PROBLEM 11.20
Solve Prob. 11.19, assuming that
40 mm.d=
PROBLEM 11.19 The beam shown is made of a nylon for which the
allowable stress is 24 MPa in tension and 30 MPa in compression.
Determine the largest couple M that can be applied to the beam.
SOLUTION
2
, mmA
0
, mmy
3
0, mmAy
PROBLEM 11.21
Knowing that for the beam shown the allowable stress is 12 ksi in tension and
16 ksi in compression, determine the largest couple M that can be applied.
SOLUTION
consent of McGraw–Hill Education.
PROBLEM 11.22
Straight rods of 0.30–in. diameter and 200–ft length are sometimes
used to clear underground conduits of obstructions or to thread wires
through a new conduit. The rods are made of high–strength steel and,
for storage and transportation, are wrapped on spools of 5–ft
diameter. Assuming that the yield strength is not exceeded,
determine (a) the maximum stress in a rod, when the rod, which is
initially straight, is wrapped on a spool, (b) the corresponding
bending moment in the rod. Use
6
29 10 psiE= ×
.
SOLUTION
Radius of cross section:
11
(0.30) 0.15 in.
22
rd= = =
Moment of inertia:
4 4 64
(0.15) 397.61 10 in
44
Ir
ππ
−
= = = ×
1
5 ft 60 in. 30 in.
2
0.15 in.
DD
cr
r
= = = =
= =
(a)
63
max
(29 10 )(0.15) 145.0 10 psi
30
Ec
sr
×
= = = ×
max
145.0 ksi
s
=
(b)
66
(29 10 )(397.61 10 )
30
EI
M
r
−
××
= =
384 lb in.M= ⋅
consent of McGraw–Hill Education.
PROBLEM 11.23
A 60–N ⋅ m couple is applied to the steel bar shown. (a) Assuming that
the couple is applied about the z axis as shown, determine the maximum
stress and the radius of curvature of the bar. (b) Solve part a, assuming
that the couple is applied about the y axis. Use
200 GPa.
E=
SOLUTION
(a) Bending about z–axis.
3 3 3 4 94
11
(12)(20) 8 10 mm 8 10 m
12 12
20 10 mm 0.010 m
2
−
== =×=×
= = =
I bh
c
6
9
(60)(0.010) 75.0 10 Pa
8 10
Mc
I
σ
−
= = = ×
×
75.0 MPa
σ
=
31
99
1 60 37.5 10 m
(200 10 )(8 10 )
M
EI
ρ
−−
−
= = = ×
××
26.7 m
ρ
=
(b) Bending about y–axis.
3 3 3 4 94
6
9
11
(20)(12) 2.88 10 mm 2.88 10 m
12 12
12 6 mm 0.006 m
2(60)(0.006) 125.0 10 Pa
2.88 10
I bh
c
Mc
I
σ
−
−
== =×=×
= = =
= = = ×
×
125.0 MPa
σ
=
31
99
1 60 104.17 10 m
(200 10 )(2.88 10 )
M
EI
ρ
−−
−
= = = ×
××
9.60 m
ρ
=
consent of McGraw–Hill Education.
PROBLEM 11.24
A couple of magnitude M is applied to a square bar of side a. For
each of the orientations shown, determine the maximum stress and
the curvature of the bar.
SOLUTION
4
33
11
12 12 12
2
a
I bh aa
a
c
= = =
=
max 42
12
a
M
Mc
Ia
σ
= =
max 3
6M
a
σ
=
4
1
12
MM
EI a
E
ρ
= =
4
1 12M
Ea
ρ
=
For one triangle, the moment of inertia about its base is
( )
34
3
1
4
21
4
12
11
2
12 12 24
2
24
12
aa
I bh a
a
II
a
II I

= = =


= =
=+=
max 43
/2 62
/12
2
a Mc Ma M
cIaa
σ
= = = =
max 3
8.49M
a
σ
=
4
1
12
ρ
= =
MM
EI a
E
4
1 12M
Ea
ρ
=
consent of McGraw–Hill Education.
PROBLEM 11.25
A bar having the cross section shown has been formed by securely
bonding brass and aluminum stock. Using the data given below,
determine the largest permissible bending moment when the
Aluminum Brass
Modulus of elasticity 70 GPa 105 GPa
Allowable stress 100 MPa 160 MPa
SOLUTION
Use aluminum as the reference material.
1.0 in aluminum
/ 105/70 1.5 in brass
=
= = =
ba
n
n EE
For the transformed section,
1
32
1
1 11 1 1
3 3 34
12
1.5 (30)(6) (1.5)(30)(6)(18) 88.29 10 mm
12
= +
=+=×
n
I bh nAd
3 3 34
2
2 22
34
31
34
123
1.0 (30)(30) 67.5 10 mm
12 12
88.29 10 mm
244.08 10 mm
= = = ×
= = ×
=++= ×
n
I bh
II
III I
94
244.08 10 m
−
= ×
= =
nMy I
M
I ny
σ
σ
Aluminum:
6
1.0, 15 mm 0.015 m, 100 10 Pa= = = = ×ny
σ
69
3
(100 10 )(244.08 10 ) 1.627 10 N m
(1.0)(0.015)
M
−
××
= =×⋅
Brass:
6
1.5, 21 mm 0.021 m, 160 10 Pa= = = = ×ny
σ
69
3
(160 10 )(244.08 10 ) 1.240 10 N m
(1.5)(0.021)
M−
××
= =×⋅
Choose the smaller value
3
1.240 10 N m=×⋅M
1.240 kN m⋅
PROBLEM 11.26
A bar having the cross section shown has been formed by securely
bonding brass and aluminum stock. Using the data given below,
determine the largest permissible bending moment when the
composite bar is bent about a horizontal axis.
Aluminum Brass
Modulus of elasticity 70 GPa 105 GPa
Allowable stress 100 MPa 160 MPa
SOLUTION
consent of McGraw–Hill Education.
PROBLEM 11.27
For the composite bar indicated, determine the largest permissible
bending moment when the bar is bent about a vertical axis.
PROBLEM 11.27. Bar of Prob. 11.25.
SOLUTION
Use aluminum as reference material.
1.0 in aluminum
/ 105/70 1.5 in brass
=
= = =
ba
n
n EE
For transformed section,
3
1
1 11
3 34
12
1.5 (6)(30) 20.25 10 mm
12
=
= = ×
n
I bh
3
2
2 22
3 34
12
1.0 (30)(30) 67.5 10 mm
12
=
= = ×
n
I bh
34
31
20.25 10 mm= = ×II
3 4 94
123
108 10 mm 108 10 mIII I
−
=++= × = ×
σ
σ
= ∴=
nMy I
M
I ny
Aluminum:
6
1.0, 15 mm 0.015 m, 100 10 Pa= = = = ×ny
σ
69
(100 10 )(108 10 ) 720 N m
(1.0)(0.015)
M
−
××
= = ⋅
Brass:
6
1.5, 15 mm 0.015 m, 160 10 Pa= = = = ×ny
σ
69
(160 10 )(108 10 ) 768 N m
(1.5)(0.015)
−
××
= = ⋅M
Choose the smaller value.
720 N mM= ⋅
Aluminum Brass
Modulus of elasticity 70 GPa 105 GPa
Allowable stress 100 MPa 160 MPa
PROBLEM 11.28
For the composite bar indicated, determine the largest permissible
bending moment when the bar is bent about a vertical axis.
PROBLEM 11.28 Bar of Prob. 11.26.
Aluminum
Brass
Modulus of elasticity
70 GPa
105 GPa
Allowable stress
100 MPa
160 MPa
SOLUTION
PROBLEM 11.29
Wooden beams and steel plates are securely bolted together to form the composite
member shown. Using the data given below, determine the largest permissible bending
moment when the member is bent about a horizontal axis.
Wood Steel
Modulus of elasticity:
6
2 10 psi×
6
29 10 psi×
Allowable stress: 2000 psi 22 ksi
SOLUTION
PROBLEM 11.21
Knowing that for the beam shown the allowable stress is 12 ksi in tension and
16 ksi in compression, determine the largest couple M that can be applied.
SOLUTION
consent of McGraw–Hill Education.
PROBLEM 11.22
Straight rods of 0.30–in. diameter and 200–ft length are sometimes
used to clear underground conduits of obstructions or to thread wires
through a new conduit. The rods are made of high–strength steel and,
for storage and transportation, are wrapped on spools of 5–ft
diameter. Assuming that the yield strength is not exceeded,
determine (a) the maximum stress in a rod, when the rod, which is
initially straight, is wrapped on a spool, (b) the corresponding
bending moment in the rod. Use
6
29 10 psiE= ×
.
SOLUTION
Radius of cross section:
11
(0.30) 0.15 in.
22
rd= = =
Moment of inertia:
4 4 64
(0.15) 397.61 10 in
44
Ir
ππ
−
= = = ×
1
5 ft 60 in. 30 in.
2
0.15 in.
DD
cr
r
= = = =
= =
(a)
63
max
(29 10 )(0.15) 145.0 10 psi
30
Ec
sr
×
= = = ×
max
145.0 ksi
s
=
(b)
66
(29 10 )(397.61 10 )
30
EI
M
r
−
××
= =
384 lb in.M= ⋅
consent of McGraw–Hill Education.
PROBLEM 11.23
A 60–N ⋅ m couple is applied to the steel bar shown. (a) Assuming that
the couple is applied about the z axis as shown, determine the maximum
stress and the radius of curvature of the bar. (b) Solve part a, assuming
that the couple is applied about the y axis. Use
200 GPa.
E=
SOLUTION
(a) Bending about z–axis.
3 3 3 4 94
11
(12)(20) 8 10 mm 8 10 m
12 12
20 10 mm 0.010 m
2
−
== =×=×
= = =
I bh
c
6
9
(60)(0.010) 75.0 10 Pa
8 10
Mc
I
σ
−
= = = ×
×
75.0 MPa
σ
=
31
99
1 60 37.5 10 m
(200 10 )(8 10 )
M
EI
ρ
−−
−
= = = ×
××
26.7 m
ρ
=
(b) Bending about y–axis.
3 3 3 4 94
6
9
11
(20)(12) 2.88 10 mm 2.88 10 m
12 12
12 6 mm 0.006 m
2(60)(0.006) 125.0 10 Pa
2.88 10
I bh
c
Mc
I
σ
−
−
== =×=×
= = =
= = = ×
×
125.0 MPa
σ
=
31
99
1 60 104.17 10 m
(200 10 )(2.88 10 )
M
EI
ρ
−−
−
= = = ×
××
9.60 m
ρ
=
consent of McGraw–Hill Education.
PROBLEM 11.24
A couple of magnitude M is applied to a square bar of side a. For
each of the orientations shown, determine the maximum stress and
the curvature of the bar.
SOLUTION
4
33
11
12 12 12
2
a
I bh aa
a
c
= = =
=
max 42
12
a
M
Mc
Ia
σ
= =
max 3
6M
a
σ
=
4
1
12
MM
EI a
E
ρ
= =
4
1 12M
Ea
ρ
=
For one triangle, the moment of inertia about its base is
( )
34
3
1
4
21
4
12
11
2
12 12 24
2
24
12
aa
I bh a
a
II
a
II I

= = =


= =
=+=
max 43
/2 62
/12
2
a Mc Ma M
cIaa
σ
= = = =
max 3
8.49M
a
σ
=
4
1
12
ρ
= =
MM
EI a
E
4
1 12M
Ea
ρ
=
consent of McGraw–Hill Education.
PROBLEM 11.25
A bar having the cross section shown has been formed by securely
bonding brass and aluminum stock. Using the data given below,
determine the largest permissible bending moment when the
Aluminum Brass
Modulus of elasticity 70 GPa 105 GPa
Allowable stress 100 MPa 160 MPa
SOLUTION
Use aluminum as the reference material.
1.0 in aluminum
/ 105/70 1.5 in brass
=
= = =
ba
n
n EE
For the transformed section,
1
32
1
1 11 1 1
3 3 34
12
1.5 (30)(6) (1.5)(30)(6)(18) 88.29 10 mm
12
= +
=+=×
n
I bh nAd
3 3 34
2
2 22
34
31
34
123
1.0 (30)(30) 67.5 10 mm
12 12
88.29 10 mm
244.08 10 mm
= = = ×
= = ×
=++= ×
n
I bh
II
III I
94
244.08 10 m
−
= ×
= =
nMy I
M
I ny
σ
σ
Aluminum:
6
1.0, 15 mm 0.015 m, 100 10 Pa= = = = ×ny
σ
69
3
(100 10 )(244.08 10 ) 1.627 10 N m
(1.0)(0.015)
M
−
××
= =×⋅
Brass:
6
1.5, 21 mm 0.021 m, 160 10 Pa= = = = ×ny
σ
69
3
(160 10 )(244.08 10 ) 1.240 10 N m
(1.5)(0.021)
M−
××
= =×⋅
Choose the smaller value
3
1.240 10 N m=×⋅M
1.240 kN m⋅
PROBLEM 11.26
A bar having the cross section shown has been formed by securely
bonding brass and aluminum stock. Using the data given below,
determine the largest permissible bending moment when the
composite bar is bent about a horizontal axis.
Aluminum Brass
Modulus of elasticity 70 GPa 105 GPa
Allowable stress 100 MPa 160 MPa
SOLUTION
consent of McGraw–Hill Education.
PROBLEM 11.27
For the composite bar indicated, determine the largest permissible
bending moment when the bar is bent about a vertical axis.
PROBLEM 11.27. Bar of Prob. 11.25.
SOLUTION
Use aluminum as reference material.
1.0 in aluminum
/ 105/70 1.5 in brass
=
= = =
ba
n
n EE
For transformed section,
3
1
1 11
3 34
12
1.5 (6)(30) 20.25 10 mm
12
=
= = ×
n
I bh
3
2
2 22
3 34
12
1.0 (30)(30) 67.5 10 mm
12
=
= = ×
n
I bh
34
31
20.25 10 mm= = ×II
3 4 94
123
108 10 mm 108 10 mIII I
−
=++= × = ×
σ
σ
= ∴=
nMy I
M
I ny
Aluminum:
6
1.0, 15 mm 0.015 m, 100 10 Pa= = = = ×ny
σ
69
(100 10 )(108 10 ) 720 N m
(1.0)(0.015)
M
−
××
= = ⋅
Brass:
6
1.5, 15 mm 0.015 m, 160 10 Pa= = = = ×ny
σ
69
(160 10 )(108 10 ) 768 N m
(1.5)(0.015)
−
××
= = ⋅M
Choose the smaller value.
720 N mM= ⋅
Aluminum Brass
Modulus of elasticity 70 GPa 105 GPa
Allowable stress 100 MPa 160 MPa
PROBLEM 11.28
For the composite bar indicated, determine the largest permissible
bending moment when the bar is bent about a vertical axis.
PROBLEM 11.28 Bar of Prob. 11.26.
Aluminum
Brass
Modulus of elasticity
70 GPa
105 GPa
Allowable stress
100 MPa
160 MPa
SOLUTION
PROBLEM 11.29
Wooden beams and steel plates are securely bolted together to form the composite
member shown. Using the data given below, determine the largest permissible bending
moment when the member is bent about a horizontal axis.
Wood Steel
Modulus of elasticity:
6
2 10 psi×
6
29 10 psi×
Allowable stress: 2000 psi 22 ksi
SOLUTION