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PROBLEM 13.21
A beam having the cross section shown is subjected to a vertical shear V. Determine (a) the
horizontal line along which the shearing stress is maximum, (b) the constant k in the following
expression for the maximum shearing stress
max
V
kA
τ
=
where A is the cross-sectional area of the beam.
SOLUTION
42
and
4
I c Ac
ππ
= =
For semicircle,
24
23
sc
Acy
π
π
= =
23
42
233
sc
Q Ay c c
π
π
= = ⋅=
(a)
max
τ
occurs at center where
2tc=
(b)
3
2
3
max 42
4
44
3
23
Vc
VQ V V
It A
cc c
π
τπ
⋅
= = = =
⋅
41.333
3
k= =
consent of McGraw-Hill Education.
PROBLEM 13.22
A beam having the cross section shown is subjected to a vertical shear V. Determine (a) the
horizontal line along which the shearing stress is maximum, (b) the constant k in the
following expression for the maximum shearing stress
max
V
kA
τ
=
where A is the cross-sectional area of the beam.
SOLUTION
For a thin-walled circular section,
2
mm
A rt
π
=
23 3
1
2, 2
m mm mm
J Ar rt I J rt
ππ
= = = =
For a semicircular arc,
2m
r
y
π
=
2
22
s mm
m
s mm mm
A rt
r
Q Ay rt rt
π
ππ
=
= = =
(a)
2
m
tt
=
at neutral axis where maximum occurs.
(b)
2
max 3
(2 ) 2
( )(2 )
mm
mm
mm m
VQ V r t V V
It r t A
rt t
τπ
π
= = = =
2.00k=
consent of McGraw-Hill Education.
PROBLEM 13.23
A beam having the cross section shown is subjected to a vertical shear V. Determine (a) the
horizontal line along which the shearing stress is maximum, (b) the constant k in the
following expression for the maximum shearing stress
max
V
kA
τ
=
where A is the cross-sectional area of the beam.
SOLUTION
33
1 11
22
2 12 6
A bh bh I bh bh
= = = =
For a cut at location y, where
,yh≤
2
23
1
() 22
2
() 3
() 23
()
by by
Ay y
hh
yy h y
by by
Q y Ay h
by
ty h
= =
= −
= = −
=
2
23
3
6
() 3 2
23
VQ h by by V y y
yV
It by h bh h h
bh
τ
= = ⋅⋅ − = −
(a) To find location of maximum of
,
τ
set
0.
d
dy
τ
=
2
[3 4 ] 0
m
dV y
dy h
bh
τ
= −=
3,
4
m
y
h=
i.e.,
1
4h±
from neutral axis.
(b)
2
339
( ) 3 2 1.125
448
m
V VV
ybh bh A
τ
= −==
1.125k=
consent of McGraw-Hill Education.
PROBLEM 13.24
A beam having the cross section shown is subjected to a vertical shear V. Determine
(a) the horizontal line along which the shearing stress is maximum, (b) the constant k in
the following expression for the maximum shearing stress
max
V
kA
τ
=
where A is the cross-sectional area of the beam.
SOLUTION
3
11
2 36
A bh I bh= =
For a cut at location y,
2
2
1
() 22
22
() 33
() ( )
3
()
by by
Ay y
hh
yy h y
by
Q y Ay h y
by
ty h
= =
= −
= = −
=
2
2
333
3
1
36
()
12 ( ) 12
() ( )
()
by
by
h
V hy
VQ Vy h y V
y hy y
It bh bh
bh
τ
−−
= = = = −
(a) To find location of maximum of
,
τ
set
0.
d
dy
τ
=
3
12 ( 2) 0
m
dV
hy
dy bh
τ
= −=
1,
2
m
yh=
i.e., at mid-height
(b)
2
22
33
12 12 1 1 3 3
()22 2
m mm
V V VV
hy y h h bh A
bh bh
τ
= −= − ==
31.500
2
k= =
consent of McGraw-Hill Education.
PROBLEM 13.25
The built-up beam shown is made by gluing together two 20 × 250-mm
plywood strips and two 50 × 100-mm planks. Knowing that the allowable
average shearing stress in the glued joints is 350 kPa, determine the largest
permissible vertical shear in the beam.
SOLUTION
3 3 64
11
(140)(250) (100)(150) 154.167 10 mm
12 12
I=−=×
64
154.167 10 m
−
= ×
33
(100)(50)(100) 500 10 mmQ Ay= = = ×
63
500 10 m
−
= ×
3
50 mm 50 mm 100 mm 100 10 mt
−
=+= =×
VQ
It
t
=
6 33
6
(154.167 10 )(100 10 )(350 10 )
500 10
It
VQ
t
−−
−
×××
= = ×
3
10.79 10 N= ×
10.79 kNV=
consent of McGraw-Hill Education.
PROBLEM 13.26
The built-up timber beam shown is subjected to a vertical shear of
1200 lb. Knowing that the allowable shearing force in the nails is 75 lb,
determine the largest permissible spacing s of the nails.
SOLUTION
32
1 11 11
1
12
I bh Ad= +
32 4
1(2)(2) (2)(2)(4) 65.333 in
12
=+=
3 34
2 22
11
(2)(10) 166.67 in
12 12
I bh= = =
4
12
4 428 inI II= +=
3
1 11
(2)(2)(4) 16 in
Q Q Ay= = = =
(1200)(16) 44.86 lb/in.
428
VQ
qI
= = =
nail
F qs=
nail
75 1.672 in.
44.86
F
sq
= = =
consent of McGraw-Hill Education.
PROBLEM 13.27
The built-up beam was made by gluing together several wooden planks.
Knowing that the beam is subjected to a 1200-lb vertical shear, determine
the average shearing stress in the glued joint (a) at A, (b) at B.
SOLUTION
33 2
11
2 (0.8)(4.8) (7)(0.8) (7)(0.8)(2.0)
12 12
I
= ++
4
60.143 in=
(a)
2
(1.5)(0.8) 1.2 in
a
A= =
2.0 in.
a
y=
3
2.4 in
a aa
Q Ay= =
0.8 in.
a
t=
(1200)(2.4)
(60.143)(0.8)
a
aa
VQ
It
t
= =
59.9 psi
a
t
=
(b)
2
(4)(0.8) 3.2 in
b
A= =
2.0 in.
b
y=
3
(3.2)(2.0) 6.4 in
b bb
Q Ay= = =
(2)(0.8) 1.6 in.
b
t= =
(1200)(6.4)
(60.143)(1.6)
b
bb
VQ
It
t
= =
79.8 psi
b
t
=
consent of McGraw-Hill Education.
PROBLEM 13.28
Knowing that a W360 × 122 rolled-steel beam is subjected to a 250-kN vertical shear,
determine the shearing stress (a) at point A, (b) at the centroid C of the section.
SOLUTION
For
W360 122,×
363 mm,d=
257 mm,
F
b=
21.70 mm,
F
t=
13.0 mm
w
t=
6 4 64
367 10 mm 367 10 mI
−
=×=×
(a)
2
(105)(21.70) 2278.5 mm
a
A= =
363 21.70 170.65 mm
22 2 2
F
a
dt
y=−= − =
3 3 63
388.8 10 mm 388.8 10 m
a aa
Q Ay
−
==×=×
3
21.70 mm 21.7 10 m
aF
tt
−
= = = ×
36 6
63
(250 10 )(388.8 10 ) 12.21 10 Pa
(367 10 )(21.7 10 )
a
aa
VQ
It
t
−
−−
××
= = = ×
××
12.21MPa
a
t
=
(b)
2
1
(257)(21.70) 5577 mm
FF
A bt= = =
1
363 21.70 170.65 mm
22 2 2
F
dt
y=−= − =
2
2(13.0)(159.8) 2077 mm
2
wF
d
At t
= −= =
2179.9 mm
22 F
d
yt
= −=
33
(5577)(170.65) (2077)(79.9) 1117.7 10 mm
c
Q Ay=Σ= + = ×
63
1117.7 10 m
−
= ×
3
13.0 mm 13 10 m
cw
tt
−
= = = ×
36
6
63
(250 10 )(1117.7 10 ) 58.6 10 Pa
(367 10 )(13 10 )
c
cc
VQ
It
t
−
−−
××
= = = ×
××
58.6 MPa
c
t
=
consent of McGraw-Hill Education.
PROBLEM 13.29
An extruded aluminum beam has the cross section shown.
Knowing that the vertical shear in the beam is 150 kN, determine
consent of McGraw-Hill Education.
PROBLEM 13.21
A beam having the cross section shown is subjected to a vertical shear V. Determine (a) the
horizontal line along which the shearing stress is maximum, (b) the constant k in the following
expression for the maximum shearing stress
max
V
kA
τ
=
where A is the cross-sectional area of the beam.
SOLUTION
42
and
4
I c Ac
ππ
= =
For semicircle,
24
23
sc
Acy
π
π
= =
23
42
233
sc
Q Ay c c
π
π
= = ⋅=
(a)
max
τ
occurs at center where
2tc=
(b)
3
2
3
max 42
4
44
3
23
Vc
VQ V V
It A
cc c
π
τπ
⋅
= = = =
⋅
41.333
3
k= =
consent of McGraw-Hill Education.
PROBLEM 13.22
A beam having the cross section shown is subjected to a vertical shear V. Determine (a) the
horizontal line along which the shearing stress is maximum, (b) the constant k in the
following expression for the maximum shearing stress
max
V
kA
τ
=
where A is the cross-sectional area of the beam.
SOLUTION
For a thin-walled circular section,
2
mm
A rt
π
=
23 3
1
2, 2
m mm mm
J Ar rt I J rt
ππ
= = = =
For a semicircular arc,
2m
r
y
π
=
2
22
s mm
m
s mm mm
A rt
r
Q Ay rt rt
π
ππ
=
= = =
(a)
2
m
tt
=
at neutral axis where maximum occurs.
(b)
2
max 3
(2 ) 2
( )(2 )
mm
mm
mm m
VQ V r t V V
It r t A
rt t
τπ
π
= = = =
2.00k=
consent of McGraw-Hill Education.
PROBLEM 13.23
A beam having the cross section shown is subjected to a vertical shear V. Determine (a) the
horizontal line along which the shearing stress is maximum, (b) the constant k in the
following expression for the maximum shearing stress
max
V
kA
τ
=
where A is the cross-sectional area of the beam.
SOLUTION
33
1 11
22
2 12 6
A bh bh I bh bh
= = = =
For a cut at location y, where
,yh≤
2
23
1
() 22
2
() 3
() 23
()
by by
Ay y
hh
yy h y
by by
Q y Ay h
by
ty h
= =
= −
= = −
=
2
23
3
6
() 3 2
23
VQ h by by V y y
yV
It by h bh h h
bh
τ
= = ⋅⋅ − = −
(a) To find location of maximum of
,
τ
set
0.
d
dy
τ
=
2
[3 4 ] 0
m
dV y
dy h
bh
τ
= −=
3,
4
m
y
h=
i.e.,
1
4h±
from neutral axis.
(b)
2
339
( ) 3 2 1.125
448
m
V VV
ybh bh A
τ
= −==
1.125k=
consent of McGraw-Hill Education.
PROBLEM 13.24
A beam having the cross section shown is subjected to a vertical shear V. Determine
(a) the horizontal line along which the shearing stress is maximum, (b) the constant k in
the following expression for the maximum shearing stress
max
V
kA
τ
=
where A is the cross-sectional area of the beam.
SOLUTION
3
11
2 36
A bh I bh= =
For a cut at location y,
2
2
1
() 22
22
() 33
() ( )
3
()
by by
Ay y
hh
yy h y
by
Q y Ay h y
by
ty h
= =
= −
= = −
=
2
2
333
3
1
36
()
12 ( ) 12
() ( )
()
by
by
h
V hy
VQ Vy h y V
y hy y
It bh bh
bh
τ
−−
= = = = −
(a) To find location of maximum of
,
τ
set
0.
d
dy
τ
=
3
12 ( 2) 0
m
dV
hy
dy bh
τ
= −=
1,
2
m
yh=
i.e., at mid-height
(b)
2
22
33
12 12 1 1 3 3
()22 2
m mm
V V VV
hy y h h bh A
bh bh
τ
= −= − ==
31.500
2
k= =
consent of McGraw-Hill Education.
PROBLEM 13.25
The built-up beam shown is made by gluing together two 20 × 250-mm
plywood strips and two 50 × 100-mm planks. Knowing that the allowable
average shearing stress in the glued joints is 350 kPa, determine the largest
permissible vertical shear in the beam.
SOLUTION
3 3 64
11
(140)(250) (100)(150) 154.167 10 mm
12 12
I=−=×
64
154.167 10 m
−
= ×
33
(100)(50)(100) 500 10 mmQ Ay= = = ×
63
500 10 m
−
= ×
3
50 mm 50 mm 100 mm 100 10 mt
−
=+= =×
VQ
It
t
=
6 33
6
(154.167 10 )(100 10 )(350 10 )
500 10
It
VQ
t
−−
−
×××
= = ×
3
10.79 10 N= ×
10.79 kNV=
consent of McGraw-Hill Education.
PROBLEM 13.26
The built-up timber beam shown is subjected to a vertical shear of
1200 lb. Knowing that the allowable shearing force in the nails is 75 lb,
determine the largest permissible spacing s of the nails.
SOLUTION
32
1 11 11
1
12
I bh Ad= +
32 4
1(2)(2) (2)(2)(4) 65.333 in
12
=+=
3 34
2 22
11
(2)(10) 166.67 in
12 12
I bh= = =
4
12
4 428 inI II= +=
3
1 11
(2)(2)(4) 16 in
Q Q Ay= = = =
(1200)(16) 44.86 lb/in.
428
VQ
qI
= = =
nail
F qs=
nail
75 1.672 in.
44.86
F
sq
= = =
consent of McGraw-Hill Education.
PROBLEM 13.27
The built-up beam was made by gluing together several wooden planks.
Knowing that the beam is subjected to a 1200-lb vertical shear, determine
the average shearing stress in the glued joint (a) at A, (b) at B.
SOLUTION
33 2
11
2 (0.8)(4.8) (7)(0.8) (7)(0.8)(2.0)
12 12
I
= ++
4
60.143 in=
(a)
2
(1.5)(0.8) 1.2 in
a
A= =
2.0 in.
a
y=
3
2.4 in
a aa
Q Ay= =
0.8 in.
a
t=
(1200)(2.4)
(60.143)(0.8)
a
aa
VQ
It
t
= =
59.9 psi
a
t
=
(b)
2
(4)(0.8) 3.2 in
b
A= =
2.0 in.
b
y=
3
(3.2)(2.0) 6.4 in
b bb
Q Ay= = =
(2)(0.8) 1.6 in.
b
t= =
(1200)(6.4)
(60.143)(1.6)
b
bb
VQ
It
t
= =
79.8 psi
b
t
=
consent of McGraw-Hill Education.
PROBLEM 13.28
Knowing that a W360 × 122 rolled-steel beam is subjected to a 250-kN vertical shear,
determine the shearing stress (a) at point A, (b) at the centroid C of the section.
SOLUTION
For
W360 122,×
363 mm,d=
257 mm,
F
b=
21.70 mm,
F
t=
13.0 mm
w
t=
6 4 64
367 10 mm 367 10 mI
−
=×=×
(a)
2
(105)(21.70) 2278.5 mm
a
A= =
363 21.70 170.65 mm
22 2 2
F
a
dt
y=−= − =
3 3 63
388.8 10 mm 388.8 10 m
a aa
Q Ay
−
==×=×
3
21.70 mm 21.7 10 m
aF
tt
−
= = = ×
36 6
63
(250 10 )(388.8 10 ) 12.21 10 Pa
(367 10 )(21.7 10 )
a
aa
VQ
It
t
−
−−
××
= = = ×
××
12.21MPa
a
t
=
(b)
2
1
(257)(21.70) 5577 mm
FF
A bt= = =
1
363 21.70 170.65 mm
22 2 2
F
dt
y=−= − =
2
2(13.0)(159.8) 2077 mm
2
wF
d
At t
= −= =
2179.9 mm
22 F
d
yt
= −=
33
(5577)(170.65) (2077)(79.9) 1117.7 10 mm
c
Q Ay=Σ= + = ×
63
1117.7 10 m
−
= ×
3
13.0 mm 13 10 m
cw
tt
−
= = = ×
36
6
63
(250 10 )(1117.7 10 ) 58.6 10 Pa
(367 10 )(13 10 )
c
cc
VQ
It
t
−
−−
××
= = = ×
××
58.6 MPa
c
t
=
consent of McGraw-Hill Education.
PROBLEM 13.29
An extruded aluminum beam has the cross section shown.
Knowing that the vertical shear in the beam is 150 kN, determine
consent of McGraw-Hill Education.
