PROBLEM 13.40
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.
PROBLEM 13.41
A plate of 4–mm thickness is bent as shown and then used as a beam. For a
vertical shear of 12 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.
SOLUTION
20
tan 22.62
48
aa
= = °
Slanted side:
2
3 34
(4 sec )(48) 208 mm
1(4 sec )(48) 39.936 10 mm
12
s
s
A
I
a
a
= =
= = ×
Top:
3 2 34
1(50)(4) (50)(4)(24) 115.46 10 mm
12
T
I=+=×
Bottom:
34
115.46 10 mm
BT
II= = ×
3 4 94
2 310.8 10 mm 310.8 10 m
sT B
I III
−
= ++= × = ×
(a)
3 3 63
3
(25)(4)(24) 2.4 10 mm 2.4 10 m
4mm 4 10 m
A
Q
t
−
−
= =×=×
= = ×
36 6
93
(12 10 )(2.4 10 ) 23.2 10 Pa
(310.8 10 )(4 10 )
A
A
VQ
It
t
−
−−
××
= = = ×
××
23.2 MPa
A
t
=
(b) Maximum shearing occurs at point M, 24 mm above the bottom
3 3 33
63
(4 sec )(24)(12) 2.4 10 1.248 10 3.648 10 mm
3.648 10 m
MA
QQ
a
−
=+ =×+ ×= ×
= ×
36 6
93
(12 10 )(3.648 10 ) 35.2 10 Pa
(310.8 10 )(4 10 )
M
M
VQ
It
t
−
−−
××
= = = ×
××
35.2 MPa
M
t
=
23.2 MPa
B AB A
QQ
tt
= = =
Copyright © McGraw–Hill Education. All rights reserved. No reproduction or distribution without the prior written
consent of McGraw–Hill Education.
PROBLEM 13.42
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.
SOLUTION
333 2
34 94
3 93
3 93
33
111
2 (2)(48) (2)(52) (20)(2) (20)(2)(25)
12 12 12
133.75 10 mm 133.75 10 mm
(2)(24)(12) 576 mm 576 10 mm
0
(12)(2)(25) 600 mm 600 10 m
(2)(24)(12) 1.176 10 mm 1.176 1
a
a
cb
dc
I
Q
Q
QQ
QQ
−
−
−
= +++
=×=×
= = = ×
=
=− =− =−×
=− =−× =−×
63
3 93
0m
(2)(26)(13) 600 mm 500 10 m
ed
QQ
−
−
=+ =− =−×
39 6
93
(5 10 )(576 10 ) 10.77 10 Pa
(133.75 10 )(2 10 )
a
a
VQ
It
t
−
−−
××
= = = ×
××
10.76 MPa
a
t
=
b
b
VQ
It
t
=
0
b
t
=
39 6
93
(5 10 )(600 10 ) 11.21 10 Pa
(133.75 10 )(2 10 )
c
c
VQ
It
t
−
−−
××
= = = ×
××
11.21MPa
c
t
=
36 6
93
(5 10 )(1.176 10 ) 22.0 10 Pa
(133.75 10 )(2 10 )
d
d
VQ
It
t
−
−−
××
= = = ×
××
22.0 MPa
d
t
=
39 6
93
(5 10 )(500 10 ) 9.35 10 Pa
(133.75 10 )(2 10 )
e
e
VQ
It
t
−
−−
××
= = = ×
××
9.35 MPa
e
t
=
consent of McGraw–Hill Education.
PROBLEM 13.43
A plate of
1
4
-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
22 2
(1.2) (1.6) 2.0 in. (0.25)(2.0) 0.5 in
BD BD
LA= += = =
Locate neutral axis and compute moment of inertia.
Part
2
(in )A
(in.)y
3
(in )Ay
(in.)d
24
(in )Ad
4
(in )I
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 0.8 0.320 0.2134
2 24
24
*1 1 0.8
(0.5)(1.6) 0.1067 in 0.4 in.
12 12 2.0
0.5334 in
BD
Ay
Ah Y A
I Ad I
Σ
= = = = =
Σ
=Σ +Σ =
(a)
3
3
3
(2)(0.25)(0.4) 0.2 in
(0.5)(0.25)(0.2) 0.025 in
0.225 in
m AB BC
AB
BC
m
QQ Q
Q
Q
Q
= +
= =
= =
=
(1.2)(0.225)
(0.5334)(0.25)
m
mVQ
It
t
= =
2.025 ksi
m
t
=
(b)
3
0.2 in
B AB
QQ= =
(1.2)(0.2)
(0.5334)(0.25)
B
B
VQ
It
t
= =
1.800 ksi
B
t
=
0
D
t
=
Copyright © McGraw–Hill Education. All rights reserved. No reproduction or distribution without the prior written
consent of McGraw–Hill Education.
PROBLEM 13.44
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
consent of McGraw–Hill Education.
PROBLEM 13.45
For a beam made of two or more materials with different moduli of elasticity, show that Eq. (13.6)
ave
VQ
It
t
=
remains valid provided that both Q and I are computed by using the transformed section of the beam
(see Sec. 11.3), and provided further that t is the actual width of the beam where
ave
t
is computed.
SOLUTION
PROBLEM 13.46
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
consent of McGraw–Hill Education.
PROBLEM 13.47
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. 13.45.)
SOLUTION
PROBLEM 13.47 (Continued)
PROBLEM 13.48
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
6
29 10
×
psi for the steel and
6
10.6 10×
psi for the
aluminum, determine (a) the average shearing stress at the bonded surface, (b)
the maximum shearing stress in the beam. (Hint: Use the method indicated in
Prob. 13.45.)
SOLUTION
1n=
in aluminum.
6
6
29 10 psi 2.7358
10.6 10 psi
n×
= =
×
in steel.
Part
2
(in )nA
(in.)y
3
(in )
nAy
(in.)d
22
(in )nAd
4
(in )nI
Alum. 3.0 2.0 6.0 0.8665 2.2525 1.0
Steel 4.1038 0.5 2.0519 0.6335 1.6469 0.3420
Σ 7.1038 8.0519 3.8994 1.3420
24
8.0519 1.1335 in.
7.1038
5.2414 in
nAy
YnA
I nAd nI
Σ
= = =
Σ
=Σ +Σ =
(a) At the bonded surface,
3
(1.5)(2)(0.8665) 2.5995 inQ= =
(4)(2.5995)
(5.2414)(1.5)
VQ
It
t
= =
1.323 ksi
t
=
(b) At the neutral axis,
3
1.8665
(1.5)(1.8665) 2.6129 in
2
Q
= =
max
(4)(2.6129)
(5.2814)(1.5)
VQ
It
t
= =
max
1.329 ksi
t
=
consent of McGraw–Hill Education.
PROBLEM 13.41
A plate of 4–mm thickness is bent as shown and then used as a beam. For a
vertical shear of 12 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.
SOLUTION
20
tan 22.62
48
aa
= = °
Slanted side:
2
3 34
(4 sec )(48) 208 mm
1(4 sec )(48) 39.936 10 mm
12
s
s
A
I
a
a
= =
= = ×
Top:
3 2 34
1(50)(4) (50)(4)(24) 115.46 10 mm
12
T
I=+=×
Bottom:
34
115.46 10 mm
BT
II= = ×
3 4 94
2 310.8 10 mm 310.8 10 m
sT B
I III
−
= ++= × = ×
(a)
3 3 63
3
(25)(4)(24) 2.4 10 mm 2.4 10 m
4mm 4 10 m
A
Q
t
−
−
= =×=×
= = ×
36 6
93
(12 10 )(2.4 10 ) 23.2 10 Pa
(310.8 10 )(4 10 )
A
A
VQ
It
t
−
−−
××
= = = ×
××
23.2 MPa
A
t
=
(b) Maximum shearing occurs at point M, 24 mm above the bottom
3 3 33
63
(4 sec )(24)(12) 2.4 10 1.248 10 3.648 10 mm
3.648 10 m
MA
QQ
a
−
=+ =×+ ×= ×
= ×
36 6
93
(12 10 )(3.648 10 ) 35.2 10 Pa
(310.8 10 )(4 10 )
M
M
VQ
It
t
−
−−
××
= = = ×
××
35.2 MPa
M
t
=
23.2 MPa
B AB A
QQ
tt
= = =
Copyright © McGraw–Hill Education. All rights reserved. No reproduction or distribution without the prior written
consent of McGraw–Hill Education.
PROBLEM 13.42
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.
SOLUTION
333 2
34 94
3 93
3 93
33
111
2 (2)(48) (2)(52) (20)(2) (20)(2)(25)
12 12 12
133.75 10 mm 133.75 10 mm
(2)(24)(12) 576 mm 576 10 mm
0
(12)(2)(25) 600 mm 600 10 m
(2)(24)(12) 1.176 10 mm 1.176 1
a
a
cb
dc
I
Q
Q
QQ
QQ
−
−
−
= +++
=×=×
= = = ×
=
=− =− =−×
=− =−× =−×
63
3 93
0m
(2)(26)(13) 600 mm 500 10 m
ed
QQ
−
−
=+ =− =−×
39 6
93
(5 10 )(576 10 ) 10.77 10 Pa
(133.75 10 )(2 10 )
a
a
VQ
It
t
−
−−
××
= = = ×
××
10.76 MPa
a
t
=
b
b
VQ
It
t
=
0
b
t
=
39 6
93
(5 10 )(600 10 ) 11.21 10 Pa
(133.75 10 )(2 10 )
c
c
VQ
It
t
−
−−
××
= = = ×
××
11.21MPa
c
t
=
36 6
93
(5 10 )(1.176 10 ) 22.0 10 Pa
(133.75 10 )(2 10 )
d
d
VQ
It
t
−
−−
××
= = = ×
××
22.0 MPa
d
t
=
39 6
93
(5 10 )(500 10 ) 9.35 10 Pa
(133.75 10 )(2 10 )
e
e
VQ
It
t
−
−−
××
= = = ×
××
9.35 MPa
e
t
=
consent of McGraw–Hill Education.
PROBLEM 13.43
A plate of
1
4
-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
22 2
(1.2) (1.6) 2.0 in. (0.25)(2.0) 0.5 in
BD BD
LA= += = =
Locate neutral axis and compute moment of inertia.
Part
2
(in )A
(in.)y
3
(in )Ay
(in.)d
24
(in )Ad
4
(in )I
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 0.8 0.320 0.2134
2 24
24
*1 1 0.8
(0.5)(1.6) 0.1067 in 0.4 in.
12 12 2.0
0.5334 in
BD
Ay
Ah Y A
I Ad I
Σ
= = = = =
Σ
=Σ +Σ =
(a)
3
3
3
(2)(0.25)(0.4) 0.2 in
(0.5)(0.25)(0.2) 0.025 in
0.225 in
m AB BC
AB
BC
m
QQ Q
Q
Q
Q
= +
= =
= =
=
(1.2)(0.225)
(0.5334)(0.25)
m
mVQ
It
t
= =
2.025 ksi
m
t
=
(b)
3
0.2 in
B AB
QQ= =
(1.2)(0.2)
(0.5334)(0.25)
B
B
VQ
It
t
= =
1.800 ksi
B
t
=
0
D
t
=
Copyright © McGraw–Hill Education. All rights reserved. No reproduction or distribution without the prior written
consent of McGraw–Hill Education.
PROBLEM 13.44
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
consent of McGraw–Hill Education.
PROBLEM 13.45
For a beam made of two or more materials with different moduli of elasticity, show that Eq. (13.6)
ave
VQ
It
t
=
remains valid provided that both Q and I are computed by using the transformed section of the beam
(see Sec. 11.3), and provided further that t is the actual width of the beam where
ave
t
is computed.
SOLUTION
PROBLEM 13.46
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
consent of McGraw–Hill Education.
PROBLEM 13.47
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. 13.45.)
SOLUTION
PROBLEM 13.47 (Continued)
PROBLEM 13.48
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
6
29 10
×
psi for the steel and
6
10.6 10×
psi for the
aluminum, determine (a) the average shearing stress at the bonded surface, (b)
the maximum shearing stress in the beam. (Hint: Use the method indicated in
Prob. 13.45.)
SOLUTION
1n=
in aluminum.
6
6
29 10 psi 2.7358
10.6 10 psi
n×
= =
×
in steel.
Part
2
(in )nA
(in.)y
3
(in )
nAy
(in.)d
22
(in )nAd
4
(in )nI
Alum. 3.0 2.0 6.0 0.8665 2.2525 1.0
Steel 4.1038 0.5 2.0519 0.6335 1.6469 0.3420
Σ 7.1038 8.0519 3.8994 1.3420
24
8.0519 1.1335 in.
7.1038
5.2414 in
nAy
YnA
I nAd nI
Σ
= = =
Σ
=Σ +Σ =
(a) At the bonded surface,
3
(1.5)(2)(0.8665) 2.5995 inQ= =
(4)(2.5995)
(5.2414)(1.5)
VQ
It
t
= =
1.323 ksi
t
=
(b) At the neutral axis,
3
1.8665
(1.5)(1.8665) 2.6129 in
2
Q
= =
max
(4)(2.6129)
(5.2814)(1.5)
VQ
It
t
= =
max
1.329 ksi
t
=
consent of McGraw–Hill Education.