PROBLEM 11.40
Solve Prob. 11.39, assuming that the 300–mm width is increased to
350 mm.
PROBLEM 11.39 The reinforced concrete beam shown is subjected to a
positive bending moment of 175 kN ⋅ m. Knowing that the modulus of
elasticity is 25 GPa for the concrete and 200 GPa for the steel, determine
(a) the stress in the steel, (b) the maximum stress in the concrete.
SOLUTION
22
32
32
200 GPa 8.0
25 GPa
4 (4) (25)
44
1.9635 10 mm
15.708 10 mm
s
c
s
s
E
nE
Ad
nA
ππ
= = =
= =
= ×
= ×
Locate the neutral axis.
3
23 6
350 (15.708 10 )(480 ) 0
2
175 15.708 10 7.5398 10 0
− × −=
+ × − ×=
x
xx
xx
Solve for x.
3 32 6
15.708 10 (15.708 10 ) (4)(175)(7.5398 10 )
(2)(175)
167.48 mm, 480 312.52 mm
x
xx
− ×+ × + ×
=
= −=
3 32
3 32
9 4 34
1(350) (15.708 10 )(480 )
3
1(350)(167.48) (15.708 10 )(312.52)
3
2.0823 10 mm 2.0823 10 m
σ
−
= +× −
= +×
=×=×
= −
Ix x
nMy
I
(a) Steel:
312.52 mm 0.31252 my=−=−
36
3
(8.0)(175 10 )( 0.31252) 210 10 Pa
2.0823 10
σ
−
×−
=−=×
×
210 MPa
σ
=
(b) Concrete:
167.48 mm 0.16748 my= =
36
3
(1.0)(175 10 )(0.16748) 14.08 10 Pa
2.0823 10
σ
−
×
=− =−×
×
14.08 MPa
σ
= −
consent of McGraw–Hill Education.
PROBLEM 11.41
A concrete slab is reinforced by
5
8
–in.–diameter steel rods
placed on 5.5-in. centers as shown. The modulus of elasticity is
3 × 106 psi for the concrete and 29 × 106 psi for the steel.
Using an allowable stress of 1400 psi for the concrete and
20 ksi for the steel, determine the largest bending moment in a
portion of slab 1 ft wide.
SOLUTION
6
6
29 10 9.6667
3 10
s
c
E
nE
×
= = =
×
Consider a section 5.5 in. wide.
2
22
50.3068 in
4 48
= = =
ss
Ad
ππ
2
2.9657 in
s
nA =
Locate the natural axis.
5.5 (4 )(2.9657) 0
2
x
xx−− =
2
2.75 2.9657 11.8628 0xx+−=
Solve for x.
1.6066 in. 4 2.3934 in.= −=xx
32
3 24
1(5.5) (2.9657)(4 )
3
1(5.5)(1.6066) (2.9657)(2.3934) 24.591in
3
=+−
=+=
Ix x
nMy I
M
I ny
σ
σ
= =
Concrete:
1, 1.6066 in., 1400 psi= = =ny
σ
3
(24.591)(1400) 21.429 10 lb in.
(1.0)(1.6066)
M= = ×⋅
consent of McGraw–Hill Education.
SOLUTION Continued
Steel:
3
9.6667, 2.3934 in., 20 ksi=20 10 psi= = = ×ny
σ
33
(24.591)(20 10 ) 21.258 10 lb in.
(9.6667)(2.3934)
M×
= = ×⋅
Choose the smaller value as the allowable moment for a 5.5 in. width.
3
21.258 10 lb in.M= ×⋅
For a 1 ft = 12 in. width,
33
12 (21.258 10 ) 46.38 10 lb in.
5.5
M= ×= × ⋅
46.38 kip in.= ⋅M
3.87 kip ft⋅
consent of McGraw–Hill Education.
PROBLEM 11.42
Solve Prob. 11.41, assuming that the spacing of the
5
8
–in.–
diameter steel rods is increased to 7.5 in.
PROBLEM 11.41 A concrete slab is reinforced by
5
8
–in.–
diameter steel rods placed on 5.5–in. centers as shown. The
modulus of elasticity is 3 × 106 psi for the concrete and 29 × 106
psi for the steel. Using an allowable stress of 1400 psi for the
concrete and 20 ksi for the steel, determine the largest bending
moment in a portion of slab 1 ft wide.
SOLUTION
6
6
29 10 psi 9.667
3 10 psi
×
= = =
×
s
c
E
nE
Number of rails per foot:
12 in. 1.6
7.5 in.
= =
Area of
5-in.-
8
diameter bars per foot:
22
5
1.6 0.4909 in
48
π
=
s
A
Transformed section, all concrete.
First moment of area:
12 4.745(4 ) 0
2
− −=
x
xx
1.4266 in.x=
2
9.667(0.4909) 4.745 in= =
s
nA
3 24
1(12)(1.4266) 4.745(4 1.4266) 43.037 in
3
= +− =
NA
I
For concrete:
all
1400 psi 1.4266 in.= = =cx
s
4
43.037 in
(1400 psi) 1.4266 in.
= =
I
Mc
s
42.24 kip.in.M=
For steel:
all
20 ksi 4 4 1.4266 2.5734 in.
s
= =−=− =cx
4
steel
20 ksi 43.042 in
9.667 2.5734 in.
= ⋅= ⋅
I
Mnc
s
34.60 kip in.= ⋅M
We choose the smaller M.
34.60 kip in.M= ⋅
Steel controls.
2.88 kip ft= ⋅M
consent of McGraw–Hill Education.
PROBLEM 11.43
Knowing that the bending moment in the reinforced concrete beam is
+100 kip ⋅ ft and that the modulus of elasticity is
6
3.625 10 psi×
for the
concrete and
6
29 10 psi×
for the steel, determine (a) the stress in the
steel, (b) the maximum stress in the concrete.
SOLUTION
6
6
22 2
29 10 8.0
3.625 10
(4) (1) 3.1416 in 25.133 in
4
s
c
ss
E
nE
A nA
π
×
= = =
×
= = =
Locate the neutral axis.
(24)(4)( 2) (12 ) (25.133)(17.5 4 ) 0
2
x
xx x
+ + − −− =
22
96 192 6 339.3 25.133 0 or 6 121.133 147.3 0xx x x x++− + = + − =
Solve for x.
2
121.133 (121.133) (4)(6)(147.3) 1.150 in.
(2)(6)
x−+ +
= =
317.5 4 12.350 in.dx= −−=
32 3 2 4
1 11 11
3 34
22
2 24
3 33
4
123
11
(24)(4) (24)(4)(3.150) 1080.6 in
12 12
11
(12)(1.150) 6.1 in
33
(25.133)(12.350) 3833.3 in
4920 in
= += + =
= = =
= = =
=++=
I bh Ad
I bx
I nA d
III I
nMy
I
σ
= −
where
100 kip ft 1200 kip in.M= ⋅= ⋅
(a) Steel:
8.0n=
12.350 in.y= −
(8.0)(1200)( 12.350)
4920
s
σ
−
= −
24.1 ksi
s
σ
=
(b) Concrete:
1.0, 4 1.150 5.150 in.ny= =+=
(1.0)(1200)(5.150)
4920
c
σ
= −
1.256 ksi
c
σ
= −
consent of McGraw–Hill Education.
PROBLEM 11.44
A concrete beam is reinforced by three steel rods placed as shown. The
modulus of elasticity is
6
3 10 psi×
for the concrete and
6
29 10 psi×
for the
steel. Using an allowable stress of 1350 psi for the concrete and 20 ksi for the
steel, determine the largest allowable positive bending moment in the beam.
SOLUTION
6
6
2
2 22
29 10 9.67
3 10
7
3 (3) 1.8040 in 17.438 in
4 48
s
c
ss
E
nE
A d nA
ππ
×
= = =
×
= = = =
Locate the neutral axis:
2
8 (17.438)(14 ) 0
2
4 17.438 244.14 0
− −=
+ −=
x
xx
xx
Solve for x.
2
17.438 17.438 (4)(4)(244.14) 5.6326 in.
(2)(4)
x−+ +
= =
14 8.3674 in.x−=
3 2 3 24
11
8 (14 ) (8)(5.6326) (17.438)(8.3674) 1697.45 in
33
s
I x nA x= + −= + =
σ
σ
= ∴=
nMy I
M
I ny
Concrete:
1.0, 5.6326 in., 1350 psiny
σ
= = =
3
(1350)(1697.45) 406.835 10 lb in. 407 kip in.
(1.0)(5.6326)
= = × ⋅= ⋅M
Steel:
3
9.67, 8.3674 in., 20 10 psiny
σ
= = = ×
3
(20 10 )(1697.45) 419.72 lb in. 420 kip in.
(9.67)(8.3674)
×
= = ⋅= ⋅M
Choose the smaller value.
407 kip in.= ⋅M
33.9 kip ftM= ⋅
PROBLEM 11.45
Five metal strips, each a 0.5 × 1.5–in. cross section, are bonded together to
form the composite beam shown. The modulus of elasticity is 30 × 106 psi
for the steel, 15 × 106 psi for the brass, and 10 × 106 psi for the aluminum.
Knowing that the beam is bent about a horizontal axis by a couple of
moment
12 kip in.,⋅
determine (a) the maximum stress in each of the three
metals, (b) the radius of curvature of the composite beam.
SOLUTION
PROBLEM 11.46
Five metal strips, each a 0.5 × 1.5–in. cross section, are bonded together to
form the composite beam shown. The modulus of elasticity is 30 × 106 psi
for the steel, 15 × 106 psi for the brass, and 10 × 106 psi for the aluminum.
Knowing that the beam is bent about a horizontal axis by a couple of
moment 12 kip · in., determine (a) the maximum stress in each of the
three metals, (b) the radius of curvature of the composite beam.
SOLUTION
Use aluminum as the reference material.
6
6
6
6
30 10 3.0 in steel
10 10
15 10 1.5 in brass
10 10
1.0 in aluminum
×
= = =
×
×
= = =
×
=
s
a
b
a
E
nE
E
nE
n
For the transformed section,
32
1
1 11 1 11
32
4
12
3.0 (1.5)(0.5) (3.0)(0.75)(1.0)
12
2.2969 in
= +
= +
=
n
I bh n Ad
32 3 2 4
2
2 22 2 22
3 34
3
3 33
44
42 51
54
1
1.0 (1.5)(0.5) (1.0)(0.75)(0.5) 0.2031in
12 12
1.5 (1.5)(0.5) 0.0234 in
12 12
0.2031in 2.2969 in
5.0234 in
=+= + =
= = =
= = = =
= =
∑i
n
I bh nAd
n
I bh
II II
II
(a) Steel:
(3.0)(12)(1.25) 8.96 ksi
5.0234
nMy
I
s
= = =
Aluminum:
(1.0)(12)(0.75) 1.792 ksi
5.0234
nMy
I
s
= = =
Brass:
(1.5)(12)(0.25) 0.896 ksi
5.0234
nMy
I
s
= = =
(b)
361
6
1 12 10 238.89 10 in.
(10 10 )(5.0234)
ρ
−−
×
= = = ×
×
a
M
EI
4186 in. 349 ft= =
ρ
consent of McGraw–Hill Education.
PROBLEM 11.47
The composite beam shown is formed by bonding together a brass rod and an
aluminum rod of semicircular cross sections. The modulus of elasticity is
6
15 10 psi×
for the brass and
6
10 10 psi×
for the aluminum. Knowing that the
composite beam is bent about a horizontal axis by couples of moment
8 kip in.,⋅
determine the maximum stress (a) in the brass, (b) in the aluminum.
SOLUTION
For each semicircle,
22
0.8 in. 1.00531 in
2
r Ar
π
= = =
,
44
0 base
4 (4)(0.8) 0.33953 in. 0.160850 in
33 8
π
ππ
= = = = =
r
y Ir
2 24
base 0
0.160850 (1.00531)(0.33953) 0.044953 in=−= − =I I Ay
Use aluminum as the reference material.
6
6
1.0 in aluminum
15 10 1.5 in brass
10 10
b
a
n
E
nE
=
×
= = =
×
Locate the neutral axis.
0
0.17067 0.06791 in.
2.51327
= =Y
The neutral axis lies 0.06791 in.
above the material interface.
12
2 24
1 1 11
2 24
2 2 22
12
0.33953 0.06791 0.27162 in., 0.33953 0.06791 0.40744 in.
(1.5)(0.044957) (1.5)(1.00531)(0.27162) 0.17869 in
(1.0)(0.044957) (1.0)(1.00531)(0.40744) 0.21185 in
dd
I n I n Ad
I n I n Ad
III
=−= =+=
=+= + =
=+= + =
= +
4
0.39054 in=
(a) Brass:
1.5, 0.8 0.06791 0.73209 in.ny==−=
(1.5)(8)(0.73209)
0.39054
nMy
I
σ
=−=−
22.5 ksi
σ
= −
(b) Aluminium:
1.0, 0.8 0.06791 0.86791 in.ny= =−− =−
(1.0)(8)( 0.86791)
0.39054
nMy
I
σ
−
=−=−
17.78 ksi
σ
=
A, in2
nA, in2
0
, in.y
3
0, innAy
1.00531
1.50796
0.33953
0.51200
1.00531
1.00531
−0.33953
−0.34133
Σ
2.51327
0.17067
PROBLEM 11.48
A steel pipe and an aluminum pipe are securely bonded together to form
the composite beam shown. The modulus of elasticity is 200 GPa for the
steel and 70 GPa for the aluminum. Knowing that the composite beam is
bent by a couple of moment 500 N ⋅ m, determine the maximum stress
(a) in the aluminum, (b) in the steel.
SOLUTION
consent of McGraw–Hill Education.
PROBLEM 11.41
A concrete slab is reinforced by
5
8
–in.–diameter steel rods
placed on 5.5-in. centers as shown. The modulus of elasticity is
3 × 106 psi for the concrete and 29 × 106 psi for the steel.
Using an allowable stress of 1400 psi for the concrete and
20 ksi for the steel, determine the largest bending moment in a
portion of slab 1 ft wide.
SOLUTION
6
6
29 10 9.6667
3 10
s
c
E
nE
×
= = =
×
Consider a section 5.5 in. wide.
2
22
50.3068 in
4 48
= = =
ss
Ad
ππ
2
2.9657 in
s
nA =
Locate the natural axis.
5.5 (4 )(2.9657) 0
2
x
xx−− =
2
2.75 2.9657 11.8628 0xx+−=
Solve for x.
1.6066 in. 4 2.3934 in.= −=xx
32
3 24
1(5.5) (2.9657)(4 )
3
1(5.5)(1.6066) (2.9657)(2.3934) 24.591in
3
=+−
=+=
Ix x
nMy I
M
I ny
σ
σ
= =
Concrete:
1, 1.6066 in., 1400 psi= = =ny
σ
3
(24.591)(1400) 21.429 10 lb in.
(1.0)(1.6066)
M= = ×⋅
consent of McGraw–Hill Education.
SOLUTION Continued
Steel:
3
9.6667, 2.3934 in., 20 ksi=20 10 psi= = = ×ny
σ
33
(24.591)(20 10 ) 21.258 10 lb in.
(9.6667)(2.3934)
M×
= = ×⋅
Choose the smaller value as the allowable moment for a 5.5 in. width.
3
21.258 10 lb in.M= ×⋅
For a 1 ft = 12 in. width,
33
12 (21.258 10 ) 46.38 10 lb in.
5.5
M= ×= × ⋅
46.38 kip in.= ⋅M
3.87 kip ft⋅
consent of McGraw–Hill Education.
PROBLEM 11.42
Solve Prob. 11.41, assuming that the spacing of the
5
8
–in.–
diameter steel rods is increased to 7.5 in.
PROBLEM 11.41 A concrete slab is reinforced by
5
8
–in.–
diameter steel rods placed on 5.5–in. centers as shown. The
modulus of elasticity is 3 × 106 psi for the concrete and 29 × 106
psi for the steel. Using an allowable stress of 1400 psi for the
concrete and 20 ksi for the steel, determine the largest bending
moment in a portion of slab 1 ft wide.
SOLUTION
6
6
29 10 psi 9.667
3 10 psi
×
= = =
×
s
c
E
nE
Number of rails per foot:
12 in. 1.6
7.5 in.
= =
Area of
5-in.-
8
diameter bars per foot:
22
5
1.6 0.4909 in
48
π
=
s
A
Transformed section, all concrete.
First moment of area:
12 4.745(4 ) 0
2
− −=
x
xx
1.4266 in.x=
2
9.667(0.4909) 4.745 in= =
s
nA
3 24
1(12)(1.4266) 4.745(4 1.4266) 43.037 in
3
= +− =
NA
I
For concrete:
all
1400 psi 1.4266 in.= = =cx
s
4
43.037 in
(1400 psi) 1.4266 in.
= =
I
Mc
s
42.24 kip.in.M=
For steel:
all
20 ksi 4 4 1.4266 2.5734 in.
s
= =−=− =cx
4
steel
20 ksi 43.042 in
9.667 2.5734 in.
= ⋅= ⋅
I
Mnc
s
34.60 kip in.= ⋅M
We choose the smaller M.
34.60 kip in.M= ⋅
Steel controls.
2.88 kip ft= ⋅M
consent of McGraw–Hill Education.
PROBLEM 11.43
Knowing that the bending moment in the reinforced concrete beam is
+100 kip ⋅ ft and that the modulus of elasticity is
6
3.625 10 psi×
for the
concrete and
6
29 10 psi×
for the steel, determine (a) the stress in the
steel, (b) the maximum stress in the concrete.
SOLUTION
6
6
22 2
29 10 8.0
3.625 10
(4) (1) 3.1416 in 25.133 in
4
s
c
ss
E
nE
A nA
π
×
= = =
×
= = =
Locate the neutral axis.
(24)(4)( 2) (12 ) (25.133)(17.5 4 ) 0
2
x
xx x
+ + − −− =
22
96 192 6 339.3 25.133 0 or 6 121.133 147.3 0xx x x x++− + = + − =
Solve for x.
2
121.133 (121.133) (4)(6)(147.3) 1.150 in.
(2)(6)
x−+ +
= =
317.5 4 12.350 in.dx= −−=
32 3 2 4
1 11 11
3 34
22
2 24
3 33
4
123
11
(24)(4) (24)(4)(3.150) 1080.6 in
12 12
11
(12)(1.150) 6.1 in
33
(25.133)(12.350) 3833.3 in
4920 in
= += + =
= = =
= = =
=++=
I bh Ad
I bx
I nA d
III I
nMy
I
σ
= −
where
100 kip ft 1200 kip in.M= ⋅= ⋅
(a) Steel:
8.0n=
12.350 in.y= −
(8.0)(1200)( 12.350)
4920
s
σ
−
= −
24.1 ksi
s
σ
=
(b) Concrete:
1.0, 4 1.150 5.150 in.ny= =+=
(1.0)(1200)(5.150)
4920
c
σ
= −
1.256 ksi
c
σ
= −
consent of McGraw–Hill Education.
PROBLEM 11.44
A concrete beam is reinforced by three steel rods placed as shown. The
modulus of elasticity is
6
3 10 psi×
for the concrete and
6
29 10 psi×
for the
steel. Using an allowable stress of 1350 psi for the concrete and 20 ksi for the
steel, determine the largest allowable positive bending moment in the beam.
SOLUTION
6
6
2
2 22
29 10 9.67
3 10
7
3 (3) 1.8040 in 17.438 in
4 48
s
c
ss
E
nE
A d nA
ππ
×
= = =
×
= = = =
Locate the neutral axis:
2
8 (17.438)(14 ) 0
2
4 17.438 244.14 0
− −=
+ −=
x
xx
xx
Solve for x.
2
17.438 17.438 (4)(4)(244.14) 5.6326 in.
(2)(4)
x−+ +
= =
14 8.3674 in.x−=
3 2 3 24
11
8 (14 ) (8)(5.6326) (17.438)(8.3674) 1697.45 in
33
s
I x nA x= + −= + =
σ
σ
= ∴=
nMy I
M
I ny
Concrete:
1.0, 5.6326 in., 1350 psiny
σ
= = =
3
(1350)(1697.45) 406.835 10 lb in. 407 kip in.
(1.0)(5.6326)
= = × ⋅= ⋅M
Steel:
3
9.67, 8.3674 in., 20 10 psiny
σ
= = = ×
3
(20 10 )(1697.45) 419.72 lb in. 420 kip in.
(9.67)(8.3674)
×
= = ⋅= ⋅M
Choose the smaller value.
407 kip in.= ⋅M
33.9 kip ftM= ⋅
PROBLEM 11.45
Five metal strips, each a 0.5 × 1.5–in. cross section, are bonded together to
form the composite beam shown. The modulus of elasticity is 30 × 106 psi
for the steel, 15 × 106 psi for the brass, and 10 × 106 psi for the aluminum.
Knowing that the beam is bent about a horizontal axis by a couple of
moment
12 kip in.,⋅
determine (a) the maximum stress in each of the three
metals, (b) the radius of curvature of the composite beam.
SOLUTION
PROBLEM 11.46
Five metal strips, each a 0.5 × 1.5–in. cross section, are bonded together to
form the composite beam shown. The modulus of elasticity is 30 × 106 psi
for the steel, 15 × 106 psi for the brass, and 10 × 106 psi for the aluminum.
Knowing that the beam is bent about a horizontal axis by a couple of
moment 12 kip · in., determine (a) the maximum stress in each of the
three metals, (b) the radius of curvature of the composite beam.
SOLUTION
Use aluminum as the reference material.
6
6
6
6
30 10 3.0 in steel
10 10
15 10 1.5 in brass
10 10
1.0 in aluminum
×
= = =
×
×
= = =
×
=
s
a
b
a
E
nE
E
nE
n
For the transformed section,
32
1
1 11 1 11
32
4
12
3.0 (1.5)(0.5) (3.0)(0.75)(1.0)
12
2.2969 in
= +
= +
=
n
I bh n Ad
32 3 2 4
2
2 22 2 22
3 34
3
3 33
44
42 51
54
1
1.0 (1.5)(0.5) (1.0)(0.75)(0.5) 0.2031in
12 12
1.5 (1.5)(0.5) 0.0234 in
12 12
0.2031in 2.2969 in
5.0234 in
=+= + =
= = =
= = = =
= =
∑i
n
I bh nAd
n
I bh
II II
II
(a) Steel:
(3.0)(12)(1.25) 8.96 ksi
5.0234
nMy
I
s
= = =
Aluminum:
(1.0)(12)(0.75) 1.792 ksi
5.0234
nMy
I
s
= = =
Brass:
(1.5)(12)(0.25) 0.896 ksi
5.0234
nMy
I
s
= = =
(b)
361
6
1 12 10 238.89 10 in.
(10 10 )(5.0234)
ρ
−−
×
= = = ×
×
a
M
EI
4186 in. 349 ft= =
ρ
consent of McGraw–Hill Education.
PROBLEM 11.47
The composite beam shown is formed by bonding together a brass rod and an
aluminum rod of semicircular cross sections. The modulus of elasticity is
6
15 10 psi×
for the brass and
6
10 10 psi×
for the aluminum. Knowing that the
composite beam is bent about a horizontal axis by couples of moment
8 kip in.,⋅
determine the maximum stress (a) in the brass, (b) in the aluminum.
SOLUTION
For each semicircle,
22
0.8 in. 1.00531 in
2
r Ar
π
= = =
,
44
0 base
4 (4)(0.8) 0.33953 in. 0.160850 in
33 8
π
ππ
= = = = =
r
y Ir
2 24
base 0
0.160850 (1.00531)(0.33953) 0.044953 in=−= − =I I Ay
Use aluminum as the reference material.
6
6
1.0 in aluminum
15 10 1.5 in brass
10 10
b
a
n
E
nE
=
×
= = =
×
Locate the neutral axis.
0
0.17067 0.06791 in.
2.51327
= =Y
The neutral axis lies 0.06791 in.
above the material interface.
12
2 24
1 1 11
2 24
2 2 22
12
0.33953 0.06791 0.27162 in., 0.33953 0.06791 0.40744 in.
(1.5)(0.044957) (1.5)(1.00531)(0.27162) 0.17869 in
(1.0)(0.044957) (1.0)(1.00531)(0.40744) 0.21185 in
dd
I n I n Ad
I n I n Ad
III
=−= =+=
=+= + =
=+= + =
= +
4
0.39054 in=
(a) Brass:
1.5, 0.8 0.06791 0.73209 in.ny==−=
(1.5)(8)(0.73209)
0.39054
nMy
I
σ
=−=−
22.5 ksi
σ
= −
(b) Aluminium:
1.0, 0.8 0.06791 0.86791 in.ny= =−− =−
(1.0)(8)( 0.86791)
0.39054
nMy
I
σ
−
=−=−
17.78 ksi
σ
=
A, in2
nA, in2
0
, in.y
3
0, innAy
1.00531
1.50796
0.33953
0.51200
1.00531
1.00531
−0.33953
−0.34133
Σ
2.51327
0.17067
PROBLEM 11.48
A steel pipe and an aluminum pipe are securely bonded together to form
the composite beam shown. The modulus of elasticity is 200 GPa for the
steel and 70 GPa for the aluminum. Knowing that the composite beam is
bent by a couple of moment 500 N ⋅ m, determine the maximum stress
(a) in the aluminum, (b) in the steel.
SOLUTION