Applied Statics And Strength of Materials, 7th Edition By George Limbrunner, Craig D'Allaird SOLUTIO

Solutions Manual for

Applied Statics and Strength of Materials Seventh Edition George F. Limbrunner (Inactive) Craig T. D’Allaird

Contents Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18 Chapter 19 Chapter 20

1 12 20 49 71 115 158 177 202 221 232 267 285 330 361 398 428 474 502 513

iii Copyright © 2021 Pearson Education, Inc.

INSTRUCTOR’S MANUAL FOR

APPLIED STATICS AND STRENGTH OF MATERIALS Seventh Edition George F. Limbrunner, P.E., (INACTIVE) Craig T. D’Allaird, P.E.

NOTES: 1. The solutions presented herein are, in general, somewhat abbreviated to conserve space.

Very little explanation is furnished. Sketches are kept to a minimum. Few checks are shown.

2. The solutions follow the procedures developed in the examples in the text.

3. The solutions are based on the limited tables furnished in the text and/or the appendices.

The tables furnished are for the purposes of this text only and should not be used for design.

4. The solutions for the design problems are generally not the only solution nor are they necessarily the most economical solutions.

5. Please note that problem numbers in the solution manual are depicted with both dashes (‐) and periods (.) between the chapter and problem numbers. These are interchangeable.

6. It should be noted that the previous editions of the text used lowercase s to denote stresses.

This has changed in the 7th edition, but due to time constraints this solution manual has not been completely updated to reflect such.

7. If you find errors in this manual or in the text, please forward them to me at c.dallaird@hvcc.edu.

Craig T. D’Allaird Troy, NY January 2021

iv Copyright © 2021 Pearson Education, Inc.

Prob. 1.1 (a) 𝑐 = √10 + 7 = 12.21 ft (b) 𝑏 = √20 − 16 = 12.00 m -----------------------------------------------------------Prob. 1.2 (a) 𝑎 = 25 sin 48° = 18.58 ft (b) 𝑏 = √25 − 18.58 = 16.73 ft (c) ℎ = b sin 48° = 16.73 sin 48° = 12.43 ft -----------------------------------------------------------Prob. 1.3 𝑎 = √72 − 67.3 = 25.59 ft A = cos–1(67.3/72) = 20.8° B = sin–1(67.3/72) = 69.2° -----------------------------------------------------------Prob. 1.4 AB = 28 sin 70° = 26.3 ft -----------------------------------------------------------Prob. 1.5 𝑐=

10 + 6 = 11.6 ft

𝜃 = tan (6⁄10) = 31.0° -----------------------------------------------------------Prob. 1.6 𝐴𝐵 =

12 + 16 = 20 ft

𝐵𝐶 = 12 + 32 = 34.2 ft 𝐴 = tan (12⁄16) = 36.9° 𝐶 = tan (12⁄32) = 20.6° -----------------------------------------------------------Prob. 1.7 θ = sin (5⁄6) = 56.4° 𝑥=

12 + 10 = 6.63 ft

-----------------------------------------------------------1-1

Prob. 1.8 Assume all angles to be 45° 𝑅 = +2 + 3 cos 45° + 0 − 6 sin 45° = 0.1213 mi 𝑅 = 0 − 3 sin 45° + 6 − 6 sin 45° = −0.364 mi. 𝑅=

(−0.1213) + (−0.364) = 0.384 mi.

-----------------------------------------------------------Prob. 1.9 (a) c2 = 112 + 132 – 2(11)(13)cos 80° = 15.50 ft sin 80° sin A sin B = = 15.50 11 13 →A = 44.3°, B = 55.7° (b) a2 = 782 + 852 – 2(78)(85)cos 72° = 96.0 ft sin 72° sin C sin B = = a 85 78 →A = 57.4°, B = 50.6° (c) Right Triangle Check Check a2 + b2 = c2 𝐴 = tan (7⁄24) = 16.26° 𝐵 = tan (24⁄7) = 73.7° -----------------------------------------------------------Prob. 1.10 C = 180° – 55° – 63° = 62° 𝑎 100 𝑐 = = sin 63° sin 55° sin 62° sin 63° (100) = 108.8 ft ∴𝑎= sin 55° sin 62° (100) = 107.8 ft &𝑐 = sin 55° Perimeter = 𝑎 + 𝑏 + 𝑐 = 317 ft -----------------------------------------------------------1-2

Prob. 1.11 S = weight of one shock P = weight of one set of brake pads Eq1: 8S + 10P = 101.6 lb Eq2: 10S + 6P = 106.2 lb Multiply Eq2 by 8/10: Eq3: 8S + 4.8P = 84.96 lb 8S + 10P = 101.6 lb - 8S – 4.8P = -84.96 lb 5.2P = 16.64 lb P = 3.20 lb ∴

S = 8.70 lb

----------------------------------------------------------------Prob. 1.12

20 26 = → ∴ 𝐵 = 41.77° sin B sin 60° D1= 78.23° and D2= 101.77° 𝐴𝐵 20 = →∴ 𝐴𝐵 = 29.39 ft sin 78.23° sin 41.77° 𝐵𝐶 =

(𝐴𝐵) + 50 − 2(𝐴𝐵)(50) cos 60°

𝐵𝐶 = 43.52 ft 50 43.5 = → ∴ 𝐵 = 84.25° sin B sin 60° C = 180° – 60° – 84.25° = 35.75° ------------------------------------------------------------

1-3

Prob. 1.13

-----------------------------------------------------------Prob. 1.14

-----------------------------------------------------------Prob. 1.15 AB = 32(tan 55°) = 45.7 m -----------------------------------------------------------Prob. 1.16

------------------------------------------------------------

1-4

Prob. 1.17

--------------------------------------------------------------Prob. 1.18

-----------------------------------------------------------Prob. 1.19

-----------------------------------------------------------1-5

Prob. 1.20

-----------------------------------------------------------Prob. 1.21

------------------------------------------------------------

1-6

Prob. 1.22

-----------------------------------------------------------Prob. 1.23

--------------------------------------------------------------Prob. 1.24 (a)0.015 ton × 2000 lb⁄ton = 30.0 lb (b)30.0 lb × 16 oz.⁄lb = 480 oz.

-----------------------------------------------------------Prob. 1.25 (a)5 mi × 5280 ft⁄mi × 1 yd⁄3 ft = 8800 yd (b)5 mi × 5280 ft⁄mi = 26,400 ft -----------------------------------------------------------1-7

Prob. 1.26 60

mi ft 1hr 1 min ft × 5280 × × = 88 hr mi 60 min 60 sec sec

-----------------------------------------------------------Prob. 1.27 43,560

ft 1 yd × acre 3 ft

×

1 rod 5.5 yd

= 160

rod acre

----------------------------------------------------------------Prob. 1.28 (a) 125,000,000,000 gal = 384 × 10 acre − ft gal ft × 43,560 7.481 acre ft (b) 125,000,000,000 gal × 62.4 7.481

gal lb × 2000 ton ft

lb ft = 521 × 10 tons

-----------------------------------------------------------Prob. 1.29 3 7.75 in. (a)27′ − 7  → = 0.646 ft → 27.65 ft in 4 12 ft (b)1.815 ft → in.

0.815 ft × 12

in = 9.78 in. ft

0.78 in.× 32 = 24.96 ∴ 1.815 ft = 1 − 9

25  32

-----------------------------------------------------------Prob. 1.30 Volume=Area × length π(2 in. ) ft (0.25 mi) 5280 mi 4 = in.

1728 ft

12

in. ft

1-8

=28.80 ft3 Flushing Water=2(28.8 ft3)(7.481 gal/ft3) = 431 gal -----------------------------------------------------------Prob. 1.31

-----------------------------------------------------------Prob. 1.32

----------------------------------------------------------------Prob. 1.33

-----------------------------------------------------------Prob. 1.34

------------------------------------------------------------

1-9

Prob. 1.36

-----------------------------------------------------------Prob. 1.37

-----------------------------------------------------------Prob. 1.38

-------------------------------------------------------------------Prob. 1.39

------------------------------------------------------------

1-10

Prob. 1.40

-----------------------------------------------------------Prob. 1.41

-----------------------------------------------------------Prob. 1.42

------------------------------------------------------------

1-11

Prob. 2.1 (a) R = 502 + 602

(b) R = 220 2 + 180 2

= 78.1 lb

= 284 N

θx = tan–1 (60/50)

θx = tan–1 (220/180)

= 50.2°

= 50.7°

(c) R = 4 2 + 1.5 2 = 4.27 k θx = tan–1 (1.5/4) =20.6° -----------------------------------------------------------Prob. 2.3 (a) Fx = +600 cos 30° = +520 lb Fy = −600 sin 30° = −300 lb (b) Fx = −5 cos 45° = −3.54 k Fy = −5 sin 45° = −3.54 k (c) Fx = −1200 sin 30° = −600 lb Fy = +1200 cos 30° = +1039 lb -----------------------------------------------------------Prob. 2.4

(a) Fx = + Fy = +

3 13 2 13

(425) = +354 lb (425) = +236 lb

5 (110) = −42.3 lb 13 12 Fy = − (110) = −101.5 lb 13

(b) Fx = −

2-12

3 (c) Fx = − (3) = −1.80 k 5 4 Fy = + (3) = +2.40 k 5 -----------------------------------------------------------Prob. 2.5 (a) Px = +320 cos 20° = +301 lb Py = −320 sin 20° = −109.4 lb (b) Px = +640 cos 30° = +554 lb Py = −640 sin 30° = −320 lb (c) Px = +320 cos 40° = +245 lb Py = −320 sin 40° = −206 lb (d) Px = +320 cos 88° = +11.17 lb Py = −320 sin 88° = −320 lb -----------------------------------------------------------Prob. 2.6

F = 1.32 + 0.130 2 = 1.304 kN  1.3  θ x = tan −1    0.103  = 85.5° -----------------------------------------------------------Prob. 2.7 R = 175/(cos 45°) = 247 lb θx = tan–1(175/175) = 45° -----------------------------------------------------------Prob. 2.8 Px = 120 cos 50° = 77.1 lb Py = 120 sin 50° = 91.9 lb ------------------------------------------------------------

2-13

Prob. 2.9 Fx = +300 sin 42° = +201 lb Fy = −300 cos 42° = −223 lb -----------------------------------------------------------Prob. 2.10 (a) Fy = 650 cos 20° = −611 lb Fx = 650 sin 20° = +222 lb (b)

Fx Fy 300 = = 12 5 13 Fx = (12/13)(300) = +277 lb Fy = (95/13)(300) = −115 lb

-----------------------------------------------------------Prob. 2.11 (a) Px = 120 cos 30° = +103.9 kN Py = 120 sin 30° = −60.0 kN

(b) Px = 120 cos 75° = +31.1 kN Py = 120 sin 75° = −115.9 kN

(c) Px = 120 sin 5° = +10.46 kN Py = 120 cos 5° = +119.5 kN ----------------------------------------------------------Prob. 2.12 (a) Fy = 800cos30° = −693 lb Fx = 800sin 30° = −400 lb

2-14

(b)

Fy 4

=

Fx 15 50 = 3 5

4 Fy = (150) = −120 1 lb 5 3 Fx = (150) = −90 9 lb 5 -------------------------------------------------------------Prob.. 2.13

FH = 1000 sin 40° = −643 lb 0° = −766 lb FV = 1000 cos 40 -------------------------------------------------------------Prob.. 2.14 (a) Py = 80cos20°° = −75.2 k Px = 80sin20° = −27.4 k (b) Px = 60 sin 30 0° = −30.0 kN N Py = 60 cos 30 0° = +52.0 kN -------------------------------------------------------------Prob.. 2.15 FV = 500sin42° = +335 lb FH = 500cos42° = +372 lb -------------------------------------------------------------Prob.. 2.16 FH = 443 lb = 500 0 cos θ

 443   = 27.6°  500 

θ = cos c −1 

--------------------------------------------------------------

2-15

Prob.. 2.17

Fx Fy 120 = = 2 1 5 2 120 = 10 Fx = 07.3kN 5 120 = 53.7 N Fy = 5 -------------------------------------------------------------Prob.. 2.18 W = weight w of thee skiers W = 38(175) 3 = 66 650 lb T = Wx = 6650 sin n 17° = 1944 1 lb -------------------------------------------------------------Prob.. 2.19

6N F1x = F2x = 10 cos 40° = 7.66

Hypo otenuse of slope triangle =

s2 +1

By siimilar triangles: 1 = 7.66

s2 +1 → s = 1.68 84 15

Resultant: F1 y

F1x 1.684 4 1 F1 y = 1.684(7.66) = 12.90 N =

F2y = 10 sin 40° = 6.43 N R = 6.43 6 + 12.90 = 19.33 N ↑ 2-16

-------------------------------------------------------------Prob.. 2.20

Fy

Fx 9500 = 4 8.06 7 7 Fy = (9500) = −8250 lb 8.06 4 Fx = (9500) = −4710 lb 8.06 =

---------------------------------------------------------------Prob.. 2.21 (a) P = 300 2 + 200 2 = 361 lb θx = tan–1(200 0/300) = 33.7 7° (b) P = 500 2 + 300 2 = 583 lb θx = tan–1(300 0/500) = 31.0 0° (c) P = 2402 + 3602 = 433 lb θx = tan–1(360 0/240) = 56.3 3° (d) P = 2502 + 4602 = 524 lb θx = tan–1(46 60/250) = 61.5° ---------------------------------------------------------------Prob.. 2.22 P1x = 7.78 cos 80 0° = +1.351 k P1y = 7.78 sin 80°° = +7.66 k P2x = 10 sin 40° = + 6.43 k P2y = 10 cos 40° = −7.66 k --------------------------------------------------------------

2-17

Prob.. 2.23 R = 1502 + 44.12 = 156.3 kN N  150   = 73.6°  44.1 

θ = tan −1 

1 = 0.294 m 6° tan 73.6 x1 = 1.5 + 0.2 294 = 1.794 m x=

-------------------------------------------------------------Prob.. 2.24

TV = 2000 sin 70° = 1879 lb TH = 2000 cos 70 0° = 684 lb ----------------------------------------------------------------Prob.. 2.25 For th he 2-kip load ds:

FV FH 2 = = 2 1 5 2(2) = 1.789 k FV = 5 1(2) = 0.894 4k FH = 5

For the 4-kip load: FV = 1.789(2) = 3.58 k FH = 0.894(2) = 1.789 k

----------------------------------------------------------------

2-18

Prob. 2.26

θ = tan–1(4/5) = 38.66° 6-lb force: Fy = 6 cos 19.66° = 5.65 lb Fx = 6 sin 19.66° = 2.02 lb 9-lb force: Py = 9 sin 63.66° = 8.07 lb Px = 9 cos 63.66° = 3.99 lb

2-19

Prob. 3.1

Using triangle OAC: R2 = 252 + 452 − 2(25)(45) cos 130° R = 64.0 lb 64.0 45 = sin 130° sin θ

  45   θ = sin −1 sin130°    = 32.59°  64.0    θx = 60° − 32.59° = 27.4°

-----------------------------------------------------------Prob. 3.2

θ1= tan−1(4/3) = 53.13° θ2 = tan−1(5/12) = 22.62° φ = 180° − θ1 − θ2 = 104.25° α = 180° − φ = 75.75° R2 = 122 + 102 − 2(12)(10) cos 75.75° R = 13.60 k φ1 = sin−1(sin75.75°(10/13.60)) = 45.45° 3-20

θx = θ2 +φ1 = 22.62° + 45.45° = 68.07°

-----------------------------------------------------------Prob. 3.3

θ1 = tan−1(1/2) = 26.57° θ = 180° − 26.57°−30° = 123.43° φ = 180° − θ = 56.57° R2 = 752 + 602 −2(75)(60)cos 56.57° R = 65.32 lb

R 60 = sin 56.57° sin θ2   60   θ2 = sin −1 sin 56.57°    = 50.05°  65.32    θ x = θ1 + θ2 = 76.6° -----------------------------------------------------------Prob. 3.4

Rx = 45 cos 10° + 25 cos 60° = +56.82 lb Ry = 45 sin 10° + 25 sin 60° = +29.46 lb

3-21

R = 56.82 2 + 29.46 2 R = 64.0 lb  29.46    56.82  = 27.4°

θ x = tan −1 

-----------------------------------------------------------Prob. 3.5

Rx = +12(12/13)−10(3/5) = +5.08 k Ry = +12(5/13) + 10(4/5) = +12.62 k

R = 5.082 + 12.622 = 13.60 k  12.62  θ x = tan −1    5.08  = 68.1° -----------------------------------------------------------Prob. 3.6

 2  R x = −75  + 60 cos 30° = −15.12 lb  5  1  R y = −75  − 60 sin 30° = −63.54 lb  5

R = (−15.12) 2 + ( −63.5) 2 R = 65.3 lb  63.54  θ x = tan −1    15.12  = 76.6°

------------------------------------------------------------

3-22

Prob. 3.7

Px = −75 cos 40° = −57.45 lb Py = +75 sin 40° = + 48.21 lb Rx = −150 cos 65° = −63.39 lb Ry = +150 sin 65° = + 135.95 lb Fy + Py = Ry Fy + 48.21 lb = 135.95 lb Fy = + 87.74 lb Fx + Px = Rx Fx − 57.45 lb = −63.39 lb

F = 5.942 + 87.742 = 87.9 lb  5.94  θ = tan −1    87.74  = 3.87°

Fx = −5.94 lb -----------------------------------------------------------Prob. 3.8

Force

Horiz. (N)

Vert. (N)

100 N

90.6

42.3

120 N

75.0

93.7

Σ

165.6

136.0

3-23

R = 165.62 + 136.02 = 214 N  136.0  θ = tan −1    165.6  = 39.4° -----------------------------------------------------------Prob. 3.9 Force (N) Horiz. (N)

Vert. (N)

160

+120.4

−105.4

200

−120.4

−159.7

Σ

0

−265

R = 265 N ↓ -----------------------------------------------------------Prob. 3.10

↑→ (+)

Use components: For sloping force Slope triang. hypt = 5 2 + 3 2 = 5.83 Fx Fy 200 = = 3 5 5.83 Fx = 102.9; Fy = 171.5 Total horiz. = −200 −102.9 = − 302.9 lb

R = 302.9 2 + 171.52 = 348 lb  171.5  θ = tan −1   = 29.5°  302.9 

------------------------------------------------------------

3-24

Prob. 3.11 Force (lb)

Fx (lb)

Fy (lb)

400

−400

0

300

−212.1

−212.1

200

+68.4

−187.9

Σ

−543.7

−400

R = 543.7 2 + 4002 = 675 lb  400  θ x = tan −1   = 36.3°  543.7 

-----------------------------------------------------------Prob. 3.12 (a)

R1, 2 = 50 2 + 70 2 − 2(50)(70) cos100° = 92.82 N sinθ 1 sin 100° = 50 92.82  50 sin 100°  θ 1 = sin −1   = 32.04°  92.82  θ x = θ 1 − 10° = 22.04°

3-25

(b)

R3,4 = 602 + 202 − 2(60)(20) cos 76.33° = 58.59 N sin θ1 sin 76.33° = → θ1 = 84.30° 60 58.59 θ x = 84.30° − 36.87° = 47.43° (more) (c)

R = R3,4 2 + R1,2 2 − 2 R3,4 R1,2 cos 25.39° = 47.14 N sinθ1 sin 24.39° = → θ1 = 57.59°(122.41°) 92.82 47.14 θ x = 10.16°

-----------------------------------------------------------Prob. 3.13

→↑ +

(a) Method of components Force (N)

Horiz. (N)

Vert. (N)

50

+17.10

+46.98

70

+68.94

−12.16

20

−16.00

+12.00

60

−23.63

−55.15

Σ +46.41

−8.33

3-26

R = 46.412 + (−8.33) 2 = 47.2 N  8.33  θ x = tan −1   = 10.18°  46.41  -----------------------------------------------------------→↑ +

Prob. 3.14

Rx = ΣFx = +200 cos 30° +100 cos 45° −400 −300 cos 60° = −306.1 lb Ry = ΣFy = +200 sin 30° −100 sin 45° −50 +300 sin 60° = +239.1 lb

R = (−306.1) 2 + 239.12 = 388 lb  239.1  θ x = tan −1   = 38.0°  306.1  --------------------------------------------------------→↑ +

Prob. 3.15 R = Ry = +300 lb

Rx = ΣFx = 0 = −500 + F1x + 240 cos 30°

∴ F1x = +292.2 lb Ry = ΣFy = +300 = +F1y −240 sin 30°

∴ F1y = +420 lb F1 = 4202 + 292.22

= 512 lb  420  θ x = tan −1   = 55.2°  292.2  --------------------------------------------------------Prob. 3.16

→↑ +

Rx = ΣFx = +1000 + 600 sin30° = +1300 lb

3-27

Ry = ΣFy = +900 −600 cos30° = +380.4 lb R = 380.4 2 + 1300 2 = 1355 lb  380.4   = 16.3°  1300 

θ x = tan −1 

-----------------------------------------------------------↑→ +

Prob. 3.17

Rx = ΣFx = +100 cos 20° − 95 cos 80°

− 110 cos 70° = + 39.9 N → Ry = ΣFy = +95 sin 80° − 110 sin 70°

− 100 sin 20° = − 44.0 N ↓ R = 39.92 + (−44.0) 2 = 59.5 N  44.0  θ x = tan −1   = 47.8°  39.9  -----------------------------------------------------------Prob. 3.18

↑→ +

For the 200-lb resultant force: Rx = 200 cos 60° = +100 lb Ry = 200 sin 60° = +173.2 lb Rx = ΣFx = +100 = − 160 + F2x ∴ F2x = 260 lb Ry = ΣFy = +173.2 = F2y F2 = 173.22 + 26.02 = 312 lb  173.2  θ x = tan −1   = 33.7°  260  ------------------------------------------------------------

3-28

Prob. 3.19

↑→ +

For the resultant: Rx = 300 cos45° = − 212.1 lb Ry = 300 sin45° = 212.1 lb Also: Rx = ΣFx = −212.1 = +F1 cos60° − F2 cos 30° ∴

−212.1 = 0.5F1 − 0.866F2 F1 = −424.2 + 1.732F2

(I.)

Ry = ΣFy = +212.1 = F1 sin 60° − F2 sin 30° ∴

+212.1 = +0.866F1 − 0.5F2 F1 = +244.9 + 0.577F2

(II.)

Solve simultaneous eqns. (I.) and (II.): F2 = 579 lb F1 = 579 lb -----------------------------------------------------------Prob. 3.20

P1 P2 30 = = sin 84.09° sin14.04° sin 81.87° from which: P1 = 30.1 kN P2 =

7.35 kN

-----------------------------------------------------

3-29

↑→ +

Prob. 3.21 For the resultant:

Rx = +150 cos 60° = +75 lb Ry = −150 sin 60° = −129.9 lb Rx = ΣFx = +75 = +F1 cos 30° − F2 cos 67° +75 = +0.866 F1 − 0.39 F2

∴

F1 = +86.6 + 0.45F2

(I.)

Ry = ΣFy = −129.9 = +F1 sin 30° − F2 sin 67° −129.9= +0.5F1 − 0.92F2

∴

F1 = −259.8 + 1.84F2

(II.)

Solve simultaneous eqns. (I.) and (II.): F2 = 249.2 lb F1 = 198.7 lb -----------------------------------------------------------↑→ +

Prob. 3.22 Force

Horiz.

Vert.

(lb)

(lb)

(lb)

20

−5.176

−19.32

30

−9.271

+28.53

Σ

−14.45

+9.21

R = (−14.45) 2 + 9.212 = 17.14 lb  9.21  θ = tan −1   = 32.5°  14.45  ----------------------------------------------------------

3-30

↑→ +

Prob. 3.23

Rx= +20 cos 30° + 50 cos 70° − 40 cos 15° = − 4.216 lb Ry = +20 sin 30° − 50 sin 70° − 40 sin 15° = − 47.34 lb

R = ( −4.216) 2 + (−47.34) 2 = 47.5 lb  47.34  θ x = tan −1   = 84.9°  4.216  -----------------------------------------------------------↑→ +

Prob. 3.24

Rx = +75 cos 45° − 250 cos 70° −200 − 90 cos 30° = −310.4 lb Ry = +75 sin45° − 250 sin 70° + 150 + 90 sin 30° = +13.11 lb R = (−310.4) 2 + 13.112 = 311 lb  13.11  θ x = tan −1   = 2.42°  310.4 

-----------------------------------------------------------Prob. 3.25

↑→ +

For the resultant: Rx = 100 cos 20° = +93.97 lb Ry = 100 sin 20° = +34.20 lb Also: Rx = ΣFx = +93.97 = +F2 cos 30° − F1 cos 45° 3-31

∴

+93.97 = +0.866F2 − 0.707F1 F1 = +1.225F2 −132.91

(I.)

Ry = ΣFy = +34.20 = +F2 sin30° + F1 sin45°−200 +34.20 = + 0.5F2 +0.707F1 − 200

∴

F1 = − 0.707F2 + 331.26

(II.)

Solve simultaneous eqns. (I.) and (II.): F2 = 240.3 lb F1 = 161.4 lb -----------------------------------------------------------Prob. 3.26 FH = Fsin18° FH = AH AH = 5000 sin 25° = 2113 lb

F=

FH 2113 = = 6840 lb sin18° 0.309

-----------------------------------------------------------Prob. 3.27 Solve the force triangle: F2 = 60002 + 50002 −2(6000)(5000) cos 25° F = 2570 lb

sin θ sin 25° = 5000 2570 θ = 55.3° -----------------------------------------------------------Prob. 3.28

Refer to Prob. 2.8.

CC moment: +

W = weight of barrel and contents  100  W = 60 +  (62.4) = 894 lb  7.481 

3-32

ΣMo = +(894 − 120 sin 50°)(1.25) − (120 cos 50°)(2.75) = +791 lb-ft --------------------------------------------------------Prob. 3.29 (a) MA : (counterclockwise: +) Force

Moment

Moment

(lb)

arm (ft)

(lb-ft)

20

4

−80

40

9

−360

50

15

+750

30

23

−690

−−

−380

Σ

(b) ΣMA = −380 lb-ft (clockwise) ------------------------------------------------------------Prob. 3.30 [CC moment: +]; [↑→ +] (a) [F1] Mo = 0 [F2] Mo = 50(6) = +300 lb-ft [F3] Mo = 75 sin 15°(6) = +116.5 lb-ft (b) ΣMo = +300 + 116.5 = + 417 lb-ft (c) Rx = ΣFx = +25 − 75 cos 15° = −47.4 lb Ry = ΣFy = +50 + 75 sin 15° = +69.4 lb (more) R = 69.42 + (−47.4) 2 = 84.0 lb  69.4  θ x = tan −1   = 55.7°  47.4 

3-33

(d) Mo = 84.0(6 sin55.7°) = + 416 lb-ft (Say OK) -----------------------------------------------------------Prob. 3.31

[CC moment is +]

(a) [F1] Mo = 30 sin60°(4) = +103.9 lb-ft [F2] Mo = 40 sin45°(4) = −113.1 lb-ft [F3] Mo = 10(4) = −40 lb-ft [F4] Mo = 0 (b) ΣMo = +103.9 − 113.1 − 40 = − 49.2 lb-ft -----------------------------------------------------------↑→ +

Prob. 3.32

Rx = ΣFx = 30 cos 60° + 40 cos 45° − 50 = −6.716 lb Ry = ΣFy = 30 sin 60° −40 sin 45° −10 = −12.304 lb R = (−6.716) 2 + (−12.304)2 = 14.02 lb  12.304  θ x = tan −1   = 61.4°  6.716 

Moment of resultant: ΣMo= 14.02 sin 61.4°(4) = 49.2 lb-ft (checks) -----------------------------------------------------------Prob. 3.33

CC moment: +

For 30 kN force: Fx = 26.49 kN, Fy = 14.08 kN (a) ΣMA Force (kN) Arm (m)

Moment (kN⋅m)

70

1.50

−105.0

26.49

2.00

+52.98

14.08

1.70

−23.94 3-34

60 Σ

2.35

−141.0

−−

−217

(b) ΣMB Force (kN) Arm (m)

Moment (kN⋅m)

70

1.50

+105.0

26.49

2.00

+52.98

14.08

1.30

+18.30

60

0.65

+39.0

−−

+215

Σ

-----------------------------------------------------------Prob. 3.34

CC moment: +

For the 900-lb forces: Px Py 900 lb = = 4 3 5

Px = 720 lb Py = 540 lb 3-35

ΣMA = −540(2)(20) + 720(78) + 720(68) + 400(48) = +102,700 lb-ft -------------------------------------------------------Prob. 3.35

CC moment: +

ΣMB = −12(3) − 14(10) + 25(4) = −76.0 k-ft -------------------------------------------------------Prob. 3.36 Resultant of dist. force = 10(6) = 60 N R = ΣFy = −50 − 60 + 30 − 60 = −140 N ΣMA = −50(2) − 60(6) + 30(11) − 60(14) = −970 N⋅m x is measured to the right from point A:

x=

ΣM A 970 = = 6.93 m R 140

------------------------------------------------------------Prob. 3.37

CC moment: +

b = 5/(tan40°) = 5.959 ft a = 10 − 5.959 = 4.041 ft d = 4.041 sin 40° = 2.598 ft Mo = 550(2.598) = −1429 lb-ft Using Varig. theorem w/ components at A: ΣMo = −550 sin 40°(10) + 550 cos 40°(5) = −1429 lb-ft

(checks)

-----------------------------------------------------------3-36

Prob. 3.38

CC moment: +

Components at B: ΣMo = 550 sin40° = −1429 lb-ft (checks) Components at C: ΣMo = 550 cos 40°(4.041 tan40°) = −1429 lb-ft (checks) --------------------------------------------------------------Prob. 3.39 Fx = 2 cos20° = 1.879 k Fy = 2 sin20° = 0.684 k ΣMA = −1.879(3 sin 50°) − 0.684(3 cos 50°) = −5.64 k-ft --------------------------------------------------------------Prob. 3.40 CC moment: + Fx = 150 cos 60°= 75.0 lb Fy = 150 sin 60°=129.9 lb ΣMA= +129.9(5) + 75.0(20) = +2150 lb-ft ------------------------------------------------------------Prob. 3.41

CC moment: +

ΣMA = −3(5) − (4/5)(5)(10) = −55.0 k-ft ------------------------------------------------------------Prob. 3.42

CC moment: +; ↑→ +

R = −300 − 450 − 200 = −950 lb (↓) ΣMA = −300(3) − 450(8) − 200(6) = −7700 lb-ft x = 7700 / 950 = 8.11 ft

-----------------------------------------------------------3-37

Prob. 3.43

CC moment: +; ↑→ +

R = +5 + 3 − 8 + 12 − 2 = +10 k (↑) ΣMA = +3(3) − 8(6) + 12(11) − 2(13) = +67 k-ft

x = 67 / 10 = 6.7 ft -----------------------------------------------------------Prob. 3.44

CC moment: +; ↑→ +

R = + 40 − 20 − 20 + 80 = +80 lb (↑) ΣMA = − 20(2) − 20(10) + 80(12) = +720 lb-ft x = 720 / 80 = 9.0 ft

-----------------------------------------------------------Prob. 3.45

CC moment: +; ↑→ +

R = + 4 − 5 − 22 = −3 k (↓) ΣMA = −4(7) + 5(3) − 2(2) = −17 k-ft ∴ R is located to the right of A x = 17 / 3 = 5.67 ft

-----------------------------------------------------------Prob. 3.46

CC moment: +; ↑→ +

R = −8 − 32 − 32 = − 72 k(↓) ΣMA = −32(14 − 32(28) = − 1344 k-ft x = 1344 / 72 = 18 .67 ft

-----------------------------------------------------------Prob. 3.47

CC moment: +; ↑→ +

Note: F1 (assumed ↑) is located x1 to the right of A. + F1 − 10 − 20 = − 60 k F1 = −30 k (↓) ΣMA = −10(4) − 20(9) − 30x1 = −60(10) = −600 k-ft ∴ x1 = 12.67 ft -----------------------------------------------------------3-38

Prob. 3.48

CC moment: +; ↑→ +

Note: F1 (assumed ↑)is located x1 to the right of A. +F1 + 9 + 16 = +10 kN F1 = − 15 kN (↓) ΣMA: +10(5) = +9(1) +16(8) −15x1 ∴ x1 = 5.80 m -------------------------------------------------------------Prob. 3.49

R = −10 − 28 = − 38 k (↓) ΣMA = −10(5) − 28(17) = − 526 k-ft

x=

526 = 13.84 ft 38

--------------------------------------------------------Prob. 3.50

R = −1000 − 7000 − 1600 = −9600 lb (↓) ΣMA = −7000(8) − 1600(12) = −75,200 lb-ft

x=

75,200 = 7.83 ft 9600

-------------------------------------------------------------

3-39

Prob. 3.51 (a) pmax = 62.4(18) = 1123 psf (b) R = (1123/2)(1)(18) = 10,100 lb (c) y =18 / 3 = 6 ft (up from base) ------------------------------------------------------------Prob. 3.52 Resultant of the triangular load is 36 k (↓) located 3 ft right of A. R = 36 + 5 = 41 k ↓ ΣMA = −36(3) − 5(13) = − 173 k-ft

x=

173 = 4.22 ft 41

------------------------------------------------------------Prob. 3.53

CC moment: +; ↑→ +

R = − 21 − 15 − 10 = −46 kN ΣMA = −21(7) − 15(11.5) − 10(14) = − 459.5 kN⋅m

x=

459.5 = 9.99 m 46

-----------------------------------------------------------Prob. 3.54

CC moment: +; ↑→ +

Hydrostatic force F = (1/2)(62.4)(18)2 =10,100 lb (located 18/3 = 6 ft up from point A) For the resultant: Rx = +10,100 lb 3-40

Ry = − 19,800 lb R = 10,1002 + (−19,800) 2 = 22, 200 lb  119,800  θ x = tan −1   = 63°  10,100 

ΣMA = − 19,800(4) − 10,100(6) = −139,800 lb-ft Resultant intersects the base of the dam x to the right of point A:

x=

ΣM A 139,800 = = 7.06 ft Ry 19,800

(Within middle 1/3: O.K.) -------------------------------------------------------Prob. 3.55

CC moment: +; ↑→ +

F1 = 0.5(900 − 350)(15) = 4125 lb F2 = 350(15) = 5250 lb F3 = 0.5(350)(15) = 2275 R = ΣFy = + 11,650 lb ΣMA = 4125(5) + 5250(7.5) + 2275(28.8.67) = 104,000 lb-ft x=

ΣM A 104,000 = = 8.93 ft 11,650 R

------------------------------------------------------------Prob. 3.56

CC moment: +; ↑→ +

R = ΣFy = − 900 − 600 = −1500 lb ΣMA = − 900(2) − 600(15) = −10,800 lb-ft x=

ΣM A 10,800 = = 7.20 ft R 1500

---------------------------------------------------------

3-41

Prob. 3.57

CC moment: +; ↑→ +

Ry = −53 − 70 = −123 kN Rx = −21 kN R = (−123) 2 + (−21) 2 = 124.8 kN

(

)

θ = tan -1 123 21 = 80.3° ΣMA = −53(1) − 70(2) + 21(1) = − 172 kN⋅m x is measured to the right from point A:

ΣM A 172 = = 1.398 m R 123

x=

-----------------------------------------------------------Prob. 3.58 (a) 312 + 499.2 (3)(3) = 3650 lb 2 R1 = 312(3)(3) = 2808 lb Ra =

187.2 (3)(3) = 842.4 lb 2 R (1.5) + R2 (1) y= 1 = 1.385 ft R1 + R2

R2 =

124.8 ( 2)(3) = 374.4 lb 2 Rb = −3650 + 374.4 = 3276 lb R3 =

y=

3650(1.385) − 374.4(0.667) = 1.467 ft 3276

-----------------------------------------------------------Prob. 3.59 1. R = (52/2)(0.833)(5) = 108.3 lb

y=

0.833 = 0.278 ft 3

2. R1 = 52(0.833)(5) = 216.7 lb

3-42

R2 = 108.3 lb R = R1 + R2 = 325 lb  0.833  216 .7  + 108 .3(0.278) 2   = 0.370 ft y= 325

3. R1 = 104(0.833)(5) = 433 lb R2 = 108.3 lb R = R1 + R2 = 541 lb  0.833  433  + 108.3(0.278) 2   = 0.389 ft y= 541

------------------------------------------------------------Prob. 3.60 R = ΣFy = −25(4) −12−40 = −77 k ΣMA = −25(4)−12(14)−40(28) = −1388 k-ft x=

ΣM A 1388 = = 18.03 ft Ry 77

------------------------------------------------------------Prob. 3.61 F1 + F2 = 150+100 = +250 lb F1 = 250 − F2

I.

ΣMA = +220 = + F1(4) + F2(8) −100(12) 4F1 + 8F2 = +1420 F1 = −2F2 + 355

II.

Solve simultaneous equations I. & II.: F2 = 105 lb ↑ F2 = 145 lb ↑ -------------------------------------------------------------

3-43

Prob. 3.62

CC moment: +

M = M1 + M2 = − 4(7) +10(21) = +182 k-ft -----------------------------------------------------------Prob. 3.63

CC moment: +

M = M1 + M2 = − 150(10) + 80(5) = − 1100 lb-in.

----------------------------------------------------------Prob. 3.64

CC moment: +

M = M1 + M2 = +50(6) + 30(6) = + 480 lb-ft -----------------------------------------------------------Prob. 3.65

CC moment: +

M = M1 + M2 + M3 = +30(3) + 20(6) − 10(5) = +160 lb-ft Since M = Fd = +160 lb-ft and d = 2.5 in.: F = M/d = 160/2.5 = 64 lb ------------------------------------------------------------Prob. 3.66

Rx = +15 cos 75° − 25(5/13) = −5.733 k ← Ry = −10 − 15 sin75° − 25(12/13) = −47.57 k ↓

3-44

R = ( −5.733) 2 + (−47.57) 2 = 47.9 k  47.57  θ x = tan −1    5.733  = 83.1°

ΣMo = −10(8) − 15 sin75°(14) − 25(12/13)(19)= −721.3 k-ft

x=

ΣM O 721.3 = = 15.16 ft Ry 47.57

------------------------------------------------------------Prob. 3.67

CC moment: +; ↑→ +

For the resultant: Rx = ΣFx = +500 − 100 = +400 lb → Ry = ΣFy = −400 + 250 = −150 lb ↓ R = 4002 + (−150) 2 = 427 lb

 150  θ x = tan −1   = 20.6°  400  ΣMo = +100(4) − 400(5) + 250(17) − 500(12) = − 3350 lb-in.

Note: resultant pierces base of block x to the right of point O.

x=

ΣM O 3350 = = 22.3 in.

Ry 150

3-45

------------------------------------------------------------Prob. 3.68

CC moment: +; ↑→ +

For the resultant: Rx = ΣFx = + 200 N → Ry = ΣFy = −80 − 50 = −130 N ↓ R = 2002 + (−130) 2 = 239 N

 130  θ x = tan −1   = 33.0°  200  ΣMA = −200(1.333) + 50(2) + 80(1) = − 86.6 N⋅m

x=

ΣM A 86.6 = = 0.666 m (as shown) Ry 130

------------------------------------------------------------Prob. 3.69

CC moment: +; ↑→ +

For the resultant: Rx = ΣFx = +15 k → Ry = ΣFy = −50 − 20 = −70 k ↓ R = 152 + (−70) 2 = 71.6 k

 70  θ x = tan −1   = 77.9°  15  3-46

ΣMA = −15(6) + 50(6) +20(5) = +310 k-ft.

Note: resultant pierces bottom of footing x to the left of point A.

x=

ΣM A 310 = = 4.43 ft Ry 70

------------------------------------------------------------Prob. 3.70 (a) ΣM A = −1200(3) − 500(4 / 5)(12) − 500(3 / 5)(9) − 400( 20) = −19,100 lb - ft

(b) Rx=ΣFx = +400+500(4/5) = +800 lb Ry = ΣFy = −1200−500(3/5) = −1500 lb R = 800 2 + (−1500) 2 = 1700 lb x=

ΣM A 19,100 = = 23.9 ft Rx 800

(Above point A) ------------------------------------------------------------Prob. 3.71 Units: Newtons and meters: (a) ΣMA = −(0.12×106)(1)−980(6) −(8×103) sin 40°(13) = −192.7 × 103 N⋅m = −192.7 kN⋅m ΣMB = +(0.12×106)(15)+980(10) +(8×103) sin40°(3) = 1.825 × 106 N.m = 1.825 MN⋅m (b) Rx = ΣFx = +(8×103)cos40° = +6.128×103 N

3-47

Ry = ΣFy = −(0.12×106)−980 −(8×103) sin 40° = −126.1×103 N R = (6.128 × 103 ) 2 + (−126.1× 103 ) 2 = 126.3 × 103 N  126.1  θ x = tan −1   = 87.2°  6.128  x is measured to the right from point A:

x=

ΣM A 192.7 × 10 3 = = 1.53 m Ry 126.1 × 10 3

------------------------------------------------------------Prob. 3.72

CC moment: +; ↑→ +

Rx = ΣFx = +25 cos45° −40(4/5) = −14.32 lb Ry = ΣFy = + 25 sin45° + 40(3/5) − 20 = + 21.68 lb R = (−14.32) 2 + 21.682 = 26.0 lb

 21.68  θ x = tan −1   = 56.6°  14.32  ΣMA = + 40(4/5)(3) + 25 sin45°(5) − 80 = + 104.4 lb-ft Resultant intersects the bottom of the shape at x to the right of point A:

x=

ΣM A 104.4 = = 4.82 ft Ry 21.68

-------------------------------------------------------------

3-48

Prob 4.11

-----------------------------------------------------------Prob. 4.12

------------------------------------------------------------

4-49

Prob. 4.13

-----------------------------------------------------------Prob. 4.14

-----------------------------------------------------------Prob. 4.15

------------------------------------------------------------

4-50

Prob. 4.16

-----------------------------------------------------------Prob. 4.17

-----------------------------------------------------------Prob. 4.18

4-51

-----------------------------------------------------------Prob. 4.19

-----------------------------------------------------------Prob. 4.20

------------------------------------------------------------

4-52

Prob. 4.21

-----------------------------------------------------------Prob. 4.22

--------------------------------------------------------------Prob. 4.23

------------------------------------------------------------

4-53

Prob. 4.24

-----------------------------------------------------------Prob. 4.25

-----------------------------------------------------------Prob. 4.26

---------------------------------------------------------

4-54

Prob. 4.27

-----------------------------------------------------------Prob. 4.28

-----------------------------------------------------------Prob. 4.29

------------------------------------------------------------

4-55

Prob. 4.30

----------------------------------------------------------Prob. 4.31

-----------------------------------------------------------Prob. 4.32

4-56

-----------------------------------------------------------Prob. 4.33

--------------------------------------------------------Prob. 4.34

------------------------------------------------------------

4-57

Prob. 4.35

--------------------------------------------------------Prob. 4.36

--------------------------------------------------------------

4-58

Prob. 4.37

------------------------------------------------------------

4-59

Prob 4.38

------------------------------------------------------------

4-60

Prob. 4.39

(Neglect Boom Weight)

-----------------------------------------------------------Prob. 4.40

------------------------------------------------------------

4-61

Prob. 4.41

-----------------------------------------------------------Prob. 4.42

-----------------------------------------------------------4-62

Prob. 4.43

-----------------------------------------------------------Prob 4.44

---------------------------------------------------------

4-63

Prob. 4.45

-----------------------------------------------------------Prob. 4.46

-----------------------------------------------------------Prob. 4.47

-----------------------------------------------------------4-64

Prob. 4.48

-----------------------------------------------------------Prob. 4.49

-----------------------------------------------------------Prob. 4.50

ΣMA = −1.0(10) −1.0(15) − 2.4(12) + 7.0(15) +5(5) + RB(24) = 0 RBV = − 3.18 k ↓ (downward) ΣMB = −RA(24) −1.0(10) + 2.4(24) −1.0(15) + 2.4(12) + 7(15) + 5(5) = 0 RA = + 7.98 k ↑ ΣFx = + 1.0 + 1.0 −7 −5 + RBH = 0 RBH = +10 k → -----------------------------------------------------------4-65

Prob. 4.51

ΣMA = −130(10) − 200(17.5) + RBH (34.64) = 0 RBH = +138.6 lb ← ΣFx = + RAH − RBH = 0 RAH = +138.6 lb → ΣFy = +RAV −130 − 200 = 0 RAV = + 330 lb ↑ -----------------------------------------------------------Prob. 4.52

ΣMA = +CDV (6) −16(8) = 0 CDV = + 21.33 k ↑ CDV = 33.2 k sin 40° CDV = 25.42 k ← CDH = tan 40° CD =

4-66

ΣFx = + RAH − 25.42 = 0 RAH = 25.4 k → ΣFy = + RAV + 21.33 − 16 = 0 RAV = −5.33 k ↓ -----------------------------------------------------------Prob. 4.53

ΣMA = +RBH (15) −600(8) −3000(16) −300(6) = 0 RBH = + 3640 lb ← ΣFx = +RAH −3640 = 0 RAH = + 3640 lb → ΣFy = +RAV − 400 −600 −300− 3000 = 0 RAV = + 4300 lb ↑ -----------------------------------------------------------Prob. 4.54

ΣMA =−12,800(8) + BDV (12) − 500(16) = 0 BDV = + 9200 lb↑ 4-67

BDH = (9/12)BDV = + 12,270 lb → BD = (15/9)BDV = 15,330 lb (as shown) ΣFx = −RAH + BDH = 0 RAH = 12,270 lb ← ΣFy = +RAV − 12,800 + 9200 − 500 = 0 RAV = 4100 lb ↑ -----------------------------------------------------------Prob. 4.55

TH TV T = = 12 5 13 TH = (12/13)T; TV = (5/13)T

 12  ΣM A = + T (15)  13  5 +  T (25) − 300(34) = 0  13  T = 435 k --------------------------------------------------------Prob. 4.56

4-68

h = 6 sin77.5° = 5.858 ft F=

62 .4(5.858 ) 2 = 1071 lb 2

Fy = 1017 cos 77.5° = 232 lb Fx = 1071 sin 77.5° = 1046 lb BV = BH = BD sin45° = (0.707)BD x1 = 3.5 cos77.5° = 0.758 ft y1 = 3.5 sin77.5° = 3.42 ft ΣMC = −1071(2.0) + BH y1 + BV x1 = 0 (by substitution) BD = 725 lb BV = BH = 0.707(725) = 513 lb ΣFx = −CH + 1046 −513 = 0 CH = +533 lb ← ΣFy = +CV −232 + 513 = 0 CV = −281 lb ↓ -----------------------------------------------------------Prob. 4.57

ΣFx = 0 T = 2(6.43) = 12.86 k ΣMB = −7.66(3.72) − 7.66(10.25) + W2 (10) +12.86(1) = 0 W2 = 9.42 k 4-69

ΣFy = +W1 + 9.42 − 2(7.66) = 0 W1 = 5.90 k --------------------------------------------------------------Prob. 4.58 Cylinder A: 75(9.81) = 736 N Cyliner B: 50(9.81) = 491 N

Solve the force triangle: BH =

491 = 740 N tan 33.56°

C = 7402 + 4912 = 888 N

Solve the force triangle: AH = 888 cos33.56° = 740 N AV − 736 = 888 sin33.56° AV = 1227 N ---------------------------------------------------------4-70

Prob. 5.1 By symmetry: RAV = RCV = 5 k RAH = 0 ΣFV = +5 −AB sin45° AB = +7.07 k (comp.) ΣFH = +AC − AB cos45° = 0 AC = 0.707(0.707) = +5.0 k (tens.)

-----------------------------------------------------------Prob. 5.2

ΣMA = −5(10)−10(8)+RCV(24) = 0 RCV = +5.42 k ↑ ΣMC = −RAV(24) + 10(16) −5(10) = 0 RAV = +4.58 k ↑ ΣFH = −RAH + 5 = 0 RAH = +5 k ←

 5  ΣFV = +4.58−AB   =0  6.40  AB = 5.86 k (comp.)

 4   =0  6.40 

ΣFH = −5+AC−5.86 

AC = +8.66 k (tens.)

 4   8   −BC  =0  6.40   9.434 

ΣFH = +5+5.86 

BC = +10.21 k (comp.)

5-71

-----------------------------------------------------------Note: Probs. 5.3 through 5.7, only summary solutions are shown.

-----------------------------------------------------------Prob. 5.3 Units are kips

-----------------------------------------------------------Prob. 5.4

Units are kN

-----------------------------------------------------------Prob. 5.5

------------------------------------------------------------

5-72

Prob. 5.6

------------------------------------------------------------Prob. 5.7 Units are kN

-------------------------------------------------------------Note: Probs. 5.8 through 5.15: Only summary solutions shown.

-------------------------------------------------------------Prob. 5.8

--------------------------------------------------------------

5-73

Prob. 5.9

-------------------------------------------------------------Prob. 5.10 (units are kN)

-------------------------------------------------------------Prob. 5.11 (units are kips)

--------------------------------------------------------------

5-74

Prob 5.12

--------------------------------------------------------Prob. 5.13

-------------------------------------------------------------Prob. 5.14

---------------------------------------------------------5-75

Prob. 5.15

-------------------------------------------------------------Prob. 5.16

ΣMA = 0  RFV = +15.75 k↑ ΣMF = 0  RAV = +8.25 k↑ Cut section a-a, use left portion:

5-76

ΣMH = −8.25(20) + CD(9) = 0 CD = +18.33 k (comp.) ΣMD = −8.25(25) + 10(5) + HI(9) = 0 HI = +17.36 k (tens.) ΣFV = + 8.25 −10 + HD(9/10.3) = 0 HD = +2.0 k (tens.)

-------------------------------------------------------------Prob. 5.17

ΣMA = 0  RGH = +1949 lb → ΣMG = 0  RAH = + 2699 lb ← ΣFV = 0  RAV = +1299 lb ↑ Cut section a-a, use right portion:

ΣME = +BC(2.67) −1299(4) −750(2.67) = 0 BC = + 2697 lb (tens.) ΣMB = −1299(8) + FE(8 sin33.69°) = 0 FE = +2342 lb (comp.) ΣFV = −1299 + 2342(sin33.69°) + BE sin33.69° = 0 BE = 0 ------------------------------------------------------------5-77

Prob. 5.18

ΣME = 0  RA = 72.19 kN↑ Cut section a-a, use the left portion. (BC assumed tensile.) ΣMH = −72.19(8) + 45(4) + BC(3) = 0 BC = +132.5 kN (tens.)

Cut section b-b, use the left portion. (CH assumed compressive.) ΣFV = +72.19 −45 −60 +CH = 0 CH = 32.8 kN (comp.)

Cut section c-c, use left portion. (CG assumed tensile.) ΣFV = +72.19 −45 −60 +CGV = 0 CGV = 32.81 kN (tens.)

CGV CG = 3 5  CG = 54.7 kN (tens.)

-------------------------------------------------------------Prob. 5.19

CC moment: +; ↑→ +

Cut section a-a, use the left portion: ΣMA = 0  CI = 0 ΣMB = −4000(8) + IJ(4) = 0 5-78

IJ = + 8000 lb (tens.)

 2   (4) = 0 ΣMH = −4000(8) + BC   5 BC = +8944 lb (comp.)

-------------------------------------------------------------Prob. 5.20

ΣMA = 0  RC = 24.17 kN↑ ΣMC = 0  RAV = 9.167 kN ↓ ΣFH = 0  RAH = 10 kN ← Cut section a-a , use portion to the left. (BC is assumed compressive.) ΣMD = −10(2) + 9.167(3) − BC(2) = 0 BC = 3.75 kN (comp.)

(Assume DE tensile:) ΣMC = +9.167(6) −DE(2) = 0 DE = 27.5 kN (tens.)

Cut section b-b, use portion to the left. (CE assumed compressive:) ΣFV = −9.167 + 24.17 − CE CE = +15.0 kN (comp.)

--------------------------------------------------------------

5-79

Prob. 5.21

Cut section a-a, use the right-hand portion. (BC & BG assumed tensile, FG assumed compressive.) ΣMB = −12(10) + FG(10) = 0 FG = +12 k (comp.) ΣMG = −12(5) + BC(10) = 0 BC = + 6 k (tens.)

 2   = 0 ΣFV = −12 + BG   5 BG = +13.4 k (tens.)

-------------------------------------------------------------Prob. 5.22

CC moment: +; ↑→ +

Cut section a-a, use the portion to the left. (Assume CD compressive, BE &BD tensile.) ΣMB = 0  CD = 15 kN (comp.) ΣFV = 0  BD = 0 ΣFH = 0  BE = 20 kN (tens.)

5-80

Cut section b-b, use the portion to the left.

(Assume CB tensile.) ΣFH = 0

CBH = 20 kN→ CB 2

=

CBH 1

CB = 2CBH CB = 28.3 kN (tens.)

-------------------------------------------------------------Prob. 5.23

ΣML = 0  RIV = +26.7 k ↑ ΣMI = 0  RLV = +38.3 k ↑ Cut section a-a, use the left portion:

5-81

ΣMK = 0  CD = +14.5 k (tens.) ΣFV = 0  CK = +13.3 k (comp.)

For member EM, cut section b-b, use the right-hand portion:

 4  ΣFV = −10 + EM  = 0  41  EM = +16 k (tens) ------------------------------------------------------------Prob. 5.24

ΣMA = 0  RDV = +1667 lb↑ ΣMD = 0  RAV = + 1333 lb ↑

5-82

Cut section a-a, use the left portion:

ΣMB = 0  EF = +1778 lb (tens) ΣMF = −1333(16) +1000(8)

4 + BC  (12) = 0 5 BC = +1389 lb (comp.) ΣFy = +1333 −1000−1389(3/5) + BF (3/5) = 0 BF = +833 lb (comp.)

-------------------------------------------------------------Prob. 5.25

ΣMA = 0  REH +45.83 k → ΣME = 0  RAH = +5.83 k ← ΣFV = 0  RAV = + 35 k ↑ 5-83

For member AF:cut section a-a, use the left portion:

 6  =0 ΣFV = +35 −AF   61  AF = +45.6 k (tens.)

For members BG & FG: cut section b-b , use the right-hand portion:

 6  =0 ΣFV = −20 + BG  61  BG = +26 k (tens.) ΣMB = 0  FG = +16.7 k (comp.)

---------------------------------------------------------------

5-84

Prob. 5.26

ΣMA = 0  RGV = +36 k↑ ΣMG = 0  RAV = +4 k↑ Cut section a-a, use left portion:

Resolve force BH into components at B:

 1   (40) = 0 ΣMD = −4(40) + BH   5 BH = + 8.94 k (tens.)

5-85

Cut section b-b, use the left portion:

ΣMG = 0  CD = +24 k (tens.) ΣMC = 0  HG = +8 k (tens.)

-------------------------------------------------------------Prob. 5.27

ΣMA = 0  RDV = +8.67 k↑ ΣMD = 0  RAV = +7.33 k↑

5-86

Cut section a-a, use left portion:

ΣFV = 0  BE = +1.33 k (tens.) (more) For member CE, cut section b-b, use left portion:

 5  ΣFV = +7.33 − 6 − CE =0  41  CE = +1.703 k (comp.)

Cut section c-c around joint F. ΣFv = CF = 0

-----------------------------------------------------

5-87

Prob. 5.28

ΣMA = 0  RJV = +26 k↑ ΣMJ = 0  RAV = +14 k ↑ ΣFH = 0  RAH = +10 k → Cut section a-a, use right-hand portion: Resolve CE into components at E

 1  ΣMH = −16(8) +10(8) + CE =0  2 CE = +8.49 k (tens.)

 1  ΣFv = −24 −16 −8.49   + EH = 0  2 EH = +46 k (comp.)

5-88

Cut section b-b, use lower portion:

ΣMH = 0  BC = +6 k (tens.)

Cut section c-c, use lower portion:

ΣMI = 0  AB = + 4 k (comp.)

-------------------------------------------------------------Prob. 5.29

By symmetry: RCV = RFV = 3600 lb ↑ Cut section a-a, use lower portion:

5-89

Resolve force CB into components at B: ΣME = 0  CB = +10,182 lb (comp.)

 1   1  =0 ΣFV = 3600−10182   + CD  2  5 CD = +8050 lb (tens.)

Cut section b-b, use right-hand portion:

Resolve force CD into components at C: ΣMF = 0  AD = +10,800 lb (tens.)

--------------------------------------------------------Prob. 5.30

5-90

ΣMA = 0  EV = +833 lb ↑ ΣME = 0  AV = +1167 lb ↑

ΣMB = 0  DV = +667 lb ↑ ΣMD = 0  BV = +1333 lb ↑

ΣFV = 0  CV = + 166 lb↑ ΣMB = 0  CH = +751 lb ← ΣFH = 0  BH = +751 lb→

ΣFH = 0  DH = +751 lb ← Pin Reactions (lb) Pin

V

H

React.

A

1167

−

1167

B

1333

751

1530

5-91

C

166

751

769

D

667

751

1004

E

833

0

833

-------------------------------------------------------------Prob. 5.31

ΣMC = 0

 BH = 600 N← ΣFH = 0

CH = 100 N→ ΣFV = 0

 BV = CV

ΣFH = 0  AH = 600 N ← ΣMA = 0  BV = 100 N ↑ ΣFV = 0  AV = 100 N↑ 5-92

(from above) BV = CV  CV = 100 N↑ Pin Reactions (N) Pin

V

H

React.

A

100

600

608

B

100

600

608

C

100

100

141.4

-------------------------------------------------------------Prob. 5.32

ΣMA = 0  FV = +517lb↑ ΣMF = 0  AV = +27 lb ↓ ΣH = 0  AH = 80 lb←

5-93

ΣMB = 0  EV = + 440 lb ↑ ΣFV = 0  BV = +80 lb ↓ ΣFH = 0  BH = EH

ΣMC = 0 BH = +160 lb → ∴EH = 160 lb ΣFH = 0  CH = +80 lb ← ΣFV = 0  CV = +53 lb↓ Pin Reactions (lb) Pin

V

H

Reaction

A

27

80

84.4

B

80

160

178.9

C

53

80

96.0

D

130

80

*

E

440

160

468

F

517

0

517

* V and H are applied loads.

--------------------------------------------------------------

5-94

Prob. 5.33

ΣMA = 0  DH = +6855 lb → ΣMD = 0  AH = 6855 lb ←

ΣMD = 0  BV = +4500 lb ↑ ΣFV = 0  DV = +1500 lb ↓ ΣFH = 0  +6855 lb ←

ΣFV = 0  AV = +7000 lb ↑ Pin Reactions (lb)

5-95

Pin

V

H

Reaction

A

7000

6855

9800

B

4500

6855

8200

D

1500

6855

7020

-------------------------------------------------------------Prob. 5.34

ΣMB = 0  AH =+11.6 k → ΣMA = 0  BH = 11.6 k ←

By symmetry: BV = CV =2 k ↑ ΣFH = +11.6 k →

5-96

ΣFV = 0  AV = +8 k ↑ Pin Reactions (kips) Pin

V

H

Reaction

A

8

11.6

14.1

B

2

11.6

11.8

C

2

11.6

11.8

-------------------------------------------------------------Prob. 5.35 W = mg = 500(9.81) = 4.91 kN

5-97

ΣMC = 0  AB = 2.34 kN ↑ ΣFV = 0  CV = 2.57 kN ↑ ΣFH = 0  CH = 2.84 kN → Pin reaction at C : = 2.57 2 + 2.84 2 = 3.83 kN

-------------------------------------------------------------Prob. 5.36

5-98

ΣMC = 0  BH = +3.25 k ← ΣMB = 0  CH = +0.05 k →

ΣMA = 0  BV = +1 k ↑ ΣFV = 0  AV = +1 k ↑ ΣFH = 0  AH = +3.25 k →

ΣMA = 0  CV = +3.4 k ↑ Pin reactions (kips) Pin

V

H

Reaction

A

1.0

3.25

3.40

B

1.0

3.25

3.40

C

3.4

0.05

3.40

--------------------------------------------------------------

5-99

Prob. 5.37

From Prob. 5.7:

EH = 30 kN ←,

EV = 46.7 kN ↑, FV = 153.3 kN ↑ Cut section a-a, use lower portion:

ΣFH = 0  ED = +50 kN (tens.) ΣMD = 0  CE = +86.7 kN (comp.)

-------------------------------------------------------------Prob. 5.38

5-100

ΣMF = 0  AV = +31.88 kN ↑ ΣMA = 0  FV = +28.12 kN ↑ ΣFH = 0  AH = +10 kN

Resolve force BD into components at D: ΣME = − 31.88(8) + 20(4) + BD(4/5)(6) =0 BD = +36.5 kN (comp.)

Resolve force BE into components at E: ΣMA = −20(4) + BE(3/5)(8) = 0 BE = +16.7 kN (comp.)

-------------------------------------------------------------Prob. 5.39

5-101

ΣMA = 0  BH = +3150 lb → ΣFH = 0  AH = +3150 lb← ΣFV = 0  AV = +7100 lb lb ↑

ΣMC = 0  EV = +8100 lb ↑ ΣFV = 0  CV = +2300 lb ↓ ΣFH = 0  CH = EH

ΣMD = 0  EH = +6300 lb → ∴CH = 6300 lb ΣFV = 0  DV = +8700 lb ↑ ΣFH = 0  DH = 6300 lb ←

5-102

Pin Reactions (lb) Pin

V

H

Reaction

A

7100

3150

7767

B

−

3150

3150

C

2300

6300

6707

D

8700

6300

10742

E

8100

6300

10262

------------------------------------------------------------Prob. 5.40

ΣMA = 0  DV = +1000 lb ↑ ΣFV = 0  AV = +600 lb ↓

5-103

ΣFV = 0  BV = +1000 lb↑ ΣMA = 0  BH = +800 lb → ΣFH = +800-AH=0  AH = 800 lb←

ΣFH = 0  DH = +200 lb→

5-104

Pin Reactions Pin

V

H

Reaction

A

600

800

1000

B

1000

800

1281

D

1000

200

1020

-------------------------------------------------Prob. 5.41 Resultant FH = 29/2(3) = 43.5 kN

ΣMA = 0  RE = 29.0 kN ↑ ΣFV = 0  AV = 29.0 kN↓ ΣFH = 0  AH = 43.5 kN ←

5-105

ΣMC = 0  BD = 32.6 kN → ΣFH = 0  CH = 32.6 kN ← ΣFV = 0  CV = 29.0 kN↑ Pin Reactions (kN) Pin

V

H

Reaction

A

29.0

43.5

52.3

B

0

32.6

32.6

C

29.0

32.6

43.6

-----------------------------------------------------Prob. 5.42

ΣMC = 0  AH = +606 lb← ΣFH = 0  CH = +606 lb→

5-106

ΣMB = 0  AV = +950 lb↑ ΣFV = 0  BV = +250 lb ↑ ΣFH = 0  BH = +606 lb →

ΣFV = 0  CV = +1850 lb ↑ Pin Reactions (lb) Pin

V

H

Reaction

A

950

606

1127

B

250

606

656

C

1850

606

1947

---------------------------------------------------

5-107

Prob. 5.43

By symmetry:DV = EV = 140 lb ↑ ΣFH = DH = 0

By symmetry: AV = BV = 140 lb↑ ΣFH = 0  BH = AH

5-108

ΣFV = 0  CV = 0 ΣMB = 0  CH = +280 lb → ΣFH = 0  BH = +280 lb ← and ∴ AH = 280 lb Pin Reactions (lb) Pin

V

H

Reaction

A

140

280

313

B

140

280

313

C

0

280

280

D

140

0

140

E

140

−

140

----------------------------------------------------Prob. 5.44

ΣMA = 0  EV = +4.1 kN↑ ΣME = 0  AV = +1.1 kN ↓ ΣFH = 0  AH = +2 kN ←

5-109

ΣMB = 0  DV = +2.40 kN↑ ΣMD = 0 BV = + 0.6 kN↑ ΣFH = 0  DH = BH

ΣMC = 0  BH = +1.25 kN→ and ∴ DH = 1.25 kN ΣFV = 0  CV = +1.7 kN↑ ΣFH = 0  CH = +1.25 kN← (more) Pin reactions (kN) Pin

V

H

Reaction

A

1.1

2.0

2.28

B

0.6

1.25

1.39

5-110

C

1.7

1.25

2.11

D

2.4

1.25

2.71

E

4.1

−

4.1

-----------------------------------------------------Prob. 5.45

ΣMA = 0 C = 40 lb ΣFV = 0  RA = 52 lb ----------------------------------------------------Prob. 5.46

↑→+

β = cos-1 1/5 = 78.46° α = cos-11/2 = 60°

ΣFV = −P +Q cos78.46° + R cos60° = 0

 −10 + 0.20Q + 0.50R = 0

(I)

ΣFH = +R sin 60°− Q sin 78.46 = 0 5-111

+0.866R − 0.980Q = 0

(II)

Solve eqns. I and II simultaneously: Q = 13.05 lb ΣFH = 0 − F +Q sin 78.46° = 0

 F = 12.80 lb -----------------------------------------------Prob. 5.47

ΣFH = 0

 QH = 100 lb Q=

100 cos 10°

=101.5 lb ΣFV = 0 −P + 2(101.5 sin10°) = 0  P = 35.3 lb

--------------------------------------------------------------

5-112

Prob. 5.48

CC moment:+, ↑→ +

(a) ΣMC = −30(15)+AV(3) = 0  AV = 150 lb

ΣFV = 0  AV = BV

Force on can = 150 lb @ 90° (b)

Cosine law: 92 = x2 + 32 − 2(3)(x)cos 45° x2 −4.243x −72 = 0  x = 10.85 in. sin 45° sin A sin B = = 9 10.85 3  A = 121.52°  B = 13.63°

5-113

ΣMC = +BAy (3) −30(15) = 0  BAy = 150 lb BA =

BAy cos 31.52°

= 176 lb

BAV = BA cos13.63° BAV = 171 lb

-------------------------------------------------------------NOTES

5-114

Prob. 6.1

ΣFy = 0  N = 130 lb ΣFx = 0  F = 34.6 lb (when P = 40 lb) Max. available resistance: F = μsN = 0.30(130) = 39 lb

34.6 < 39 ∴ block will not move.

---------------------------------------------------Prob. 6.2

(a) θ = 0° ΣFy = 0  N = 200 lb Max. available resistance: F = μsN = 0.50(200) = 100 lb

ΣFx = 0  P = 100 lb (b) θ = 20° ΣFy = +N − 200 + P sin 20° = 0 N = 200 − P sin 20°

ΣFx = +P cos 20° − F = 0 F = P cos 20° F = μsN

6-115

P cos 20° = 0.50(200 − P sin 20°)

 P = 90 lb

---------------------------------------------------Prob. 6.3 W = mg = 140(9.81) = 1373 N

μs = 0.38

ΣFy = 0  N = 1373 N F = μsN = 0.38(1373) = 522 N

ΣFx = 0  P = 522 N -----------------------------------------------------Prob. 6.4 (a) Motion up the plane:

ΣFy = 0  N = 80 lb Max. available resistance: F = μsN = 0.40(80) = 32 lb

ΣFx = +P −32 − 100(3/5) = 0 P = 92 lb

6-116

(b) Motion M down n the plane:

ΣFy = +N −100(4 4/5) = 0 N = 80 lb F = μsN 0.40(80) = 32 lb ΣFx = 0  P = 28 2 lb ---------------------------------------------------------------Prob.. 6.5 From m Prob 6-4, th he body willl be at rest fo or 28#≤P P≤98# (a) When W P=40# (assume mo otion is down n the plane) ΣF Σ x = −40 − F+100(3/5)= F =0  F=20 0# # (b) When W P=60 ΣF Σ x=60 – F – 100(3/5)=0 0  F=0 (c) When W P=70# (Assume mo otion is up th he plane) # ΣF Σ x=70 – F – 100(3/5)=0 0  F=10 ----------------------------------------------------------------Prob.. 6.6 On th he verge of sliding s FA=0 0.5(98.1)=49.1 N FB=0 0.15(98.1+19 96.2)=44.1 N Sincee FB<FA, Mo otion will im mpend when P reaches 444.1 N. The blocks will sllide as a unitt.

-----------------------------------------------------------------

6-117

Prob.. 6.7

ΣFy=N – 100 cos 30°-P sin 30°=0 N=0.5P + 86.6 ΣFx=–F – 100 sin n 30+-P cos 30°=0 F=0.866P – 50 F=µsN 0.866 6P – 50=0.25 5(0.6P+86.6 6)  P=96.7 7# ----------------------------------------------------------------Prob.. 6.8

ΣFy=N – 175 Coss 25°-P Sin 25°=0 2 N=0.4226P+ N +158.6 ΣFx = −F−P cos 25°+175 2 sin 25°=0 F=74.0 F – 0.90 063P F= =µsN 74.0 – 0.9063P=0 0.65(0.4226P P+158.6) P = –24.6# Negaative indicatees that P actss to the lect. Therefore nno force P accting to the right is needeed to preven nt the block from f sliding down the pllane.

-----------------------------------------------------------------

6-118

Prob. 6.9

--------------------------------------------------------------Prob. 6.10

ΣFy = 0  N = 45 lb ΣFx = 0  F = 18 lb F = μsN

μs = F/N = 18/45 = 0.40 ------------------------------------------------------------6-119

Prob. 6.11

μs = 0.33

ΣFy = 0  N = 90 lb F = μsN = 0.33(90) = 30 lb

ΣFx = 0  P = 30 lb ------------------------------------------------------------Prob 6.12

--------------------------------------------------------------6-120

Prob. 6.13

-------------------------------------------------------------

6-121

Prob. 6.14

-------------------------------------------------------------

6-122

Prob. 6.15

---------------------------------------------------------------

6-123

Prob.. 6.16

--------------------------------------------------------------Prob.. 6.17

---------------------------------------------------------------

6-124

Prob. 6.18

------------------------------------------------------------Prob. 6.19

--------------------------------------------------------------6-125

Prob. 6.20

---------------------------------------------------------------

6-126

Prob. 6.21

---------------------------------------------------------------

6-127

Prob. 6.22

----------------------------------------------------------

6-128

Prob. 6.23

-------------------------------------------------------------

6-129

Prob. 6.24

---------------------------------------------------------------

6-130

Prob. 6.25

--------------------------------------------------------------

6-131

Prob. 6.26

---------------------------------------------------------------

6-132

Prob. 6.27

---------------------------------------------------------------

6-133

Prob. 6.28

6-134

--------------------------------------------------------------Prob. 6.29

6-135

--------------------------------------------------------------Prob. 6.30

6-136

--------------------------------------------------------------Prob. 6.31

6-137

---------------------------------------------------------------

6-138

Prob. 6.32

6-139

---------------------------------------------------------------

6-140

Prob. 6.33

---------------------------------------------------------6-141

Prob.. 6.34

-----------------------------------------------------------------

6-142

Prob. 6.35

--------------------------------------------------------------6-143

Prob. 6.36 (see prob. 3.32)

Wedge See wedge free-body-digram in prob. 6.32

6-144

-------------------------------------------------------------Prob. 6.37

6-145

---------------------------------------------------------------

6-146

Prob. 6.38

--------------------------------------------------------------6-147

Prob. 6.39

--------------------------------------------------------------Prob. 6.40

------------------------------------------------------------Prob 6.41

--------------------------------------------------------------6-148

Prob. 6.42

---------------------------------------------------------Prob. 6.43

--------------------------------------------------------------Prob. 6.44

--------------------------------------------------------------6-149

Prob. 6.45

--------------------------------------------------------------Prob. 6.46

--------------------------------------------------------------Prob. 6.47

--------------------------------------------------------------6-150

Prob. 6.48

------------------------------------------------------------Prob. 6.49

------------------------------------------------------------6-151

Prob. 6.50

-------------------------------------------------------------Prob. 6.51

6-152

--------------------------------------------------------------Prob. 6.52

---------------------------------------------------------6-153

Prob. 6.53

↑→ +

r = 11 mm, μs = 0.14, p = 6 mm.

Qa = Pr

P=

Qa 225(250) = = 5114 N r 11

 6   p   = 4.96°  = tan −1  2 2 ( 11 ) π r π    

θ = tan −1 

ΣFx = +5114 − F cosθ − N sinθ = 0 (F = 0.14N)  N = 22.6 kN

ΣFy = − C + N cosθ −F sinθ = 0 (N = 22.6 kN; F= 0.14 N)  C = 22.2 kN

-------------------------------------------------Prob. 6.54 r = 1.50 in.., μs = 0.15, p = 0.25 in., a = 20 in.

 0.25   p   = 1.519°  = tan −1   2π (1.50)   2π r 

θ = tan −1 

6-154

ΣFy = −4000 + N cosθ −F sinθ = 0 −4000 + N(0.9996) −0.15N (0.02651) = 0  N = 4018 lb

(more)

ΣFx = +P − N sinθ − Fcosθ = 0 +P − 4018(0.02651) − 0.15(4018)(0.9996) = 0  P = 709 lb

Qa = Pr

Q=

709(1.50) = 53.2 lb 20

-------------------------------------------------Prob. 6.55

-------------------------------------------------------------6-155

Prob. 6.56

ΣFx = + 2541 − N sinθ − Fcosθ = 0 +2541 −N(0.0746) − 0.13N (0.9972) = 0 N = 12,440 lb

ΣFy = −W + N cosθ −F sin θ = 0 − W +12,440(0.9972) − 0.13(12,440)(0.0746) = 0 W = 12,280 lb

(b) μs = 0.18 (same sketch as part (a)) ΣFx = +2541 − N sinθ − F cosθ = 0 +2541 − N (0.0746) − 0.18N (0.9972) = 0 N = 9996 lb

ΣFy = −W + N cosθ − F sinθ = 0 − W + 9996(0.9972) − 0.18(9996)(0.0746) = 0 W = 9834 lb

6-156

Force is decreased: 12,280 − 9834 = 2446 lb − W + 9996(0.9972) − 0.18(9996)(0.0746) = 0 W = 9834 lb

Force is decreased: 12,280 − 9834 = 2446 lb -----------------------------------------------------

6-157

Prob. 7.1

C.I.: 450 pcf Consider W2 to be negative.

W1 is the weight of the solid 1′×5′ shaft.

W1 = 0.7854(1)2 (5)(450) = 1767 lb W2 = − 0.7854(0.5)2(3)(450) = −265 lb

x=

1767(2.5) − 265(1.5) = 2.68 ft 1767 − 265

----------------------------------------------------Prob. 7.2 Magnesium alloy: 112 pcf Weight W3 = 0.7854(0.5)2(3)(12) = 66.0 lb

x=

1767(2.5) − 265(1.5) + 66(1.5) = 2.63 ft 1767 − 265 + 66.0

------------------------------------------------------------Prob. 7.3

7-158

Wt. of bar: W1 =

0.7854 (1.5) 2 ( 24 ) (165) = 4.05 lb 1728

Wt. of sphere: W2 =

4

3

π (2.5) 2 1728

( 490 ) = 18.56 lb

R.A. on the left side:

x=

4.05(1) + 18.56(26.5) = 23.9 in.

4.05 + 18.56

---------------------------------------------Prob. 7.4

Comp.

d (in.)

L (in.)

Wt. (lb)

1

5

6

33.4

2

3

8

16.04

3

1

4

0.891

Total 50.3 33.4(3) + 16.04(10) + 0.891(16) 50.3 = 5.46 in.

x=

-----------------------------------------------------

7-159

Prob. 7.5

W1 =

0.7854 (1.0) 2 l ( 40 ) = 0.0182 l lb 1728

W2 =

0.7854 (3.0) 2 (6) ( 40) = 0.9817 lb 1728

𝑙 0.9817(𝑙 + 1.5) + 0.0182 2 𝑥̅ = 0.9817 + 0.0182𝑙

also: x = l − 6 Solve for l : l = 35.1 in.

--------------------------------------------------Prob. 7.6

W460 × 60:

wt. = 60(9.81) = 589 N/m d = 455 mm tw = 8.00 mm

Steel: mass: 490×16.02 = 7.85×103 kg/m3 unit wt.: 7.85×103 (9.81) = 77.0 kN/m3 7-160

Consider hole to be a negative volume: Wt. = 0.7854(300)2(8)×10-9(77×103 N/m3) = (−) 43.5 N

3(589)(6.5) + 8(589)(4) − 43.5(1) 3(589) + 8(589) − 43.5 = 4.71 m 5(589)(228) − 43.5(228) + 3(589)(683) y= 8(589) − 43.5 + 3(589) = 353 mm

x=

--------------------------------------------------Prob. 7.7

63 (490) = 61.25 lb W1 (steel) = 1728 0.7854(2) 2 (5) (490) = 4.45 lb W2 (steel) = 1728 0.7854(2) 2 (5) (710) = 6.45 lb W3 (lead) = 1728

y1 = 3 in., y2 = 3.5 in.

61.25(3) − 4.45(3.5) + 6.45(3.5) 61.25 − 4.45 + 6.45 = 3.02 in.

y=

------------------------------------------------------

7-161

Prob. 7.8

y=

(Refer to Prob. 7.18)

ΣWy 61.25(3) − 4.45(3.5) = 2.96 in. = 61.25 − 4.45 ΣW

-----------------------------------------------------Prob. 7.9 (Neglect attachment of handle to head.)

0.7854 (3) 2 (5) ( 490 ) = 10 .02 lb 1728 0.7854 (1) 2 (36) W2 = (80 ) = 1.31 lb 1728 W1 =

ΣW = 11.33 lb x1 = 37.5 in., x2 = 18 in.

x=

10.02(37.5) + 1.31(18) = 35.2 in.

11.33

---------------------------------------------------------Prob. 7.10

7-162

y=

60(5)(14.5) + 12(12)(6) = 11.74 in.

60(5) + 12(12)

---------------------------------------------------Prob. 7.11

Comp.

A (in.2)

y (in.)

Ay (in.3)

1

25.0

3.33

83.25

2

30.0

5.00

150.0

3

−6.28

3.00

−18.84

4

−3.14

7.00

−22.0

Σ45.6

y=

Σ192.4

192.4 = 4.22 in.

45.6

-------------------------------------------------

7-163

Prob. 7.12

3 × 10(S4S) A = 23.1 in.2 4 × 4(S4S) A = 12.3 in.2

2(23.1)(4.625) + 12.3(1.75) 2(23.1) + 12.3 = 4.02 in.

y=

-----------------------------------------------Prob. 7.13 (a)

A1 = 50(30) = 1500 mm2 A2 = ½ π (15)2 = 353 mm2

y=

1500(25) + 353(56.37) = 31.0 mm 1500 + 353

7-164

(b)

W4 × 13: A = 3.83 in.2 d = 4.16 in.

PL ¾ × ¾ : A = 0.563 in.2

y=

3.83(2.08) + 0.563(4.54) = 2.40 in.

3.83 + 0.563

------------------------------------------------Prob. 7.14 (a)

Σ:

A

y

x

Ay

Ax

(in.2)

(in.)

(in.)

(in.3)

(in.3)

1

4

11.5

2

46

8

2

9

6.5

0.5

58.5

4.5

3

12

1

3

12

36

116.5

48.5

25

7-165

𝑥=

48.5 = 1.94 in.

25

𝑦=

116.5 = 4.66 in.

25

(b)

A

y

x

Ay

Ax

(in.2)

(in.)

(in.)

(in.3)

(in.3)

1

18

2.25

2

40.5

36

2

4.5

1.5

4.67

6.75

21.0

3

−3.14

0.849

0.849

−2.67

−2.67

Σ

19.36

44.6

54.3

𝑥=

54.3 = 2.80 in.

19.36

𝑦=

44.6 = 2.30in.

19.36

(c)

7-166

A

y

x

Ay

Ax

(in.2)

(in.)

(in.)

(in.3)

(in.3)

1

72

4.0

4.5

288

324

2

−9

6.5

1.5

−58.5

−13.5

3

−12.57

3.0

5.0

−37.7

−62.9

Σ

50.4

191.8

248

𝑥=

248 = 4.92 in.

50.4

𝑦=

191.8 = 3.81 in.

50.4

(d)

A

y

x

Ay

Ax

(in.2)

(in.)

(in.)

(in.3)

(in.3)

1

36

15.33

6.0

552

216

2

252

7.0

9.0

1764

2270

3

−54

2.0

9.0

−108

−486

Σ

234

2210

2000

𝑥=

2000 = 8.55 in.

234

𝑦=

2210 = 9.44 in.

234

--------------------------------------------------7-167

Prob. 7.15

A (in.2)

y (in.)

Ay (in.3)

1

14.7

9.0

132.3

2

6.0

18.25

109.5

Σ

20.7

Σ

241.8

y=

ΣAy 241.8 = = 11.68 in. ΣA 20.7

-----------------------------------------------------Prob.7.16 (a)

a1 = ½ (10)(12) = 60 in.2 a2 = ½ (0.7854)(4)2 = −6.28 in.2 Σa = 53.72 in.2 x1 = 8 in., x2 = 10 in., y1 = 3.33 in.

y2 =

4(2) = 0.849 in.

3π

7-168

x=

60(8) − 6.28(10) = 7.77 in.

53.72

y=

60(3.33) − 6.28(0.849) = 3.62 in.

53.72

(b)

a1 = 6 in.2, a2 = 11 in.2, a3 = 6 in.2 x1 = 3 in., x2 = 5.5 in., x3 = 8 in. y1 = 12.5 in., y2 = 6.5 in., y3 = 0.5 in.

6(3) + 11(5.5) + 6(8) = 5.5 in.

6 + 11 + 6 6(12.5) + 11(6.5) + 6(0.5) y= = 6.5 in.

6 + 11 + 6

x=

----------------------------------------------------Prob. 7.17

W1 = 3.5×103(9.81) = 34.3 kN W2 = 12×103(9.81) = 117.7 kN W3 = 14×103(9.81) = 137.3 kN 7-169

ΣW = 289 kN 𝑥=

34.3(0) + 117.7(4.5) + 137.3(14.5) = 8.72 m 289

----------------------------------------------------Prob. 7.18

y=

ΣAy 9100(261.5) + 42100(532.85) = ΣA 9100 + 42100 = 485 mm

----------------------------------------------------Prob. 7.19

a1 = 18.3 in.2, a2 = 7.34 in.2, Σa = 25.64 in.2 y1 = 21.0 /2 = 10.5 in. y2 = 21.0 + 0.387 − 0.674 = 20.7 in.

y=

18.3(10.5) + 7.34(20.7) = 13.42 in.

25.64

-----------------------------------------------------

7-170

Prob. 7.20

a (m2)

y (m)

ay (m3)

1

0.13

0.45

0.0585

2

0.120

0.20

0.024

3

0.040

0.267

0.0107

Σ 0.290

y=

0.0932

Σay 0.0932 = = 0.321 m Σa 0.290

----------------------------------------------------Prob. 7.21

7-171

14,100(180.5) + 3610(180.5) + 2170(226.5) 14,100 + 3610 + 2170 = 185.5 mm

y=

3610(5) + 14,100(138.5) + 2170(245.7) 3610 + 14,100 + 2170 = 126 mm

x=

-----------------------------------------------------------Prob. 7.22 (a)

A

y

x

Ay

Ax

(in.2)

(in.)

(in.)

(in.3)

(in.3)

1

288

6

12

1728

3456

2

144

16

16

2304

2304

3

226

12

29.09

2712

6574

Σ

658

6744 12330

ΣAx 12,330 = = 18.74 in. ΣA 658 ΣAy 6744 y= = = 10.25 in. ΣA 658 x=

7-172

(b)

2.96(2.50) + 2(1.19)(0.711) 2.96 + 2(1.19) = 1.703 in.

y=

----------------------------------------------------------Prob. 7.23

a1 = 12h, a2 = ½ (10)(12) = 60 in.2 y1 = 10 + h/2 y2 = 6.67 in. Σay Σa 12h(10 + h / 2) + 60(6.67) 10 = 60 + 12h y=

Solve for h: h = 5.77 in.

------------------------------------------------------

7-173

Prob. 7.24 (a)

A

y 2

x

Ay

Ax 3

(in. )

(in.)

(in.)

(in. )

(in.3)

1

20

7

1

140

20

2

16

1

4

16

64

Σ

36

x=

156 84 = 2.33 in.; y = = 4.33 in.

36 36

156

84

(b)

A

y 2

x

Ay

Ax 3

(in. )

(in.)

(in.)

(in. )

(in.3)

1

98

3.5

7

343

686

2

30

2.5

17

75

510

Σ

128

418

1196

7-174

x=

196 = 9.34 in.; 128

y=

418 = 3.27 in.

128

(c)

A

y

x

Ay

Ax

(in.2)

(in.)

(in.)

(in.3)

(in.3)

1

100

5

5

500

500

2

30

12

3.33

360

100

3

−3.14

5

5

−15.7

−15.7

Σ

126.9

844

584

𝑥=

584 = 4.60 in.; 126.9

𝑦=

844 = 6.65 in.

126.9

--------------------------------------------------------------Prob. 7.25

(a) a1 = 14.1 in.2, a2 = 11.7 in.2 y1 = 13.8/2 = 6.90 in., y2 = 13.94 in.

y=

14.1(6.90) + 11.7(13.94) = 10.09 in.

14.1 + 11.7 7-175

(b)

A

y

x

Ay

Ax

(in.2)

(in.)

(in.)

(in.3)

(in.3)

1

24.4

10.70

4.18

261.1

102.0

2

8.81

14.73

9.44

129.8

83.2

Σ

33.21

390.9

185.2

x=

185.2 = 5.58 in.; 33.21

y=

390.9 = 11.77 in.

33.21

(c)

A

y

x

Ay

Ax

(in.2)

(in.)

(in.)

(in.3)

(in.3)

1

5.87

5.0

2.13

29.4

12.5

2

11.0

1.86

4.60

20.5

50.6

Σ

16.87

49.9

63.1

x=

63.1 = 3.74 in.; 16.87

y=

49.9 = 2.96 in.

16.87

--------------------------------------------------------------7-176

Prob. 8.1 (a)

bh 3 12( 24) 3 = = 13824 in. 4 12 12 π d 4 π (8) 4 I x2 = = = 201 in. 4 64 64 I x = 13,824 − 201 = 13623 in. 4 I x1 =

(b)

Ix =

π 64

(d 4 − d14 ) = 527 in.4

(c)

Ix =

100 4 40(70) 3 − = 7.19 × 10 6 mm 4 12 12

----------------------------------------------------

8-177

Prob. 8.2 8(12) 3 = 384 in. 4 36 8(12) 3 I xb = 12 = 1152 in. 4 Ix =

----------------------------------------------------Prob. 8.3 I x1 = 1170 in.4 1(15) 3 12 = 563 in.4

I x 2 = (2)

Ix = 1170 + 563 = 1733 in.4 ----------------------------------------------------Prob. 8.4

Ix =

1 1 (390)(190) 3 − 2 (95)(110) 3 12  12  1 − 2 (40)(110) 3 = 193.0 × 10 6 mm 4  12 

--------------------------------------------------------------

8-178

Prob. 8.5 (a) a = 18 in.2 Ixb = aΣy2 Ixb = 18(10.52 + 7.52 + 4.52 + 1.52) = 3402 in.4 (b) a = 12 in.2 Ixb = aΣy2 Ixb = 12(112 + 92 + 72 + 52 + 12 = 3432 in.4 -----------------------------------------------------Prob. 8.6 I xb =

6(12) 3 = 3456 in. 4 3

% Error: 3402 − 3456 × 100% = −1.56% 3456 3432 - 3456 8 - 5(b) : × 100% = −0.69% 3456 8 - 5(a) :

----------------------------------------------------Prob. 8.7

Ix =

Nominal 300 × 610

292(597) 3 = 5.178 × 10 9 mm 4 12

Tabulated value: App. E-2 (SI): 5180×106 mm4 (O.K.) ----------------------------------------------------8-179

Prob. 8.8

W18 × 71: tw = 0.495 in.

Iy = 60.3 in.4

I Y = 60.3 + (2)

15(1) 3 + 2(15)(0.748) 2 12

= 79.6 in.4 ---------------------------------------------------Prob. 8.9 (a)

a1 = 10(8) = 80 in.2 a2 = 0.7854(8)2(0.5) = 25.1 in.2 (each) 10(8) 3 = 427 in.4 12 π (8) 4 (0.5) = 100.5 in 4 I x2 = 64 I x1 =

Ix = 427 + 100.5(2) = 628 in.4 I y1 =

8(10) 3 = 667 in.4 12

Iy2 = 2(0.1098(4)4 + 25.1(6.7)2) = 2310 in.4 Iy = 667 + 2310 = 2980 in.4

8-180

(b)

a1 = 5 in.2, a2 = 8 in.2 5(1) 3 + 5(4.5) 2 = 101.7 in.4 12 1(8) 3 I x2 = = 42.7 in.4 12 I x1 =

Ix = 2(101.7) + 42.7 = 246 in.4 1(5) 3 + 5(2) 2 = 30.4 in.4 12 8(1) 3 I y2 = = 0.67 in. 4 12 I y1 =

Iy = 2(30.4) + 0.67 = 61.5 in.4 (c)

a1 =2(10) = 20 in.2 , y1 = 7 in. a2 = 2(14) = 28 in.2, y2 = 1 in.

y=

Σay 20(7) + 28(1) = = 3.5 in. Σa 20 + 28

8-181

2(10) 3 I x1 = + 20(3.5) 2 = 412 in.4 12 14(2) 3 I x2 = + 28(2.5) 2 = 184.3 in.4 12 Ix = 412 + 184.3 = 596 in.4 10(2) 3 = 6.67 in.4 12 2(14) 3 I y2 = = 457 in.4 12 I y1 =

Iy = 6.67 + 457 = 464 in.4 (d)

a1 = 100 in.2, a2 = 50 in.2 y=

100(5) + 2(50)(3.33) = 4.17 in.

100 + 2(50)

10(10) 3 + 100(0.83) 2 = 902 in.4 12 10(10) 3 I x2 = + 50(0.83) 2 = 313 in.4 36 I x1 =

Ix = 902 + 2(313) = 1528 in.4 10(10) 3 I y1 = = 833 in.4 12 10(10) 3 I y2 = + 50(8.33) 2 = 3747 in.4 36 Iy = 833 + 2(3747) = 8327 in.4 ----------------------------------------------------------

8-182

Prob. 8.10 (a)

For the semi-circular cutout: 4(45 / 2) = 30.45 mm 3π Ac = 0.7854(45) 2 / 2 = 795 mm 2 y c = 40 −

I ox = 0.1098 R 2 = 28.1 × 10 3 mm 4 I oy = π R 4 / 8 = 100.6 × 10 3 mm 4 80 4 − 2(28.1 × 10 3 ) − 2(795)(30.45) 2 IX = 12 = 1.883 × 10 6 mm 4 IY =

80 4 − 2(100.6 × 10 3 ) = 3.21 × 10 6 mm 4 12

(b)

For the hole: A = 7854 mm2 , I = 4.91×106 mm4 8-183

y=

300(600)(300) − 7854(420) = 294.5 mm 300(600) − 7854

Ix =

300(600) 3 + 300(600)(5.5) 2 − 4.91× 10 6 12 − 7854 × 125.5) 2 = 5.28 × 10 9 mm 4

600(300) 3 − 4.91 × 10 6 12 = 1.345 × 10 9 mm 4

Iy =

-------------------------------------------------------------Prob. 8.11

a1 = 12(24) = 288 in.2, y1 = 6 in. a2 = −16 in.2 , y2 = 4 in.

288(6) − 16(4) = 6.12 in.

288 − 16 X - X axis : y=

24(12) 3 + 288(0.12) 2 = 3460 in.4 12 4(4) 3 I x2 = + 16(2.12) 2 = −93 in.4 12 I x1 =

Ix = 3460 − 93 = 3370 in.4 X'-X' axis: 24(12) 3 = 13,820 in.4 3 4(4) 3 I x '2 = + 16(4) 2 = −277 in.4 12 I x '1 =

Ix' = 13,820 − 277 = 13,540 in.4 ------------------------------------------------------

8-184

Prob. 8.12

a1 = 27 in.2,

y1 = 15.5 in.

2

a2 = 11.8 in. , y2 = 7.5 in.

y=

27(15.5) + 2(11.8)(7.5) = 11.77 in.

27 + 2(11.8)

27(1) 3 + 27(3.73) 2 = 378 in.4 12 I x 2 = 348 + 11.8(4.27) 2 = 563 in.4 I x1 =

Ix = 378 + 2(563) = 1504 in.4 1(27) 3 = 1640 in.4 12 I y 2 = 9.17 + 11.8(8.78) 2 = 919 in.4 I y1 =

Iy = 1640 + 2(919) = 3480 in.4 ----------------------------------------------------Prob. 8.13 (refer to Prob. 7.18)

a1: (W530 × 72)

a2: (W840 × 329)

A = 9100 mm2

A = 42,100 mm2

Ix = 399×106 mm4

Ix = 5370×106 mm4

Iy = 16.1×106 mm4

Iy = 350×106 mm4

d = 523 mm

tw = 19.7 mm 8-185

Ix1 = 399×106 + 9100(223.5)2 = 854×106 mm4 Ix2 = 350×106 + 42,100(47.85)2 = 446×106 mm4 Ix = 854×106 + 446×106 = 1.300×109 mm4 Iy1 = 16.1×106 mm4 Iy2 = 5370×106 mm4 Iy = 16.1×106 + 5370×106 = 5.39×109 mm4 -----------------------------------------------------------Prob. 8.14

a1 = 24 in.2, y1 = 13 in. a2 = 24 in.2, y2 = 6 in. y=

24(13) + 2( 24)(6) = 8.33 in.

24 + 2(24)

12(2)3 + 24(4.76) 2 = 531 in.4 12 2(12) 3 I x2 = + 24(2.33) 2 = 418 in.4 12 I x1 =

Ix = 531 + 2(418) = 1367 in.4 --------------------------------------------------------Prob. 8.15

8-186

C12 × 25: A = 7.34 in.2

x = 0.674 in.

Ix = 144 in.4 Iy = 4.45 in.4 For 2 channels: IX =2(144) = 288 in.4 For Y-Y axis: 144 = 4.45 + 7.34x2  x = 4.360 in.

s/2 = x − 0.674  s = 7.37 in.

------------------------------------------------------Prob. 8.16

W150X24: A = 3060 mm2 d = 160 mm Ix = 13.4×106 mm4 Iy = 1.84×106 mm4 y=

3060(99) + 19(150)(9.5) = 55.84 mm 3060 + 19(150)

d1 = 99 − 55.84 = 43.16 mm d2 = 55.84 − 9.5 = 46.34 mm

150(19) 3 I x = 13.4 ×10 + 3060(43.12) + 12 2 + 19(150)(46.34) = 25.3 ×10 6 mm 4 6

2

8-187

19(150) 3 I y = 1.84 × 10 + = 7.18 × 10 6 mm 4 12 6

-----------------------------------------------------Prob. 8.17

a1 = 8(3/4) = 6.0 in.2, x1 = 0.375 in., y1 = 4 in. a2 = 5.25(3/4) = 3.94 in.2 x2 = 3.375 in., y2 = 0.375 in. Σa = 9.94 in.2 6( 4) + 3.94(0.375 = 2.56 in.

9.94 6(0.375) + 3.94(3.375) x= = 1.56 in.

9.94 y=

X-X axis: 0.75(8) 3 + 6(1.44) 2 = 44.4 in.4 12 5.25(0.75) 3 I x2 = + 3.94(2.185) 2 = 19.0 in.4 12 I x1 =

Ix = 44.4 + 19.0 = 63.4 in.4 Y-Y axis: 8(0.75) 3 + 6(1.185) 2 = 8.71 in.4 12 0.75(5.25) 3 I y2 = + 3.94(1.185) 2 = 22.0 in.4 12 I y1 =

Iy = 8.71 + 22.0 = 30.7 in.4 -----------------------------------------------------8-188

Prob. 8.18

a (in.2)

y (in.)

ay (in.3)

1

84

21.5

1806

2

120

10

1200

3

8

18.67

149.4

Σa = 84+ 120 + 2(8) = 220 in.2 Σay = 1806 + 1200 + 2(149.4) = 3305 in.3

y=

3305 = 15.02 in.

220

28(3) 3 + 84(6.48) 2 = 3590 in.4 12 6(20) 3 I x2 = + 120(5.02) 2 = 7024 in.4 12 4(4) 3 36 I x3 = + 8(3.65) 2 = 113.7 in.4 I x1 =

Ix = 3590 + 7024 + 2(113.7) = 10,840 in.4 -----------------------------------------------------Prob. 8.19

8-189

a1 = 24(0.375) = 9 in.2 , a2 = 3.61 in.2

0.375(24) 3 = 432 in.4 12 I x 2 = [4.86 + 3.61(11.067) 2 ](4) = 1788 in.4 I x1 =

Ix = 432 + 1788 = 2220 in.4 -------------------------------------------------------------Prob. 8.20

a1 = 13.3 in.2, y1 = 5.05 in. a2 = 6.08 in.2, y2 = 9.684 in.

y=

13.3(5.05) + 6.08(9.684) = 6.504 in.

13.3 + 6.08

Ix1 = 248 + 13.3(1.454)2 = 276.1 in.4 Ix2 = 3.86 + 6.08(3.180)2 = 65.3 in.4 Ix = 276.2 + 65.5 = 341.5 in.4 Iy1 = 53.4 in.4 Iy2 = 129 in.4 Iy = 53.4 + 129 = 182.4 in.4 -----------------------------------------------------

8-190

Prob.. 8.21

W8 × 40: A = 11 1.7 in.2, tw = 0.360 in., bf = 8.07 8 in., Ix = 146 in.4, Iy = 49.1 in.4

11 1 .7(4.04) + 3.00(7.22) + 2.50(10.47) 11.7 + 3.00 + 2.50 = 5.53 in.

y=

Ix = 49.1 4 + 11.7(1 1.49)2 +

0.5(6) 3 12

= 153.7 1 in.4 I y = 146 +

6(0.5) 3 0.5(5) 3 + = 151.3 in.4 12 12

------------------------------------------------------------Prob.. 8.22 (a)

C6 × 10.5 A = 3.07 in.2 8-191

bf = 2.03 in. x = 0.500 in.

Ix = 15.1 in.4 Iy = 0.860 in.4 X-X axis:

tw = 0.314 in.

a (in.2)

y (in.)

ay (in.3)

1

4.50

6.405

28.82

2

3.00

4.03

12.09

3

3.09

1.53

4.73

Σ

10.59

y=

45.69 = 4.310 in.

10.59

45.64

d (in.)

Ad2 (in.4)

Ixo (in.4)

1

2.09

19.75

0.211

2

0.280

.235

4.00

3

2.78

23.88

0.860

Σ 43.87

5.071

Ix = 43.85 + 5.077 = 48.9 in.4 Y-Y axis: 0.75(6)3 4(0.75)3 Iy = + + 15.2 = 28.8in.4 12 12 (b)

8-192

Pipe 4 x-strong: O.D. = 4.50 in.

A = 4.14 in.2 I = 9.12 in.4 X-X axis:

y=

4.14(7.25) + 3(3) + 6(0.5) = 3.20 in.

4.14 + 3 + 6 d (in.)

Ad2 (in.4)

Ixo (in.4)

1

4.05

69.9

9.12

2

0.20

0.12

4.00

3

2.70

43.74

0.50

Σ 111.76

13.62

Ix = 111.76 + 13.62 = 125.4 in.4 Y-Y axis: I y = 9.12 +

4(0.75) 2 1(6) 3 + = 27.3 in.4 12 12

----------------------------------------------------------Prob. 8.23 (a) W10 × 54: A = 15.8 in.2, Ix = 303 in.4 Iy = 103 in.4, rx = 4.37 in., ry = 2.56 in. rx =

303 = 4.38 in. ≈ 4.37 in. (O.K.) 15.8

ry =

103 = 2.55 in. ≈ 2.56 in. (O.K.) 15.8

(b) C10 × 15.3: A = 4.48 in.2 , Ix = 67.3 in.4, Iy = 2.27 in.4, rx = 3.88 in., ry = 0.711 in.

rx =

67.3 = 3.88 in. (O.K.) 4.48

ry =

2.28 = 0.712 in. (O.K.) 4.49

-------------------------------------------------------8-193

Prob. 8.24 C10 × 15.3: A = 4.48 in.2 Ix = 67.3 in.4 Iy = 2.27 in.4 rx =3.88 in. ry = 0.711 in. x = 0.634 in.

Ix = 2(67.3) = 134.6 in.4 Iy = 2(2.27 + 4.48(1.966)2) = 39.1 in.4 rx =

134.6 = 3.87 in. (same as single L ) 2(4.48)

ry =

39.2 = 2.09 in.

2(4.48) (0.713 in. for single L )

----------------------------------------------------Prob. 8.25 (Refer to Prob. 8.12) Atotal = 27 + 2(11.8) = 50.6 in.2

rx =

1506 = 5.46 in.

50.6

ry =

3480 = 8.29 in.

50.6

----------------------------------------------------Prob. 8.26

8-194

a (in.2)

y (in.)

ay (in.3)

1

0.25

1.875

0.469

2

0.50

1.00

0.500

3

0.125

0.125

0.01563

Σa = 0.25 + 2(0.50 + 0.125) = 1.500 in.2 Σay = 0.469 +2(0.50 + 0.01563) =1.500 in.3

y=

1.500 = 1.00 in.

1.500 dx (in.) dy (in.)

Ixo (in.4)

Iyo (in.4)

1

0.875

0

0.0013

0.0208

2

0

0.625

0.1667

0.0026

3

0.875

1.00

0.0007

0.0026

Ix = ΣIxo + Σadx2 = 0.0013 + 0.25(0.875)2 + 2(0.1667 + 0) + 2(0.0007 + 0.125(0.875)2) = 0.719 in.4

rx =

0.719 = 0.692 in.

1.500

Iy = ΣIyo + Σady2 = 0.0208 + 0 + 2(0.0026 + 0.5(0.625)2) + 2(0.0026 + 0.125(1)2) = 0.672 in.4

ry =

0.672 = 0.669 in.

1.500

-------------------------------------------------------------Prob. 8.27 (Refer to Probs. 8.3 and 8.8) Atotal = 20.9 + 2(15) = 50.9 in.2

rx =

Ix 1733 = = 5.84 in.

A 50.9

8-195

ry =

Iy A

=

79.6 = 1.25 in.

50.9

-------------------------------------------------------------Prob. 8.28 (Refer to Prob. 8.14) Atotal = 3(24) = 72 in.2 X-X axis:

rx =

1386 = 4.36 in.

72

Y-Y axis: 2(12) 3 = 288 in.4 12 12(2) 3 I y2 = + 24(5) 2 = 608 in.4 12 I y1 =

Iy = 288 + 2(608) =1504 in.4

ry =

1504 = 4.57 in 4 72

----------------------------------------------------------Prob. 8.29 (a)

8-196

X-X axis a

d 4

×10

ad2

Iox

6

×10

×106

mm2

(mm)

(mm4)

(mm4)

1

2.40

115

317

12.80

2

1.20

0

0

22.5

Ix = Σ(Io + Ad2) = 2[(12.80 + 22.5)×106 + 317×106] = 705×106 mm4 Y-Y axis a

d

×104

ad2

Ioy

×106

×106

mm2

(mm)

(mm4)

(mm4)

1

2.40

0

0

180

2

1.20

90

97.2

6.40

Iy = 2[(180 + 6.40)×106 + 97.2×106] = 567×106 mm4 (controls) ry =

567 × 10 6 = 88.7 mm 2(2.40 + 1.20) × 10 4

(b)

8-197

W530 × 101: A = 12 900 mm2 Ix = 616×106 mm4 Iy = 26.9×106 mm4 bf = 210 mm Ix = 2(616×106) = 1232×106 mm4 Iy = 2(26.9×106) + 2(12 900)(105)2 = 338×106 mm4 (controls) ry =

338 ×10 6 = 114.5 mm 2(12 900)

---------------------------------------------------Prob. 8.30 W8 × 40: A = 11.7 in.2 Ix = 146 in.4 Iy = 49.1 in.4 (a) Ix = 2[146 + 11.7(8.25/2)2] = 690.2 in.4 Iy = 2(49.1) = 98.2 in.4 (b)

rx =

690.2 = 5.43 in.

2(11.7)

ry =

98.2 = 2.05 in.

2(11.7)

---------------------------------------------------Prob. 8.31 (Solid shaft) J=

π d 4 π (3) 4 = = 7.95 in.4 32 32

-------------------------------------------------Prob. 8.32 (Hollow circular shaft)

J=

π 4 π (d − d14 ) = (34 − 2.54 ) = 4.12in.4 32 32 8-198

--------------------------------------------------------Prob. 8.33 (Refer to Prob. 8.9 (a) & (b)) (a) J = 628 + 2980 = 3610 in.4 (b) J = 246 + 61.5 = 308 in.4 --------------------------------------------------------Prob. 8.34 (a) Pipe 152 mm (Stn'd): I = 11.0×106 mm4 J = Ix + Iy = 2(11.7×106) = 22.0×106 mm4 (b) W530 × 101: Ix = 616×106 mm4 Iy = 26.9×106 mm4 J = Ix + Iy= 643×106 mm4 (c) 200 × 200 timber (nominal dimensions) Ix = Iy = d4/12 = 133.3×106 mm4 J = Ix + Iy = 267×106 mm4 ---------------------------------------------------Prob. 8.35 (Refer to Prob. 8.35) (a) J = Ix + Iy = (705 + 567)×106 = 1272×106 mm4 (b) J = Ix + Iy = (1232 + 338)×106 = 1570×106 mm4 ---------------------------------------------------Prob. 8.36 (Refer to Prob. 8.32) Atotal = 19.39 in.2 Ix = 341.7 in.4, Iy = 182.4 in.4

rx =

341.5 = 4.20 in.

19.38

ry =

182.4 = 3.07 in.

19.38

--------------------------------------------------

8-199

Prob. 8.37

a1 = 6(5.196)(2) = 62.35 in.2 a2 = ½ (5.196)(2)(3) = 15.59 in.2 Σa = 62.35 + 2(15.59) = 93.53 in.2 X-X axis: 6(5.196)(2) 3 I x1 = = 561.1 in.4 12  3(5.196)(2) 3   = 140.3 in.4 I x 2 = 4 12   Ix = 561.1 + 140.3 = 701 in.4 Y-Y axis: 5.196(2)(6) 3 I y1 = = 187.1 in.4 12  5.196(3) 3 15.59 2  I y 2 = 4 + (4)  = 514.5 in.4 2  36  Iy = 187.1 + 514.5 = 702 in.4 J = Ix + Iy = 701 + 702 = 1403 in.4 ---------------------------------------------------Prob. 8.38

8-200

L2½ × 2½ × ¼ A = 1.19 in.2 Ix = Iy = 0.692 in.4 x = y = 0.711 in.

Pipe 4 Std. :

A = 2.96 in.2 I = 6.82 in.4 O.D. = 4.50 in.

I = Σ(Io + Ad2) Ix = Iy = 6.82 + 4(0.692) + 4(1.19)(1.789)2 = 24.82 in.4 rx = ry =

24.82 = 1.793 in.

2.96 + 4(1.19)

J = Ix + Iy = 2(24.82) = 49.6 in.4 ----------------------------------------------------

8-201

Prob. 9.2 σ = P/A A = πd2/4 = 0.7854d2 = 28.27 in.2 σ = 113.1/28.27 = 4.00 ksi = 4000 psi -------------------------------------------------------------Prob. 9.3 AB: A = 4 in.2 BC: A = 0.7854(1.75)2 = 2.41 in.2 σAB = 75/4 = 18.75 ksi σBC = 75/2.41 = 31.1 ksi -------------------------------------------------------------Prob. 9.4 P = 70 kN = 70 ×103 N (a) A = 25(50) = 1250 mm2 σt =

P 70 × 103 = = 56 MPa A 1250

(b) A = 0.7854(150)2 = 17.68×10−3 m2 70 × 103 σt = = 3.96 × 106 Pa −3 17.68 × 10 = 3.96 MPa

(c) A = 0.7854(25)2 = 490.9 mm2 σt =

70 × 103 = 142.6 MPa 490.9

------------------------------------------------------------Prob. 9.5 (a) σt =

16, 000 = 4000 psi 4

16, 000 = 4178 psi 3.83 --------------------------------------------------------------

(b) σt =

9-202

Prob. 9.6 Segment A: P = 10 k; A = 0.7854(1.25)2 = 1.23 in.2

σc =

10 = 8.13 ksi 1.23

Segment B P = 22 k; A = 2(2) = 4 in.2

σc =

22 = 5.5 ksi 4

--------------------------------------------------------Prob. 9.7 Total load = 16,000 lb Load per rod = 16,000/3 = 5333 lb A = 0.7854(0.625)2 = 0.3068 in.2 5333 = 17,380 psi σt = 0.3068 --------------------------------------------------------------Prob. 9.8 Req'd A = Req'd d =

P σ t (all)

=

35, 000 = 1.75 in.2 20, 000

A = 1.49 in.2 0.7854

Use a 1½″ diam. rod --------------------------------------------------------------Prob. 9.9

P = 60 k

Column: σ = 60/11.7 = 5.13 ksi Base Plate: σ = 60/196 = 0.306 ksi Pedestal: σ = 60/452.4 = 0.133 ksi Footing: σ = 60/16 = 3.75 ksf ---------------------------------------------------------------

9-203

Prob. 9.10 Req' d A =

P

σ p (all)

=

60 = 24 ft 2 2 .5

Req' d side = 24 = 4.90 ft

Use a square footing: 5′-0″×5′-0″ --------------------------------------------------------------Prob. 9.11 Load/rod = 75/3 = 25 kN = 25×103 N A = 0.7854(19)2 = 283.5 mm2 P 25 ×103 σt = = = 88.2 MPa A 283.5 ---------------------------------------------------------Prob. 9.12 A = 25(100) = 2500 mm2 σt(all) = 140 MPa = 140 N/mm2 P(all) = A σt(all) = 2500(140) = 350,000 N = 350 kN --------------------------------------------------------------Prob. 9.13 w = wt. of wire per foot

0.7854(0.1875) 2 (12) w= (490) 1728 = 0.09396 lb/ft Total wire wt. = 0.09396(400) = 37.58 lb

(a) σ t =

P 37.58 = = 1362 psi A (0.7854)(0.1875) 2

(b) P=Aσt = (0.7854)(0.1875)2(24,000 – 136.2) = 625 lb ---------------------------------------------------------------

9-204

Prob. 9.14 W12 × 40 A = 11.7 in.2 tf = 0.515 in. tw = 0.295 in.

An = Ag – Ah = 11.7 – 4(0.75)(0.515) – 2(0.75)(0.295) = 9.71 in.2

σt =

P 190 = = 19.56 ksi An 9.71

--------------------------------------------------------------Prob. 9.15 Floor load = 10.5 kPa = 10.5×103 N/m2 Load to each rod = P P = 7(10.5×103) = 73.5×103 N 73.5 × 10 3 Req' d A = = = 525 mm 2 140 σ t (all) P

525 = 25.9 mm 0.7854

Req' d d =

---------------------------------------------------------Prob. 9.16 P = 20 k, L = 30 ft, δ(limit) = 0.25 in. σt(all) = 24 ksi For tension: Req’d A = P/ σt(all) = 20/24 = 0.833 in.2 For elongation: Req' d A =

PL

δ (limit) E

=

20(30)(12) 0.25(30,000)

= 0.96 in. 2 ⇐ controls

9-205

Req' d d =

0.96 = 1.106 in.

0.7854

Use a 1 1/8″ rod --------------------------------------------------------------Prob. 9.17

P A A = 0.7854d 2 = 0.7854(32) 2 = 804mm 2

τ (ult) =

τ (ult) =

300 ×103 = 373 MPa 804

--------------------------------------------------------------Prob. 9.18 AB = shear area per bolt AB = 0.7854(0.875)2 = 0.6013 in.2

P = τ (all) AB = 14,500(2)(0.6013) = 17,440 lb --------------------------------------------------------------Prob. 9.19 A = shear area A = πdt = π(1)(0.5) = 1.571 in.2

P = Aτ (ult) = 1.571(42,000) = 65,980 lb --------------------------------------------------------------Prob. 9.20 (a) A = 0.7854(0.75)2 = 0.4418 in.2 P 6000 = 13,580 psi σt = = A 0.4418

9-206

(b) A = πdt = π(0.75)(0.375) = 0.8836 in.2

τ=

6000 P = = 6790 psi A 0.8836

--------------------------------------------------------------Prob. 9.21

(2 bolts)

Ab = bolt cross-sectional area Ab = 0.7854(25)2 = 491 mm2 Total area A = 2(491) = 982 mm2 P 120 × 103 τ= = = 122.2 MPa A 982 --------------------------------------------------------------Prob. 9.22

Max. AH = τ(all)A =1050(2)(2) = 4200 lb ΣFx = 0  BH = 4200 lb ΣMA = –1400(8) + BH(x) = 0  x = 2.67 ft

--------------------------------------------------------------Prob. 9.23 Washer: I.D. = 27 mm, O.D. = 64 mm t = 5 mm Resisting shear area: A = πt (I.D. + O.D.) = π(5)(27+64) = 1.429×103 mm2 P = τ(ult) A = 400(1.429×103) = 572 kN --------------------------------------------------------------

9-207

Prob. 9.24 A = π d t = π (0.75)(0.5) = 1.178 in.2

τ=

P 60,000 = = 50,930 psi A 1.178

--------------------------------------------------------------Prob. 9.25 ΣMo : 0.5P = 40(12) P= 960 lb

τ=

P 960 = A (1 / 4)(7 / 8) = 4390 psi

--------------------------------------------------------------Prob. 9.26 P = Force (lb) acting on key ΣMo : 0.5P = 60(12) P= 1440 lb

Area of key:

1440 4000 = 0.36 in.2

Req' d A =

Width of key b: A = b×length Req’d b = 0.36/1.5 = 0.24 in.

Use ¼” width --------------------------------------------------------------Prob. 9.27

9-208

PV = PH = 0.707(20) = 14.14 kN (a) Comp. stress in diagonal:

σc =

P 20 ×10 3 = 0.889 MPa A 150 2

(b) Bearing stress on plane a-b:

σ=

14.14 ×103 = 2.36 MPa 40(150)

(c) Shear stress on plane a-c :

τ=

14.14 ×103 = 0.589 MPa 160(150)

--------------------------------------------------------------Prob. 9.28 Refer to Appendix G. Req’d shear area = 150× mm2 P 14.14 × 103 Req'd area = = = 27,200 mm 2 τ (all) 0.52 Req’d x = (27 200)/150 = 181.3 mm --------------------------------------------------------------Prob. 9.30 (a) ε =

δ 1.2 = = 0.0010 in./in.

L 100(12)

(b) L =

δ 0.5(12) = = 20, 000 in. = 1667 ft ε 0.00030

(c) δ = εL = 0.00042(1000) = 0.42in.

--------------------------------------------------------------Prob. 9.31

ε=

δ 8 − 7.85 = = 0.01875 in./in.

L 8

--------------------------------------------------------------Prob. 9.32 Total elongation = 0.80 in.

2δ 2(0.80) = = 0.001333 in./in.

L 100(12) --------------------------------------------------------------Max. ε =

9-209

Prob. 9.33 δ=

PL σL 15, 000(2.5) = = = 0.0125 in.

AE E 30, 000, 000

--------------------------------------------------------------Prob. 9.34

E = 30,000 ksi

(a) A = 0.7854d2 = 0.442 in.2

σ= (b) ε =

P 15, 000 = = 33,900 psi A 0.442 s 33,900 = = 0.00113 E 30, 000, 000

(c) δ = EL = 0.00113(25)(12) = 0.340 in.

--------------------------------------------------------------Prob. 9.35

E = 70×103 MPa

(a) A = 0.7854d2 = 491 mm2

σ=

P 65 ×10 3 = = 132.4 MPa A 491

(b) ε =

s 132.4 = = 1.89 × 10 3 3 E 70 × 10

(c) δ = εL = 1.89 ×10 −3 (4.0 ×10 3 ) = 7.56 mm ---------------------------------------------------------------

9-210

Prob. 9.36

Segment 1: A = 0.7854(0.75)2 = 0.442 in.2 5000(5)(12) PL δ1 = = = 0.0226 in.

AE 0.442(30,000,000) Segment 2: A = 0.7854(0.50)2 = 0.1963 in.2 5000(5)(12) PL δ2 = = = 0.0509 in.

AE 0.1963(30,000,000) Total δ = δ1 + δ2 = 0.0735 in.

--------------------------------------------------------------Prob. 9.37

(P.L. = 125 ksi)

δ = 0.500 in.,

P = 100 k,

L = 16(12) = 192 in., A = 2.25 in.2

σt =

P 100 = = 44.4 ksi A 2.25

44.4 < 125 O.K.

E=

PL sL 44.4(192) = = = 17,050 ksi Aδ δ (0.500)

--------------------------------------------------------------Prob. 9.38 P = 67 kN = 67×103 N A = 0.7854(25)2 = 491 mm2 (a) σ t =

P 67 × 10 3 = = 136.5 MPa A 491

136.5 MPa < 207 MPa O.K.

9-211

(b) ε =

σ E

=

136.5 = 1.95 ×10 −3 3 70 ×10

(c) δ = εL = (1.95 × 10 −3 )(3)(1000) = 5.85 mm --------------------------------------------------------------Prob. 9.39 8” × 8” nominal timber: Dressed A = 56.3 in.2

P 40,000 = = 710 psi 56.3 A

(a) σ t =

710 psi < 6000 psi O.K. (b) ε =

σ E

=

710 = 0.000418 1,700,000

(c) δ = εL = (0.000418)(8.5)(12) = 0.0426 in.

--------------------------------------------------------------Prob. 9.40 PL 15(100 = = 0.01 ft AE 0.005(30,000,000) P 15 (b) σ t = = = 3000 psi A 0.005

(a ) δ =

--------------------------------------------------------------Prob. 9.41 A = cross-sectional area = 0.7854(1.25)2 = 1.227 in.2 (a) σ t = (b) ε =

P 12,000 = = 9780 psi A 1.227

σ E

=

9780 = 0.000326 30,000,000

(c) δ = εL = (0.000326)(18) = 0.00587 in.

--------------------------------------------------------------9-212

Prob. 9.42

σ PA E= =  P = AEε ε ε P = 0.7854( 26) 2 ( 207 × 10 3 )(0.0007) = 76.9 × 10 3 N = 76.9 kN

--------------------------------------------------------------Prob. 9.43 δ = 45 mm, d = 5 mm, A = 0.7854(5)2 = 19.64 mm2 L = 18 m = 18×103 mm E = 207,000 MPa

P=

δAE

=

45(19.64)(207 000)

L 18 × 10 3 = 10.16 × 10 3 N

σt =

P 10.16 ×103 = = 517 MPa A 19.64

--------------------------------------------------------------Prob. 9.44 A = 0.7854(1.5)2 = 1.767 in.2

(a) δ =

30,000(20)(12) PL = = 0.1358 in.

AE 1.767(30,000,000)

(b) Req' d A =

30,000(20)(12) PL = δE 0.10(30,000,000)

= 2.40 in. 2 Req' d d =

A = 1.748 in.

0.7854

Use a 1 ¾” diam. Rod ---------------------------------------------------------------

9-213

Prob. 9.45

(a) Req' d A =

50,000(24)(12) PL = δE 0.20(30,000,000)

= 2.40 in.2 Req' d A Req' d w = = 3.20 in.

3 4 P 50,000 (b) σ t = = = 20,800 psi 2.40 A --------------------------------------------------------------Prob. 9.46 A = 0.7854(0.75)2 = 0.442 in.2 Due to own wt.: P = 0.442(100)(12)(1/1728)(490) = 150.4 lb

δ=

PL (150.4 / 2)(100)(12) = = 0.00681in.

AE 0.442(30,000,000)

Due to applied load:

δ=

8000(100)(12) PL = = 0.724 in.

AE 0.442(30,000,000)

Total δ = δ1 + δ2 = 0.731 in.

--------------------------------------------------------------Prob. 9.47

ΣMB = 0  AV = 80 k 9-214

Cut section 1-1 and use left portion:

ΣFV = +80 – 100 + CD(4/5) = 0 CD = +25 k (comp.)

δ=

PL 25(5)(12) = = 0.02 in.

AE 2.5(30,000)

Check stress level: σ = 25/2.5 = 10 ksi < 34 ksi O.K. --------------------------------------------------------------Prob. 9.48

A = 0.25 in.2 E = 30,000,000 psi

δ=

PL AE

Section

Load

L

δ (in.)

(lb; tens.)

(in.)

A

6000

72

0.0576

B

2000

48

0.0128

C

4000

24

0.0128

Total:

0.0832

Check max. stress level: σ = 6000/0.25 = 24,000 psi < 34,000 psi OK 9-215

--------------------------------------------------------------Prob. 9.49 A = 0.25 in.2; E = 30,000,000 psi

Segmnt Load (lb)

L

δ (in.)

(in.)

A

6000 (T)

72

+0.0576

B

8000 (C)

48

–0.0512

C

4000 (T)

24

+0.0128

Total δ = +0.0192 (elongation) Check max. stress level: σ = 8000/0.25 = 32,000 psi < 34,000 psi OK --------------------------------------------------------------Prob. 9.50

9-216

AC BC 45 = = sin 51.34° sin 101.31° sin 27.35°  BC = 96.0 kN (tens.)  AC = 76.5 kN (comp.)

For member BC: A = 0.7854(17)2 = 227 mm2

δ=

PL 96 × 103 (6400) = = 13.08 mm AE 227(207 000)

For member AC:

δ=

PL 76 500(5100) = = 1.396 mm AE 1350(207 000)

--------------------------------------------------------------Prob. 9.51

ΣFV = 2 (0.7071T ) – 2000 = 0 T = 1414 lb A = 0.7854(0.25)2 = 0.0491 in.2

δ=

1414(10)(12) PL = = 0.1152 in.

AE 0.0491(30,000,000)

Check stress level: σ = 1414/0.0491 = 28,800 psi < 34,000 psi OK y = drop of the hook

y=

0.1152 sin 45°

= 0.163 in.

9-217

--------------------------------------------------------------Prob. 9.52

For the tie-back: L=

3 2 + 2 2 = 3.61 m

A = 0.7854(33)2 = 855 mm2 Wt. = 6000(9.81) = 58.9 kN max P = σt(ult) A = 170(855) = 145.4 kN

δAE PL  max P = L AE 2.5(855)( 207 000) max P = = 122.6 kN ⇐ 3610

δ=

(controls) PV = 2 (122.6)/3.61 = 67.9 kN ΣMA : 67.9(3) = 58.9L Max L = 3.46 m --------------------------------------------------------------Prob. 9.53

9-218

PH = 29 cos30° = 25.11 lb PV = 29 sin30° = 14.5 lb FH = F cos 20° = 0.9397F ΣMo = –14.5(6) –25.11(11.5) + 0.9397F(4) = 0 F = 100 lb Area of rod = 0.7854(5/16)2 = 0.0767 in.2 σ = 100/0.0767=1304 psi --------------------------------------------------------------Prob. 9.54

ΣMo = 0  F = 63.75 lb Cable: A = 0.7854(0.047)2(7) = 0.01214 in.2 σ = 63.75/0.01214 = 5250 psi --------------------------------------------------------------Prob. 9.55

Piston area: A A = 0.7854(400)2 = 125 700 mm2 Total axial force P = sA 9-219

= 1.72 (125 700) = 216 kN For the piston rod: Req' d AP = Req' d d =

P

σ (all)

=

216 000 N = 3130 mm 2 69.0 MPa

Req' d A = 63.1 mm 0.7854

--------------------------------------------------------------NOTES

9-220

Prob. 10.1 A = 0.7854(0.5625)2 = 0.2485 in.2

6500 P = = 26,160 psi A 0.2485

(a) σ t =

δ

(b) ε =

L

=

0.00715 = 0.000894 8

(c) 26,160 psi < 34,000 psi OK:

E=

26,160 σ = = 29,300,000 psi ε 0.000894

--------------------------------------------------------------Prob. 10.2 A = 0.7854(150)2 = 17.67×103 mm2

P 89 ×103 σc = = = 5.04 MPa A 17.67 ×103

ε=

δ

E=

L s

ε

=

0.074 = 0.000247 300

=

5.04 = 20.4 × 103 MPa 0.000247

--------------------------------------------------------------Prob. 10.3 Pt.

Stress (psi)

Strain

1

10,000

0.000343

2

25,000

0.000858

3

40,000

0.0350

Point 1:

E=

10,000 σ = = 29,200,000 psi ε 0.000343

Point 2:

E=

25,000 = 29,100,000 psi 0.000858

E is essentially the same at points 1 and 2. At point 3, E > P.L. and therefore not valid.

10-221

-------------------------------------------------------Prob. 10.4 A = 0.7854(0.505)2 = 0.2003 in.2 Data Load (k) δ (in.)

σ (k/in.2)

ε = δ/2

0

0

0

0

2.0

0.0006

9.99

0.0003

4.0

0.0013

19.97

0.00065

6.0

0.0019

30.0

0.00095

(more) Data (cont.)

Load (k)

δ (in.)

σ (k/in.2)

ε = δ/2

8.0

0.0025

39.9

0.00125

7.83

0.0043

39.1

0.00215

7.80

0.0120

38.9

0.006

8.0

0.0361

39.9

0.0181

10.0

0.1100

49.9

0.055

13.0

0.2320

64.9

0.116

13.9

0.3200

69.4

0.160

14.0

0.3600

69.9

0.180

12.9

0.4000

64.4

0.200

10.8

0.4200

53.9

0.210

Stress-strain diagram not shown.

39.9 = 31,900,000 psi Point 5: E = 0.00125 Upper yield point, point 5, highest stress before the drop: 39.9 ksi Lower yield point, point 7: 38.9 ksi 10-222

Ultimate strength: 69.9 ksi Rupture strength: 53.9 ksi % reduction in area: =

Orig. A − Final A × 100 Orig. A

=

0.2003 − 0.7854 (0.397 ) 2 (100 ) = 38.2 % 0.2003

--------------------------------------------------------------Prob. 10.5 % Elongation:

.

. .

(100) = 21.0%

% Reduction in area: Af = 0.7854(0.422)2 = 0.1399 in.2 Ao = 0.7854(0.50)2 = 0.1964 in.2 %Reduction =

0.1964 − 0.1399 (100) = 28.8% 0.1964

-----------------------------------------------------------Prob. 10.6 δ = 14 mm = 14.0 × 10

m

P = 320 N, L = 4 m 𝐴 = 0.7854(1.0 × 10 ) = 785.4 × 10 m 𝐸=

𝑃𝐿 320(4.0) = 𝐴𝐸 14.0 × 10 (785.4 × 10 )

= 116.4 × 10 Pa = 116.4 × 10 MPa

-----------------------------------------------------------Prob. 10.7 A = 0.7854(6)2 = 28.27 in.2 𝐸=

𝑃𝐿 (20,000)(12) = = 2,930,000 psi 𝐴δ (28.27)(0.0029)

------------------------------------------------------------

10-223

Prob. 10.8 A = 1/2(2) = 1 in.2, P = 16,000 lb σ=

𝑃 16,000 = = 16,000 psi < 21,000 psi ∴ OK 𝐴 1

-----------------------------------------------------------Prob. 10.9

𝐸=

∆σ 134 − 35 = = 215 ×× 10 MPa ∆ϵ 0.00063 − 0.00017

Expected Stress: σ = 𝐸ϵ = 215 × 10 (0.00025) = 53.8 MPa 53.8 MPa < 200 MPa O.K.

-----------------------------------------------------------Prob. 10.10 % Elongation =

δ = 60.7 – 50 = 10.7 mm 10.7 (100) = 21.4% 50

% Reduction in Area = 𝐴 = (0.7854)(10.7) = 89.9 mm 𝐴 = (0.7854)(12.5) = 122.7 mm % Reduction =

122.7 − 89.9 (100) = 26.7% 122.7

-----------------------------------------------------------Prob. 10.11 𝑃(

) = 13,100 lb, δ = 0.52 in.

𝑑 = 0.50 in. , 𝑔 = 2 in. 𝐴 = 0.7854(0.50) = 0.1964 in

10-224

𝑃( ) 13,100 = = 66,700 psi 𝐴 0.1964 0.52 (100) = 26.0% b) %Elongation = 2 a) σ(

) =

--------------------------------------------------------------Prob. 10.12 d = 0.505 in., δ = 0.002 in.

A = 0.7854(0.505)2 = 0.2003 in.2 𝑃=

δ𝐴𝐸 0.02(0.2003)(30,000) = = 6.01 k 𝐿 2

Check Stress: σ=

6.01 = 30 ksi < 34 ksi ∴ O. K. 0.2003

-----------------------------------------------------------Prob. 10.13

σt(all) = 60 ksi, δlimit = 0.05 in., P.L. = 120 ksi Based on stress: σt = P/A, A = 0.7854d2 Req' d d 2 =

24 = 0.509 in. 2 60(0.7854)

Based on δ: δ=

PL AE

Req'd d 2 = =

Req’d d =

PL 0.7854 Eδlimit

24(18) = 0.667 in.2 0.7854(16,500)(0.05) (controls)

0.667 = 0.817 in.

--------------------------------------------------------------10-225

Prob. 10.14 P = load to each rod = 75(250) = 18,750 lb req d 𝐴 =

P σ(

=

)

18,750 = 0.852 in.

22,000

0.852 = 1.04 in.

0.7854

req d 𝑑 =

--------------------------------------------------------------Prob. 10.15 Stress: req d 𝐴 =

𝑃 σ(

)

=

55 × 10 = 392.9 × 10 140 × 10

m

Elongation req d 𝐴 =

𝑃𝐿 55 × 10 (0.45) = = 598 × 10 δ𝐸 0.2x10 (207 × 10 )

m

Elongatoin Controls: 598 × 10 0.7854

req d 𝑑 =

= 27.6 × 10

m = 27.6 mm

Check Stress: σ=

𝑃 55 × 10 = 𝐴 589 × 10

= 92.0 × 10 Pa

92.0 MPa < 175 MPa (O.K.) --------------------------------------------------------------Prob. 10.16 (a) 𝑃(

) =σ (

(b) σ =

=

, .

) A = 22,000(8.62) = 189,600 lb

= 17,400 psi

--------------------------------------------------------------Prob. 10.17

Each Plate: 225 mm × 32 mm

Area per plate: A = 225(32) = 7200 mm2 10-226

P(all) per plate = σt(all)A = 165(7200) = 1188 kN # of plates, N = 6800/1188 = 5.72 plates (use 6) Comp’d Stress: σ = P/A = 6800/(6(7200)) = 157.4 MPa -----------------------------------------------------------Prob. 10.18

∑𝑀 = 𝑅(0.5) + 35(8.5) = 0

R = 595 lb Req d 𝐴 =

Req d 𝑑 =

𝑃 τ(

)

595 = 0.0496 in.

12,000

=

0.0496 = 0.251 in.

0.7854

Use 9/32 in. diameter pin.

-----------------------------------------------------------Prob. 10.19 Req d 𝐴 =

P = 15,000 lb, σp(all) = 300 psi 𝑃 σ (

)

=

15,000 = 50 in.

300

Contact area = 9.5x Req’d x = 5.26 in.

(Use 6 in. bearing length)

-----------------------------------------------------------Prob.10.20 P = 67.5 kN, L = 3 m (a) Steel: Fy = 310 MPa 310 σ( )= = 124 MPa 2.5 67.5 × 10 Req d 𝐴 = = 544 mm 124

10-227

𝑅𝑒𝑞 𝑑 𝑑 =

544 = 26.3 mm 0.7854

(b) Aluminum

Fy = 240 MPa

σt(all) = 240/2.5 = 96.0 MPa Req d 𝐴 =

67.5 × 10 = 703 mm 96.0

Req d 𝑑 =

703 = 29.9 mm 0.7854

--------------------------------------------------------------Prob. 10.21 Gray C.I. σult = 80 ksi σ(

) = 80⁄6 = 13.33 psi

Req d 𝐴 = Req d 𝑑 =

100 = 7.50 in.

13.33 7.50 = 3.09 in.

0.7854

--------------------------------------------------------------Prob. 10.22 A = 0.7854(0.16)2 = 0.02011 in.2 Failure load = σ (

) 𝐴 = 1470(0.02011)(230,000)

= 6,799,000 lb F. S. =

6,799,000 = 2.62 2,600,000

--------------------------------------------------------------Prob. 10.23 (a) Based on yield: F.S. = 350/210 = 1.67 (b) Based on tensile strength: F.S.=827/210 = 3.94 ---------------------------------------------------------------

10-228

Prob. 10.24

TMax(allowable) = 252/4 = 63 lb ∑ 𝐹 = +48 − 2(𝑇 cos θ/2) θ 48 48 cos = = = 0.38095 2 2T 2(63)

Max. θ/2 = 67.6°  θmax = 135.2° --------------------------------------------------------------Prob. 10.25 P = 60,000 lb, σp(all) = 300 psi Req’d contact area A = 60,000/300 = 200 in.2 Di = 4 + 2(1/32) = 4.0625 in. 𝐴 = 0.7854(𝐷 − 𝐷 ) Req d 𝐷 =

200 − (4.0625) = 16.47 in.

0.7854

-----------------------------------------------------------Prob 10.26 P = 200,000 lb, σult = 95,000 psi a) F.S. = 5.0: Req d 𝐴 =

200,000 (5.0) = 10.53 in.

95,000

b) δmax = 0.10 in.

Req d 𝐴 =

𝑃𝐿 200,000(10)(12) = = 8.00 in. 𝐴𝐸 0.10(30,000,000)

------------------------------------------------------------

10-229

Prob. 10.27

A = 1.2 in.2, σt(all) = 22,000 psi, Fy = 36,000 psi, F.S. = 1.85 Elastic: 𝑃𝐿 σ𝐿 σ 𝐿 σ 𝐿 =  = 𝐸 𝐸 𝐴𝐸 𝐸 σ 𝐿 6 σ ∴σ =σ( σ = = 𝐿 8 δ=

)

σ = 22,000 psi σ = (6⁄8)(22,000) = 16,500 psi 𝑃 = 2σ 𝐴 + σ 𝐴 = 2(16,500)(1.2) + 22,000(1.2) = 66,000 lb

(a) Ultimate Strength: (All rods stressed to Fy.) 1.85P = 3FyA ∴𝑃=

3(36,000)(1.2) = 70,100 lb 1.85

-----------------------------------------------------------Prob. 10.28 a) δ =

→ δ = δ ; 𝐸 is constant.

σALA = σBLB → σB = LAσA/LB = (4/3)σA →σB reaches allowable (185 MPa) first σB = 185 MPa = 4/3 σA→σA = 138.8 MPa P=σAAA + σBAB = 138.8(580) + 185(700) = 210 kN

10-230

b) Both bars will be stressed to 275 MPa: 𝑃=

275(700) + 275(580) = 195.5 kN 1.80

-----------------------------------------------------------Prob. 10.29 a) Load to each rod: ∑𝑀 = −20,000(3) + 𝑃 (4) = 0 ∴ 𝑃 ∑𝐹 = 0 ∴ 𝑃

= 15 k

=5k

Elongation of each Rod: δ

=

5000(2)(12) = 0.0040 in.

1(30,000,000)

δ

=

15,000(3)(12) = 0.0090 in.

2(30,000,000)

b) Ratio of areas for δAB = δCD: 𝑃 𝐿 𝑃 𝐿 = 𝐴 𝐸 𝐴 𝐸 ∴

𝐴 𝐴

or:

= 𝐴 𝐴

𝑃 𝐿 𝑃 𝐿

=

5000(2) = 0.222 15,000(3)

= 4.5

------------------------------------------------------------

10-231

Prob. 11.1 (a) ε axial =

δ 0.48 = = 0.0040 L 10(12)

(b) ε transverse =

δ 0.0024 = = 0.0012 2 L

εtransverse = μ εaxial μ = 0.0012/0.0040 = 0.300 -------------------------------------------------------------Prob. 11.2 σt =

P 300, 000 = A 12 = 25, 000 psi < 34,000 psi (O.K.)

ε axial =

σt 25, 000 = = 0.000833 E 30, 000, 000

εtransverse =μ εaxial = 0.25(0.000833) = 0.000208 Dimensional change in 6-in. transverse direction: δ = εtransverse L = 0.000208(6) = 0.00125 in. (decrease) --------------------------------------------------------------Prob. 11.3 μ=

E 110 GPa −1 = − 1 = 0.250 2G (2)44 GPa

--------------------------------------------------------------Prob. 11.4 30, 000, 000 = 11, 280, 000 psi 2(1 + 0.33) 16, 000, 000 (b) μ = − 1 = 0.379 2(5,800, 000) (a) G =

---------------------------------------------------------------

11-232

Prob. 11.5

E = 207 × 103 MPa; μ = 0.25

Stresses

σy =

P 600 ×103 = = 53.3 MPa (Comp.)

A 450(25) = 53.3 MPa < 230 MPa (O.K.)

σx =

265 ×103 = 106.0 ×106 MPa (Tens) 100(25)

= 106 MPa < 230 MPa (O.K.) Strains: εtransverse = μεaxial 1 (−μσ x + μσ y ) E −0.25(106) + 0.25(53.3) = = −63.6 ×10 −6 (decrease) 3 207 × 10 1 ε y = (−σ y − μσ x ) E −53.3 − 0.25(106) = = −386 × 10 −6 (decrease) 3 207 ×10 εz =

1 (+σ x + μσ y ) E 106 + 0.25(53.3) = = 576 ×10−6 (increase) 3 207 ×10

εx =

Dimensional changes: δz = 63.6 × 10−6(25) = 1.590 × 10−3 mm (decrease) δy = 386 × 10−6(100) = 0.0386 mm (decrease) δx = 576 × 10−6(450) = 0.259 mm (increase) --------------------------------------------------------------11-233

Prob. 11.6 (a) σ t =

P 279 × 103 = = 137.5 × 106 Pa A 1963 ×10−6

= 137.5 MPa < 234 MPa (O.K.) σt 137.5 = = 664 × 10 −6 3 E 207 × 10 ε transv = με axial = 0.25(664 × 10 −6 ) ε axial =

= 166.0 × 10 −6

(b) δ = εaxial L = 664 × 10−6(1 m) =0.664 mm --------------------------------------------------------------Prob. 11.7

σt =

P 64, 000 = = 16, 000 psi A 4

Longitudinal (axial) ε axial =

σt 16, 000 = = 5.33 × 10−4 E 30, 000, 000

δ axial = εL = 5.33 × 10 −4 (4)(12) = 0.0256 in. (increase)

Transverse (lateral) εtransv = μεaxial = 0.25(5.33 × 10−4) = 1.33 × 10−4 δtransv = εtransvL = 1.33 × 10−4(2) = 2.66 × 10−4 in. (decrease) ------------------------------------------------------------

11-234

Prob. 11.8 d = 150 mm A = 0.7854d2 = 17.67 × 103 mm2

0.275 300 = 0.000917 0.0125 εy = 150 = 0.0000833 εz =

μ=

εy εz

=

0.0000833 = 0.0908 0.000917

From Table G-2 (SI): E = 21.5 × 103 MPa

δ z AE 0.275(17.67 ×103 )(21.5 ×103 ) = 300 L = 348 kN

P=

--------------------------------------------------------------Prob. 11.9 d = 1.5 in.

A = 1.767 in.2 εz = σz =

δ z 0.0023 = = 0.00115 2 Lz P 60, 000 = = 33,960 psi A 1.767 < 34,000 psi (O.K.)

E=

σ z 33,960 = = 29,530, 000 psi ε z 0.00115

εx =

δ x 0.00043 = = 0.000287 Lx 1.5

μ=

ε x 0.000287 = = 0.2496 εz 0.00115

---------------------------------------------------------------

11-235

Prob. 11.10

s 30, 000 = 0.001 ε axial = c = E 30, 000, 000 ε transv = με axial = 0.25(0.001) = 0.00025 δ transv = ε transv L = 0.00025(3) = 0.00075 in. (increase) --------------------------------------------------------------Ly = 12 in., μ = 0.25

Prob. 11.11

Stresses: 90, 000 = 30, 000 psi < 34,000 psi (OK) 3(1) 288, 000 = 24, 000 psi < 34,000 psi (OK) σx = 12(1) 648, 000 = 18, 000 psi < 34,000 psi (OK) σz = 3(12) σy =

Strains: εx = εy = εz =

σ x + μσ y − μσ z

E σ y + μσ x + μσ z

E σ z + μσ y − μσ x E

= 0.0009 (decrease) = 0.00135 (increase) = 0.00065 (decrease)

New dimensions of bar: δx = 0.0009(3) = 0.0027 in. (decrease) 3 − 0.0027 = 2.9973 in.

11-236

δy = 0.00135(12) = 0.0162 in. (increase) 12 + 0.0162 = 12.0162 in. δz = 0.00065(1) = 0.00065 (decrease) 1 − 0.00065 = 0.99935 in.

--------------------------------------------------------------Prob. 11.12

Stresses: P 20 = = 20 ksi < 34 ksi (O.K.) A 2(0.5) 40 σx = = 8 ksi < 34 ksi (O.K.) 10(0.5) σy =

Strain: εz =

μσ x + μσ y

E 0.25(8) + 0.25(20) = = 0.000233 (decrease) 30, 000

Dimensional change in thickness: δz = εzL = 0.000233 (0.5) = 0.000117 in. (decrease) --------------------------------------------------------------Prob. 11.13 δ = αL(ΔT) = 0.0000065(1000)(70 − 32) = 0.247 ft (Distance measured 0.247 ft too long.)

-------------------------------------------------------------11-237

Prob. 11.14 Initial temp. = 68°F (a) At 32°: wire contracts, stress increases. ΔT = 68 – 32 = 36° σ = αE(ΔT) =13.1 × 10−6(10,000) (36) = 4.72 ksi Final stress = 5.00 + 4.72 = 9.72 ksi (b) At 90°F: wire expands, stress decreases. σ = 13.1 × 10−6(10,000)(22) = 2.88 ksi Final stress = 5.00 − 2.88 = 2.12 ksi -------------------------------------------------------------Prob. 11.15

E = 103 × 103 MPa α = 16.7 × 10−6 1/°C ΔT to close gap: δ = αLΔT = 0.05 mm ΔT =

0.5 = 29.9°C 16.7 ×10−6 (1000)

ΔT for stress development = 50–29.9 = 20.1 °C σ = αEΔT = 16.7 × 10−6 (103 × 103) 20.1 = 34.6 MPa --------------------------------------------------------------Prob. 11.16

L = 60 ft @ 30°: ΔT = – 40° δ = αLΔT = 6.5 × 10−6 (60)(12)(40) = −0.187 in.

Total gap = 0.50 + 0.187 = 0.687 in. @ 110° ΔT = + 40°; δ = + 0.187 in.

Total gap = 0.50 – 0.187 in. = 0.313 in.

--------------------------------------------------------------11-238

Prob. 11.17 α = 11.7 × 10−6 1/°C

E = 207 × 103 MPa ΔT = +34°C σ = αEΔT = 11.7 × 10−6(207 × 103)(34) =82.3 MPa (<235 MPa O.K.) --------------------------------------------------------------Prob. 11.18 (a) δ = αL (ΔT) ΔT =

δ 0.01 = = 46.6° αL 0.0000065(33)

Temp. @ touching = 46.6 + 15 = 61.6°F (b) Temp. rise to 110°F: ΔT = 110 − 61.6 = 48.4°F σ = Eα(ΔT) = 30,000(0.0000065)(48.4) = 9.44 ksi --------------------------------------------------------------Prob. 11.19

δB + δS = 0.030 in. αBLB(ΔT) + αSLS(ΔT) = 0.030 in. ΔT =

0.030 α B LB + αS LS

0.03 in.

[10.4 ×10 (3) + 6.5 × 10−6 (6)](12 in./ft) = 35.6°F =

−6

---------------------------------------------------------------

11-239

Prob. 11.20 ΔT to expand bar a length of 0.38(2) = 0.76 mm: ΔT =

δ 0.76 = = 21.65°C αL 0.0000117(3000)

ΔT to cause thermal stress: ΔT = 27 − 21.65 = 5.35°C σ = EαΔT = 207 × 103(11.7 × 10−6)(5.35) = 12.96 MPa --------------------------------------------------------------Prob. 11.21 (a) @ 112°F, error per measured distance: δ = αL (ΔT) = 0.0000065(2208.56)(112 − 68) = 0.63 ft Tape is too long; distance is short. Distance corrected for temperature: 2208.56 + 0.63 = 2209.19 ft (b) @ 20°F δ = 0.0000065(2208.56)(68 − 20) = 0.69 ft Tape is too short; distance is long. Distance corrected for temperature: 2208.56 − 0.69 = 2207.87 ft --------------------------------------------------------------Prob. 11.22 Cut section a-a as shown.

CD = 72 kN (Comp.)

Force δ: δ=

PL AE

δ=

72 000 N(4000 mm) 2500 mm 2 (207 ×103 MPa)

= 0.556 mm (shortening) 11-240

Temp. δ δ = αL (ΔT) = 11.7 × 10−6(4000)(28) = 1.310 mm (shortening) Total δ = 1.310 + 0.556 = 1.866 mm (shortening) -----------------------------------------------------Prob. 11.23 Refer to Prob. 5-25 ΔT = +40°F

L

A

P

PL AE

(ft)

2

(in. )

(k)

(in.)

1.00

13.33

0.80

0.0468

(T)

(+)

(+)

10.67

0.0171

0.0374

(C)

(–)

(+)

CD 15

FD

12

3.00

αLΔT (in.)

Total δ: δCD = 0.080 + 0.0468 = +0.1268 in. δFD = – 0.0171 + 0.0374 = +0.0203 in.

--------------------------------------------------------------Prob. 11.24 δ = αL (ΔT) = 0.0000065(100)(70-25) = 0.02925 ft PL δAE  P= AE L 0.02925(12)(0.016)(30, 000, 000) P= (100 − 0.02925)(12) = 140.4lb

δ=

Ptotal = 140.4 + 15 = 155.4 lb --------------------------------------------------------------11-241

Prob. 11.25 1-in. diam. rod: A = 0.7854 in.2 Due to tensioning:

σ=

P 10,000 = = 12,730 psi (tension) A 0.7854

Due to temp. rise of 30°F: σ = Eα(ΔT) = 30,000,000(0.0000065)(30) = 5850 psi (comp.)

Stress in rod = 12,730 – 5850 = 6880 psi (tension) Due to temp. fall of 50°F: σ = 30,000,000(0.0000065)(50) = 9750 psi (tension) Stress in rod = 12,730 + 9750 = 22,480 psi (tension) --------------------------------------------------------------Prob. 11.26 α = 0.0000093, E = 15,000,000 psi (a) σ = Eα(ΔT)  ΔT = s/(Eα) ΔT =

15, 000 = 107.5°F 15, 000, 000(0.0000093)

(b) In addition to (a): δ = αL (ΔT)  ΔT = δ/(αL) ΔT =

0.05 = 22.4 F° 0.0000093(20)(12)

ΔTtotal = 107.5 + 22.4 = 129.9°F ---------------------------------------------------------------

11-242

Prob. 11.27 ΔT to cause stress of 70 MPa: ΔT =

σ 70 = = 28.9°C 3 E α 207 × 10 (11.7 × 10−6 )

Initial temp to which member must be heated: 20°C + 28.9°C = 48.9°C --------------------------------------------------------------Prob. 11.28

Bar remains horizontal. δ for wire due to 100°F temp increase: δt1 = αL1 (ΔT) = 0.0000065(60)(100) = 0.0390 in. δt2 = αL2 (ΔT) = 0.0000065(30)(100) = 0.0195 in. δ with load applied:

δ P1 =

PL P1 (60) 1 1 = = 0.00002 P1 A1 E 0.1(30, 000, 000)

δP2 =

P2 L2 P2 (30) = = 0.000005P2 A2 E 0.2(30, 000, 000)

δt1 + δP1 = δt2 + δP2 ------- Eqn. I. ΣFy = 2P1 + P2 − 5000 = 0

P2 = 5000 − 2P1 ------------ Eqn. II Solve eqns. I & II simultaneously:  P1 = 183.3 lb  P2 = 4633 lb

--------------------------------------------------------------11-243

Prob. 11.29

δt2 + δP2 = δt1 ----------Eqn. I. δt1 = αL1(ΔT) = 0.0000065(60)(ΔT) δt2 = αL2(ΔT) = 0.0000065(30)(ΔT)

δ P2 =

P2 L2 5000(30) = = 0.025 in.

A2 E 0.2(30,000,000)

Subs. into Eqn. I and solve for ΔT: ΔT = 128.2°F Therefore, must heat wires to 68°F + 128.2 °F = 196.2°F --------------------------------------------------------------Prob. 11.30 (a) Unrestrained contraction: δst = αstLst (ΔT) = 0.0000065(4)(12)(100) = 0.0312 in. δcu = αcuLcu(ΔT) = 0.0000093(2)(12)(100) = 0.0223 in.

Total δ = 0.0312 + 0.0223 = 0.0535 in. (b) Force P applied to restore member to its original length:

 PL   PL  δ=  +   AE st  AE cu (more)

11-244

δ

P=

 L   L    +   AE st  AE cu 0.0535 = = 22,300 lb  4(12) 2(12)   (1)(30 × 106 ) + 2(15 × 106 )   

22,300 = 22, 300 psi 1 22,300 σ cu = = 11,150 psi 2 σst =

--------------------------------------------------------------Prob. 11.31 Wood: S.Pine σc(all) = 1250 psi

E = 1700 ksi Steel: A36 σc(all) = 20 ksi

Es = 30,000 ksi n = Es/Ew = 17.65 Assume steel at allowable: σw =

σ st 20, 000 = = 1133 psi < 1250 psi (O.K.) n 17.65

Aw = 32 in.2 As = 2(8)(0.25) + 2(4)(0.125) = 5.0 in.2 Pmax = Asσst + Awσw = 5.0(20,000) + 32(1133) = 136,300 lb --------------------------------------------------------------Prob. 11.32 Post: (Doug. fir) 150 mm × 150 mm

Aw = 1502 = 22.5 × 103 mm2 Ew = 12 × 103 MPa 11-245

σc(all)w = 7.24 MPa Steel plates:

Ast = 2(150)(10) = 3000 mm2 Est = 207 MPa σc(all)st = 147 MPa

n = Est/Ew = 17.25 Assume wood at allowable: σst =nσw = 17.25(7.24) = 124.9 MPa (<147 MPa (O.K.))

P = Astσst + Awσw = 3000(124.9) + 22.5 × 103(7.24) = 538 kN --------------------------------------------------------------Prob. 11.33

T = 8000 lb Ecu = 15,000 ksi Est = 30,000 ksi N = Es/Ecu = 2.0 σst = nσcu

Ast = 0.7854(0.5)2 = 0.1964 in.2 Acu = 0.7854(0.625)2 − Ast = 0.1104 in.2 σcu Acu + σst Ast = 8000 lb σcu(0.1104) + 2σst(0.1964) = 8000 lb σcu = 15,900 psi σst = 31,800 psi --------------------------------------------------------------Prob. 11.34

ADF = AHF = 11.5(5.5) = 63.25 in.2 EDF = 1700 ksi; EHF = 1400 ksi

n =EDF/EHF = 1.2143 σDF = nσHF = 1.2143 σHF 11-246

P = (Aσ)DF + (Aσ)HF 70,000 = 63.25(1.2143σHF) + 63.25σHF σHF = 500 psi

PHF = AHFσHF = 63.5(500) = 31,625 lb P = PHF + PDF PDF = 70,000 – 31,625 = 38,375 lb --------------------------------------------------------------Prob. 11.35

AS = 4.0 in.2, LS = 12 in.

ACI = 9.0 in.2 , LCI = 14 in.

ES = 30,000 ksi, ECI = 15,000 ksi

 PL   PL  (a)   +  = 0.01 in.  AE  S  AE  CI   12 14   = 0.01 in. +  4(30,000) 9(15,000)  P = 49.1 k

( b) σ =

P 49.1 = = 12.28 ksi A 4.0

--------------------------------------------------------------Prob. 11.36

Copper: Acu = 2(100)(3.2) = 640 mm2 Ecu = 103 × 103 MPa Steel:

Ast = 100(12.7) = 1270 mm2 Est = 207 × 103 MPa

n = Est/Ecu = 2.01 11-247

σst = nσcu = 2.01σcu P = Astσst + Acuσcu 225 × 103 N = 1270 × 10−6 m2(2.01σcu) + 640 × 10−6 m2(σcu) σcu = 70.5MPa; σst = 2.01(70.5) = 141.7MPa εcu = εst = σst/E =141.7/(207 × 103) = 684 × 10−6 --------------------------------------------------------------Prob. 11.37 L5 × 5 × ½ : A = 4.79 in.2 Redwoood: E = 1300 ksi, σc(all) = 1050 psi n = Est/Ew = 30,000/1300 = 23.1 Assume steel at allowable: σw =20,000/23.1= 866 psi < 1050 psi (O.K.) P = σstAst + σwAw = 20,000(4)(4.79) + 866(15.5)2 = 591 k --------------------------------------------------------------Prob. 11.38

Steel Pipe: Ast = 18.06 in.2, Est = 30,000 ksi Concrete: Ac = 88.0 in.2, Ec = 3120 ksi C.I.: ACI = 7.07 in.2, ECI = 15,000 ksi σstAst + σcAc + σCI ACI = 400 k (Eqn. I) εst = εc = εCI 

σ st Est

=

σc Ec

=

σ CI ECI

11-248



σ st 30,000

=

σc 3120

=

σ CI 15,000

30,000σ CI = 2σ CI 15,000 3120σ CI σc = = 0.208σ CI 15,000

σ st =

Sub. into Eqn. I: 2σCI (18.06) + 0.208σCI (88.0) + σCI(7.07) = 400 k σCI = 6.50 ksi σst = 2(6.50) = 13.0 ksi σc = 0.208(6.50) = 1.352 ksi --------------------------------------------------------------Prob. 11.39

Steel bars: Ast = 4(0.7854)(0.875)2 = 2.41 in.2 Est = 30,000 ksi Concrete: Ac = 142 – 2.41 = 193.6 in.2 Ec = 3605 ksi n = Est/Ec = 8.322 σst = nσc = 8.322σc P = Astσst + Acσc 100 = 2.41(8.322σc) + 193.6σc σc = 0.468 ksi σst = 8.322(0.468) = 3.89 ksi 11-249

Load carried by each material: Pc = 0.468(193.6) = 90.6 k 90.6 (100) = 90.6% 100 Pst = 3.89(2.41) = 9.4 k 9.4 (100) = 9.4% 100 --------------------------------------------------------------Prob. 11.40 Est = 207 × 103 MPa, Ast = 1250 mm2 Ecu = 103 × 103 MPa, Acu = 1800 mm2 The steel rod will initially carry the load until the 0.06 mm gap closes:

σ=

δE 0.06(207 ×103 ) = = 33.11 MPa L 375.06

The two materials then jointly resist the remainder of the load: 112 – 41.39 = 70.61 kN 70.61 kN = σstAst + σcuAcu

( Eqn. I)

n=Est/Ecu = 2.01 σst = nσcu = 2.01 σcu Subs. into Eqn. I 2.01σcu (1250) + σcu(1800) = 70.61 kN σcu = 16.37 MPa  σst = 2.01(16.37) + 33.11 = 66.0 MPa

--------------------------------------------------------------Prob. 11.41

11-250

Eal = 10,000 ksi; Est = 30,000 ksi ΣFy = Pst + Pal – 24,000 = 0

Eqn. I

ΣMA = Pal(10) – 24,000a = 0 Eqn. II Bar remains horizontal:  PL   PL   =   AE st  AE al Pst (12)(12) P (6)(12) = al (1)(30,000) 1.5(10,000)

δ st = δ al  

 Pst = Pal

Subs. into Eqn. I:  Pst = Pal = 12,000 lb

Subs. into Eqn. II:  a = 5.00 ft

Stresses: 12,000 = 12,000 psi 1 12,000 σ al = = 8,000 psi 1.5

σ st =

--------------------------------------------------------------Prob. 11.42

Steel: Ast = 5162 × 10−6 m2 Est = 207 × 109 Pa Brass: Abr = 3226 × 10−6 m2 Ebr = 97 × 109 Pa n = Est/Ebr = 207/97 = 2.13 σst = nσbr = 2.13σbr P = Astσst + Abrσbr

11-251

580 × 103 = 5162 × 10−6(2.13σbr) + 3226 × 10−6σbr  σbr =40.8 × 106 Pa = 40.8 MPa

σst = 2.13(40.8) = 86.9 MPa --------------------------------------------------------------Prob. 11.43

Each wire: 1/4″ diam., A = 0.49 in.2, L=20′ ΣFy = 2Pst + Pbr – 2000 = 0 ----- Eqn. I Box remains horizontal: δst = δbr  PL   PL    =   AE st  AE br

Pst (20)(12) Pbr (20)(12) = 0.049(30,000,000) 0.049(12,000,000)

 Pst = 2.5Pbr

Subs. into Eqn. I: 2(2.5Pbr) + Pbr = 2000 lb  Pbr = 333.3 lb Pst = 2.5(333.3) = 833.3 lb (more) Stresses: Bronze: σst = Steel : σst = (b) δst =

333.3 = 6800 psi 0.049

833.3 = 17, 010 psi 0.049

PL σL 17,101(20)(12) = = = 0.136 in.

AE E 30, 000, 000

---------------------------------------------------------------

11-252

Prob. 11.44 (Refer to Prob. 11.57)

Each wire: 1/4″ diam., A = 0.49 in.2, L=20′ ΣFy = 2Pal + Pst – 1500 = 0 ----- Eqn. I Box remains horizontal: δal = δst  PL   PL    =  AE  al  AE st

Pal (20)(12) Pst (20)(12) = 0.049(10,000,000) 0.049(30,000,000)

 Pst = 3.0Pal

Subs. into Eqn. I: 2Pal + 3.0Pal = 1500 lb  Pal = 300 lb Pst = 3.0(300) = 900 lb (a) Stresses: Aluminum : σ al =

300 = 6120 psi 0.049

Steel : σ st =

900 = 18,370 psi 0.049

(b) δst =

PL σL 18370(20)(12) = = = 0.147 in.

AE E 30, 000, 000

--------------------------------------------------------------Prob. 11.45

Steel: Diam. = 25 mm Ast = 490.9 × 10−6 m2 Lst = 1050 mm Est = 207 × 103 MPa 11-253

Brass: Diam. = 38 mm Abr = 1134 × 10−6 m2 Lbr = 500 mm Ebr = 97 × 103 MPa ΣFy = 2Pst + Pbr – (180 × 103) = 0

Eqn. I

δst = δbr  PL   PL    =   AE st  AE br

 Pst = 0.44Pbr

Subs. into Eqn. I: 2(0.44Pbr) + Pbr = 180×103 Pbr = 95.7×103 N 2Pst = 180×103 – 95.7×103 Pst = 42.2×103 N -------------------------------------------------------------Prob. 11.46

Anet = (4–1)(0.5) = 1.5 in.2

σ t (avg) =

P 30,000 = = 20,000 psi Anet 1.5

r/d = 0.5/3 = 0.167 Fig. 11.14: k = 2.35  P   = 2.35( 20,000) = 47,000psi  Anet 

σ t (max) = k 

---------------------------------------------------------------

11-254

Prob. 11.47

(a) r = 8 mm; r/d = 8/104 = 0.077 Anet = 13(120 − 2(8)) = 1352 mm2 Fig. 11.14: k = 2.4

σ t (max) =

kP 2.4(50 000) = = 88.8 MPa Anet 1352

(b) r = 21 mm, r/d = 21/78 = 0.27 Fig. 11.14: k = 1.85 Anet = 13(120 − 2(21)) = 1014 mm2

σ t (max) =

kP 1.85(50 000) = = 91.2 MPa Anet 1014

(c) r = 38 mm, r/d = 38/44 = 0.86 Fig. 11.14: k = 1.25 Anet = 13(120 − 2(38)) = 572 mm2

σ t (max) =

kP 1.25(50 000) = = 109.3 MPa Anet 572

--------------------------------------------------------------Prob. 11.48

(a) σ t =

P 12,000 = = 8000 psi A 4(0.375)

(b) σ t =

P 12,000 = = 10,700 psi Anet 3(0.375)

11-255

(c) r/d = 0.5/3 = 0.167: Fig 11.14: k = 1.7

σ t (max) =

kP = 1.7(10,700) = 18,190 psi Anet

--------------------------------------------------------------Prob. 11.49

Anet = (5 – 2)(0.375) = 1.125 in.2 r/d = 1/(5 − 2) = 0.333 Fig. 11.14: k = 2.22

σ t (max) = P=

σ t (all) Anet kP P= Anet k

22,000(1.125) = 11,150 lb 2.22

--------------------------------------------------------------Prob. 11.50

σt =

P 12,000 = = 9600 psi Anet 2.5(0.5)

σt(all) = 15,000 psi  P  15,000   k = = 1.56 A 9600 net  

σ t = k 

Fig. 11.14: Req’d r/d = 0.30 Req’d rmin = 0.30d = 0.30(2.5) = 0.75 in.

---------------------------------------------------------------

11-256

Prob. 11.51

d=75 – 19 = 56 mm r = 19/2 = 9.5 mm Ag = 75(10) = 750 mm2 = 750 × 10−6 m2 Ah = 19(10) = 190 mm2 = 190 × 10−6 m2 Anet = 560 × 10−6 m2 r/d = 9.5/56 = 0.17; Fig. 11.14: k = 2.35

σ avg =

P 180 × 10 3 = = 32.14 × 10 6 Pa −6 Aavg 560 × 10  P   = 2.35(32.14) = 75.5 MPa  Anet 

σ max = k 

--------------------------------------------------------------Prob. 11.52

(a) W = 2.0 in., D = 0.25 in.

Ag = 2(0.375) = 0.75 in.2 Ah = 0.25(0.375) = 0.094 in.2 d = W – D = 2.0 – 0.25 = 1.75 in. r/d = 0.125/1.75 = 0.071 Fig. 11.14: k = 2.56  P  4000   = 2.56   0.75 − 0.094   Anet 

σ t = k 

= 15,610 psi (b) W = 2.5 in., D = 0.50 in.

Ag = 2.5(0.375) = 0.938 in.2 Ah = 0.50(0.375) = 0.1875 in.2 11-257

Anet = 0.938 – 0.1875 = 0.75 in.2 d = W – D = 2.5 – 0.50 = 2.00 in. r/d = 0.25/2.00 = 0.125 Fig. 11.14: k = 2.43  P  4000   = 2.43   0.75   Anet 

σ t = k 

= 12,960 psi (c) W = 3.0 in., D = 1.00 in.

Ag = 3.0(0.375) = 1.125 in.2 Ah = 1.00(0.375) = 0.375 in.2 Anet= 1.125 – 0.375 = 0.75 in.2 d = W – D = 3.0 – 1.00 = 2.00 in. r/d = 0.50/2.00 = 0.25 Fig. 11.14: k = 2.28  P  4000   = 2.28   0.75   Anet 

σ t = k 

= 12,160 psi --------------------------------------------------------------Prob. 11.53

d = 125 – 38 = 87 mm r = 38/2 = 19 mm Ag = 125(10) = 1250 mm = 1250 × 10−6 m2 Ah = 38(10) = 380 mm2 = 380 × 10−6 m2 Anet = 870 × 10−6 m2 r/d = 19/87 = 0.22 Fig. 11.14: k = 2.3 11-258

 P    Anet 

σ max = k 

σ (all) Anet

150 ×10 6 (870 ×10 − 6 ) P= = k 2.3 3 = 56.7 ×10 N = 56.7 kN --------------------------------------------------------------Prob. 11.54

P sin 2θ 2A 12,000 = sin 96° = 30,400 psi 2(0.1963)

(a) τ ' =

P cos 2 θ A 12,000 = cos 2 0° = 61,100 psi 0.1963 P (c) σ n = cos 2 θ A 12,000 = cos 2 48° = 27,400 psi 0.1963 (b) σ n (max) =

--------------------------------------------------------------Prob. 11.55

A = 50 mm × 75 mm = 3.75 × 103 mm2 (a) τ’ max. on 45° plane:

11-259

τ' =

P 500 ×103 = = 66.7 MPa 2 A 2(3.75 ×103 )

P sin 2θ 2A 500 × 10 3 = sin 140° = 42.9 MPa 2(3.75 × 10 3 ) P 500 ×10 3 σ n = cos 2 θ = cos 2 70° = 15.60 MPa 3 A 3.75 ×10 (b) τ ' =

--------------------------------------------------------------Prob. 11.56 P sin 2θ 2A 8000 = sin 120 ° 2(1) = 3460 psi

τ' =

P cos 2 θ A 8000 = cos 2 60° = 2000 psi 1

σn =

--------------------------------------------------------------Prob. 11.57 P = 67 k A = 0.7854(6)2 = 28.3 in.2

11-260

P sin 2θ 2A 67 = sin 90° 2(28.3) = 1.184 ksi

τ' =

--------------------------------------------------------------Prob. 11.58 A = 4.0 in.2 τ(ult) = 1200 psi

τ' = P=

P sin 2θ 2A τ (ult) (2 A)

sin 2θ 1200(2)(4.0) = sin(2 × 80°) = 28,100 psi

--------------------------------------------------------------Prob. 11.59 A = 490.9 × 10−6 m2

(a) θ = 30°

τ' =

(more)

80 ×10 3 P sin 2θ = sin 60° 2A 2(490.9 × 10 − 6 )

= 70.6 × 10 6 Pa = 70.6 MPa 80 ×103 P 2 cos 2 30° σ n = cos θ = −6 A 490.9 ×10 6 = 122.2 ×10 Pa = 122.2 MPa 11-261

(b) Max. normal stress (θ = 0°) P 80 ×10 3 cos 2 θ = cos 2 0° A 490.9 ×10 − 6 = 163.0 × 10 6 Pa = 163.0 MPa

σ n (max) =

Max. shear stress (θ = 45°) ' = τ (max)

80 ×103 P sin 2θ = sin 90° 2A 2(490.9 ×10 − 6 )

= 81.5 ×10 6 Pa = 81.5 MPa --------------------------------------------------------------Prob. 11.60 A = 1.0 in.2 P = Aσ = 1.0(2000) = 2000 lb P 2000 sin 2θ = sin 60° 2A 2(1) = 866 psi

τ' =

σn =

2000 P cos 2 θ = cos 2 60° = 500 psi 1 A

--------------------------------------------------------------Prob. 11.61 On plane CD: τ' = 35 MPa

P sin 2θ = 35 MPa 2A P 2(35 ×10 6 ) = = 80.83 ×10 6 Pa A sin 60° P σ n = cos 2 θ = 80.83 ×10 6 cos 2 30° A = 60.6 ×10 6 Pa = 60.6 MPa

τ' =

11-262

----------------------------------------------------------------Prob.. 11.62 A = 50(75) 5 = 3750 mm2 = 37 750 × 10−6 m2 On 50 0° plane: τ' = 138 MPa (θ = 40°) 4 P sin 2θ = 138 MPaa 2A 138 1 × 10 6 ( 2)((3750 × 10 − 6 ) P= sin( 2 × 40 °)

(a) τ ' =

= 1.051 × 10 6 N = 1.051 MN M

P 1.051×100 6 cos 2 θ = cos 2 40° −6 A 3750 × 10 0 6 = 164.5 × 10 Pa = 164..5 MPa

(b) σ n =

----------------------------------------------------------------Prob.. 11.63 (a)

(b) σ n = τ sin 2θ = 10,000 sin n 70° = 9400 psi p -----------------------------------------------------------------

11-263

Prob. 11.64 (a) See element of Prob. 11.24:

τ ' = τ cos 2θ 6000 = 10,000 cos 2θ  6000  2θ = cos −1   = 53.13°  10,000  θ = 26.57° (b)

Use element of unit thickness and use diagonal d = 1. Therefore, h = sinθ and w = cosθ. ΣF parallel to the diagonal plane: –6000d(1) – 10,000sin26.57°h(1) + 10,000cos26.57°w(1) = 0 –6000(1)(1) – 10,000sin226.57°(1) +10,000cos226.57°(1) = 0 –6000 – 2000 + 8000 = 0 (O.K.) --------------------------------------------------------------Prob. 11.65 6” diam. compression member: A = 28.27 in.2 Based on shear stress (max. when θ = 45°): P sin 2θ 2A τ '(max) (2 A) 5000(2)(28.27) Pmax = = sin 2θ sin 90°

τ' =

11-264

Pmax = 283,000 lb Based on comp. stress (max when θ = 0°): Pmax = σc(max)A = 12,000(28.27) = 339,000 lb  Controlling Pmax = 283,000 lb

--------------------------------------------------------------Prob. 11.66 A = 150(200) = 30 × 103 mm2 = 30 × 10−3 m2 σc(all) = 6 MPa = 6 × 106 Pa τ(all) = 1.6 MPa = 1.6 × 106 Pa Based on comp. stress (max when θ = 0°): Pmax = σc(all)A = 6 × 106(30 × 103) = 180 × 103 N Based on shear stress (max. when θ = 45°): P sin 2θ 2A τ (all) (2 A) 1.6 × 10 6 ( 2)(30 × 10 −3 ) = Pmax = sin 2θ sin 90° = 96.0 × 10 3 N (Controls)

τ' =

--------------------------------------------------------------Prob. 11.67 Tens. and/or comp. stress (max. at 45°) (a) σn = τsin2θ = 5000(sin 2 × 45°) = 5000 psi (b)

---------------------------------------------------------------

11-265

Prob. 11.68 Tens. and/or comp. stress (max. at 45°) (a) σn = τ sin 2θ = 9200(sin 2 × 45°) = 9200 psi (b)

---------------------------------------------------------------

11-266

Prob. 12.1

------------------------------------------------------------Prob. 12.2 Tint(A) = Fd = 1(18) = 18 k-in.

Tint(B) = 18 + 100 = 118 k-in.

-------------------------------------------------------------

12-267

Prob. 12.3

--------------------------------------------------------------Prob. 12.4 J=

T=

π (854 − 404 ) = 4.87 × 106 mm 4 32 = 4.87 × 10−6 m 4

τ (all) J c

(68 × 106 Pa)(4.87 × 10-6 m 4 ) 85 / 2 × 10−3 m = 7.79 × 103 N ⋅ m = 7.79 kN ⋅ m =

--------------------------------------------------------------Prob. 12.5 π d 4 π (5) 4 = = 61.36 in.4 32 32 τ J 10, 000(61.36) TR = (all) = = 245, 000 lb-in. c 2.5 J=

--------------------------------------------------------------Prob.12.6 TR =

τ (all) J c

12-268

π (d 4 − d14 ) π (1004 − 754 ) = = 6.71× 106 mm 4 32 32 62 MPa(6.71× 106 mm 4 ) TR = = 8.32 × 106 N ⋅ mm 50 mm Tc 8.32 × 106 N ⋅ mm(37.5 mm) τ= = J 6.71 × 106 mm 4 = 46.5 MPa J=

--------------------------------------------------------------Prob. 12.7

--------------------------------------------------------------Prob. 12.8

-----------------------------------------------------------Prob. 12.9

--------------------------------------------------------------12-269

Prob. 12.10

-------------------------------------------------------------Prob. 12.11 J=

π d 4 π(3.5)4 = = 14.73 in.4 32 32

τ=

Tc 5000(12)(3.5 / 2) = = 7130 psi 14.73 J

-----------------------------------------------------Prob. 12.12

-----------------------------------------------------------12-270

Prob. 12.13

--------------------------------------------------------------Prob. 12.14

---------------------------------------------------------------

12-271

Prob. 12.15

--------------------------------------------------------------Prob. 12.16

--------------------------------------------------------------Prob. 12.17

--------------------------------------------------------------12-272

Prob. 12.18

--------------------------------------------------------------Prob. 12.19

-----------------------------------------------------------Prob. 12.20

---------------------------------------------------------------

12-273

Prob. 12.21

--------------------------------------------------------------Prob. 12.22

--------------------------------------------------------------Prob. 12.23

--------------------------------------------------------------12-274

Prob. 12.24

--------------------------------------------------------------Prob. 12.25

--------------------------------------------------------------Prob. 12.26

---------------------------------------------------------------

12-275

Prob. 12.27

--------------------------------------------------------------Prob. 12.28

--------------------------------------------------------------12-276

Prob. 12.29

--------------------------------------------------------------Prob. 12.30

--------------------------------------------------------------Prob. 12.31

---------------------------------------------------------------

12-277

Prob. 12.32

--------------------------------------------------------------Prob. 12.33

--------------------------------------------------------------

12-278

Prob. 12.34

--------------------------------------------------------------Prob. 12.35

---------------------------------------------------------12-279

Prob. 12.36

--------------------------------------------------------------Prob. 12.37

------------------------------------------------------------Prob. 12.38

--------------------------------------------------------------12-280

Prob. 12.39

--------------------------------------------------------------Prob. 12.40

--------------------------------------------------------------Prob. 12.41

---------------------------------------------------------------

12-281

Prob.. 12.42

----------------------------------------------------------------Prob.. 12.43

----------------------------------------------------------------Prob.. 12.44

----------------------------------------------------------------12-282

Prob. 12.45

--------------------------------------------------------------Prob. 12.46

--------------------------------------------------------------Prob. 12.47 π (1.5) 2 = 0.497 in.4 Tc 1500(1.5 / 2) 32 τ= = = 2260 psi J 0.497 1500(1000) Tnr HP = = = 23.8 hp 63,025 63,025 J=

---------------------------------------------------------------

12-283

Prob. 12.48 (a) Eqn. 12.7. (nr = 3000 rpm) T=

63, 025(120) 2520 lb-in.

3000

Req'd d = 3

16T 16(2520) =3 = 1.171 in. π τ(all) π(8000)

Use a 1 163 ″ solid shaf t (b) nr = 300 rpm T=

63, 025(120) 25, 200 lb-in.

300

Req'd d = 3

16T 16(25, 200) =3 = 2.52 in. π τ(all) π(8000)

Use a 2 85 ″ solid shaf t (c) For 1 163 ″ solid shaft: π d 4 π (1.188) 4 = = 0.1956 in.4 32 32 TL 2520(10)(12) = = 0.1288 rad θ= JG 0.1956(12,000,000) = 7.38° J=

For 2 85 ″ solid shaft: π d 4 π (2.625) 4 = = 4.66 in.4 32 32 25,200(10)(12) TL = = 0.0541 rad θ= JG 4.66(12,000,000) = 3.10° J=

---------------------------------------------------------------

12-284

Prob. 13.1

CC moment: +

(a)

ΣMA = +RB(24) – 6(24)(12) = 0 RB = +72.0 k ↑ ΣMB = – RA (24) + 6(24)(12) = 0 RA = + 72.0 k ↑ (b)

ΣMA = +RB(16) – 12(8) – 10(3) = 0 RB = +7.88 k ↑ ΣMB = –RA (16) +12(8) + 10(13) = 0 RA = + 14.13k ↑ ----------------------------------------------------Prob. 13.2

CC moment: +

(a)

ΣMA = –54(9) – 9(13) +RB(18) = 0 RB = +33.5 k ↑ ΣMB = +54(9) + 9(5) – RA(18) = 0 RA = + 29.5 k ↑

13-285

(b)

ΣMA = –3(4) – 5(10) – 36(9) + RB(18) = 0 RB = +21.44 k ↑ ΣMB = −RA (18) + 3 (14) +36(9) +5(8) = 0 RA = + 22.56 k ↑ ----------------------------------------------------Prob. 13.3

CC moment: +

(a)

ΣMA = −24(4) +RB(20) = 0 RB = +4.80 k ↑ ΣMB = −RA (20) + 24(16) = 0 RA = + 19.2 k ↑ (b)

ΣMA = −30(3) − 6(7.5) + RB(14) = 0 RB = +9.64 k ↑ ΣMB = −RA (14) + 30(11) +6(6.5) = 0 RA = + 26.4 k ↑ -----------------------------------------------------

13-286

Prob. 13.4 (a)

ΣMA = −325(6.5) − 75(16) + RB(20) = 0 RB = +165.6 kN ↑ ΣMB = − RA (20) + 325(13.5) + 75(4) = 0 RA = + 234 kN ↑ (b)

ΣMA = −40(2) −140(6) − 60(11) + RB(12) = 0 RB = +131.7 kN ↑ ΣMB = −RA (12) + 40(10) + 140(6) +60(1) = 0 RA = + 108.3 kN ↑ ----------------------------------------------------Prob. 13.5

CC moment: +

(a) ΣMA = +6(4) −10(8) + RB = 0 RB = 4.00 k ↑ ΣMB = +6(18) −RA(14) +10(6) = 0 RA = 12.00 k ↑

13-287

(b)

ΣMA = −20(10) + RB(12) − 10 0(20) = 0 RB = 33.3 k ↑ ΣMB = −RA(12) + 20(12) − 10 0(8) = 0 RA = −3.33 k RA = 3.33 k ↓ ------------------------------------------------------Prob.. 13.6

ΣMA = −54(9) − 16.5(12.5) 1 +R + B(13) = 0 RB = 52.3 k ↑ ΣMB = −RA(13) +54(4) + + 16.5 5(0.5) = 0 RA = 17.25 k ↑ (b)

ΣMA = +RB(18) − 24(14) − 63 3(9) = 0 RB = 50.2 k ↑ ΣMB = −RA(18) + 63(9) + 24((4) = 0 RA = 36.8 8k↑ ------------------------------------------------------13-288

Prob. 13.7 (a)

ΣMA = −10(5) −40(10) +RB(20) = 0 RB = +22.5 k ↑ ΣMB = − RA (20) +10(15) +40(10) = 0 RA = +27.5 k ↑ (b)

ΣMA = −16(8) + RB (20) = 0 RB = +6.4 k ↑ ΣMB = −RA (20) +16(12) = 0 RA = +9.6 k ↑ (c)

ΣMA = −20(4) − 20(13) + RB(22) = 0 RB = + 15.45 k ↑ ΣMB = −RA(22) +20(18) + 20(9) = 0 RA = +24.5 k ↑

13-289

(d)

ΣMA = −28(7) − 10(20) − 12(26)+ RB(30) = 0 RB = + 23.6 k ↑ ΣMB = −RA(30) +28(23) +10(10) +12(4) = 0 RA = +26.4 k ↑ -----------------------------------------------------Prob. 13.8 (a)

ΣMA = −400(5) − 200(13) + RB(10) = 0 RB = + 460 lb ↑ ΣMB = −RA(10) +400(5) + 200(3) = 0 RA = +140 lb ↑ (b)

ΣMA = +4000(2) + RB(10) −6000(11) = 0 RB = + 5800 lb ↑ ΣMB = +4000(12) − RA(10) − 6000(1) = 0 RA = +4200 lb ↑

13-290

(c)

ΣMA = −9000(4) − 10,000(13) +RB(14) = 0 RB = +11,860 lb ↑ ΣMB = −RA(14) +9000(10) + 10,000(1) = 0 RA = +7140 lb ↑ (d)

ΣMA = −48(6) + RB(12) −18(14) = 0 RB = + 45 k ↑ ΣMB = −RA(12) +48(6) −18(2) = 0 RA = +21 k ↑ ---------------------------------------------------------------Prob. 13.9 (a)

ΣMA = −200(4) − 315(12.5) + RB(17) = 0 RB = +279 kN ↑ ΣMB = +200(13) +315(4.5) − RA(17) = 0 RA = +236 kN ↑

13-291

(b)

ΣMA = −42(3.5) − 30(4) − 20(9) + RB(10) = 0 RB = + 44.7 kN ↑ ΣMB = −RA(10) +42(6.5) +30(6) − 20(1) = 0 RA = +47.3 kN ↑ -----------------------------------------------------Prob. 13.10 (a)

ΣMA = +80(3) − 180(3) + RB(10) = 0 RB = +30.0 kN ↑ ΣMB = +80(13) − RA(10) +180(7) = 0 RA = +230 kN ↑ (b)

ΣMA = +80(2) − 120(5.33) + RB(12) − 40(14)= 0 RB = +86.6 kN ↑ ΣMB = +80(14) − RA(12) +120(6.67) − 40(2) = 0

RA = +153.4 kN ↑ --------------------------------------------------------------13-292

Prob. 13.11 (a)

V = +6.55 kN M = +6.55(4) = 26.2 kN∙m

V = +6.55 − 12 = −5.45 kN M = +6.55(7) − 12(2) = + 21.9 kN∙m V = +11.11 − 10 = + 1.11 kN M = +11.11(4) − 10(1) = +34.4 kN∙m (b)

@ 7m (left of the point load):

13-293

V = +11.11 + − 10 = +1.11 kN N M = +11.11(7) + − 10(4) = +37 7.8 kN∙m @ 7m m (right of po oint load):

V = +11.11 + − 10 − 20 = −18.8 89 kN M = +11.11(7) + − 10(4) = +37 7.8 kN∙m ------------------------------------------------------Prob.. 13.12 (a)

w = 2 k/ft, L = 20 ft RA = RB = ½ (2)(20) = 20 2 k↑

V & M 3 ft from left: V= +2 20 − 6 = +14 + k M = +20(3) + − 6(1 1.5) = + 51 k-ft V&M M 8 ft from left:

V = +20 + − 16 = + 4 k M = +20(8) + − 16((4) = + 96 k--ft

13-294

(b)

V&M at 3 ft from left: V = +14.25 − 0 = +14.25 k M = +14.25(3) = +42.75 k-ft V&M 8 ft from left:

V = +14.25 − 15 = −0.75 k M = +14.25(8) − 15(2) = +84 k-ft -----------------------------------------------------

13-295

Prob. 13.13 (a)

ΣMA = − 100(6.67) + RB(20) = 0 RB = 33.3 k ↑ ΣMB = −RA (20) + 100(13.33) = 0 RA = 66.7 k ↑

V = +66.7 − 25 − 50 = −8.3 k M = +66.7(10) − 25(6.67) − 50(5) = +250 k-ft (b)

At midspan, just left of point load:

13-296

V= +39 − 30 = +9 k M = +39(10) − 30(5) = +240 k-ft At midspan, just right of point load:

V = +39 − 30 − 18 = −9 k M = +39(10) − 30(5) = +240 k-ft ----------------------------------------------------Prob. 13.14 (a) RA = RB = ½ (30)(2) = 30 k↑ At 5 ft from left, just left of support: V = −10 k M = −10(2.5) = −25 k-ft At 5 ft from left , just right of support:

13-297

V = −10 + 30 =+ 20 k M = −10(2.5) = −25 k-ft At 15 ft:

V = + 30 − 30 = 0 M = +30(10) − 30(7.5) = +75.0 k-ft (b)

ΣMA = 0  RB = +57.6 k ↑ ΣMB = 0  RA = +9.4 k ↑ At 5 ft from left: V = +9.4 − 15 = −5.6 k M = +9.4(5) − 15(2.5) = + 9.5 k-ft At 15 ft from left, just left of 10-k load: 13-298

V = +9.4 − 45 + 57.6 = +22 k M = +9.4(15) − 45(7.5) + 57.6(3) = −23.7 k-ft At 15 ft from left, just right of 10-k load:

V = +9.4 − 45 +57.6 − 10 = +12 k M = = +9.4(15) − 45(7.5) + 57.6(3) = −23.7 k-ft -------------------------------------------------------------Prob. 13.15 (a)

At 5m from left:

13-299

V = +47.2 − 25 − 20 = +2.2 kN M = +47.2(5) − 25(2.5) − 20(1) = +153.5 kN∙m At 10 m from left:

V = +47.2 − 20 − 50 = −22.8 kN M = +47.2(10) − 20(6) − 50(5) = +102 kN∙m b)

ΣMA = +10(4) − 20(6) − 32(10) +RB(10) = 0 RB = +40.0 kN ↑ ΣMB = 10(14) − RA(10) +20(4) = 0 RA = 22 kN ↑ At 5 m from left:

13-300

V = −10 + 22 = +12 kN M = −10(5) + 22(1) = −28 kN∙m At 10 m from left:

V = −10 + 22 = +12 kN M = −10(10) +22(6) = +32 kN∙m ----------------------------------------------------Prob. 13.16 (a)

At 2 m from left: V = −80 kN M = −80(2) = −160 kN∙m At 8 m from left:

13-301

V = − 80 + 230 − 150 = 0 M = − 80(8) + 230(5) − 150(2.5) = +135 kN∙m (b)

At 2 m from left: V = –40 kN M = –40(1) = –40 kN∙m

At 8 m from left:

V = − 80 + 153.4 − 30 = + 43.4 kN M = − 80(6) + 153.4(4) − 30(1.33 ) = 93.7 kN∙m ----------------------------------------------------------------

13-302

Prob. 13.17

ΣMA = + 16(4) − 80(6) +RB(16) = 0 RB = +26.0 k ↑ ΣMB = +16(20) − RA(16) + 80(10) = 0 RA = +70.0 k ↑ At 6 ft from left:

V = +70 − 16 − 24 = + 30.0 k M = − 16(6) − 24(3) + 70(2) = –28 k-ft At 16 ft from left:

V = − 16 + 70 − 64 = –10 k M = − 16(16) + 70(12) − 64(8) = +72.0 k-ft -----------------------------------------------------Prob. 13.18 (Refer to Prob. 13.8) (a) At 4 ft from left: V = + 140 lb M = +140(4) = +560 k-ft At 10 ft from left (just left of RB): 13-303

V = +140 + – 400 = –260 lb M = +140(10) + – 400(5) 4 = –60 00 lb-ft At 10 0 ft from leftt (just right of o RB): V = +140 + – 400 +460 = +200 lb M = –600 – k-ft (b) At 4 ft from left: V = +4200 + – 4000 0 = +200 + lb M = +4200(2) + – 4000(4) 4 = –7600 – lb-ft At 10 0 ft from leftt:

V = +4200 + – 4000 0 – 1200 = –1000 – lb M = +4200(8) + – 4000(10) 4 – 1200(1) 1 = –7600 – lb-ft -------------------------------------------------------------

13-304

Prob. 13.19

ΣMA = 0  RB = 106.7 kN ↑ ΣMB = 0  RA = 43.3 kN ↑ At 2 m from left:

V = +43.3 – 40 – 40 = –36.7 kN M = +43.3(2) – 40(1) – 40(1) = +6.6 kN∙m At 3.5 m from left:

V = + 43.3 – 40 – 70 + 106.7 = +40 kN M = +43.3(3.5) – 40(2.5) – 70 (1.75) + 106.7(0.5) = –17.6 kN∙m ------------------------------------------------------13-305

Prob. 13.20 (Refer to Prob. 13.4(a))

At 10 m from left:

V = +234.4 – 250 = –15.6 kN M = +234.4(10) – 250(5) = +1094 kN∙m At 16 m from left (just left of 75 kN load):

V = +234.4 – 325 = –90.6 kN M = +234.4(16) – 325(9.5) = +663 kN∙m At 16 m from left (just right of 75 kN load):

V = +234.4 – 325 – 75 = –165.6 kN M = +234.4(16) – 325(9.5) = +663 kN∙m ---------------------------------------------------------------13-306

Prob. 13.21

---------------------------------------------------------------Prob. 13.22

---------------------------------------------------------------

13-307

Prob. 13.23

----------------------------------------------------------------Prob. 13.24

---------------------------------------------------------------

13-308

Prob. 13.25

---------------------------------------------------------------Prob. 13.26

13-309

-----------------------------------------------------Prob. 13.27

--------------------------------------------------------------13-310

Prob. 13.28

---------------------------------------------------------------

13-311

Prob. 13.29

---------------------------------------------------------------

13-312

Prob. 13.30

---------------------------------------------------------------Prob. 13.31

---------------------------------------------------------------13-313

Prob. 13.32

--------------------------------------------------------------Prob. 13.33

--------------------------------------------------------------13-314

Prob. 13.34

--------------------------------------------------------------Prob. 13.35

--------------------------------------------------------------13-315

Prob. 13.36

--------------------------------------------------------------Prob. 13.37

--------------------------------------------------------------13-316

Prob. 13.38

--------------------------------------------------------------Prob. 13.39 (Refer to Prob. 13.1(b)

---------------------------------------------------------------

13-317

Prob. 13.40 (Refer to Prob. 13.8(a))

---------------------------------------------------------------Prob. 13.41 (Refer to Prob. 13.2(a))

---------------------------------------------------------------

13-318

Prob. 13.42 (Refer to Prob. 13.2(b)

---------------------------------------------------------------Prob. 13.43 (Refer to Prob. 13.3(a))

---------------------------------------------------------------

13-319

Prob. 13.44 (Refer to Prob. 13.3(b))

---------------------------------------------------------------Prob. 13.45 (Refer to Prob. 13.4(b))

--------------------------------------------------------------13-320

Prob. 13.46 (Refer to Prob. 13.8(b))

---------------------------------------------------------------Prob. 13.47 (Refer to Prob. 13.10(a))

---------------------------------------------------------------

13-321

Prob. 13.48 (Refer to Prob. 13.10(b))

---------------------------------------------------------------Prob. 13.49 (Refer to Prob. 13.11(a))

--------------------------------------------------------------13-322

Prob. 13.50

--------------------------------------------------------------Prob. 13.51

--------------------------------------------------------------13-323

Prob. 13.52

-----------------------------------------------------Prob. 13.53 For abs. max. V: Max. V = RA ΣMB = –RA(30) + 20(30) + 20(20) = 0 RA = +33.3 k ↑ For abs. max. M:

Max. M @ C: M = +16.67(12.5) =208 k-ft --------------------------------------------------------------13-324

Prob. 13.54 ΣMA: 38x = 12(16) x = 5.052 ft For abs. max. V:

Max. V = RB ΣMC = –RB (40) +26(40) +12(24) = 0 RB = 33.2 k ↑ For abs. max. M:

Max. M occurs @ point D: MD = +16.6(17.475) = 290 k-ft ------------------------------------------------------Prob. 13.55

x=

142(4.30) + 142(13.30) = 7.81 m 320

13-325

For max. M:

Under the left 142 kN load: M = +138.3(11.25) – 36(4.3) = +1401 kN∙m For max. V:

V = RR =

12.7 17.0 (36) + (142) + 142 = 252 kN 26.0 26.0

-----------------------------------------------------Prob. 13.56 For absolute max. V:

Max. V = RA ΣMB = –RA (15) + 200(15) + 200(10) = 0 RA = 333 kN ↑ For absolute max. M:

13-326

ΣMB = –RA (15) +400(6.25) = 0 RA = 166.7 kN Max. M occurs at point C: MC = +166.7(6.25) = +1042 kN∙m -----------------------------------------------------Prob. 13.57

ΣM L : 12 x = 1(8) + 4(16)  x = 6.0 ft For absolute max. V:

Max. V = RA ΣMB = –RA (40) + 7(40) + 1(32) + 4(24) = 0 RA = 10.2 k ↑ For absolute max. M:

13-327

ΣMB = –RA (40) +12(17) = 0 RA = 5.1 k ↑ Max. M occurs at point C: MC = +5.1(17) = +86.7 k-ft ---------------------------------------------------------------Prob. 13.58

R = 23 k, x =6.78 ft (a) Max. M under load C

RL = 11.85 k ↑ MC = +11.85(20.61) – 5(8) – 5(4) = 184.2 k-ft (b) Max. M under load D:

13-328

RR = 10.0 k ↑, MD = 10(17.39) = 173.9 k-ft (c) Absolute max. V:

Max. V = RR = 20 k -----------------------------------------------------

13-329

Prob. 14.1 (a) bh 2 5.5(9.5) 2 = 6 6 3 = 82.7 in.

Sx =

(b) π (d 4 − d14 ) 64 π (1144 − 1004 ) = 64 = 3.38 × 106 mm 4

I=

I 3.38 ×106 = 59.3 ×103 mm3 S= = c 114 / 2

(c) Refer to Example 14.16 I y = Σ( I o + Ad 2 ) = 2( 112)(2)(16) 3 + 112 (24)(1) 3 = 1367 in. 4 Sy =

Iy c

=

1367 = 170.9 in.3 8

-------------------------------------------------------------Prob. 14.2 (a)

14-330

a1 = 36 in.2, a2 = 40 in.2, Σa = 76 in.2 36(11.5) + 40(5) = 8.079 in.

76 12(3) 3 I x1 = + 36(3.421) 2 = 448.3 in. 4 12 y=

4(10) 3 + 40(3.079) 2 = 712.5 in. 4 12 ΣI x = 1160.8 in. 4 I x2 =

S x (top) = S x (bott) =

Ix c(top) Ix c(bott)

=

1160.8 = 236 in.3 4.921

=

1160.8 = 143.7 in.3 8.079

(b)

W24 × 162: Ix = 5170 in.4

18(1)3  I x (PL) = 2  + 18(13)2  = 6090 in.4  12  ΣIx = 5170 + 6090 = 11,260 in.4 𝑆 =

𝐼 11,260 = = 834 in. 𝑐 13.5

--------------------------------------------------------------Prob. 14.3 W18 × 71: Sx = 127 in.3 wL2 (0.40 + 0.071)( 28) 2 = = 46.16 k - ft 8 8 M 46.16(12) = = 4.36 ksi fb = Sx 127

M =

--------------------------------------------------------------14-331

Prob. 14.4 W18 × 71: Fy = 15.8 in.3 M = 46.16 k - ft fb =

M 46.16(12) = = 35.1 ksi Sy 15.8

--------------------------------------------------------------Prob. 14.5 See Prob. 14.2(a) wL2 500 (12) 2 = 9000 lb - ft 8 8 M 9000 (12 ) = = 752 psi Max. f b = Sx 143 .7 M =

-------------------------------------------------------------Prob. 14.6 See Prob. 14.2(b) wL2 3( 40) 2 = = 600 k - ft 8 8 M 600 (12) = = 8.63 ksi Max. f b = Sx 834 M =

-------------------------------------------------------------Prob. 14.7 Steel rod: 25 mm diam. π d3

π (25 ×10 −3 ) 3

= 1.534 ×10 −6 m 3 32 32 160 M = = 104.3 ×106 N/m 2 fb = −6 S 1.534 ×10 = 104.3 MPa

S=

=

-------------------------------------------------------------Prob. 14.8 Square steel bar: 38 mm x 38 mm bh 2 38(38) 2 = = 9.15 ×103 mm3 6 6 = 9.15 ×10 −6 m 3 M 460 N ⋅ m fb = = = 50.3 ×10 6 N/m 2 S x 9.15 ×10 -6 m 3

Sx =

= 50.3 MPa

-------------------------------------------------------------14-332

Prob. 14.9 A992: Fy = 50 ksi, W36 × 302: Zx = 1280 in.3 (Consider only the strong axis) MR = 0.6FyZx = 0.6(50)(1280) = 38,400 k-in. = 3200 k-ft -------------------------------------------------------------Prob. 14.10 (a) Beam: 6”x16” (nominal) Sx = bh2/6 = 6(16)2/6 = 256 in.3 MR = FbSx = 1000(256) = 256,000 lb-in. = 256 k-in. (b) Beam 6”x16” (S4S) MR = FbSx = 1000(220) = 220,000 lb-in. = 220 k-in.

-------------------------------------------------------------Prob. 14.11 (a)

  10(2) 3 I x1 = 2 + 20(6) 2  = 1453.3 in.4   12  2(10) 2   = 333.3 in.4 I x 2 = 2 12   Total I x = 1786.6 in.4

Sx = Ix/c = 1786.6/7 = 255 in.3 14-333

(b)

a1 = 12 in.2, a2 = 24 in.2, a3 = 16 in.2 Total a = 52 in.2 12(11) + 24(1) + 16(6) = 4.846 in.

52 6( 2 ) 3 I x1 = + 12(6.154) 2 = 458.5 in.4 12 y=

I x2 =

12(2) 3 + 24(3.846) 2 = 363.0 in.4 12

I x3 =

2(8) 3 + 16(1.154) 2 = 106.6 in.4 12

Total Ix = 928.1 in.4 Sx = Sx =

Ix c(bott) Ix c(top)

=

928.1 = 191.5 in.3 4.846

=

928.1 = 129.7 in.3 7.154

-------------------------------------------------------------Prob. 14.12 (See Prob. 14.29(a)) wL2 = 20,250 lb - ft 8 M 20,250 (12 ) = = 953 psi fb = Sx 255

M =

---------------------------------------------------------------

14-334

Prob. 14.13

100(150) 3 (a) I x = = 28.13 × 10 6 mm 4 12 = 28.13 × 10 −6 m 4

fb =

Mc 2600 N ⋅ m(75 ×10 -3 m) = I 28.13 ×10 -6 m 4 = 6.93 ×10 6 N/m 2 = 6.93 MPa

(b) fb =

My 2600 N ⋅ m(50 × 10 −3 m) = I 28.13 ×10 − 6 m 4 = 4.62 × 10 6 N/m 2 = 4.62 MPa

--------------------------------------------------------------Prob. 14.14 (Refer to Prob. 14.13) Ix =

150(100) 3 = 12.5 × 10 6 mm 4 12 = 12.5 × 10 −6 m 4

(a) fb =

Mc 2600(50 ×10 −3 ) = = 10.4 ×10 6 N/m 2 −6 12.5 × 10 I = 10.4 MPa

(b) fb =

My 2600(25 × 10 −3 ) = = 5.20 ×10 6 N/m 2 12.5 ×10 −6 I = 5.20 MPa

---------------------------------------------------------------

14-335

Prob. 14.15

M = 12(2) = 24 lb-ft

S=

π d3

fb =

32

=

π (0.5) 3 32

= 0.01227 in.3

24(12) M = = 23,500 psi S 0.01227

--------------------------------------------------------------Prob. 14.16

W530 × 101: Sx = 2290×103 mm3 wt. = 9.81(101) = 991 N/m 991(13) 2 M = + 54 × 10 3 ( 4.33) 8 = 255 × 10 3 N ⋅ m fb =

M 255 × 10 6 N ⋅ mm = 111.4 MPa = Sx 2290 × 10 3 mm 3

-------------------------------------------------------------Prob. 14.17

a1 = 4 in.2 a2 = 3 in.2 Σa = 7 in.2 14-336

y=

Σay 4(3.5) + 3(1.5) = = 2.643 in. Σa 7

4(1) 3 + 4(0.857) 2 = 3.271 in.4 12 1(3) 3 I x2 = + 3(1.143) 2 = 6.169 in.4 12 Total I x = 9.44 in.4 I x1 =

M = 450(6) = 2700 lb-ft f b (top) = f b (bott) =

Mc top Ix

=

Mc(bott) Ix

2700(1.357 )(12) = 4660 psi 9.44 =

2700(12)(2.643) = 9070 psi 9.44

--------------------------------------------------------------Prob. 14.18 W920 × 390

Fy = 345 MPa Zx = 18,000 × 103 mm3

MR = 0.60FyZx = 0.60(345)(18,000 × 103) = 3.73 × 109 N∙mm = 3.73 × 103 kN∙m --------------------------------------------------------------Prob. 14.19

f b (max) =

M 13,080(12) = 15,840 psi 9.91 Sx

14-337

-------------------------------------------------------------Prob. 14.20 (a) f v =

3V 3(20 ×103 ) = = 3.13 MPa 2 A 2(80)(120)

(b)

A = 0.7854d 2 = 0.7854(120) 2 = 11.31×103 mm 2 fv =

4V 4(20 ×103 ) = = 2.36 MPa 3 A 3(11.31×103 )

--------------------------------------------------------------Prob. 14.21

Vmax = RA = RB = 16,000/2 = 8000 lb (a) f v (max) =

1.5V 1.5(8000) = = 100 psi 6(20 A

(b) 4” above and below the N.A.;

6(20) 3 12 = 4000 in.4

I NA =

y = 7 in. a1 = 36 in.2

Q = a1 y = 36(7) = 252 in.3 f v ( 4) =

VQ 8000(252) = = 84 psi 4000(6) Ib

8” above & below N.A.:

14-338

y = 9 in. a2 = 12 in.2 Q = a2 y = 12(9) = 108 in.3

f v (8) =

VQ 8000(108) = = 36 psi 4000(6) Ib

--------------------------------------------------------------Prob. 14.22

ΣMA = –40(19) + RB(16) –1.03(19)(9.5) = 0 RB = +59.1 k ↑ ΣFy = +59.1 + RA – 40 –1.03(19) = 0 RA = +0.47 k ↑

Vmax = 43.1 k (a)

W14 × 30: A = 8.85 in.2

Ix = 291 in.4 0.385   Qflange = 6.73(0.385) 6.90 −  2   = 17.38 in.3

 6.90 − 0.385  Qweb = (6.90 − 0.385)(0.270)  2   3 = 5.73 in.

Total Q = 17.4 8+ 5.77 = 23.11 in.3 14-339

fv =

VQ 43.1(23.11) = = 12.72 ksi Ib 291(0.270)

(b) f v =

V 43.1 = = 11.57 ksi dt w 13.8(0.270)

--------------------------------------------------------------Prob. 14.23 W460 × 60:

d = 455 mm tw = 8.0 mm

Max. V = Fvdtw = 100(455)(8.0) = 364 kN --------------------------------------------------------------Prob. 14.24 Steel pin: 1½″ diam.

A = 0.7854(1.5)2 = 1.767 in.2 fv =

4V 4(10,000) = = 7550 psi 3 A 3(1.767)

--------------------------------------------------------------Prob. 14.25

A = 0.7854(10)2 = 78.54 in.2 I=

π (10) 4

64 = 490.9 in.4

Vmax = 300 lb Mmax = 300(20) = 6000 lb-ft

fb =

Mc 6000(12)(5) = = 733 psi 490.9 I

fv =

4V 4(300) = = 5.09 psi 3 A 3(78.54)

---------------------------------------------------------------

14-340

Prob. 14.26

Vmax = RB = (15/20)(1000) = 750 lb

1(4) 3 = 5.33 in.4 12 1(2) 3  4 I x 2 = 2  = 1.34 in.

12   I x1 =

Total Ix = 6.67 in.4 On plane x-x:

Q1 = 2(1)(1) = 2 in.3 Q2 = 2(1)(1)(0.5) = 1 in.3 Total Q = 3 in.3

fv =

VQ 750(3.0) = = 112.4 psi 6.67(3) Ib

On plane x1-x1: Q = (1)(1)(1.5) = 1.5 in.3

fv =

VQ 750(1.5) = = 168.7 psi 6.67(1) Ib

-------------------------------------------------------------Prob. 14.27 Rect. Beam: 100 mm x 250 mm Ix =

100( 250) 3 = 130.2 × 10 6 mm 4 12 = 130.2 × 10 −6 m 4

14-341

Q = 100(125)(62.5) = 781.3×103 mm3 = 781.3×10-6 m3 fv =

VQ (140 × 10 3 N)(781.3 × 10 -6 ) = (130.2 × 10 − 6 )(100 × 10 −3 ) Ib = 8.40 × 10 6 N/m 2 = 8.40 MPa

--------------------------------------------------------------Prob. 14.28 (See Prob. 14.29(a))

Ix = 1786.6 in.4, Vmax = 4500 lb Max. fv occurs at N.A. (axis x-x):

Q1 = 10(2)(6)

= 120 in.3

Q2 = 5(2)(2.5)(2) = 50 Total Q = 170 in.3

fv =

VQ 4500(170) = = 107.0 psi Ib 1786.6(4)

--------------------------------------------------------------Prob. 14.29

10(2) 3  I x1 = 2 + 20(5) 2  = 1013 in.4  12  I x2 =

2(8) 3 = 85.3 in.4 12

Total Ix = 1098 in.4 Shear stress on x-x plane = 120 psi

Qx1 = 10(2)(5) = 100 in.3 Qx2 = 4(2)(2) = 16 14-342

Total Qx = 116 in.3 fv =

f Ib 120(1098)(2) VQ V = v = = 2270 lb 116 Q Ib

Shear stress in plane of junction between web and flange:

Q = 10(2)(5) = 100 in.3 Web: f v =

VQ 2270(100) = = 103.4 psi 1098(2) Ib

Flange: f v =

2270(100) = 20.7 psi 1098(10)

------------------------------------------------------------Prob. 14.30 W36 × 150, Iy = 270 in.4

V = 140 k

Plane y-y; (neutral axis)

Qy(flng) = 5.675(0.940)(3.15)(2) = 33.61 in.3

 0.625  3 Q y (web) = 35.85 (0.1560) = 1.75 in.  2  Total Qy = 35.36 in.3

fv =

140(35.36) = 0.511 ksi 270(35.85)

Web-flange junction:

Qy = 5.675(0.940)(3.15)(2) = 33.61 in.3 Flange : f v =

140(33.61) = 9.27 ksi 270(0.940)(2)

14-343

Web : f v =

140(33.61) = 0.486 ksi 270(35.85)

2 in. from tip of flange:

Qy = 2(0.940)(4.988)(2) = 18.75 in.3 140(18.75) fv = = 5.17 ksi 270(0.940)(2) 4 in. from tip of flange:

Qy = 4(0.940)(3.988)(2) = 29.99 in.3 140( 29.99) fv = = 8.27 ksi 270(0.940)(2) -------------------------------------------------------------Prob. 14.31 W10 × 45: L = 14 ft

Zx = 54.9 in.3, d = 10.10 in., tw = 0.350 in.

0.6(50)(54.9) = 137.3 k - ft 12 8M 8(137.3) = 5.60 k/ft w( total) = 2 R = L 14 2 (a) M R =

superimposed w(all) = 5.60 – 0.045 = 5.56 k/ft (b)

Vmax =5.60(14)/2 = 39.2 k Qx(flng) = 8.02(0.620)(4.74) = 23.57 in.3 Qx(web) = 4.43(0.350)(2.22) = 3.44 in.3 Total Qx = 27.01 in.3

fv =

39.2(27.01) = 12.20 ksi 248(0.350)

--------------------------------------------------------------

14-344

Prob. 14.32

W610 × 113: Ix = 864×106 mm4,

d = 607 mm, tf = 17.3 mm, tw = 11.20 mm 286.2 2 (11.20) (a) Qx = 228(17.3)(294.85) + 2 6 3 = 1.622 × 10 mm fv =

VQ 525 × 10 3 (1.622 × 10 6 ) = 87.9 MPa = 864 × 10 6 (11.20) Ib

(b) Avg. web shear:

525×103 V fv = = = 77.2 MPa dtw 607(11.20) --------------------------------------------------------------Prob. 14.33

W30 × 108: L = 10 ft

Ix = 4470 in.4, d = 29.8 in., Sx = 299 in.3 42(10) 2 = 525 k - ft 8 M 525(12) = = 21.1 ksi fb = Sx 299

M max =

42(10) 2 = 210 k

Vmax =

14-345

Qx(flng) =10. 5(0.760)(14.52) = 115.9 in.3 Qx(web) = 14.14(0.545)(7.07) = 54.5 in.3 Qx(total) = 170.4 in.3

fv =

210(170.4) = 14.7 ksi 4470(0.545)

--------------------------------------------------------------Prob. 14.34

W6 × 12: A = 3.55 in.2

bf = 4.00 in., d = 6.03 in., tf = 0.28 in.

Ix = 22.1 in.4

tw = 0.23 in.

Bar: A = (0.75 in.)2 = 0.5625 in.2

y=

ΣAy 3.55(3.015) + 0.5625(6.405) = = 3.479 in. ΣA 3.55 + 0.5625

Ix = 22.1 + 3.55(0.464)2 + 0.5625(2.926)2 = 27.68 in.4 Based on ss(max) at bottom of bar: Q = ΣAy = 0.5625(2.926) = 1.646 in.3 f v (max) Ib VQ fv =  Vmax = Ib Q 4.0( 27.68)(10 .75) = = 50 .0 k 1.646 Based on fv(max) = 20 ksi at the N.A.: Q = 1.646 + 4.00(0.28)(2.411) +

Vmax =

2.27112 (0.23) = 4.939 in.3 2

20(27.68)(0.23) = 25.8 k ⇐ Controls 4.939

--------------------------------------------------------------14-346

Prob. 14.35

Mmax = 400(8)/4 = 800 lb-ft Vmax = 400/2 = 200 lb (a) Not glued:  bh 2  4(6)(1)  = = 4 in.3 S x = 4 6 6   M 800(12) = = 2400 psi fb = 4 Sx

(b) Glued: bh 2 6( 4) 2 = = 16 in.3 Sx = 6 6 M 800 (12) = = 600 psi fb = Sx 16

(c) Shear stress at NA (plane x-x) 6( 4) 3 = 32 in.4 12 Q x = 6( 2)(1) = 12 in.3 Ix =

fv =

VQ 200(12) = = 12.5 psi Ib 32(6)

------------------------------------------------------------Prob. 14.36

a1 = 4 in.2, a2 = 4 in.2, Σa = 8 in.2

14-347

4( 4.5) + 4(0.25) = 2.375 in.

8 0.5(8) 3 I x1 = + 4( 2.125) 2 = 39.4 in.4 12 8(0.5) 3 I x2 = + 4(2.125) 2 = 18.15 in.4 12 y=

Total Ix = 57.55 in.4

M=

WL 10,000(6) = = 7500 lb - ft 8 8 Mc 7500 (12)(6.125) = 57.55 I = 9580 psi (Compressi on)

f b (top) =

f b (bott) =

7500 (12)( 2.375) = 3710 psi (Tension) 57.55

Vmax = 10,000/2 = 5000 lb At N.A.:

 6.125  3 Q = 6.125(0.5)  = 9.38 in.  2  VQ 5000(9.38) fv = = = 1630 psi Ib 57.55(0.5) At web-flange junction:

Q = 8(0.5)(2.125) = 8.5 in.3 5000 (8.5) = 1477 psi 57.55(0.5) 5000 (8.5) f v (flng) = = 92.3 psi 57.55(8) f v (web) =

-------------------------------------------------------------Prob. 14.37 (a)

14-348

Elastic:

y=

ΣAy 8(1.5)(6.75) + 6(2)(3) = = 4.875 in. ΣA 8(1.5) + 6(2)

I xx =

8(1.5) 3 2( 6) 3 + 12(1.875) 2 + + 12 12 12(1.875) 2 = 122.6 in.4

I 122.6 = = 25.15 in.3 c 4.875

S x (bott) =

Fully plastified section (N.A. is at the bottom of the top flange):

Zx = 8(1.5)(0.75) + 6(2)(3) = 45 in.3 Shape factor = Z/S = 45/25.15 = 1.79 (b) Elastic: 8(12) 3 6(10) 3 I xx = + = 652 in.4 12 12 I 652 Sx = = = 108.7 in.3 c 6

Fully plastified section (N.A. coincides with the centroidal axis):

Zx = 2(6)(8)(3) – 2(5)(6)(2.5) = 138 in.3 Shape factor = Z/S = 138/108.7 = 1.27 --------------------------------------------------------------Prob. 14.38 (Refer to Prob. 14.17)

L = 14 ft, Fy = 36 ksi (a) My = FySx = 36(25.15)/12 = 75.5 k-ft

wy =

8M y 2

L

=

8(75.5) = 3.08 k/ft 14 2

Mp = FyZx = 36(45)/12 = 135 k-ft

wp =

8M p 2

L

=

8(135) = 5.51 k/ft 14 2

(b) My = FySx = 36(108.7)/12 = 326 k-ft

wy =

8M y 2

L

=

8(326) = 13.31 k/ft 142 14-349

Mp = FyZx = 36(138)/12 = 414 k-ft

wp =

8M p 2

L

=

8(414) = 16.90 k/f t 142

-----------------------------------------------------------Prob. 14.39

Fy = 40,000 psi

 0.7854(3) 2  4(1.5)   Z x = 2(6)(3)(1.5) − 2  2   3π  = 49.5 in.3

Mp = FyZx = 40(49.5)/12 = 165 k-ft Max. M = Pa

P = Mp/a = 165/3 = 55 k --------------------------------------------------------------Prob. 14.40 (1.50 kN/m = 1.50 N/m)

bh 2 300(50) 2 = = 125 × 103 mm 3 6 6 A = 50(300) = 15 000 mm 2

Sx =

Based on moment:

fb =

M wL2  = Fb S x Sx 8

14-350

Fb S x 8 11.0(125 ×103 )(8) = 1.5 w 3 = 2.71×10 mm = 2.71 m ⇐ controls

max L =

Based on shear:

3V 3  wL  =   2A 2A  2  4 AFv 4(15 000)(0.83) max L = = 3w 3(1.5) fv =

= 11.07 ×103 mm = 11.07 m --------------------------------------------------------------Prob. 14.41 Fv = 60 psi

Ix =

4(6)3 = 72 in.4 12

In glued joint: Q = 4(2)(2) = 16 in.3 (a) f v =

F Ib 60(72)( 4) VQ  VR = v = 16 Ib Q = 1080 lb

Since max V = P: P =1080 lb (b) Mmax = 1080(3) = 3240 lb-ft

fb =

Mc 3240(12)(3) = = 1620 psi 72 I

------------------------------------------------------------Prob. 14.42

14-351

Fv=1.5 MPa, Fb = 10 MPa (more) 250(300) 3 200(250) 3 − 12 12 6 4 = 302.1× 10 mm = 302.1 × 10 - 6 m 4

Ix =

 125  Qx = 250(25)(137.5) + 2(125)(25)   2  = 1250 × 103 mm3 = 1.25 × 10 −3 m 3

Moment: Fb I 10 × 10 6 (302.1× 10 −6 ) MR = = c 0.15 = 21.14 × 10 3 N ⋅ m

PL wL2 + 4 8 4 wL2   max P =  M − 8  L M=

4 2.7 × 10 3 ( 4) 2   =  20.14 × 10 3 − 4 8  max P = 14.74 × 10 3 N = 14.74 kN Shear: 1.5 × 10 6 (302.1× 10 −6 )(0.050) VR = 1.25 × 10 −3 = 18.13 × 10 3 N

14-352

P wL + 2 2 wL   max P = 2V −  2  

V=

 2.7 × 10 3 (4)  3   = 218.13 ×10 − 2   3 = 25.45 ×10 N = 25.45 kN

Smaller controls: P = 14.74 kN --------------------------------------------------------------Prob. 14.43

a1 = 16 in.2 a2 = 12 in.2 Σa = 28 in.2 Σay Σa = 5.286 in.

y=

14-353

8( 2) 3 I x1 = + 16(1.714 ) 2 = 52.33 in. 4 12 2(6) 3 I x2 = + 12( 2.286 ) 2 = 98.71 in. 4 12

Total Ix = 151.04 in.4 25(12)(2.714) = 5.39 ksi (tens.)

151.04 25(12)(5.286) f b (bott) = = 10.50 ksi (comp.)

151.04 f b (top) =

Max. shear stress at the N.A.: 2(5.286 ) 2 = 27 .94 in.3 2 8.7( 27 .94) f v (max) = = 0.805 ksi 151 .04( 2)

Q=

--------------------------------------------------------------Prob. 14.44

Vmax = 920 lb a1 = 6 in.2 a2 = 6 in.2  6(1) 3  + 6(3.5) 2  = 148 in. 4 I x1 = 2   12   1(6) 3   = 36 in. 4 I x 2 = 2  12  Total I x = 184 in. 4

Shear stress at plane A:

Q = 6(1)(3.5) = 21.0 in.3 fv =

VQ 920(21) = = 52.5 psi 184( 2) Ib

Allowable shear load per screw = 250 lb 250/52.5 = 4.76 in.2/screw Boards are 1″ thick, therefore max. screw spacing = 4.76 in.

--------------------------------------------------------------14-354

Prob. 14.45 6 × 12 (S4S): wt. = 17.5 lb/ft 1800 (121) = 18,150 lb - ft 12 17.5(18) 2 M beam wt. = = 709 lb - ft 8

M R = Fb S x =

Allowable M for superimposed loads: 18,150 – 709 = 17,441 lb-ft

M max = P (6 ft)  P =

17,441 = 2907 lb 6

Fv A 140(63.3) = = 5908 lb 1 .5 1 .5 17.5(18) Vbeam wt. = = 158 lb 2 VR =

Allowable V for superimposed load: 5908 – 158 = 5750 lb Vmax = P = 5750 lb Smaller controls, therefore Allowable P = 2907 lb --------------------------------------------------------------Prob. 14.46 6 × 18 (S4S): wt. = 26.7 lb/ft 1800 (281) = 42,150 lb - ft 12 26.7(18) 2 M beam wt. = = 1081 lb - ft 8 MR =

Allowable M for superimposed loads: 42,150 – 1081 = 41,069 lb-ft

M max = P (6 ft)  P =

41,069 = 6845 lb 6

Fv A 140(96.3) = = 8990 lb 1 .5 1 .5 26.7(18) Vbeam wt. = = 240 lb 2 VR =

14-355

Allowable V for superimposed load: 8990 – 240 = 8750 lb Vmax = P = 8750 lb Smaller controls, therefore Allowable P = 6845 lb --------------------------------------------------------------Prob. 14.47

ΣAy 16(13) + 10(7) + 24(1) = = 6.04 in. ΣA 16 + 10 + 24 8(2) 3 1(10) 3 + 16(6.96) 2 + + 10(0.96) 2 I xx = 12 12 3 12(2) + + 24(5.04) 2 = 1490.6 in.4 12 I 1409.6 = 187.3 in.3 S xx = xx = c top 7.96 y=

Plastic N.A. is located 1” above the top of the bottom flange (for equal areas).

Zx = 16(10) + 9(4.5) + 1(0.5) + 24(2) = 249 in.3 Shape factor = Z/S = 249/187.3 = 1.33 ---------------------------------------------------------------

14-356

Prob. 14.48 4 × 12 (S4S) Hem-fir

A = 39.4 in.2 Sx = 73.8 in.3 Wt. = 10.9 lb/ft

Fb = 1000 psi Fv = 150 psi

wL2 (325 + 10.9)(12) 2 = = 6046 lb - ft 8 8 M 6046(12) fb = = = 983 psi Sx 73.8

M=

983 psi < 1000 psi

(O.K.)

wL (325 + 10.9)12 = = 2015 lb 2 2 3V 3(2015) fv = = = 76.7 psi 2 A 2(39.4) 76.7 psi < 150 psi (O.K.)

Max. V = R =

--------------------------------------------------------------Prob. 14.49 A992: Fy = 50 ksi W16 × 36:

Sx = 56.5 in.3, d = 15.9 in., tw = 0.295 in.

Zx = 64.0 in.3

w =1.25 + 0.036 = 1.286 k/ft Moment:

14-357

1.286(24) 2 10(24) M= + = 152.6 k - ft 8 4 0.6(50)(64) M R = 0.6 Fy Z x = =160 k - ft 12 160 k - ft > 152.6 k - ft (O.K.) Shear : V =R=

1.286(24) + 10 / 2 = 20.4 k 2

VR = 0.4Fy dtw = 0.4(50)(15.9)(0.295) = 93.8 k 93.8 k > 20.4 k (O.K.) -------------------------------------------------------------Prob. 14.50 Calculate all. load w (kN/m) A992 W250 × 115

Fy =345 MPa Zx = 1600×103 mm3 d = 269 mm, tw = 13.5 mm MR = 0.6Fy Zx = 0.6(345)(1600×103) = 331×106 N∙mm = 331×103 N∙m wL2 = MR 8 8M 8(331×10 3 ) w = 2R = = 106 ×10 3 N/m 2 L 5 = 106 kN/m

VR = 0.4Fydtw = 0.4(345)(269)(13.5) = 501×103 N = 501 kN wL = VR 2 2V 2(501)  w= R = = 200 kN/m L 5

Therefore, all. load = 106 kN/m --------------------------------------------------------------14-358

Prob. 14.51 (Refer to Example 14.13) Douglas fir: Fb = 900 psi, Fv = 180 psi 3×12 (S4S): Sx = 52.7 in.3, A = 28.1 in.2 wt. = 7.81 lb/ft

M R = Fb S x =

900(52.7) = 3950 lb - ft 12

8M 8(3950) = = 123.4 lb/ft L2 16 2 F A 180(28.1) = 3370 lb VR = v = 1.5 1.5 2V 2(3370) w= R = = 4.21 lb/ft L 16 w=

(a) All. total w = 123.4 lb/ft All. superimposed w = 123.4 – 7.81 = 115.6 lb/ft (b) All. superimposed load (psf): 115.6 = 86.7 psf  16     12 

-------------------------------------------------------------Prob. 14.52 Calculate all. load w (lb/ft) A992, W14 × 34 (L = 6.5 ft) Fy =50 ksi,

Zx = 54.6 in.3

d = 14.0 in., tw = 0.285 in.

0.6(50)(54.6) = 136.5 k - ft 12 8M 8(136.5) = 25.8 k/ft w = 2R = L 6.5 2 VR = 0.4 Fy dt w = 0.4(50)(13.98)(0.285) M R = 0.6 Fy Z x =

= 79.7 k w=

2VR 2(79.7) = = 24.5 k 6.5 L

14-359

Therefore, shear controls: All. load = 24.5 – 0.034 = 24.5 k/ft --------------------------------------------------------------Prob. 14.53 Given: Fb =1600 psi, Fv= 100 psi 3 × 12 (S4S): A = 28.1 in.2 bh 3 11.25(2.5) 3 = = 14.65 in.4 12 12 I 14.65 = 11.72 in.3 S= = c 1.25 1600 (11.72) = 1563 lb - ft MR = 12 8M 8(1563) = 195.4 lb/ft w= 2 = 82 L F A 100( 28.1) = 1873 lb VR = v = 1.5 1.5 2V 2(1873) = 468 lb/ft w= R = L 8 allowable w = 195.4 lb/ft I=

--------------------------------------------------------------Prob. 14.54 L = 30 ft, Fy = 50 ksi W18 × 50: Sx = 88.9 in.3, Zx = 101 in.3 M y = Fy S x = wmax =

8(370) = 3.29 k/ft (at yield) 30 2

M p = Fy Z x = wmax =

50(88.9) = 370 k - ft 12

50(101) = 421 k - ft 12

8(421) = 3.74 k/ft (at full plasification) 30 2

---------------------------------------------------------------

14-360

Prob. 15.1 Ec R 10,000(0.125) = = 20.0 ksi (125 / 2)

fb =

--------------------------------------------------------------Prob. 15.2 Ec R 30,000( 164) = = 46.9 ksi ((20 / 2))

fb =

--------------------------------------------------------------Prob. 15.3

L 60° = 2π R 360° 360 L 360(500) R= = = 477.5 mm 2π 60 2π 60 Ec 207 ×10s (0.50) fb = = = 217 MPa R 477.5 --------------------------------------------------------------Prob. 15.4 E ( d / 2) D/2 ( D / 2)(30,000) = 10,000,00( d / 2) Min. D = 333d (in.) fb =

---------------------------------------------------------------

15-361

Prob. 15.5 3((0.5) 3 = 0.0313 in. 4 12 EI RM 62(700) = R= E = 0.0313 M I E = 1,389,000 psi I=

-------------------------------------------------------------Prob. 15.6 6(14) 3 = 1372 in. 4 12 EI 1,700,000(1372 ) R= = M 200,000 11,660 in. = 972 ft I=

-------------------------------------------------------------Prob. 15.7

sb =

Ec Ec(2) = R D

 0.08  30,000,00 (2) 2   60,000 = D  D = 40 in.

--------------------------------------------------------------Prob. 15.8

1.5(1.5) 3 = 0.4219 in.4 I= 12 EI 30,000,000(0.4219) = M= R 15,000 = 844 lb - in. (= 70.3 lb - ft) --------------------------------------------------------------Prob. 15.9 E = 207×103 MPa, d = 38 mm, L = 800 mm, P = 3 kN

15-362

π (38) 4

= 102.4 × 103 mm4 64 PL3 3 × 103 (800) 3 Δ= = 48EI 48(207 × 103 )(102.4 × 103 ) I=

= 1.510 mm --------------------------------------------------------------Prob. 15.10 (Refer to Prob. 15.7) d = 25 mm L = 500 mm

I= Δ=

π (25) 4 64

= 19.17 × 10 3 mm 4

3 × 10 3 (500) 3 48(207 × 10 3 )(19.17 × 10 3 ) = 1.969 mm

--------------------------------------------------------------Prob. 15.11 Redwood: E = 1300 ksi 10 × 14 (S4S): I = 1950 in.4 5wL4 5(1.0)(15) 4 (1728) = = 0.449 in.

384 EI 384(1300)(1950) 15(12) Δ all = = 0.60 in.

300 0.449 in. < 0.60 in. (O.K.) Δ max =

--------------------------------------------------------------Prob. 15.12 Δ all =

15(12) = 0.60 in Prob. 15.10 (cont.)

300

Δ max =

PL3 5(15) 3 (1728) = = 0.240 in..

48 EI 48(1300)(1950) 0.240 in. < 0.60 in. (O.K.)

---------------------------------------------------------------

15-363

Prob. 15.13 W16 × 45: Ix = 586 in.4 Pa (3L2 − 4a 2 ) 24 EI 12(8)(12) = [3(24) 2 − 4(8) 2 ](144) 24(30,000,000)(586) = 0.579 in.

Δ max =

---------------------------------------------------------------------------

Prob. 15.14 4-in. diam. shaft: I = πd4/64 = 12.57 in.4 wt. =

π r 2 (12) 1728

(490) = 42.8 lb/ft

Δlimit = 0.50 in.

5wL4 5(42.8)(12) 4 (1728) Δ wt. = = 384 EI 384(30,000,000)(12.57) = 0.0530 in.

Allowable Δ for P = 0.50 − 0.0530 = 0.447 in.

48 EIΔ 48(30,000)(12.57)(0.447) = L3 (12) 3 (1728) = 2.71 k

P max =

--------------------------------------------------------------Prob. 15.15 Δall at center of span = 0.010L 5wL4 384 EI π d 4 π (3) 4 I= = = 3.98 in.4 64 64 π r 2 (12) w= (490) = 24.00 lb/ft 1728 5wL4 0.010(384) EI = 0.010 L  L = 3 384 EI 5w(1728)

Δ=

15-364

L=3

0.010(384)(30,000,000)(3.98) 5(24)(1728)

= 13.03 ft --------------------------------------------------------------Prob. 15.16

W = ½ (5)(12) = 30 k

12(12) = 0.60 in.

240 WL3 WL3 Δ= I = 15EI 15EΔ all Δ all =

I min =

30(12) 3 (1728) = 332 in.4 15(30,000)(0.60)

--------------------------------------------------------------Prob. 15.17 Δ max =

PL3 3EI

(a) Douglas fir Δ =

5(6) 3 (1728) = 0.385 in.

3(1700)(951)

(b) W10 × 22 Δ =

5(6) 3 (1728) = 0.1757 in.

3(30,000)(118)

(c) Pipe 8 x-strong Δ =

5(6) 3 (1728) = 0.1956 in.

3(30,000)(106)

---------------------------------------------------------------

15-365

Prob. 15.18 W16 × 36, Ix = 448 in.4 PL3 5wL4 + 384 EI 48 EI 5(1.0)(20) 4 (1728) 10(20) 31728 = + 384(30,000)(448) 48(30,000)( 448) = 0.482 in.

(a) Δ max =

1( 20 ) 2 10 ( 20 ) + = 10 k - ft 8 4 M 100 (12 ) = = 21 .2 ksi sb = 56 .5 Sx

(b) M max =

--------------------------------------------------------------Prob. 15.19 W410 × 100, Ix = 397×106 mm4 Wt. = 100(9.81) = 981 N/m = 0.981 kN/m Total w = 4 + 0.981 = 4.98 kN/m E = 207×103 MPa, P = 40 kN L = 9×103 mm, a = 3×103 mm 5wL4 Pa + (3L2 − 4a 2 ) 384 EI 24 EI Δ max = 17.77 mm Δ max =

--------------------------------------------------------------Prob. 15.20

W10 × 22, Ix = 118 in.4 Neglect beam weight.

Free Δ (no center support):

15-366

5wL4 5(1000)( 20) 4 (1728) = 384 EI 384(30,000,000)(118) = 1.017 in.

Δ=

Apply force RC to create Δ1 where Δ1 = 1.017 − 0.50 = 0.517 in.

RC L3 48 EI 0.517 ( 48)(30,000,000)(118)  RC = ( 20) 3 (1728 ) = 6350 lb Δ1 =

--------------------------------------------------------------Prob. 15.21

Δmax = 0.20 mm = 0.20×103 m

Δ=

Pa (3L2 − 4a 2 ) 24EI

I=

π d4 64

req' d d = 4 =4

64 Pa(3L2 − 4a 2 ) π (24) EΔ

64(3 ×10 3 )(0.2333)[3(0.7) 2 − 4(2.333) 2 ] π (24)(207 ×109 )(0.20 ×10 −3 )

req'd d = 65.1×10–3 m = 65.1 mm

π (65.1× 10 −3 ) 4 = 882 × 10 −9 m 4 64 π (65.1× 10 −3 ) 2 A= = 3.33 × 10 −3 m 2 4

I=

M = Pa = 3×103(0.2333) = 700 N⋅m

15-367

 65.1× 10 3   700 2 Mc   = 25.8 × 10 6 Pa sb = = −9 I 882 × 10 = 25.8 MPa < 165 MPa (O.K.) 4V 4(3 × 103 ) = = 1.201×10 6 Pa 3 A 3(3.33 ×10 −3 ) = 1.201 MPa < 100 MPa (O.K.)

ss =

----------------------------------------------------Prob. 15.22 W10 × 22: Ix = 118 in.4

AM 1 2 (8)(40,000)(144) = 30,000,000(118) EI = 0.00651 rad θ 1 = θ 2 = 0.00651(57.3) = 0.373°

θ1 = θ 2 =

-----------------------------------------------------

15-368

Prob. 15.23 W24 × 84 Ix = 2370 in.4

AM 2 3 (15)(337.5)(144) = 30,000(2370) EI = 0.00684 rad θ 1 = θ 2 = 0.00684(57.3) = 0.392°

θ1 = θ 2 =

--------------------------------------------------------------Prob. 15.24

π (76.0 2 − 62.7 2 ) I= = 879 × 10 3 mm 4 64 AM 2 × 10 3 ( 2.1 × 10 6 ) = EI 70 × 10 3 (879 × 10 3 ) = 0.0683 rad θ = 0.0683(57.3) = 3.91°

θ=

--------------------------------------------------------------Prob. 15.25 1.5(1)3 12 = 0.125 in.4

Ix =

15-369

AM 60(4000) = = 0.064 rad EI 30,000,000(0.125) θ = 0.064(57.3) = 3.67°

θ=

--------------------------------------------------------------Prob. 15.26 (Refer to Prob. 15.18)

θ1 = θ 2 =

AM 4000(30) + 1 2 (40)(4000) = 30,000,000(0.125)) EI = 0.0533 rad

θ 1 = θ 2 = 0.0533(57.3) = 3.06° --------------------------------------------------------------Prob. 15.27 Refer to Prob. 15.9 Redwood E = 1300 ksi

x1 = 5 8 (7.5) = 4.688 ft AM x 2 3 (7.5)(28.125)(4.688)(1728) = EI 1300(1950) = 0.449 in.

y AB =

Δ B = y AB = 0.449 in.

---------------------------------------------------------------

15-370

Prob. 15.28 (Refer to Prob. 15.12 (& 15.11)) Redwood: E = 1300 ksi

x1 = 2 3 (7.5) = 5.0 ft AM x 1 2 (7.5)(18.75)(5.0)(1728 ) = EI 1300 (1950 ) = 0.240 in. Δ B = y AB = 0.240 in. y AB =

--------------------------------------------------------------Prob. 15.29 Doug Fir 8 × 12 (S4S)

ΔA = yAB = 0.75 in.

AM = ½(10)(−10P) = − 50P ft2-lb x = 2 3 (10) = 6.67 ft AM x (−50 P)(6.67)(1728) = = −0.75 in.

1700(951) EI  P = 2.10 k y AB =

--------------------------------------------------------------15-371

Prob. 15.30 Refer to Prob. 15.29

AM = ½(5)(− 5P) = − 12.5P ft2-lb

x = 2 3 (5) + 5 = 8.33 ft AM x − 12.5P(8.33)(1728) = = 0.75 in.

EI 1700(951) P = 6.74 k y AB =

--------------------------------------------------------------Prob. 15.31 Refer to Prob. 15.15

A1 = ½(8)(96) = 384 ft2-k A2 = 96(4) = 384 ft2-kProb. 15.24 (cont.) x1 = 2 3 (8) = 5.33 ft x2 = 2 + 8 = 10.0 ft AM x [384(5.33) + 384(10)](1728) = 30,000(586) EI = 0.579 in. Δ B = y AB = 0.579 in.

= y AB =

---------------------------------------------------------------

15-372

Prob. 15.32 W12 × 30: Ix = 238 in.4 5wL4 PL3 + 384 EI 48 EI 5(675)(12 4 )(1728) = 384(30,000,000)( 238)

Δ=

14,00(12 3 )(1728) + 48(30,000,000)( 238) = 0.1661 in.

Δ all =

12(12) = 0.40 in.

360 0.1661in. < 0.40 in. (O.K.)

--------------------------------------------------------------Prob. 15.33 Refer to Prob. 15.23 AM x 2 3 (337.5)( 5 8 )(15)(1728) = 30,000(2370) EI = 0.769 in.

y AB =

Δ B = y AB = 0.769 in.

--------------------------------------------------------------Prob. 15.34 See text Fig. 15.16 (Max. M @ conc. loads =88 kN⋅m)

---------------------------------------------------------------

15-373

Prob. 15.35 See text Fig. 15.4 (Max. M at supports = −2700 lb-in.)

--------------------------------------------------------------Prob. 15.36 See text Fig. 15.24 (Max. M in center portion = 7.10 kN⋅m)

--------------------------------------------------------------Prob. 15.37

---------------------------------------------------------------

15-374

Prob. 15.38

A1 = ½(3)(360) = 540 ft2-lb A2 = 360(2) = 720 ft2-lb x1 = 2 3 (3) = 2 ft, x 2 = 1 + 3 = 4 ft Ix =

3(1) 3 = 0.25 in.4 12

Max. deflection @ midspan: AM x [540(2) + 720(4)](1728) = EI 30,000,000(0.25) = 0.9124 in. Δ E = y AE = 0.9124 in. y AE =

Deflection at point C: 720(1)(1728) = 0.1659 in.

30,000,000(0.25) Δ C = y AE − yCE = 0.9124 − 0.1659 = 0.747 in. yCE =

---------------------------------------------------------------

15-375

Prob. 15.39 (See Example 15.22(b))

From Example 15.22, yBA = +1.853 in.

By proportion: 30 (1.853) = + 2.316 in.

24 1 A x  1 [ 2 (24)(276)(14) yCA = M =  EI  30,000(394)  yC =

− 1 3 (24)(288)(12) − 1 2 (6)(12)(4)](1728) = +2.716 in. Δ C = yCA − yC = 2.716 − 2.316 = 0.400 in. (upward) --------------------------------------------------------------Prob. 15.40

For 1.72″ × 1.75″ section: 15-376

Ix =

bh 3 1.72(1.753 ) = = 0.7682 in.4 12 12

Max. defl. = ΔB = yAB = 0.74 in.

AM x 1 2 (14)(1400)( 2 3 )(14) = EI E (0.7682)  E = 1,609,000 psi y AB =

-------------------------------------------------------------Prob. 15.41 W530 × 138: Ix = 862×106 mm4 S

3 3 x = 3150×10 mm

Wgt.= 138(9.81) = 1.354 kN/m

sb =

M wL2  sb S x = Sx 8

8sb S x 8(152)(3150 ×103 ) = L2 (12 ×103 ) 2 = 26.6 N/mm = 26.6 kN/m

w=

Total load = 26.6 + 1.354 = 28.0 kN/m M=

wL2 28.0(12 ×103 ) 2 = = 504 ×10 6 N ⋅ mm 8 8

AM x EI 2 AM = 3 bh = 2 3 (6000)(499 ×10 6 ) y AB = Δ B =

= 2.02 ×1012 N ⋅ mm 2 2.02 ×1012 (3.75 ×10 3 ) y AB = = 42.4 mm 207 ×10 3 (862 ×10 6 )

--------------------------------------------------------------15-377

Prob. 15.42

Max. defl. between supports: [ 12 (120)(242 )( 13 ) − 12 (144)(122 )( 13 )]1728 y BA = 30,000(228) = 2.037 in.

tan θ = θ =

θ=

2.037 = 0.007074 rad 24(12)

AM 1 2 m(5m)(144) = = 0.007074 EI 30,000 (228) m = 11.59 ft

(11.59) 2 (5)(11.59)( 2 3 )(1728) 30,000(228) = 0.655 in. Δ E = Δ max − y AE = 0.655 in. y AE =

1

2

Defl. @ point D: yD =

30 ( 2.037 ) = 2.546 in.

24

15-378

 1 1 [ 2 (120)(24)(14) y DA =   30,000(228)  − 1 2 (144)(12)(10) − 1 2 (24)(6) 2 ( 2 3 )]1728

= 2.838 in.

ΔD = yDA − yD = 2.838 − 2.546 = 0.292 in. ↑ --------------------------------------------------------------Prob. 15.43 Aluminum beam, Ix = 200 in.4

  1 1 [− 2 (12000)(6 2 )( 2 3 ) y AB =   10,000,000(200)  − 13 (4500)(3)(5.25)]1728 = −0.144 in.

ΔA = −yAB = 0.144 in.

--------------------------------------------------------------Prob. 15.44

15-379

 1 1 [ 2 (800 )( 40) 2 ( 1 3 ) y BA =   30,000 ( 4080 )  − 1 2 (600 )( 20) 2 ( 1 3 ) − 1 2 ( 200 )(10) 2 ( 1 3 )]1728 = 2.40 in.

θ = tan θ =

2.40 = 0.005 rad 40(12)

(Assume 20′ < m < 30′):

θ=

1 AM  1 [ 2 m(20m) =  EI  30,000(4080)  − 1 2 ( m − 20) 2 (30)]144

Subst. for θ, solve for m: m = 20.63 ft 1  1 [ 2 ( 20)(20.63) 2 ( 2 3 )(20.63) y AE =    30,000(4080)  − 1 2 (30)(0.65) 2 ( 2 3 {0.63} + 20)]1728 = 0.825 in.

ΔE = yAE = 0.825 in.

--------------------------------------------------------------Prob. 15.45

Δmax= 0.10 in. w = weight (lb/ft) − 13 (10)(50 w)( 3 4 )(10)(1728) = 0.10 in.

30,000,000( I ) 0.072 w = = 0.10 in.

I  I = 0.72 w y AB =

15-380

Steel wt.: 490 lb/ft3 = 3.40 lb/in.2/ft Round rod w = 3.40A = 3.40(0.7854d2)

π d4

I=

64

π d4

and I = 0.72 w

= 0.72(3.40)(0.7854 d 2 )

64  Req' d d = 6.26 in.

--------------------------------------------------------------Prob. 15.46 Note M/I diagram.

θ =

1 AM  1 =  [ 2 (10)(12)(960) EI  30,000,000 

+ 15(12)(342.86) + 1 2 (15)(12)(514.28) + 1 2 (15)(12)(771.43)] = 0.00783 rad = 0.00783(57.3) = 0.45° Max. deflection: y AB =

AM x  1  1 = [− 2 (960)(10) 2 (12) 2 ( 2 3 ) EI  30,000,000 

− (15)(342.86)(12) 2 (17.5) − 1 2 (514.28)(15)(12) 2 (20) − 1 2 (771.43)(15)(12) 2 (20)]

= − 1.511 in. ΔA = yAB = 1.511 in.

--------------------------------------------------------------15-381

Prob. 15.47

y BA =

1

2

(2160)(9 2 )(12) 2 + 12 (2880)(6)(4)(12) 2 30,000,000

= 0.586 in.

0.586 = 0.00326 rad 15(12) 1 m( 240m)(12) A θ= M = 2 = 0.00326 30,000,000 EI  m = 8.24 ft 1 (8.24)(12)(240)(8.24)( 2 )(8.24)(12) 3 y AD = 2 30,000,000

θ = tan θ =

= 0.215 in. ΔD = yAD = 0.215 in.

---------------------------------------------------------------

15-382

Prob. 15.48

I1 = 15.2 in.4, I2 = 7.23 in.4

y DA =

1 [−394.7(15)(17.5)(12) 2 30,000,000

− 829.9(5)(7.5)(12) 2 − 1 2 (829.9)(5)(12)( 2 3 )(5)(12)] = 0.680 in.

ΔD = −yDA = 0.680 in. (to the right) --------------------------------------------------------------Prob. 15.49

15-383

1 1 [− 2 L( PL)( 2 3 ) L − 1 2 ( L 2 )( PL2 )( 23 ( L2 ) + L2 )] EI 7 PL3 =− 16 EI 7 PL3 Δ A = − y AB = 16 EI y AB =

---------------------------------------------------------Prob. 15.50

( PL3 )( L3 ) 2 ( 2 3 ) + ( PL3 )( L6 )( 125 L) y AE = EI 3 23PL = = ΔE 648EI 1

2

--------------------------------------------------------------Prob. 15.51 (Refer to Prob. 15.24)

15-384

y AB =

− 12 (6)(30)( 2 3 )(6)(1728) 622,000 =− EI EI

(a) E = 1700 ksi, Ix = 951 in.4 Δ A = − y AB =

622,000 = 0.385 in.

1700(951)

(b) E = 30,000 ksi, Ix = 118 in.4 Δ A = − y AB =

622,000 = 0.1757 in.

30,000(118)

(c) E = 30,000 ksi, Ix = 106 in.4 Δ A = − y AB =

622,000 = 0.1956 in.

30,000(106)

---------------------------------------------------Prob. 15.52

Steel wt.: 490 lb/ft3 = 3.40 lb/in.2/ft w = 3.40(πd2/4) = 2.67 lb/ft

π d4 = 0.0491 in.4 I= 64 y AB =

− 13 (12.5)(208.6)( 3 4 )(12.5)(1728) 30,000,000(0.0491)

= −9.56 in. ΔA = −yAB = 9.56 in.

---------------------------------------------------------------

15-385

Prob. 15.53 Pipe 102 mm - Std., Ix = 3.00×106 mm4

Δ C = y AC =

AM x EI

  1 × =  3 6  207 10 ( 3 .

00 10 ) × ×   6 1 [ 2 (2000)(12 × 10 )(1333) − 13 (1000)(3 × 10 6 )(1750)] = 22.9 mm --------------------------------------------------------------Prob. 15.54 (Refer to Prob. 15.31)

I = 879×103 mm4 AM 2 × 10 3 (2.1× 10 6 )(1000) = EI 70 × 10 3 (879 × 10 3 ) = 68.3 mm

y AC =

---------------------------------------------------------------

15-386

Prob. 15.55 Redwood 8 × 12 (S4S), E = 1300 ksi Sx = 165 in.3 1350(165) = 18.56 k - ft 12,000 PL ML 4(18.56) P= = = 3.71 k M= 4 4 20 M = sb (all) S x =

At midspan: PL3 3.71(20) 3 (1728) Δ= = = 0.864 in.

48EI 48(1300)(951)

At quarter points:

Px (3L2 − 4 x 2 ) 48EI  3.71(5)  = (3(20) 2 − 4(5) 2 )(1728)   48(1300)(951) 

Δ=

= 0.594 in.

--------------------------------------------------------------Prob. 15.56 I=

π d4 64

Δ max =

= 3.98 in.4

PL3 500(20) 31728 = = 1.206 in.

48EI 48(30,000,000)(3.98)

--------------------------------------------------------------Prob. 15.57

--------------------------------------------------------------15-387

Prob. 15.58

--------------------------------------------------------------Prob. 15.59

15-388

Max. deflection: point D: y BA =

[ 12 (2100)(14)(10.67) + 12 (2100)(6) 2 ( 2 3 )]1728 30,000,000(3.976)

=2.637 in.

θ = tan θ =

2.637 = 0.01099 rad 20(12)

1 (150 m )( m )(144) AM = 2 = 0.01099 EI 30,000,000(3.976)  m = 11.017 ft

θ=

11.017 (2.637) = 1.4526 in.

20 1 / 2 (150)(11.017) 2 ( 1 )(11.017)(1728) 3 y DA = 30,000,000(3.976 = 0.4843 in. Δ D = y D − y DA = 0.968 in. yD =

Defl. @ midspan: (point E) yE = 2.637/2 = 1.3185 in. y EA =

1

2

(1500)(10) 2 ( 13 )(1728) = 0.3623 in.

30,000,000(3.976)

ΔE = yE − yEA = 1.3185−0.3623 = 0.956 in.

Defl. under load (point C) yC = (14/20)(2.637)=1.8459 in.

(2100)(14) 2 ( 13 )(1728) = 0.9938 in.

30,000,000(3.976) Δ C = yC − yCA = 1.8459 − 0.9938 = 0.852 in.

1

yCA = 2

---------------------------------------------------------------

15-389

Prob. 15.60 W27 × 114, Ix = 4080 in.4

Slope at free end: [ 1 2 ( 200 )( 20) + 13 (100 )(10) + 20(10)]12 2 30,000 ( 4080 ) = 0.005086 rad = 0.005086(5 7.3) = 0.291 °

θ=

Deflection at free end:   1 1 [ − 2 ( 20) 2 (200)( 2 3 ) y AB =  30 , 000 ( 4080 )   2 − 13 (10) (100)( 3 4 ) − 100(10)(15)

− 1 2 ( 200)(10)(16.67)]1728 = −0.857 in.

ΔA = −yAB = 0.859 in.

Deflection at point C:   1 1 [− 2 (100)(10) 2 ( 2 3 ) yCB =   30,000(4080)  − 100(10)(5) − 100(10)(5)

− 1 2 (200)(10) 2 ( 2 3 )]1728 = −0.282 in. ΔC = −yCB = 0.283 in.

---------------------------------------------------------------

15-390

Prob. 15.61 [6 × 10(S4S), Hem-fir, Ix = 393 in.4]

 1 1 [ 2 (8) 2 (17,000)( 1 3 ) y BA =  1 , 400 , 000 ( 393 )   2 − 1 3 (12,800)(8) ( 1 4 ) − 1 2 (6000)(2) 2 ( 1 3 )]1728 = +0.3425 in. yC = (6/8)(0.3425) = 0.2569 in. yCA =

[ 12 (12,750)(6) 2 ( 13 ) − 13 (7200)(6) 2 ( 14 )]1728 1,400,000(393)

= 0.1724 in. ΔC = yC − yCA = 0.2569−0.1724 = 0.0845 in. yE = (4/5)(0.3425) = 0.1713 in. y EA =

[ 12 (8500)(4) 2 ( 13 ) − 13 (3200)(4) 2 ( 14 )]1728 1,400,000(393)

yEA = 0.0578 in.

ΔE = yE − yEA = 0.1713 − 0.0578 = 0.1135 in.

--------------------------------------------------------------15-391

Prob. 15.62 E = 207×103 MPa, Ix = 861×106mm4

  1 × y AB =  3 6   207 × 10 (861× 10 )  [ 1 2 (2.97 × 109 )(12 × 103 ) 2 ( 13 ) − 13 (2.88 × 109 )(12 × 103 ) 2 ( 1 4 ) − 1 2 (90 × 106 )(3 × 103 ) 2 ( 13 )] = 205.3 mm

At point C:

  1 × yCB =  3 6   207 × 10 (861×10 )  [ 1 2 (2.228 ×10 9 )(9 ×10 3 ) 2 ( 1 3 ) − 1 3 (1.62 ×10 9 )(9 ×10 3 ) 2 ( 1 4 )] = 107.4 mm yC = (9/12)(205.3) = 154.0 mm

ΔC = yC − yCB = 46.6 mm (more)

15-392

At midspan (point D):

  1 × y DB =  3 6   207 × 10 (861× 10 )  [ 1 2 (1.485 × 10 9 )(6 × 10 3 ) 2 ( 1 3 ) − 13 (720 × 10 6 )(6 × 10 3 ) 2 ( 1 4 ) = 37.87 mm yD = (6/12)(205.3) = 102.7 mm

ΔD = yD − yDB = 64.8 mm --------------------------------------------------------------Prob. 15.63 3″ diam. solid steel shaft I=

π (3) 4 = 3.976 in.4 64

A = 0.7854(3)2 = 7.07 in.2

Wt/ft = 3.4(7.07) = 24.0 lb/ft (See Prob. 15.38)

wL4 Δ max = 8EI 24.0 L4 (1728) 0.2 = 8(30,000,000)(3.976)  L = 8.24 ft --------------------------------------------------------------Prob. 15.64 W10 × 33: Ix = 170 in.4 (a) Δ at free end: wL4 800(10) 4 (1728) Δ= = = 0.339 in.

8 EI 8(30,000,000)(170)

(b) If Δ = 0.25 in.: By proportion: Total all. w = (0.25/0.339)(800) = 590 lb/ft All. superimposed load: 590 − 33 = 557 lb/ft ---------------------------------------------------------------

15-393

Prob. 15.65

Assuming P =0: ΔD =

PL3 10(12) 3 (1728) = = 0.0526 in.

48EI 48(30,000)(394)

Total Δ due to P loads: ΔD = 0.0526 + 0.05 = 0.1026 in.

For one P load: ΔD =

Pax 2 ( L − x 2 ) (AISC 9/e case #26) 6 EIL ( where a = 4 ft, x = 6 ft, L = 12 ft)

P(4)(6)(1728) 0.1026 (12 2 − 6 2 ) = 2 6(30,000)(394)(12)  P = 9.75 k

--------------------------------------------------------------Prob. 15.66 W36 × 194: Ix = 12,100 in.4

15-394

 1 1 [ 2 (600)(30) 2 ( 1 3 ) y BA =   30,000(12,100)  − 1 3 (900)(30) 2 ( 1 4 )]1728 = 0.1071 in.

yD = 0.1071/2 = 0.05355 in.

(more)  1 1 [ 2 (300)(15) 2 ( 1 3 ) y DA =   30,000(12,100)  − 1 3 ( 225)(15) 2 ( 1 4 )](1728) = 0.03347 in.

ΔD =yD − yDA = 0.5355 − 0.03347 = 0.0201 in. (down) yC = (40/30)(0.1071) = 0.1428 in.

 1 1 [ 2 (800)(40) 2 ( 13 ) yCA =   30,000(12,100)  − 13 (1600)(40) 2 ( 1 4 ) + 1 2 (800)(10) 2 ( 1 3 )]1728 = 0.06347 in. ΔC = yC − yCA = 0.0793 in. (down) -----------------------------------------------------Prob. 15.67

15-395

 1 1 y BA =  [ 2 (0.45)(12)(16)  30,000  + 0.288(12)(6) + 1 2 (0.288)(12)( 4) − 1 2 (1.728)(12)( 4)]144 = 0.1410 in.

Deflection at point C: yC = ½(0.1410) = 0.0705 in. yCA =

1

2

(0.45)(12)(4)(144) = 0.0518 in.

30,000

ΔC = yC − yCA = 0.0187 in. (down) Deflection at point D: yD = (36/24)(0.1410) = 0.2115 in.

 1 1 y DA =  [ 2 (0.45)(12)(28)  30,000  + 0.288(24)(12) + 1 2 (0.576)(24)(8) − 1 2 (3.456)(24)(8) + 1 2 (2.592)(12)(4)]144 = −0.2675 in.

Negative sign indicates that deflected point D lies below the tangent line from point A. Therefore: ΔD = yD + yDA = 0.2115 + 0.2675 = 0.479 in. (down) (Note that the elastic curve diagram is incorrect.)

--------------------------------------------------------------Prob. 15.68

15-396

 1  y AB =  [− 1 2 PL3 ( 2 3 )  EI  − 1 2 ( 2 3 L)( 2 3 PL)( 7 9 L) − 1 2 ( L / 3)( PL / 3)( 8 9 L)] = − Δ A = − y AB =

5PL3 9 EI

5 PL3 9 EI

--------------------------------------------------------------Prob. 15.69

 wb 2  b(a + 3 4 b) − 13   2  y AB = EI

wb 3 ( a + 3 4 b) 6 = EI 3 wb ( L − b / 4) − 6 = EI 3 wb =− ( 4 L − b) 24 EI wb 3 Δ A = − y AB = ( 4 L − b) 24 EI −

---------------------------------------------------------------

15-397

Prob. 16.1

--------------------------------------------------------------Prob. 16.2 (Neglect beam wt.)

16-398

--------------------------------------------------------------Prob. 16.3

--------------------------------------------------------------Prob. 16.4

16-399

--------------------------------------------------------------Prob. 16.5

---------------------------------------------------------------

16-400

Prob. 16.6

--------------------------------------------------------------Prob. 16.7

16-401

(more)

--------------------------------------------------------------Prob. 16.8

16-402

--------------------------------------------------------------Prob. 16.9

--------------------------------------------------------------Prob. 16.10

16-403

--------------------------------------------------------------Prob. 16.11

16-404

--------------------------------------------------------------Prob. 16.12

16-405

--------------------------------------------------------------Prob. 16.13

16-406

--------------------------------------------------------------Prob. 16.14

16-407

16-408

--------------------------------------------------------------Prob. 16.15

16-409

--------------------------------------------------------------Prob. 16.16

---------------------------------------------------------------

16-410

Prob. 16.17

--------------------------------------------------------------Prob. 16.18 See problem 16.10

16-411

--------------------------------------------------------------Prob. 16.19

--------------------------------------------------------------16-412

Prob. 16.20

16-413

--------------------------------------------------------------Prob. 16.21

16-414

--------------------------------------------------------------Prob. 16.22

16-415

--------------------------------------------------------------Prob. 16.23

16-416

--------------------------------------------------------------Prob. 16.24

16-417

--------------------------------------------------------------Prob. 16.25

---------------------------------------------------------------

16-418

Prob. 16.26

--------------------------------------------------------------Prob. 16.27

16-419

--------------------------------------------------------------Prob. 16.28

16-420

--------------------------------------------------------------Prob. 16.29

----------------------------------------------------------16-421

Prob. 16.30

--------------------------------------------------------------Prob. 16.31

-----------------------------------------------------------16-422

Prob. 16.32

--------------------------------------------------------------Prob. 16.33

16-423

--------------------------------------------------------------Prob. 16.34

16-424

--------------------------------------------------------------Prob. 16.35

---------------------------------------------------------------

16-425

Prob. 16.36

--------------------------------------------------------------Prob. 16.37

Prob. 16.41

16-426

16-427

Prob. 17.1

---------------------------------------------------------------

17-428

Prob. 17.2

--------------------------------------------------------------Prob. 17.3

17-429

--------------------------------------------------------------Prob. 17.4

17-430

----------------------------------------------------------------Prob.. 17.5

----------------------------------------------------------------Prob.. 17.6

17-431

Prob.. 17.7

----------------------------------------------------------------Prob.. 17.8

-----------------------------------------------------------------

17-432

Prob.. 17.9

----------------------------------------------------------------Prob.. 17.10

-----------------------------------------------------------------

17-433

Prob. 17.11

--------------------------------------------------------------Prob. 17.12

17-434

---------------------------------------------------------------

17-435

Prob.. 17.13

----------------------------------------------------------------Prob.. 17.14

----------------------------------------------------------------Prob.. 17.15

-----------------------------------------------------------------

17-436

Prob. 17.16

--------------------------------------------------------------Prob. 17.17

--------------------------------------------------------------Prob. 17.18

--------------------------------------------------------------17-437

Prob. 17.19

--------------------------------------------------------------Prob. 17.20

--------------------------------------------------------------Prob. 17.21

---------------------------------------------------------------

17-438

Prob. 17.22

--------------------------------------------------------------Prob. 17.23

--------------------------------------------------------------Prob. 17.24

--------------------------------------------------------------17-439

Prob.. 17.25

----------------------------------------------------------------Prob.. 17.26

----------------------------------------------------------------Prob.. 17.27

----------------------------------------------------------------17-440

Prob.. 17.28

----------------------------------------------------------------Prob.. 17.29

17-441

--------------------------------------------------------------Prob. 17.30

---------------------------------------------------------------

17-442

Prob. 17.31

--------------------------------------------------------------Prob. 17.32

--------------------------------------------------------------17-443

Prob. 17.33

--------------------------------------------------------------Prob. 17.34

17-444

(b)

17-445

(c)

17-446

-------------------------------------------------------------Prob. 17.35 (a)

17-447

17-448

(b)

17-449

--------------------------------------------------------------Prob. 17.36

17-450

---------------------------------------------------------------

17-451

Prob. 17.37

---------------------------------------------------------------

17-452

Prob. 17.38

--------------------------------------------------------------Prob. 17.39

17-453

--------------------------------------------------------------Prob. 17.40

17-454

17-455

--------------------------------------------------------------Prob. 17.41

17-456

--------------------------------------------------------------Prob. 17.42

--------------------------------------------------------------17-457

Prob.. 17.43

-----------------------------------------------------------------

17-458

Prob. 17.44

--------------------------------------------------------------17-459

Prob. 17.45

17-460

--------------------------------------------------------------Prob. 17.46

17-461

17-462

(b)

17-463

(c)

17-464

----------------------------------------------------------------Prob.. 17.47

17-465

(b)

17-466

--------------------------------------------------------------Prob. 17.48

17-467

--------------------------------------------------------------17-468

Prob. 17.49

17-469

--------------------------------------------------------------Prob. 17.50

17-470

(b)

---------------------------------------------------------------

17-471

Prob. 17.51

17-472

---------------------------------------------------------------

17-473

Prob. 18.1 fe =

L = 12 ft = 144 in.

π 2E

(L r )

2

K = 1.0, A = 6.49 in.2, ry = 1.04 in.

L/r = 144/1.04 = 138.5

π 2 30,000 fe = = 15.44 ksi < 48 ksi (O.K.) 138.5 2 Pe = feA = 15.44(6.49) = 100.2 k ----------------------------------------------------Prob. 18.2 (a) L = 50 ft

π 2 (30,000)

= 7.109 ksi < 34 ksi (O.K.) 2  50(12)     2.94  Pe = 7.109(8.40) = 59.7 k fe =

(b) L = 35 ft

π 2 (30,000)

= 14.5 ksi < 34 ksi (O.K.) 2 35 ( 12 )     2 .

94   Pe = 14.5(8.40) = 121.8 k fe =

(c) L = 20 ft fe =

π 2 (30,000)

= 44.4 ksi > 34 ksi 2  20(12)     2.94  Euler not applicable.

(d) L = 15 ft 15 ft < 20 ft (part c) therefore, Euler not applicable.

-----------------------------------------------------

18-474

Prob. 18.3 E = 69.0×103 MPa, P.L. = 220 MPa

π 2E

max f e =

min

 L   r

2

= 220 MPa

L π 2E π 2 (69 ×103 ) = = = 55.6 220 220 r

----------------------------------------------------Prob. 18.4 Steel bar - 1”×2”, A = 2.00 in.2 weak axis r = fe =

1 = 0.2887 in.

12

π 2E

 min L = r 2

(L r )

π 2E fe

π 2 (30,000,000) min L = 0.2887 = 28.7 in.

30,000 For 5-ft long bar:

fe =

π 2 (30,000,000) 2

 5(12)     0.2887  = 6855 psi < 30,000 psi (O.K.)

Pe = 6855(2) = 13,710 lb ----------------------------------------------------Prob. 18.5 E = 207×103 MPa, P.L. = 331 MPa, fe =

π 2 (207 × 10 3 )

L/r 0 50 100

(L r )2

fe (MPa) – 817 204 18-475

150 200

90.8 51.1

-----------------------------------------------------Prob. 18.6 Aluminum alloy A = 0.7854(1.5)2 = 1.767 in.2 r = 0.25(1.5) = 0.375 in. (a) L = 24 in.

Fe =

π 2E

= 2

π 2 (10,000) 2

L  24      r  0.375  = 24.1 ksi < 32 ksi (O.K.) Pe = 24.1(1.767 ) = 42.6 k

(b) L = 60 in.

Fe =

π 2E

= 2

π 2 (10,000) 2

L  60      r  0.375  = 3.855 ksi < 32 ksi (O.K.) Pe = 3.855(1.767) = 6.81 k

18-476

(c) L = 96 in. fe =

π 2E

= 2

π 2 (10,000) 2

L  96      r  0.375  = 1.506 ksi < 32 ksi (O.K.) Pe = 1.506(1.767) = 2.66 k

----------------------------------------------------Prob. 18.7 A = 0.7854(0.5)2 = 0.1964 in.2 r = 0.25(0.5) = 0.125 in. fe =

π 2E

= 2

π 2 (30,000) 2

L  60      r  0.125  = 1.285 ksi < 48 ksi (O.K.) Pe = 1.285(0.1964) = 0.252 k

-----------------------------------------------------Prob. 18.8 For minimum length for Euler, se = P.L. fe =

L=

π 2E

π 2 Er 2

L   r

se

 L= 2

π 2 (30,000)(1.94) 2 48

= 152.4 in. = 12.70 ft

----------------------------------------------------Prob. 18.9 d = 25 mm, A = 490.9 mm2, r = 0.25(25) = 6.25 mm - steel (a) L = 1 m

π 2E

π 2 (207 ×10 3 ) fe = = 2 2 L  1000      r  6.25  = 79.81 MPa < 331 MPa (O.K.) Pe = 79.81(490.9) = 39.18 ×10 3 N = 39.2 kN

18-477

(b) L = 2 m fe =

π 2E

π 2 (207 ×103 ) 2 L  2000      r  6.25  = 19.95 MPa < 331 MPa (O.K.) = 2

Pe = 19.95(490.9) = 9.79 ×10 3 N = 9.79 kN

(c) L = 4 m fe =

π 2E

π 2 (207 ×103 ) 2 L  4000      r  6.25  = 4.99 MPa < 331 MPa (O.K.) = 2

Pe = 4.99(490.9) = 2.45 ×10 3 N = 2.45 kN

----------------------------------------------------Prob. 18.10 (Rework Prob. 18.34) d = 25 mm, A = 490.9 mm2, r = 0.25(25) = 6.25 mm - aluminum E = 70×103 MPa, P.L. = 220 MPa (a) L = 1 m fe =

π 2E

π 2 (70 ×10 3 ) 2 L  1000      r  6.25  = 27.0 MPa < 220 MPa (O.K.) = 2

Pe = 27.0( 490.9) = 13.25 ×10 3 N = 13.25 kN

(b) L = 2 m fe =

π 2E

π 2 (70 ×103 ) 2 L  2000      r  6.25  = 6.75 MPa < 220 MPa (O.K.) = 2

Pe = 6.75(490.9) = 3.31×10 3 N = 3.31 kN

18-478

(c) L = 4 m fe =

π 2E

= 2

π 2 (70 ×10 3 ) 2

L  4000      r  6.25  = 1.687 MPa < 220 MPa (O.K.) Pe = 1.687(490.9) = 828 N

-----------------------------------------------------Prob. 18.11 W12×22: A = 6.48 in.2, ry = 0.848

KL 1.0(14)(12) = = 198.1 0.848 r π 2 (30,000) = 7.54 ksi < 48 ksi (O.K.) fe = (198.1) 2 f A 7.54(6.48) = 24.4 k Pa = e = F.S. 2 ----------------------------------------------------Prob. 18.12 P.L. = 40 ksi fe =

min

π 2E

(KL r )

2

L = r

= 40 ksi

π 2E

40 = 86.04 K K

K

Min L/r

(a)

1.0

86.0

(b)

0.5

172.1

(c)

0.7

122.9

(d)

2.0

43.0

-----------------------------------------------------

18-479

wProb. 18.13 P.L.= 30,000 psi A = 2(1) – 0.8(1.8) = 0.560 in.2

rx = fe =

2(1) 3 − 1.8(0.8) 3 = 0.4006 in.

12(0.56)

π 2 (10,000,000)

= 6874 psi < 30,000 psi 2  1.0(48)     0.4006  f A 6874(0.56) Pa = e = = 1925 lb F.S. 2.0

-----------------------------------------------------Prob. 18.14 L = 7 m, K = 1.0 P.L. = 331 MPa, E = 207×103 MPa W200×59: A = 7550 mm2, Iy = 20.4×106 mm4, ry = 51.8 mm

Fa = f e F.S. fe =

π 2E

= 2

(KL r )

π 2 (207 ×10 3 )  1.0(7000)     51.8 

2

= 111.9 MPa

111.9 MPa < 331 MPa (O.K.) Fa = 111.9/2.5 = 44.8MPa Pa = FaA = 45.1(7550) = 337×103 N Pa = 337 kN -----------------------------------------------------

18-480

Prob. 18.15 Try Eqn. 18.1 req' d I =

Pe ( KL) 2 2.5(350)(34 ×12) 2 = = 492 in.4 2 2 π E π (30,000)

Try W14×132: A = 38.8 in.2, ry = 3.76 in., Iy = 548 in.4 Check Euler applic.:

π 2 (30,000) fe = = 25.1 ksi < 48 ksi (O.K.) 2  34 ×12     3.76  Check:

Pa =

25.1(38.8) = 389 k > 350 k (O.K.) 2.5

Use W14×132 -----------------------------------------------------Prob. 18.16

A = 14.1 + 2(16) = 46.1 in.2 2

 13.80  I x = 486.7 + 2(16) + 0.5  = 2239 in. 4  2 

I y = 51.4 + Iy

1(16) 3 (2) = 734 in.4 12

734 = 3.99 in.

A 46.1 KL 1.0( 22)(12) max = = 66.2 r 3.99 ry =

=

------------------------------------------------------

18-481

Prob. 18.17 2” diam. standard weight pipe, fixed-pinned, L = 84 in., P.L. = 34 ksi

π 2E

fe =

= 2

π 2 (30,000) 2

 KL   0.8(84)       r   0.787  = 40.6 ksi > 34 ksi

Euler formula is not applicable.

-----------------------------------------------------Prob. 18.18 P ( KL ) 2 ( F.S .) π 2E 20(1.0 × 30 × 12) 2 ( 2.0) = = 17.51 in. 4 2 π (30,000 )

req' d I =

Try W8×24: Iy = 18.3 in.4, ry = 1.61 in., Check Euler applic.: fe =

π 2 (30,000) = 5.92 ksi < 48 ksi (O.K.) 2  1.0 × 30 ×12    1.61  

Use W8×24 --------------------------------------------------------------Prob. 18.19 W14×90: A = 26.5 in.2, ry = 3.70 in.

Trans.

30,000 KL E = 4.71 = 4.71 = 115.4 50 r Fy

KL 1.0(16)(12) = = 51.9 < 115.4 (short/int.)

3.70 r Use Eqn. 18.8:

fe =

Pa =

π 2E

= 2

(KL r )

π 2 (30,000) 51.9 2

= 109.9

 50     109.9 

0.658 (50) (26.5) = 656 k 1.67

-----------------------------------------------------18-482

Prob. 18.20 K

KL/r

fe (ksi)

Pa (k)

(a)

0.65

33.7

260.7

732

(b)

0.80

41.5

171.9

702

-----------------------------------------------------Prob. 18.21 W250 × 115: A = 14 600 mm2, ry = 66.0 mm, Fy = 345 MPa, E = 207 × 103 MPa P = 1300 kN Transition.

KL 207 × 10 3 = 4.71 = 115.4 r 345

(a) K = 1.0, L = 6 m

KL 1.0(6000) π 2E π 2 (207 ×103 ) = = 90.9 < 115.4 Use (18 - 8) f e = = = 247 MPa 2 r 66.0 90.6 2  KL     r   345     247 

345 N/mm 2 (14,600 mm 2 ) Pa = 1.67 = 1680 kN 0.658

1300 kN < 1680 kN (O.K.) (b) K = 1.0, L = 9 m

KL 1.0(9000) π 2E π 2 (207 ×103 ) = = 136 > 115.4  Use (18 - 7) f e = = = 109.8 MPa 2 r 66.0 136 2  KL     r  Pa = 0.525(109.8)(14,600) = 842 kN 1300 kN > 842 kN (N.G.) -----------------------------------------------------

18-483

Prob. 18.22 Steel pipe: A = 11.9 in.2, r = 3.67 in Fy = 35 ksi Trans.

30,000 KL E = 4.71 = 4.71 = 137.9 35 r Fy

KL 1.0(20)(12) = = 65.4 < 137.9 (short/int.)

3.67 r Use Eqn. 18.8:

fe =

Pa =

π 2E

= 2

(KL r )

π 2 (30,000) 65.4 2

= 69.2

 35     69.2 

0.658 (35) (11.9) = 202 k 1.67

----------------------------------------------------Prob. 18.23 W12×50: A= 26.6 in.2, Fy = 36 ksi, L = 32 ft

 0.5(12) 3   = 200.3 in.4 I y = 56.3 + 2  12  2

  12.20 I x = 391.3 + 2(12)(0.5) + 0.25  = 875 in.4   2 Y axis controls:

ry =

Iy A

=

Transition

200.3 = 2.744 in.

26.6

KL 30,000 = 4.71 = 136 r 36

KL 1.0(32)(12) = = 140.1 > 136  Use (18 - 7) 2.74 r 18-484

π 2E

π 2 (30,000) fe = = = 15.09 ksi 2 140.12  KL     r  Pa = 0.525(15.09)(26.6) = 211 k -----------------------------------------------------Prob. 18.24 Select standard wt. steel pipe.

Fy = 240 MPa, P = 100 kN, L = 5.5 m, K = 1.0 Try Fa = Fy /3 = 80 MPa

req' d A =

100,000 N = 1250 mm 2 2 80 N/mm

Try 102 mm diam. pipe: A = 2040 mm2, r = 38.3 mm KL 207 × 10 3 = 4.71 = 138.3 r 240 KL 1.0(5500) = = 143.6 (use Eqn. 18 - 7) 38.3 r

Trans.

fe =

π 2E

= 2

π 2 (207 ×103 )

143.6  KL     r  Pa = 0.525(99.1)(2040)

2

= 106 kN > 100 kN

= 99.1 MPa

(O.K.)

Use 102 mm diam. Pipe ----------------------------------------------------Prob. 18.25 Select W14, Fy = 50 ksi, E = 30,000 ksi, P = 360 k, L = 24 ft, K = 1.0 Try Fa = Fy /3 = 16.7 ksi

req' d A =

360 = 21.6 in.2 16.7

Try W14×74: A = 21.8 in.2, ry = 2.48 in.

18-485

KL 30,000 = 4.71 = 115.4 r 50 KL 1.0(24)(12) = = 116.1 (use Eqn.18 - 7) r 2.48

Trans.

fe =

π 2 (30,000)

= 22.0 ksi 116.12 Pa = 0.525(22.0)(21.8) = 252 k < 360 k (N.G.) Need a larger shape.

Try W14×90: A =26.5 in.2, ry = 3.70 in.

Trans.

KL 1.0(24)(12) 30,000 KL E = = 77.8 < 115.4 = 4.71 = 4.71 = 115.4 r 3.70 50 r Fy

Use Eqn. 18.8: fe =

Pa =

π 2E

=

(KL r )2

π 2 (30,000) 77.8 2

= 48.9

 50     48.9 

0.658 (50) (26.5) 1.67 = 517 k > 360 k (O.K.)

Use W14×94 ----------------------------------------------------Prob. 18.26 W250×73: A = 9290 mm2, ry = 64.6 mm A992: Fy = 345 MPa, E = 207 × 103 MPa

trans.

KL 207 ×103 = 4.71 = 115.4 r 345

K

L

KL/r

Eqn

(m)

fe

Pa

(MPa)

(kN)

a

1.0

5

77.5

18-8

340

1255

b

1.0

10

155

18-7

85.0

415

c

0.8

10

124

18-7

132.9

648

---------------------------------------------------18-486

Prob. 18.27 K = 1.0, L = 20 ft W14×132: A = 38.8 in.2, ry = 3.76 in.

A992: Fy = 50 ksi, E = 30,000 ksi Trans. KL/r = 115.4 KL 1.0(20)(12) = = 63.8 Use Eqn. (18 - 8) r 3.76 π 2E = 72.7 ksi fe = 2  KL     r  Pa =

 50     72.7 

0.658 50 (38.8) = 871 k 1.67

----------------------------------------------------Prob. 18.28 (A36 steel)

ΣA = 38.8 + 2(14.70)(0.5) = 53.5 in.2  0.5(14.70) 3   = 1794 .7 in. 4 I xx = 1530 + 2 12  

Iyy = 551 + 2(0.5)(14.70)(7.60)2 = 1400 in.4 y-y axis controls ry =

1400 = 5.12 in.

53.50

Trans.

30,000 KL E = 4.71 = 4.71 = 136.0 36 r Fy

KL 1.0(19)(12) = = 44.6 < 136.0 (short/int.) r 5.12 18-487

Use Eqn. 18.8: fe =

π 2E  KL     r 

= 2

π 2 (30,000) 45 2

= 148.7 ksi

 36   

0.658 148.7  36 Pa = (53.50) = 1042 k 1.67

----------------------------------------------------Prob. 18.29 Select a W10 shape.

A992: Fy = 50 ksi, E = 30,000 ksi P = 280 k, K = 1.0, L = 18 ft trans.

(30,000) KL E = 4.71 = 4.71 = 115.4 50 r sY

Try Fa = Fy/3 = 16.7 ksi Req’d A = 280/16.7 = 16.8 in.2 Try W10×68: A = 19.9 in.2, ry = 2.59 in.

KL 1.0(18)(12) = = 83.4 ( Use Eqn.18 - 8) r 2.59 π 2E = 42.6 ksi fe = 2  KL     r  Pa =

 50     42.6 

0.658 (50) (19.9) = 365 k 1.67

365 k > 280 k (O.K. – over by 31%) Try a smaller shape; try W10×54: A = 15.8 in.2, ry = 2.56 in.

KL 1.0(18)(12) = = 84.4 ( Use Eqn. 18 - 8) r 2.56 π 2E fe = = 41.6 ksi 2  KL     r 

18-488

 50   

0.658  41.6  (50) Pa = (15.8) = 286 k 1.67

286 k > 280 k (O.K.) Use W10×54 ----------------------------------------------------Prob. 18.30 Select Pipe X-strong Fy = 35 ksi, E = 30,000 ksi P = 90 k, K = 1.0, L = 16 ft trans.

(30,000) KL E = 4.71 = 4.71 = 137.9 35 r Fy

Try Fa = Fy/3 = 11.67 ksi Req’d A = 90/11.67 = 7.71 in.2 Try Pipe 6 X-strong (6” diam.)

A = 7.83 in.2, r = 2.20 in.

KL 1.0(16)(12) = = 87.3 ( Use Eqn. 18 - 8) r 2.20 π 2E fe = = 38.8 ksi 2  KL     r   35   

0.658 38.5  (35) Pa = (7.83) = 112 k 1.67

112 k > 90 k (O.K.) Use Pipe 6 X-strong ----------------------------------------------------Prob. 18.31 (Euler-Johnson) least r = trans.

h 1 = = 0.2887 in.

12 12

KL = r

2π 2 E = Fy

KL 1.0(20) 2π 2 (30,000) = = 69.3 = 118.7 r 0.2887 42

69.3 < 118.7, use the J. B. Johnson formula:

18-489

 42(69.3) 2  1 −  4π 2 (30,000)  42  = 9.956 ksi Fa =  3.5

Pa = FaA = 9.956(2.0) = 19.91 k -----------------------------------------------------Prob. 18.32 A = 0.7854d2, r = 0.25d Assume rod will be “intermediate” and J. B. Johnson formula applies.  Fy (KL r )2  1 −  Fy 4π 2 E    Fa = F.S. 2   1.0(20)   42     1 −  0.25d   42  4π 2 (30,000)      = 16.8 − 3.813 = 2.5 d2

Pa 3.813 6.00  16.8 − 2 = A d 0.7854d 2 from which : d = 0.8256 in.

Fa =

Try 7/8” diameter rod.

Check : KL 1.0(20) = = 91.4 r 0.25(0.875) trans.

KL = r

2π 2 E = Fy

2π 2 (30,000) = 118.7 42

91.4 < 118.7 (O.K.) Use 7/8” diam. rod ----------------------------------------------------Prob. 18.33 Assume rod to be slender.

A = 0.7854d2, r = 0.25d

18-490

Fa =

π 2E 2

 KL    F.S.  r 

=

π 2 (30,000) 2

 1.0(54)    3 .5  0.25d  = 1.8132d 2

FaA = Pa (1.8132d2)(0.7854d2) = 1.90 k d = 1.075 in.

Try 1 1/8” diam. rod Check if slender: KL 1.0(54) = = 192 r 0.25(1.125) trans.

2π 2 E = Fy

KL = r

2π 2 (30,000) = 118.7 42

192 > 118.7 (O.K.) Use 1 1/8” diam. rod ----------------------------------------------------Prob. 18.34 Assume rod to be slender.

A = 0.7854d2, r = 0.25d Fa =

π 2E 2

 KL    F.S.  r 

=

π 2 (30,000) 2

 1.0(96)    3 .0  0.25d  = 0.6693d 2

FaA = Pa (0.6693d2)(0.7854d2) = 18.0 k d = 2.419 in.

Try 2½” diam. rod Check if slender: KL 1.0(96) = = 153.6 r 0.25(2.50)

18-491

trans.

KL = r

2π 2 E = Fy

2π 2 (30,000) = 83.5 85

153.6 > 83.5 (O.K.) Use 2½” diam. rod -----------------------------------------------------

Prob. 18.35 Assume rod to be intermediate category. r = 0.25d Pa = A sa 2   1× 24   50     1 −  0.25d  50 2  4π (30,000)     10 k = 0.7854d 2  3.0

Solve for d: Req’d d = 1.074 in., try 1 1/8” diam. rod.

Check validity of J.B.Johnson formula: KL 1.0(24) = = 85.3 r 0.25(1.125) trans.

KL = r

2π 2 E = sY

2π 2 (30,000) = 108.8 85.3 < 108.8 (O.K.) 50

Use 1 1/8” diam. rod ----------------------------------------------------Prob. 18.36 19 mm diam. steel rod, K = 1.0, L = 350 mm, F.S. = 2.5 r = 0.25(19) = 4.75 mm A = 0.7854(19)2 = 284 mm2

18-492

trans.

KL = r

2π 2 E = sY

2π 2 ( 207 × 10 3 ) = 118.7 290

KL 1.0(350) = = 73.68 r 4.75

73.68 < 118.7, therefore column is intermediate and J.B.Johnson formula applies:  290(73.68) 2  1 − 4π 2 ( 207 × 10 3 )  290  Fa =  = 93.7 MPa 2. 5

Pa = 93.7(284) = 26.6×103 N = 26.6 kN ----------------------------------------------------Prob. 18.37 Cross section is d×2d K = 1.0, L = 16 in., P = 4.60 k, H.S. steel with Fy = 110,000 psi, S.F. = 3.0 A = 2d2 in.2 r = d / 12 = 0.2887 d in.

Assume bar is slender: Pa = AFa     2 π (30,000)  2 4.60 k = 2d   1(16)  2    3 .0    0.2887 d  

Solve for d: Req’d d = 0.517 in., 2d = 1.035 in. (more) trans.

KL = r

2π 2 E = sY

2π 2 (30,000) = 73.4 110

KL 1.0(16) = = 107 r 0.2887(0.517) 107 > 73.4

(O.K. – bar is slender.)

Req’d dimensions: 0.517” × 1.035” 18-493

----------------------------------------------------Prob. 18.38 8×8 (S4S) Doug. fir, E = 1700 ksi, K = 1.0, A = 56.3 in.2, Fc = 1050 psi, d = 7.5 in. (a) Le = KL = 1.0(10) = 10.0 ft Le 10(12) = = 16 < 50 (O.K.) 7 .5 d 0.3E 0.3(1.7 ×10 6 ) α= = = 1.897 2 16 2 (1050)  Le    Fc d  2

CP =

1 + 1.897  1 + 1.897  1.897 −  = 0.858  − 1.6 0.8  1 .6 

Fa = FcCP = 1050(0.858) = 901 psi Pa = FaA = 901(56.3) = 50,700 lb (b) ) Le = KL = 1.0(17) = 17.0 ft Le 17 (12 ) = = 27.2 < 50 (O.K.) d 7 .5 0.3(1.7 × 10 6 ) α= = 0.657 27.2 2 (1050 ) 2

CP =

1 + 0.657  1 + 0.657  0.657 −  = 0.534  − 1 .6 0 .8  1 .6 

Fa = FcCP = 1050(0.534) = 561 psi Pa = FaA = 561(56.3) = 31,600 lb -----------------------------------------------------Prob. 18.39 (Rework Prob. 18.38) 250 mm × 250 mm, Eastern white pine, E = 7.6×103 MPa, K = 0.8, A = 58.1×103 mm2, Fc = 5.00 MPa, d = 241 mm (a) Le = KL = 0.8(3050) = 2440 mm

18-494

Le 2440 = = 10.12 < 50 (O.K.) 241 d 0.3E 0.3(7.6 ×10 3 ) α= = = 4.45 2 10.12 2 (5.00)  Le    Fc d  2

CP =

1 + 4.45  1 + 4.45  4.45 −  = 0.949  − 1 .6 0 .8  1 .6 

Fa = FcCP = 5.00(0.949) = 4.75 MPa Pa = FaA = 4.75(58.1×103) = 276 kN (b) ) Le = KL = 0.80(5180) = 4144 mm Le 4144 = = 17.20 < 50 (O.K.) d 241 0.3(7.6 × 10 3 ) α= = 1.541 17.20 2 (5.00) 2

1 + 1.541  1 + 1.541  1.541 −  CP = = 0.816  − 1.6 0.8  1 .6 

Fa = FcCP = 5.00(0.816) = 4.08 MPa Pa = FaA = 4.08(58.1×103) = 237 kN -----------------------------------------------------Prob. 18.40 Doug. fir: E = 1700 ksi, Fc = 1050 psi K = 1.0, L = 18 ft, P = 90,000 lb Assume Fa = Fc:

req' d A =

90,000 = 85.7 in.2 1050

Try 10×10 (S4S): A = 90.3 in.2, d = 9.5 in.

2

CP =

1 + 0.943  1 + 0.943  0.943 −  = 0.670  − 1 .6 0 .8  1 .6 

Fa = FcCP = 1050(0.670) = 704 psi

90,000 704 = 127.8 in.2 > 90.3 in.2 (N.G.)

req'd A =

18-495

Try 12×12 (S4S): A = 132 in.2, d = 11.5 in.

Le 18(12) = = 18.78 < 50 (O.K.) d 11.5 0.3(1.7 × 10 6 ) α= = 1.377 18.78 2 (1050) 2

1 + 1.377  1 + 1.377  1.377 −  CP = = 0.789  − 1.6 0.8  1 .6 

Fa = FcCP = 1050(0.789) = 829 psi Pa = FaA = 829(132) = 109,400 lb 109.4 k > 90 k (O.K.) Use 12x12 (S4S) ----------------------------------------------------Prob. 18.41 Southern pine: E = 1700 ksi Fc = 1250 psi K = 1.0, L = 16 ft (a) P = 48,000 lb Assume Fa = Fc:

req' d A =

48,000 = 38.4 in.2 1250

Try 8×8 (S4S): A = 56.3 in.2, d = 7.5 in.

Le 1.0(16)(12) = = 25.6 < 50 (O.K.) 7 .5 d 0.3(1.7 × 10 6 ) α= = 0.623 25.6 2 (1250 ) 2

CP =

1 + 0.623  1 + 0.623  0.623 = 0.514 −   − 1.6 0.8  1.6 

Fa = FcCP = 1250(0.514) = 643 psi

48,000 643 = 74.7 in.2 > 56.3 in.2 (N.G.)

req'd A =

18-496

Try 8×12 (S4S): A = 86.3 in.2, d = 7.5 in.

Fa is unchanged: Pa = 643(86.3) = 55,500 lb > 48,000 lb (O.K.) Use 8×12 (S4S) (b) P = 60,000 lb From part (a), 8×12 too small.

Try 10×10: A = 90.3 in.2, d = 9.5 in.

Le 1.0(16)(12) = = 20.2 < 50 (O.K.) 9.5 d 0.3(1.7 × 10 6 ) α= = 1.00 20.2 2 (1250 ) 2

CP =

1 + 1.00  1 + 1.00  1.00 −  = 0.691  − 1.6 0.8  1.6 

Fa = FcCP = 1250(0.691) = 864 psi Pa = 864(90.3) = 78,000 lb > 60,000 lb (O.K.) Use 10×10 (S4S) ----------------------------------------------------Prob. 18.42 (See Prob. 18.41) Southern pine: E = 1700 ksi Fc = 1250 psi K = 0.8, L = 16 ft (a) P = 48,000 lb Assume Fa = Fc:

req' d A =

48,000 = 38.4 in.2 1250

Try 8×8 (S4S): A = 56.3 in.2, d = 7.5 in.

Le 0.8(16)(12) = = 20.48 < 50 (O.K.) 7 .5 d 0.3(1.7 × 10 6 ) α= = 0.973 20.48 2 (1250 )

18-497

2

CP =

1 + 0.973  1 + 0.973  0.973 −  = 0.681  − 1.6 0.8  1.6 

Fa = FcCP = 1250(0.681) =851psi

48,000 851 = 56.4 in.2 ≈ 56.3 in.2 (Say O.K.)

req'd A =

Use 8×8 (S4S) (b) P = 60,000 lb Assume an 8” nominal thickness from part (a) and Fa = 851 psi:

req' d A =

60,000 = 70.5 in.2 851

Use 8×10 (S4S) (A = 71.3 in.2) ----------------------------------------------------Prob. 18.43 Doug. fir: E = 1700 ksi, Fc = 1050 psi 10×12 (S4S): A = 109 in.2, d = 9.5 in.

K = 0.8, L = 16 ft

Le 0.8(16)(12) = = 16.17 < 50 d 9.5

α=

(O.K.)

0.3(1700 × 10 3 ) = 1.858 (16.17) 2 (1050) 2

1 + 1.858  1 + 1.858  1.858 CP = −  = 0.854  − 1.6 0.8  1.6 

Fa = FcCP = 1050(0.854) = 897 psi Pa = 897(109) = 97,800 lb = 97.8 k ----------------------------------------------------Prob. 18.44 Southern pine: E = 1700 ksi, Fc = 1250 psi Full nom. 12×12: A = 144 in.2, d = 12 in.

K = 1.0, L = 20 ft 18-498

Le 20(12) = = 20 < 50 d 12

α=

(O.K.)

0.3(1700 × 10 3 ) = 1.02 20 2 (1250) 2

1 + 1.02  1 + 1.02  1.02 CP = −  = 0.698  − 1.6 0.8  1.6 

Fa = FcCP = 1250(0.698) = 873 psi Pa = 873(144) = 125,700 lb = 125.7 k ----------------------------------------------------Prob. 18.45 Doug. fir: E = 1700 ksi, Fc = 1050 psi K = 1.0, L = 18 ft, P = 62 k Assume Fa = Fc:

req' d A =

62,000 = 59.0 in.2 1050

Try 10×10 (S4S): A = 90.3 in.2, d = 9.5 in.

Le 1.0(18)(12) = = 22.7 < 50 (O.K.) d 9.5 0.3(1.7 × 10 6 ) α= = 0.943 22.7 2 (1050 ) 2

CP =

1 + 0.943  1 + 0.943  0.943 −  = 0.670  − 1 .6 0 .8  1 .6 

Fa = FcCP = 1050(0.670) = 704 psi

62,000 704 = 88.1 in.2 < 90.3 in.2 (O.K.)

req'd A =

Use 10×10 (S4S) ----------------------------------------------------Prob. 18.46 18-499

Southern pine: E = 1700 ksi Fc = 1250 psi K = 1.0, L = 20 ft

P=

100(2000) = 50,000 lb 4

Assume Fa = Fc:

req' d A =

50,000 = 40 in.2 1250

Try 8×8 (S4S): A = 56.3 in.2, d = 7.5 in.

Le 1.0( 20)(12 ) = = 32 < 50 (O.K.) 7 .5 d 0.3(1.7 × 10 6 ) α= = 0.398 32 2 (1250 ) 2

CP =

1 + 0.398  1 + 0.398  0.398 −  = 0.358  − 1.6 0.8  1.6 

Fa = FcCP = 1250(0.358) = 448 psi

50,000 448 = 111.6 in.2 > 56.3 in.2 (N.G.)

req'd A =

Try 10×10 (S4S): A = 90.3 in.2, d = 9.5 in.

Le 1.0( 20)(12) = = 25.3 < 50 (O.K.) 9 .5 d 0.3(1.7 × 10 6 ) α= = 0.637 25.3 2 (1250 ) 2

CP =

1 + 0.637  1 + 0.637  0.637 −  = 0.523  − 1.6 0.8  1.6 

Fa = FcCP = 1250(0.523) = 654 psi Pa = 654(90.3) = 59,100 lb > 50,000 lb (O.K.) Use 10×10 (S4S) -----------------------------------------------------Prob. 18.47 Southern pine: E = 12×103 MPa , 18-500

Fc = 8.62 MPa K = 1.0, L = 6 m, P = 450 kN Assume Fa = Fc: 450 ×10 3 req' d A = = 52.2 ×10 3 mm 2 8.62 Try 250×250 (S4S) 250×250 300×300 A (mm2) 58.1×103 85.3×103 d (mm)

241

292

Le/d

24.9

20.5

α

0.674

0.994

CP

0.544

0.689

Pa (kN)

272

507

Decision N.G.

O.K.

Use 300×300 (S4S) -----------------------------------------------------

18-501

Prob. 19.1 A-36 plates, 7/8" H.S. bolts (A325-X) Ab = 0.601 in.2 Bolt shear: PS = 1.0(0.601)(34)(9) = 183.9 k Bearing on VP plate: PP = 0.875(0.5)(87)(9) = 343 k Tension: (Gross area): Pg= 12(0.5)(21.6) = 129.6 k (Net area): An = bt -NFdHt = 12(0.5) -3(1)(0.5) = 4.5 in.2 P„ = 4.50(29) = 130.5 k All. tensile load = PG= 130.5 k -------------------------------------------------------------Prob. 19.2 (see Prob. 19.1) A325-N PS = 1.0(0.601)(27)(9) = 146.0 k From Prob. 1-1: PP = 343 k PG= 129.6 k PN = 130.5 k All. tensile load = PG = 129.6 k -------------------------------------------------------------Prob. 19.3

19-502

-------------------------------------------------------------Prob. 19.4

-------------------------------------------------------------Prob. 19.5

--------------------------------------------------------------

19-503

Prob. 19.6

-------------------------------------------------------------Prob. 19.7

--------------------------------------------------------------

19-504

Prob. 19.8

--------------------------------------------------------------Prob. 19.9

--------------------------------------------------------------

19-505

Prob. 19.10

-------------------------------------------------------------Prob. 19.11

-------------------------------------------------------------19-506

Prob. 19.12

-------------------------------------------------------------Prob. 19.13

-------------------------------------------------------------19-507

Prob. 19.14

-------------------------------------------------------------Prob. 19.15

19-508

-------------------------------------------------------------Prob. 19.16 A36 Steel

-------------------------------------------------------------Prob. 19.17 A36 steel

--------------------------------------------------------------

19-509

Prob. 19.18

-------------------------------------------------------------Prob. 19.19

-------------------------------------------------------------Prob. 19.20

--------------------------------------------------------------

19-510

Prob. 19.21

-------------------------------------------------------------Prob. 19.22

--------------------------------------------------------------

19-511

Prob. 19.23

-------------------------------------------------------------Prob. 19.24

-------------------------------------------------------------Prob. 19.25

--------------------------------------------------------------

19-512

Prob. 20.1

--------------------------------------------------------------Prob. 20.2

--------------------------------------------------------------Prob. 20.3

--------------------------------------------------------------Prob. 20.4

--------------------------------------------------------------20-513

Prob. 20.5

--------------------------------------------------------------Prob. 20.6

--------------------------------------------------------------Prob. 20.7

--------------------------------------------------------------Prob. 20.8

--------------------------------------------------------------20-514

Prob.. 20.9

----------------------------------------------------------------Prob.. 20.10

----------------------------------------------------------------Prob.. 20.11

-----------------------------------------------------------------

20-515

Prob.. 20.12

----------------------------------------------------------------Prob.. 20.13

----------------------------------------------------------------Prob.. 20.14

-----------------------------------------------------------------

20-516

Prob. 20.15

--------------------------------------------------------------Prob. 20.16

--------------------------------------------------------------Prob. 20.17

---------------------------------------------------------------

20-517

Prob.. 20.18

----------------------------------------------------------------Prob.. 20.19

-----------------------------------------------------------Prob.. 20.20

----------------------------------------------------------------Prob.. 20.21

----------------------------------------------------------------20-518

Prob. 20.22

--------------------------------------------------------------Prob. 20.23

---------------------------------------------------------------

20-519