Solution Manual for Applied Statics and Strength of Materials 7th Edition by George Limbrunner.

Solution Manual for Applied Statics and Strength of Materials, 7th Edition by George Limbrunner

Solution Manual for

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

iv Copyright © 2020 Pearson Education, Inc.

Copyright © 2022 by Pearson Education, Inc. or its affiliates, 221 River Street, Hoboken, NJ 07030. All Rights Reserved. Manufactured in the United States of America. This publication is protected by copyright, and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise. For information regarding permissions, request forms, and the appropriate contacts within the Pearson Education Global Rights and Permissions department, please visit www.pearsoned.com/permissions/.

ISBN-13: 978-0-13-571672-4 ISBN-10: 0-13-571672-1

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.

Solution 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 = √102 + 72 = 12.21 ft (b) 𝑏 = √202 − 162 = 12.00 m Prob. 1.2 (a) 𝑎 = 25 sin 48° = 18.58 ft (b) 𝑏 = √252 − 18.582 = 16.73 ft (c) ℎ = b sin 48° = 16.73 sin 48° = 12.43 ft Prob. 1.3 𝑎 = √722 − 67.32 = 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 = √102 + 62 = 11.6 ft 𝜃 = tan−1(6⁄10) = 31.0° Prob. 1.6 𝐴𝐵 = √122 + 162 = 20 ft 𝐵𝐶 = √122 + 322 = 34.2 ft 𝐴 = tan−1(12⁄16) = 36.9° 𝐶 = tan−1(12⁄32) = 20.6° Prob. 1.7 θ = sin−1(5⁄6) = 56.4° 𝑥 = √122 + 102 = 6.63 ft

1-1

stumerit@gmail.com

Prob. 1.8 Assume all angles to be 45° 𝑅𝑥 = +2 + 3 cos 45° + 0 − 6 sin 45° = 0.1213 mi 𝑅𝑦 = 0 − 3 sin 45° + 6 − 6 sin 45° = −0.364 mi. 𝑅 = √(−0.1213)2 + (−0.364)2 = 0.384 mi.

Prob. 1.9 (a) c2 = 112 + 132 – 2(11)(13)cos 80° = 15.50 ft sin 80° sin A sin B = = 13 15.50 11 →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 78 85 →A = 57.4°, B = 50.6° (c) Right Triangle Check Check a2 + b2 = c2 𝐴 = tan−1(7⁄24) = 16.26° 𝐵 = tan−1(24⁄7) = 73.7° Prob. 1.10 C = 180° – 55° – 63° = 62° 100 𝑐 𝑎 = = 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

stumerit@gmail.com

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

26 20 = → ∴ 𝐵1 = 41.77° sin B1 sin 60° D1= 78.23 and D2= 101.77 20 𝐴𝐵 = →∴ 𝐴𝐵 = 29.39 ft sin 78.23° sin 41.77° 𝐵𝐶 = √(𝐴𝐵)2 + 502 − 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 mi ft 1hr 1 min ft 60 × 5280 × × = 88 sec hr mi 60 min 60 sec Prob. 1.27 43,560

ft2

1 yd 2 1 rod 2 rod2 ×( ) ×( ) = 160 acre acre 3 ft 5.5 yd

Prob. 1.28 (a) 125,000,000,000 gal gal ft2 = 384 × 103acre − ft 7.481 × 43,560 acre ft3 (b) lb 125,000,000,000 gal × 62.4 ft3 = 521 × 106 tons gal lb 7.481 × 2000 ton ft3 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. in 0.815 ft × 12 = 9.78 in. ft 0.78 in.× 32 = 24.96 25 ∴ 1.815 ft = 1′ − 9  32 Prob. 1.30 Volume=Area  length π(2 in. )2 ft in.

(0.25 mi) (5280 ) (12 4 mi ft ) = 3 in.

1728 ft3 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  2202 1802

= 78.1 lb

= 284 N

x = tan–1 (60/50)

x = tan–1 (220/180)

= 50.2

= 50.7

 

(c) R  42 1.52 = 4.27 k x = 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

(425)  354 lb (425)  236lb

13

5 (110)  42.3 lb 13 12 Fy   (110)  101.5lb 13

(b) Fx  

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  0.1302  1.304 kN    tan1  1.3    x  0.103   85.5 



  Prob. 2.7

R = 175/(cos 45) = 247 lb x = 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 12



Fy 5



300 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 4



Fx 3



150 5

F  (150)  120 lb y 5 3 F  (150)  90 lb x 5

Prob. 2.13

FH = 1000 sin 40 = 643 lb FV = 1000 cos 40 = 766 lb Prob. 2.14 (a) Py = 80cos20 = 75.2 k Px = 80sin20 = 27.4 k (b) Px = 60 sin 30 = 30.0 kN Py = 60 cos 30 = +52.0 kN Prob. 2.15 FV = 500sin42 = +335 lb FH = 500cos42 = +372 lb Prob. 2.16 FH = 443 lb = 500 cos   443   cos1 500  27.6   

2-15

Prob. 2.17 Fx Fy 120   2 1 5 2 120  107.3 kN Fx  5 120 F  53.7 N y 5 Prob. 2.18 W = weight of the skiers W = 38(175) = 6650 lb T = Wx = 6650 sin 17 = 1944 lb Prob. 2.19

F1x = F2x = 10 cos 40 = 7.66 N Hypotenuse of slope triangle =

s2 1

By similar triangles: 1  s 2 1   s 1.684 7.66 15 Resultant: F1y



F1x

1.684 1 F1y  1.684(7.66)  12.90 N

F2y = 10 sin 40 = 6.43 N R = 6.43 + 12.90 = 19.33 N 

2-16

 



Prob. 2.20 Fy



Fx 9500  7 4 8.06 7 (9500)  8250 lb Fy  8.06 F  4 (9500)  4710 lb x 8.06 

Prob. 2.21 (a) P  3002  2002  361lb x = tan–1(200/300) = 33.7 (b) P  5002  3002  583lb x = tan–1(300/500) = 31.0 (c) P  2402  3602  433lb x = tan–1(360/240) = 56.3 (d) P  2502  4602  524 lb x = tan–1(460/250) = 61.5

Prob. 2.22 P1x = 7.78 cos 80 = +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  150    tan   73.6 44.1   1 x  0.294 m tan 73.6 x1  1.5  0.294  1.794 m

Prob. 2.24

TV = 2000 sin 70 = 1879 lb TH = 2000 cos 70 = 684 lb Prob. 2.25 For the 2-kip loads:

FV



FH

 2 2 1 5 2(2) F   1.789 k V 5 1(2)  0.894 k 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

 = 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  sin130 sin   45   sin1 sin130  32.59



   64.0   x  60  32.59  27.4



Prob. 3.2

1= tan1(4/3) = 53.13 2 = tan1(5/12) = 22.62  = 180  1  2 = 104.25  = 180   = 75.75 R2 = 122 + 102  2(12)(10) cos 75.75 R = 13.60 k 1 = sin1(sin75.75(10/13.60)) = 45.45 3-20



x = 2 +1 = 22.62 + 45.45 = 68.07



 

Prob. 3.3

1 = tan1(1/2) = 26.57  = 180  26.5730 = 123.43  = 180   = 56.57 R2 = 752 + 602 2(75)(60)cos 56.57 R = 65.32 lb R 60  sin 56.57 sin 2   60    sin1 sin 56.57  50.05  65.32   x  1  2  76.6 2

 

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  56.822  29.462 R  64.0 lb 29.46  x  tan1  56.82      27.4

 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    tan1   5.08   x  68.1  

Prob. 3.6

R    2  60cos30  15.12 lb 75   x  5   1   60sin 30  63.54 lb Ry  75  5  R  (15.12)2  (63.5)2 R  65.3 lb  63.54    tan1  15.12   x  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

F  5.942  87.742

Fy + 48.21 lb = 135.95 lb Fy = + 87.74 lb

 87.9 lb  θ  tan1  5.94   87.74     3.87

Fx + Px = Rx Fx  57.45 lb = 63.39 lb





Fx = 5.94 lb  





Prob. 3.8

Force

Horiz. (N)

Vert. (N)

100 N

90.6

42.3

120 N

75.0

93.7



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



0

265

R = 265 N   

 (+)

Prob. 3.10 Use components: For sloping force

Slope triang. hypt  52  32  5.83 Fx Fy 200   3 5 5.83 Fx  102.9; Fy  171.5 Total horiz. = 200 102.9 =  302.9 lb 302.92 171.52

R θ  tan1

 348 lb  171.5   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



543.7

400

 

R  543.72  4002  675 lb  400  θ  tan1  36.3   543.7 

x

 

Prob. 3.12 (a)

R1,2  502  702  2(50)(70) cos100  92.82 N



sin1 sin100  50 50sin100  92.82   sin 1  32.04   92.82   x  1 10  22.04 1

3-25



(b) R3,4  602  202  2(60)(20) cos 76.33  58.59 N sin 1 sin 76.33   θ1  84.30 60 58.59 θx  84.30  36.87  47.43 (more)

(c)

R  R3,4 2  R1,2 2  2R3,4 R1,2 cos 25.39  47.14 N sin1 sin 24.39   θ 1  57.59(122.41) 92.82 47.14 θx  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

 +46.41

8.33

3-26

R  46.412  (8.33)2  47.2 N  8.33  θ  tan1  10.18   x  46.41   

 +

Prob. 3.14

Rx = Fx = +200 cos 30 +100 cos 45 400 300 cos 60 = 306.1 lb Ry = Fy = +200 sin 30 100 sin 45 50 +300 sin 60 = +239.1 lb R  (306.1)2  239.12  388 lb

θ  tan1 x

 239.1  38.0  306.1  

 

 +

Prob. 3.15 R = Ry = +300 lb

Rx = Fx = 0 = 500 + F1x + 240 cos 30  F1x = +292.2 lb Ry = Fy = +300 = +F1y 240 sin 30  F1y = +420 lb



F1  4202  292.22  512 lb   θ  tan1  420   55.2  292.2  x  



Prob. 3.16

 +

Rx = Fx = +1000 + 600 sin30 = +1300 lb

3-27

Ry = Fy = +900 600 cos30 = +380.4 lb



R  380.42 13002  1355 lb  380.4   x  tan1  1300  16.3  

 +

Prob. 3.17

Rx = Fx = +100 cos 20  95 cos 80  110 cos 70 = + 39.9 N  Ry = Fy = +95 sin 80  110 sin 70  100 sin 20 =  44.0 N  

R  39.92  (44.0)2  59.5 N  44.0  θ  tan1  47.8  39.9  x    

 +

Prob. 3.18 For the 200-lb resultant force: Rx = 200 cos 60 = +100 lb Ry = 200 sin 60 = +173.2 lb

Rx = Fx = +100 =  160 + F2x  F2x = 260 lb Ry = Fy = +173.2 = F2y

F2  173.22  26.02  312 lb  173.2  θ x  tan1  260   33.7 





3-28

Prob. 3.19

 +

For the resultant: Rx = 300 cos45 =  212.1 lb Ry = 300 sin45 = 212.1 lb Also: Rx = Fx = 212.1 = +F1 cos60  F2 cos 30 

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

(I.)

Ry = Fy = +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 = Fx = +75 = +F1 cos 30  F2 cos 67 

+75 = +0.866 F1  0.39 F2 F1 = +86.6 + 0.45F2

(I.)

Ry = Fy = 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



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  tan1  84.9  4.216  x    

 +

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  tan1  2.42  310.4  x  

 

Prob. 3.25

 +

For the resultant: Rx = 100 cos 20 = +93.97 lb Ry = 100 sin 20 = +34.20 lb Also: Rx = Fx = +93.97 = +F2 cos 30  F1 cos 45 3-31



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

(I.)

Ry = Fy = +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 FH 2113 F   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   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

Mo = +(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



(b) MA = 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) Mo = +300 + 116.5 = + 417 lb-ft (c) Rx = Fx = +25  75 cos 15 = 47.4 lb Ry = Fy = +50 + 75 sin 15 = +69.4 lb (more) R  69.42  (47.4)2  84.0 lb  69.4  θ  tan1  55.7  47.4  x  

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) Mo = +103.9  113.1  40 =  49.2 lb-ft  +

Prob. 3.32

Rx = Fx = 30 cos 60 + 40 cos 45  50 = 6.716 lb Ry = Fy = 30 sin 60 40 sin 45 10 = 12.304 lb R  (6.716)2  (12.304)2  14.02 lb  12.304  θ  tan1  61.4  6.716  x   Moment of resultant: Mo= 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) MA 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 

2.35

141.0

 

217

(b) MB 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



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

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

CC moment: +

MB = 12(3)  14(10) + 25(4) = 76.0 k-ft Prob. 3.36 Resultant of dist. force = 10(6) = 60 N R = Fy = 50  60 + 30  60 = 140 N MA = 50(2)  60(6) + 30(11)  60(14) = 970 Nm x is measured to the right from point A: M A  970  6.93 m x 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: Mo = 550 sin 40(10) + 550 cos 40(5) = 1429 lb-ft

(checks)

3-36

Prob. 3.38

CC moment: +

Components at B: Mo = 550 sin40 = 1429 lb-ft (checks) Components at C: Mo = 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 MA = 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 MA= +129.9(5) + 75.0(20) = +2150 lb-ft Prob. 3.41

CC moment: +

MA = 3(5)  (4/5)(5)(10) = 55.0 k-ft Prob. 3.42

CC moment: +;  +

R = 300  450  200 = 950 lb () MA = 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 () MA = +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 () MA =  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 () MA = 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() MA = 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 () MA = 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 () MA: +10(5) = +9(1) +16(8) 15x1  x1 = 5.80 m Prob. 3.49

R = 10  28 =  38 k () MA = 10(5)  28(17) =  526 k-ft 526 x  13.84 ft 38 Prob. 3.50

R = 1000  7000  1600 = 9600 lb () MA = 7000(8)  1600(12) = 75,200 lb-ft 75,200 x  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 (upfrombase) Prob. 3.52 Resultant of the triangular load is 36 k () located 3 ft right of A. R = 36 + 5 = 41 k  MA = 36(3)  5(13) =  173 k-ft x

173

 4.22 ft

41 Prob. 3.53

CC moment: +;  +

R =  21  15  10 = 46 kN MA = 21(7)  15(11.5)  10(14) =  459.5 kNm 459.5 x  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  10,100  63 



MA =  19,800(4)  10,100(6) = 139,800 lb-ft Resultant intersects the base of the dam x to the right of point A: M A  139,800  7.06 ft x 19,800 Ry (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 = Fy = + 11,650 lb MA = 4125(5) + 5250(7.5) + 2275(28.8.67) = 104,000 lb-ft x

M A  104,000  8.93 ft R 11,650

Prob. 3.56

CC moment: +;  +

R = Fy =  900  600 = 1500 lb MA =  900(2)  600(15) = 10,800 lb-ft M A  10,800  7.20 ft x 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





  tan-1 12321  80.3 MA = 53(1)  70(2) + 21(1) =  172 kNm x is measured to the right from point A: M A  172  1.398 m x R 123

Prob. 3.58 (a) R

312  499.2

(3)(3)  3650 lb 2 R1  312(3)(3)  2808 lb 187.2 R  (3)(3)  842.4 lb 2 2 R (1.5)  R2 (1) y 1  1.385 ft R1  R2 124.8 R  (2)(3)  374.4 lb 3 2 Rb  3650  374.4  3276 lb 3650(1.385)  374.4(0.667) y  1.467 ft 3276 a

Prob. 3.59 1. R = (52/2)(0.833)(5) = 108.3 lb 0.833 y  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  y  0.370 ft 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   y  0.389 ft 541

Prob. 3.60 R = Fy = 25(4) 1240 = 77 k MA = 25(4)12(14)40(28) = 1388 k-ft x

M A  1388  18.03 ft Ry 77

Prob. 3.61 F1 + F2 = 150+100 = +250 lb F1 = 250  F2

I.

MA = +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  tan1   x  5.733   83.1  

Mo = 10(8)  15 sin75(14)  25(12/13)(19)= 721.3 k-ft MO 721.3 x   15.16 ft Ry 47.57

Prob. 3.67

CC moment: +;  +

For the resultant: Rx = Fx = +500  100 = +400 lb  Ry = Fy = 400 + 250 = 150 lb  R  4002  (150)2



 427 lb  150  θ  tan1  20.6   x  400 



Mo = +100(4)  400(5) + 250(17)  500(12) =  3350 lb-in.

Note: resultant pierces base of block x to the right of point O. M O  3350  22.3 in. x 150 Ry

3-45

Prob. 3.68

CC moment: +;  +

For the resultant: Rx = Fx = + 200 N  Ry = Fy = 80  50 = 130 N  R  2002  (130)2  239 N  130  θ  tan1  33.0  200  x   

MA = 200(1.333) + 50(2) + 80(1) =  86.6 Nm MA 86.6 x   0.666 m (as shown) Ry 130

Prob. 3.69

CC moment: +;  +

For the resultant: Rx = Fx = +15 k  Ry = Fy = 50  20 = 70 k  R  152  (70)2  71.6 k  70  θ  tan1   77.9 x  15  3-46

MA = 15(6) + 50(6) +20(5) = +310 k-ft.  

Note: resultant pierces bottom of footing x to the left of point A. MA 310 x   4.43 ft Ry 70

Prob. 3.70 (a) M A  1200(3)  500(4 / 5)(12)  500(3 / 5)(9)  400(20)  19,100 lb - ft

(b) Rx=Fx = +400+500(4/5) = +800 lb Ry = Fy = 1200500(3/5) = 1500 lb R  8002  (1500)2  1700 lb M A  19,100  23.9 ft x 800 Rx (Above point A) Prob. 3.71 Units: Newtons and meters: (a) MA = (0.12106)(1)980(6) (8103) sin 40(13) = 192.7  103 Nm = 192.7 kNm MB = +(0.12106)(15)+980(10) +(8103) sin40(3) = 1.825  106 N.m = 1.825 MNm (b) Rx = Fx = +(8103)cos40 = +6.128103 N

3-47

Ry = Fy = (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  tan1   87.2 x 6.128   

x is measured to the right from point A: x

M A Ry

 192.7 10  1.53 m 126.1103

Prob. 3.72

3

CC moment: +;  +

Rx = Fx = +25 cos45 40(4/5) = 14.32 lb Ry = Fy = + 25 sin45 + 40(3/5)  20 = + 21.68 lb R  (14.32)2  21.682  26.0 lb  21.68  θ  tan1  56.6  14.32  x   

MA = + 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: M A  104.4  4.82 ft x 21.68 Ry

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

stumerit@gmail.com

Prob. 4.21

Prob. 4.22

Prob. 4.23

4-53

stumerit@gmail.com

Prob. 4.24

Prob. 4.25

Prob. 4.26

4-54

stumerit@gmail.com

Prob. 4.27

Prob. 4.28

Prob. 4.29

4-55

stumerit@gmail.com

Prob. 4.30

Prob. 4.31

Prob. 4.32

4-56

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Prob. 4.33

Prob. 4.34

4-57

stumerit@gmail.com

Prob. 4.35

Prob. 4.36

4-58

stumerit@gmail.com

Prob. 4.37

4-59

stumerit@gmail.com

Prob 4.38

4-60

stumerit@gmail.com

Prob. 4.39

(Neglect Boom Weight)

Prob. 4.40

4-61

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Prob. 4.41

Prob. 4.42

4-62

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Prob. 4.43

Prob 4.44

4-63

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Prob. 4.45

Prob. 4.46

Prob. 4.47

4-64

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Prob. 4.48

Prob. 4.49

Prob. 4.50

MA = 1.0(10) 1.0(15)  2.4(12) + 7.0(15) +5(5) + RB(24) = 0 RBV =  3.18 k  (downward) MB = RA(24) 1.0(10) + 2.4(24) 1.0(15) + 2.4(12) + 7(15) + 5(5) = 0 RA = + 7.98 k  Fx = + 1.0 + 1.0 7 5 + RBH = 0 

RBH = +10 k 

4-65

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

Prob. 4.51





MA = 130(10)  200(17.5) + RBH (34.64) = 0 RBH = +138.6 lb  Fx = + RAH  RBH = 0 RAH = +138.6 lb  Fy = +RAV 130  200 = 0 RAV = + 330 lb 

 

Prob. 4.52

MA = +CDV (6) 16(8) = 0 CDV = + 21.33 k  CDV  33.2 k sin 40 CDV CDH   25.42 k  tan 40 CD 

4-66

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Fx = + RAH  25.42 = 0 RAH = 25.4 k  Fy = + RAV + 21.33  16 = 0 

RAV = 5.33 k 



Prob. 4.53

MA = +RBH (15) 600(8) 3000(16) 300(6) = 0 RBH = + 3640 lb  Fx = +RAH 3640 = 0 RAH = + 3640 lb  Fy = +RAV  400 600 300 3000 = 0 

RAV = + 4300 lb 



Prob. 4.54

MA =12,800(8) + BDV (12)  500(16) = 0 BDV = + 9200 lb

4-67

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BDH = (9/12)BDV = + 12,270 lb  BD = (15/9)BDV = 15,330 lb (as shown) Fx = RAH + BDH = 0 RAH = 12,270 lb  Fy = +RAV  12,800 + 9200  500 = 0 

RAV = 4100 lb 



Prob. 4.55

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



 12  M A   13 T (15)   5     T (25)  300(34)  0  13  T = 435 k

 



Prob. 4.56

4-68

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h = 6 sin77.5 = 5.858 ft F 

62.4(5.858 )2 2

 1071 lb

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 MC = 1071(2.0) + BH y1 + BV x1 = 0 (by substitution) BD = 725 lb BV = BH = 0.707(725) = 513 lb Fx = CH + 1046 513 = 0 CH = +533 lb  Fy = +CV 232 + 513 = 0 

CV = 281 lb 



Prob. 4.57

Fx = 0 T = 2(6.43) = 12.86 k MB = 7.66(3.72)  7.66(10.25) + W2 (10) +12.86(1) = 0 W2 = 9.42 k 4-69

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Fy = +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

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Prob. 5.1 By symmetry: RAV = RCV = 5 k RAH = 0 FV = +5 AB sin45 AB = +7.07 k (comp.) FH = +AC  AB cos45 = 0 AC = 0.707(0.707) = +5.0 k (tens.)

Prob. 5.2

MA = 5(10)10(8)+RCV(24) = 0 RCV = +5.42 k  MC = RAV(24) + 10(16) 5(10) = 0 RAV = +4.58 k  FH = RAH + 5 = 0 RAH = +5 k   5  FV = +4.58AB  =0 6.40    AB = 5.86 k (comp.)  4  FH = 5+AC5.86  =0 6.40    AC = +8.66 k (tens.)  4   8   BC  = 0 6.40 9.434     

FH = +5+5.86 

BC = +10.21 k (comp.)

5-71

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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

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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

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Prob. 5.9

Prob. 5.10 (units are kN)

Prob. 5.11 (units are kips)

5-74

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Prob 5.12

Prob. 5.13

Prob. 5.14

5-75

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Prob. 5.15

Prob. 5.16

MA = 0  RFV = +15.75 k



MF = 0  RAV = +8.25 k

Cut section a-a, use left portion:

5-76

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MH = 8.25(20) + CD(9) = 0 CD = +18.33 k (comp.) MD = 8.25(25) + 10(5) + HI(9) = 0 HI = +17.36 k (tens.) FV = + 8.25 10 + HD(9/10.3) = 0 HD = +2.0 k (tens.)

Prob. 5.17

MA = 0  RGH = +1949 lb  MG = 0  RAH = + 2699 lb  FV = 0  RAV = +1299 lb  Cut section a-a, use right portion:

ME = +BC(2.67) 1299(4) 750(2.67) = 0 BC = + 2697 lb (tens.) MB = 1299(8) + FE(8 sin33.69) = 0 FE = +2342 lb (comp.) FV = 1299 + 2342(sin33.69) + BE sin33.69 = 0 BE = 0

5-77

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Prob. 5.18



ME = 0  RA = 72.19 kN

Cut section a-a, use the left portion. (BC assumed tensile.) MH = 72.19(8) + 45(4) + BC(3) = 0 BC = +132.5 kN (tens.)

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

Cut section c-c, use left portion. (CG assumed tensile.) FV = +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: MA = 0  CI = 0 MB = 4000(8) + IJ(4) = 0

5-78

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IJ = + 8000 lb (tens.)  2  MH = 4000(8) + BC   (4) = 0  5  BC = +8944 lb (comp.)

Prob. 5.20

MA = 0  RC = 24.17 kN MC = 0  RAV = 9.167 kN  FH = 0  RAH = 10 kN  Cut section a-a , use portion to the left. (BC is assumed compressive.) MD = 10(2) + 9.167(3)  BC(2) = 0 BC = 3.75 kN (comp.)

(Assume DE tensile:) MC = +9.167(6) DE(2) = 0 DE = 27.5 kN (tens.)

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

5-79

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Prob. 5.21

Cut section a-a, use the right-hand portion. (BC & BG assumed tensile, FG assumed compressive.) MB = 12(10) + FG(10) = 0 FG = +12 k (comp.) MG = 12(5) + BC(10) = 0 BC = + 6 k (tens.)  2  FV = 12 + BG   = 0  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.) MB = 0  CD = 15 kN (comp.) FV = 0  BD = 0 FH = 0  BE = 20 kN (tens.)

5-80

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

(Assume CB tensile.) FH = 0 CBH = 20 kN CB 2



CBH 1

CB  2CBH CB = 28.3 kN (tens.)

Prob. 5.23

ML = 0  RIV = +26.7 k 



MI = 0  RLV = +38.3 k 

Cut section a-a, use the left portion:

5-81

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MK = 0  CD = +14.5 k (tens.) FV = 0  CK = +13.3 k (comp.)

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

F = 10 + EM V

 4  =0    41 

EM = +16 k (tens) Prob. 5.24

MA = 0  RDV = +1667 lb MD = 0  RAV = + 1333 lb 

5-82

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

Cut section a-a, use the left portion:







MB = 0  EF = +1778 lb (tens)



MF = 1333(16) +1000(8)  4   BC (12)  0  5  BC = +1389 lb (comp.) Fy = +1333 10001389(3/5) + BF (3/5) = 0 BF = +833 lb (comp.)

Prob. 5.25

MA = 0  REH +45.83 k  ME = 0  RAH = +5.83 k  FV = 0  RAV = + 35 k  5-83

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

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



  

F = +35 AF V

 6 

0    61 

AF = +45.6 k (tens.)

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

 6  FV = 20 + BG 0 61   BG = +26 k (tens.) MB = 0  FG = +16.7 k (comp.)

5-84

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Prob. 5.26

MA = 0  RGV = +36 k



MG = 0  RAV = +4 k

Cut section a-a, use left portion:

Resolve force BH into components at B:  1  MD = 4(40) + BH  (40) = 0  5  BH = + 8.94 k (tens.)

5-85

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Cut section b-b, use the left portion:

MG = 0  CD = +24 k (tens.) MC = 0  HG = +8 k (tens.)

Prob. 5.27

MA = 0  RDV = +8.67 k MD = 0  RAV = +7.33 k

5-86

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

Cut section a-a, use left portion:





FV = 0  BE = +1.33 k (tens.) (more)

For member CE, cut section b-b, use left portion:

F = +7.33  6  CE V

 5  0    41 

CE = +1.703 k (comp.)

Cut section c-c around joint F. Fv = CF = 0

5-87

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Prob. 5.28

MA = 0  RJV = +26 k MJ = 0  RAV = +14 k 



FH = 0  RAH = +10 k 

Cut section a-a, use right-hand portion: Resolve CE into components at E

 1  MH = 16(8) +10(8) + CE   0  2  CE = +8.49 k (tens.)  1  Fv = 24 16 8.49   + EH = 0  2  EH = +46 k (comp.)

5-88

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Cut section b-b, use lower portion:

MH = 0  BC = +6 k (tens.)

Cut section c-c, use lower portion:

MI = 0  AB = + 4 k (comp.)

Prob. 5.29

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

5-89

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Resolve force CB into components at B:



ME = 0  CB = +10,182 lb (comp.)  1   1   CD 0 F = 360010182     V  2   5  CD = +8050 lb (tens.)

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

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

Prob. 5.30

5-90

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MA = 0  EV = +833 lb  ME = 0  AV = +1167 lb 

MB = 0  DV = +667 lb   

MD = 0  BV = +1333 lb 

FV = 0  CV = + 166 lb MB = 0  CH = +751 lb  

FH = 0  BH = +751 lb

FH = 0  DH = +751 lb  Pin Reactions (lb) Pin

V

H

React.

A

1167



1167

B

1333

751

1530

5-91

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C

166

751

769

D

667

751

1004

E

833

0

833

Prob. 5.31

MC = 0  BH = 600 N FH = 0 CH = 100 N FV = 0  BV = C V

FH = 0  AH = 600 N  MA = 0  BV = 100 N  FV = 0  AV = 100 N 5-92

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(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

MA = 0  FV = +517lb MF = 0  AV = +27 lb  

H = 0  AH = 80 lb

5-93

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MB = 0  EV = + 440 lb  FV = 0  BV = +80 lb  FH = 0  BH = EH

MC = 0 BH = +160 lb  EH = 160 lb FH = 0  CH = +80 lb 



FV = 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

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Prob. 5.33

MA = 0  DH = +6855 lb  MD = 0  AH = 6855 lb 

MD = 0  BV = +4500 lb  FV = 0  DV = +1500 lb  

FH = 0  +6855 lb 

FV = 0  AV = +7000 lb  Pin Reactions (lb)

5-95

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Pin

V

H

Reaction

A

7000

6855

9800

B

4500

6855

8200

D

1500

6855

7020

Prob. 5.34

MB = 0  AH =+11.6 k   



MA = 0  BH = 11.6 k 

By symmetry: BV = CV =2 k  FH = +11.6 k 

5-96

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

FV = 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

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MC = 0  AB = 2.34 kN  FV = 0  CV = 2.57 kN   

FH = 0  CH = 2.84 kN  Pin reaction at C :   2.572  2.842  3.83kN

Prob. 5.36

5-98

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MC = 0  BH = +3.25 k   

MB = 0  CH = +0.05 k 

MA = 0  BV = +1 k  FV = 0  AV = +1 k  

FH = 0  AH = +3.25 k 





MA = 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

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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:

FH = 0  ED = +50 kN (tens.) MD = 0  CE = +86.7 kN (comp.)

Prob. 5.38

5-100

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MF = 0  AV = +31.88 kN  MA = 0  FV = +28.12 kN  FH = 0  AH = +10 kN

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

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

Prob. 5.39

5-101

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MA = 0  BH = +3150 lb  FH = 0  AH = +3150 lb 

FV = 0  AV = +7100 lb lb 

MC = 0  EV = +8100 lb  FV = 0  CV = +2300 lb  FH = 0  CH = EH

MD = 0  EH = +6300 lb  CH = 6300 lb FV = 0  DV = +8700 lb  FH = 0  DH = 6300 lb 

5-102

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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

MA = 0  DV = +1000 lb  FV = 0  AV = +600 lb 

5-103

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

 

FV = 0  BV = +1000 lb MA = 0  BH = +800 lb   

FH = +800-AH=0  AH = 800 lb

FH = 0  DH = +200 lb

5-104

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

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

MA = 0  RE = 29.0 kN  FV = 0  AV = 29.0 kN FH = 0  AH = 43.5 kN 

5-105

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MC = 0  BD = 32.6 kN  FH = 0  CH = 32.6 kN 



FV = 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

MC = 0  AH = +606 lb FH = 0  CH = +606 lb

5-106

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

MB = 0  AV = +950 lb FV = 0  BV = +250 lb   

FH = 0  BH = +606 lb 

 

FV = 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

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Prob. 5.43

By symmetry:DV = EV = 140 lb  FH = DH = 0

By symmetry: AV = BV = 140 lb FH = 0  BH = AH

5-108

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FV = 0  CV = 0 MB = 0  CH = +280 lb  FH = 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

MA = 0  EV = +4.1 kN ME = 0  AV = +1.1 kN  FH = 0  AH = +2 kN 

5-109

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

MB = 0  DV = +2.40 kN MD = 0 BV = + 0.6 kN FH = 0  DH = BH

MC = 0  BH = +1.25 kN and  DH = 1.25 kN FV = 0  CV = +1.7 kN FH = 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

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C

1.7

1.25

2.11

D

2.4

1.25

2.71

E

4.1



4.1

Prob. 5.45

MA = 0 C = 40 lb FV = 0  RA = 52 lb Prob. 5.46

+

 = cos-1 1/5 = 78.46  = cos-11/2 = 60 

FV = P +Q cos78.46 + R cos60 = 0  10 + 0.20Q + 0.50R = 0

(I)

FH = +R sin 60 Q sin 78.46 = 0

5-111

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+0.866R  0.980Q = 0

(II)

Solve eqns. I and II simultaneously: Q = 13.05 lb FH = 0  F +Q sin 78.46 = 0  F = 12.80 lb Prob. 5.47

FH = 0  QH = 100 lb Q=

100 cos10

=101.5 lb FV = 0 P + 2(101.5 sin10) = 0  P = 35.3 lb

5-112

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Prob. 5.48

CC moment:+,  +

(a) MC = 30(15)+AV(3) = 0  AV = 150 lb

FV = 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

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





MC = +BAy (3) 30(15) = 0  BAy = 150 lb BA 

BAy cos31.52

  176 lb

BAV = BA cos13.63 BAV = 171 lb NOTES

5-114

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Prob. 6.1

Fy = 0  N = 130 lb Fx = 0  F = 34.6 lb (when P = 40 lb) Max. available resistance: F = sN = 0.30(130) = 39 lb 34.6 < 39  block will not move.

Prob. 6.2

(a)  = 0 Fy = 0  N = 200 lb Max. available resistance: F = sN = 0.50(200) = 100 lb Fx = 0  P = 100 lb (b)  = 20 Fy = +N  200 + P sin 20 = 0 N = 200  P sin 20 Fx = +P cos 20  F = 0 F = P cos 20 F = sN 6-115

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P cos 20 = 0.50(200  P sin 20)  P = 90 lb Prob. 6.3 W = mg = 140(9.81) = 1373 N s = 0.38

Fy = 0  N = 1373 N F = sN = 0.38(1373) = 522 N Fx = 0  P = 522 N Prob. 6.4 (a) Motion up the plane:

Fy = 0  N = 80 lb Max. available resistance: F = sN = 0.40(80) = 32 lb Fx = +P 32  100(3/5) = 0 P = 92 lb

6-116

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(b) Motion down the plane:

Fy = +N 100(4/5) = 0 N = 80 lb F = sN 0.40(80) = 32 lb Fx = 0  P = 28 lb Prob. 6.5 From Prob 6-4, the body will be at rest for 28#≤P≤98# (a) When P=40# (assume motion is down the plane) ΣFx = 40  F+100(3/5)=0  F=20# (b) When P=60# ΣFx=60 – F – 100(3/5)=0  F=0 (c) When P=70# (Assume motion is up the plane) ΣFx=70 – F – 100(3/5)=0  F=10# Prob. 6.6

On the verge of sliding FA=0.5(98.1)=49.1 N FB=0.15(98.1+196.2)=44.1 N

Since FB<FA, Motion will impend when P reaches 44.1 N. The blocks will slide as a unit.

6-117

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Prob. 6.7

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

ΣFy=N – 175 Cos 25°-P Sin 25°=0 N=0.4226P+158.6 ΣFx = FP cos 25°+175 sin 25°=0 F=74.0 – 0.9063P F=µsN 74.0 – 0.9063P=0.65(0.4226P+158.6) P = –24.6# Negative indicates that P acts to the lect. Therefore no force P acting to the right is needed to prevent the block from sliding down the plane.

6-118

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Prob. 6.9

Prob. 6.10

Fy = 0  N = 45 lb Fx = 0  F = 18 lb F = sN s = F/N = 18/45 = 0.40

6-119

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Prob. 6.11

s = 0.33

Fy = 0  N = 90 lb F = sN = 0.33(90) = 30 lb Fx = 0  P = 30 lb Prob 6.12

6-120

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Prob.

6.121

6-121

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Prob.

6.122

6-122

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Prob.

6.123

6-123

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Prob.

6.124

Prob. 6.17

6-124

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Prob. 6.18

Prob. 6.19

6-125

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Prob.

6.126

6-126

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Prob.

6.127

6-127

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Prob.

6.128

6-128

stumerit@gmail.com

Prob.

6.129

6-129

stumerit@gmail.com

Prob.

6.130

6-130

stumerit@gmail.com

Prob.

6.131

6-131

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Prob.

6.132

6-132

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Prob.

6.133

6-133

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Prob.

6.134

6-134

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Prob. 6.29

6-135

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Prob. 6.30

6-136

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Prob. 6.31

6-137

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6-138

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Prob. 6.32

6-139

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6-140

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Prob.

6.141

6-141

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Prob.

6.142

6-142

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Prob.

6.143

6-143

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Prob. 6.36 (see prob. 3.32)

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

6-144

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Prob. 6.37

6-145

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6-146

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Prob.

6.147

6-147

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Prob.

6.148

Prob. 6.40

Prob 6.41

6-148

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Prob. 6.42

Prob. 6.43

Prob. 6.44

6-149

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Prob. 6.45

Prob. 6.46

Prob. 6.47

6-150

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Prob. 6.48

Prob. 6.49

6-151

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Prob. 6.50

Prob. 6.51

6-152

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Prob. 6.52

6-153

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Prob. 6.53

 +

r = 11 mm, s = 0.14, p = 6 mm.

Qa = Pr Qa 225(250)   5114 N r 11 6  1 1 p       tan 2 r  tan 2 (11)   4.96     P



Fx = +5114  F cos  N sin = 0 (F = 0.14N)  N = 22.6 kN Fy =  C + N cos F sin = 0 (N = 22.6 kN; F= 0.14 N)  C = 22.2 kN Prob. 6.54 r = 1.50 in.., s = 0.15, p = 0.25 in., a = 20 in.

 p  1  0.25   tan  2 r   tan  2 (1.50)   1.519     1



6-154

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Fy = 4000 + N cos F sin = 0 4000 + N(0.9996) 0.15N (0.02651) = 0  N = 4018 lb

(more)

Fx = +P  N sin  Fcos = 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

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Prob. 6.56

Fx = + 2541  N sin  Fcos = 0 +2541 N(0.0746)  0.13N (0.9972) = 0 N = 12,440 lb Fy = W + N cos F sin  = 0  W +12,440(0.9972)  0.13(12,440)(0.0746) = 0 W = 12,280 lb (b) s = 0.18 (same sketch as part (a)) Fx = +2541  N sin  F cos = 0 +2541  N (0.0746)  0.18N (0.9972) = 0 N = 9996 lb Fy = W + N cos  F sin = 0  W + 9996(0.9972)  0.18(9996)(0.0746) = 0 W = 9834 lb 6-156

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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

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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

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Wt. of bar: W1 



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

Wt. of sphere: 4  (2.5)2 3 (490 )  18.56 lb W2  1728 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

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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 2

0.9817(𝑙 + 1.5) + 0.0182 (𝑙 ) 2 𝑥= 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

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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  4.71m

x

5(589)(228)  43.5(228)  3(589)(683) 8(589)  43.5  3(589)  353 mm

y

Prob. 7.7

W1 (steel) 

63

(490)  61.25 lb 1728 0.7854(2)2 (5)  W2 (steel)  (490)  4.45 lb 1728 0.7854(2)2 (5) W3 (lead)  (710)  6.45 lb 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

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Prob. 7.8 y

(Refer to Prob. 7.18)

Wy 61.25(3)  4.45(3.5)   2.96 in. W 61.25  4.45

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



0.7854(3)2 (5) (490)  10.02 lb 1728 0.7854(1)2 (36) (80)  1.31lb W2  1728 W1 

W = 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

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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

45.6 192.4 y  4.22 in.

45.6

192.4

7-163

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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  (15)2 = 353 mm2 y

1500(25)  353(56.37)  31.0 mm 1500  353

7-164

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(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)

:

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

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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



19.36

44.6

54.3

𝑥=

54.3 = 2.80 in.

19.36

𝑦=

44.6 = 2.30in.

19.36

(c)

7-166

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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



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



234

2210

2000

𝑥=

2000 = 8.55 in.

234

𝑦=

2210 = 9.44 in.

234

7-167

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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



20.7  241.8 241.8 y   11.68 in. A 20.7 Ay

Prob.7.16 (a)

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

4(2) y  0.849 in.

2 3

7-168

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x y

60(8)  6.28(10)  7.77 in.

53.72 60(3.33)  6.28(0.849) 53.72

 3.62 in.

(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

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W = 289 kN 𝑥=

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

Prob. 7.18

y

Ay 9100(261.5)  42100(532.85)  A 9100  42100  485 mm

Prob. 7.19

a1 = 18.3 in.2, a2 = 7.34 in.2, a = 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

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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

 0.290 0.0932 y   0.321m a 0.290

0.0932

ay

Prob. 7.21

7-171

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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

658 6744 Ax 12,330 x   18.74 in. A 658 Ay 6744 y   10.25 in. A 658

12330



7-172

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(b)

2.96(2.50)  2(1.19)(0.711) 2.96  2(1.19)  1.703in.

y

Prob. 7.23

a1 = 12h, a2 = ½ (10)(12) = 60 in.2 y1 = 10 + h/2 y2 = 6.67 in. ay a 12h(10  h / 2)  60(6.67) 10  60  12h y

Solve for h: h = 5.77 in.

7-173

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Prob. 7.24 (a)

A

y

x

Ay

Ax

(in.2)

(in.)

(in.)

(in.3)

(in.3)

1

20

7

1

140

20

2

16

1

4

16

64



36 84

 2.33 in.; y 

156

x

156

36

84

 4.33 in.

36

(b)

A

y

x

Ay

Ax

(in.2)

(in.)

(in.)

(in.3)

(in.3)

1

98

3.5

7

343

686

2

30

2.5

17

75

510



128

418

1196

7-174

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x

196

 9.34 in.;

y

128

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



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

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(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



33.21 185.2 x  5.58 in.; 33.21

390.9 185.2 390.9 y  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



16.87

49.9

63.1

x

63.1  3.74 in.; 16.87

y

49.9  2.96 in.

16.87

7-176

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Prob.

8.177 (a)

bh3 12(24)3   13824 in.4 12 12 4 d  (8)4   201in.4 I x 2  64 64 I x  13,824  201  13623in.4 I x1 

(b)

I x 

 64

(d 4  d 4 )  527in.4 1

(c)

I x 

100 4 40(70)3   7.19 10 6 mm 4 12 12

8-177

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

Prob.

8.178 8(12)3  384 in.4 I x  36 8(12)3 I xb  12  1152 in.4

Prob. 8.3 I x1  1170 in.4 1(15)3 I x 2  (2) 12  563in.4 Ix = 1170 + 563 = 1733 in.4 Prob. 8.4



1 3 1 I x  12 (390)(190)  2 12 (95)(110) 3   1 3 6 4  2 (40)(110)  193.0 10 mm  12 

8-178

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Prob. 8.5 (a) a = 18 in.2 Ixb = ay2 Ixb = 18(10.52 + 7.52 + 4.52 + 1.52) = 3402 in.4 (b) a = 12 in.2 Ixb = ay2 Ixb = 12(112 + 92 + 72 + 52 + 12 = 3432 in.4 Prob. 8.6 6(12)3 4 I xb   3456 in.

3 % Error: 3402  3456 8 - 5(a): 100%  1.56% 3456 3432 - 3456 8 - 5(b): 100%  0.69% 3456 Prob. 8.7

I x 

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

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Prob. 8.8

W18  71: tw = 0.495 in.

Iy = 60.3 in.4 IY  60.3  (2)  79.6 in.

15(1)3 12

 2(15)(0.748)

2

4

Prob. 8.9 (a)

a1 = 10(8) = 80 in.2 a2 = 0.7854(8)2(0.5) = 25.1 in.2 (each) I x1  I x2 

10(8)3  427 in.4 12  (8)4

 

4

0.5  100.5 in 64 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

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(b)

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

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 4 I y2   0.67 in.

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

ay 20(7)  28(1)   3.5 in. a 20  28

8-181

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2(10)3  20(3.5)2  412 in.4 12 14(2)3 2 4 I x2   28(2.5)  184.3 in.

12 I x1 

Ix = 412 + 184.3 = 596 in.4 10(2)3  6.67 in.4 I y1  12 2(14)3  457 in.4 I y 2  12 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 I x1   100(0.83)2  902 in.4 12 10(10)3 I x2   50(0.83)2  313in.4 36 Ix = 902 + 2(313) = 1528 in.4 10(10)3  833in.4 12 10(10)3 2 4  50(8.33)  3747 in.

I y 2  36 I y1 

Iy = 833 + 2(3747) = 8327 in.4

8-182

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Prob. 8.10 (a)

For the semi-circular cutout: 4(45 / 2) y  40   30.45 mm c 3 Ac  0.7854(45)2 / 2  795 mm2 I ox  0.1098R 2  28.1103 mm4 I oy   R 4 / 8  100.6 103 mm4 804  2(28.1103 )  2(795)(30.45)2 12  1.883106 mm4 804 3 6 4 IY   2(100.6 10 )  3.2110 mm 12 I X 

(b)

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

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300(600)(300)  7854(420)  294.5 mm 300(600)  7854 300(600)3  300(600)(5.5)2  4.91106 I x  12  7854125.5)2  5.28109 mm4 y

600(300)3  4.91106 12  1.345109 mm4

I y 

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 2 4 I x2  16(2.12)  93 in.

12 I x1 

Ix = 3460  93 = 3370 in.4 X'-X' axis: 24(12)3 4 I x'1   13,820 in.

3 4(4)3 2 4 I x'2   16(4)  277 in.

12 Ix' = 13,820  277 = 13,540 in.4

8-184

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Prob. 8.12

a1 = 27 in.2,

y1 = 15.5 in.

a2 = 11.8 in.2 , 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 x2  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

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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  531in.4 12 2(12)3 2 4 Ix 2   24(2.33)  418 in.

12 Ix1 

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

8-186

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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 12 2 19(150)(46.34)  25.3106 mm4

I x  13.410 6  3060(43.12)2 

8-187

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I y  1.84 10 6 

19(150 )3  7.18 10 6 mm 4 12

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. a = 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 2 4 I x1   6(1.44)  44.4 in.

12 5.25(0.75)3  3.94(2.185)2  19.0 in.4 I x2  12 Ix = 44.4 + 19.0 = 63.4 in.4 X-Y axis: 8(0.75)3  6(1.185)2  8.71in.4 12 0.75(5.25)3  3.94(1.185)2  22.0 in.4 I y 2  12 I y1 

Iy = 8.71 + 22.0 = 30.7 in.4

8-188

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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

a = 84+ 120 + 2(8) = 220 in.2 ay = 1806 + 1200 + 2(149.4) = 3305 in.3 3305 y  15.02 in.

220 28(3)3  84(6.48)2  3590 in.4 12 6(20)3 2 4 I x2  120(5.02)  7024 in.

12 4(4) 3 36 2 4 I x3   8(3.65)  113.7 in.

I x1 

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

8-189

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a1 = 24(0.375) = 9 in.2 , a2 = 3.61 in.2 0.375(24)3  432 in.4 12 I x2  [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

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Prob. 8.21

W8  40: A = 11.7 in.2, tw = 0.360 in., bf = 8.07 in., Ix = 146 in.4, Iy = 49.1 in.4 11.7(4.04)  3.00(7.22)  2.50(10.47) 11.7  3.00  2.50  5.53 in.

y

Ix = 49.1 + 11.7(1.49)2 +

0.5(6)3 12

= 153.7 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

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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



10.59 45.69 y  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

 43.87

5.071

Ix = 43.85 + 5.077 = 48.9 in.4 X-Y axis: I y 

0.75(6)3 4(0.75)3   15.2  28.8in.4 12 12

(b)

8-192

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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

 111.76

13.62

Ix = 111.76 + 13.62 = 125.4 in.4 X-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

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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

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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

a = 0.25 + 2(0.50 + 0.125) = 1.500 in.2 ay = 0.469 +2(0.50 + 0.01563) =1.500 in.3 1.500 y  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 = Ixo + adx2 = 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 = Iyo + ady2 = 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

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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 4  288 in.

I y1   12 12(2)3 I y 2   24(5)2  608 in.4 12 



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

ry 

1504  4.57 in 4 72

Prob. 8.29 (a)

8-196

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X-X axis a

d

104

ad2

Iox

106

106

mm2

(mm)

(mm4)

(mm4)

1

2.40

115

317

12.80

2

1.20

0

0

22.5

Ix = (Io + Ad2) = 2[(12.80 + 22.5)106 + 317106] = 705106 mm4

X-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106  88.7 mm 2(2.40 1.20) 104

(b)

8-197

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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106  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) 4 π d 4 π (3)4 J   7.95 in.

32 32

Prob. 8.32 (Hollow circular shaft) π 4 4 4 π J  (d 4  d 14 )  (3  2.5 )  4.12in.

32 32

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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

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Prob. 8.37

a1 = 6(5.196)(2) = 62.35 in.2 a2 = ½ (5.196)(2)(3) = 15.59 in.2 a = 62.35 + 2(15.59) = 93.53 in.2 X-X axis: 6(5.196)(2)3  561.1in.4 12  3(5.196)(2)3  4   140.3 in.

I x 2  4 12   Ix = 561.1 + 140.3 = 701 in.4 I x1 

X-Y axis: 5.196(2)(6)3 4  187.1in.

I y1  12 5.196(3)3 15.59 2  4  (4)   514.5 in.

I y 2  4 2  36  Iy = 187.1 + 514.5 = 702 in.4 J = Ix + Iy = 701 + 702 = 1403 in.4 Prob. 8.38

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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 = (Io + Ad2) Ix = Iy = 6.82 + 4(0.692) + 4(1.19)(1.789)2 = 24.82 in.4 rx  ry 

24.82 2.96  4(1.19)

 1.793in.

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

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Prob. 9.2  = P/A A = πd2/4 = 0.7854d2 = 28.27 in.2  = 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 AB = 75/4 = 18.75 ksi BC = 75/2.41 = 31.1 ksi Prob. 9.4 P = 70 kN = 70 ×103 N (a) A = 25(50) = 1250 mm2 σ  t

 70 10  56 MPa A 1250

P

3

(b) A = 0.7854(150)2 = 17.68×103 m2 σt 

70  103

6

17.6810

3

 3.9610 Pa  3.96 MPa

(c) A = 0.7854(25)2 = 490.9 mm2 70 103  142.6 MPa σt  490.9

Prob. 9.5 16, 000  4000 psi (a) σt  4 16, 000  4178 psi (b) σt  3.83

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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 22 σ   5.5 ksi c 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  35, 000  1.75 in.2 σt (all) 20, 000 A  1.49 in.2 0.7854

Use a 1½ diam. rod Prob. 9.9

P = 60 k

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

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Prob. 9.10 Req'd A 

P

 p (all)

Req'd side 



60

 24 ft2

2.5

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 σ  t

P

3

 2510  88.2 MPa A 283.5

Prob. 9.12 A = 25(100) = 2500 mm2 t(all) = 140 MPa = 140 N/mm2 P(all) = A t(all) = 2500(140) = 350,000 N = 350 kN Prob. 9.13 w = wt. of wire per foot w 

0.7854(0.1875) 2 (12)

1728  0.09396lb/ft

(490)

Total wire wt. = 0.09396(400) = 37.58 lb P 37.58  1362 psi (a) t   A (0.7854)(0.1875)2 (b) P=At = (0.7854)(0.1875)2(24,000 – 136.2) = 625 lb

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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 Req'd A  

P

 t (all)



73.5103  525 mm2 140

525  25.9 mm 0.7854

Req' d d 

Prob. 9.16 P = 20 k, L = 30 ft, δ(limit) = 0.25 in. t(all) = 24 ksi For tension: Req’d A = P/ t(all) = 20/24 = 0.833 in.2 For elongation: Req' d A 

PL

 (limit) E



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

 0.96 in. 2  controls

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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  804mm2 300 103 τ(ult)   373 MPa 804 τ(ult) 

Prob. 9.18 AB = shear area per bolt AB = 0.7854(0.875)2 = 0.6013 in.2 P   (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  A(ult) 1.571(42,000)  65,980 lb

Prob. 9.20 (a) A = 0.7854(0.75)2 = 0.4418 in.2 P σ   6000  13, 580 psi t A 0.4418

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(b) A = πdt = π(0.75)(0.375) = 0.8836 in.2 P 6000     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

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Prob. 9.24 A = π d t = π (0.75)(0.5) = 1.178 in.2 P 60,000    50,930 psi A 1.178 Prob. 9.25 ΣMo : 0.5P = 40(12) P= 960 lb 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 Req'd A  4000  0.36 in.2 Width of key b: A = b×length Req’d b = 0.36/1.5 = 0.24 in.

Use ¼” width Prob. 9.27

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PV = PH = 0.707(20) = 14.14 kN (a) Comp. stress in diagonal: P 20 10 3    0.889 MPa c A 150 2





(b) Bearing stress on plane a-b: 14.14103    2.36 MPa 40(150) (c) Shear stress on plane a-c : 14.14103    0.589 MPa 160(150)

Prob. 9.28 Refer to Appendix G. Req’d shear area = 150 mm2 3 2 P 14.1410   27,200 mm Req'd area  τ(all) 0.52 Req’d x = (27 200)/150 = 181.3 mm Prob. 9.30  1.2 (a) ε    0.0010 in./in.

L 100(12) δ (b) L   0.5(12)  20, 000 in.  1667 ft ε 0.00030 (c) δ  L  0.00042(1000)  0.42in.

Prob. 9.31 δ 8  7.85    0.01875 in./in.

L 8 Prob. 9.32 Total elongation = 0.80 in.

2 2(0.80) Max. ε    0.001333 in./in.

L 100(12)

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Prob. 9.33 PL σL 15, 000(2.5) δ    0.0125 in.

AE E 30, 000, 000

Prob. 9.34

E = 30,000 ksi

(a) A = 0.7854d2 = 0.442 in.2 P 15, 000 σ   33, 900 psi A 0.442 s 33, 900  0.00113 (b) ε   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 P 65 10 3     132 .4 MPa A 491 s (b)    132.4  1.89103 E 70103 (c)   L  1.89103 (4.0103 )  7.56 mm

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Prob. 9.36

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

1 AE 0.442(30,000,000) Segment 2: A = 0.7854(0.50)2 = 0.1963 in.2 PL  5000(5)(12)   0.0509 in.  2  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 P 100     44.4 ksi t A 2.25 44.4 < 125 O.K. PL sL 44.4(192) E    17,050ksi A  (0.500)

Prob. 9.38 P = 67 kN = 67×103 N



A = 0.7854(25)2 = 491 mm2 P 67 10 3      136 .5 MPa (a) t 491 A 136.5 MPa < 207 MPa O.K.

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 136.5 (b)      1.95103 3 E 7010 (c)   L  (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 (a)    t A 56.3 710 psi < 6000 psi O.K.  710  0.000418 (b)     E 1,700 ,000 (c)   L  (0.000418)(8.5)(12)  0.0426in.

Prob. 9.40 PL 15(100  (a)    0.01ft AE 0.005(30,000,000) P 15 (b) t    3000 psi A 0.005

Prob. 9.41 A = cross-sectional area = 0.7854(1.25)2 = 1.227 in.2 P 12,000  9780 psi (a)    t A 1.227  9780  0.000326 (b)     E 30,000 ,000 (c)   L  (0.000326)(18)  0.00587 in.

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Prob. 9.42



P  A  P  AE   P  0.7854(26)2 (207103 )(0.0007) E

 76.9 103 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

AE 45(19.64)(207 000)  L 18103  10.16103 N

P





P

t

A



10.16 10 3 19.64

  517 MPa

Prob. 9.44 A = 0.7854(1.5)2 = 1.767 in.2 PL (a)    30,000(20)(12)  0.1358in.

AE 1.767(30,000,000) (b) Req' d A 

PL

 30,000(20)(12) E 0.10(30,000,000)

 2.40 in. 2 Req' d d 

A  1.748 in.

0.7854

Use a 1 ¾” diam. Rod

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Prob. 9.45

(a) Req'd A 

PL

 50,000(24)(12) E 0.20(30,000,000)

 2.40 in.2 Req'd A  3.20 in.

Req'd w  3 4 P 50,000  20,800 psi (b)    t A 2.40 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



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

AE 0.442(30,000,000)

Due to applied load: PL   8000(100)(12)  0.724 in.

AE 0.442(30,000,000) Total δ = δ1 + δ2 = 0.731 in.

Prob. 9.47

ΣMB = 0  AV = 80 k

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Cut section 1-1 and use left portion:

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



25(5)(12) PL   0.02 in.

AE 2.5(30,000 )

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

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



PL AE

Section

δ (in.)

Load

L

(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:  = 6000/0.25 = 24,000 psi < 34,000 psi OK 9-215

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

Segmnt Load (lb)

δ (in.)

L (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:  = 8000/0.25 = 32,000 psi < 34,000 psi OK

Prob. 9.50

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AC BC 45   sin 51.34 sin101.31 sin 27.35  BC = 96.0 kN (tens.)  AC = 76.5 kN (comp.)

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

 



PL 96 103 (6400) 13.08 mm   AE 227(207 000)

For member AC:



PL

 76 500(5100)  1.396 mm AE 1350(207000)

Prob. 9.51

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



PL AE



1414(10)(12)  0.1152in.

0.0491(30,000,000)

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

0.1152 sin 45

= 0.163 in.

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Prob. 9.52

For the tie-back: L = 32  22  3.61m A = 0.7854(33)2 = 855 mm2 Wt. = 6000(9.81) = 58.9 kN max P = t(ult) A = 170(855) = 145.4 kN PL AE   max P  AE L 2.5(855)(207 000)  max P   122.6 kN  3610 (controls) PV = 2 (122.6)/3.61 = 67.9 kN ΣMA : 67.9(3) = 58.9L Max L = 3.46 m Prob. 9.53

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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  = 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  = 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

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= 1.72 (125 700) = 216 kN

For the piston rod: Req'd A  P  216000 N  3130 mm2 P (all) 69.0 MPa Req'd d 

Req'd A  63.1 mm 0.7854

NOTES

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Prob. 10.1 A = 0.7854(0.5625)2 = 0.2485 in.2 P 6500  26,160 psi (a)  t   A 0.2485  0.00715 (b)      0.000894 L 8 (c) 26,160 psi < 34,000 psi OK:  E    26,160  29,300,000 psi  0.000894 Prob. 10.2 A = 0.7854(150)2 = 17.67×103 mm2 3 P    89 10 5.04 MPa c A 17.67 10 3  0.074     0.000247 L 300 s E   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  29,200,000 psi  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.

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

 (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.)

 (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

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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: 2.42−2.0 (100) = 21.0% 2.0

% Reduction in area: Af = 0.7854(0.422)2 = 0.1399 in.2 Ao = 0.7854(0.50)2 = 0.1964 in.2 0.1964 − 0.1399 (100) = 28.8% %Reduction = 0.1964 Prob. 10.6 δ = 14 mm = 14.0 × 10−3 m P = 320 N, L = 4 m 𝐴 = 0.7854(1.0 × 10−3)2 = 785.4 × 109 m2 𝑃𝐿 320(4.0) 𝐸= = 𝐴𝐸 14.0 × 10−3(785.4 × 10−9) = 116.4 × 109 Pa = 116.4 × 103 MPa Prob. 10.7 A = 0.7854(6)2 = 28.27 in.2 (20,000)(12) 𝑃𝐿 = 𝐸= = 2,930,000 psi 𝐴δ (28.27)(0.0029)

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Prob. 10.8 A = 1/2(2) = 1 in.2, P = 16,000 lb 𝑃 16,000 σ= = = 16,000 psi < 21,000 psi ∴ OK 𝐴 1 Prob. 10.9

𝐸=

∆σ

=

134 − 35

∆ϵ 0.00063 − 0.00017 Expected Stress:

= 215 ×× 103 MPa

σ = 𝐸ϵ = 215 × 103(0.00025) = 53.8 MPa 53.8 MPa < 200 MPa O.K.  = 60.7 – 50 = 10.7 mm 10.7 (100) = 21.4% % Elongation = 50 % Reduction in Area = Prob. 10.10

𝐴𝑓 = (0.7854)(10.7)2 = 89.9 mm2 𝐴0 = (0.7854)(12.5)2 = 122.7 mm2 122.7 − 89.9 (100) = 26.7% % Reduction = 122.7 Prob. 10.11 𝑃(ult) = 13,100 lb, δ = 0.52 in. 𝑑 = 0.50 in. , 𝑔 = 2 in. 𝐴 = 0.7854(0.50)2 = 0.1964 in2

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𝑃(ult) 𝐴

=

13,100

= 66,700 psi 0.1964 0.52 (100) = 26.0% b) %Elongation = 2 a) σ(ult) =

Prob. 10.12 d = 0.505 in.,  = 0.002 in.

A = 0.7854(0.505)2 = 0.2003 in.2 δ𝐴𝐸 0.02(0.2003)(30,000) = 𝑃= = 6.01 k 𝐿 2 Check Stress: 6.01 σ= = 30 ksi < 34 ksi ∴ O. K. 0.2003 Prob. 10.13

t(all) = 60 ksi, δlimit = 0.05 in., P.L. = 120 ksi Based on stress: t = P/A, A = 0.7854d2 Req' d d 2 

24  0.509 in.2 60(0.7854)

Based on δ: PL AE Req'd d 2 

δ

PL

0.7854Elimit 24(18)   0.667 in.2 0.7854(16, 500)(0.05) (controls) Req’d d =

0.667  0.817 in.

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Prob. 10.14 P = load to each rod = 75(250) = 18,750 lb P 18,750 req′d 𝐴 = = = 0.852 in.2 σt(all) 22,000 req′d 𝑑 = √

0.852 0.7854

= 1.04 in.

Prob. 10.15 Stress: 𝑃

req′d 𝐴 =

σ𝑡(all)

=

55 × 103

= 392.9 × 10−6 m2

140 × 106

Elongation req′d 𝐴 =

𝑃𝐿 δ𝐸

=

55 × 103(0.45) = 598 × 10−6 m2 0.2x10−3(207 × 109)

Elongatoin Controls: 598 × 10−6 req′d 𝑑 = √ = 27.6 × 10−3 m = 27.6 mm 0.7854 Check Stress: 𝑃

55 × 103 = 92.0 × 106 Pa 𝐴 589 × 10−6 92.0 MPa < 175 MPa (O.K.) σ=

=

Prob. 10.16 (a) 𝑃(all) 𝑃= σ150,000 𝑡(all)A = 22,000(8.62) = 189,600 lb (b) σ = = = 17,400 psi 𝑡

𝐴

Prob. 10.17

8.62

Each Plate: 225 mm  32 mm

Area per plate: A = 225(32) = 7200 mm2

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P(all) per plate = t(all)A = 165(7200) = 1188 kN # of plates, N = 6800/1188 = 5.72 plates (use 6) Comp’d Stress:  = P/A = 6800/(6(7200)) = 157.4 MPa Prob. 10.18

∑𝑀o = 𝑅(0.5) + 35(8.5) = 0 R = 595 lb Req′d 𝐴 =

𝑃 τ(all)

Req′d 𝑑 = √

=

595 12,000

0.0496

= 0.0496 in.2

= 0.251 in.

0.7854 Use 9/32 in. diameter pin.

P = 15,000 lb, p(all) = 300 psi 𝑃 = 15,000 = 50 in.2 Req′d 𝐴 = 300 σ𝑝(all) Prob. 10.19

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 = 124 MPa σ𝑡(all) = 2.5 67.5 × 103 ′ Req d 𝐴 = = 544 mm2 124

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𝑅𝑒𝑞′𝑑 𝑑 = √

544 = 26.3 mm

0.7854 (b) Aluminum

Fy = 240 MPa

t(all) = 240/2.5 = 96.0 MPa 67.5 × 103 Req′d 𝐴 = = 703 mm2 96.0 Req′d 𝑑 = √

703 0.7854

= 29.9 mm

Prob. 10.21 Gray C.I. ult = 80 ksi σ𝑐(all) = 80⁄6 = 13.33 psi 100 Req′d 𝐴 = = 7.50 in.2 13.33 Req′d 𝑑 = √

7.50 0.7854

= 3.09 in.

Prob. 10.22 A = 0.7854(0.16)2 = 0.02011 in.2 Failure load = σ𝑡(ult)𝐴 = 1470(0.02011)(230,000) = 6,799,000 lb 6,799,000 F. S. = = 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

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Prob. 10.24

TMax(allowable) = 252/4 = 63 lb ∑ 𝐹v = +48 − 2(𝑇 cos θ/2) θ 48 48 cos = = = 0.38095 2 2T 2(63) 

Max. /2 = 67.6  max = 135.2



Prob. 10.25 P = 60,000 lb, p(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(𝐷2 − 𝐷2) 𝑐

𝑖

Req′d 𝐷 = √ 200 − (4.0625)2 = 16.47 in. 𝑐 0.7854

Prob 10.26 P = 200,000 lb, ult = 95,000 psi a) F.S. = 5.0: 200,000 (5.0) = 10.53 in.2 Req′d 𝐴 = 95,000 b) max = 0.10 in.

Req′d 𝐴 =

𝑃𝐿 200,000(10)(12) = = 8.00 in.2 𝐴𝐸 0.10(30,000,000)

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Prob. 10.27

A = 1.2 in.2, t(all) = 22,000 psi, Fy = 36,000 psi, F.S. = 1.85 Elastic: 𝑃𝐿 σ𝐿 σ1𝐿1 σ2𝐿2 δ= = =  𝐸 𝐴𝐸 𝐸 𝐸 6 σ2𝐿2 σ1 = = ( ) σ2 ∴ σ2 = σ𝑡(all) 8 𝐿1 σ2 = 22,000 psi σ1 = (6⁄8)(22,000) = 16,500 psi 𝑃 = 2σ1𝐴 + σ2𝐴 = 2(16,500)(1.2) + 22,000(1.2) = 66,000 lb (a) Ultimate Strength: (All rods stressed to Fy.) 1.85P = 3FyA 3(36,000)(1.2) ∴𝑃= = 70,100 lb 1.85 Prob. 10.28 𝜎𝐿 a) δ = → δ = δ ; 𝐸 is constant. 𝐸

𝐴

𝐵

ALA = BLB  B = LAA/LB = (4/3)A B reaches allowable (185 MPa) first B = 185 MPa = 4/3 AA = 138.8 MPa P=AAA + BAB = 138.8(580) + 185(700) = 210 kN

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b) Both bars will be stressed to 275 MPa: 275(700) + 275(580) 𝑃= = 195.5 kN 1.80 Prob. 10.29 a) Load to each rod: ∑𝑀0 = −20,000(3) + 𝑃𝐶𝐷(4) = 0 ∴ 𝑃𝐶𝐷 = 15 k ∑𝐹V = 0 ∴ 𝑃𝐴𝐵 = 5 k Elongation of each Rod: 5000(2)(12) δ𝐴𝐵 = = 0.0040 in.

1(30,000,000) 15,000(3)(12) δ𝐶𝐷 = = 0.0090 in.

2(30,000,000) b) Ratio of areas for AB = CD: 𝑃𝐴𝐵𝐿𝐴𝐵 𝑃𝐶𝐷𝐿𝐶𝐷 = 𝐴𝐴𝐵𝐸 𝐴𝐶𝐷𝐸 𝐴𝐴𝐵 𝑃𝐴𝐵𝐿𝐴𝐵 5000(2) ∴ = = = 0.222 𝐴𝐶𝐷 𝑃𝐶𝐷 𝐿𝐶𝐷 15,000(3) or:

𝐴𝐶𝐷 = 4.5 𝐴𝐴𝐵

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Prob. 11.1 

 0.48  0.0040 L 10(12)  0.0024 ε  0.0012 (b) transverse  L  2 εtransverse = μ εaxial

(a) ε axial 

μ = 0.0012/0.0040 = 0.300 Prob. 11.2 P 300, 000  A 12  25, 000 psi  34,000 psi (O.K.) σ   t  25, 000  0.000833 axial E 30, 000, 000 σt 



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 E 110 GPa μ 1  1  0.250 (2)44 GPa 2G

Prob. 11.4 30, 000, 000 (a) G   11, 280, 000 psi 2(1 0.33) 16, 000, 000 (b) μ  1  0.379 2(5,800, 000)

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Prob.

11.233

E = 207 × 103 MPa; μ = 0.25 Stresses

σ  y

P

3

 60010  53.3 MPa (Comp.)

A 450(25)

= 53.3 MPa < 230 MPa (O.K.) 6 265103 σx   106.010 MPa (Tens) 100(25) = 106 MPa < 230 MPa

(O.K.)

Strains: εtransverse = μεaxial 1 ε  (μσ x  μσ ) y z E 0.25(106)  0.25(53.3)   63.6106 (decrease) 3 207 10 1 ε  (σ y  μσ ) x y E 53.3  0.25(106)   386106 (decrease) 207 103 1 ε  (σ x  μσ ) y x E 106  0.25(53.3)   576106 (increase) 207 103 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

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Prob.

11.234 P 3 (a) σ   279  10  137.510 6 Pa t A 1963106 = 137.5 MPa < 234 MPa (O.K.) σt 137.5   664 106 E 207 103 ε transv  με axial  0.25(664 106 ) εaxial 

 166.0 106

(b) δ = εaxial L = 664 × 106(1 m) =0.664 mm Prob. 11.7 P 64, 000 σ   16, 000 psi t A 4 Longitudinal (axial) εaxial 

σt 16, 000   5.33104 E 30, 000, 000

δaxial  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)

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Prob. 11.8 d = 150 mm A = 0.7854d2 = 17.67 × 103 mm2 0.275 300  0.000917

εz

0.0125 150  0.0000833

ε y

μ

εy εz



0.0000833  0.0908 0.000917

From Table G-2 (SI): E = 21.5 × 103 MPa δz AE  0.275(17.67 103)(21.5103)  L 300  348 kN

P

Prob. 11.9 d = 1.5 in.

A = 1.767 in.2 δ 0.0023 ε  z   0.00115 z 2 Lz P 60, 000 σ    33, 960 psi z A 1.767  34,000 psi (O.K.) σz

33, 960  29, 530, 000 psi ε z 0.00115 δ 0.00043 ε  x   0.000287 x 1.5 Lx εx 0.000287 μ   0.2496 0.00115 εz E



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Prob. 11.10 s 30, 000 εaxial  c   0.001 E 30, 000, 000 εtransv  μεaxial  0.25(0.001)  0.00025 δtransv  εtransv L  0.00025(3)  0.00075 in. (increase)

Prob. 11.11

Ly = 12 in., μ = 0.25

Stresses: 90, 000 σ   30, 000 psi  34,000 psi (OK) y 3(1) 288, 000 σ   24, 000 psi  34,000 psi (OK) x 12(1) 648, 000 σ   18, 000 psi  34,000 psi (OK) z 3(12) Strains: εx 

σx  μσ y  μσz  0.0009 (decrease) E

εy  εz 

σ y  μσx  μσz  0.00135 (increase) E σz  μσ y  μσx E

 0.00065 (decrease)

New dimensions of bar: δx = 0.0009(3) = 0.0027 in. (decrease) 3  0.0027 = 2.9973 in.

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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 σ   20  20 ksi  34 ksi (O.K.) y A 2(0.5) σx 

40  8 ksi  34 ksi (O.K.) 10(0.5)

Strain: μσx  μσ y εz  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.)

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Prob. 11.14 Initial temp. = 68°F (a) At 32°: wire contracts, stress increases. ΔT = 68 – 32 = 36°  = 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.  = 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 T 

0.5  29.9C 16.7 106 (1000)

ΔT for stress development = 50–29.9 = 20.1 °C  = 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.

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Prob. 11.17 α = 11.7 × 106 1/°C E = 207 × 103 MPa ΔT = +34°C  = αEΔT = 11.7 × 106(207 × 103)(34) =82.3 MPa (<235 MPa O.K.) Prob. 11.18



(a) δ = αL (ΔT) δ 0.01 T    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  = 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. T 



0.030 αB LB  αS LS

0.03 in.  [10.4106 (3)  6.5106 (6)](12 in./ft)  35.6F

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Prob. 11.20 ΔT to expand bar a length of 0.38(2) = 0.76 mm: δ 0.76 T    21.65C αL 0.0000117(3000) ΔT to cause thermal stress: ΔT = 27  21.65 = 5.35°C  = 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 mm2 (207 103 MPa)

 0.556 mm (shortening) 11-240

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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)

(in.2 )

(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 δ  P  AE AE L 0.02925(12)(0.016)(30, 000, 000) P (100  0.02925)(12)  140.4lb Ptotal = 140.4 + 15 = 155.4 lb

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Prob. 11.25 1-in. diam. rod: A = 0.7854 in.2 Due to tensioning: P 10,000    12,730 psi (tension) A 0.7854 Due to temp. rise of 30°F:  = 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:  = 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)  = Eα(ΔT)  ΔT = s/(Eα) T 

15, 000  107.5F 15, 000, 000(0.0000093)

(b) In addition to (a): δ = αL (ΔT)  ΔT = δ/(αL) T 

0.05  22.4 F 0.0000093 (20)(12)



ΔTtotal = 107.5 + 22.4 = 129.9°F

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Prob. 11.27 ΔT to cause stress of 70 MPa: T 

σ 70   28.9C 3 E 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: PL  P1 (60)  0.00002P1 δP1  1 1  A1E 0.1(30, 000, 000) δP2 

P2 L2  P2 (30)   0.000005P2 A2E 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

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Prob. 11.29

δt2 + δP2 = δt1 ----------------- Eqn. I. δt1 = αL1(ΔT) = 0.0000065(60)(ΔT) δt2 = αL2(ΔT) = 0.0000065(30)(ΔT)

 P2 

P2 L2



A2E

5000(30)  0.025in.

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   AE   st  cu (more)

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  L   L   AE    AE   st  cu 0.0535   22, 300 lb   4(12) 2(12)  (1)(30106 )  2(15 6   122,300   0)  22,300 psi st 1 22,300    11,150 psi cu 2 P









Prob. 11.31 Wood: S.Pine c(all) = 1250 psi E = 1700 ksi Steel: A36 c(all) = 20 ksi Es = 30,000 ksi n = Es/Ew = 17.65 Assume steel at allowable: 20, 000 σ  1133 psi  1250 psi (O.K.) σw  st  n 17.65 Aw = 32 in.2 As = 2(8)(0.25) + 2(4)(0.125) = 5.0 in.2 Pmax = Asst + Aww = 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

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c(all)w = 7.24 MPa Steel plates: Ast = 2(150)(10) = 3000 mm2 Est = 207 MPa c(all)st = 147 MPa n = Est/Ew = 17.25 Assume wood at allowable: st =nw = 17.25(7.24) = 124.9 MPa (<147 MPa (O.K.)) P = Astst + Aww = 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 st = ncu Ast = 0.7854(0.5)2 = 0.1964 in.2 Acu = 0.7854(0.625)2  Ast = 0.1104 in.2 cu Acu + st Ast = 8000 lb cu(0.1104) + 2st(0.1964) = 8000 lb cu = 15,900 psi st = 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 DF = nHF = 1.2143 HF 11-246

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P = (A)DF + (A)HF 70,000 = 63.25(1.2143HF) + 63.25HF HF = 500 psi PHF = AHFHF = 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   0.01in. (a)  AE    AE     S

CI

  12 14   0.01in.    4(30,000) 9(15,000)  P  49.1k P 49.1  12.28 ksi b σ   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

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st = ncu = 2.01cu P = Astst + Acu cu 225 × 103 N = 1270 × 106 m2(2.01cu) + 640 × 106 m2(cu) cu = 70.5MPa; st = 2.01(70.5) = 141.7MPa εcu = εst = st/E =141.7/(207 × 103) = 684 × 106 Prob. 11.37 L5  5  ½ : A = 4.79 in.2 Redwoood: E = 1300 ksi, c(all) = 1050 psi n = Est/Ew = 30,000/1300 = 23.1 Assume steel at allowable: w =20,000/23.1= 866 psi < 1050 psi (O.K.) P = stAst + wAw = 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 stAst + cAc + CI ACI = 400 k εst = εc = εCI

(Eqn. I)

st  c  CI Est

Ec

ECI

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 st



30,000

c   CI 3120 15,000

30,000CI  2 CI 15,000 3120CI    0.208 c CI 15,000

 st 

Sub. into Eqn. I: 2CI (18.06) + 0.208CI (88.0) + CI(7.07) = 400 k CI = 6.50 ksi st = 2(6.50) = 13.0 ksi c = 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 st = nc = 8.322c P = Astst + Acc 100 = 2.41(8.322c) + 193.6c c = 0.468 ksi st = 8.322(0.468) = 3.89 ksi

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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: E 0.06(207 10 3 )     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 = stAst + cuAcu

( Eqn. I)

n=Est/Ecu = 2.01 st = ncu = 2.01 cu Subs. into Eqn. I 2.01cu (1250) + cu(1800) = 70.61 kN cu = 16.37 MPa  st = 2.01(16.37) + 33.11 = 66.0 MPa Prob. 11.41

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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  st al  AE   AE   st  al Pst (12)(12) Pal (6)(12)  (1)(30,000) 1.5(10,000) 





 Pst = Pal Subs. into Eqn. I:  Pst = Pal = 12,000 lb Subs. into Eqn. II:  a = 5.00 ft Stresses: st   al 

12,000 1 12,000

 12,000 psi  8,000 psi

1.5

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 st = nbr = 2.13br P = Astst + Abrbr

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580 × 103 = 5162 × 106(2.13br) + 3226 × 106br  br =40.8 × 106 Pa = 40.8 MPa st = 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   AE   st  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:  

333.3

 6800 psi

st

0.049 833.3  17, 010 psi Steel : st = 0.049 (b)

δ  PL  σL  17,101(20)(12)  0.136 in. st AE E 30, 000, 000

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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   AE   al  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 :  al 

300  6120 psi 0.049

Steel :  900  18,370 psi st 0.049

(b) δ  PL  σL  18370(20)(12)  0.147 in. st 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

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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   AE   st  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 P  30,000  20,000 psi t (avg)  Anet 1.5 r/d = 0.5/3 = 0.167 Fig. 11.14: k = 2.35  P   t (max)  k A   2.35(20,000)  47,000psi  net  

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

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

 t (max) 

kP 2.4(50000)   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

 t (max) 

kP 1.85(50000)   91.2 MPa 1014 Anet

(c) r = 38 mm, r/d = 38/44 = 0.86 Fig. 11.14: k = 1.25 Anet = 13(120  2(38)) = 572 mm2

 t (max) 

kP 1.25(50000)   109.3 MPa Anet 572

Prob. 11.48

(a) t 

P

(b) t 

P 12,000   10,700 psi Anet 3(0.375)

 12,000  8000 psi A 4(0.375)

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(c) r/d = 0.5/3 = 0.167: Fig 11.14: k = 1.7 kP  1.7(10,700)  18,190 psi  t (max)  A net

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

 t (max)  P

 A kP  P  t (all) net Anet k

22,000(1.125)  11,150 lb 2.22

Prob. 11.50

 t 

P 12,000   9600 psi Anet 2.5(0.5)

t(all) = 15,000 psi  P  15,000    k  t  k  A 9600  1.56  net  Fig. 11.14: Req’d r/d = 0.30 Req’d rmin = 0.30d = 0.30(2.5) = 0.75 in.

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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 P 180103     32.14  10 6 Pa avg Aavg 56010  6

 P 

 max  k A   2.35(32.14)  75.5 MPa 

net



 

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    t  kA   2.56 0.75  0.094     net   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

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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   t  kA   2.43 0.75     net   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   t  kA   2.28 0.75     net   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

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 P    max  k  A net   A 150106 (870106 ) P  (all) net  k 2.3 3  56.7 10 N  56.7 kN

Prob. 11.54

P sin 2 2A 12,000  sin 96  30,400 psi 2(0.1963)

(a)  ' 



P 2  (b) n(max)  cos 

A 12,000  cos2 0  61,100 psi 0.1963 P 2 (c)  n  cos  A 12,000  cos2 48  27,400 psi 0.1963 Prob. 11.55

A = 50 mm × 75 mm = 3.75 × 103 mm2 (a) τ’ max. on 45° plane:

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' 

P

3  50010  66.7 MPa 2A 2(3.75103 )

P sin 2 2A 500103  sin140  42.9 MPa 2(3.75103 )  P 500 10 3 2   cos2   cos 70  15.60 MPa n 3.75 10 3 A (b)  ' 

Prob. 11.56 

P sin 2 2A 8000  sin120 2(1)

  '

 3460 psi

  n



P

cos2 

A 8000

cos2 60 2000 psi

1 Prob. 11.57 P = 67 k A = 0.7854(6)2 = 28.3 in.2

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P sin 2 2A 67  sin 90 2(28.3)  1.184 ksi

 ' 

Prob. 11.58 A = 4.0 in.2 τ(ult) = 1200 psi P sin 2 2A  (ult) (2 A)

 '  P

sin 2 1200(2)(4.0)  sin(2  80)  28,100 psi

Prob. 11.59 A = 490.9 × 106 m2

(a) θ = 30° P  '  sin 2  



(more) 80103









sin 60 2A 2(490.9106 )  70.6106 Pa  70.6 MPa P 2      80103 2  cos n cos 30 A 490.9106 6  122.210 Pa 122.2 MPa 



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

(b) Max. normal stress (θ = 0°) P 2 80103    2 cos n(max)  cos 0 A 490.9 106  6  163.0 10 Pa 163.0 MPa 



Max. shear stress (θ = 45°) P 80103 '  sin 2 







(max)

2A

2(490.9106 ) 6  81.510 Pa  81.5MPa



sin 90

Prob. 11.60 A = 1.0 in.2 P = A = 1.0(2000) = 2000 lb P 2000 sin 2  sin 60 2A 2(1)  866 psi

 ' 

  n

P

cos2   A

2000

cos2 60  500 psi

1

  Prob. 11.61 On plane  CD:  MPa τ' = 35    P '   sin 2  35 MPa

2A P 2(35106 )  80.83 106 Pa   A sin 60 P   cos2   80.83106 cos2 30 n A  60.6 106 Pa  60.6 MPa 



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Prob. 11.62 A = 50(75) = 3750 mm2 = 3750 × 106 m2 On 50° plane: τ' = 138 MPa (θ = 40°) P sin 2  138 MPa 2A 138106 (2)(3750106 )

(a)   '

P

sin( 2  40)

 1.051106 N  1.051MN (b)  n 

P

2

1.051106

2

cos   cos 40 A 3750106  164.5106 Pa 164.5MPa

Prob. 11.63 (a)

(b) n   sin 2  10,000sin 70  9400 psi

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Prob. 11.64 (a) See element of Prob. 11.24:

 '   cos2 6000  10,000cos2   2  cos1  6000  53.13 10,000    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 2 A' τ (2 A) 5000(2)(28.27) (max) Pmax   sin 2θ sin 90

' 





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Pmax = 283,000 lb Based on comp. stress (max when θ = 0°): Pmax = c(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 c(all) = 6 MPa = 6 × 106 Pa τ(all) = 1.6 MPa = 1.6 × 106 Pa Based on comp. stress (max when θ = 0°): Pmax = c(all)A = 6 × 106(30 × 103) = 180 × 103 N Based on shear stress (max. when θ = 45°): P sin 2 2A  (all) (2 A) 1.6106 (2)(30103 )  Pm ax  sin 2 sin 90  96.0103 N (Controls)

 ' 

Prob. 11.67 Tens. and/or comp. stress (max. at 45°) (a) n = τsin2θ = 5000(sin 2 × 45°) = 5000 psi (b)

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Prob.

266.68 Tens. and/or comp. stress (max. at 45°) (a) n = τ sin 2θ = 9200(sin 2 × 45°) = 9200 psi (b)

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Prob.

267.1

Prob. 12.2 Tint(A) = Fd = 1(18) = 18 k-in.

Tint(B) = 18 + 100 = 118 k-in.

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Prob. 12.3

Prob. 12.4 J

T

π

(854  404)  4.87 106 mm4 32  4.87 106 m4

 (all) J c

(68 106 Pa)(4.87 10-6 m4) 85 / 2 103 m  7.79 103 N  m 

 7.79 kN  m

Prob. 12.5 π d 4 π (5)4 4 J   61.36 in.

32 32 10, 000(61.36) τ J  245, 000 lb-in.

TR  (all)  c 2.5

Prob.12.6 TR 

 (all) J c

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4 4 4 4 J  π(d  d1 )   (100  75 )  6.7110 6 mm 4 32 32 62 MPa(6.71106 mm4) TR   8.32 106 N  mm 50 mm



Tc 8.32 106 N  mm(37.5 mm)  J 6.71106 mm4  46.5 MPa

Prob. 12.7

Prob. 12.8

Prob. 12.9

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Prob. 12.10

Prob. 12.11 J

πd4 32





π(3.5)4

14.73 in.

4

32

Tc 5000(12)(3.5 / 2)   7130 psi J 14.73

Prob. 12.12

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Prob. 12.13

Prob. 12.14

12-271

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Prob. 12.15

Prob. 12.16

Prob. 12.17

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Prob. 12.18

Prob. 12.19

Prob. 12.20

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Prob. 12.21

Prob. 12.22

Prob. 12.23

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Prob. 12.24

Prob. 12.25

Prob. 12.26

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Prob. 12.27

Prob. 12.28

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Prob. 12.29

Prob. 12.30

Prob. 12.31

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Prob. 12.32

Prob. 12.33

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Prob. 12.34

Prob. 12.35

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Prob. 12.36

Prob. 12.37

Prob. 12.38

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Prob. 12.39

Prob. 12.40

Prob. 12.41

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Prob. 12.42

Prob. 12.43

Prob. 12.44

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Prob. 12.45

Prob. 12.46

Prob. 12.47 J  π(1.5) 32  0.497 in.

2

HP 

4

Tnr 1500(1000)   23.8 hp 63,025 63,025

τ

Tc J



1500(1.5 / 2) 0.497

 2260 psi

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Prob. 12.48 (a) Eqn. 12.7. (nr = 3000 rpm) 63, 025(120) T 2520 lb-in.

3000 16(2520) 16T Req'd d  3 3  1.171 in. (all) (8000) Use a 1 163  solid shaft (b) nr = 300 rpm 63, 025(120) T 25, 200 lb-in.

300 16(25, 200) 16T Req'd d  3 3  2.52 in. (all) (8000) Use a 2 58  solid shaft (c) For 1 163  solid shaft: J  π d  π(1.188)  0.1956 in.

32 32 TL 2520(10)(12)    0.1288 rad JG 0.1956(12,000,000)  7.38 4



4

4

For 2 85 solid shaft: J  π d  π(2.625)  4.66 in.

32 32 TL 25, 200(10)(12)    0.0541 rad JG 4.66(12,000,000)  3.10 4

4

4



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

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 ↑

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(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 ↑

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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 ↑

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(b)

ΣMA = 20(10) + RB(12)  10(20) = 0 RB = 33.3 k ↑ ΣMB = RA(12) + 20(12)  10(8) = 0 RA = 3.33 k RA = 3.33 k ↓ Prob. 13.6

ΣMA = 54(9)  16.5(12.5) +RB(13) = 0 RB = 52.3 k ↑ ΣMB = RA(13) +54(4) + 16.5(0.5) = 0 RA = 17.25 k ↑ (b)

ΣMA = +RB(18)  24(14)  63(9) = 0 RB = 50.2 k ↑ ΣMB = RA(18) + 63(9) + 24(4) = 0 RA = 36.8 k ↑

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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 ↑

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(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 ↑

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(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 ↑

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(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 ↑

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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):

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V = +11.11  10 = +1.11 kN M = +11.11(7)  10(4) = +37.8 kN∙m

@ 7m (right of point load):

V = +11.11  10  20 = 18.89 kN M = +11.11(7)  10(4) = +37.8 kN∙m Prob. 13.12 (a)

w = 2 k/ft, L = 20 ft RA = RB = ½ (2)(20) = 20 k ↑

V & M 3 ft from left: V= +20  6 = +14 k M = +20(3)  6(1.5) = + 51 k-ft V&M 8 ft from left:

V = +20  16 = + 4 k M = +20(8)  16(4) = + 96 k-ft

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(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

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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:

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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:

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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

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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:

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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:

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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:

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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

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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

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V = +140 – 400 = –260 lb M = +140(10) – 400(5) = –600 lb-ft At 10 ft from left (just right of RB): V = +140 – 400 +460 = +200 lb M = –600 k-ft

(b) At 4 ft from left: V = +4200 – 4000 = +200 lb M = +4200(2) – 4000(4) = –7600 lb-ft

At 10 ft from left:

V = +4200 – 4000 – 1200 = –1000 lb M = +4200(8) – 4000(10) – 1200(1) = –7600 lb-ft

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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

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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

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Prob. 13.21

Prob. 13.22

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Prob. 13.23

-Prob. 13.24

13-308

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Prob. 13.25

Prob. 13.26

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Prob. 13.27

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Prob. 13.28

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Prob.

13.312

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Prob.

13.313

Prob. 13.31

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Prob. 13.32

Prob. 13.33

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Prob. 13.34

Prob. 13.35

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Prob. 13.36

Prob. 13.37

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Prob. 13.38

Prob. 13.39 (Refer to Prob. 13.1(b)

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Prob. 13.40 (Refer to Prob. 13.8(a))

Prob. 13.41 (Refer to Prob. 13.2(a))

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Prob. 13.42 (Refer to Prob.

13.319(b)

Prob. 13.43 (Refer to Prob. 13.3(a))

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Prob. 13.44 (Refer to Prob.

13.320(b))

Prob. 13.45 (Refer to Prob. 13.4(b))

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Prob. 13.46 (Refer to Prob. 13.8(b))

Prob. 13.47 (Refer to Prob. 13.10(a))

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Prob. 13.48 (Refer to Prob. 13.10(b))

Prob. 13.49 (Refer to Prob. 13.11(a))

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Prob. 13.50

Prob. 13.51

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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

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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

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For max. M:

Under the left 142 kN load: M = +138.3(11.25) – 36(4.3) = +1401 kN∙m For max. V:

V R  R

12.7

(36) 

26.0

17.0

(142) 142  252 kN

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:

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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

ML : 12x 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:

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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:

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RR = 10.0 k ↑, MD = 10(17.39) = 173.9 k-ft (c) Absolute max. V:

Max. V = RR = 20 k

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Prob. 14.1 (a) bh 2 5.5(9.5) 2  6 6 3  82.7 in.

Sx 

(b) π (d 4  d 4 ) I

1

64 π (1144 1004 )  64  3.38106 mm4 I 3.38106 3 3 S   59.310 mm c 114 / 2 (c) Refer to Example 14.16 I y  (I o  Ad 2 )  2( 112)(2)(16)3  1 12 (24)(1)3  1367 in.4 I y 1367 S    170.9 in.3 y c 8

Prob. 14.2 (a)

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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 I x  1160.8 in. 4 I x2 

I x  1160.8  236 in.3 c(top) 4.921 1160.8 Ix   143.7 in.3 S  x(bott) c(bott) 8.079 Sx(top) 

(b)

W24  162: Ix = 5170 in.4 18(1)3 4 2 Ix(PL)  2  18(13)  6090 in.  12   ΣIx = 5170 + 6090 = 11,260 in.4 𝐼𝑥 11,260 𝑆 = = = 834 in.3 𝑥 𝑐 13.5 Prob. 14.3 W18  71: Sx = 127 in.3 wL (0.40  0.071)(28)2 2   46.16 k - ft M  8 8 M 46.16(12)  4.36 ksi f b   127 Sx

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Prob. 14.4 W18  71: Fy = 15.8 in.3 M  46.16 k - ft M 46.16(12)  35.1ksi f b   15.8 Sy

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  143.7 Sx 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.  d3

 (25103 )3  1.534106 m3 32 32 M 160 f    104.3106 N/m2 b S 1.534106  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 m3 M fb   460 N  m  50.3106 N/m2 -6 3 Sx 9.1510 m Sx 

 50.3 MPa

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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 4 2 I x1  2  20(6)   1453.3 in.  12   2(10)2  4   333.3 in.

I x 2  2  12  Total I x  1786.6 in.4

Sx = Ix/c = 1786.6/7 = 255 in.3 14-333

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(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 12(6.154)2  458.5 in.4 I x1  12 y

12(2)3  24(3.846 )2  363 .0 in.4 12 2(8)3 2 4 I x3  16(1.154 )  106 .6 in.

12 I x 2 



Total Ix = 928.1 in.4 928.1 I  191.5 in.3 S x  x  c(bott) 4.846 928.1 I  129.7 in.3 S x  x  c(top) 7.154

Prob. 14.12 (See Prob. 14.29(a)) wL2 M   20,250 lb - ft 8 M 20,250(12)  953 psi f b   255 Sx

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Prob. 14.13

(a) I x 

f  b

100(150)3  28.13106 mm4 12  28.13106 m4

 2600 N m(7510 m) I 28.1310-6 m4  6.93106 N/m2  6.93 MPa

Mc

-3

(b) fb 

My  2600 Nm(50103 m)  I 28.13106 m4  4.62106 N/m2  4.62 MPa

Prob. 14.14 (Refer to Prob. 14.13) I x 

150(100)3  12.5106 mm4 12  12.5106 m4

(a) fb 

Mc I



2600(50103 ) 12.5106

6

2

 10.410 N/m  10.4 MPa

(b) fb 

My I



2600(25103 ) 12.5106

6

 5.2010 N/m

2

 5.20 MPa

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Prob. 14.15

M = 12(2) = 24 lb-ft S 

 d3

f  b

32 M S





 (0.5) 3  0.01227 in.3 32

24(12)  23,500 psi 0.01227

Prob. 14.16

W530  101: Sx = 2290×103 mm3 wt. = 9.81(101) = 991 N/m 991(13)2 M   54103 (4.33) 8  255103 N m fb 

M 255106 N mm   111.4 MPa 2290103 mm3 Sx

Prob. 14.17

a1 = 4 in.2 a2 = 3 in.2 Σa = 7 in.2 14-336

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y

ay 4(3.5)  3(1.5)   2.643in. a 7

4(1) 3  4(0.857)2  3.271in.4 12 1(3) 3  3(1.143)2  6.169 in.4 I x 2  12 Total I x  9.44 in.4 I x1 

M = 450(6) = 2700 lb-ft f b(top)  f b(bott) 

Mctop

 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 Sx



13,080(12)

15,840 psi

9.91

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Prob. 14.20 3V 3(20103 )   3.13 MPa (a) fv  2A 2(80)(120) (b) A  0.7854d 2  0.7854(120)2  11.31103 mm2 4V 4(20103 ) fv    2.36 MPa 3A 3(11.31103 )

Prob. 14.21

Vmax = RA = RB = 16,000/2 = 8000 lb (a) fv(max) 

1.5V 1.5(8000)   100 psi A 6(20

(b) 4” above and below the N.A.; 6(20)3 I NA  12  4000 in.4 y  7 in. a1  36 in.2

Q  a1 y  36(7)  252in.3 f v(4) 

VQ 8000(252)   84 psi Ib 4000(6)

8” above & below N.A.:

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y  9 in. a 2  12 in.2 Q  a2 y  12(9)  108 in.3 f v(8) 

VQ 8000(108)   36 psi Ib 4000(6)

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  Q    6.73(0.385)   6.90  flange 2    17.38 in.3



Q

 6.90  0.385   (6.90  0.385)(0.270)  2   3  5.73 in. 

web

Total Q = 17.4 8+ 5.77 = 23.11 in.3

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VQ 43.1(23.11)   12.72 ksi Ib 291(0.270) V 43.1 (b) f    11.57 ksi v dt w 13.8(0.270) fv 



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 3A 3(1.767)

Prob. 14.25 A = 0.7854(10)2 = 78.54 in.2  (10)4 I  64  490.9 in.4 Vmax = 300 lb Mmax = 300(20) = 6000 lb-ft fb 

Mc 6000(12)(5)   733 psi I 490.9

fv 

4V 4(300)   5.09 psi 3A 3(78.54)

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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 f 

VQ

v

Ib



750(3.0)

 112.4 psi

6.67(3)

On plane x1-x1: Q = (1)(1)(1.5) = 1.5 in.3 f 

VQ

v

Ib



750(1.5)

 168.7 psi

6.67(1)

Prob. 14.27 Rect. Beam: 100 mm x 250 mm I x 

100(250)3  130.2106 mm4 12  130.2106 m4

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Q = 100(125)(62.5) = 781.3×103 mm3 = 781.3×10-6 m3 fv 

VQ Ib



(140103 N)(781.310-6) (130.2 106 )(100103 )  8.40106 N/m2

 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 4 2 I x1  2  20(5)   1013in.  12  3 2(8)  85.3 in.4 I x 2  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

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Total Qx = 116 in.3 VQ f  V  fv Ib  120(1098)(2)  2270 lb v Ib Q 116 

Shear stress in plane of junction between web and flange: Q = 10(2)(5) = 100 in.3 Web: f  VQ  2270(100)  103.4 psi v Ib 1098(2) Flange: fv 

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 Q  35.85 (0.1560)  1.75 in.3   y(web)  2  Total Qy = 35.36 in.3 f 

140(35.36)

v

270(35.85)

 0.511ksi

Web-flange junction: Qy = 5.675(0.940)(3.15)(2) = 33.61 in.3 Flange: fv 

140(33.61)  9.27 ksi 270(0.940)(2)

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Web: fv 

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)  8.27 ksi fv  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 8(137.3) 8M  5.60 k/ft w( total)  2R  2 L 14 (a) MR 

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)

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Prob. 14.32

W610  113: Ix = 864×106 mm4, d = 607 mm, tf = 17.3 mm, tw = 11.20 mm 286.22 (11.20) (a) Qx  228(17.3)(294.85)  2 6 3  1.62210 mm VQ 525103 (1.622106 ) fv   87.9 MPa  Ib 864106 (11.20) (b) Avg. web shear: f  v

V dt w



Prob. 14.33

525103 607(11.20)

77.2 MPa

W30  108: L = 10 ft

Ix = 4470 in.4, d = 29.8 in., Sx = 299 in.3 42(10)2 M max   525 k - ft 8 M  525(12)  21.1 ksi fb  Sx 299 42(10) Vmax  2  210 k

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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.

2

Bar: A = (0.75 in.) = 0.5625 in.2

y

Ay 3.55(3.015)  0.5625(6.405)   3.479 in. A 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 VQ f Ib f   V  v(max) max v 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) 2.2711 2 (0.23)  4.939 in.3  2 Vmax 

20(27.68)(0.23)  25.8 k 4.939

 Controls

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Prob. 14.35

Mmax = 400(8)/4 = 800 lb-ft Vmax = 400/2 = 200 lb (a) Not glued:  bh2  4(6)(1) 3   S x  4  4 in.

6  6  M  800(12)  2400 psi fb  4 Sx (b) Glued: 3 bh2 6(4)2 Sx    16 in.

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 VQ 200(12) f    12.5 psi v Ib 32(6) I x 

Prob. 14.36

a1 = 4 in.2, a2 = 4 in.2, Σa = 8 in.2

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4(4.5)  4(0.25)  2.375 in.

8 0.5(8)3  4(2.125)2  39.4 in.4 I x1  12 8(0.5)3 2 4 I x2   4(2.125)  18.15 in.

12 y

Total Ix = 57.55 in.4 WL 10,000(6) M   7500 lb - ft 8 8 Mc 7500(12)(6.125)  I 57.55  9580 psi(Compression)

f b(top) 

fb(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 fv(web) 

5000(8.5)  1477 psi 57.55(0.5)

fv (flng) 

5000(8.5)  92.3 psi 57.55(8)

Prob. 14.37 (a)

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Elastic: y

Ay 8(1.5)(6.75)  6(2)(3)   4.875 in. A 8(1.5)  6(2)

I xx 

8(1.5)3 12



S x(bott)

I

2

12(1.875) 



2(6)3

 12 12(1.875)2  122.6 in.4

122.6

c

 25.15 in.3

4.875

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   652 in.4 12 12 I 652 S    108.7 in.3 x c 6 I xx 

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 L2



8(75.5)

 3.08 k/ft

14 2

Mp = FyZx = 36(45)/12 = 135 k-ft wp 

8M p L2



8(135 ) 14 2

 5.51 k/ft

(b) My = FySx = 36(108.7)/12 = 326 k-ft wy 

8M y L2



8(326 ) 14 2

 13.31 k/ft

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Mp = FyZx = 36(138)/12 = 414 k-ft wp 

8M p L2



8(414 ) 14 2

 16.90 k/f t

Prob. 14.39

Fy = 40,000 psi  0.7854(3)2  4(1.5)    Zx  2(6)(3)(1.5)  2 2   3  = 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)

bh2 300(50)2   125103 mm3 Sx  6 6 A  50(300)  15000 mm2

Based on moment: fb 

M wL2   Fb Sx Sx 8

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F b Sx 8 11.0(125103 )(8)  w 1.5 3  2.7110 mm

max L 

 2.71m



controls

Based on shear: 3V f   3  wL    v 2 A 2 A  2  4 AFv 4(15000)(0.83) max L   3(1.5) 3w 



 11.07103 mm 11.07 m

Prob. 14.41 Fv = 60 psi 4(6)3 4 Ix   72 in.

12 In glued joint: Q = 4(2)(2) = 16 in.3 VQ  V  Fv Ib  60(72)(4) (a) f  v R Ib Q 16  1080 lb

Since max V = P: P =1080 lb (b) Mmax = 1080(3) = 3240 lb-ft fb 

Mc 3240(12)(3)   1620 psi I 72

Prob. 14.42

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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 125  Q  250(25)(137.5)  2(125)(25)  I x 



x

   2  1250 103 mm3  1.25 103 m3

Moment: F I 10106 (302.1106 ) M R  b  c 0.15  21.14103 N m PL

2

 wL 8 4 4 wL2   max P   M  L 8  4 3 2.7 103 (4)2     20.1410  4 8  M

max P  14.74103 N  14.74 kN

Shear: 1.5106 (302.1106 )(0.050) VR  1.25103  18.13103 N

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P wL  2 2 wL  max P  2 V     2    2.7 103 (4)  3   218.1310  2   3  25.4510 N  25.45kN

V 

Smaller controls: P = 14.74 kN Prob. 14.43

a1 = 16 in.2 a2 = 12 in.2 Σa = 28 in.2 ay a  5.286 in.

y

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8(2)3 16(1.714) 2  52.33 in. 4 12 2(6)3 I x2  12(2.286) 2  98.71in. 4 12 I x1 

Total Ix = 151.04 in.4 25(12)(2.714)  5.39 ksi(tens.) fb(top)  151.04 25(12)(5.286)  10.50 ksi(comp.) fb(bott)  151.04 Max. shear stress at the N.A.: Q 

2(5.286)2

 27.94 in.3

2 8.7(27.94)  0.805 ksi fv(max)  151.04(2)

Prob. 14.44 Vmax = 920 lb a1 = 6 in.2 a2 = 6 in.2  6(1)3 4 2 I x1  2  6(3.5)   148 in.  12  1(6)3  4   36 in.

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 VQ 920(21) f    52.5 psi v Ib 184(2) 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.

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Prob. 14.45 6  12 (S4S): wt. = 17.5 lb/ft 1800 (121)  18,150 lb - ft 12 17.5(18)2 M beamwt.   709 lb - ft 8 M R  Fb Sx 

Allowable M for superimposed loads: 18,150 – 709 = 17,441 lb-ft 17,441  2907 lb M max  P(6ft)  P  6 Fv A 140(63.3)   5908 lb 1.5 1.5 17.5(18)  158 lb Vbeamwt.  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 M  (281)  42,150 lb - ft R 12 26.7(18)2 M beamwt.   1081lb - ft 8 Allowable M for superimposed loads: 42,150 – 1081 = 41,069 lb-ft 41,069  6845 lb M max  P(6ft)  P  6 Fv A 140(96.3)   8990 lb 1.5 1.5 26.7(18)  240 lb Vbeamwt.  2 VR 

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Allowable V for superimposed load: 8990 – 240 = 8750 lb Vmax = P = 8750 lb Smaller controls, therefore Allowable P = 6845 lb

Prob. 14.47

Ay 16(13) 10(7)  24(1)   6.04 in. A 16 10  24 8(2)3 1(10)3 16(6.96)2  10(0.96)2 I xx  12 12 12(2)3    24(5.04)2 1490.6 in.4 12 I xx  1409.6  187.3 in.3 S xx  ctop 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

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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)  983 psi f b   73.8 Sx

M 



983 psi 1000 psi Max.V  R  f  v

wL



2 3(2015)

(325 10.9)12

(O.K.)  2015 lb

2

3V   76.7 psi 2 A 2(39.4) 76.7 psi 150 psi

(O.K.)

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:

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1.286(24)2

10(24)  152.6 k - ft 4 8 0.6(50)(64) 160 k - ft M R  0.6Fy Zx  12 160k - ft  152.6 k - ft (O.K.) M 



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 R 8(331103 ) 3  w  2  N/m  10610 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

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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 900(52.7)  3950 lb - ft 12 8M 8(3950)  123.4 lb/ft w 2  L 162 F A 180(28.1) VR  v   3370 lb 1.5 1.5 2(3370) 2V  4.21lb/ft w R  L 16 M R  Fb Sx 

(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 8(136.5) 8M  25.8 k/ft w  2R  L 6.52 VR  0.4Fy dtw  0.4(50)(13.98)(0.285) M R  0.6Fy Zx 

 79.7 k w

2VR 2(79.7)   24.5 k L 6.5

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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 bh3

11.25(2.5)3  14.65 in.4 12 12 I 14.65 S   11.72 in.3 c 1.25 1600 M  (11.72)  1563 lb - ft R 12 8M    195.4 lb/ft w 2 L  F A 100(28.1) VR  v   1873 lb 1.5 1.5 2(1873) 2V  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 50(88.9)  370 k - ft 12 8(370) w   3.29 k/ft (at yield) max 2 30 M y  Fy Sx 

50(101)  421k - ft 12 8(421) w   3.74 k/ft (at full plasification) max 2 30 M p  Fy Zx 

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Prob. 15.1 Ec R 10,000(0.125)   20.0 ksi (125 / 2)

fb 

Prob. 15.2 fb  

Ec R 30,000( 164)

 46.9 ksi

((20 / 2))

Prob. 15.3

L 60  2π R 360 360L  360(500)  477.5 mm R 2π 60 2π 60 fb 

Ec  207 10s (0.50)   217 MPa R 477.5

Prob. 15.4 E(d / 2) D/2 (D / 2)(30,000)  10,000,00(d / 2) fb 

Min. D  333d (in.)

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Prob. 15.5 3((0.5)3 I   0.0313in.4 12 EI RM 62(700) R E   M I 0.0313 E  1,389,000 psi

Prob. 15.6 I 

6(14)3

 1372 in.4 12 EI 1,700,000(1372) R  M 200,000 11,660 in.  972 ft

Prob. 15.7 Ec Ec(2)  R D  0.08  30,000,00 (2)    2  60,000  D  D  40 in. sb 

Prob. 15.8 1.5(1.5)3 I   0.4219 in.4 12 EI 30,000,000(0.4219) M  R 15,000  844 lb - in. ( 70.3lb - ft)

Prob. 15.9 E = 207103 MPa, d = 38 mm, L = 800 mm, P = 3 kN

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I 

 (38)4

 102.4 103 mm4 64 PL3 3 103 (800)3   48EI 48(207 103 )(102.4 103 )  1.510 mm

Prob. 15.10 (Refer to Prob. 15.7) d = 25 mm L = 500 mm I 

 (25)4 64



 19.17 103 mm4

3 103 (500)3 48(207 103 )(19.17 103 )  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.

384EI 384(1300)(1950) 15(12)  0.60 in. all  300 0.449in.  0.60 in. (O.K.)

 max 

Prob. 15.12 15(12)  0.60 in Prob. 15.10 (cont.) all  300  ma x 

PL 3 5(15)3 (1728)   0.240 in..

48EI 48(1300)(1950) 0.240in.  0.60in.

(O.K.)

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Prob. 15.13 W16  45: Ix = 586 in.4 Pa (3L2  4a 2 ) 24EI 12(8)(12)  [3(24) 2  4(8) 2 ](144) 24(30,000,000)(586)  0.579 in.

 ma x 

Prob. 15.14 4-in. diam. shaft: I = d4/64 = 12.57 in.4  r 2 (12) wt.  (490 )  42.8 lb/ft 1728



limit = 0.50 in.

5wL4 5(42.8)(12)4 (1728)  384EI 384(30,000,000)(12.57)  0.0530 in.

wt. 

Allowable  for P = 0.50  0.0530 = 0.447 in.

48EI 48(30,000)(12.57)(0.447)  L3 (12)3 (1728)  2.71k

P max 

Prob. 15.15 all at center of span = 0.010L 5wL4 384EI  d 4   (3)4  I  3.98 in.4 64 64  r 2 (12) w  (490)  24.00 lb/ft 1728 0.010(384)EI 5wL4  0.010L  L 3 384EI 5w(1728)



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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. all  240 WL3 WL3   I  15EI 15Eall 30(12)3 (1728) 4  332 in.

I min  15(30,000)(0.60)

Prob. 15.17  max 

PL3 3EI

(a) Douglas fir  

5(6)3 (1728) 3(1700)(951)

 0.385 in.

3 (b) W10  22   5(6) (1728)  0.1757 in.

3(30,000)(118)

(c) Pipe 8 x-strong  

5(6)3 (1728)  0.1956 in.

3(30,000)(106)

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Prob. 15.18 W16  36, Ix = 448 in.4 5wL4  PL3  384EI 48EI 5(1.0)(20)4 (1728)  10(20)31728   384(30,000)(448) 48(30,000)(448)  0.482 in.

(a)  m ax 

1(20) 2 10(20)   10 k - ft 8 4 M 100(12)  21.2 ksi sb   Sx 56.5

(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 2 2  max   (3L  4a ) 384EI 24EI  max  17.77 mm

Prob. 15.20

W10  22, Ix = 118 in.4 Neglect beam weight.

Free  (no center support):

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5wL4 5(1000)(20)4 (1728)  384EI 384(30,000,000)(118)  1.017 in.



Apply force RC to create 1 where 1 = 1.017  0.50 = 0.517 in.

R L3 1  C 48EI 0.517(48)(30,000,000)(118)  RC  (20)3 (1728)  6350 lb

Prob. 15.21

max = 0.20 mm = 0.20103 m 

Pa (3L2  4a2 ) 24EI

 d4 I  64 req'd d   



4



4

64Pa(3L2  4a2 )

 (24)E

64(3103 )(0.2333)[3(0.7)2  4(2.333)2 ]

 (24)(207109 )(0.20103 )

req'd d = 65.110–3 m

I  A 

= 65.1 mm  (65.1103 )4 64  (65.1103 )2 4

9

 88210 m

4

 3.3310 3 m2

M = Pa = 3103(0.2333) = 700 Nm

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 65.1103   2 Mc 700    25.8106 Pa sb   9 I 88210  25.8MPa  165 MPa (O.K.) ss 

4V



4(3103 )

6

 1.20110 Pa

3A 3(3.33103 ) 1.201MPa  100 MPa

(O.K.)

Prob. 15.22 W10  22: Ix = 118 in.4

AM 12 (8)(40,000)(144)  EI 30,000,000(118)  0.00651 rad 1   2  0.00651(57.3)  0.373

1   2 



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Prob. 15.23 W24  84 Ix = 2370 in.4

AM 23 (15)(337.5)(144)  EI 30,000(2370)  0.00684 rad 1   2  0.00684(57.3)  0.392

1   2 

 

Prob. 15.24

I 

 (76.02  62.72 ) 64



A M

EI



 879 10 3 mm 4

2 103 (2.1106 ) 70 103 (879103 )  0.0683 rad

  0.0683(57.3)  3.91

 

Prob. 15.25 1.5(1)3 Ix  12  0.125in.4

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

AM

60(4000)  0.064 rad EI 30,000,000(0.125)   0.064(57.3)  3.67 

 

Prob. 15.26 (Refer to Prob. 15.18)

1   2 

AM 4000(30)  12 (40)(4000)  EI 30,000,000(0.125))  0.0533 rad

1   2  0.0533(57.3)  3.06

 

Prob. 15.27 Refer to Prob. 15.9 Redwood E = 1300 ksi

x1  58 (7.5)  4.688 ft y AB 

AM x 23 (7.5)(28.125)(4.688)(1728)  EI 1300(1950)

 0.449 in. B  yAB  0.449 in.

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Prob. 15.28 (Refer to Prob. 15.12 (& 15.11)) Redwood: E = 1300 ksi

x1  2 3 (7.5)  5.0 ft A x 1 (7.5)(18.75)(5.0)(1728) y AB  M  2 EI 1300(1950)  0.240 in.  B  yAB  0.240 in.

Prob. 15.29 Doug Fir 8  12 (S4S)

A = yAB = 0.75 in.

AM = ½(10)(10P) =  50P ft2-lb x  2 3 (10)  6.67 ft A x (50P)(6.67)(1728) y AB  M   0.75 in.

EI 1700(951)  P  2.10 k

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Prob. 15.30 Refer to Prob. 15.29

AM = ½(5)( 5P) =  12.5P ft2-lb x  2 3 (5)  5  8.33 ft A x 12.5P(8.33)(1728) y AB  M   0.75 in.

EI 1700(951) P  6.74 k

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  y AB 

AM x [384(5.33)  384(10)](1728)  EI 30,000(586)

 0.579 in.  B  yAB  0.579 in.

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Prob. 15.32 W12  30: Ix = 238 in.4 5wL4  PL3  384EI 48EI 5(675)(124 )(1728)  384(30,000,000)(238)





14,00(123 )(1728) 48(30,000,000)(238)

 0.1661in. all 

12(12)

 0.40 in.

360 0.1661in. 0.40in. (O.K.)

Prob. 15.33 Refer to Prob. 15.23 AM x 23 (337.5)(58 )(15)(1728)  EI 30,000(2370)  0.769 in.

y AB 

 B  yAB  0.769 in.

Prob. 15.34 See text Fig. 15.16 (Max. M @ conc. loads =88 kNm)

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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

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Prob. 15.38

A1 = ½(3)(360) = 540 ft2-lb A2 = 360(2) = 720 ft2-lb x1  23 (3)  2 ft, x2  1 3  4 ft 3(1)3 4 Ix   0.25 in.

12

Max. deflection @ midspan: AM x [540(2)  720(4)](1728)  EI 30,000,000(0.25)  0.9124 in.

y AE 

 E  yAE  0.9124 in.

Deflection at point C: 720(1)(1728)  0.1659 in.

30,000,000(0.25) C  yAE  yCE  0.9124  0.1659  0.747 in. yCE 

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Prob. 15.39 (See Example 15.22(b))

From Example 15.22, yBA = +1.853 in.

By proportion: 30 y  (1.853)  2.316 in.

C 24 AM x  1  1 2 (24)(276)(14) yCA  EI   30,000(394) [   1 1  3 (24)(288)(12)  2 (6)(12)(4)](1728)  2.716 in. C  yCA  yC  2.716  2.316  0.400 in. (upward)

Prob. 15.40

For 1.72  1.75 section: 15-376

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bh3 1.72(1.753 )   0.7682 in.4 12 12

I x 

Max. defl. = B = yAB = 0.74 in. y AB 

AM x 12 (14)(1400)(23)(14)  EI E(0.7682)

 E  1,609,000 psi

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 

wL2 M  sb Sx  8 Sx

w

8sb S x  8(152)(3150103 )  L2 (12103 )2

 26.6 N/mm  26.6 kN/m Total load = 26.6 + 1.354 = 28.0 kN/m M 

wL2 8



28.0(12 10 3 )2  504 10 6 N  mm 8

AM x EI 2 AM  3 bh  2 3 (6000)(499106 ) yAB   B 

 2.021012 N mm 2 yAB 

2.021012 (3.75103 )  42.4 mm 207103 (862106 )

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Prob. 15.42

Max. defl. between supports: yBA 

[ 12 (120)(242 )(13 )  12 (144)(122 )(13 )]1728 30,000(228)

 2.037 in.

tan   



AM

1

2.037  0.007074 rad 24(12)

m(5m)(144)  0.007074 30,000(228) EI m  11.59 ft 

2

(11.59)2 (5)(11.59)(2 3 )(1728) 30,000(228)  0.655 in.

1

yAE  2

 E   max  yAE  0.655 in.

Defl. @ point D: y  D

30

(2.037)  2.546 in.

24

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 1  1 yDA   [ 2 (120)(24)(14)  30,000(228)   1 2 (144)(12)(10)  1 2 (24)(6) 2 ( 2 3 )]1728  2.838 in. 

D = yDA  yD = 2.838  2.546 = 0.292 in. 



Prob. 15.43 Aluminum beam, Ix = 200 in.4

 1  1 yAB   [ 2 (12000)(62 )(23) 10,000,000(200)   13 (4500)(3)(5.25)]1728  0.144 in.

A = yAB = 0.144 in.

Prob. 15.44

15-379

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 1  1 [ 2 (800)(40)2 ( 13 ) yBA    30,000(4080)   12 (600)(20)2 ( 13 )  1 2 (200)(10) 2 ( 1 3 )]1728  2.40 in.

  tan 

2.40  0.005 rad 40(12)

(Assume 20 < m < 30): AM  1 [ 1 2 m(20m)   EI   30,000(4080)     12 (m  20)2 (30)]144 Subst. for , solve for m: m = 20.63 ft 

1

2 2  1 [ 2 (20)(20.63) ( 3)(20.63)

yAE   30,000(4080)  12 (30)(0.65)2 (23 {0.63}  20)]1728  0.825 in.

E = yAE = 0.825 in.

Prob. 15.45

max= 0.10 in. w = weight (lb/ft)  13 (10)(50w)(34)(10)(1728)  0.10 in.

30,000,000(I ) 0.072w   0.10 in.

I  I  0.72w yAB 

15-380

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Steel wt.: 490 lb/ft3 = 3.40 lb/in.2/ft Round rod w = 3.40A = 3.40(0.7854d2) I

 d4

and I  0.72w

64

 d4

 0.72(3.40)(0.7854d 2 ) 64  Req'd d  6.26 in.

Prob. 15.46 Note M/I diagram. 

A  1 1 (10)(12)(960)   EIM   30,000,000 [  2

 15(12)(342.86)  12 (15)(12)(514.28)  12 (15)(12)(771.43)]  0.00783 rad  0.00783(57.3)  0.45



Max. deflection: 

  1  2 2 2 1  [ 30,000,000 2 (960)(10) (12) ( 3) EI    (15)(342.86)(12)2 (17.5)  12 (514.28)(15)(12)2 (20) 

y

AM x







AB

 12 (771.43)(15)(12)2 (20)]

=  1.511 in. A = yAB = 1.511 in.

15-381

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Prob.

15.382

1

yBA  2

(2160)(92 )(12)2  12 (2880)(6)(4)(12)2 30,000,000

= 0.586 in.

0.586  0.00326 rad   tan  15(12) AM 12 m(240m)(12)    0.00326 30,000,000 EI  m  8.24 ft y AD 

1 2

(8.24)(12)(240)(8.24)(2 3 )(8.24)(12) 30,000,000

= 0.215 in. D = yAD = 0.215 in.

15-382

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Prob.

15.383

I1 = 15.2 in.4, I2 = 7.23 in.4

yDA 

1 [394.7(15)(17.5)(12)2 30,000,000

 829.9(5)(7.5)(12)2  12 (829.9)(5)(12)(2 3 )(5)(12)]  0.680 in.

D = yDA = 0.680 in. (to the right) Prob. 15.49

15-383

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y AB

 1 [ 1 L(PL)(2 )L  1 (L )( PL )(2 ( L )  L )] 2 3 2 2 2 3 2 2 EI 

7PL3 16EI 7PL3

 A  yAB 

16EI

Prob. 15.50

( PL3 )( L3)2 (2 3)  ( PL3 )( L6)( 512L) EI 3 23PL   E 648EI 1

yAE  2

Prob. 15.51 (Refer to Prob. 15.24)

15-384

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yAB 

 12 (6)(30)(23)(6)(1728) 622,000  EI EI

(a) E = 1700 ksi, Ix = 951 in.4  A   y AB 

622 ,000  0.385 in.

1700 (951)

(b) E = 30,000 ksi, Ix = 118 in.4  A   y AB 

622 ,000  0.1757 in.

30,000 (118 )

(c) E = 30,000 ksi, Ix = 106 in.4  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(d2/4) = 2.67 lb/ft I 

 d4 64

y AB 

 0.0491 in.4

 13 (12.5)(208 .6)( 34)(12.5)(1728 ) 30,000 ,000 (0.0491 )

= 9.56 in. A = yAB = 9.56 in.

15-385

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Prob. 15.53 Pipe 102 mm - Std., Ix = 3.00106 mm4

 C  y AC 

AM x EI

   1   207  103 (3.00  106 )    6 1 [ 2 (2000)(12  10 )(1333)  13 (1000)(3  106 )(1750)]  22.9 mm

Prob. 15.54 (Refer to Prob. 15.31)

I = 879103 mm4 3 6 A y  M  2 10 (2.110 )(1000) AC EI 70103 (879103 )  68.3 mm

15-386

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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) M  P   3.71k 4 4 20 M  s b(all)Sx 

At midspan: 

PL3 3.71(20)3 (1728)    0.864 in.

48EI 48(1300)(951)

At quarter points: Px (3L2  4x 2 ) 48EI   3.71(5) (3(20)2  4(5)2 )(1728)   48(1300)(951) 



  0.594 in.



Prob. 15.56  d4 I  3.98 in.4 64 PL3 500(20)31728   max   1.206 in.

48EI 48(30,000,000)(3.98)

Prob. 15.57

15-387

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Prob. 15.58

Prob. 15.59

15-388

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Max. deflection: point D: yBA 

[12 (2100)(14)(10.67)  12 (2100)(6)2 (23)]1728 30,000,000(3.976)

=2.637 in.

  tan  

AM



2.637  0.01099 rad 20(12) 1

(150m)(m)(144)  0.01099 30,000,000(3.976) 2

EI  m  11.017 ft

y 

11.017

D

(2.637)  1.4526 in.

20 (150)(11.017)2 ( 13 )(11.017)(1728) yDA  30,000,000(3.976  0.4843in.  D  yD  yDA  0.968 in.

1/ 2

Defl. @ midspan: (point E) yE = 2.637/2 = 1.3185 in. yEA  2 1

(1500)(10)2 ( 1 3)(1728)  0.3623in.

30,000,000(3.976)

E = yE  yEA = 1.31850.3623 = 0.956 in.

Defl. under load (point C) yC = (14/20)(2.637)=1.8459 in.

1 (2100)(14)2 ( 1 )(1728) 2 3  0.9938 in. yCA  30,000,000(3.976) C  yC  yCA  1.8459  0.9938  0.852 in.

15-389

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Prob. 15.60 W27  114, Ix = 4080 in.4

Slope at free end: [ 12 (200)(20)  13 (100)(10)  20(10)]122 30,000(4080)  0.005086 rad  0.005086(57.3)  0.291

  

Deflection at free end: yAB   [  30,000(4080)   2 (20) 2 (200)( 2 3) 1 1

   13 (10)2 (100)(3 4) 100(10)(15)  12 (200)(10)(16.67)]1728  0.857 in. A = yAB = 0.859 in.

Deflection at point C:  1 1 yCB  [ 2 (100)(10)2 (2 3)  30,000(4080)  100(10)(5) 100(10)(5)  12 (200)(10)2 (2 3 )]1728  0.282 in. C = yCB = 0.283 in.

15-390

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Prob. 15.61 [6  10(S4S), Hem-fir, Ix = 393 in.4]

 1  1 yBA   [ 2 (8)2 (17,000)(13 ) 1,400,000(393)   13 (12,800)(8)2 ( 14)  12 (6000)(2)2 ( 13)]1728  0.3425 in. yC = (6/8)(0.3425) = 0.2569 in. [ 1 (12,750)(6)2 ( 13)  13 (7200)(6)2 ( 1 4)]1728 yCA  2 1,400,000(393)



= 0.1724 in. C = yC  yCA = 0.25690.1724 = 0.0845 in. yE = (4/5)(0.3425) = 0.1713 in. yEA 

2 1 [ 1 2 (8500)(4) ( 3 )  13 (3200)(4)2 ( 1 4)]1728  1,400,000(393)

yEA = 0.0578 in. E = yE  yEA = 0.1713  0.0578 = 0.1135 in.

15-391

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Prob. 15.62 E = 207103 MPa, Ix = 861106mm4

yAB  

 1  207 103(861106 )    9 1 [ 2 (2.97 10 )(12103)2 ( 13)  13 (2.88109 )(12 103)2 ( 14)

 12 (90106 )(3103)2 ( 13 )]  205.3 mm

At point C:  1 yCB    207103 (861106 )    [ 12 (2.228109 )(9 103 )2 ( 13 )  13 (1.62109 )(9 103 )2 ( 14)]  107.4 mm yC = (9/12)(205.3) = 154.0 mm C = yC  yCB = 46.6 mm (more)

15-392

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At midspan (point D):  1  yDB   207 103 (861106 )    9 3 2 1 1 [ 2 (1.48510 )(6 10 ) ( 3 )  13 (720106 )(6 103 )2 ( 14)  37.87 mm yD = (6/12)(205.3) = 102.7 mm D = yD  yDB = 64.8 mm Prob. 15.63 3 diam. solid steel shaft  (3)4 I   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)  max 

wL4 8EI 24.0L4 (1728)

0.2  8(30,000,000)(3.976)  L  8.24 ft



Prob. 15.64 W10  33: Ix = 170 in.4 (a)  at free end: 

wL4 800(10)4 (1728)   0.339 in.

8EI 8(30,000,000)(170)

(b) If  = 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

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Prob. 15.65

Assuming P =0: D 

PL3 10(12)3 (1728)   0.0526in.

48EI 48(30,000)(394)

Total  due to P loads: D = 0.0526 + 0.05 = 0.1026 in.

For one P load: Pax 2 2 (L  x ) (AISC9/e case#26) 6EIL ( where a  4 ft, x  6 ft, L  12 ft) 0.1026  P(4)(6)(1728)  (122  62 ) 6(30,000)(394)(12) 2  P  9.75 k  D 

Prob. 15.66 W36  194: Ix = 12,100 in.4

15-394

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[ yBA    30,000(12,100) 1  2 (600)(30) 2 (1 3) 1



 2 1 1  3 (900)(30) ( 4)]1728  0.1071in.

yD = 0.1071/2 = 0.05355 in. (more) [ yDA    30,000(12,100) 1  2 (300)(15) 2 (1 3) 1



 2  13 (225)(15) ( 14)](1728)  0.03347 in.

D =yD  yDA = 0.5355  0.03347 = 0.0201 in. (down) yC = (40/30)(0.1071) = 0.1428 in.  1  1 yCA   [ 2 (800)(40)2 ( 13 )  30,000(12,100)   13 (1600)(40)2 ( 14)  12 (800)(10)2 ( 13 )]1728  0.06347 in. C = yC  yCA = 0.0793 in. (down) Prob. 15.67

16-395

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 1  1 [ 2 (0.45)(12)(16) y BA  30,000    0.288(12)(6)  12 (0.288)(12)(4)  12 (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

C = yC  yCA = 0.0187 in. (down) Deflection at point D: yD = (36/24)(0.1410) = 0.2115 in.  1  1 [ 2 (0.45)(12)(28) y DA  30,000    0.288(24)(12)  12 (0.576)(24)(8)  12 (3.456)(24)(8)  12 (2.592)(12)(4)]144  0.2675 in.

Negative sign indicates that deflected point D lies below the tangent line from point A. Therefore: D = yD + yDA = 0.2115 + 0.2675 = 0.479 in. (down) (Note that the elastic curve diagram is incorrect.)

Prob. 15.68

15-396

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

y

  

AB

 1 

[ 1 PL3 (2 )   2 3 EI1 2  2 ( 3 L)(2 3 PL)(7 9 L)

5PL3  12 (L / 3)(PL/ 3)(89 L)]   9EI 5PL3    yAB  A 9EI

Prob. 15.69



 wb2  b(a  34 b)    2  yAB  EI 1 3



wb3

(a  34 b)

6



EI 



wb3

(L  b / 4)

6 EI 3

wb (4L  b) 24EI wb3 (4L  b)  A  yAB  24EI 

15-397

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Prob. 16.1

Prob. 16.2 (Neglect beam wt.)

16-398

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Prob. 16.3

Prob. 16.4

16-399

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Prob. 16.5

16-400

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Prob. 16.6

Prob. 16.7

16-401

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(more)

Prob. 16.8

16-402

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Prob. 16.9

Prob. 16.10

16-403

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Prob. 16.11

16-404

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Prob. 16.12

16-405

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Prob. 16.13

16-406

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Prob. 16.14

16-407

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16-408

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Prob. 16.15

16-409

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Prob. 16.16

16-410

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Prob. 16.17

Prob. 16.18 See problem 16.10

16-411

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Prob. 16.19

16-412

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Prob. 16.20

16-413

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Prob. 16.21

16-414

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Prob. 16.22

16-415

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Prob. 16.23

16-416

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Prob. 16.24

16-417

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Prob. 16.25

16-418

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Prob. 16.26

Prob. 16.27

16-419

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Prob. 16.28

16-420

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Prob. 16.29

16-421

stumerit@gmail.com

Prob. 16.30

Prob. 16.31

16-422

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Prob. 16.32

Prob. 16.33

16-423

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Prob. 16.34

16-424

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Prob. 16.35

16-425

stumerit@gmail.com

Prob. 16.36

Prob. 16.37

Prob. 16.41

16-426

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16-427

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Prob.

17.428

17-428

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Prob.

17.429

Prob. 17.3

17-429

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Prob. 17.4

17-430

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Prob. 17.5

Prob. 17.6

17-431

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Prob. 17.7

Prob. 17.8

17-432

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Prob. 17.9

Prob. 17.10

17-433

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Prob. 17.11

Prob. 17.12

17-434

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17-435

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Prob. 17.13

Prob. 17.14

Prob. 17.15

17-436

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Prob. 17.16

Prob. 17.17

Prob. 17.18

17-437

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Prob. 17.19

Prob. 17.20

Prob. 17.21

17-438

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Prob. 17.22

Prob. 17.23

Prob. 17.24

17-439

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Prob. 17.25

Prob. 17.26

Prob. 17.27

17-440

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Prob. 17.28

Prob. 17.29

17-441

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Prob. 17.30

17-442

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Prob. 17.31

Prob. 17.32

17-443

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Prob. 17.33

Prob. 17.34

17-444

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(b)

17-445

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(c)

17-446

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Prob. 17.35 (a)

17-447

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17-448

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(b)

17-449

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Prob. 17.36

17-450

stumerit@gmail.com

17-451

stumerit@gmail.com

Prob.

17.452

17-452

stumerit@gmail.com

Prob.

17.453

Prob. 17.39

17-453

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Prob. 17.40

17-454

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17-455

stumerit@gmail.com

Prob. 17.41

17-456

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Prob. 17.42

17-457

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Prob.

17.458

17-458

stumerit@gmail.com

Prob.

17.459

17-459

stumerit@gmail.com

Prob.

17.460

17-460

stumerit@gmail.com

Prob. 17.46

17-461

stumerit@gmail.com

17-462

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(b)

17-463

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(c)

17-464

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Prob. 17.47

17-465

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(b)

17-466

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Prob. 17.48

17-467

stumerit@gmail.com

17-468

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Prob. 17.49

17-469

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Prob. 17.50

17-470

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(b)

17-471

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Prob. 17.51

17-472

stumerit@gmail.com

17-473

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Prob. 18.1  2E fe  2 L r

L = 12 ft = 144 in.

 

K = 1.0, A = 6.49 in.2, ry = 1.04 in.

L/r = 144/1.04 = 138.5  2 30,000  15.44 ksi  48 ksi (O.K.) fe  138 .52 Pe = feA = 15.44(6.49) = 100.2 k Prob. 18.2 (a) L = 50 ft  2 (30,000) fe   50(12) 2  7.109 ksi 34 ksi (O.K.)     2.94  Pe  7.109(8.40)  59.7 k

(b) L = 35 ft  2 (30,000) fe   35(12) 2  14.5 ksi 34 ksi(O.K.)     2.94  Pe  14.5(8.40)  121.8k

(c) L = 20 ft  2 (30, 000)

2  44.4 ksi  34 ksi  20(12)  2.94    Euler not applicable.

fe 

(d) L = 15 ft 15 ft < 20 ft (part c) therefore, Euler not applicable.

18-474

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Prob. 18.3 E = 69.0×103 MPa, P.L. = 220 MPa  2E max fe  2  220 MPa  L    r  2 2 3 L min   E   (6910 )  55.6 220 220 r

Prob. 18.4 Steel bar - 1”×2”, A = 2.00 in.2 weak axis r 

fe 

 2E

L r  2

1 12

 0.2887 in.

 min L  r

 2E fe

min L  0.2887

 2 (30,000,000)  28.7 in.

30,000

For 5-ft long bar:  2 (30,000,000)  fe   5(12) 2   0.2887     6855 psi 30,000psi (O.K.) 

Pe = 6855(2) = 13,710 lb Prob. 18.5 E = 207×103 MPa,  2 (207103 ) fe  L r 2 L/r 0 50 100

P.L. = 331 MPa,

fe (MPa) – 817 204 18-475

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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.

2 2 Fe   E2  (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   E2  (10,000) 2  L   60   r 0.375       3.855 ksi  32 ksi (O.K.) Pe  3.855(1.767)  6.81k 2

2

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(c) L = 96 in.  2 E  2 (10,000) fe   L 2   96 2     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.

2 2 fe   E2  (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.  2E  2 Er2 fe  L 2  L    se    r   L 

 2 (30,000)(1.94)2  152.4in. 12.70ft 48

Prob. 18.9 d = 25 mm, A = 490.9 mm2, r = 0.25(25) = 6.25 mm - steel (a) L = 1 m fe 

 2 E  2 (207103 ) 2 2   1000  L 6.25    r    79.81 MPa  331MPa (O.K.)

Pe  79.81(490.9)  39.18103 N  39.2 kN

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(b) L = 2 m  2 E  2 (207103 )  2 2  fe   2000  L   6.25  r    19.95 MPa  331MPa (O.K.) 

Pe  19.95(490.9)  9.79103 N  9.79 kN (c) L = 4 m  2 E  2 (207103 )  2 2  fe   4000  L   6.25  r    4.99 MPa  331MPa (O.K.) 

Pe  4.99(490.9)  2.45103 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

 2 E  2 (70103 ) fe  2  2  1000  L 6.25    r    27.0 MPa  220 MPa (O.K.) Pe  27.0(490.9)  13.25103 N  13.25 kN (b) L = 2 m  2 E  2 (70103 )  2 2  fe   2000   L   6.25  r    6.75MPa  220 MPa (O.K.) 

Pe  6.75(490.9)  3.31103 N  3.31kN

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(c) L = 4 m  2 E  2 (70103 )  2 2 fe   4000   L   6.25  r    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 r 0.848  2 (30,000)  7.54 ksi  48 ksi (O.K.) fe  (198.1)2 f A 7.54(6.48) Pa  e   24.4 k F.S. 2

Prob. 18.12 P.L. = 40 ksi  2E fe  2  40 ksi KL r

 

 2E L min  40  86.04 r 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

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wProb. 18.13 P.L.= 30,000 psi A = 2(1) – 0.8(1.8) = 0.560 in.2

2(1)3 1.8(0.8)3  0.4006 in.

12(0.56)  2 (10,000,000) fe   6874 psi  30,000psi 2 1.0(48)  0.4006    fe A 6874(0.56) Pa    1925 lb F.S. 2.0 rx 

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  fe F.S. fe  

 2E

 2 (207103 )  1.0(7000) 2  111.9 MPa   KL 2   r  51.8 

 

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

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Prob. 18.15 Try Eqn. 18.1 req'd I 

Pe (KL)2

2

 2.5(350)(3412)  492 in.4 2 2  (30,000)  E

Try W14×132: A = 38.8 in.2, ry = 3.76 in., Iy = 548 in.4 Check Euler applic.:  2 (30,000) fe   3412 2  25.1 ksi  48 ksi (O.K.)    3.76  Check: 25.1(38.8) P   389 k  350 k (O.K.) a 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 3  1(16) 4 I y  51.4  (2)  734 in.

12 I y  734  ry   3.99 in.

A 46.1 KL 1.0(22)(12) max   66.2 r 3.99

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Prob. 18.17 2” diam. standard weight pipe, fixed-pinned, L = 84 in., P.L. = 34 ksi

 2E  2 (30,000) fe  KL 2   0.8(84) 2      r   0.787   40.6 ksi  34 ksi Euler formula is not applicable.

Prob. 18.18 P(KL)2 (F.S.)  req'd I   2E 20(1.0 3012)2 (2.0)    17.51in.4  2 (30,000) 

Try W8×24: Iy = 18.3 in.4, ry = 1.61 in.,



Check Euler applic.:  2 (30,000) fe   1.0  3012 2  5.92 ksi  48 ksi (O.K.)    1.61  Use W8×24 Prob. 18.19 W14×90: A = 26.5 in.2, ry = 3.70 in.

KL

 4.71 E  4.71 30,000  115.4 Fy r 50 KL 1.0(16)(12)   51.9  115.4 (short/int.) r 3.70

Trans.

Use Eqn. 18.8:  2E  2 (30,000) fe    109.9 KL r 2  51.92 

 50  109.9    

0.658  (50) Pa  (26.5)  656 k 1.67

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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 KL

 4.71

Transition. r



207 103  115.4 345

 (a) K = 1.0, L = 6 m  2 (207103 )  2E KL 1.0(6000)    90.9 115.4 Use (18- 8) fe  KL 2   247 MPa r 66.0 90.62      r  





 345  247



0.658  

2

345 N/mm Pa  (14,600 mm2 ) 1.67  1680 kN 1300 kN < 1680 kN (O.K.)



  

(b) K = 1.0, L = 9 m KL 1.0(9000)   2E  2 (207103 )   136 115.4 Use (18- 7) fe  KL 2   109.8 MPa r 66.0 1362      r  



Pa = 0.525(109.8)(14,600) = 842 kN 1300 kN > 842 kN (N.G.)

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Prob. 18.22 Steel pipe: A = 11.9 in.2, r = 3.67 in Fy = 35 ksi KL

 4.71 E  4.71 30,000  137.9 35 Fy r KL 1.0(20)(12)   65.4  137.9 (short/int.) r 3.67

Trans.

Use Eqn. 18.8:  2E  2 (30,000) fe    69.2 KL r 2  65.42 Pa 



 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  4   200.3 in.

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



200.3  2.744 in.

26.6

KL

 4.71 30,000  136 36 r KL 1.0(32)(12)   140.1 136 Use (18 - 7) r 2.74

Transition

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 2E  2 (30,000) fe  KL 2  140.12  15.09 ksi      r  Pa  0.525(15.09)(26.6)  211k

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 mm2 2 80 N/mm

Try 102 mm diam. pipe: A = 2040 mm2, r = 38.3 mm KL Trans.

 4.71

207103  138.3 240

r KL 1.0(5500)   143.6 (use Eqn.18 - 7) r 38.3  2E  2 (207103 )  99.1 MPa fe   143.62  KL 2    r  Pa  0.525(99.1)(2040)

 106 kN 100 kN

(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 360 req'd A   21.6 in.2 16.7 Try W14×74: A = 21.8 in.2, ry = 2.48 in.

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KL

 4.71 30,000  115.4 50 r KL 1.0(24)(12)   116.1 (useEqn.18 - 7) r 2.48

Trans.

 2 (30,000)  22.0 ksi 116.12 Pa  0.525(22.0)(21.8)  252 k  360 k (N.G.) fe 

Need a larger shape.

Try W14×90: A =26.5 in.2, ry = 3.70 in.

KL 1.0(24)(12) KL   77.8  115.4 Trans.  4.71 E  4.71 30,000  115.4 r 3.70 r Fy 50 Use Eqn. 18.8:  2E  2 (30,000) fe    48.9 KL r 2  77.82 Pa 

 50  48.9  

(50) 0.658  (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 r

K

 4.71 L

207103  115.4 345 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

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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 UseEqn.(18- 8) r 3.76  2E fe   KL 2  72.7 ksi    r  Pa 

 50  72.7  

50 0.658  (38.8)  871k 1.67 

Prob. 18.28 (A36 steel)

ΣA = 38.8 + 2(14.70)(0.5) = 53.5 in.2  0.5(14.70)3  4   1794.7 in.

I xx  1530  2 12   Iyy = 551 + 2(0.5)(14.70)(7.60)2 = 1400 in.4 y-y axis controls ry  Trans.

1400  5.12 in.

53.50 KL

 4.71 E  4.71 30,000  136.0 Fy 36 r

KL 1.0(19)(12)   44.6  136.0 (short/int.) r 5.12

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Use Eqn. 18.8:  2E  2 (30,000) fe  KL 2   148.7 ksi 452      r   Pa 

36   148.7 

36 0.658  (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.

KL

 4.71 E  4.71 (30,000)  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  2E fe  KL 2  42.6 ksi      r  Pa 

 50  42.6  

(50) 0.658  (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  2E fe   KL 2  41.6 ksi    r 

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 50   41.6  

0.658  (50) (15.8)  286 k 1.67 

P  a

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.

KL r

 4.71 E  4.71 (30,000)  137.9 35 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  2E fe  KL 2 38.8 ksi      r  

 35  38.5 

0.658  (35) (7.83)  112 k Pa  1.67

112 k > 90 k (O.K.) Use Pipe 6 X-strong

Prob. 18.31 (Euler-Johnson) h

least r  trans.

12

KL

 

r



1  0.2887 in.

12 

 2 2 E  2 2 (30,000)  118.7 KL 1.0(20)   69.3 42 r 0.2887 Fy

69.3 < 118.7, use the J. B. Johnson formula:

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 42(69.3)2  1  4 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.

Fa 

 F KL r 2  1 y F y 2 4 E   F.S. 2  1.0(20)     142   0.25d  42 2  4 (30,000)     





3.813  16.8  d2

2.5

F 

Pa

a

A from which:

3.813

6.00 d 0.7854d 2 d  0.8256 in.

 16.8 

2



Try 7/8” diameter rod.

Check : KL  1.0(20)   91.4 0.25(0.875 ) r trans.

 



KL  2 2 E 2 2 (30,000)   118.7  42 r Fy

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

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Fa  

 2E  2 (30,000) 2 2   KL  F.S. 1.0(54)  3.5 





   r 

   0.25d  2  1.8132d

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.

 

 KL  2 2 E 2 2 (30,000)   118.7  r 42 Fy

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  

 2E  2 (30,000) 2 2   KL  F.S. 1.0(96)  3.0 



   r 



   0.25d  2  0.6693d

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

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trans.

KL



r

2 2 E  2 2 (30,000)  83.5 85 Fy

153.6 > 83.5 (O.K.) Use 2½” diam. rod

Prob. 18.35 Assume rod to be intermediate category. r = 0.25d Pa  As a

50 1 24  2   0.25d    50 1  2  4 (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

2 2 E  2 2 (30,000)  108.8 85.3 < 108.8 (O.K.) 50 sY

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

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trans.

KL

2 2 E  2 2 (207103 )  118.7 290 sY



r

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 4 2 (207103 ) 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  4.60 k  2d 2   (30,000)  2  1(16)   0.2887d  3.0     Solve for d: Req’d d = 0.517 in., 2d = 1.035 in. (more) trans.

KL r





2 2 E sY





2 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

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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.) d 7.5 0.3E 0.3(1.7 106 )   L 2   1.897  e  162 (1050)   Fc  d  2

C P 

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 106 )     0.657 27.22 (1050) C P 

1 0.657 1.6

2

1 0.657  0.657    0.534   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

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Le



2440

 10.12  50 (O.K.)

d

241 0.3(7.6 103 )  0.3E    L 2  10.122 (5.00)  4.45  e  Fc  d  

2

1 4.45   1 4.45  4.45    0.949 C P    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 103 )     1.541 17.202 (5.00) C P 

11.541 1.6

2

11.541 1.541    0.816   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: 90,000 req'd A   85.7 in.2 1050 Try 10×10 (S4S): A = 90.3 in.2, d = 9.5 in.

C P 

1 0.943 1.6

2

1 0.943  0.943    0.670   0.8  1.6 

Fa = FcCP = 1050(0.670) = 704 psi 90,000 req'd A  704  127.8 in.2 90.3 in.2 (N.G.)

18-495

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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 106 )    1.377 18.782 (1050) C P 

11.377 1.6

2

11.377  1.377    0.789   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: 48,000 req'd A   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.) d 7.5 0.3(1.7 106 )    0.623 25.62 (1250) CP 

1 0.623 1.6

2

1 0.623  0.623    0.514   0.8  1.6 

Fa = FcCP = 1250(0.514) = 643 psi 48,000 req'd A  643  74.7 in.2  56.3 in.2 (N.G.)

18-496

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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.) d 9.5 0.3(1.7 106 )    1.00 20.22 (1250) CP 

11.00 1.6

2

11.00  1.00    0.691    1.6  0.8

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: 48,000 req'd A   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.) d 7.5 0.3(1.7 106 )    0.973 20.482 (1250)

18-497

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2

CP 

1 0.973   1 0.973  0.973      0.681 0.8 1.6  1.6 

Fa = FcCP = 1250(0.681) =851psi 48,000 req'd A  851  56.4 in.2  56.3 in.2 (Say O.K.) Use 8×8 (S4S) (b) P = 60,000 lb Assume an 8” nominal thickness from part (a) and Fa = 851 psi: 60,000 req'd A   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



(O.K.)

0.3(1700103 )   1.858 (16.17)2 (1050)

CP 

11.858 1.6

2

11.858  1.858    0.854   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

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Le

20(12)



d

 20  50

(O.K.)

12



0.3(1700103 )   1.02 202 (1250)

CP 

11.02 1.6

2

11.02  1.02    0.698    1.6  0.8

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: 62,000 req'd A   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 106 )     0.943 22.72 (1050) C P 

1 0.943 1.6

2

1 0.943  0.943    0.670   0.8  1.6 

Fa = FcCP = 1050(0.670) = 704 psi 62,000 req'd A  704  88.1in.2  90.3 in.2 (O.K.) Use 10×10 (S4S)

Prob. 18.46 18-499

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Southern pine: E = 1700 ksi Fc = 1250 psi K = 1.0, L = 20 ft 100(2000) P  50,000 lb 4 Assume Fa = Fc: 50,000 req'd A   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.) d 7.5 0.3(1.7 106 )    0.398 322 (1250) CP 

1 0.398 1.6

2

1 0.398  0.398    0.358   0.8  1.6 

Fa = FcCP = 1250(0.358) = 448 psi 50,000 req'd A  448  111.6 in.2  56.3 in.2 (N.G.) Try 10×10 (S4S): A = 90.3 in.2, d = 9.5 in.

Le 1.0(20)(12)   25.3  50 (O.K.) d 9.5 0.3(1.7 106 )    0.637 25.32 (1250) CP 

1 0.637 1.6

2

1 0.637  0.637    0.523   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

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Fc = 8.62 MPa K = 1.0, L = 6 m, P = 450 kN Assume Fa = Fc: 450 10 3 3 2 req'd A   52.2 10 mm 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

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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

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Prob. 19.4

Prob. 19.5

19-503

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Prob. 19.6

Prob. 19.7

19-504

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Prob. 19.8

Prob. 19.9

19-505

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Prob. 19.10

Prob. 19.11

19-506

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Prob. 19.12

Prob. 19.13

19-507

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Prob. 19.14

Prob. 19.15

19-508

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Prob. 19.16 A36 Steel

Prob. 19.17 A36 steel

19-509

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Prob. 19.18

Prob. 19.19

Prob. 19.20

19-510

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Prob. 19.21

Prob. 19.22

19-511

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Prob. 19.23

Prob. 19.24

Prob. 19.25

19-512

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Prob. 20.1

Prob. 20.2

Prob. 20.3

Prob. 20.4

20-513

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Prob. 20.5

Prob. 20.6

Prob. 20.7

Prob. 20.8

20-514

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Prob. 20.9

Prob. 20.10

Prob. 20.11

20-515

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Prob. 20.12

Prob. 20.13

Prob. 20.14

20-516

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Prob. 20.15

Prob. 20.16

Prob. 20.17

20-517

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Prob. 20.18

Prob. 20.19

Prob. 20.20

Prob. 20.21

20-518

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Prob. 20.22

Prob. 20.23

20-519

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Prob. 20.22

20-520

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