Mechanics of materials 7th edition beer solutions manual by john.hook262

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Mechanics of Materials 7th Edition Beer Solutions Manual

C C H H A A P P T T E E R R 6 6

PROBLEM 6.1

50 mm

50 mm 100 mm

SOLUTION 113364 (100)(150)28.12510mm 1212  I bh 28.12510m64   2 (100)(50)5000mm A 1 50mm  y 3363 1 25010mm25010mQAy 6 3 6 (1500)(25010) 13.333310N/m 28.12510    VQ q I nail 2  qsF 3 nail 3 2(2)(400) 60.010 m 13.333310   F s q 60.0mm  s 

PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.

893

50 mm s

Three full-size 50  100-mm boards are nailed together to form a beam that is subjected to a vertical shear of 1500 N. Knowing that the allowable shearing force in each nail is 400 N, determine the largest longitudinal spacing s that can be used between each pair of nails.

s s

50 mm

100 mm

PROBLEM 6.2

For the built-up beam of Prob. 6.1, determine the allowable shear if the spacing between each pair of nails is s  45 mm.

PROBLEM 6.1 Three full-size 50  100-mm boards are nailed together to form a beam that is subjected to a vertical shear of 1500 N. Knowing that the allowable shearing force in each nail is 400 N, determine the largest longitudinal spacing s that can be used between each pair of nails.

SOLUTION

I bh

113364 (100)(150)28.12510mm 1212

28.12510m64

2 (100)(50)5000mm

1 50mm

3363 1 25010mm25010m

Eliminating ,q nail 2 VQF I

Solving for V,

3 nail 63 2(2)(28.12510)(400) 2.0010N (25010)(4510)

894



 



 y



QAy  VQ q I nail 2  qsF



   IF

2.00kN 

◄

V Qs

PROBLEM 6.3

Three boards, each 2 in. thick, are nailed together to form a beam that is subjected to a vertical shear. Knowing that the allowable shearing force in each nail is 150 lb, determine the allowable shear if the spacing s between the nails is 3 in.

895 2 in. 2 in. 6 in. s s s 2 in. 4 in.

SOLUTION 32 1 324 334 2 4 31 4 123 1 12 1 (6)(2)(6)(2)(3)112 in 12 11 (2)(4)10.6667 in 1212 112 in 234.67 in IbhAd I bh II IIII      3 11 nail (6)(2)(3)36 in (1) (2) QAy qsF VQ q I    

Eq. (2) by Eq. (1), nail 1 VQ s FI  nail (150)(234.67) (36)(3) FI V Qs  326 lb V  

Dividing

s s s

20 mm

80 mm

20 mm

(

120 mm

PROBLEM 6.4

A square box beam is made of two 2080-mm  planks and two 20120-mm  planks nailed together as shown. Knowing that the spacing between the nails is 30 s  mm and that the vertical shear in the beam is 1200 N, V  determine (a) the shearing force in each nail, (b) the maximum shearing stress in the beam.

896

33 2211 3364 64 11 1212 11 1212(120)(120)(80)(80)13.866710mm 13.866710m Ibhbh   

a) 2 1 1 3363 111 (120)(20)2400 mm 50 mm 12010 mm12010m A y QAy    6 3 6 nail (1200)(12010) 10.384610N/m 13.866710 2     VQ q I qsF 33 nail (10.384610)(3010) 22   qs F nail 155.8 N F  

SOLUTION

(

b) 1 3333 63

12010321015210mm

QQ   6 max 63 (1200)(15210) (13.866710)(22010) VQ It     3 32910Pa max 329 kPa   

(2)(20)(40)(20)

15210m

S310

PROBLEM 6.5

The American Standard rolled-steel beam shown has been reinforced by attaching to it two 16200-mm  plates, using 18-mm-diameter bolts spaced longitudinally every 120 mm. Knowing that the average allowable shearing stress in the bolts is 90 MPa, determine the largest permissible vertical shearing force. SOLUTION

Calculate moment of inertia:

200 mm

2 26464 * * (mm) (mm)(10mm)(10mm)

Top plate320082.430.07 160.5 S310526650 095.3 Bot. plate320082.430.07 160.5 164.8695.44 d AAdI   26 46 4 33 63 plateplate 23 26 2 boltbolt 663 boltallbolt * 30516 160.5 mm 22 260.310mm260.310m (3200)(160.5)513.610mm513.610m

44 (9010)(254.4710)22.9010N d IAdI QAd Ad FA qs        3 3 bolt bolt 3 63 3 6 2(2)(22.9010) 2381.710N/m 12010 (260.310)(381.710) 193.510N 513.610 F Fq s qVVQIq IQ       193.5 kN V  

Part

(1810)254.4710m

C12

PROBLEM 6.6

The beam shown is fabricated by connecting two channel shapes and two plates, using bolts of 3 4 -in. diameter spaced longitudinally every 7.5 in. Determine the average shearing stress in the bolts caused by a shearing force of 25 kips parallel to the y axis.

20.7 16 in. in. C z y 1 2

4 C1220.7:12.00 in.,129 in x dI 

top plate, 3 24 12.0011 6.25 in. 222 111 (16)(16)(6.25)312.667 in 1222 t y I        

bottom plate, 4 312.667 in b I  Moment of inertia of fabricated beam: 4 3 plateplate bolt 2 22 boltbolt (2)(129)312.667312.667 883.33 in 1 (16)(6.25)50 in 2 (25)(50) 1.41510 kips/in. 883.33 11 (1.41510)(7.5)5.3066 kips 22 3 ()0.44179 in 444 I QAy VQ q I Fqs Ad                  bolt bolt bolt 5.3066 12.01 ksi 0.44179 F A  bolt 12.01 ksi   

SOLUTION

For

PROBLEM 6.7

A columm is fabricated by connecting the rolled-steel members shown by bolts of 3 4 -in. diameter spaced longitudinally every 5 in. Determine the average shearing stress in the bolts caused by a shearing force of 30 kips parallel to the y axis

Geometry:

Determine moment of inertia.

899

25.4

13.7 C z y

S10

SOLUTION

10.0

      

() 2

0.3035.303in. 2 0.534in. 5.3030.5344.769in. wC s

ft x yfx

2244

(in)(in)

Part C813.74.044.76991.881.52 S1025.47.4500123 C813.74.044.76991.881.52 183.76126.04 AAdI d     24 3 1 bolt 2 22 boltbolt 183.76126.04309.8in (4.04)(4.769)19.2668in

309.8 11

22 3

444 IAdI QAy VQ q I Fqs Ad             bolt bolt bolt 4.6643 10.56ksi 0.44179   F A bolt 10.56ksi   

(in.)

(in)

(30)(19.2668) 1.86573kip/in.

(1.86573)(5)4.6643kips

0.44179in

PROBLEM 6.8

The composite beam shown is fabricated by connecting two W620  rolled-steel members, using bolts of 5 8 -in. diameter spaced longitudinally every 6 in. Knowing that the average allowable shearing stress in the bolts is 10.5 ksi, determine the largest allowable vertical shear in the beam.

900

SOLUTION 24 W620:5.87 in,6.20 in.,41.4 in 1 3.1 in. 2 x AdI yd   Composite: 2 4 3 2[41.4(5.87)(3.1)] 195.621 in (5.87)(3.1)18.197 in I QAy    Bolts: all 2 2 bolt boltallbolt bolt 5 in.,10.5 ksi,6 in. 8 5 0.30680 in 48 (10.5)(0.30680)3.2214

2(2)(3.2214) 1.07380 kips/in. 6 ds A FA F q s          

(195.621)(1.0780) 18.197 qVVQIq IQ 11.54 kips V  

kips

Shear:

PROBLEM 6.9

For the beam and loading shown, consider section n-n and determine (a) the largest shearing stress in that section, (b) the shearing stress at point a

901 90 120 15 1515 15 30 20 20 20 40 20 72 kN n n Dimensions in mm 1.5 m 0.5 m 0.8 m a

SOLUTION 0:2.3(1.5)(72)0 46.957 kN B MA   A At section n-n, 46.957 kN VA Calculate moment of inertia: 333 6464 111 2(15)(40)2(15)(80)(30)(120) 121212 5.7610mm5.7610m I      At a, 33 63 36 6 6 30 mm 0.030 m (3020)(50)3010mm 3010m (46.95710)(3010)

8.1510 Pa = 8.15 MPa a a a a a t Q VQ It        

(5.7610)(0.030)

36 6 6 60 mm0.060 m (6020)(30)301036106610mm6610m (46.95710)(6610) 8.9710Pa8.97 MPa

b ba b b b t QQ VQ It       At NA, NA

NA 36 6 NA NA 6 NA 90 mm0.090 m (9020)(10)661018108410mm8410m (46.95710)(8410) 7.6110Pa7.61 MPa

b t QQ VQ It      

a) max occurs at b. max 8.97 MPa   

b) 8.15 MPa a  

333364

(5.7610)(0.060)

333363

(5.7610)(0.090)

(

PROBLEM 6.10

For the beam and loading shown, consider section n-n and determine (a) the largest shearing stress in that section, (b) the shearing stress at point a.

902 1.5 m 100 mm 200 mm 40 mm 12 mm 12 mm 150 mm 0.3 m 10 kN n a n

SOLUTION

section n-n, 10 kN. V  12 332 11 2222 332 666 64 64 4 11 4 1212 11 (100)(150)4(50)(12)(50)(12)(69) 12 12 28.1251040.0072102.856610 39.5810mm39.5810m III bhbhAd              

a) 1122 33 63 2 (100)(75)(37.5)(2)(50)(12)(69) 364.0510mm364.0510m 100 mm = 0.100 m QAyAy t     36 3 max 6 (1010)(364.0510) 92010Pa (39.5810)(0.100) VQ It     max 920 kPa   

b) 1122 3363 2 (100)(40)(55)(2)(50)(12)(69) 302.810mm302.810m 100 mm0.100 m QAyAy t     36 3 6 (1010)(302.810) 76510Pa (39.5810)(0.100) a VQ It     765 kPa a  

(

PROBLEM 6.11

For the beam and loading shown, consider section n-n and determine (a) the largest shearing stress in that section, (b) the shearing stress at point a

SOLUTION

112333 (8in.)(9in.)(7.25in.)(8.5in.)(0.25in.)(3in.)

121212  I

4 113.8in  I = 25 kips V (25 kips)(18 in.) = 450 kipin  M (a)

: At neutral axis, thickness  0.25 in 2(3 in.0.25 in.)(3 in.)(7.5 in.)(0.25 in.)(4.375 in.)(4.25 in.)(0.25 in.)(2.125 in.)

3333 4.5 in8.203 in2.258 in14.96 in 0.25 in. 

(25kips)(14.96in) (113.8in)(0.25in.)

3 in. 3 in. 3 in. 8 in. 25 in. 18 in. a n n t

0.25 in. t

25 kips







 

◄

Q t

Q m V It 13.15ksi   m

(b)

PROBLEM 6.11 (Continued)

a: At point a, = 0.25 in. t See sketch above.

333 4.5 in8.203 in12.70 in  Qa

  a a VQ It

3 4 (25kips)(12.70in) (113.8in)(0.25in.)

11.16ksi   a ◄

904