# PROBLEM 4.1

advertisement ```PROBLEM 4.1
Knowing that the couple shown acts in a vertical plane, determine the
stress at (a) point A, (b) point B.
SOLUTION
For rectangle:
I=
1 3
bh
12
For cross sectional area:
I = I1 + I 2 + I 3 =
1
1
1
(2)(1.5)3 +
(2)(5.5)3 +
(2)(1.5)3 = 28.854 in 4
12
12
12
(a)
y A = 2.75 in.
σA = −
My A
(25)(2.75)
=−
I
28.854
σ A = −2.38 ksi
(b)
yB = 0.75 in.
σB = −
MyB
(25)(0.75)
=−
I
28.854
σ B = −0.650 ksi
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PROBLEM 4.7
Two W4 &times; 13 rolled sections are welded together as shown. Knowing that for the steel
alloy used σ Y = 36 ksi and σ U = 58 ksi and using a factor of safety of 3.0, determine
the largest couple that can be applied when the assembly is bent about the z axis.
SOLUTION
Properties of W4 &times; 13 rolled section.
(See Appendix C.)
Area = 3.83 in 2
Depth = 4.16 in.
I x = 11.3 in 4
For one rolled section, moment of inertia about axis a-a is
I a = I x + Ad 2 = 11.3 + (3.83)(2.08) 2 = 27.87 in 4
For both sections,
I z = 2 I a = 55.74 in 4
c = depth = 4.16 in.
M all
σU
58
= 19.333 ksi
F .S . 3.0
σ I (19.333) (55.74)
= all =
c
4.16
σ all =
=
σ=
Mc
I
M all = 259 kip ⋅ in
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PROBLEM 4.9
Two vertical forces are applied to a beam of the cross section shown. Determine
the maximum tensile and compressive stresses in portion BC of the beam.
SOLUTION
A1 =
π
2
r2 =
π
2
(25) 2 = 981.7 mm 2
A2 = bh = (50)(25) = 1250 mm 2
y =
4r
(4)(25)
=
= 10.610 mm
3π
3π
h
25
y2 = − = −
= −12.5 mm
2
2
y1 =
A1 y1 + A2 y2
(981.7)(10.610) + (1250)(−12.5)
=
= −2.334 mm
A1 + A2
981.7 + 1250
I1 = I x1 − A1 y12 =
π
r 4 − A1 y12 =
π
(25) 4 − (981.7)(10.610) 2 = 42.886 &times; 106 mm 4
8
8
d1 = y1 − y = 10.610 − ( −2.334) = 12.944 mm
I1 = I1 + A1d12 = 42.866 &times; 103 + (981.7)(12.944)2 = 207.35 &times; 103 mm 4
1 3 1
bh = (50)(25)3 = 65.104 &times; 103 mm 4
12
12
d 2 = y2 − y = −12.5 − (−2.334) = 10.166 mm
I2 =
I 2 = I 2 + A2 d 22 = 65.104 &times; 103 + (1250)(10.166)2 = 194.288 &times; 103 mm 4
I = I1 + I 2 = 401.16 &times; 103 mm 4 = 401.16 &times; 10−9 m 4
ytop = 25 + 2.334 = 27.334 mm = 0.027334 m
ybot = −25 + 2.334 = −22.666 mm = −0.022666 m
M − Pa = 0 :
σ top =
σ bot =
−Mytop
M = Pa = (4 &times; 103 )(300 &times; 10−3 ) = 1200 N ⋅ m
(1200)(0.027334)
= −81.76 &times; 106 Pa
401.16 &times; 10−9
σ top = −81.8 MPa
−Mybot
(1200)(−0.022666)
=−
= 67.80 &times; 106 Pa
I
401.16 &times; 10−9
σ bot = 67.8 MPa
I
=−
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PROBLEM 4.14
Knowing that a beam of the cross section shown is bent about a horizontal axis
and that the bending moment is 50 kip ⋅ in., determine the total force acting
(a) on the top flange, (b) on the shaded portion of the web.
SOLUTION
The stress distribution over the entire cross-section is given by the bending stress formula:
σx = −
My
I
where y is a coordinate with its origin on the neutral axis and I is the moment of
inertia of the entire cross sectional area. The force on the shaded portion is
calculated from this stress distribution. Over an area element dA, the force is
dF = σ x dA = −
My
dA
I
The total force on the shaded area is then
F = dF = −
My
M
dA = −
I
I
ydA = −
M * *
y A
I
where y * is the centroidal coordinate of the shaded portion and A* is its area.
Calculate the moment of inertia.
1
1
(6 in.)(7 in.)3 − (4 in.)(4 in.)3 = 150.17 in 4
12
12
M = 50 kip ⋅ in
I =
(a)
Top flange:
A* = (6 in.)(1.5 in.) = 9 in 2
F =
(b)
Half web:
50 kip ⋅ in
(9 in 2 )(2.75 in.) = 8.24 kips
150.17 in 4
A* = (2 in.)(2 in.) = 4 in 2
F =
y * = 2 in. + 0.75 in. = 2.75 in.
F = 8.24 kips
y * = 1 in.
50 kip ⋅ in
(4 in 2 )(1 in.) = 1.332 kips
150.17 in 4
F = 1.332 kips
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PROBLEM 4.24
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 E = 200 GPa.
SOLUTION
(a)
Bending about z-axis.
1 3 1
bh = (12)(20)3 = 8 &times; 103 mm4 = 8 &times; 10−9 m 4
12
12
20
c=
= 10 mm = 0.010 m
2
I=
σ=
Mc (60)(0.010)
=
= 75.0 &times; 106 Pa
I
8 &times; 10−9
1
M
60
=
= 37.5 &times; 10−3 m −1
EI (200 &times; 109 )(8 &times; 10 −9 )
ρ
(b)
=
σ = 75.0 MPa
ρ = 26.7 m
Bending about y-axis.
1 3 1
bh = (20)(12)3 = 2.88 &times; 103 mm4 = 2.88 &times; 10−9 m4
12
12
12
c=
= 6 mm = 0.006 m
2
Mc (60)(0.006)
=
= 125.0 &times; 106 Pa
σ=
I
2.88 &times; 10−9
I=
1
ρ
=
M
60
=
= 104.17 &times; 10−3 m −1
−9
9
EI (200 &times; 10 )(2.88 &times; 10 )
σ = 125.0 MPa
ρ = 9.60 m
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PROBLEM 4.34
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
70 GPa
105 GPa
100 MPa
160 MPa
Modulus of elasticity
Allowable stress
SOLUTION
Use aluminum as the reference material.
For aluminum, n = 1.0
For brass, n = Eb /Ea = 105/70 = 1.5
Values of n are shown on the sketch.
For the transformed section,
n1
1.5
(8)(32)3 = 32.768 &times; 103 mm 4
b1h13 =
12
12
n
1.0
I 2 = 2 b2 H 23 − h23 =
(32)(323 − 163 ) = 76.459 &times; 103 mm 4
12
12
I 3 = I1 = 32.768 &times; 103 mm 4
I1 =
(
)
I = I1 + I 2 + I 3 = 141.995 &times; 103 mm 4 = 141.995 &times; 10−9 m 4
|σ | =
Aluminum:
M=
σI
ny
n = 1.0, | y | = 16 mm = 0.016 m, σ = 100 &times; 106 Pa
M=
Brass:
nMy
I
(100 &times; 106 )(141.995 &times; 10−9 )
= 887.47 N ⋅ m
(1.0)(0.016)
n = 1.5, | y | = 16 mm = 0.016 m, σ = 160 &times; 106 Pa
M=
Choose the smaller value.
(160 &times; 106 )(141.995 &times; 10−9 )
= 946.63 N ⋅ m
(1.5)(0.016)
M = 887 N ⋅ m
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PROBLEM 4.41
The 6 &times; 12-in. timber beam has been strengthened by bolting to it the steel
reinforcement shown. The modulus of elasticity for wood is 1.8 &times; 106 psi and for
steel, 29 &times; 106 psi. Knowing that the beam is bent about a horizontal axis by a
couple of moment M = 450 kip ⋅ in., determine the maximum stress in (a) the
wood, (b) the steel.
SOLUTION
Use wood as the reference material.
For wood,
n =1
For steel,
n = Es / Ew = 29 /1.8 = 16.1111
= wood
Transformed section:
421.931
112.278
= 3.758 in.
Yo =
= steel
A, in 2
nA, in 2
72
72
2.5
40.278
yo
6
−0.25
112.278
nA yo , in 3
432
−10.069
421.931
The neutral axis lies 3.758 in. above the wood-steel interface.
n1
1
b1h13 + n1 A1d12 = (6)(12)3 + (72)(6 − 3.758)2 = 1225.91 in 4
12
12
n2
16.1111
I 2 = b2 h23 + n2 A2 d 22 =
(5) (0.5)3 + (40.278)(3.578 + 0.25)2 = 647.87 in 4
12
12
I = I1 + I 2 = 1873.77 in 4
I1 =
σ =−
M = 450 kip ⋅ in
(a)
Wood:
n = 1,
y = 12 − 3.758 = 8.242 in
σw = −
(b)
Steel:
(1) (450) (8.242)
= −1.979 ksi
1873.77
n = 16.1111,
σs = −
nMy
I
σ w = −1.979 ksi
y = −3.758 − 0.5 = −4.258 in
(16.1111) (450) (−4.258)
= 16.48 ksi
1873.77
σ s = 16.48 ksi
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PROBLEM 4.42
The 6 &times; 12-in. timber beam has been strengthened by bolting to it the steel
reinforcement shown. The modulus of elasticity for wood is 1.8 &times; 106 psi
and for steel, 29 &times; 106 psi. Knowing that the beam is bent about a
horizontal axis by a couple of moment M = 450 kip ⋅ in., determine the
maximum stress in (a) the wood, (b) the steel.
SOLUTION
Use wood as the reference material.
For wood,
n =1
For steel,
n=
Es 29 &times; 106
=
= 16.1111
Ew 1.8 &times; 106
For C8 &times; 11.5 channel section,
A = 3.38 in 2 , t w = 0.220 in., x = 0.571 in., I y = 1.32 in 4
For the composite section, the centroid of the channel (part 1) lies 0.571 in. above the bottom of the section.
The centroid of the wood (part 2) lies 0.220 + 6.00 = 6.22 in. above the bottom.
Transformed section:
A, in2
3.38
Part
1
72
2
Σ
Y0 =
nA, in2
54.456
y , in.
0.571
nAy , in 3
31.091
d, in.
3.216
72
6.22
447.84
2.433
478.93
126.456
478.93 in 3
= 3.787 in.
126.456 in 2
d = y0 − Y0
The neutral axis lies 3.787 in. above the bottom of the section.
I1 = n1 I1 + n1 A1d12 = (16.1111)(1.32) + (54.456)(3.216)2 = 584.49 in 4
n2
1
b2 h23 + n2 A2 d 22 = (6)(12)3 + (72)(2.433)2 = 1290.20 in 4
12
12
4
I = I1 + I 2 = 1874.69 in
I2 =
M = 450 kip ⋅ in
(a)
Wood:
n = 1,
σw = −
(b)
Steel:
n My
I
y = 12 + 0.220 − 3.787 = 8.433 in.
σ =−
(1)(450)(8.433)
= −2.02 ksi
1874.69
n = 16.1111,
σs = −
σ w = −2.02 ksi
y = −3.787 in.
(16.1111) (450) (−3.787)
= 14.65 ksi
1874.67
σ s = 14.65 ksi
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PROBLEM 4.51
A concrete beam is reinforced by three steel rods placed as shown. The
modulus of elasticity is 3 &times; 106 psi for the concrete and 29 &times; 106 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
n=
Es 29 &times; 10 6
=
= 9.67
Ec
3 &times; 106
As = 3
π
4
d 2 = (3)
π
4
7
8
2
= 1.8040 in 2
nAs = 17.438 in 2
x
− (17.438)(14 − x) = 0
2
4 x 2 + 17.438 x − 244.14 = 0
Locate the neutral axis:
8x
Solve for x.
x=
−17.438 + 17.4382 + (4)(4)(244.14)
= 5.6326 in.
(2)(4)
14 − x = 8.3674 in.
I =
1 3
1
8x + nAs (14 − x)2 = (8)(5.6326)3 + (17.438)(8.3674) 2 = 1697.45 in 4
3
3
σ =
nMy
I
∴ M=
σI
ny
n = 1.0,
Concrete:
M =
M =
Choose the smaller value.
σ = 1350 psi
(1350)(1697.45)
= 406.835 &times; 103 lb ⋅ in = 407 kip ⋅ in
(1.0)(5.6326)
n = 9.67,
Steel:
y = 5.6326 in.,
y = 8.3674 in., σ = 20 &times; 103 psi
(20 &times; 103 )(1697.45)
= 419.72 lb ⋅ in = 420 kip ⋅ in
(9.67)(8.3674)
M = 407 kip ⋅ in
M = 33.9 kip ⋅ ft
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PROBLEM 4.52
Knowing that the bending moment in the reinforced concrete beam is
+100 kip ⋅ ft and that the modulus of elasticity is 3.625 &times; 106 psi for the
concrete and 29 &times; 106 psi for the steel, determine (a) the stress in the
steel, (b) the maximum stress in the concrete.
SOLUTION
n=
Es
29 &times; 106
=
= 8.0
Ec
3.625 &times; 106
As = (4)
π
4
(1) 2 = 3.1416 in 2
nAs = 25.133 in 2
Locate the neutral axis.
(24)(4)( x + 2) + (12 x)
96 x + 192 + 6 x 2 − 339.3 + 25.133x = 0
x=
Solve for x.
x
− (25.133)(17.5 − 4 − x) = 0
2
or
6 x 2 + 121.133x − 147.3 = 0
−121.133 + (121.133) 2 + (4)(6)(147.3)
= 1.150 in.
(2)(6)
d3 = 17.5 − 4 − x = 12.350 in.
1
b1h13 +
12
1
I 2 = b2 x3 =
3
I1 =
A1d12 =
1
(24)(4)3 + (24)(4)(3.150) 2 = 1080.6 in 4
12
1
(12)(1.150)3 = 6.1 in 4
3
I 3 = nA3d32 = (25.133)(12.350) 2 = 3833.3 in 4
I = I1 + I 2 + I 3 = 4920 in 4
σ =−
(a)
Steel:
nMy
I
n = 8.0
y = −12.350 in.
σs = −
(b)
Concrete:
n = 1.0,
σc = −
where M = 100 kip ⋅ ft = 1200 kip ⋅ in.
(8.0)(1200)(−12.350)
4920
σ s = 24.1 ksi
y = 4 + 1.150 = 5.150 in.
(1.0)(1200)(5.150)
4920
σ c = −1.256 ksi
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PROBLEM 4.55
Five metal strips, each 40 mm wide, are bonded together to form the
composite beam shown. The modulus of elasticity is 210 GPa for the
steel, 105 GPa for the brass, and 70 GPa for the aluminum. Knowing that
the beam is bent about a horizontal axis by a couple of moment 1800 N ⋅
m, 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.
n = 1 in aluminum.
n = Es / Ea = 210 / 70 = 3 in steel.
n = Eb / Ea = 105 / 70 = 1.5 in brass.
Due to symmetry of both the material arrangement and the geometry, the
neutral axis passes through the center of the steel portion.
For the transformed section,
n1
1
b1h13 + n1 A1d12 = (40)(10)3 + (40)(10)(25) 2 = 253.33 &times; 103 mm 4
12
12
n2
1.5
3
2
(40)(10)3 + (1.5)(40)(10)(15) 2 = 140 &times; 103 mm 4
I 2 = b2 h2 + n2 A2 d 2 =
12
12
n
3.0
(40)(20)3 = 80 &times; 103 mm 4
I 3 = 3 b3 h33 =
12
12
I 4 = I 2 = 140 &times; 103 mm 4
I 5 = I1 = 253.33 &times; 103 mm 4
I1 =
I=
(a)
= 866.66 &times; 103 mm 4 = 866.66 &times; 10−9 m 4
nMy
where M = 1800 N ⋅ m
I
n = 1.0
y = −30 mm = 0.030 m
Aluminum:
σ =−
σa =
(1.0)(1800)(0.030)
= 62.3 &times; 106 Pa
−9
866.66 &times; 10
Brass:
σb =
Steel:
σs =
(b)
I
n = 1.5
y = −20 mm = −0.020 m
(1.5)(1800)(0.020)
= 62.3 &times; 106 Pa
−9
866.66 &times; 10
n = 3.0
σ b = 62.3 MPa
y = −10 mm = −0.010 m
(3.0)(1800)(0.010)
= 62.3 &times; 106 Pa
−9
866.66 &times; 10
Radius of curvature.
σ a = 62.3 MPa
1
ρ
=
M
1800
=
= 0.02967 m −1
9
Ea I (70 &times; 10 )(866.66 &times; 10−9 )
σ s = 62.3 MPa
ρ = 33.7 m
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PROBLEM 4.57
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
15 &times; 106 psi for the brass and 10 &times; 106 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,
yo =
r = 0.8 in.
π
A=
2
4r
(4)(0.8)
=
= 0.33953 in.
3π
3π
r 2 = 1.00531 in 2 ,
I base =
π
8
r 4 = 0.160850 in 4
I = I base − Ayo2 = 0.160850 − (1.00531)(0.33953)2 = 0.044953 in 4
Use aluminum as the reference material.
n = 1.0 in aluminum
n=
Eb 15 &times; 106
=
= 1.5 in brass
Ea 10 &times; 106
Locate the neutral axis.
A, in2
nA, in2
yo , in.
1.00531
1.50796
0.33953
0.51200
1.00531
1.00531
−0.33953
−0.34133
Σ
2.51327
nAyo , in 3
Yo =
0.17067
= 0.06791 in.
2.51327
The neutral axis lies 0.06791 in.
above the material interface.
0.17067
d1 = 0.33953 − 0.06791 = 0.27162 in., d 2 = 0.33953 + 0.06791 = 0.40744 in.
I1 = n1I + n1 Ad12 = (1.5)(0.044957) + (1.5)(1.00531)(0.27162)2 = 0.17869 in 4
I 2 = n2 I + n2 Ad 22 = (1.0)(0.044957) + (1.0)(1.00531)(0.40744)2 = 0.21185 in 4
I = I1 + I 2 = 0.39054 in 4
(a)
Brass:
n = 1.5,
y = 0.8 − 0.06791 = 0.73209 in.
σ =−
(b)
Aluminium:
n = 1.0,
nMy
(1.5)(8)(0.73209)
=−
I
0.39054
σ = −22.5 ksi
y = −0.8 − 0.06791 = −0.86791 in.
σ =−
nMy
(1.0)(8)(−0.86791)
=−
I
0.39054
σ = 17.78 ksi
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