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Chapter 6_Shearing stress in beams
Mechanics of Materials (Concordia University)
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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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PROBLEM 6.5
16 ⫻ 200 mm
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 (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
*d  305  16  160.5 mm
2
2
I  Ad  I  260.3  106 mm 4  260.3  106 m 4
2
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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15 15
30
PROBLEM 6.9
15 15
20
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
a
20
n
40
120
n
20
1.5 m
20
0.8 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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10
10
30
40
Dimensions in mm
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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PROBLEM 6.19
P
L/2
b
L/2
C
A
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
3PL
M max

S
2bh 2
h
m


 m 2L
(a)
(b)
h
Solving for 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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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.
44.86
q

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0.8
PROBLEM 6.31
0.8
1.5
1.5
4
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 
PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
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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
(6  103 )(375  106 )(0.225)
VQs

 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 

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