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=
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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.