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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 s 50 mm 50 mm 100 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 3 3 6 4 1 1 (100)(150) 28.125 10 mm 12 12     I bh 6 4 28.125 10 m   2 (100)(50) 5000 mm   A 1 50 mm y 3 3 6 3 1 250 10 mm 250 10 m      Q Ay 6 3 6 (1500)(250 10 ) 13.3333 10 N/m 28.125 10         VQ q I nail 2  qs F nail 3 3 2 (2)(400) 60.0 10 m 13.3333 10       F s q 60.0 mm  s

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50 mm s s 50 mm 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 3 3 6 4 1 1 (100)(150) 28.125 10 mm 12 12     I bh 6 4 28.125 10 m   2 (100)(50) 5000 mm   A 1 50 mm y 3 3 6 3 1 250 10 mm 250 10 m      Q AyVQ q I qs 2Fnail Eliminating ,q VQ 2Fnail Is Solving for V, 6 3 nail 6 3 2 (2)(28.125 10 )(400) 2.00 10 N (250 10 )(45 10 )           IF V Qs 2.00 kN  V ◄

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

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.

SOLUTION 3 2 1 3 2 4 3 3 4 2 4 3 1 4 1 2 3 1 12 1 (6)(2) (6)(2)(3) 112 in 12 1 1 (2)(4) 10.6667 in 12 12 112 in 234.67 in I bh Ad I bh I I I I I I               3 1 1 nail (6)(2)(3) 36 in (1) (2) Q A y qs F VQ q I      Dividing Eq. (2) by Eq. (1),

nail 1 VQ sF I nail (150)(234.67) (36)(3) F I V Qs   V 326 lb 

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s s s 120 mm 80 mm 20 mm 20 mm

PROBLEM 6.4

A square box beam is made of two 20 80-mm planks and two 20 120-mm planks nailed together as shown. Knowing that the spacing between the nails is s30 mm and that the vertical shear in the beam is V 1200 N, determine (a) the shearing force in each nail, (b) the maximum shearing stress in the beam.

SOLUTION 3 3 2 2 1 1 3 3 6 4 6 4 1 1 12 12 1 1 (120)(120) (80)(80) 13.8667 10 mm 12 12 13.8667 10 m I b h b h          (a) 2 1 1 3 3 6 3 1 1 1 (120)(20) 2400 mm 50 mm 120 10 mm 120 10 m A y Q A y          6 3 6 nail (1200)(120 10 ) 10.3846 10 N/m 13.8667 10 2          VQ q I qs F 3 3 nail (10.3846 10 )(30 10 ) 2 2    qsF Fnail155.8 N  (b) 1 3 3 3 3 6 3 (2)(20)(40)(20) 120 10 32 10 152 10 mm 152 10 m Q Q            6 max 6 3 (1200)(152 10 ) (13.8667 10 )(2 20 10 ) VQ It         3 329 10 Pa   max 329 kPa 

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897 16 ⫻ 200 mm

S310 ⫻ 52

PROBLEM 6.5

The American Standard rolled-steel beam shown has been reinforced by attaching to it two 16 200-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:

2 2 6 4 6 4 * * (mm) (mm ) (10 mm ) (10 mm ) Part Top plate 3200 160.5 82.43 0.07 S310 52 6650 0 95.3 Bot. plate 3200 160.5 82.43 0.07 164.86 95.44 d A Ad I   2 6 4 6 4 3 3 6 3 plate plate 2 3 2 6 2 bolt bolt 6 6 3

bolt all bolt

* 305 16 160.5 mm 2 2 260.3 10 mm 260.3 10 m (3200)(160.5) 513.6 10 mm 513.6 10 m (18 10 ) 254.47 10 m 4 4 (90 10 )(254.47 10 ) 22.90 10 N d I Ad I Q A d A d F A qs                                     3 3 bolt bolt 3 6 3 3 6 2 (2)(22.90 10 ) 2 381.7 10 N/m 120 10 (260.3 10 )(381.7 10 ) 193.5 10 N 513.6 10 F F q s VQ Iq q V I Q                   193.5 kN V  

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C12 ⫻ 20.7 16 in. ⫻ in. C z y 1 2

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.

SOLUTION

4 C12 20.7: d12.00 in., Ix 129 in For top plate,

3 2 4 12.00 1 1 6.25 in. 2 2 2 1 1 1 (16) (16) (6.25) 312.667 in 12 2 2 t y I                       

For bottom plate, 312.667 in4

b

I  Moment of inertia of fabricated beam:

4 3 plate plate bolt 2 2 2 bolt bolt (2)(129) 312.667 312.667 883.33 in 1 (16) (6.25) 50 in 2 (25)(50) 1.41510 kips/in. 883.33 1 1 (1.41510)(7.5) 5.3066 kips 2 2 3 ( ) 0.44179 in 4 4 4 I Q A y VQ q I F qs A d                                     bolt bolt bolt 5.3066 12.01 ksi 0.44179 F A    bolt 12.01 ksi 

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899 S10 ⫻ 25.4 C8 ⫻ 13.7 C z y

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.

SOLUTION Geometry: 1 ( ) 2 10.0 0.303 5.303 in. 2 0.534 in. 5.303 0.534 4.769 in. w C s d f t x y f x                 

Determine moment of inertia.

2 (in.) 2 4 4

(in ) (in ) (in )

Part C8 13.7 4.04 4.769 91.88 1.52 S10 25.4 7.45 0 0 123 C8 13.7 4.04 4.769 91.88 1.52 183.76 126.04 A d Ad I     2 4 3 1 bolt 2 2 2 bolt bolt 183.76 126.04 309.8 in (4.04)(4.769) 19.2668 in (30)(19.2668) 1.86573 kip/in. 309.8 1 1 (1.86573)(5) 4.6643 kips 2 2 3 0.44179 in 4 4 4 I Ad I Q Ay VQ q I F qs A d                                 bolt bolt bolt 4.6643 10.56 ksi 0.44179  F   A bolt10.56 ksi 

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PROBLEM 6.8

The composite beam shown is fabricated by connecting two W6 20 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.

SOLUTION 2 4 W6 20: 5.87 in , 6.20 in., 41.4 in 1 3.1 in. 2 x A d I y d       Composite: 2 4 3 2[41.4 (5.87)(3.1) ] 195.621 in (5.87)(3.1) 18.197 in I Q Ay       Bolts: all 2 2 bolt

bolt all bolt bolt 5

in., 10.5 ksi, 6 in.

8 5 0.30680 in 4 8 (10.5)(0.30680) 3.2214 kips 2 (2)(3.2214) 1.07380 kips/in. 6 d s A F A F q s                     Shear: (195.621)(1.0780) 18.197  VQIqq V I Q 11.54 V  kips 

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

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. SOLUTION 0: 2.3 (1.5)(72) 0 46.957 kN B M A        A At section n-n, 46.957 V  A kN

Calculate moment of inertia:

3 3 3 6 4 6 4 1 1 1 2 (15)(40) 2 (15)(80) (30)(120 ) 12 12 12 5.76 10 mm 5.76 10 m I                At a, 3 3 6 3 3 6 6 6 30 mm 0.030 m (30 20)(50) 30 10 mm 30 10 m (46.957 10 )(30 10 ) (5.76 10 )(0.030) 8.15 10 Pa = 8.15 MPa a a a a a t Q VQ It                    At b, 3 3 3 3 6 4 3 6 6 6 60 mm 0.060 m (60 20)(30) 30 10 36 10 66 10 mm 66 10 m (46.957 10 )(66 10 ) 8.97 10 Pa 8.97 MPa (5.76 10 )(0.060) b b a b b b t Q Q VQ It                          At NA, NA 3 3 3 3 6 3 NA 3 6 6 NA NA 6 NA 90 mm 0.090 m (90 20)(10) 66 10 18 10 84 10 mm 84 10 m (46.957 10 )(84 10 ) 7.61 10 Pa 7.61 MPa (5.76 10 )(0.090) b t Q Q VQ It                         

(a) max occurs at b. max 8.97 MPa 

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1.5 m 100 mm 200 mm 40 mm 12 mm 12 mm150 mm 0.3 m 10 kN n a n

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. SOLUTION At section n-n, V 10 kN. 1 2 3 3 2 1 1 2 2 2 2 3 3 2 6 6 6 6 4 6 4 4 1 1 4 12 12 1 1 (100)(150) 4 (50)(12) (50)(12)(69) 12 12 28.125 10 4 0.0072 10 2.8566 10 39.58 10 mm 39.58 10 m I I I b h b h A d                               (a) 1 1 2 2 3 3 6 3 2 (100)(75)(37.5) (2)(50)(12)(69) 364.05 10 mm 364.05 10 m 100 mm = 0.100 m Q A y A y t           3 6 3 max 6 (10 10 )(364.05 10 ) 920 10 Pa (39.58 10 )(0.100) VQ It           max  920 kPa  (b) 1 1 2 2 3 3 6 3 2 (100)(40)(55) (2)(50)(12)(69) 302.8 10 mm 302.8 10 m 100 mm 0.100 m Q A y A y t           

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903 3 in. 3 in. 3 in. 8 in. 25 in. 18 in. a n n t t t = 0.25 in. t 25 kips

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

3 3 3

1 1 2

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

12 12 12    I 4 113.8 in  I = 25 kips V

(25 kips)(18 in.) = 450 kip in.

 

M

(a) m: 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.)

    Q 3 3 3 3 4.5 in 8.203 in 2.258 in 14.96 in 0.25 in.      Q t 3 4 (25 kips)(14.96 in ) (113.8 in )(0.25 in.)   Qm V Itm 13.15 ksi ◄

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PROBLEM 6.11 (Continued)

(b) a: At point a, t = 0.25 in. See sketch above.

3 3 3 4.5 in 8.203 in 12.70 in    a Q 3 4 (25 kips)(12.70 in ) (113.8 in )(0.25 in.)   aa VQ Ita 11.16 ksi ◄

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905 8 in.

16 in. 12 in. 16 in.

4 in. 4 in. n 10 kips 10 kips n a in. 1 2 in. 1 2

PROBLEM 6.12

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 By symmetry, RARB. 0: 10 10 0 10 kips y A B A B F R R R R        

From the shear diagram, V 10 kips at - .n n

3 3 2 2 1 1 3 3 4 1 1 12 12 1 1 (4)(4) (3)(3) 14.5833 in 12 12      I b h b h (a) 1 1 2 2 (3) 1 (1.75) (2) 1 (2)(1) 4.625 in3 2 2 QA yA y              1 1 1 in. 2 2 t    max (10)(4.625) (14.5833)(1)  VQIt max 3.17 ksi  (b) (4) 1 (1.75) 3.5 in3 2 1 1 1 in. 2 2 Q Ay t             (10)(3.5) (14.5833)(1)   VQIt 2.40 a  ksi 

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Dimensions in mm 40 40

40

10 10

30

PROBLEM 6.13

For a beam having the cross section shown, determine the largest allowable vertical shear if the shearing stress is not to exceed 60 MPa.

SOLUTION

Calculate moment of inertia.

3 3 6 4 6 4 6 4 1 1 2 (10 mm)(120 mm) (30 mm)(40 mm) 12 12 2[1.440 10 mm ] 0.160 10 mm 3.04 10 mm I           6 4 3.04 10 m I   

Assume that m occurs at point a.

3 3 6 3 10 mm 0.01 m (10 mm 40 mm)(40 mm) 16 10 mm 16 10 m t Q Q         

For all60 MPa, m all VQ

It    6 3 6 6 4 (16 10 m ) 60 10 Pa (3.04 10 m )(0.01m) V       114.0 kNV  

Check  at neutral axis:

3 3 6 3 50 mm 0.05 m 2[(10 60)(30)] (30 20)(10) 42 10 m 42 10 m t Q           

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907 10 10 10 30 10 Dimensions in mm 30 30 40

PROBLEM 6.14

For a beam having the cross section shown, determine the largest allowable vertical shear if the shearing stress is not to exceed 60 MPa.

SOLUTION

Calculate moment of inertia.

3 4 2 6 4 6 4 6 4 6 4 1 1 (50 mm)(120 mm) 2 (30 mm) (30 mm 30 mm)(35 mm) 12 12 7.2 10 mm 2[1.170 10 mm ] 4.86 10 mm 4.86 10 m I I                 

Assume that m occurs at point a.

3 3 6 3 2(10 mm) 0.02 m (10 mm 50 mm)(55 mm) 2[(10 mm 30 mm)(35 mm)] 48.5 10 mm 48.5 10 m t Q           

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PROBLEM 6.14 (Continued)

For all60 MPa, m all VQ

It    6 3 6 6 4 (48.5 10 m ) 60 10 Pa (4.86 10 m )(0.02 m) V       120.3 kNV  

Check  at neutral axis: t 50 mm 0.05 m

3 3 6 3 6 4 (50 60)(30) (30 30)(35) 58.5 10 mm (120.3 kN)(58.5 10 m ) 29.0 MPa 60 MPa OK (4.86 10 m )(0.05 m) Q VQ It               

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909 1.5 in. 2 in. 2 in. 1.5 in. 4 in. w = 2.5 in.

PROBLEM 6.15

For a timber beam having the cross section shown, determine the largest allowable vertical shear if the shearing stress is not to exceed 150 psi.

SOLUTION 3 3 1 (1.5 8 2(0.5)(4) ) 12    I 4 69.333 in  I all 150 psi  

At point a: Q(1.5 in.)(2 in.)(3 in.) 9 in ;3 t 1.5 in. ; mVQ It 3 4 (9 in ) 150 psi ; (69.333 in )(1.5 in.) VV 1733 lb  At neutral axis: 3 (1.5 in.)(2 in.)(3 in.) (2.5 in.)(2 in.)(1in.) 14 in

   Q , t  2.6 in. ; mVQ It 3 4 (14 in ) 150 psi ; (69.333 in )(2.5 in.)  V V 1857 lb 

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220 mm 12 mm 12 mm 252 mm W250 3 58

PROBLEM 6.16

Two steel plates of 12  220-mm rectangular cross section are welded to the W250  58 beam as shown. Determine the largest allowable vertical shear if the shearing stress in the beam is not to exceed 90 MPa.

SOLUTION

Calculate moment of inertia.

Part A(mm )2 d(mm) Ad2(10 mm )6 4 (10 mm ) I 6 4 Top plate 2640 *132 45.999 0.032 W250 × 58 7420 0 0 87.3 Bot. plate 2640 *132 45.999 0.032  91.999 87.364 252 12 * 2 2 d   2 179.363 10 mm6 4 179.363 10 m6 4 I  Ad   I     max

 occurs at neutral axis. t 8.0 mm8.0 10 m 3

Part A(mm )2 y(mm) Ay(10 mm )3 3

Top plate 2640 132 348.48

Top flange 2740.5 119.25 326.805

Half web 900 56.25 50.625  725.91 Dimensions in mm:

12220 ;

13.5203 ;

8.0 112.5

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911

A

(a) (b)

B A B

PROBLEM 6.17

Two W8  31 rolled-steel sections may be welded at A and B in either of the two ways shown in order to form a composite beam. Knowing that for each weld the allowable shearing force is 3000 lb per inch of weld, determine for each arrangement the maximum allowable vertical shear in the composite beam.

SOLUTION 2 9.12 in  A W8 31 Ix 110 in4 4 37.1 in  y I (a) I 2[IxAd2] 4 2 2 2[110 in (9.12 in )(4 in.) ]   4 511.84 in  I 2 3 (9.12 in )(4 in.) 36.4 in    Q Ay

For two welds each with allowable shearing force of 3 kips/in.,

2(3 kips/in.) 6 kips/in.   q ;  VQ q I 3 4 (36.4 in ) 6 kips/in. (511.84 in )  V V 84.2 kips  (b) 2 2[ ]  yI I Ad 4 2 2 2[37.1 in (9.12 in )(4 in.) ]   4 366.04 in  I 2 3 (9.12 in )(4 in.) 36.4 in    Q Ay

6 kips/in. (same as in part )  q a ;  VQ q I 3 4 (36.4 in ) 6 kips/in. (366.04 in )  V V  60.2 kips 

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2.4 kN 4.8 kN 7.2 kN 1 m 1 m 1 m 0.5 m 150 mm A E b B C D

PROBLEM 6.18

For the beam and loading shown, determine the minimum required width b, knowing that for the

grade of timber used, all 12 MPa and

all 825 kPa.   SOLUTION 0: 3 (2)(2.4) (1)(4.8) (0.5)(7.2) 0 2 kN D M A          A

Draw the shear and bending moment diagrams.

3 max 3 max | | 7.2 kN 7.2 10 N | | 3.6 kN m 3.6 10 N m V M         Bending: M S   max min 3 6 6 3 3 3 | | 3.6 10 12 10 300 10 m 300 10 mm M S          

For a rectangular section, 1 2

6 Sbh 3 2 2 6 (6)(300 10 ) 80 mm (150) S b h    

Shear: Maximum shearing stress occurs at the neutral axis of bending for a rectangular section. 2 3 2 1 1 1 1 , , 2 4 8 1 12 ( )       A bh y h Q Ay bh I bh t b V bh

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913 B b h C L/2 L/2 A P

PROBLEM 6.19

A timber beam AB of length L and rectangular cross section carries a single concentrated load P at its midpoint C. (a) Show that the ratio

/

m m

  of the maximum values of the shearing and normal stresses in the beam is equal to 2 / ,h L where h and L are, respectively, the depth and the length of the beam. (b) Determine the depth h and the width b of the beam, knowing that L 2 m, 40 kN,P  960 kPa,m  and

12 MPa. m   SOLUTION Reactions: RARBP/2 (1) max 2 A P VR

(2) Abh for rectangular section.

(3) 3 max 3

2 4

m

V P

A bh

   for rectangular section.

(4) max 4 PL M  (5) 1 2 6

Sbh for rectangular section.

(6) max 2 3 2 m M PL S bh    (a) 2 m m h L     (b) Solving for h, 3 3 6 2 (2)(2)(960 10 ) 320 10 m 12 10 m m L h          320 mmh  

Solving Eq. (3) for b,

3 3 3 3 3 (3)(40 10 ) 97.7 10 m 4 m (4)(320 10 )(960 10 ) P b h          97.7 mmb  

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B b h A C D w L/2 L/4 L/4

PROBLEM 6.20

A timber beam AB of length L and rectangular cross section carries a uniformly distributed load w and is supported as shown. (a) Show that the ratio  m/ m of the maximum values of the shearing and normal stresses in the beam is equal to 2 / ,h L where h and L are, respectively, the depth and the length of the beam. (b) Determine the depth h and the width b of the beam, knowing that L 5 m, w kN/m, 8

1.08 MPa, m   and m 12 MPa. SOLUTION 2 A B wL RR

From shear diagram, | |

4

m

wL

V  (1)

For rectangular section, Abh (2)

3 3 2 8 m m V wL A bh    (3)

From bending moment diagram, | | 2

32

m

wL

M  (4)

For a rectangular cross section,

2 1 6 Sbh (5) 2 2 | | 3 16 m m M wL S bh    (6)

(a) Dividing Eq. (3) by Eq. (6), m 2

m h L     (b) Solving for h, 6 3 6 (5)(1.08 10 ) 225 10 m 2 (2)(12 10 ) m m L h          225 mmh  

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915 B b a A 10 in. 20 in. 20 in. 25 kips 25 kips n 7.25 in. in. 1.5 in. 1.5 in. 3 4 8 in. in. 3 4 in. 3 4 n

PROBLEM 6.21

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

SOLUTION

25 kipsRARB

At section n-n, 25 kipsV

Locate centroid and calculate moment of inertia.

Part A(in )2 y(in.) Ay(in )3 d(in.) Ad2(in )4 I(in )4

 4.875 6.875 33.52 2.244 24.55 0.23  10.875 3.625 39.42 1.006 11.01 47.68  15.75 72.94 35.56 47.91 2 4 72.94 4.631 in. 15.75 35.56 47.91 83.47 in Ay Y A I Ad I             (a) 3 (1.5)(4.631 0.75) 4.366 in3 4 3 0.75 in. 4 a Q Ay t            (25)(4.366) (83.47)(0.75) a VQ It    1.744 ksia   (b) 3 (3)(4.631 1.5) 7.045 in3 4 0.75 in. b Q Ay t           (25)(7.045) (83.47)(0.75) b VQ It    2.81 ksib 

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180 kN 500 mm 500 mm 100 mm 160 mm 30 mm 30 mm 30 mm 20 mm 20 mm A B b a n n

PROBLEM 6.22

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

SOLUTION

Draw the shear diagram. | |Vmax90 kN

Part A(mm )2 y(mm) Ay(10 mm )3 3 d(mm) Ad2(10 mm )6 4 I(10 mm )6 4  3200 90 288 25 2.000 0.1067  1600 40 64 25 1.000 0.8533  1600 40 64 25 1.000 0.8533  6400 416 4.000 1.8133 3 2 6 4 6 4 6 4 416 10 65 mm 6400 (4.000 1.8133) 10 mm 5.8133 10 mm 5.8133 10 m Ay Y A I Ad I                   (a) A(80)(20) 1600 mm2 3 3 6 3 25 mm 40 10 mm 40 10 m a y Q Ay        3 6 6 6 3 (90 10 )(40 10 ) 31.0 10 Pa (5.8133 10 )(20 10 ) a a VQ It             a 31.0 MPa  (b) 2 3 3 6 3 (30)(20) 600 mm 65 15 50 mm 30 10 mm 30 10 m b A y Q Ay           

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917 B b a A 10 in. 20 in. 20 in. 25 kips 25 kips n 7.25 in. in. 1.5 in. 1.5 in. 3 4 8 in. in. 3 4 in. 3 4 n

PROBLEM 6.23

For the beam and loading shown, determine the largest shearing stress in section n-n.

SOLUTION

25 kipsRARB

At section n-n, V 25 kips

Locate centroid and calculate moment of inertia.

Part A(in )2 y(in.) Ay(in )3 d(in.) Ad2(in )4 I(in )4

 4.875 6.875 33.52 2.244 24.55 0.23  10.875 3.625 39.42 1.006 11.01 47.68  15.75 72.94 35.56 47.91 2 4 72.94 4.631 in. 15.75 35.56 47.91 83.47 in Ay Y A I Ad I            

Largest shearing stress occurs on section through centroid of entire cross section.

3 3 4.631 (4.631) 8.042 in 4 2 3 0.75 in. 4 (25)(8.042) (83.47)(0.75) Q Ay t VQ It                  3.21 ksim 

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180 kN 500 mm 500 mm 100 mm 160 mm 30 mm 30 mm 30 mm 20 mm 20 mm A B b a n n

PROBLEM 6.24

For the beam and loading shown, determine the largest shearing stress in section n-n.

SOLUTION

Draw the shear diagram. | |Vmax90 kN

Part A(mm )2 x Ay(10 mm )3 3 d(mm) Ad2(10 mm )6 4 I(10 mm )6 4  3200 90 288 25 2.000 0.1067  1600 40 64 25 1.000 0.8533  1600 40 64 25 1.000 0.8533  6400 416 4.000 1.8133 3 2 6 4 6 4 6 4 416 10 65 mm 6400 (4.000 1.8133) 10 mm 5.8133 10 mm 5.8133 10 m Ay Y A I Ad I                   Part A(mm )2 y(mm) Ay(10 mm )3 3  3200 25 80  300 7.5 2.25  300 7.5 2.25  84.5 3 3 6 3 3 3 6 max 6 3 84.5 10 mm 84.5 10 m (2)(20) 40 mm 40 10 m (90 10 )(84.5 10 ) (5.8133 10 )(40 10 ) Q Ay t VQ It                      

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919

c

PROBLEM 6.25

A beam having the cross section shown is subjected to a vertical shear V. Determine (a) the horizontal line along which the shearing stress is maximum, (b) the constant k in the following expression for the maximum shearing stress

max

V k

A

 

where A is the cross-sectional area of the beam.

SOLUTION 4 and 2 4 I  c Ac For semicircle, 2 4 2 3 s c Ac y    2 4 2 3 2 3 3 s c Q A yc c     

(a) maxoccurs at center where t 2c 

(b) 3 2 3 max 4 2 4 4 4 3 2 3 V c VQ V V Itc c c A         4 1.333 3 k   

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rm

tm

PROBLEM 6.26

A beam having the cross section shown is subjected to a vertical shear V. Determine (a) the horizontal line along which the shearing stress is maximum, (b) the constant k in the following expression for the maximum shearing stress

max

V k

A

 

where A is the cross-sectional area of the beam.

SOLUTION

For a thin-walled circular section, A 2r tm m

2 2 3 , 1 3

2

m m m m m

JAr  r t IJ r t

For a semicircular arc, y 2rm

  2 2 2 s m m m s m m m m A r t r Q A y r t r t       

(a) 2ttm at neutral axis where maximum occurs. 

(b) 2 max 3 (2 ) 2 ( )(2 ) m m m m m m m VQ V r t V V It r t t r t A        2.00k  

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921

h

h

b

PROBLEM 6.27

A beam having the cross section shown is subjected to a vertical shear V. Determine (a) the horizontal line along which the shearing stress is maximum, (b) the constant k in the following expression for the maximum shearing stress

max

V k

A

 

where A is the cross-sectional area of the beam.

SOLUTION 3 3 1 1 1 2 2 2 12 6 Abhbh Ibhbh    

For a cut at location y, where y h ,

2 2 3 1 ( ) 2 2 2 ( ) 3 ( ) 2 3 ( ) by by A y y h h y y h y by by Q y Ay h by t y h             2 2 3 3 6 ( ) 3 2 2 3 VQ h by by V y y y V It bh by h bh h h                       

(a) To find location of maximum of , set d 0.

dy 2[3 4 m] 0 d V y dy bh h 3 , 4 m y h  i.e., 1 4h

 from neutral axis. 

(b) 2 3 3 9 ( ) 3 2 1.125 4 4 8 m V V V y bh bh A                     1.125k  

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b

h

PROBLEM 6.28

A beam having the cross section shown is subjected to a vertical shear V. Determine (a) the horizontal line along which the shearing stress is maximum, (b) the constant k in the following expression for the maximum shearing stress

max

V k

A

 

where A is the cross-sectional area of the beam.

SOLUTION

3

1 1

2 36

Abh Ibh

For a cut at location y,

2 2 1 ( ) 2 2 2 2 ( ) 3 3 ( ) ( ) 3 ( ) by by A y y h h y y h y by Q y Ay h y by t y h             2 2 3 3 3 3 1 36 ( ) 12 ( ) 12 ( ) ( ) ( ) by by h V h y VQ Vy h y V y hy y It bh bh bh        

(a) To find location of maximum of , set d 0.

dy 3 12 ( 2 m) 0 d V h y dy bh 1 , 2 m yh i.e., at mid-height  (b) 2 2 2 3 3 12 12 1 1 3 3 ( ) 2 2 2 m m m V V V V hy y h h bh A bh bh                3 1.500 2 k   

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923 2 in. 2 in. 2 in. 10 in. s s s

PROBLEM 6.29

The built-up timber beam shown is subjected to a vertical shear of 1200 lb. Knowing that the allowable shearing force in the nails is 75 lb, determine the largest permissible spacing s of the nails.

SOLUTION 3 2 1 1 1 1 1 1 12   I b h A d 3 2 4 1 (2)(2) (2)(2)(4) 65.333 in 12    3 3 4 2 2 2 1 1 (2)(10) 166.67 in 12 12    I b h 4 1 2 4 428 in    I I I 3 1 1 1 (2)(2)(4) 16 in     Q Q A y (1200)(16) 44.86 lb/in. 428 VQ q I    nail  F qs nail 75 1.672 in. 44.86 F s q    

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20 mm 100 mm 20 mm 50 mm 50 mm 150 mm

PROBLEM 6.30

The built-up beam shown is made by gluing together two 20  250-mm plywood strips and two 50  100-mm planks. Knowing that the allowable average shearing stress in the glued joints is 350 kPa, determine the largest permissible vertical shear in the beam.

SOLUTION 3 3 6 4 1 1 (140)(250) (100)(150) 154.167 10 mm 12 12     I 6 4 154.167 10 m   3 3 (100)(50)(100) 500 10 mm     Q Ay 6 3 500 10 m   3 50 mm 50 mm 100 mm 100 10 m      t   VQ It 6 3 3 6 (154.167 10 )(100 10 )(350 10 ) 500 10           It V Q 3 10.79 10 N   V 10.79 kN 

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925 0.8 0.8 0.8 0.8 B A 1.5 1.5 4 3.2 Dimensions in inches

PROBLEM 6.31

The built-up beam was made by gluing together several wooden planks. Knowing that the beam is subjected to a 1200-lb vertical shear, determine the average shearing stress in the glued joint (a) at A, (b) at B.

SOLUTION 3 3 2 1 1 2 (0.8)(4.8) (7)(0.8) (7)(0.8)(2.0) 12 12        I 4 60.143 in  (a) Aa (1.5)(0.8) 1.2 in2 ya  2.0 in. 3 2.4 in   a a a Q A y 0.8 in.  a t (1200)(2.4) (60.143)(0.8)   aa a VQ It 59.9 psia   (b) (4)(0.8) 3.2 in2 b A yb  2.0 in. 3 (3.2)(2.0) 6.4 in    b b b Q A y (2)(0.8) 1.6 in.   b t (1200)(6.4) (60.143)(1.6)   bb b VQ It  b 79.8 psi 

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A B 60 mm 20 mm 20 mm 20 mm 20 mm 20 mm 30 mm 30 mm

PROBLEM 6.32

Several wooden planks are glued together to form the box beam shown. Knowing that the beam is subjected to a vertical shear of 3 kN, determine the average shearing stress in the glued joint (a) at A, (b) at B.

SOLUTION 3 2 3 2 6 4 3 3 6 4 3 3 6 4 6 4 6 4 1 1 (60)(20) (60)(20)(50) 12 12 3.04 10 mm 1 1 (60)(20) 0.04 10 mm 12 12 1 1 (20)(120) 2.88 10 mm 12 12 2 2 11.88 10 mm 11.88 10 m A B C A B C I bh Ad I bh I bh I I I I                       3 3 6 3 3 (60)(20)(50) 60 10 mm 60 10 m 20 mm 20 mm 40 mm 40 10 m A Q Ay t              (a) 3 6 3 6 3 (3 10 )(60 10 ) 379 10 Pa (11.88 10 )(40 10 ) A A VQ It             0 B Q  A 379 kPa  (b) B 0 B VQ It    B   0

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927 300 100 200 400 50 50 50 50 B B A x A A A C Dimensions in mm

PROBLEM 6.33

The built-up wooden beam shown is subjected to a vertical shear of 8 kN. Knowing that the nails are spaced longitudinally every 60 mm at A and every 25 mm at B, determine the shearing force in the nails (a) at A, (b) at B. (Given: 1.504 10 mm .)Ix   9 4 SOLUTION 9 4 6 4 1.504 10 mm 1504 10 m 60 mm 0.060 m 25 mm 0.025 m x A B I s s          (a) 1 1 1 3 3 6 3 (50)(100)(150) 750 10 mm 750 10 m A Q Q A y         3 6 1 6 (8 10 )(750 10 )(0.060) 1504 10 A A A A F q s VQ s I         FA  239 N  (b) 2 2 2 3 3 3 3 1 2 6 3 (300)(50)(175) 2625 10 mm 2 4125 10 mm 4125 10 m B Q A y Q Q Q            3 6 6 (8 10 )(4125 10 )(0.025) 1504 10 B B B B B VQ s F q s I         549 NFB  

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A C

105 mm

PROBLEM 6.34

Knowing that a W360  122 rolled-steel beam is subjected to a 250-kN vertical shear, determine the shearing stress (a) at point A, (b) at the centroid C of the section.

SOLUTION For W360 122, d  363 mm, bF 257 mm, tF  21.70 mm, tw 13.0 mm 6 4 6 4 367 10 mm 367 10 m     I (a) (105)(21.70) 2278.5 mm2 a A 363 21.70 170.65 mm 2 2 2 2   F    a d t y 3 3 6 3 388.8 10 mm 388.8 10 m      a a a Q A y 3 21.70 mm 21.7 10 m     a F t t 3 6 6 6 3 (250 10 )(388.8 10 ) 12.21 10 Pa (367 10 )(21.7 10 )           a a a VQ Ita 12.21 MPa  (b) 2 1  F F  (257)(21.70) 5577 mm A b t 1 363 21.70 170.65 mm 2 2 2 2  dtF    y 2 2 w 2 F (13.0)(159.8) 2077 mm d Att     2 1 79.9 mm 2 2      Fd y t 3 3 (5577)(170.65) (2077)(79.9) 1117.7 10 mm       c Q Ay 6 3 1117.7 10 m   3 13.0 mm 13 10 m     c w t t 3 6 6 6 3 (250 10 )(1117.7 10 ) 58.6 10 Pa (367 10 )(13 10 )          c c c VQ It 58.6 MPac  

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929 b 12 12 40 80 150 Dimensions in mm 6 6 a

PROBLEM 6.35

An extruded aluminum beam has the cross section shown. Knowing that the vertical shear in the beam is 150 kN, determine the shearing stress at (a) point a, (b) point b.

SOLUTION 3 3 1 1 (150)(80) (126)(68) 12 12   I 6 4 6 4 3.098 10 mm 3.0985 10 m     (a) QaA y1 12A y 2 2 (126)(6)(37) (2)(12)(40)(20)   3 3 6 3 47.172 10 mm 47.172 10 m     (2)(12) 24 mm 0.024 m    a t 3 6 6 6 (150 10 )(47.172 10 ) 95.2 10 Pa (3.0985 10 )(0.024)          a a a VQ It 95.2 MPa a    (b) 1 1 3 3 6 3 (126)(6)(37) 27.972 10 mm 27.972 10 m       b Q A y (2)(6) 12 mm 0.012 m    b t 3 6 6 6 (150 10 )(27.972 10 ) 112.8 10 Pa (3.0985 10 )(0.012)          b b b VQ It 112.8 MPa b  

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b 12 12 40 80 80 Dimensions in mm 6 6 a

PROBLEM 6.36

An extruded aluminum beam has the cross section shown. Knowing that the vertical shear in the beam is 150 kN, determine the shearing stress at (a) point a, (b) point b.

SOLUTION 3 3 6 4 6 4 1 1 (80)(80) (56)(68) 1.9460 10 mm 12 12 1.946 10 m I        (a) 1 1 2 2 3 3 6 3 2 (56)(6)(37) (2)(12)(40)(20) 31.632 10 mm 31.632 10 m (2)(12) 24 mm 0.024 m a a Q A y A y t             3 6 6 6 (150 10 )(31.632 10 ) 101.6 10 Pa (1.946 10 )(0.024) a a a VQ It          a101.6 MPa  (b) 1 1 3 3 6 3 (56)(6)(37) 12.432 10 mm 12.432 10 m b Q A y        3 6 6 6 (2)(6) 12 mm 0.012 m (150 10 )(12.432 10 ) 79.9 10 Pa (1.946 10 )(0.012) b b b b t VQ It              b79.9 MPa 

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931 40 30 30 40 10 10 160 120 50 50 20 20 c b a Dimensions in mm

PROBLEM 6.37

Knowing that a given vertical shear V causes a maximum shearing stress of 75 MPa in an extruded beam having the cross section shown, determine the shearing stress at the three points indicated.

SOLUTION VQ It   τ is proportional to Q/t. Point c: 3 3 2 (30)(10)(75) 22.5 10 mm 10 mm / 2250 mm c c c c Q t Q t      Point b: 3 3 2 (20)(50)(55) 77.5 10 mm 20 mm / 3875 mm b c b b b Q Q t Q t       Point a: 3 3 2 2 (120)(30)(15) 209 10 mm 120 mm / 1741.67 mm a b a a a Q Q t Q t       ( / )Q t m occurs at b. 75 m b  MPa / / / a b c a a b b c c Q t Q t Q t  2 2 2 75 MPa 1741.67 mm 3875 mm 2250 mm a a  33.7 MPa a    75.0 MPa b    43.5 MPa c   

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b d d 8 in. 0.5 in. 0.5 in. 5 in. 4 in. a

PROBLEM 6.38

The vertical shear is 1200 lb in a beam having the cross section shown. Knowing that d  4 in., determine the shearing stress at (a) point a, (b) point b.

SOLUTION 3 2 4 1 1 (4)(0.5) (4)(0.5)(3.75) 28.167 in 12    I 3 4 2 1 (5)(4) 106.67 in 3   I 4 1 2 4 2 326 in    I I I (a) Qa  2A y1 1A y 2 2 3 (2)(4)(0.5)(3.75) (5)(4)(2) 55 in    5 in.  a t (1200)(55) 40.5 psi (326)(5)   a   a a VQ It(b) 4 1 1 (4)(0.5)(3.75) 7.5 in    b Q A y 0.5 in.  b t (1200)(7.5) 55.2 psi (326)(0.5)   b   b b VQ It  

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933 b d d 8 in. 0.5 in. 0.5 in. 5 in. 4 in. a

PROBLEM 6.39

The vertical shear is 1200 lb in a beam having the cross section shown. Determine (a) the distance d for which a  b, (b) the corresponding shearing stress at points a and b.

SOLUTION 2 1 0.5 in , Ad y1 3.75 in., tb  0.5 in. 2 2 (5)(4)  20 in , A y2  2 in., ta 5 in. 3 1 1 1.875 in   b Q A y d 1.875 3.75 0.5   b   b b VQ V d Vd It I I 2 2 2 (20)(2) (2)(1.875 )     a b Q A y Q d 40 3.75d 5 in.ta(a) (40 3.75 ) 8 0.75 3.75 (5)   a       a b a VQ V d V Vd Vd It I I I I 80.75d 3.75d 8 2.6667 in. 3   d 2.67 in  d . (b) 1 1 (2.6667)(0.5)3 (2.6667)(0.5)(3.75)2 18.780 in4 12    I 3 4 2 1 (0.5)(4) 106.667 in 3   I 4 1 2 4 2 288.46 in    I I I (3.75)(1200)(2.6667) 3.75 288.46 a bVdI 41.6 psia  

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1.25 in. 1.25 in. 1.25 in. 1.25 in. b c a e d

PROBLEM 6.40

The extruded aluminum beam has a uniform wall thickness of 1

8in. Knowing that the vertical shear in the beam is 2 kips, determine the corresponding shearing stress at each of the five points indicated.

SOLUTION 3 3 4 1 1 (2.50)(2.50) (2.125)(2.25) 1.23812 in 12 12    I  0.125 in. 

t at all sections.

2 kips  V  0 a Qa a VQ It   a0  3 1.25 (0.125)(1.25) 0.097656 in 2       b Q  (2)(0.097656) (1.23812)(0.125)   bb VQ It b 1.262 ksi  2 (1.0625)(0.125)(1.1875) 0.25537 in. c b QQ    (2)(0.25537) (1.23812)(0.125)   cc VQ It  3.30 ksic   2 2 (0.125) (1.1875) 0.52929 d c QQ    (2)(0.52929) (1.23812)(0.125)   dd VQ It d 6.84 ksi  1.125 (0.125)(1.125) 0.60839 2 e d QQ      (2)(0.60839) (1.23812)(0.125) eVQIte 7.86 ksi 

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935 1.25 in. 1.25 in. 1.25 in. 1.25 in. b c a e d

PROBLEM 6.41

The extruded aluminum beam has a uniform wall thickness of 1

8in. Knowing that the vertical shear in the beam is 2 kips, determine the corresponding shearing stress at each of the five points indicated.

SOLUTION 3 3 4 1 1 (2.50)(2.50) (2.125)(2.25) 1.23812 in 12 12    I

Add symmetric points c’, b’, and a’. 0 e Q   3 1.125 (0.125)(1.125) 0.079102 in 2       d Q 0.125 in.td   2 4 (0.125) (1.1875) 0.097657 in    c d Q Q 0.25 in.tc  3 (2)(1.0625)(0.125)(1.1875) 0.41309 in    b c Q Q 0.25 in.tb  3 1.25 (2)(0.125)(1.25) 0.60840 in 2        a b Q Q 0.25 in.ta  (2)(0.60840) (1.23812)(0.25)   aa a VQ It 3.93 ksia   (2)(0.41309) (1.23812)(0.25)   bb b VQ It 2.67 ksib  (2)(0.097657) (1.23812)(0.25)   cc c VQ It 0.631 ksic   (2)(0.079102) (1.23812)(0.125)   dd d VQ It 1.02 ksid   e e e VQ It   e  0

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40 mm 30 mm 50 mm 30 mm 10 mm 10 mm 12 mm 40 mm b c a

PROBLEM 6.42

Knowing that a given vertical shear V causes a maximum shearing stress of 50 MPa in a thin-walled member having the cross section shown, determine the corresponding shearing stress (a) at point a, (b) at point b, (c) at point c.

SOLUTION 3 3 3 3 3 3 3 3 (12)(30)(25 10 15) 18 10 mm (40)(10)(25 5) 12 10 mm 2 (12)(10)(25 5) 45.6 10 mm 25 (12)(25) 49.35 10 mm 2 12 mm 10 mm a b c a b m c a c m b Q Q Q Q Q Q Q t t t t                            50 MPam(a) 18 12 0.36474 49.35 12  mamama    Q t Q t 18.23 MPaa   (b) 12 12 0.29179 49.35 10  mbmbmb    Q t Q t 14.59 MPab  (c) 45.6 12 0.92401 49.35 12  mcmcmc    Q t Q t 46.2 MPac 

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937 2 in. 2 in. 10 in. 10 in. 4 in.

PROBLEM 6.43

Three planks are connected as shown by bolts of 3

8-in. diameter spaced every 6 in. along the longitudinal axis of the beam. For a vertical shear of 2.5 kips, determine the average shearing stress in the bolts.

SOLUTION Locate neutral axis.

2 (2)(2)(10) (10)(4) 80 in  A   3 (2)(2)(10)(5) (10)(4)(8) 520 in Ay    520 6.5 in. 80 Ay Y A      3 2 3 2 4 1 2 (2)(10) (2)(10)(1.5) 12 1 (10)(4) (10)(4)(1.5) 566.67 in 12         I 3 (2)(10)(1.5) 30 in   Q (2.5)(30)(6) 0.79411 kips 566.67   VQs   F qs I 2 2 2 bolt bolt 3 0.110447 in 4 4 8           A d bolt bolt 0.79411 7.19 ksi 0.110447   F   A 

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100 mm 100 mm 50 mm 100 mm 50 mm 25 mm 25 mm

PROBLEM 6.44

A beam consists of three planks connected as shown by steel bolts with a longitudinal spacing of 225 mm. Knowing that the shear in the beam is vertical and equal to 6 kN and that the allowable average shearing stress in each bolt is 60 MPa, determine the smallest permissible bolt diameter that can be used. SOLUTION Part A(mm )2 y(mm) Ay2(10 mm )6 4 I(10 mm )6 4  7500 50 18.75 14.06  7500 50 18.75 14.06  15,000 50 37.50 28.12 Σ 75.00 56.24 2 6 4 6 4 3 3 1 1 6 3

bolt bolt bolt

131.25 10 mm 131.25 10 m (7500)(50) 375 10 mm 375 10 m I Ay I Q A y VQs F A qs I                     3 6 6 2 bolt 6 6 bolt 2 (6 10 )(375 10 )(0.225) 64.286 10 m (6 10 )(131.25 10 ) 64.286 mm VQs A I             bolt bolt 4A (4)(64.286) d     dbolt 9.05 mm 

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939 6 in.

1 in. 1 in.

PROBLEM 6.45

A beam consists of five planks of 1.5 6-in. cross section connected by steel bolts with a longitudinal spacing of 9 in. Knowing that the shear in the beam is vertical and equal to 2000 lb and that the allowable average shearing stress in each bolt is 7500 psi, determine the smallest permissible bolt diameter that can be used.

SOLUTION

Part A(in )2 y0(in.) Ay0(in )3 y(in.) Ay2(in )4 I(in )4

 9 5 45 0.8 5.76 27  9 4 36 0.2 0.36 27  9 3 27 1.2 12.96 27  9 4 36 0.2 0.36 27  9 5 45 0.8 5.76 27  45 189 25.20 135 0 2 4 189 4.2 in. 45 160.2 in Ay Y A I Ad I           Between  and : 3 12  1  1 (9)(0.8) 7.2 in Q Q Ay Between  and : 3 23 1 2 7.2 (9)( 0.2) 5.4 in QQAy     VQ q IMaximum q is based on Q12 7.2 in .3 bolt (2000)(7.2) 89.888 lb/in. 160.2 (89.888)(9) 809 lb      q F qs 2 bolt bolt bolt bolt bolt bolt 2 bolt

bolt bolt bolt

809 0.1079 in 7500 4 (4)(0.1079) 4 F F A A A A d d             dbolt 0.371 in.

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12 mm 400 mm

PROBLEM 6.46

Four L102  102  9.5 steel angle shapes and a 12  400-mm plate are bolted together to form a beam with the cross section shown. The bolts are of 22-mm diameter and are spaced longitudinally every 120 mm. Knowing that the beam is subjected to a vertical shear of 240 kN, determine the average shearing stress in each bolt. SOLUTION For one L102  102  9.5, 2 1845 mm  A 6 2 1.815 10 mm   I 2 (1845 mm )(171 mm)  Q 3 3 315.5 10 mm   6 4 315.5 10 m  

For 12-mm 400-mm plate and four angle,

3 6 4 2 2 1 (12 mm)(400 mm) 4[1.815 10 mm (1845 mm )(171 mm) ] 12     I 6 4 6 4 287.06 10 mm 287.06 10 m     One angle: 6 3 6 4 (240 kN)(315.5 10 m ) 263.78 kN/m 287.06 10 m        VQ q I (263.78 kN/m)(0.120 m) 31.65 kN    F qs Diam. bolt  22 mm 2 31.65 kN ; (0.022 m) 4   F A  83.3 MPa ◄

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941 1.6 in. 2 in. 2 in. 1.2 in. 1.2 in. A B D E F

PROBLEM 6.47

A plate of 1

4-in. thickness is corrugated as shown and then used as a beam. For a vertical shear of 1.2 kips, determine (a) the maximum shearing stress in the section, (b) the shearing stress at point B. Also, sketch the shear flow in the cross section.

SOLUTION

2 2 2

(1.2) (1.6) 2.0 in. (0.25)(2.0) 0.5 in

BD BD

L    A  

Locate neutral axis and compute moment of inertia.

Part A(in )2 y(in.) Ay(in )3 (in.)d Ad2(in )4 I(in )4

AB 0.5 0 0 0.4 0.080 neglect BD 0.5 0.8 0.4 0.4 0.080 *0.1067 DE 0.5 0.8 0.4 0.4 0.080 *0.1067 EF 0.5 0 0 0.4 0.080 neglect Σ 2.0 0.8 0.320 0.2134 2 2 4 2 4 * 1 1 0.8 (0.5)(1.6) 0.1067 in 0.4 in. 12 12 2.0 0.5334 in BD Ay A h Y A I Ad I             (a) 3 3 3 (2)(0.25)(0.4) 0.2 in (0.5)(0.25)(0.2) 0.025 in 0.225 in m AB BC AB BC m Q Q Q Q Q Q        (1.2)(0.225) (0.5334)(0.25) m m VQ It    2.02 m  ksi  (b) QBQAB  0.2 in3 (1.2)(0.2) (0.5334)(0.25) B B VQ It    B 1.800 ksi  0 D   

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d a e b c 50 mm 10 mm 10 mm 22 mm

PROBLEM 6.48

A plate of 2-mm thickness is bent as shown and then used as a beam. For a vertical shear of 5 kN, determine the shearing stress at the five points indicated and sketch the shear flow in the cross section.

SOLUTION 3 3 3 2 3 4 9 4 3 9 3 3 9 3 3 3 1 1 1 2 (2)(48) (2)(52) (20)(2) (20)(2)(25) 12 12 12 133.75 10 mm 133.75 10 mm (2)(24)(12) 576 mm 576 10 mm 0 (12)(2)(25) 600 mm 600 10 m (2)(24)(12) 1.176 10 mm 1.176 1                                    a a c b d c I Q Q Q Q Q Q 6 3 3 9 3 0 m (2)(26)(13) 600 mm 500 10 m          e d Q Q 3 9 6 9 3 (5 10 )(576 10 ) 10.77 10 Pa (133.75 10 )(2 10 ) a a VQ It           a 10.76 MPa  b b VQ It   b   0 3 9 6 9 3 (5 10 )(600 10 ) 11.21 10 Pa (133.75 10 )(2 10 ) c c VQ It           11.21 MPac   3 6 6 9 3 (5 10 )(1.176 10 ) 22.0 10 Pa (133.75 10 )(2 10 ) d d VQ It           22.0 MPad  

References

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