CHAPTER 6
s
PROBLEM 6.1
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.
50 mm
50 mm
50 mm
100 mm
SOLUTION
I 
1 3
1
(100)(150)3  28.125  106 mm 4
bh 
12
12
 28.125  106 m 4
A  (100)(50)  5000 mm2
y1  50 mm
Q  A y1  250  103 mm3  250  106 m3
VQ
(1500)(250  106 )

 13.3333  103 N/m
6
I
28.125  10
2 Fnail
(2)(400)
qs  2Fnail
s 

 60.0  103 m
q
13.3333  103
q 
s  60.0 mm 
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893
s
PROBLEM 6.2
s
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.
50 mm
50 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.
50 mm
100 mm
SOLUTION
I 
1 3
1
bh 
(100)(150)3  28.125  106 mm 4
12
12
 28.125  106 m 4
A  (100)(50)  5000 mm 2
y1  50 mm
Q  A y1  250  103 mm3  250  106 m3
q 
Eliminating q,
Solving for V,
VQ
I
qs  2 Fnail
VQ
2 Fnail

I
s
V 
2IFnail
(2)(28.125  106 )(400)

 2.00  103 N
Qs
(250  106 )(45  103 )
V  2.00 kN ◄
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894
PROBLEM 6.3
s
s
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.
s
2 in.
4 in.
2 in.
2 in.
6 in.
SOLUTION
1 3
bh  Ad 2
12
1
(6)(2)3  (6)(2)(3) 2  112 in 4

12
1 3
1
(2)(4)3  10.6667 in 4
I2 
bh 
12
12
I1 
I 3  I1  112 in 4
I  I1  I 2  I 3  234.67 in 4
Q  A1 y1  (6)(2)(3)  36 in 3
qs  Fnail
(1)
VQ
I
(2)
q
Dividing Eq. (2) by Eq. (1),
1
VQ

s
Fnail I
V 
Fnail I
(150)(234.67)

(36)(3)
Qs
V  326 lb 
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895
PROBLEM 6.4
s
s
s
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.
20 mm
80 mm
20 mm
120 mm
SOLUTION
1
1
b2 h23  b1h13
12
12
1
1
 (120)(120)3  (80)(80)3  13.8667  106 mm 4
12
12
6 4
 13.8667  10 m
I
(a)
A1  (120)(20)  2400 mm 2
y1  50 mm
Q1  A1 y1  120  103 mm3  120  106 m3
VQ (1200)(120  106 )

 10.3846  103 N/m
6
I
13.8667  10
qs  2 Fnail
q
Fnail 
(b)
qs (10.3846  103 )(30  103 )

2
2
Fnail  155.8 N 
Q  Q1  (2)(20)(40)(20)
 120  103  32  103  152  103 mm3
 152  106 m3
 max 
VQ
(1200)(152  106 )

It
(13.8667  106 )(2  20  103 )
 max  329 kPa 
 329  103 Pa
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896
16 ⫻ 200 mm
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.
S310 ⫻ 52
SOLUTION
Calculate moment of inertia:
A (mm 2 )
3200
6650
3200
Part
Top plate
S310  52
Bot. plate

*d

d (mm)
*
160.5
0
*
160.5
Ad 2 (106 mm 4 )
82.43
82.43
164.86
I (106 mm 4 )
0.07
95.3
0.07
95.44
305 16

 160.5 mm
2
2
I  Ad 2  I  260.3  106 mm 4  260.3  106 m 4
Q  Aplate d plate  (3200)(160.5)  513.6  103 mm3  513.6  106 m3
Abolt 

4
2

d bolt

4
(18  103 )2  254.47  106 m 2
Fbolt   all Abolt  (90  106 )(254.47  106 )  22.90  103 N
qs  2 Fbolt
q
VQ
I
q
V 
2Fbolt
(2)(22.90  103 )

 381.7  103 N/m
3
s
120  10
Iq (260.3  106 )(381.7  103 )

 193.5  103 N
Q
513.6  106
V  193.5 kN 
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897
y
16 in. ⫻
1
2
in.
C12 ⫻ 20.7
z
C
PROBLEM 6.6
The beam shown is fabricated by connecting two channel shapes and two
plates, using bolts of 34 -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
C12  20.7: d  12.00 in., I x  129 in 4
For top plate,
y
12.00  1  1 
     6.25 in.
2
 2  2 
3
1
1
1
I t  (16)    (16)   (6.25) 2  312.667 in 4
12
2
2
For bottom plate,
I b  312.667 in 4
Moment of inertia of fabricated beam:
I  (2)(129)  312.667  312.667
 883.33 in 4
1
Q  Aplate yplate  (16)   (6.25)  50 in 3
2
VQ (25)(50)
q

 1.41510 kips/in.
I
883.33
1
1
Fbolt  qs    (1.41510)(7.5)  5.3066 kips
2
2
 bolt

(d bolt ) 2 
 3
2
 0.44179 in 2
4
4  4 
F
5.3066
 bolt 
 12.01 ksi
Abolt 0.44179
Abolt 
 bolt  12.01 ksi 
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898
PROBLEM 6.7
y
A columm is fabricated by connecting the rolled-steel members shown by
bolts of 34 -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.
C8 ⫻ 13.7
z
C
S10 ⫻ 25.4
SOLUTION
Geometry:
d
f     (tw )C
 2 s
10.0
 0.303  5.303 in.
2
x  0.534 in.

y1  f  x  5.303  0.534  4.769 in.
Determine moment of inertia.
d (in.)
4.769
0
4.769
A(in 2 )
Part
C8  13.7
4.04
S10  25.4 7.45
C8  13.7
4.04

Ad 2 (in 4 )
91.88
0
91.88
183.76
I (in 4 )
1.52
123
1.52
126.04
I  Ad 2  I  183.76  126.04  309.8 in 4
Q  A y1  (4.04)(4.769)  19.2668 in 3
VQ (30)(19.2668)

 1.86573 kip/in.
I
309.8
1
1
 qs    (1.86573)(5)  4.6643 kips
2
2
q
Fbolt
Abolt 
 bolt 

4
2

d bolt
 3
2
2
   0.44179 in
4 4
4.6643
Fbolt

 10.56 ksi
Abolt 0.44179
 bolt  10.56 ksi 
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899
PROBLEM 6.8
The composite beam shown is fabricated by connecting two W6  20 rolled-steel
members, using bolts of 85 -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
W6  20: A  5.87 in 2 , d  6.20 in., I x  41.4 in 4
y 
Composite:
1
d  3.1 in.
2
I  2[41.4  (5.87)(3.1) 2 ]
 195.621 in 4
Q  A y  (5.87)(3.1)  18.197 in 3
Bolts:
d 
Shear:
 5
2
2
   0.30680 in
4 8
  all Abolt  (10.5)(0.30680)  3.2214 kips
Abolt 
Fbolt
5
in.,  all  10.5 ksi, s  6 in.
8
q
2Fbolt
(2)(3.2214)

 1.07380 kips/in.
6
s
q
VQ
I
V 
Iq (195.621)(1.0780)

Q
18.197
V  11.54 kips 
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900
15 15
30
PROBLEM 6.9
15 15
20
a
0.5 m
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.
72 kN
20
n
40
120
n
20
20
0.8 m
1.5 m
90
Dimensions in mm
SOLUTION
M B  0: 2.3 A  (1.5)(72)  0
A  46.957 kN 
V  A  46.957 kN
At section n-n,
Calculate moment of inertia:
1

1
 1
I  2  (15)(40)3   2  (15)(80)3   (30)(1203 )
12

12
 12
 5.76  106 mm 4  5.76  106 m 4
ta  30 mm  0.030 m
At a,
Qa  (30  20)(50)  30  103 mm3
 30  106 m3
a 
VQa (46.957  103 )(30  106 )

Ita
(5.76  106 )(0.030)
 8.15  106 Pa = 8.15 MPa
tb  60 mm  0.060 m
At b,
Qb  Qa  (60  20)(30)  30  103  36  103  66  103 mm3  66  106 m 4
b 
VQb (46.957  103 )(66  106 )

 8.97  106 Pa  8.97 MPa
Itb
(5.76  106 )(0.060)
t NA  90 mm  0.090 m
At NA,
QNA  Qb  (90  20)(10)  66  103  18  103  84  103 mm3  84  106 m3
 NA 
(a)
VQNA (46.957  103 )(84  106 )

 7.61  106 Pa  7.61 MPa
It NA
(5.76  106 )(0.090)
 max occurs at b.
 max  8.97 MPa 
 a  8.15 MPa 
(b)
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901
PROBLEM 6.10
0.3 m
n
10 kN
40 mm
a
100 mm
12 mm
150 mm
12 mm
n
200 mm
1.5 m
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.
I  I1  4I 2

1
 1

b1h13  4  b2h23  A2d 22 
12
12



 1 

1
(100)(150)3  4   (50)(12)3  (50)(12)(69)2 
12
 12 

 28.125  106  4 0.0072  106  2.8566  106 
 39.58  106 mm 4  39.58  106 m 4
(a)
Q  A1 y1  2 A2 y2
 (100)(75)(37.5)  (2)(50)(12)(69)
 364.05  103 mm3  364.05  106 m3
t  100 mm = 0.100 m
 max 
(b)
VQ (10  103 )(364.05  106 )

 920  103 Pa
It
(39.58  106 )(0.100)
 max  920 kPa 
Q  A1 y1  2 A2 y2
 (100)(40)(55)  (2)(50)(12)(69)
 302.8  103 mm3  302.8  106 m3
t  100 mm  0.100 m
a 
VQ (10  103 )(302.8  106 )

 765  103 Pa
It
(39.58  106 )(0.100)
 a  765 kPa 
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902
PROBLEM 6.11
t
3 in.
18 in.
a
t
t
n
25 kips
n
3 in.
3 in.
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.
t = 0.25 in.
25 in.
8 in.
SOLUTION
I 
1
1
2
(8 in.)(9 in.)3  (7.25 in.)(8.5 in.)3  (0.25 in.)(3 in.)3
12
12
12
I  113.8 in 4
V = 25 kips
M  (25 kips)(18 in.) = 450 kip  in.
(a)
 m:
At neutral axis,
thickness  0.25 in.
Q  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  4.5 in 3  8.203 in 3  2.258 in 3  14.96 in 3
t  0.25 in.
V
(25 kips)(14.96 in 3 )
m  Q 
It
(113.8 in 4 )(0.25 in.)
 m  13.15 ksi ◄
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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903
PROBLEM 6.11 (Continued)
(b)
 a: At point a,
t = 0.25 in.
See sketch above.
Qa  4.5 in 3  8.203 in 3  12.70 in 3
a 
VQa
(25 kips)(12.70 in 3 )

It
(113.8 in 4 )(0.25 in.)
 a  11.16 ksi ◄
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904
1
2
10 kips 10 kips
8 in.
PROBLEM 6.12
in.
a
n
1
2
4 in.
in.
n
16 in.
12 in.
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.
4 in.
16 in.
SOLUTION
By symmetry, RA  RB .
Fy  0: RA  RB  10  10  0
RA  RB  10 kips
From the shear diagram,
V  10 kips at n-n.
1
1
b2h23  b1h13
12
12
1
1

(4)(4)3  (3)(3)3  14.5833 in 4
12
12
I 
(a)
1
1
Q  A1 y1  A2 y2  (3)   (1.75)  (2)   (2)(1)  4.625 in 3
2
2
t 
 max 
(b)
1 1
  1 in.
2 2
(10)(4.625)
VQ

(14.5833)(1)
It
 max  3.17 ksi 
1
Q  Ay  (4)   (1.75)  3.5 in 3
2
1 1
t    1 in.
2 2
 
(10)(3.5)
VQ

(14.5833)(1)
It
 a  2.40 ksi 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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905
10
40
Dimensions in mm
30
10
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.
40
40
SOLUTION
Calculate moment of inertia.
1
 1
I  2  (10 mm)(120 mm)3   (30 mm)(40 mm)3
12
 12
 2[1.440  106 mm 4 ]  0.160  106 mm 4
 3.04  106 mm 4
I  3.04  106 m 4
Assume that  m occurs at point a.
t  10 mm  0.01 m
Q  (10 mm  40 mm)(40 mm)
 16  103 mm3 Q  16  106 m3
For
 all  60 MPa,
60  106 Pa 
 m   all 
V (16  106 m3 )
(3.04  106 m 4 )(0.01m)
VQ
It
V  114.0 kN 
Check  at neutral axis:
t  50 mm  0.05 m
Q  2[(10  60)(30)]  (30  20)(10)  42  103 m3  42  106 m3
 NA 
VQ (114.0 kN)(42  106 m3 )

 31.5 MPa  60 MPa
It
(3.04  106 m 4 )(0.05 m)
OK
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906
10
30
PROBLEM 6.14
10
For a beam having the cross section shown, determine the largest
allowable vertical shear if the shearing stress is not to exceed 60 MPa.
10
30
40
Dimensions in mm
30
10
SOLUTION
Calculate moment of inertia.
I 
1
1

(50 mm)(120 mm)3  2  (30 mm)4  (30 mm  30 mm)(35 mm)2 
12
12

I  7.2  106 mm 4  2[1.170  106 mm 4 ]  4.86  106 mm 4
 4.86  106 m 4
Assume that  m occurs at point a.
t  2(10 mm)  0.02 m
Q  (10 mm  50 mm)(55 mm)  2[(10 mm  30 mm)(35 mm)]
 48.5  103 mm3  48.5  106 m3
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907
PROBLEM 6.14 (Continued)
For  all  60 MPa,
 m   all 
60  106 Pa 
Check  at neutral axis:
VQ
It
V (48.5  106 m3 )
(4.86  106 m 4 )(0.02 m)
V  120.3 kN 
t  50 mm  0.05 m
Q  (50  60)(30)  (30  30)(35)  58.5  103 mm3
 
VQ (120.3 kN)(58.5  106 m3 )

 29.0 MPa  60 MPa
It
(4.86  106 m 4 )(0.05 m)
OK
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908
PROBLEM 6.15
1.5 in.
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.
2 in.
4 in.
w = 2.5 in.
2 in.
1.5 in.
SOLUTION
I 
1
(1.5  83  2(0.5)(4)3 )
12
I  69.333 in 4
 all  150 psi
Q  (1.5 in.)(2 in.)(3 in.)  9 in 3;
At point a:
m 
VQ
;
It
150 psi 
t  1.5 in.
V (9 in 3 )
;
(69.333 in 4 )(1.5 in.)
V  1733 lb 
At neutral axis:
Q  (1.5 in.)(2 in.)(3 in.)  (2.5 in.)(2 in.)(1 in.)  14 in 3, t  2.6 in.
m 
VQ
;
It
150 psi 
V (14 in 3 )
;
(69.333 in 4 )(2.5 in.)
V  1857 lb 
V  1733 lb ◄
We choose smaller shear.
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909
PROBLEM 6.16
220 mm
12 mm
W250 3 58
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.
252 mm
12 mm
SOLUTION
Calculate moment of inertia.
Part
Top plate
A(mm 2 )
2640
W250 × 58
7420
0
Bot. plate
2640
*132
Ad 2 (106 mm 4 )
45.999
d (mm)
*132
0

*d 
I (106 mm 4 )
0.032
87.3
45.999
0.032
91.999
87.364
Ay (103 mm3 )
252 12

2
2
I  Ad 2  I  179.363  106 mm 4  179.363  106 m 4
 max occurs at neutral axis. t  8.0 mm  8.0  103 m
Part
 Top plate
A(mm 2 )
y (mm)
2640
132
348.48
 Top flange
2740.5
119.25
326.805
56.25
50.625
 Half web
900
725.91

Dimensions in mm:  12  220 ;  13.5  203 ; 
8.0  112.5
Q  A y  725.91  103 mm3  725.91  106 m3
 
VQ
It
V 
It
(179.363  106 )(8.0  103 )(90  106 )

Q
725.91  106
 177.9  103 N
V  177.9 kN 
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910
PROBLEM 6.17
A
B
A
(a)
B
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.
(b)
SOLUTION
A  9.12 in 2
I x  110 in 4
W8  31
I y  37.1 in 4
I  2[ I x  Ad 2 ]
(a)
 2[110 in 4  (9.12 in 2 )(4 in.)2 ]
I  511.84 in 4
Q  Ay  (9.12 in 2 )(4 in.)  36.4 in 3
For two welds each with allowable shearing force of 3 kips/in.,
q  2(3 kips/in.)  6 kips/in.
q
VQ
;
I
6 kips/in. 
V (36.4 in 3 )
(511.84 in 4 )
V  84.2 kips 
(b)
I  2[ I y  Ad 2 ]
 2[37.1 in 4  (9.12 in 2 )(4 in.) 2 ]
I  366.04 in 4
Q  Ay  (9.12 in 2 )(4 in.)  36.4 in 3
q  6 kips/in. (same as in part a)
q
V (36.4 in 3 )
VQ
; 6 kips/in. 
I
(366.04 in 4 )
V  60.2 kips 
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911
2.4 kN
4.8 kN
PROBLEM 6.18
7.2 kN
b
B
C
D
A
150 mm
E
1m
1m
1m
For the beam and loading shown, determine the
minimum required width b, knowing that for the
and
grade of timber used,  all  12 MPa
 all  825 kPa.
0.5 m
SOLUTION
M D  0: 3 A  (2)(2.4)  (1)(4.8)  (0.5)(7.2)  0
A  2 kN 
Draw the shear and bending moment diagrams.
|V |max  7.2 kN  7.2  103 N
|M |max  3.6 kN  m  3.6  103 N  m

Bending:
S min 
M
S
|M |max

3.6  103

12  106
 300  106 m3
 300  103 mm3
1
S  bh 2
6
6 S (6)(300  103 )
b 2 
 80 mm
h
(150)2
For a rectangular section,
Shear: Maximum shearing stress occurs at the neutral axis of bending for a rectangular section.
1
1
1
bh, y  h, Q  Ay  bh 2
2
4
8
1 3
I  bh t  b
12
V ( 18 bh 2 )
3V
VQ



3
1
It
( 12 bh )(b) 2 bh
A
b
3V
(3)(7.2  103 )

 87.3  103 m
2h (2)(150  103 )(825  103 )
b  87.3 mm 
The required value of b is the larger one.
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912
PROBLEM 6.19
P
L/2
A
b
L/2
C
B
h
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 2h/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, P  40 kN,  m  960 kPa, and
 m  12 MPa.
SOLUTION
RA  RB  P/2 
Reactions:
P
2
(1)
Vmax  RA 
(2)
A  bh for rectangular section.
(3)
m 
(4)
M max 
(5)
S 
(6)
m 
3 Vmax
3P

for rectangular section.
2 A
4bh
PL
4
1 2
bh for rectangular section.
6
M max
3PL

S
2bh 2
m
h


 m 2L
(a)
(b)
Solving for h,
h
2 L m
(2)(2)(960  103 )
 320  103 m
6
12  10
h  320 mm 
3P
(3)(40  103 )

 97.7  103 m
4h m
(4)(320  103 )(960  103 )
b  97.7 mm 
m

Solving Eq. (3) for b,
b
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913
PROBLEM 6.20
w
b
A
B
C
L/4
D
L/2
L/4
h
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 2h/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  8 kN/m,
 m  1.08 MPa, and  m  12 MPa.
SOLUTION
RA  RB 
From shear diagram,
For rectangular section,
|V |m 
wL
4
(1)
A  bh
m 
From bending moment diagram,
wL
2
(2)
3 Vm
3wL

2 A
8bh
|M |m 
(3)
wL2
32
(4)
1 2
bh
6
(5)
|M |m
3wL2

S
16 bh 2
(6)
For a rectangular cross section,
S 
m 
(a)
Dividing Eq. (3) by Eq. (6),
(b)
Solving for h,
h
L m
(5)(1.08  106 )

 225  103 m
6
2 m
(2)(12  10 )
m
2h


L
m
h  225 mm 
Solving Eq. (3) for b,
b
3wL
(3)(8  103 )(5)

8h m
(8)(225  103 )(1.08  106 )
 61.7  103 m
b  61.7 mm 
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914
25 kips
PROBLEM 6.21
25 kips
n
3
4
7.25 in.
in.
b
a
B
A
n
20 in.
10 in.
3
4
20 in.
1.5 in.
1.5 in.
3
4
in.
For the beam and loading shown,
consider section n-n and determine the
shearing stress at (a) point a, (b) point b.
in.
8 in.
SOLUTION
RA  RB  25 kips
V  25 kips
At section n-n,
Locate centroid and calculate moment of inertia.
Part
A(in 2 )
y (in.)
A y (in 3 )
d(in.)
Ad 2 (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
35.56
47.91
72.94
Ay 72.94

 4.631 in.
A 15.75
I  Ad 2  I  35.56  47.91  83.47 in 4
Y 
(a)
3
Qa  Ay    (1.5)(4.631  0.75)  4.366 in 3
4
3
t   0.75 in.
4
a 
(b)
VQ
(25)(4.366)

It
(83.47)(0.75)
 a  1.744 ksi 
3
Qb  Ay    (3)(4.631  1.5)  7.045 in 3
4
t  0.75 in.
b 
VQ
(25)(7.045)

It
(83.47)(0.75)
 b  2.81 ksi 
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915
PROBLEM 6.22
160 mm
180 kN
a
n
A
20 mm
100 mm
B
b
n
500 mm
30 mm
For the beam and loading
shown, consider section n-n
and determine the shearing
stress at (a) point a, (b) point b.
500 mm
30 mm
30 mm
20 mm
SOLUTION
|V |max  90 kN
Draw the shear diagram.
Part
A(mm 2 )
y (mm)
A y (103 mm3 )

3200
90
288

1600
40

1600
40

6400
Ad 2 (106 mm 4 )
I (106 mm 4 )
25
2.000
0.1067
64
25
1.000
0.8533
64
25
1.000
0.8533
4.000
1.8133
d(mm)
416
Ay
416  103

 65 mm
A
6400
Y 
I  Ad 2  I  (4.000  1.8133)  106 mm 4
 5.8133  106 mm 4  5.8133  106 m 4
(a)
A  (80)(20)  1600 mm 2
y  25 mm
Qa  Ay  40  103 mm3  40  106 m3
a 
VQa
(90  103 )(40  106 )

 31.0  106 Pa
It
(5.8133  106 )(20  103 )
 a  31.0 MPa 
(b)
A  (30)(20)  600 mm 2
y  65  15  50 mm
Qb  Ay  30  103 mm3  30  106 m3
b 
VQb
(90  103 )(30  106 )

 23.2  106 Pa
It
(5.8133  106 )(20  103 )
 b  23.2 MPa 
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916
25 kips
PROBLEM 6.23
25 kips
n
7.25 in.
3
4
in.
b
a
B
A
n
20 in.
10 in.
3
4
20 in.
1.5 in.
1.5 in.
3
4
in.
For the beam and loading shown,
determine the largest shearing stress
in section n-n.
in.
8 in.
SOLUTION
RA  RB  25 kips
V  25 kips
At section n-n,
Locate centroid and calculate moment of inertia.
Part
A(in 2 )
y (in.)
Ay (in 3 )
d(in.)
Ad 2 (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
35.56
47.91
72.94
I (in 4 )
A y 72.94

 4.631 in.
A 15.75
I  Ad 2  I  35.56  47.91  83.47 in 4
Y 
Largest shearing stress occurs on section through centroid of entire cross section.
3
 4.631 
 8.042 in 3
Q  Ay    (4.631) 

4
 2 
3
t   0.75 in.
4
VQ
(25)(8.042)


It
(83.47)(0.75)
 m  3.21 ksi 
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Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted
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917
PROBLEM 6.24
160 mm
180 kN
a
n
A
For the beam and loading
shown, determine the largest
shearing stress in section n-n.
100 mm
B
b
n
500 mm
20 mm
30 mm
500 mm
30 mm
30 mm
20 mm
SOLUTION
|V |max  90 kN
Draw the shear diagram.
Part
A (mm 2 )
x

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
4.000
1.8133
Ay (103 mm3 ) d(mm) Ad 2 (106 mm 4 ) I (106 mm 4 )
416
Y 
Ay
416  103

 65 mm
A
6400
I  Ad 2  I  (4.000  1.8133)  106 mm 4
 5.8133  106 mm 4  5.8133  106 m 4
Part
A(mm 2 )

3200

300
7.5
2.25

300
7.5
2.25
y (mm)
25

Ay (103 mm3 )
80
84.5
Q  Ay  84.5  103 mm3  84.5  106 m3
t  (2)(20)  40 mm  40  103 m
 max 
VQ
(90  103 )(84.5  106 )

It
(5.8133  106 )(40  103 )
 32.7  106 Pa
 m  32.7 MPa 
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
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918
PROBLEM 6.25
c
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  k
V
A
where A is the cross-sectional area of the beam.
SOLUTION
I 
For semicircle,
As 

4

2
c4
c2
Q  As y 
(a)
 max occurs at center where
(b)
 max 
and
A   c2
y 

2
c2 
4c
3
4c
2
 c3
3
3
t  2c 
V  2 c3
VQ
4V
4V
  43


2
It
3A
c  2c 3 c
4
k 
4
 1.333 
3
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919
PROBLEM 6.26
tm
rm
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  k
V
A
where A is the cross-sectional area of the beam.
SOLUTION
A  2 rmtm
For a thin-walled circular section,
J  Arm2  2 rm3tm ,
For a semicircular arc,
y 
I 
1
J   rm3tm
2
2rm

As   rmtm
Q  As y   rmtm
(a)
t  2tm
(b)
 max 
2rm

 2rm2tm
at neutral axis where maximum occurs. 
VQ
V (2rm2tm )
V
2V



3
It
A
( rmtm )(2tm )  rmtm
k  2.00 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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920
PROBLEM 6.27
h
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
h
 max  k
b
V
A
where A is the cross-sectional area of the beam.
SOLUTION
1 
A  2  bh   bh
2 
 1
 1
I  2  bh3   bh3
 12
 6
For a cut at location y, where y  h,
A( y) 
1  by 
by 2
 y 
2 h 
2h
y ( y)  h 
2
y
3
Q( y)  Ay 
t ( y) 
 ( y) 
(a)
by 2 by 3

2
3h
by
h
2
VQ
6
h by 2 by 3
V   y
 y 
V 3 



3    2   
It
3h
bh   h 
bh by 2
 h  
To find location of maximum of  , set
d
 0.
dy
d
V
y
 2 [3  4 m ]  0
dy
h
bh
(b)
 ( ym ) 
2
V  3
V
3  9 V
 1.125
3    2    
bh   4 
A
 4   8 bh
ym
3
1
 , i.e.,  h from neutral axis. 
h
4
4
k  1.125 
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921
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
b
h
 max  k
V
A
where A is the cross-sectional area of the beam.
SOLUTION
A
1
bh
2
I 
1 3
bh
36
For a cut at location y,
A( y) 
1  by 
by 2

y
 
2 h 
2h
y ( y) 
2
2
h y
3
3
Q( y)  Ay 
t ( y) 
by 2
(h  y )
3
by
h
2
by
VQ V 3 (h  y) 12Vy(h  y) 12V



(hy  y 2 )
 ( y) 
3
3
1 bh3 ) by
It
bh
bh
( 36
h
(a)
To find location of maximum of  , set
d
 0.
dy
d
12V
( h  2 ym )  0

dy
bh3
(b)
m
2
12V
12V  1 2  1   3V
3V
2
(hym  ym ) 


 h   h  
3
3
2A
bh
bh  2
 2   bh
ym 
1
h, i.e., at mid-height 
2
k 
3
 1.500 
2
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922
PROBLEM 6.29
2 in.
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.
10 in.
2 in.
s
s
s
2 in.
SOLUTION
1
b1h13  A1d12
12
1
(2)(2)3  (2)(2)(4)2  65.333 in 4

12
1
1
(2)(10)3  166.67 in 4
I2 
b2h23 
12
12
I1 
I  4 I1  I 2  428 in 4
Q  Q1  A1 y1  (2)(2)(4)  16 in 3
q
(1200)(16)
VQ

 44.86 lb/in.
I
428
Fnail  qs
s 
75
Fnail

 1.672 in.
q
44.86

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923
PROBLEM 6.30
50 mm
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.
150 mm
50 mm
20 mm
20 mm
100 mm
SOLUTION
I 
1
1
(140)(250)3  (100)(150)3  154.167  106 mm 4
12
12
 154.167  106 m 4
Q  Ay  (100)(50)(100)  500  103 mm3
 500  106 m3
t  50 mm  50 mm  100 mm  100  103 m
 
VQ
It
V 
It
(154.167  106 )(100  103 )(350  103 )

Q
500  106
 10.79  103 N
V  10.79 kN 
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924
1.5
0.8
0.8
4
PROBLEM 6.31
1.5
0.8
A
B
3.2
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.
0.8
Dimensions in inches
SOLUTION
1
1

I  2  (0.8)(4.8)3  (7)(0.8)3  (7)(0.8)(2.0) 2 
12
12

 60.143 in 4
(a)
Aa  (1.5)(0.8)  1.2 in 2
Qa  Aa ya  2.4 in
ya  2.0 in.
3
ta  0.8 in.
a 
(b)
VQa
(1200)(2.4)

Ita
(60.143)(0.8)
Ab  (4)(0.8)  3.2 in 2
 a  59.9 psi 
yb  2.0 in.
Qb  Ab yb  (3.2)(2.0)  6.4 in 3
tb  (2)(0.8)  1.6 in.
b 
VQb
(1200)(6.4)


Itb
(60.143)(1.6)
 b  79.8 psi 
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925
PROBLEM 6.32
20 mm 60 mm 20 mm
A
20 mm
30 mm
B
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.
20 mm
30 mm
20 mm
SOLUTION
IA 
1 3
1
(60)(20)3  (60)(20)(50)2
bh  Ad 2 
12
12
 3.04  106 mm 4
1 3
1
(60)(20)3  0.04  106 mm 4
bh 
12
12
1 3
1
(20)(120)3  2.88  106 mm 4
IC 
bh 
12
12
IB 
I  2 I A  I B  2 I C  11.88  106 mm 4  11.88  106 m 4
QA  Ay  (60)(20)(50)  60  103 mm3  60  106 m3
t  20 mm  20 mm  40 mm  40  103 m
(a)
A 
VQA
(3  103 )(60  106 )

 379  103 Pa
It
(11.88  10 6 )(40  103 )
 A  379 kPa 
QB  0
(b)
B 
VQB
0
It
B  0 
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926
PROBLEM 6.33
50
300
50
B
A 100
A
50
C
400
x
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: I x  1.504  109 mm 4.)
50
A
A
200
B
Dimensions in mm
SOLUTION
I x  1.504  109 mm 4  1504  106 m 4
s A  60 mm  0.060 m
sB  25 mm  0.025 m
(a)
QA  Q1  A1 y1  (50)(100)(150)  750  103 mm3
 750  106 m3
FA  q As A

VQ1s A
(8  103 )(750  106 )(0.060)

I
1504  106
FA  239 N 
(b)
Q2  A2 y2  (300)(50)(175)  2625  103 mm3
QB  2Q1  Q2  4125  103 mm3
 4125  106 m3
FB  qB sB 
VQB sB
(8  103 )(4125  106 )(0.025)

I
1504  106
FB  549 N 
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927
105 mm
PROBLEM 6.34
A
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.
C
SOLUTION
For W360  122, d  363 mm, bF  257 mm, t F  21.70 mm, tw  13.0 mm
I  367  106 mm 4  367  106 m 4
(a)
Aa  (105)(21.70)  2278.5 mm 2
ya 
d tF
363 21.70



 170.65 mm
2
2
2
2
Qa  Aa ya  388.8  103 mm3  388.8  106 m3
ta  t F  21.70 mm  21.7  103 m
a 
(b)
VQa
(250  103 )(388.8  106 )

 12.21  106 Pa
Ita
(367  106 )(21.7  103 )
 a  12.21 MPa 
A1  bF t F  (257)(21.70)  5577 mm 2
y1 
d tF
363 21.70



 170.65 mm
2
2
2
2
d

A2  tw   t F   (13.0)(159.8)  2077 mm 2
2

y2 
1d

  t F   79.9 mm
2 2

Qc  Ay  (5577)(170.65)  (2077)(79.9)  1117.7  103 mm3
 1117.7  106 m3
tc  t w  13.0 mm  13  103 m
c 
VQc
(250  103 )(1117.7  106 )

 58.6  106 Pa
6
3
Itc
(367  10 )(13  10 )
 c  58.6 MPa 
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928
PROBLEM 6.35
b
12
12
6
a
80
6
40
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.
150
Dimensions in mm
SOLUTION
I 
1
1
(150)(80)3  (126)(68)3
12
12
 3.098  106 mm 4  3.0985  106 m 4
(a)
Qa  A1 y1  2 A2 y2
 (126)(6)(37)  (2)(12)(40)(20)
 47.172  103 mm3  47.172  106 m3
ta  (2)(12)  24 mm  0.024 m
a 
VQa
(150  103 )(47.172  106 )

 95.2  106 Pa
6
Ita
(3.0985  10 )(0.024)
 a  95.2 MPa 
(b)
Qb  A1 y1  (126)(6)(37)  27.972  103 mm3
 27.972  106 m3
tb  (2)(6)  12 mm  0.012 m
b 
VQb
(150  103 )(27.972  106 )

 112.8  106 Pa
Itb
(3.0985  106 )(0.012)
 b  112.8 MPa 
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929
6
PROBLEM 6.36
b
12
12
80
a
6
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.
40
80
Dimensions in mm
SOLUTION
I
1
1
(80)(80)3  (56)(68)3  1.9460  106 mm 4
12
12
 1.946  106 m 4
(a)
Qa  A1 y1  2 A2 y2
 (56)(6)(37)  (2)(12)(40)(20)  31.632  103 mm3
 31.632  106 m3
ta  (2)(12)  24 mm  0.024 m
a 
VQa (150  103 )(31.632  106 )

 101.6  106 Pa
6
Ita
(1.946  10 )(0.024)
 a  101.6 MPa 
(b)
Qb  A1 y1  (56)(6)(37)  12.432  103 mm3
 12.432  106 m3
tb  (2)(6)  12 mm  0.012 m
b 
VQb
(150  103 )(12.432  106 )

 79.9  106 Pa
Itb
(1.946  106 )(0.012)
 b  79.9 MPa 
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930
PROBLEM 6.37
120
50
50
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.
10
c
b
40
30
a
160
30
40
10
20
20
Dimensions in mm
SOLUTION
 
VQ
It
τ is proportional to Q/t.
Qc  (30)(10)(75)
Point c:
 22.5  103 mm3
tc  10 mm
Qc /tc  2250 mm 2
Qb  Qc  (20)(50)(55)
Point b:
 77.5  103 mm3
tb  20 mm
Qb /tb  3875 mm 2
Qa  2Qb  (120)(30)(15)
Point a:
 209  103 mm3
ta  120 mm
Qa /ta  1741.67 mm 2
 m   b  75 MPa
(Q/t )m occurs at b.
a
Qa /ta
a
1741.67 mm
2


b
Qb /tb

c
Qc /tc
75 MPa
a

2
3875 mm
2250 mm 2
 a  33.7 MPa 
 b  75.0 MPa 
 c  43.5 MPa 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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931
0.5 in.
d
5 in.
PROBLEM 6.38
d
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.
b
8 in.
a
4 in.
0.5 in.
SOLUTION
1
(4)(0.5)3  (4)(0.5)(3.75) 2  28.167 in 4
12
1
I 2  (5)(4)3  106.67 in 4
3
I1 
I  4 I1  2I 2  326 in 4
(a)
Qa  2 A1 y1  A2 y2
 (2)(4)(0.5)(3.75)  (5)(4)(2)  55 in 3
ta  5 in.
a 
(b)
VQa
(1200)(55)

 40.5 psi
Ita
(326)(5)

Qb  A1 y1  (4)(0.5)(3.75)  7.5 in 4
tb  0.5 in.
b 
VQb
(1200)(7.5)

 55.2 psi 
Itb
(326)(0.5)

PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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932
d
0.5 in.
5 in.
PROBLEM 6.39
d
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.
b
a
8 in.
4 in.
0.5 in.
SOLUTION
A1  0.5d in 2 ,
y1  3.75 in., tb  0.5 in.
A2  (5)(4)  20 in 2 ,
y2  2 in., ta  5 in.
Qb  A1 y1  1.875d in 3
b 
VQb V 1.875d
Vd

 3.75
Itb
I 0.5
I
Qa  A2 y2  2Qb  (20)(2)  (2)(1.875d )
 40  3.75d
ta  5 in.
(a)
a 
VQa
V (40  3.75d )
V
Vd
Vd

 8  0.75
  b  3.75
Ita
I (5)
I
I
I
8  0.75d  3.75d
d 
8
 2.6667 in.
3
d  2.67 in .
(b)
1
(2.6667)(0.5)3  (2.6667)(0.5)(3.75) 2  18.780 in 4
12
1
I 2  (0.5)(4)3  106.667 in 4
3
I1 
I  4 I1  2 I 2  288.46 in 4
 a   b  3.75
Vd
(3.75)(1200)(2.6667)

288.46
I
 a  41.6 psi 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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933
c
PROBLEM 6.40
b
d
1.25 in.
a
e
1.25 in.
1.25 in.
The extruded aluminum beam has a uniform wall thickness of 18 in. Knowing
that the vertical shear in the beam is 2 kips, determine the corresponding shearing
stress at each of the five points indicated.
1.25 in.
SOLUTION
I
1
1
(2.50)(2.50)3 
(2.125)(2.25)3  1.23812 in 4 
12
12
t  0.125 in.at all sections.
V  2 kips 
Qa  0
a 
VQa
It
 a  0 
 1.25 
3
Qb  (0.125)(1.25) 
  0.097656 in 
 2 
b 
VQb
(2)(0.097656)

It
(1.23812)(0.125)
 b  1.262 ksi 
Qc  Qb  (1.0625)(0.125)(1.1875)  0.25537 in.2 
c 
VQc
(2)(0.25537)


It
(1.23812)(0.125)
 c  3.30 ksi 
Qd  2Qc  (0.125)2 (1.1875)  0.52929 
d 
VQd
(2)(0.52929)

It
(1.23812)(0.125)
  d  6.84 ksi 
 1.125 
Qe  Qd  (0.125)(1.125) 
  0.60839 
 2 
e 
VQ
(2)(0.60839)

It
(1.23812)(0.125)
 e  7.86 ksi 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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934
c
PROBLEM 6.41
b
d
1.25 in.
e
1.25 in.
1.25 in.
a
The extruded aluminum beam has a uniform wall thickness of 18 in. Knowing
that the vertical shear in the beam is 2 kips, determine the corresponding shearing
stress at each of the five points indicated.
1.25 in.
SOLUTION
I
1
1
(2.50)(2.50)3  (2.125)(2.25)3  1.23812 in 4
12
12
Add symmetric points c’, b’, and a’.
Qe  0 
 1.125 
3
Qd  (0.125)(1.125) 
  0.079102 in
 2 
td  0.125 in. 
Qc  Qd  (0.125)2 (1.1875)  0.097657 in 4
tc  0.25 in. 
Qb  Qc  (2)(1.0625)(0.125)(1.1875)  0.41309 in 3
tb  0.25 in. 
 1.25 
3
Qa  Qb  (2)(0.125)(1.25) 
  0.60840 in
2


ta  0.25 in. 
a 
VQa
(2)(0.60840)

Ita
(1.23812)(0.25)
 a  3.93 ksi 
b 
VQb
(2)(0.41309)

Itb
(1.23812)(0.25)
 b  2.67 ksi 
c 
VQc
(2)(0.097657)

Itc
(1.23812)(0.25)
 c  0.631 ksi 
d 
VQd
(2)(0.079102)

Itd
(1.23812)(0.125)
 d  1.02 ksi 
e 
VQe
Ite
 e  0 
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935
PROBLEM 6.42
40 mm 12 mm 40 mm
30 mm
a
10 mm
b
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.
c
50 mm
10 mm
30 mm
SOLUTION
Qa  (12)(30)(25  10  15)  18  103 mm3
Qb  (40)(10)(25  5)  12  103 mm3
Qc  Qa  2Qb  (12)(10)(25  5)  45.6  103 mm3
 25 
Qm  Qc  (12)(25)    49.35  103 mm3
 2 
ta  tc  tm  12 mm
tb  10 mm
 m  50 MPa
(a)
 a Qa tm
18 12




 0.36474
49.35 12
 m Qm ta
 a  18.23 MPa 
(b)
 b Qb tm
12
12




 0.29179
49.35 10
 m Qm tb
 b  14.59 MPa 
(c)
c
Q t
45.6 12
 c  m 

 0.92401
49.35 12
 m Qm tc
 c  46.2 MPa 
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
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936
2 in.
PROBLEM 6.43
2 in.
10 in.
4 in.
10 in.
Three planks are connected as shown by bolts of 83 -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.
A  (2)(2)(10)  (10)(4)  80 in 2
Ay  (2)(2)(10)(5)  (10)(4)(8)  520 in 3
Y 
Ay
520

 6.5 in.
A
80
1

I  2  (2)(10)3  (2)(10)(1.5) 2 
12


1

(10)(4)3  (10)(4)(1.5) 2  566.67 in 4
12
Q  (2)(10)(1.5)  30 in 3
F  qs 
Abolt 

4
VQs
(2.5)(30)(6)

 0.79411 kips
566.67
I
2
d bolt

  3
2
2
   0.110447 in
4 8
 bolt 
F
0.79411

 7.19 ksi 
Abolt
0.110447
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.
937
PROBLEM 6.44
100 mm
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.
25 mm
25 mm
100 mm
50 mm 100 mm 50 mm
SOLUTION
Part
A(mm 2 )
y (mm)
Ay 2 (106 mm 4 )
I (106 mm 4 )

7500
50
18.75
14.06

7500
50
18.75
14.06

15,000
50
37.50
28.12
75.00
56.24
Σ
I   Ay 2   I  131.25  106 mm 4  131.25  106 m 4
Q  A1 y1  (7500)(50)  375  103 mm3
 375  106 m3
Fbolt   bolt Abolt  qs 
Abolt 
VQs
I
VQs
(6  103 )(375  106 )(0.225)

 64.286  106 m 2
 bolt I
(6  106 )(131.25  106 )
 64.286 mm 2
d bolt 
4 Abolt


(4)(64.286)
d bolt  9.05 mm 

PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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938
PROBLEM 6.45
6 in.
1 in.
1 in.
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.)
Ay 2 (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
25.20
135
189
Y0 
Ay 189

 4.2 in.
A
45
I  Ad 2  I  160.2 in 4
Between  and :
Q12  Q1  Ay1  (9)(0.8)  7.2 in 3
Between  and :
Q23  Q1  Ay2  7.2  (9)(0.2)  5.4 in 3
q
VQ
I
Maximum q is based on Q12  7.2 in 3.
(2000)(7.2)
 89.888 lb/in.
160.2
 qs  (89.888)(9)  809 lb
q
Fbolt
 bolt 
Abolt 
Fbolt
Abolt

4
2
d bolt
Abolt 
Fbolt
 bolt
d bolt 

4 Abolt

809
 0.1079 in 2
7500

(4)(0.1079)

d bolt  0.371 in. 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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on a website, in whole or part.
939
PROBLEM 6.46
400 mm
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.
12 mm
SOLUTION
For one L102  102  9.5,
A  1845 mm 2
I  1.815  106 mm 2
Q  (1845 mm 2 )(171 mm)
 315.5  103 mm3
 315.5  106 m 4
For 12-mm  400-mm plate and four angle,
I 
1
(12 mm)(400 mm)3  4[1.815  106 mm 4  (1845 mm 2 )(171 mm) 2 ]
12
 287.06  106 mm 4  287.06  106 m 4
One angle:
q
VQ
(240 kN)(315.5  106 m3 )

 263.78 kN/m
I
287.06  106 m 4
F  qs  (263.78 kN/m)(0.120 m)  31.65 kN
Diam. bolt  22 mm
 
F
31.65 kN

;

A
(0.022 m) 2
4
  83.3 MPa ◄
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940
PROBLEM 6.47
D
1.6 in.
A
B
2 in.
E
1.2 in. 1.2 in.
F
2 in.
A plate of 14 -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
LBD 
(1.2) 2  (1.6) 2  2.0 in.
ABD  (0.25)(2.0)  0.5 in 2
Locate neutral axis and compute moment of inertia.
Part
A(in 2 )
d (in.)
Ad 2 (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
0.320
0.2134
*
y (in.)
Ay (in 3 )
0.8
1
1
ABD h 2 
(0.5)(1.6)2  0.1067 in 4
12
12
Y 
Ay
0.8

 0.4 in.
2.0
A
I  Ad 2  I  0.5334 in 4
(a)
Qm  QAB  QBC
QAB  (2)(0.25)(0.4)  0.2 in 3
QBC  (0.5)(0.25)(0.2)  0.025 in 3
Qm  0.225 in 3
m 
(b)
VQm
(1.2)(0.225)

It
(0.5334)(0.25)
 m  2.02 ksi 
QB  QAB  0.2 in 3
B 
VQB
(1.2)(0.2)

It
(0.5334)(0.25)
 B  1.800 ksi 
D  0 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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941
22 mm
PROBLEM 6.48
e
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.
a
d 50 mm
b c
10 mm 10 mm
SOLUTION
1
1
1

I  2  (2)(48)3  (2)(52)3  (20)(2)3  (20)(2)(25) 2 
12
12
12

 133.75  103 mm 4  133.75  109 mm 4
Qa  (2)(24)(12)  576 mm3  576  109 mm3
Qa  0
Qc  Qb  (12)(2)(25)  600 mm3  600  109 m3
Qd  Qc  (2)(24)(12)  1.176  103 mm3  1.176  106 m3
Qe  Qd  (2)(26)(13)  600 mm3  500  109 m3
a 
VQa
(5  103 )(576  109 )

 10.77  106 Pa
It
(133.75  109 )(2  103 )
b 
VQb
It
c 
VQc
(5  103 )(600  109 )

 11.21  106 Pa
It
(133.75  109 )(2  103 )
 c  11.21 MPa 
d 
VQd
(5  103 )(1.176  106 )

 22.0  106 Pa
9
3
It
(133.75  10 )(2  10 )
 d  22.0 MPa 
e 
VQe
(5  103 )(500  109 )

 9.35  106 Pa
It
(133.75  109 )(2  103 )
 e  9.35 MPa 
 a  10.76 MPa 
b  0 

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942
60 mm
PROBLEM 6.49
A
An extruded beam has the cross section shown and a uniform wall
thickness of 3 mm. For a vertical shear of 10 kN, determine (a) the
shearing stress at point A, (b) the maximum shearing stress in the beam.
Also, sketch the shear flow in the cross section.
30 mm
16 mm
28 mm
16 mm
SOLUTION
tan  
16
30
  28.07
A  (3 sec  )(30)  102 mm 2
1
I  (3 sec  )(30)3  7.6498  103 mm 4
12
Side:
A y (103 mm3 )
Ad 2 (103 mm 4 )
I (103 mm 4 )
Part
A (mm 2 )
y0 (mm)
Top
180
30
5.4
11.932
25.627
neglect
Side
102
15
1.53
3.077
0.966
7.6498
Side
102
15
1.53
3.077
0.966
7.6498
Bot.
84
0
18.077
27.449
neglect
55.008
15.2996
Σ
468
0
d (mm)
8.46
A y 8.46  103

 18.077 mm
468
A
I  Ad 2  I  70.31  103 mm 4  70.31  109 m 4
Y0 
(a)
QA  (180)(11.932)  2.14776  103 mm3  2.14776  10 6 m3
t  (2)(3  103 )  6  103 m
A 
(b)
VQ (10  103 )(2.14776  106 )

 50.9  106 Pa
9
6
It
(70.31  10 )(6  10 )
 A  50.9 MPa 
1

Qm  QA  (2)(3 sec  )(11.932)   11.932 
2

 2.14776  103  484.06  2.6318  103 mm3
 2.6318  106 m3
t  6  103 m
m 
VQm (10  103 )(2.6318  106 )

 62.4  106 Pa
It
(70.31  109 )(6  103 )
 m  62.4 MPa 
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.
943
PROBLEM 6.49 (Continued)
QB  (28)(3)(18.077)  1.51847  103 mm3
QB
1.51847  103
A 
(50.9)
QA
2.14776  103
 36.0 MPa
B 
Multiply shearing stresses by t (3 mm  0.003 m) to get shear flow.
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
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944
PROBLEM 6.50
6 in.
E
D
4.8 in.
A
G
B F
3 in.
2 in.
A plate of thickness t is bent as shown and then used as a beam. For a
vertical shear of 600 lb, determine (a) the thickness t for which the
maximum shearing stress is 300 psi, (b) the corresponding shearing
stress at point E. Also, sketch the shear flow in the cross section.
3 in.
SOLUTION
LBD  LEF 
4.82  22  5.2 in.
Neutral axis lies at 2.4 in. above AB.
Calculate I.
I AB  (3t )(2.4) 2  17.28t
I BD 
1
(5.2t )(4.8)2  9.984t
12
I DE  (6t )(2.4)2  34.56t
I EF  I DB  9.984t
I FG  I AB  17.28t
I  I  89.09t
(a)
At point C,
QC  QAB  QBC  (3t )(2.4)  (2.6t )(1.2)  10.32t
 
(b)
VQC
It
 t 
VQ
(600)(10.32t )

 0.23168 in.
(300)(89.09t )
I
t  0.232 in. 
I  (89.09)(0.23168)  20.64 in 4
QE  QEF  QFG
 0  (3)(0.23168)(2.4)  1.668 in 3
E 
VQE
(600)(1.668)

It
(20.64)(0.23168)
 E  209 psi 
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945
3
8
3
8
in.
PROBLEM 6.51
in.
2 in.
2 in.
1
2
in.
a
The design of a beam calls for connecting two vertical rectangular 83  4 -in. plates
by welding them to horizontal 12  2 -in. plates as shown. For a vertical shear V,
determine the dimension a for which the shear flow through the welded surface is
maximum.
a
2 in.
1
2
in.
SOLUTION
3
 1  3 
 1  1
1
I  (2)    (4)3  (2)   (2)    (2)(2)   a 2
 12  8 
 12   2 
2
 4.041667  2a 2 in 4
1
Q  (2)   a  a in 3
2
VQ
Va
q

I
4.041667  2a 2
Set
dq
 0.
da
dq  (4.041667  2a 2 )  (a)(4a) 

V  0
da 
(4.041667  2a 2 )2

2a 2  4.041667
a  1.422 in. 

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946
PROBLEM 6.52
The cross section of an extruded beam is a hollow square of side a  3 in. and
thickness t  0.25 in. For a vertical shear of 15 kips, determine the maximum
shearing stress in the beam and sketch the shear flow in the cross section.
a
a
SOLUTION
Iu  I v 
1
(3.254  2.754 )
12
 4.53125 in 4
Since products of inertia  0,
I x  I y  Iu  I v
I x  4.53125 in 4
V  15 kips
QNA  2[(3 in.  0.25 in.)(1.0607 in.)]  1.59105 in 3
 m   NA 
VQNA
(15 kips)(1.59105 in 3 )

I (2t )
(4.53125 in 4 )(2)(0.25 in.)
 m  10.53 ksi ◄

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947
PROBLEM 6.53
a
a
(a)
An extruded beam has a uniform wall thickness t. Denoting by V the vertical
shear and by A the cross-sectional area of the beam, express the maximum
shearing stress as  max  k (V/A) and determine the constant k for each of the
two orientations shown.
(b)
SOLUTION
(a)
3
a
2
A1  A2  at
h
(b)
3 3
at
4
1
1 3
1
I 2  A2 h 2  at a 2  a 3t
3
3 4
4
5 3
I  2 I1  4 I 2  a t
2
A2 
I1  A1h 2  ath 2 
2
1
a h
 ath 2  at   
12
2 2
1 3
9
7

a t  a 3t  a 3t
48
16
12
3 2
a t
2
3 2
h
Q2  A2 
a t
2
4
Qm  Q1  2Q2  3a 2 t

k
3
1 a
1 3
I2  t   
at
3 2
24
5
I  4 I1  4 I 2  a3t
2
a h 3 2
Q1  at     a t
2 2 4
VQ
V 3a 2 t
3V


3
5
I (2t )
5 at
a t 2t
2


 1  a  1
Q2   at    a 2t
 2  4  8
7
Q  2Q1  2Q2  a 3t
4
V  74 a3t
VQ

m 
5 3
I (2t )
a t (2t )
2
V
6 3 V
6 3V

k
A
5 6at
5 A
6 3
5
1
at
2
I1  I1  A1d 2
Q1  A1h 
m 
a
2
A1  at
h
k  2.08 



7 V 42 V
21 V


20 at 20 6at 10 A
k
V
A
k
21
 2.10 
10
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948
PROBLEM 6.54
u
P
C
rm
t
(a) Determine the shearing stress at point P of a thin-walled pipe of the cross section
shown caused by a vertical shear V. (b) Show that the maximum shearing stress occurs for
  90 and is equal to 2V/A, where A is the cross-sectional area of the pipe.
SOLUTION
A  2 rm t
r
sin 

J  Arm2  2 rm3t
I
1
J   rm3t
2
for a circular arc.
AP  2r t
QP  AP r  2rt sin 
(a)
P 
(b)
m 
VQP (V )(2rt sin  )

I (2t )
 rm3t (2t )

P 

2V sin 2
V sin 

 rm t
m 
2 rm t
2V

A
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949
PROBLEM 6.55
For a beam made of two or more materials with different moduli of elasticity, show that Eq. (6.6)
 ave 
VQ
It
remains valid provided that both Q and I are computed by using the transformed section of the beam
(see Sec. 4.4), and provided further that t is the actual width of the beam where  ave is computed.
SOLUTION
Let Eref be a reference modulus of elasticity.
n1 
E1
E
, n2  2 , etc.
Eref
Eref
Widths b of actual section are multiplied by n’s to obtain the transformed section. The bending stress
distribution in the cross section is given by
x  
nMy
I
where I is the moment of inertia of the transformed cross section and y is measured from the centroid of the
transformed section.
The horizontal shearing force over length  x is

H   ( x ) dA 

n(M ) y
(M )
Q(M )
dA 
ny dA 
I
I
I


Q  ny dA  first moment of transformed section.
Shear flow:
q
 H M Q VQ


x
x I
I
q is distributed over actual width t, thus
q
t
VQ

It
 .
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950
PROBLEM 6.56
90 mm
A composite beam is made by attaching the timber and steel portions shown with
bolts of 12-mm diameter spaced longitudinally every 200 mm. The modulus of
elasticity is 10 GPa for the wood and 200 GPa for the steel. For a vertical shear of
4 kN, determine (a) the average shearing stress in the bolts, (b) the shearing stress
at the center of the cross section. (Hint: Use the method indicated in Prob. 6.55.)
84 mm
90 mm
6 mm
140 mm
6 mm
SOLUTION
Let steel be the reference material.
ns  1.0
Ew
10 GPa

 0.05
Es
200 GPa
nw 
Depth of section:
d  90  84  90  264 mm
For steel portion,
Is  2
For the wooden portion,
Iw 
For the transformed section,
1 3
 1 
bd  (2)   (6)(264)3  18.400  106 mm 4
12
 12 


1
1
(140)(2643  843 )  207.75  106 mm 4
b d13  d 23 
12
12
I  ns I s  nw I w
I  (1.0)(18.400  106 )  (0.05)(207.75  106 )  28.787  106 mm 4  28.787  106 m 4
(a)
Shearing stress in the bolts.
For the upper wooden portion,
Qw  (90)(140)(42  45)  1.0962  106 mm3
For the transformed wooden portion,
Q  nwQw  (0.05)(1.0962  106 )  54.81  103 mm3  54.81  106 m3
Shear flow on upper wooden portion:
q
VQ (4000)(54.81  106 )

 7616 N/m
I
28.787  106
Fbolt  qs  (7616)(0.200)  1523.2 N
Abolt 
Double shear:

4
2
d bolt


 bolt 
4
(12) 2  113.1 mm 2  113.1  106 m 2
Fbolt
1523.2

2 Abolt
(2)(113.1  106 )
 bolt  6.73 MPa 
 6.73  106 Pa


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951
PROBLEM 6.56 (Continued)
(b)
Shearing stress at the center of the cross section.
For two steel plates,
Qs  (2)(6)(90  42)(90  42)  76.032  103 mm3  76.032  106 m3
For the neutral axis,
Q  54.81  106  76.032  106  130.842  106 m3
Shear flow across the neutral axis:
q
VQ (4000)(130.842  106 )

 18.181  103 N/m
6
I
28.787  10
Double thickness:
2t  12 mm  0.012 m
Shearing stress:
 
q
18.181  103

 1.515  106 Pa
2t
0.012
  1.515 MPa 
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952
PROBLEM 6.57
150 mm
12 mm
A composite beam is made by attaching the timber and steel portions shown with
bolts of 12-mm diameter spaced longitudinally every 200 mm. The modulus of
elasticity is 10 GPa for the wood and 200 GPa for the steel. For a vertical shear of
4 kN, determine (a) the average shearing stress in the bolts, (b) the shearing stress at
the center of the cross section. (Hint: Use the method indicated in Prob. 6.55.)
250 mm
12 mm
SOLUTION
Eref  Es  200 GPa
Let
ns  1
nw 
Ew 10 GPa
1


Es 200 GPa 20
Widths of transformed section:
bs  150 mm
 1 
bw    (150)  7.5 mm
 20 
1
 1
I  2  (150)(12)3  (150)(12)(125  6)2   (7.5)(250)3
12
 12
 2[0.0216  106  30.890  106 ]  9.766  106
 71.589  106 mm 4  71.589  106 m 4
(a)
Q1  (150)(12)(125  6)  235.8  103 mm3  235.8  106 m3
VQ1 (4  103 )(235.8  106 )

 13.175  103 N/m
I
71.589  106
 qs  (23.187  103 )(200  103 )  2.635  103 N
q
Fbolt
Abolt 
 bolt 
(b)

 
2
   (12)2  113.1 mm 2  113.1  106 m 2
d bolt
4
4
Fbolt 2.635  103

 23.3  106 Pa
Abolt 113.1  106
 bolt  23.3 MPa 
Q2  Q1  (7.5)(125)(62.5)  235.8  103  58.594  103  294.4  103 mm3  294.4  106 m3
t  150 mm  150  103 m
c 
VQ2
(4  103 )(294.4  106 )

 109.7  103 Pa
It
(71.589  106 )(150  103 )
 c  109.7 kPa 
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953
PROBLEM 6.58
2 in.
Aluminum
1 in.
Steel
1.5 in.
A steel bar and an aluminum bar are bonded together as shown to form a
composite beam. Knowing that the vertical shear in the beam is 4 kips and that
the modulus of elasticity is 29  106 psi for the steel and 10.6  106 psi for the
aluminum, determine (a) the average stress at the bonded surface, (b) the
maximum shearing stress in the beam. (Hint: Use the method indicated in
Prob. 6.55.)
SOLUTION
n  1 in aluminum.
n
29  106 psi
 2.7358 in steel.
10.6  106 psi
Part
nA (in 2 )
Alum.
3.0
2.0
Steel
4.1038
0.5
Σ
7.1038
y (in.)
nA y (in 3 )
d (in.)
nAd 2 (in 2 )
6.0
0.8665
2.2525
1.0
2.0519
0.6335
1.6469
0.3420
3.8994
1.3420
8.0519
Y 
nI (in 4 )
nA y
8.0519

 1.1335 in.
nA
7.1038
I  nAd 2  nI  5.2414 in 4
(a)
At the bonded surface,
Q  (1.5)(2)(0.8665)  2.5995 in 3
 
(b)
At the neutral axis,
VQ
(4)(2.5995)

It
(5.2414)(1.5)
  1.323 ksi 
 1.8665 
3
Q  (1.5)(1.8665) 
  2.6129 in
2


 max 
VQ
(4)(2.6129)

It
(5.2814)(1.5)
 max  1.329 ksi 
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954
PROBLEM 6.59
2 in.
Steel
1 in.
Aluminum
1.5 in.
A steel bar and an aluminum bar are bonded together as shown to form a
composite beam. Knowing that the vertical shear in the beam is 4 kips and
that the modulus of elasticity is 29  106 psi for the steel and 10.6  106 psi for
the aluminum, determine (a) the average stress at the bonded surface, (b) the
maximum shearing stress in the beam. (Hint: Use the method indicated in
Prob. 6.55.)
SOLUTION
n  1 in aluminum.
n
29  106 psi
 2.7358 in steel.
10.6  106 psi
Part
nA (in 2 )
y (in.)
nA y (in 3 )
d (in.)
nAd 2 (in 2 )
nI (in 4 )
Steel
8.2074
2.0
16.4148
0.2318
0.4410
2.7358
Alum.
1.5
0.5
0.75
1.2682
2.4125
0.1250
Σ
9.7074
2.8535
2.8608
17.1648
nA y 17.1648

 1.7682 in.
9.7074
A
I  nAd 2  nI  5.7143 in 4
Y 
(a)
At the bonded surface,
Q  (1.5)(1.2682)  1.9023 in 3

(b)
At the neutral axis,
VQ
(4)(1.9023)

It
(5.7143)(1.5)
  0.888 ksi 
 1.2318 
Q  (2.7358)(1.5)(1.2318) 
 3.1133 in 3

 2 
 max 
VQ
(4)(3.1133)

It
(5.7143)(1.5)
 max  1.453 ksi 
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955
PROBLEM 6.60
P
Consider the cantilever beam AB discussed in Sec. 6.5 and the
portion ACKJ of the beam that is located to the left of the transverse
section CC and above the horizontal plane JK, where K is a point at
a distance y  yY above the neutral axis. (See figure.) (a) Recalling
that  x   Y between C and E and  x  ( Y /yY ) y between E and K,
show that the magnitude of the horizontal shearing force H exerted
on the lower face of the portion of beam ACKJ is
Plastic
C
A
J
E
yY
K
B
C'
E'
H
y

1
y2 
b Y  2c  yY 

2
yY 

x
Neutral axis
(b) Observing that the shearing stress at K is
H
1 H 1  H
 lim

 A 0  A
 x 0 b  x
b x
 xy  lim
and recalling that yY is a function of x defined by Eq. (6.14), derive
Eq. (6.15).
SOLUTION
Point K is located a distance y above the neutral axis.
The stress distribution is given by
y
  Y
for 0  y  yY and    Y
yY
for
yY  y  c.
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956
PROBLEM 6.60 (Continued)
For equilibrium of horizontal forces acting on ACKJ,

H   dA 

yY
 Y yb

c
 Y b dy
yY
yY
 b  y2  y2 
 Y  Y
   y b(c  yY )
yY 
2 

1
y2 
H  b  Y  2c  yY 

yY 
2

y
dy 
(a)
Note that yY is a function of x.
y 2 dyY 
1  H 1   yY
 Y  


b  x 2   x
yY 2 dx 
1 
y 2  dyY
   Y 1 

2 
yY 2  dx
 xy 
But
Differentiating,
M  Px 
 1 yY2 
3
M y 1 
 3 c 2 
2


dM
3
 2 yY dyY 
 P  MY  

2
dx
2
 3 c dx 
dyY
Pc 2
Pc 2
3 P



2
2
dx
yY M Y
2  Y byY
yY 3  Y bc
Then
1
2



 xy   Y 1 
y2
yY 2
3 P
3P


 2  Y b Y 4byY


y2 
1 


yY 2 

(b)
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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957
a
A
PROBLEM 6.61
B
a
D
E
Determine the location of the shear center O of a thin-walled beam of uniform
thickness having the cross section shown.
a
O
e
F
G
a
H
J
SOLUTION
2
1
9
 3a 
I AB  I HJ  at    at 3  ta 3
2
12
4
 
2
1
1
a
I DE  I FG  at    at 3  ta 2
4
 2  12
1
9
I AH  t (3a )3  ta 3
12
4
29 3
I  I 
ta
4
Part AB:
3a
3
Q  atx
2
2
3
VQ V  2 atx
6Vx



3
29
It
ta t
29a 2 t
4
A  tx
y


F1   dA 
Part DE:
a
0
6Vx
6V
t dx 
2
29a t
29a 2

x dx 
3
V
29
x dx 
1
V
29
a
0
a
1
Q  atx
2
2
1
2Vx
VQ V  2 atx



3
29
It
29 a 2t
ta t
4
A  tx

y
F2   dA 
M K 

a
0
2Vx
2V
t dx 
2
29a t
29a 2

a
0
M K : Ve  F1 (3a )  F2 (a )

e
9
1
10
Va  Va  Va
29
29
29
10
a
29
e  0.345a 
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958
PROBLEM 6.62
A
D
a
B
a
Determine the location of the shear center O of a thin-walled beam of uniform
thickness having the cross section shown.
O
e
a
F
E
a
G
2a
SOLUTION
2
1 3
7
 3a 
ta  (ta)    ta 3
12
2
3
 
1
 I EF  (2at ) a 2  (2a ) t 3  2a3t
12
1
2
28 3
I  I 
ta
 t (2a )3  ta 3
12
3
3
I AB  I FG 
I DB
I DE
Part AB:
A  t (2a  y );
Q  Ay 
y
2a  y
3
1
t (2a  y )(2a  y )
2
1
t (4a 2  y 2 )
2
VQ V
(4a 2  y 2 )


2I
It
2a V
F1   dA 
(4a 2  y 2 ) t dy
a 2I
Vt  2
y 3  2 a Vta 3 
(2)3
 1 

 (4)(1)    
 4a y   
(4)(2) 
2I 
3 a
2I 
3
 3 




5 Vta 3
5
 V
6 I
56
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959
PROBLEM 6.62 (Continued)
3a
 txa
2
 3a

 ta   x 
2


Q  (ta )
Part DB:

VQ Va  3a


 x
It
I  2


F2   dA 

Vta
 3a

 x  t dx 

I  2
I

2 a Va
0
Vta  3ax x 2


I  2
2
5
MH 




2a
0


2a 
0
3a

 2  x  dx


Vta 3  (3)(2) (2)2 



I  2
2 
3
Vta
15
 V
I
28
 M H:
Ve  F2 (2a )  2 F1 (2a )

30
20
5
Va  Va  Va
28
56
7
e
5
a  0.714a 
7
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960
D
PROBLEM 6.63
B
a
A
G
O
e
Determine the location of the shear center O of a thin-walled beam of uniform
thickness having the cross section shown.
a
E
2a
F
SOLUTION
I AB  I FG 
1 3
ta
3
1
2att 3  2ta3
12
I DB  I CP  2ata3 
1
2
16 3
t (2a)3  ta 3 I  I 
ta
12
3
3
y
1
A  ty y 
Q  ty 2
2
2
I DE 
Part AB:
2
VQ V  12 ty
3Vy 2
 
 16 3 
It
ta t
32a3t
3
a
F1    dA   0  t dy 
Part BD:
Q  QB  txa 
 
VQ

It
3V a 2
1
y dt 
V
3 0
32
32a
1 2
ta  tax
2
Vt  1 2

 a  ax 
a3t  2

16
3
3 V
( a  2 x)
32 a 2
2 a 3V
F2    dA   0
(a  2 x)dx
32a 2
2a
3V

(ax  x 2 )
2
0
32a
3V
9

(2a 2  4a 2 )  V
16
32a 2

M H 
M H :
Ve  (2a)(2 F1)  (2a)( F2 )

1
9
5
Va  Va  Va
8
8
4
e
5
a  1.250a 
4
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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on a website, in whole or part.
961
a
PROBLEM 6.64
b
D
A
B
h
O
Determine the location of the shear center O of a thin-walled beam of uniform
thickness having the cross section shown.
e
E
G
F
SOLUTION
2
Part AD:
1
1
h
I AB  I EF  (a  b)t    (a  b)t 3  t (a  b)h 2
4
 2  12
1
1
I DG  th3
I  I  t (6a  6b  h)h 2
12
12
h 1
Q  tx  thx
2 2
VQ Vhx


It
2I
a Vhx
Vht a
F1   dA 
t dx 
x dx
0 2I
2I 0


Part BD:

Vht x 2
2I 2

a

0
Vhta 2
4I
h 1
 thx
2 2
VQ Vhx


2I
It
b Vhx
Vht
F2   dA 
t dx 
0 2I
2I
Q  tx


Vht x 2

2I 2
M H 
b

0

b
0
x dx
Vhtb 2
4I
M H :
Vh 2 t (b 2  a 2 )
4I
2
2
2
3V (b 2  a 2 )
Vh t (b  a )


2
6a  6b  h
4  121 t (6a  6b  h)h
Ve  F2 h  F1h 
e
3(b 2  a 2 )

6(a  b)  h
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962
PROBLEM 6.65
6 mm
B
A
An extruded beam has the cross section shown. Determine (a) the location of the
shear center O, (b) the distribution of the shearing stresses caused by the vertical
shearing force V shown applied at O.
12 mm
O
C
192 mm
e
V ⫽ 110 kN
6 mm
E
D
72 mm
SOLUTION
2
 1 
 192   1
3
I  2   (72)(6)3  (72)(6) 
   12 (12)(192)
12
2

 
 
 15.0431  106 mm 4  15.0431  106 m 4
Part AB:
 192 
Q  Ay  (6 x) 
  576 x
 2 
VQ 576Vx

q
I
I
A  6x
x  0 at point A. x  l AB  72 mm at point B.
F1 

MC 
(a)
(b)

xB
xA
q dx 

72
0
576Vx
576V (72)2
dx 
2
I
I
2
(288)(72)
V  0.099247 V
15.0431  106
M C : Ve  (0.099247 ) V (192)
e  19.0555 mm
Point A: x  0,
e  19.06 mm 
Q  0,
Point B in part AB:
A  0 
q0
x  72 mm
Q  (576)(72)  41.472  103 mm3  41.472  106 m3
t  6 mm  0.006 m
B 
VQ (110  103 )(41.472  106 )

It
(15.0431  106 )(0.006)
 B  50.5 MPa 
 50.5  106 Pa
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on a website, in whole or part.
963
PROBLEM 6.65 (Continued)
Part BD:
Point B:
y  96 mm
Q  41.472  103 mm3  41.472  106 m3
t  12 mm  0.012 m
B 
VQ (110  103 )(41.472  106 )

It
(15.0431  106 )(0.012)
 B  25.3 MPa 
 25.271  106 Pa
Point C:
y  0, t  0.012 m
 96 
Q  41.472  103  (12)(96)    96.768  103 mm3  96.768  106 m3
 2 

VQ (110  103 )(96.768  106 )

 58.967  106 Pa
It
(15.0431  106 )(0.012)
 C  59.0 MPa 
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964
PROBLEM 6.66
4.0 in.
D
B
O
A
G
e
6.0 in.
An extruded beam has the cross section shown. Determine (a) the
location of the shear center O, (b) the distribution of the shearing stresses
caused by the vertical shearing force V shown applied at O.
V ⫽ 2.75 kips
E
F
t⫽
1
8
in.
SOLUTION
1
I AB  (0.125)(3)3  1.125 in 4
3
1
I BD  (4)(0.125)3  (4)(0.125)(3)2  4.50065 in 4
12
1
I DE  (0.125)(6)3  2.25 in 4
12
I EF  I BD  4.50065 in 4
I FG  I AB  1.125 in 4
I   I  13.50 in 4
(a)
Part AB:
y
 0.5ty 2
2
VQ ( y ) 0.5Vt 2
q( y ) 
y

I
I
3
0.5Vt
FAB  q ( y ) dy 
0
I
Q ( y )  ty

Its moment about H is
4 FAB  18

3
0
y 2 dy  4.5
Vt

I
Vt
I
QB  (0.5)(t )(3) 2  4.5 t
Part BD:
Q( x)  QB  xt (3)  (4.5  3x) t
Vq ( x) Vt
 (4.5  3 x)
I
I
4
t 4
Vt
 q ( x) dx 
(4.5  3x) dx  42 
0
I 0
I
q( x) 
FBD
Its moment about H : 3FBD  126


Vt
I
QD  [4.5  (3)(4)] t  16.5 t
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965
PROBLEM 6.66 (Continued)
By symmetry with part BD, FEF  42
Part EF:
Its moment about H is 3FEF  126
Vt

I
Vt
I
By symmetry with part AB, FFG  4.5
Part FG:
Its moment about H is 4 FFG  18
VT

I
Vt
I
Moment about H of force in part DE is zero.
Vt
144Vt
(18  126  0  126  18) 
I
I
144 t (2.88)(0.125)

e
13.50
I
Ve   M H 
(b)
e  2.67 in. 
 A  G  0 
QA  QG  0
QB  QF  4.5t
B F 
VQB (2.75)(4.5 t )

13.50 t
It

QD  QE  16.5t 

D E 
 B   F  0.917 ksi 
VQ0 (2.75)(16.5t )


13.50 t
It
At H (neutral axis),
 D   E  3.36 ksi 
QH  QD  t (3)(1.5)  21t
H 
VQH (2.75)(21t )

13.50t
It
 H  4.28 ksi 
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966
PROBLEM 6.67
A
2 in.
B
D
O
6 in.
An extruded beam has the cross section shown. Determine (a) the location of
the shear center O, (b) the distribution of the shearing stresses caused by the
vertical shearing force V shown applied at O.
e
V 5 2.75 kips
F
E
2 in.
G
4 in.
t5
1
8
in.
SOLUTION
Part
A (in2)
d (in.)
Ad 2 (in 4 )
I (in 4 )
BD
0.50
3
4.50
0
ABEG
1.25
0
0
EF
0.50
3
4.50
0

2.25
9.00
10.417
10.417
I   Ad 2   I  19.417 in 4
(a)
Q( x)  3 tx
VQ ( x) V
q( x) 
 (3tx)
I
I
Part BD:
3Vt
24Vt
(8) 
0
I
I
72Vt
( M BD ) H  3FBD 
I
FBD 
Its moment about H:
Part EF:
By same method,
3Vt
I

4
x dx 
FEF 
24Vt
I
( M EF ) H 
72Vt
I
Moments of FAB , FBE , and FEG about H are zero.
Ve   M H 
e
72Vt 72Vt 144Vt


I
I
I
144 t (144)(0.125)

19.417
I
e  0.927 in. 
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967
PROBLEM 6.67 (Continued)
(b)
 A   D   F  G  0 
At A, D, F , and G,
Q0
Just above B :
Q1  QAB  (2t )(4)  8 t
1 
Just to the right of B:
VQ2 (2.75)(12t )

(19.417) t
It
 2  1.700 ksi 
Q3  Q1  Q2  20t
3 
At H (neutral axis),
1  1.133 ksi 
Q2  QBD  (3) t (4)  12t
2 
Just below B:
VQ1 (2.75)(8t )

(19.417) t
It
VQ3 (2.75)(20t )

(19.417) t
It
 3  2.83 ksi 
QH  Q3  QBH  20t  t (3)(1.5)  24.5 t
H 
VQH (2.75)(24.5t )

(19.417) t
It
 H  3.47 ksi 
 4   3  2.83 ksi 
By symmetry,
 5   2  1.700 ksi 

 6  1  1.133 ksi 
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968
PROBLEM 6.68
6 mm
A
B
6 mm
z
30 mm
4 mm
D
O
E
An extruded beam has the cross section shown. Determine (a) the location
of the shear center O, (b) the distribution of the shearing stresses caused by
the vertical shearing force V shown applied at O.
30 mm
4 mm
e
F
N = 35 kN
G
30 mm
6 mm
H
J
30 mm
Iz = 1.149 × 106 mm4
SOLUTION
1
(30)(6)3  (30)(6)(45)2  0.365  106 mm 4
12
1
 (30)(4)3  (30)(4)(15) 2  0.02716  106 mm 4
12
I AB  I HJ 
I DE  I FG
1
(6)(90)3  0.3645  106 mm 4
12
I  I  1.14882  106 mm 4
I AH 
(a)
For a typical flange,
A( s )  ts
Q( s )  yts
VQ( s ) Vyts

I
I
b
Vytb 2
F  q( s )ds 
0
2I
q(s) 

Flange AB:
FAB 
V (45)(6)(302 )
 0.10576V 
(2)(1.14882  106 )
Flange DE:
FDE 
V (15)(4)(30)2
 0.023502V 
(2)(1.14882  106 )
Flange FG:
FFG  0.023502V 
Flange HJ:
FHJ  0.10576V 
M K 
M K : Ve  45 FAB  15FDE  15 FFG
 45 FHJ  10.223V
e  10.22 mm 
Dividing by V,
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969
PROBLEM 6.68 (Continued)
(b)
Calculation of shearing stresses.
I  1.14882  106 m 4
V  35  103 N
 0 
At B, E, G, and J,
At A and H,
Q  (30)(6)(45)  8.1  103 mm3  8.1  106 m3
t  6  103 m

VQ
(35  103 )(8.1  106 )

 41.1  106 Pa
It
(1.14882  106 )(6  103 )
  41.1 MPa 
Just above D and just below F:
Q  8.1  103  (6)(30)(30)  13.5  103 mm3  13.5  106 m3
t  6  103 m

VQ
(35  103 )(13.5  106 )

 68.5  106 Pa
It
(1.14882  106 )(6  103 )
  68.5 MPa 
Just to right of D and just to the right of F:
Q  (30)(4)(15)  1.8  103 mm3  1.8  106 m3

t  4  103 m
VQ
(35  103 )(1.8  106 )

 13.71  106 Pa
It
(1.14882  106 )(4  103 )
  13.71 MPa 
Just below D and just above F:
Q  13.5  103  1.8  103  15.3  103 mm3  15.3  106 m3
t  6  103 m

At K,
VQ
(35  103 )(15.3  106 )

 77.7  106 Pa
It
(1.14882  106 )(6  103 )
  77.7 MPa 
Q  15.3  103  (6)(15)(7.5)  15.975  103 mm3  15.975  106 m3

VQ (35  103 )(15.975  106 )

 81.1  106 Pa
It
(1.14882  106 )(6  103 )
  81.1 MPa 
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970
PROBLEM 6.69
A
1
4
2 in.
in.
B
O
D
3 in.
Determine the location of the shear center O of a thin-walled beam of uniform
thickness having the cross section shown.
3 in.
r
E
2 in.
F
4 in.
SOLUTION
LDB 
42  32  5 in.
1
ADB  LDBt  (5)    1.25 in 2
4
I DB 
1
1
AAB h 2    (1.25)(3)2  3.75 in 4
3
3
I AB 
1 1 3 1
2
4
  (2)    (2)(4)  8.1667 in
12  4 
4
I  (2)(3.75)  (2)(8.1667)  23.833 in 4
Part AB:
1
(5  y) in 2
4
1
y  (5  y) in.
2
1
1
Q  Ay  (5  y)(5  y)  (25  y 2 )
8
8
A
 
VQ
V (25  y 2 )
V (25  y 2 )


It
(8)(23.833)(0.25)
47.667
 y2) 1
 dy
(47.667) 4
5 V (25
F1    dA   3
5
1 
V

25 y  y 3 


190.667 
3 3

MD 
V
1
1


(25)(5)  (5)3  (25)(3)  (3)3   0.09091V
190.667 
3
3

M D : Ve  2F1(4)  0.7273V
e  0.727 in. 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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971
PROBLEM 6.70
B
6 mm
35 mm
60⬚
O
35 mm
D
e
Determine the location of the shear center O of a thin-walled beam of
uniform thickness having the cross section shown.
A
F
60⬚
E
SOLUTION
1
I DB  (6)(35)3  85.75  103 mm 4
3
LAB  70 mm
AAB  (70)(6)  420 mm 2
I AB 
1
1
AAB h 2    (420)(35)2  171.5  103 mm 4
3
 3
I  (2)(85.75  103 )  (2)(171.5  103 )  514.5  103 mm 4
A  ts  6 s
1
1
y  s sin 30  s
2
4
3 2
Q  Ay  s
2
VQ 3Vs 2


It
It
Part AB:

F1   dA 

M D 

70 3Vs 2
0
2 It
t ds 
3V
I

70
0
s 2 ds
(3)(70)3
1
V V
(2)(3) I
3
 M D:
Ve  2[( F1 cos 60°)(70 sin 60°)]
 20.2V
e  20.2 mm 
Dividing by V,
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972
PROBLEM 6.71
4 in.
A
Determine the location of the shear center O of a thin-walled beam of uniform
thickness having the cross section shown.
3 in.
B
O
5 in.
D
3 in.
e
E
SOLUTION
LAB  42  32  5 in.
AAB  5t
1
1
AAB h 2  AAB d 2  (5t )(3)2  (5t )(4)2  83.75t in 4
12
12
1
I BD  (t )(5)3  10.417t in 4
12
I  2 I AB  I BD  177.917t in 4
I AB 
Q  QAB  QBY
In part BD,
1
Q  (5t )(4)  (2.5  y )t   (2.5  y )
2
1
1 

 20t  3.125t  ty 2   23.125  y 2  t
2
2 


VQ
It

FBD   dA 
2.5
V (23.125  12 y 2 )t
2.5
It

 t dy
2.5
1 
Vt 
1 

23.125  y 2  dy   23.125 y  y 3 

 2.5 
2 
I 
6  2.5

Vt
I

Vt 
(2.5)3  Vt (110.417)
 2 (23.125)(2.5) 
 0.62061V

I
6 
177.917t


2.5
M K 
e
10
 10

M K : V   e    (0.62061V )
3
 3

10
[1  0.62061]
3
e  1.265 in. 
Note that the lines of action of FAB and FDE pass through point K. Thus, these forces have zero moment
about point K.
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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973
PROBLEM 6.72
A
B
60 mm
O
D
Determine the location of the shear center O of a thin-walled beam of
uniform thickness having the cross section shown.
60 mm
e
E
F
80 mm
40 mm
SOLUTION
I AB  (40t )(60)2  144  103t
LDB  802  602  100 mm
ADB  100t
1
1
ADB h 2  (100t )(60)2  120  103t
3
3
I  2 I AB  2 I DB  528  103t
I DB 
A  tx y  60 mm
Part AB:
Q  A y  60tx mm3
VQ V (60tx) 60Vx



It
It
I
40 60Vx
60Vt
F1   dA 
t dx 
0
I
I


M D 

60Vt x 2
I 2
 M D:
30
0


40
0
x dx
(60)(30)2 Vt
 0.051136V
(2)(528  103 ) t
Ve  (0.051136V )(120)
e  6.14 mm 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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974
PROBLEM 6.73
a
O
t
A
B
Determine the location of the shear center O of a thin-walled beam of uniform
thickness having the cross section shown.
e
SOLUTION
For whole cross section,
A  2 at
J  Aa 2  2 a 3t
I
1
J   a 3t
2
Use polar coordinate  for partial cross section.
A  st  a t
s  arc length
sin 
1
r a
where   

2
2
sin 
y  r sin   a

sin 2 
Q  Ay  a t a
 a 2t 2sin 2


But
M C  Ve,
2
 a 2 t 2 sin 2 
 a 2 t (1  cos  )
VQ Va 2

(1  cos  )
It
I
M C  a  dA 




2
0
Va3
Va 4t
(1  cos  ) tad 
(  sin  )
I
I
2
0
2 Va t
 2aV
 a 3t
4
e  2a 
hence
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975
A
a
O
PROBLEM 6.74
Determine the location of the shear center O of a thin-walled beam of uniform thickness
having the cross section shown.
t
B
e
SOLUTION
For a thin-walled hollow circular cross section, A  2 at
J  a 2 A  2 a 3t
I
For the half-pipe section,

2
I
1
J   a 3t
2
a 3t
Use polar coordinate  for partial cross section.
A  st  a  t
r a
sin 

s  arc length
where  
y  r cos   a
Q  Ay  a t a

2
sin  cos 

sin  cos 

 a 2t (2 sin  cos  )
 a 2t sin 2  a 2t sin 

VQ Va 2

sin 
It
I

M H  a  dA 
2


0
Va 2
Va 4 t
sin  ta d 
 cos 
a
I
I

0
4
Va t 4
 Va
I

But M H  Ve, hence
e
4

a  1.273a 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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976
3
4
PROBLEM 6.75
in.
3
4
in.
A thin-walled beam has the cross section shown. Determine the
location of the shear center O of the cross section.
1
2 in.
O
8 in.
6 in.
e
8 in.
SOLUTION
I
1 3 1
t1h1  t2 h23
12
12
1

A   h2  y  t2
2


11

y   h2  y  t2
22

Q  Ay
Right flange:
11
 1

h2  y  h2  y  t2

22
 2

11

  h22  y 2  t2
24

VQ
V 1 2



h2  y 2  t2

It2 2 It2  4


Vt2
Vt2  1 2
2
F2   dA 
 h2  y  t2 dy  2 I
 h2 / 2 2 I t  4

2



Vt2
2I
h2 / 2
1 2
y3 
 h2 y 

3 
4
h2 / 2
h2 / 2
 1 2 h 1  h 3 1 2 h 1  h 3  Vt h3
Vt2 h23
2
2
2
2
2 2





h
h
 2

2
2 3  2  4
2 3  2   12 I
t1h13  t2 h23
 4
MH 
M H:
 Ve   F2 b  V
e
t2 h23b

t1h13  t2 h23
t2 h23b
t1h13  t2 h23
(0.75)(6)3(8)
(0.75)(8)3  (0.75)(6)3
e  2.37 in. 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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977
50 mm
PROBLEM 6.76
50 mm
A
6 mm
D
O
80 mm
A thin-walled beam has the cross section shown. Determine the
location of the shear center O of the cross section.
F
60 mm
40 mm
G
E
B
e
SOLUTION
Let
h1  AB  h, h2  DE ,
I 
Part AB:

1
t h13  h23  h33
12
h3  FG

1

A   h1  y  t
2


y 
11

 h1  y 
22

Q  Ay 

 
1 1
 1

t  h1  y  h1  y 
2 2
 2

1 1 2

t  h1  y 2 
2 4

VQ
V 1 2
2

 h1  y 
It
2I  4

1h
F1    dA   2 1 h1
1
2
V
2I
1 2
2
 h1  y  t dy
4


1
h1
Vt  1 2
y3  2

 h1 y 

2I  4
3  1
 h1
2

Vt  1 2 1
1  h1 
h1 h1   
3 2 
I 4 2


h13V
h13  h23  h33
3
Vth13


12 I

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
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978
PROBLEM 6.76 (Continued)
Likewise, for Part DE,
F2 
h23V
h13  h23  h33
and for Part FG,
F3 
h33V
h13  h23  h33
M H 
e
M H : Ve  bF2  2bF3 
bh23  2bh33
V
h13  h23  h33
h23  2h33
(60)3  (2)(40)3
b
(50)
3
3
 h2  h3
(80)3  (60)3  (40)3
h13
 21.7 mm
e  21.7 mm ◄
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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979
PROBLEM 6.77
A
60 mm
B
60 mm
60 mm
D
O
F
A thin-walled beam of uniform thickness has the cross section shown. Determine the
dimension b for which the shear center O of the cross section is located at the point
indicated.
E
G
b
SOLUTION
Part AB:
1
s
2
1
Q( s )  A( s ) y ( s)  ty A s  ts 2
2
VQ( s ) Vt 
1 
q(s) 
  yAs  s2 
I
I 
2 
A( s )  ts
FAB 

y ( s)  y A 
l AB
0
q( s ) ds
2
l3 
Vt  y Al AB
 AB  

6 
I  2
1 2
QB  ty Al AB  tl AB
2
FFG  FAB

At B,
By symmetry,
Part BD:
A( x)  tx
Q( x)  QB  yB A( x)  QB  tyB x
VQ( x) V
 (QB  tyB x)
I
I
b
V
1

 q ( x) dx   QB b  tyB b 2  
0
I
2

q( x) 
FBD
By symmetry,

FEF  FBD
FDE is not required, since its moment about O is zero.
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980
PROBLEM 6.77 (Continued)
 M O  0 : b( FAB  FFG )  yB FBD  yF FEF  0
2b FAB  2 yB FBD  0
2b 
2
l3 
Vt  y Al AB
V
 AB   2 yB


I  2
I
6 
1

2
 QB b  2 tyB b   0


2Vt  1
1 3 
2Vt 
1 2 
1 2 2
2
 y A l AB  l AB  b 
 y Al AB  l AB  yB b  yB b  0
I 2
I 
6
2
2


Dividing by
2Vt
and substituting numerical data,
I
1
1
1
1

2
3
2
2 2
 (90)(60)  (60)  b  (90)(60)  (60)  (30)b  (30) b  0
6
2
2
2



126  103b  108  103b  450b 2  0
18  103b  450b 2  0
b  0 and b  40.0 mm 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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981
A
PROBLEM 6.78
B
1 in.
D
A thin-walled beam of uniform thickness has the cross section shown.
Determine the dimension b for which the shear center O of the cross section
is located at the point indicated.
E
8 in.
10 in.
O
F
G
1 in.
H
b
J
3 in.
SOLUTION
Part AB:
A  tx
 
y  5 in.
VQ V .5 tx 5Vx


It
It
I
3 5Vx
F1    dA   0

Part DE:
Q  Ay  5tx
I
tdx 
5Vt 3
 x dx
I 0
2
(5)(3) Vt
Vt
 22.5
2
I
I
A  tx
y  4 in.
Q  Ay  4tx
VQ V 4tx
4Vx


It
It
I
a 4Vx
F2    dA   0
t dx
I
4Vt a

 x dx
I 0
 

M O 
a2 
M O :
2Vta 2
I
Vt 
2Vta 2

O  (10)  22.5   (8)
I 
I

(10)(22.5)
 14.0625 in 2
(8)(2)
a  3.75 in. 
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982
PROBLEM 6.79

For the angle shape and loading of Sample Prob. 6.6, check that q dz  0 along the horizontal leg of the angle
and q dy  P along its vertical leg.

SOLUTION
Refer to Sample Prob. 6.6.
3P(a  z )(a  3 z )
3P 2

(a  4az  3z 2 )
4ta 3
4ta3
a
3P a
q dz   f t dz  3 (a 2  4az  3z 2 )dz
0
4a 0
a
3P 
z2
z2 
 3  a 2 z  4a  3 
2
3 0
4a 
f 
Along horizontal leg:



3P 3
( a  2a 3  a 3 )  0
4a 3
3P(a  y )(a  5 y )
3P 2

(a  4ay  5 y 2 )
e 
3
3
4ta
4ta
a
a
3P
q dy   e t dy  3 (a 2  4ay  5y 2 )dy
0
4a 0

Along vertical leg:






3P  2
y2
y3 
a
y
a


4
5


2
3 
4a3 
a
0
3P  3
5 3  3P 4 3
3
 a  2a  3 a   3  3 a  P
4a 3 
 4a

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983
PROBLEM 6.80
For the angle shape and loading of Sample Prob. 6.6, (a) determine the points where the shearing stress is
maximum and the corresponding values of the stress, (b) verify that the points obtained are located on the
neutral axis corresponding to the given loading.
SOLUTION
Refer to Sample Prob. 6.6.
(a)
e 
Along vertical leg:
3P(a  y )(a  5 y )
3P 2

(a  4ay  5 y 2 )
3
4ta
4ta 3
d e
3P
(4a  10 y )  0

dy 4ta 3
2

2 
 2   3P  9 2 
a 
 a 2  (4a )  a   (5)  a   
3 
5 
 5   4ta  5 

3P(a  z )(a  3 z )
3P 2

(a  4az  3z 2 )
f 
3
4ta
4ta3
m 
Along horizontal leg:
d f
dz

m 

At the corner:
(b)
1
I y  ta 3
3
3P
4ta3
3P
(4a  6 z )  0
4ta 3
3P
4ta 3
2
 2
2 
 2   3P  5 2 
a
a
a
a


 a 
(4
)
(3)

3 
3  
3 



  4ta  3 

y  0, z  0,
I z 
tan  
y
m 
2
a 
5
27 P

20 ta
z
2
a 
3
m  
1 P

4 ta

3P

4 ta
1 3
ta   45
12
I z
1
tan  
  14.036
4
I y
    45  14.036  30.964
Ay at (a/2) 1

 a
A
2at
4
Az at (a/2) 1

 a
z
A
2at
4
y
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984
PROBLEM 6.80 (Continued)
Neutral axis intersects vertical leg at
y  y  z tan 30.964
1 1

   tan 30.964  a  0.400a
4 4

y
2
a 
5
z
2
a 
3
Neutral axis intersects horizontal leg at
z  z  y tan (45 +  )
1 1

   tan 59.036  a  0.667a
4 4

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985
PROBLEM 6.81
P
D'
Determine the distribution of the shearing stresses along line DB
in the horizontal leg of the angle shape for the loading shown. The
x and y  axes are the principal centroidal axes of the cross
section.
D
a
A'
B'
B
A
2a
0.596a
y'
D'
B'
0.342a
C'
2
3
y
a
6
a
A'
15.8⬚
x'
Ix' ⫽ 1.428ta3
Iy' ⫽ 0.1557ta3
x
SOLUTION
  15.8
Vx  P cos 
A( y )  (2a  y )t
Coordinate transformation.
In particular,
y
Vy   P sin 
1
(2a  y ),
2
x 0
2 
1 


y    y  a  cos    x  a  sin 
3 
6 


1 
2 


x   x  a  cos    y  a  sin 
6 
3 


2 
1 


y    y  a  cos    x  a  sin 
3 
6 


1 
1
 1 
  y  a  cos     a  sin 
3 
2
 6 
 0.48111y  0.36612a
1 
2 


x   x  a  cos    y  a  sin 
6
3 



1 
 1 
1
   a  cos    y  a  sin 
3 
 6 
2
 0.13614 y  0.06961a
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986
PROBLEM 6.81 (Continued)
  
Vx Ax Vy A y 

I yt
I x t
( P cos  )(2a  y )(t )(0.13614 y  0.06961a)
(0.1557ta 3 )(t )
( P sin  )(2a  y )(0.48111 y  0.36612a)

(1.428 a 3t )(t )
P(2a  y )(0.750 y  0.500a)

ta3

y (a )
P
  
 at 

0
1
3
2
3
1
4
3
5
3
2
1.000
0.417
0
0.250
0.333
0.250
0
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987

PROBLEM 6.82
P
D'
For the angle shape and loading of Prob. 6.81, determine the
distribution of the shearing stresses along line DA in the vertical
leg.
D
a
A'
B'
B
A
2a
0.596a
PROBLEM 6.81* Determine the distribution of the shearing
stresses along line DB in the horizontal leg of the angle shape for
the loading shown. The x and y  axes are the principal centroidal
axes of the cross section.
y'
D'
B'
0.342a
C'
2
3
y
a
6
a
A'
15.8⬚
x'
Ix' ⫽ 1.428ta3
Iy' ⫽ 0.1557ta3
x
SOLUTION
  15.8
Vx  P cos 
Vx   P sin 
x
1
(a  x),
2
A( x)  (a  x)t
y 0
Coordinate transformation.
2 
1 


y    y  a  cos    x  a  sin 
3 
6 


1 
2 


x   x  a  cos    y  a  sin 
6
3 



In particular,
2 
1 


y    y  a  cos    x  a  sin 
3 
6 


1 
 2 
1
   a  cos    x  a  sin 
3 
 3 
2
 0.13614 x  0.73224a
1 
2 


x   x  a  cos    y  a  sin 
6 
3 


1 
1
 2 
  x  a  cos     a  sin 
3 
2
 3 
 0.48111x  0.13922a
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988
PROBLEM 6.82* (Continued)

Vx A( x) x Vy A( x) y 

I y t
I x t
( P cos  )(a  x)(t )(0.48111x  0.13922a)
(0.1557ta3 )(t )
( P sin  )(a  x)(t )(0.13614 x  0.73224a)

(1.428ta 3 )(t )
P(a  x)(3.00 x  1.000a )

ta 3

x( a )
P
  
 at 
0
1
6
1
3
1
2
2
3
5
6
1
1.000
1.250
1.333
1.250
1.000
0.583
0

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989

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.
B
100 mm
A
D
E
P ⫽ 15 kN
30 mm
SOLUTION
Use results of Example 6.06 with
e
b  30 mm,
h  100 mm, and t  8 mm.
b
30

 9.6429 mm  9.6429  103 m
100
h
2  3b 2  (3)(30)
1 2
1
th (6b  h)  (8)(100)2 [(16)(30)  100]  1.86667  106 mm 4  1.86667  106 m 4
12
12
3
V  15  10 N
I
(a)
T  Ve  (15  103 )(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
 (8)(100) 100  (4)(30)  22  103 mm3  22  106 m3
8
t  8  103 m
V 
VQ
(15  103 )(22  106 )

 22.10  106 Pa  22.10 MPa
6
3
It
(1.86667  10 )(8  10 )
Stress due to T :
a  2b  h  160 mm  0.160 m
1
t  1
8 
 0.3228
c1  1  0.630   1  (0.630)
3
a 3
160 
T
144.64
V 

 43.76  106 Pa  43.76 MPa
3 2
2
c1at
(0.3228)(0.160)(8  10 )
(b)
By superposition,
 max   V   T
 max  65.9 MPa 
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990
PROBLEM 6.84*
Solve Prob. 6.83, assuming that a 6-mm-thick plate is bent to form the
channel shown.
B
100 mm
A
D
E
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.
P ⫽ 15 kN
30 mm
SOLUTION
Use results of Example 6.06 with b  30 mm, h  100 mm, and t  6 mm.
e
b
30

 9.6429 mm  9.6429  103 m
100
h
2  3b 2  (3)(30)
1 2
1
th (6b  h)  (6)(100) 2 [(6)(30)  100]  1.400  106 mm 4  1.400  106 m 4
12
12
3
V  15  10 N
I
(a)
T  Ve  (15  103 )(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 m3
8
t  6  103 m
V 
VQ (15  103 )(16.5  106 )

 29.46  106 Pa  29.46 MPa
It
(1.400  106 )(6  106 )
Stress due to T :
a  2b  h  160 mm  0.160 m
1
t  1
 6 
c1  1  0.630   1  (0.630) 
   0.32546
3
a 3
 160  
T
144.64

 77.16  106 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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991
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  a 2 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
M O  Vd 
Data: P  500 lb
a = 1.25 in.
2

Pa
V  500 lb 
2
M O    (500)(1.25)
 

M O  398 lb  in. 


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992
PROBLEM 6.85* (Continued)
(b)
Shearing stresses.
(1)
Due to V :  V 
V 


(2)
VQmax

It
( P)(a 2 t )
2P
(2)(500)


 679 psi 
 at  (1.25)(0.375)
 3 
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 torque M O is
M 
Data:
c1lt 2
1
t
where c1   1  0.630 
l
3
l   a   (1.25)  3.927 in. t  0.375 in. c1  0.31328
M 
By superposition,
MO
397.9
 2300 psi
(0.31328)(3.927)(0.375) 2
   V   M  679 psi  2300 psi
  2980 psi 
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993
PROBLEM 6.86
B
Solve Prob. 6.85, assuming that the thickness of the
beam is reduced to 14 in.
1.25 in.
A
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
4
e a



Q  a 2 t sin 
Qmax  a 2t
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 
M O  398 lb  in. 

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994
PROBLEM 6.86* (Continued)
(b)
Shearing stresses.
(1)
Due to V ,  V 
V 
(2)
VQmax
It
( P)(a 2t )
 3 
 2 a t (t ) 



2P
(2)(500)

 1019 psi
 at  (1.25)(0.250)
Due to the torque M O :
For a long rectangular section of length l and width t, the shearing stress due to torque M O is
M 
Data:
MO
c1lt 2
1
t
where c1 = 1  0.630 
3
l
l   a   (1.25)  3.927 in. t  0.250 in. c1  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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995
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
Vx  V sin  Vy  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
I yt
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  
2
4


2 

1
I xt
 
(166.3)  
4
 0.0166665(6  x)[6.69132  0.19134 x]
Vy A y 
Due to Vy :
1   2  (6  x)[0.07141  0.035396 x]
Total:
x (in.)
6
5
4
3
2
1
0
 (ksi)
0
0.105
0.140
0.104
0.003
0.180
0.428


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996
PROBLEM 6.88
y'
3 kips
y
A'
B'
x'
A'
B'
A
C'
22.5⬚
D'
E'
x
B
12 in.
D'
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 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 .
SOLUTION
V  3 kips   22.5
Vx  V sin  Vy  V cos 
For part AB’,
For part B′Y,
Due to Vx :
1
A    (6)  1.5 in 2
4
x  3 in., y  6 in.
1
(6  y )
4
1
x  0 y  (6  y )
2
x  x cos   y sin 
y   y cos   x sin 
A
1 
1 

 )
Vx ( AAB x AB  ABY xBY
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  0.047835 y 2 ]
 0.3404  0.01614 y 2
3.4025
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997
PROBLEM 6.88* (Continued)
2 
Due to Vy :
2 

Vy ( AAB y AB  ABY y )
I x t
(V cos  )[(1.5)(6 cos   3 sin  )  14 (6  y ) 12 (6  y ) cos  ]
(166.3) 14 
(V cos  )[10.037  4.1575  0.11548y 2 ]
 0.9463  0.00770 y 2
(166.3) 14 
1   2  1.2867  0.02384 y 2
Total:
y (in.)
0
2
4
6
 (ksi)
1.287
1.191
0.905
0.428



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998
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  75mm 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
A (mm 2 )
d (mm)
7200
Middle Plank
Bottom Plank
Top Plank
Ad 2 (106 mm 4 )
I (106 mm 4 )
60
25.92
2.16
12,000
0
0
3.60
7200
60
25.92
2.16
51.84
7.92

I  Ad 2   I  59.76  106 mm 4  59.76  106 m 4
Q  (7200)(60)  432  103 mm3  432  106 m3
VQ
I
F
q  nail
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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999
PROBLEM 6.90
180
160 kN
16
12
a
0.6 m
80
100
16
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
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 mm 4
12
1
(180)(16)3  (180)(16)(42)2  5.14176  106 mm 4
I2 
12
1
(16)(68)3  419.24  103 mm 4
I3 
12
I1 
I  4I1  2 I 2  2 I 3  44.274  106 mm 4
 44.274  106 m 4
(a)
Calculate Q at neutral axis.
Q1  (12)(80)(90)  86.4  103 mm 4
Q2  (180)(16)(42)  120.96  103 mm 4
Q3  (16)(34)(17)  9.248  103 mm 4
Q  2Q1  Q2  2Q3  312.256  103 mm3  312.256  106 m3
 
(b)
At point a,
VQ
(80  103 )(312.256  106 )

 17.63  106 Pa
It
(44.274  106 )(2  16  103 )
  17.63 MPa 
Q  Q1  86.4  103 mm 4  86.4  106 m 4
 
VQ
(80  103 )(86.4  106 )

 13.01  106 Pa
It
(44.274  106 )(12  103 )
  13.01 MPa 
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1000
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:  15RA  qP  0
RA  0.6 P
Draw shear and bending moment diagrams.
V
 0.6P
max
M
max
 0.6PLAB
LAB  6 ft  72 in.
S  258 in 3
For W24  104,
Bending.
S 
P
M
max
 all

0.6 PLAB
 all
(24)(258)
 all S

 143.3 kips
0.6LAB
(0.6)(72)
Aweb  dtw
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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1001
12 kips
n
A
PROBLEM 6.92
1 in.
12 kips
4 in.
B
1 in.
1 in.
a
b
n
16 in.
10 in.
For the beam and loading shown, consider
section n-n and determine the shearing stress
at (a) point a, (b) point b.
2 in.
16 in.
4 in.
SOLUTION
RA  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.
12
A
I  Ad 2  I  32 in 4
(a)
A  1 in 2
y  3.5 in. Qa  Ay  3.5 in 3
t  1 in.
a 
(b)
VQa
(12)(3.5)

(32)(1)
It
A  2 in 2
 a  1.313 ksi 
y  3 in. Qb  A y  6 in 3
t  1 in.
b 
VQb
(12)(6)

(32)(1)
It
 b  2.25 ksi 
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1002
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  2I 2
I2 
 237.333 in 4
Q  A1 y1  (2)(4)(3)  24 in 3
q
VQ (1500)(24)

 151.685 lb/in.
I
237.333
2Fnail  qs
Fnail 
1
1
qs    (151.685)(2.5)
2
2
Fnail  189.6 lb 
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1003
PROBLEM 6.94
40 mm
b
6 mm
60 mm
4 mm
6 mm
Knowing that a given vertical shear V causes a maximum shearing
stress of 75 MPa in the hat-shaped extrusion shown, determine the
corresponding shearing stress at (a) point a, (b) point b.
14 mm
a
4 mm
20 mm 28 mm 20 mm
SOLUTION
Neutral axis lies 30 mm above bottom.
c 
VQc
It
 a Qatc

 c Qcta
a 
VQa
Ita
b 
VQb
Itb
 b Qbtc

 c Qctb
Qc  (6)(30)(15)  (14)(4)(28)  4260 mm3
tc  6 mm
Qa  (14)(4)(28)  1568 mm3
ta  4 mm
Qb  (14)(4)(28)  1568 mm3
tb  4 mm
 c  75 MPa
(a)
a 
Qa tc
1568 6
 c 
  75
Qc ta
4260 4
(b)
b 
Qb tc
1568 6
 c 
  75
Qc tb
4260 4
 a  41.4 MPa 
 b  41.4 MPa 
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1004
125 mm
100 mm
125 mm
PROBLEM 6.95
Three planks are connected as shown by bolts of 14-mm diameter
spaced every 150 mm along the longitudinal axis of the beam. For a
vertical shear of 10 kN, determine the average shearing stress in the
bolts.
100 mm
250 mm
SOLUTION
Locate neutral axis and compute moment of inertia.
Part
A(mm 2 )
y (mm)

12,500
200

25,000

12,500
Σ
50,000
Ay (mm3 )
d (mm)
Ad 2 (mm 4 )
I (mm 4 )
2.5  106
37.5
17.5781  106
10.4167  106
125
3.125  106
37.5
35.156  106
200
2.5  106
37.5
17.5781  106
8.125  106
Y 
70.312  106
130.208  106
10.4167  106
151.041  106
Ay
8.125  106

 162.5 mm
A
50  103
I  Ad 2  I  221.35  106 mm 4
 221.35  106 m 4
Q  A1 y1  (12,500)(37.5)  468.75  103 mm3
 468.75  106 m3
q
VQ (10  103 )(468.75  106 )

I
221.35  106
 21.177  103 N/m
Fbolt  qs  (21.177  103 )(150  103 )  3.1765  103 N
Abolt 
 bolt 

4
(14) 2  153.938 mm 2  153.938  106 m 2
Fbolt
3.1765  103

 20.6  106 Pa
Abolt 153.938  106
 bolt  20.6 MPa 
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1005
PROBLEM 6.96
1 in.
1 in.
C
x
18 in.
1 in.
Three 1  18-in. steel plates are bolted to four L6  6  1 angles to form a beam
with the cross section shown. The bolts have a 78 -in. diameter and are spaced
longitudinally every 5 in. Knowing that the allowable average shearing stress in
the bolts is 12 ksi, determine the largest permissible vertical shear in the beam.
(Given:I x  6123 in 4.)
18 in.
SOLUTION
Flange:
If 
1
(18)(1)3  (18)(1)(9.5) 2  1626 in 4
12
Web:
Iw 
1
(1)(18)3  486 in 4
12
I  35.5 in 4 ,
Angle:
y  1.86 in.
A  11.0 in 2
d  9  1.86  7.14 in.
I a  I  Ad 2  596.18 in 4
I  2I f  I w  4I a  6123 in 4 , which agrees with the given value.
Flange:
Q f  (18)(1)(9.5)  171 in 3
Angle:
Qa  Ad  (11.0)(7.14)  78.54 in 3
Q  Q f  2Qa  328.08 in 3
Fbolt
qall 
q
 7
2
2
   0.60132 in
48
 2 bolt Abolt  (2)(12)(0.60132)  14.4317 kips
Abolt 
Fbolt 14.4317

 2.8863 kip/s
s
5
VQ
I
Vall 
qall I
(2.8863)(6123)

Q
328.08
Vall  53.9 kips 
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1006
PROBLEM 6.97
a
112 mm
The composite beam shown is made by welding C 200  17.1 rolled-steel channels to the
flanges of a W250  80 wide-flange rolled-steel shape. Knowing that the beam is
subjected to a vertical shear of 200 kN, determine (a) the horizontal shearing force per
meter at each weld, (b) the shearing stress at point a of the flange of the wide-flange
shape.
SOLUTION
For W250  80, d  257 mm, t f  15.6 mm, I x  126  106 mm 4
For C200  17.1,
A  2170 mm 2 , b f  57.4 mm, t f  9.91 mm
I y  0.545  106 mm 4 , x  14.5 mm
For the channel in the composite beam,
yc 
257
 57.4  14.5  171.4 mm
2
For the composite beam,
I  126  106 +2 0.545  106  (2170)(171.4) 2 
 254.59  106 mm 4  254.59  106 m 4
(a) For the two welds,
Qw  Ayc  (2170) (171.4)  371.94  103 mm3  371.94  106 m3 
q
VQ (200  103 ) (371.94  106 )

 292.2  103 N/m
I
254.59  106
For one weld,
q
 146.1  103 N m
2
Shearing force per meter of weld:
(b)
146.1 kN m 
For cuts at a and a ' together,
Aa  2(112)(15.6)  3494.4 mm 2
ya 
257 15.6

 120.7 mm
2
2
Qa  371.94  103  (3494.4)(120.7)  793.71  103 mm3  793.71  106 m3
Since there are cuts at a and a ', t  2t f  (2)(15.6)  31.2 mm  0.0312 m.
a 
VQa
(200  103 )(793.71  106 )

 19.99  106 Pa
6
It
(254.59  10 )(0.0312)
 a  19.99 MPa 
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1007
PROBLEM 6.98
0.5 in.
2.5 in.
The design of a beam requires welding four horizontal plates to a
vertical 0.5  5-in. plate as shown. For a vertical shear V,
determine the dimension h for which the shear flow through the
welded surface is maximum.
h
0.5 in.
2.5 in.
h
4.5 in.
4.5 in.
0.5 in.
SOLUTION
Horizontal plate:
1
(4.5)(0.5)3  (4.5)(0.5)h 2
12
 0.046875  2.25h 2
Ih 
Vertical plate:
Iv 
1
(0.5)(5)3  5.2083 in 4
12
Whole section:
I  4 I h  I v  9h 2  5.39583 in 4
For one horizontal plate,
Q  (4.5)(0.5)h  2.25 h in 3
q
To maximize q, set
VQ
2.25Vh
 2
I
9h  5.39583
dq
 0.
dh
2.25V
(9h 2  5.39583)  18h 2
0
(9h 2  5.39583) 2
h  0.774 in. 
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1008
A
PROBLEM 6.99
B
D
E
60 mm
45 mm
F
O
A thin-walled beam of uniform thickness has the cross section shown. Determine
the dimension b for which the shear center O of the cross section is located at the
point indicated.
45 mm
60 mm
H
J
G
K
30 mm
b
SOLUTION
Part AB:
A  tx
y  60 mm
Q  Ay  60tx mm3
 
VQ
60Vx

It
I

F1   dA 
60Vt x 2

I 2
Part DE:
A  tx

60 Vx
60Vt
t dx 
I
I
30
0
30

0

30
0
x dx
(60)(30) 2 Vt
Vt
 27  103
I
I
2
y  45 mm
Q  Ay  45tx
 
VQ
45Vx

It
I
b
F2    dA   O
 MO 
45Vx
45Vt b
45b 2Vt
t dx 
x
dx


2I
I
I O
 M O : 0  (2)(45) F2  (2)(60) F1
(45) 2 b 2  (2)(60)(27  103 )  Vt  0

 I
b2 
(2)(60)(27  103 )
 1600 mm 2
452
b  40 mm 
Note that the pair of F1 forces form a couple. Likewise, the pair of F2 forces form a couple. The lines of
action of the forces in BDOGK pass through point O.
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1009
PROBLEM 6.100
B
1
4
in.
60⬚
O
D
e
1.5 in.
A
F
60⬚
Determine the location of the shear center O of a thin-walled beam of
uniform thickness having the cross section shown.
1.5 in.
E
SOLUTION
11
I AB    (1.5)3  0.28125 in 4
3 4 
1
LBD  3 in. ABD  (3)    0.75 in 2
4
1
1
I BD  ABD h 2  (0.75)(1.5) 2  0.5625 in 4
3
3
I  (2)(0.28125)  (2)(0.5625)  1.6875 in 4
Part AB:
1
1
1
y y  y Q  Ay  y 2
4
2
8
2
VQ
Vy
Vy 2



(8)(1.6875)(0.25) 3.375
It
A

F1   dA 


1.5
0
(0.25)V y 3
3.375 3
Vy 2
(0.25dy )
3.375
1.5
0

(0.25)(1.5)3
(3.375)(3)
 0.083333V
MD 
M D : Ve  2 F1 (3 sin 60)
Ve  (2)(0.083333) V (3 sin 60)
e  (2)(0.083333)(3 sin 60)
e  0.433 in. 
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on a website, in whole or part.
1010
PROBLEM 6.C1
x4
x3
x1
x2
w
P1
P2
t
h
A
B
L
a
b
A timber beam is to be designed to support a distributed load and
up to two concentrated loads as shown. One of the dimensions of
its uniform rectangular cross section has been specified and the
other is to be determined so that the maximum normal stress and
the maximum shearing stress in the beam will not exceed given
allowable values  all and  all . Measuring x from end A and using
either SI or U.S. customary units, write a computer program to
calculate for successive cross sections, from x  0 to x  L and
using given increments x, the shear, the bending moment, and
the smallest value of the unknown dimension that satisfies in that
section (1) the allowable normal stress requirement and (2) the
allowable shearing stress requirement. Use this program to solve
Prob. 5.65, assuming  all  12 MPa and  all  825 kPa and using
x  0.1 m.
SOLUTION
See solution of Prob. 5.C2 for the determination of RA , RB , V ( x), and M ( x)
We recall that
V ( x)  RA STPA  RB STPB  P1 STP1  P2 STP2
w( x  x3 ) STP3  w( x  x4 ) STP4
M ( x)  RA ( x  a ) STPA  RB ( x  a  L) STPB  P1 ( x  x1 ) STP1
1
1
w( x  x3 ) 2 STP3  w( x  x4 )2 STP4
2
2
where STPA, STPB, STP1, STP2, STP3, and STP4 are step functions defined in Problem 5.C2.
P2 ( x  x2 ) STP2 
(1)
To satisfy the allowable normal stress requirement: If unknown dimension is h:
S min  |M |/ all .
From
1
S  th 2 , we have h  h  6 S /t
6
If unknown dimension is t:
S min  |M |/ all .
From
(2)
1
S  th 2 , we have t  t  6S /h 2
6
To satisfy the allowable shearing stress requirement:
We use Equation (6.10), Page 378:
If unknown dimension is h:
If unknown dimension is t:
3 |V | 3 |V |

2 A 2 th
3|V |
h  h 
2t all
3M
t  t 
2h all
 max 
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1011
PROBLEM 6.C1 (Continued)
Program Outputs
Problem 5.65
RA  2.40 kN
RB  3.00 kN
X
m
V
kN
M
kN  m
HSIG
mm
HTAU
mm
0.00
0.10
0.20
0.30
0.40
0.50
0.60
0.70
0.80
0.90
1.00
1.10
1.20
1.30
1.40
1.50
1.60
1.70
1.80
1.90
2.00
2.10
2.20
2.30
2.40
2.40
2.40
2.40
2.40
2.40
2.40
2.40
2.40
0.60
0.60
0.60
0.60
0.60
0.60
0.60
0.60
–3.00
–3.00
–3.00
–3.00
–3.00
–3.00
–3.00
–3.00
0.00
0.000
0.240
0.480
0.720
0.960
1.200
1.440
1.680
1.920
1.980
2.040
2.100
2.160
2.220
2.280
2.340
2.400
2.100
1.800
1.500
1.200
0.900
0.600
0.300
0.000
0.00
54.77
77.46
94.87
109.54
122.47
134.16
144.91
154.92
157.32
159.69
162.02
164.32
166.58
168.82
171.03
173.21
162.02
150.00
136.93
122.47
106.07
86.60
61.24
0.05
109.09
109.09
109.09
109.09
109.09
109.09
109.09
109.09
27.27
27.27
27.27
27.27
27.27
27.27
27.27
27.27
136.36
136.36
136.36
136.36
136.36
136.36
136.36
136.36
0.00









The smallest allowable value of h is the largest of the values shown in the last two columns.
h  h  173.2 mm 
For Problem 5.65,
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1012
PROBLEM 6.C2
P
b
w
B
A
8b
L
A cantilever timber beam AB of length L and of uniform rectangular
section shown supports a concentrated load P at its free end and a
uniformly distributed load w along its entire length. Write a
computer program to determine the length L and the width b of the
beam for which both the maximum normal stress and the maximum
shearing stress in the beam reach their largest allowable values.
Assuming  all  1.8 ksi and  all  120 psi, use this program to
determine the dimensions L and b when (a) P  1000 lb and w  0,
(b) P  0 and w  12.5 lb/in., (c) P  500 lb and w  12.5 lb/in.
SOLUTION
Both the maximum shear and the maximum bending moment occur at A. We have
VA  P  wL
M A  PL 
1 2
wL
2
To satisfy the allowable normal stress requirement:
 all 
MA
MA
3M A


2
1
S
(8
)
32b3
b
b
6
 3 MA 
b  b  

 32  all 
1/3
To satisfy the allowable shearing stress requirement:
We use Equation (6.10), Page 378.
 all 
3V
3V 3 VA

 A2
2 A 2 b(8b) 16b
 3 VA 
b  b  

16  all 
1/ 2
Program
For L  0, VA  P and b > 0, while M A  0 and b  0.
Starting with L  0 and using increments L  0.001 in., we increase L until b and b become equal. We
then Print L and b.
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1013
PROBLEM 6.C2 (Continued)
Program Outputs
For P  1000 lb, w  0.0 lb/in.
For P  0 lb, w  12.5 lb/in.
Increment  0.0010 in.
Increment  0.0010 in.
L  37.5 in., b  1.250 in.
L  70.3 in., b  1.172 in.
For P  500 lb, w  12.5 lb/in.
Increment  0.0010 in.
L  59.8 in., b  1.396 in.
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1014
bn
PROBLEM 6.C3
b2
A beam having the cross section shown is subjected to a vertical
shear V. Write a computer program that, for loads and dimensions
expressed in either SI or U.S. customary units, can be used to
calculate the shearing stress along the line between any two
adjacent rectangular areas forming the cross section. Use this
program to solve (a) Prob. 6.10, (b) Prob. 6.12, (c) Prob. 6.22.
hn
h2
V
h1
b1
SOLUTION
1.
Enter V and the number n of rectangles.
2.
For i  1 to n, enter the dimensions bi and hi .
3.
Determine the area Ai  bi hi of each rectangle.
4.
Determine the elevation of the centroid of each rectangle:
yi 
i
h
k
 0.5hi
k 1
and the elevation y of the centroid of the entire section:

y 


5.


i i
i
i
i
Determine the centroidal moment of inertia of the entire section:
I
1
 12 b h
3
i i
i
6.

 A y    A 

 A i ( yi  y ) 2 

For each surface separating two rectangles i and i  1, determine Qi of the area below that surface:
Qi 
i
 A (y
k
k
 y)
k 1
7.
Select for ti the smaller of bi and bi 1.
The shearing stress on the surface between the rectangles i and i  1 is
i 
VQi
Iti





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1015
PROBLEM 6.C3 (Continued)
Program Outputs
Problem 6.10
Shearing force  10 kN
y  75.000 mm above base
I  39.580  106 mm 4
Between Elements 1 and 2:
  418.39 kPA
Between Elements 2 and 3:
  919.78 kPA
(a)
  765.03 kPA
(b)
Between Elements 3 and 4:
Between Elements 4 and 5:
  418.39 kPA
Problem 6.12
Shearing force  10 kips
y  2.000 in.
I  14.58 in 4
Between Elements 1 and 2:
  2.400 ksi
Between Elements 2 and 3:
  3.171 ksi
(a)
  2.400 ksi
(b)
Between Elements 3 and 4:
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1016
PROBLEM 6.C3 (Continued)
Program Outputs (Continued)
Problem 6.22
Shearing force  90 kN
y  65.000 mm
I  58.133  106 mm 4
Between Elements 1 and 2:
  23.222 MPA
(b)
  30.963 MPA
(a)
Between Elements 2 and 3:
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1017
PROBLEM 6.C4
y
xn
x
y2
y1
x2
x1
A plate of uniform thickness t is bent as shown into a shape
with a vertical plane of symmetry and is then used as a beam.
Write a computer program that, for loads and dimensions
expressed in either SI or U.S. customary units, can be used to
determine the distribution of shearing stresses caused by a
vertical shear V. Use this program (a) to solve Prob. 6.47, (b) to
find the shearing stress at a Point E for the shape and load of
Prob. 6.50, assuming a thickness t  14 in.
SOLUTION
For each element on the right-hand side, we compute (for i  1 to n):
Length of element  Li  ( xi  xi 1 ) 2  ( yi  yi 1 )2
Area of element  Ai  tLi
where
t
1
in.
4
1
( yi  yi 1 )
2
Distance from x axis to centroid of element  yi 
Distance from x axis to centroid of section:
y    Ai yi   Ai
Note that yn  0 and that xn 1  yn 1  0.
Moment of inertia of section about centroidal axis:
1

I  2 Ai  ( yi  yi 1 ) 2  ( yi  y )2 
12

Computation of Q at Point P where stress is desired:
Q   Ai ( yi  y ) where sum extends to the areas located between one end of section and Point P.
Shearing stress at P:

VQ
It
Note:  max occurs on neutral axis, i.e., for yP  y .
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1018
PROBLEM 6.C4 (Continued)
Program Outputs
Part (a):
I  0.5333 in 4
 max  2.02 ksi

 B  1.800 ksi

Part (b):
I  22.27 in 4
 E  194.0 psi

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1019
PROBLEM 6.C5
y
x1
x2
yn
y1
tn
O
y2
x
e
t2 t
1
The cross section of an extruded beam is symmetric with respect to
the x axis and consists of several straight segments as shown. Write
a computer program that, for loads and dimensions expressed in
either SI or U.S. customary units, can be used to determine (a) the
location of the shear center O, (b) the distribution of shearing
stresses caused by a vertical force applied at O. Use this program to
solve Prob. 6.70.
V
SOLUTION
Since section is symmetric with x axis, computations will be
done for top half.
For i  1 to n  1: (Note: n  1 is the origin)
Enter ti , xi ,
yi
Compute length of each segment.
For i  1 to n:
 xi  xi 1  xi
yi  yi 1  yi
L  ( xi2  yi2 )1/ 2
Calculate moment of inertia I x .
Consider each segment as made of 100 equal parts.
For i  1 to n:
 Area  Li ti /100
For j  1 to 100:
y  yi  yi ( j  0.5)/100
I  ( Area) y 2
Ix  Ix  I
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.
1020
PROBLEM 6.C5 (Continued)
Since only top half was used,
I x  2I x
Calculate shearing stress at ends of segments and shear forces in segments.
For i  1 to n:
 Area  Li ti /100,
 new   next
For j  1 to 100:
y  yi  yi ( j  0.5)/100
Q  ( Area) y
 old   new ,
Q  Q  Q
 new  VQ /I x ti
 ave  0.5( old   new )
     ave
Force i   (∆Area)
 i  VQ/I x ti
( adjacent )i  VQ/I x ti 1
Qi  Q
 next  ( adjacent )i
Location of shear center.
Calculate moment of shear forces about origin.
For L  1 to n,
( Fx )i  Forcei ( xi )/Li
( Fy )i  Forcei (yi )/Li
Moment i  ( Fx )i yL  ( Fy )i xi
Moment  Moment  Moment i
For whole section, moment  2(moment),
Shear center is at
e  Moment/V
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.
1021
PROBLEM 6.C5 (Continued)
Program Output
Problem 6.70
T(K) mm
X(K) mm
Y(K) mm
L(K) mm
1
6.00
60.62
0.00
70.00
2
6.00
0.00
35.00
35.00
3
6.00
0.00
0.00
Moment of inertia: I x  514487 mm 4
Junction of
Segments
Q
mm3
Shear  1000.000 N
Tau Before
MPa
Tau After
MPa
Force in
Segment kN
1 and 2
7350.00
2.38
2.38
335.01
2 and 3
11,025.00
3.57
3.57
666.27
Moment of shear forces about origin: M  20.309 N  m  counterclockwise
Distance from origin to shear center: e  20.309 mm
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.
1022
tn
t2
t1
ti
a1
t0
an
a2
ai
O
a1
PROBLEM 6.C6
A thin-walled beam has the cross section shown. Write a computer program
that, for loads and dimensions expressed in either SI or U.S. customary units,
can be used to determine the location of the shear center O of the cross
section. Use the program to solve Prob. 6.75.
ai
a2
an
b2
e
bi
bn
SOLUTION
Distribution of shearing stresses in element i.
Let
V  Shear in cross section
I  Centroidal moment of inertia of section
We have for shaded area
Q  Ay  ti (ai  y )
ai  y
2
1
ti (ai2  y 2 )
2
QV V 2


(ai  y 2 )
Iti
2I

Force exerted on element i.
Fi 

ai
ai
Vti
2I
Vti

I
Vti

I

 (ti dy )
 a
 a
ai
ai
ai
0

2
i
 y 2 dy
2
i
 y 2 dy

 3 1 3 2V 3
 ai  3 ai   3 I ti ai


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.
1023
PROBLEM 6.C6 (Continued)
The system of the forces Fi must be equivalent to V at shear center.
F  F :
M A  M A :
Divide (2) by (1):
e
2V
ti ai3  V
3 I
2V
ti ai3bi  eV
3 I
(1)
(2)
ti ai3bi

ti ai3
Program Output
Problem 6.75
For Element 1:
t  0.75 in., a  4 in., b  0
For Element 2:
t  0.75 in., a  3 in., b  8 in.
Answer:
e  2.37 in.

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.
1024