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CHAPTER 6 s PROBLEM 6 s For the beam of Prob. 6, determine the allowable shear if the spacing between each pair of nails is s 45 mm. 50 mm 50 mm PROBLEM 6 Three 50 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 ...

Description

PROBLEM 6.84* Solve Prob. 6.83, assuming that a 6-mm-thick plate is bent to form the channel shown.

B 100 mm

PROBLEM 6.83* A steel plate, 160 mm wide and 8 mm thick, is bent to form the channel shown. Knowing that the vertical load P acts at a point in the midplane of the web of the channel, determine (a) the torque T that would cause the channel to twist in the same way that it does under the load P, (b) the maximum shearing stress in the channel caused by the load P.

A

D E P ⫽ 15 kN 30 mm

SOLUTION Use results of Example 6.06 with b 30 mm, h  100 mm, and t  6 mm.

e

30 b    9.6429 mm  9.6429  10 3 m h 100 2  3 b 2  (3)(30)

I

1 2 1  th (6b  h )  (6)(100)2 [(6)(30)  100]  1.400 106 mm4  1.400 10 6 m4 12 12

V  15  10 3 N (a )

T  Ve  (15  10 3 )(9.6429  10 3 )

T  144.6 N  m 

Stress at neutral axis due to V:

Q  bt

h  h  h  1 t   th (h  4b) 2  2   4  8

1  (6)(100) 100  (4)(30)  16.5 103 mm3 16.5 10 6 m 3 8 

t  6 10 3 m

V 

VQ (15 10 3)(16.5 10 6 )   29.46 10 6 Pa  29.46 MPa It (1.400 10 6 )(6 10 6 )

Stress due to T :

a  2b  h  160 mm  0.160 m  6  t  1 1 c1  1  0.630   1  (0.630)     0.32546 3 a 3  160   T 144.64   77.16 10 6 Pa  77.16 MPa V   2 c1a t (0.32546)(0.160)(6 10 3 ) 2

(b )

By superposition,

 max  V   T

 max  106.6 MPa 

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PROBLEM 6.85

B

The cantilever beam AB, consisting of half of a thinwalled pipe of 1.25-in. mean radius and 83-in. wall thickness, is subjected to a 500-lb vertical load. Knowing that the line of action of the load passes through the centroid C of the cross section of the beam, determine (a) the equivalent force-couple system at the shear center of the cross section, (b) the maximum shearing stress in the beam. (Hint: The shear center O of this cross section was shown in Prob. 6.74 to be located twice as far from its vertical diameter as its centroid C.)

1.25 in.

A

A C

a

O t

B 500 lb

e

SOLUTION From the solution to Prob. 6.74, 

I

 

e



 2 4



Q  a 2t sin 

a 3t

Qmax  a2 t

a

For a half-pipe section, the distance from the center of the semi-circle to the centroid is

 

x



2

a



At each section of the beam, the shearing force V is equal to P. Its line of action passes through the centroid C. The moment arm of its moment about the shear center O is

d  e x  (a )

4



a

2



a

2



a

Equivalent force-couple system at O.

V P

MO  Vd 

Data: P  500 lb

a = 1.25 in.

2



Pa

V  500 lb   2 MO    (500)(1.25)  



M O  398 lb  in. 





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PROBLEM 6.85* (Continued)

(b )

Shearing stresses. (1)

Due to V :  V 

V 





(2)

VQ max  It ( P )(a2 t )

  a3 t  ( t) 2   



2P (2)(500)   679 psi   at  (1.25)(0.375)

Due to the torque M O : For a long rectangular section of length l and width t, the shearing stress due to torqueM Ois

M  Data:

t 1 where c1   1  0.630  3 l

l   a   (1.25)  3.927 in. t  0.375 in. c1  0.31328

M  By superposition,

MO c1 lt 2

397.9 (0.31328)(3.927)(0.375) 2

 2300 psi

  V   M  679 psi 2300 psi

  2980 psi 

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PROBLEM 6.86

B A

Solve Prob. 6.85, assuming that the thickness of the beam is reduced to 41 in.

1.25 in.

A C

a

O t

B

500 lb

e

PROBLEM 6.85 The cantilever beam AB, consisting of half of a thin-walled pipe of 1.25-in. mean radius and 83 -in. wall thickness, is subjected to a 500-lb vertical load. Knowing that the line of action of the load passes through the centroid C of the cross section of the beam, determine (a) the equivalent force-couple system at the shear center of the cross section, (b) the maximum shearing stress in the beam. (Hint: The shear center O of this cross section was shown in Prob. 6.74 to be located twice as far from its vertical diameter as its centroid C.)

SOLUTION 

From the solution to Prob. 6.74,



I   a 3t



e



4



a

Q  a 2t sin  Qmax  a2 t

For a half-pipe section, the distance from the center of the semi-circle to the centroid is



x 

2



a

At each section of the beam, the shearing force V is equal to P. Its line of action passes through the centroid C. The moment arm of its moment about the shear center O is

d ex  (a )



a

2



a

2



a

Equivalent force-couple system at O.

V P 

4

M O  Vd 

P  500 lb

Data:



2



Pa

a  1.25 in. V  500 lb  MO  398 lb  in. 



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PROBLEM 6.86* (Continued)

(b)

Shearing stresses. (1)

Due to V ,  V 

V 

(2)

VQ max It ( P)( a2 t) 2P (2)(500)   1019 psi  3   at  (1.25)(0.250) a t t ( ) 2   

Due to the torque M O : For a long rectangular section of length l and width t, the shearing stress due to torqueM Ois

M  Data:

MO c1lt 2

t 1 where c 1=  1  0.630  3 l

l   a   (1.25)  3.927 in. t  0.250 in. c 1  0.31996

M 

397.9  5067 psi (0.31996)(3.927)(0.250)2

By superposition,   V   M  1019 psi  5067 psi

  6090 psi 

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3 kips

PROBLEM 6.87

y

y' A'

B' x'

A' B'

A

C'

22.5⬚

D'

E'

B

12 in.

D' D

E'

E

6 in. 6 in. (a)

x

The cantilever beam shown consists of a Z shape of 14 -in. thickness. For the given loading, determine the distribution of the shearing stresses along line A  B in the upper horizontal leg of the Z shape. The x  and y axes are the principal centroidal axes of the cross section and the corresponding moments of inertia are I x  166.3 in 4 and I y  13.61 in 4 .

(b)

SOLUTION V  3 kips   22.5 V x  V sin  V y  V cos  In upper horizontal leg, use coordinate x : ( 6 in⭐ x ⭐ 0) 1 (6  x ) in. 4 1 x  ( 6  x) in. 2 y  6 in. x   x cos  + y sin  y   y cos   x sin  A

1

Due to Vx  :

Vx  Ax  Iyt

1 1  (V sin  )   (6  x )  (6  x ) cos   6sin   4 2     1   1 (13.61)   4  0.084353(6  x )( 0.47554  0.46194 x )

1 1   (V cos  )   (6  x ) 6 cos   ( 6  x ) sin   4 2     2  Ix t  1 (166.3)    4  0.0166665(6  x)[6.69132  0.19134 x] V y Ay 

Due to V y :

 1   2  (6  x )[0.07141  0.035396x ]

Total: x (in.)

6

5

4

3

2

1

 (ksi)

0

0.105

0.140

 0.104

0.003

0.180

 0.428 

0

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PROBLEM 6.88 y'

3 kips

y

A'

B' x'

A' B'

A

22.5⬚

C'

x

B

12 in.

D' D

E'

D'

E'

E

6 in. 6 in. (a)

(b)

For the cantilever beam and loading of Prob. 6.87, determine the distribution of the shearing stress along line BD in the vertical web of the Z shape. PROBLEM 6.87 The cantilever beam shown consists of a Z shape of 41 -in. thickness. For the given loading, determine the distribution of the shearing stresses along line A B  in the upper horizontal leg of the Z shape. The x  and y axes are the principal centroidal axes of the cross section and the corresponding moments of inertia are I x 166.3 in 4 and I y   13.61 in 4 .

SOLUTION V  3 kips   22.5 Vx   V sin  Vy   V cos  For part AB’,

For part B′Y,

 1 2 A    (6)  1.5 in  4 x  3 in., y  6 in.

1 (6  y ) 4 1 x  0 y  (6  y ) 2 x   x cos   y sin 

A

y   y cos   x sin  Due to V x :

1  1  

Vx ( AAB xAB  ABY xBY ) I y t (V sin )[(1.5)( 3cos  6sin )  14 (6  y ) 12 (6  y ) sin  ] (13.61) 14 

(V sin )[ 0.7133 1.7221 0.047835y 2 ] 2  0.3404  0.01614y 3.4025

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PROBLEM 6.88* (Continued)

2 

Due to V y :

2  

  ABY y) Vy  ( AAB yAB I xt ( V cos  )[(1.5)(6 cos   3 sin  )  14 (6  y) 21 (6  y)cos  ] (166.3) 14 

( V cos  )[10.037  4.1575  0.11548 y2 ] 2  0.9463  0.00770y (166.3)  14 

1  2  1.2867  0.02384 y2

Total: y (in.)

0

2

4

6

 (ksi)

1.287

1.191

0.905

0.428







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s s s

PROBLEM 6.89 Three boards are nailed together to form a beam shown, which is subjected to a vertical shear. Knowing that the spacing between the nails is s  75 mm and that the allowable shearing force in each nail is 400 N, determine the allowable shear when w  120 mm.

60 mm 60 mm 60 mm

w 200 mm

SOLUTION Part Top Plank

A (mm 2 )

d (mm)

Ad 2 (106 mm 4 )

I (10 6 mm 4 )

7200

60

25.92

2.16

Middle Plank

12,000

0

0

3.60

Bottom Plank

7200

60

25.92

2.16

51.84

7.92



I  Ad 2   I  59.76  10 6 mm 4  59.76  10 6 m 4 Q  (7200)(60)  432  10 3 mm 3  432  10 6 m3 VQ I Fnail q s

q

V 

Fnail  qs V 

Iq IFnail  Q Qs

(59.76  106 )(400) (432  106 )(75  103 )

V  738 N 

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PROBLEM 6.90

180 16

12 a

0.6 m 80

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.

n

100

16

160 kN

n

80

0.9 m

0.9 m

Dimensions in mm

SOLUTION

V  80 kN

At section n-n,

Consider cross section as composed of rectangles of types , , and . 1 (12)(80)3  (12)(80)(90)2  8.288  106 mm4 12 1 (180)(16)3  (180)(16)(42)2  5.14176  106 mm4 I2  12 1 (16)(68) 3  419.24  10 3 mm 4 I3  12

I1 

I  4I1  2I 2  2I 3  44.274  106 mm4  44.274  106 m4 (a )

Calculate Q at neutral axis. Q 1  (12)(80)(90)  86.4  10 3 mm 4 Q2  (180)(16)(42)  120.96  10 3 mm 4 Q3  (16)(34)(17)  9.248  103 mm4

Q  2Q1  Q2  2Q3  312.256  103 mm3  312.256  106 m3

 (b )

At point a,

(80  103 )(312.256  106 ) VQ   17.63  106 Pa It (44.274  106 )(2  16  103 )

  17.63 MPa 

Q  Q1  86.4  10 3 mm 4  86.4  10 6 m 4

 

(80  103 )(86.4  10 6 ) VQ   13.01  106 Pa It (44.274  10 6)(12  10 3)

  13.01 MPa 

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PROBLEM 6.91

P W24 × 104 A

C B 6 ft

9 ft

For the wide-flange beam with the loading shown, determine the largest load P that can be applied, knowing that the maximum normal stress is 24 ksi and the largest shearing stress using the approximation  m  V/Aweb is 14.5 ksi.

SOLUTION

 M C  0:  15R A  qP  0 RA  0.6 P

Draw shear and bending moment diagrams. V max  0.6 P

M max  0.6PLAB

LAB  6 ft  72 in.

For W 24  104,

Bending.

S  P

M

max

 all



S  258 in 3

0.6PLAB

 all

 all S (24)(258)   143.3 kips 0.6L AB (0.6)(72)

Aweb  dt w

Shear.

 (24.1)(0.500) 12.05 in 2

  P

V

max

Aweb

 Aweb 0.6



0.6P Aweb



(14.5)(12.05)  291 kips 0.6

The smaller value of P is the allowable value.

P 143.3 kips

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12 kips

PROBLEM 6.92

1 in.

12 kips

n

1 in. 1 in.

a

A

4 in.

B

b

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

n 16 in.

10 in.

2 in.

16 in.

4 in.

SOLUTION R A  RB  12 kips Draw shear diagram.

V  12 kips Determine section properties. Part

A(in 2 )

y (in.)

Ay (in 3)

d(in.)

Ad 2(in 4)

I (in 4)



4

4

16

2

16

5.333



8

1

8

1

8

2.667



12

24

8.000

24 Y 

24 Ay   2 in. A 12

I   Ad 2   I  32 in4

(a )

A  1 in 2

y  3.5 in.

Qa  Ay  3.5 in 3

t  1 in.

a  (b )

(12)(3.5) VQa  It (32)(1)

A  2 in 2

 a  1.313 ksi 

y  3 in. Qb  A y  6 in3

t  1 in.

b 

(12)(6) VQb  It (32)(1)

b  2.25 ksi 

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PROBLEM 6.93

2 in. 4 in. 6 in.

The built-up timber beam is subjected to a 1500-lb vertical shear. Knowing that the longitudinal spacing of the nails is s  2.5 in. and that each nail is 3.5 in. long, determine the shearing force in each nail.

4 in. 4 in.

2 in.

2 in.

2 in. 2 in.

SOLUTION I1 

1 (2)(4)3  (2)(4)(3) 2 12

 82.6667 in 4 1 (2)(6) 3  36 in 4 12 I  2 I1  2 I 2

I2 

 237.333 in 4 Q  A1y ...