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•4–1. The ship is pushed through the water using an A-36
steel propeller shaft that is 8 m long, measured from the
propeller to the thrust bearing D at the engine. If it has an
outer diameter of 400 mm and a wall thickness of 50 mm,
determine the amount of axial contraction of the shaft
when the propeller exerts a force on the shaft of 5 kN. The
bearings at B and C are journal bearings.
A
Internal Force: As shown on FBD.
B
C
D
5 kN
Displacement:
8m
dA =
-5.00 (103)(8)
PL
= p
2
2
9
AE
4 (0.4 - 0.3 ) 200(10 )
= -3.638(10 - 6) m
= -3.64 A 10 - 3 B mm
Ans.
Negative sign indicates that end A moves towards end D.
4–2. The copper shaft is subjected to the axial loads
shown. Determine the displacement of end A with respect
to end D. The diameters of each segment are dAB = 3 in.,
dBC = 2 in., and dCD = 1 in. Take Ecu = 1811032 ksi.
50 in.
A
p
The cross-sectional area of segment AB, BC and CD are AAB = (32) = 2.25p in2,
4
p
p
ABC = (22) = p in2 and ACD = (12) = 0.25p in2.
4
4
Thus,
PCD LCD
PiLi
PAB LAB
PBC LBC
=
+
+
AiEi
AAB ECu
ABC ECu
ACD ECu
2.00 (75)
6.00 (50)
=
(2.25p) C 18(10 ) D
3
+
p C 18(10 ) D
3
-1.00 (60)
+
(0.25p) C 18(103) D
= 0.766(10 - 3) in.
Ans.
The positive sign indicates that end A moves away from D.
122
60 in.
2 kip
6 kip
The normal forces developed in segment AB, BC and CD are shown in the
FBDS of each segment in Fig. a, b and c respectively.
dA>D = ©
75 in.
B 2 kip
1 kip
C
3 kip
D
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4–3. The A-36 steel rod is subjected to the loading shown.
If the cross-sectional area of the rod is 50 mm2, determine
the displacement of its end D. Neglect the size of the
couplings at B, C, and D.
1m
A 9 kN B
The normal forces developed in segments AB, BC and CD are shown in the FBDS
of each segment in Fig. a, b and c, respectively.
The
cross-sectional
areas
of
all
2
1
m
b = 50.0(10 - 6) m2.
A = A 50 mm2 B a
1000 mm
dD = ©
the
segments
are
PiLi
1
=
a PAB LAB + PBC LBC + PCD LCD b
AiEi
A ESC
1
=
50.0(10 ) C 200(109) D
-6
c -3.00(103)(1) + 6.00(103)(1.5) + 2.00(103)(1.25) d
= 0.850(10 - 3) m = 0.850 mm
Ans.
The positive sign indicates that end D moves away from the fixed support.
123
1.5 m
1.25 m
C
4 kN
D
2 kN
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*4–4. The A-36 steel rod is subjected to the loading
shown. If the cross-sectional area of the rod is 50 mm2,
determine the displacement of C. Neglect the size of the
couplings at B, C, and D.
1m
1.5 m
1.25 m
C
A 9 kN B
4 kN
D
2 kN
The normal forces developed in segments AB and BC are shown the FBDS of each
segment in Fig. a and b, respectively. The cross-sectional area of these two segments
2
1m
are A = A 50 mm2 B a
b = 50.0 (10 - 6) m2. Thus,
10.00 mm
dC = ©
PiLi
1
=
A P L + PBC LBC B
AiEi
A ESC AB AB
1
=
50.0(10 - 6) C 200(109) D
c -3.00(103)(1) + 6.00(103)(1.5) d
= 0.600 (10 - 3) m = 0.600 mm
Ans.
The positive sign indicates that coupling C moves away from the fixed support.
4–5. The assembly consists of a steel rod CB and an
aluminum rod BA, each having a diameter of 12 mm. If the rod
is subjected to the axial loadings at A and at the coupling B,
determine the displacement of the coupling B and the end
A. The unstretched length of each segment is shown in the
figure. Neglect the size of the connections at B and C, and
assume that they are rigid. Est = 200 GPa, Eal = 70 GPa.
dB =
C
18 kN
6 kN
3m
12(103)(3)
PL
= 0.00159 m = 1.59 mm
= p
2
9
AE
4 (0.012) (200)(10 )
dA = ©
A
B
2m
Ans.
18(103)(2)
12(103)(3)
PL
+ p
= p
2
9
2
9
AE
4 (0.012) (200)(10 )
4 (0.012) (70)(10 )
= 0.00614 m = 6.14 mm
Ans.
4–6. The bar has a cross-sectional area of 3 in2, and
E = 3511032 ksi. Determine the displacement of its end A
when it is subjected to the distributed loading.
x
w ⫽ 500x1/3 lb/in.
A
4 ft
P(x) =
L0
x
x
w dx = 500
x3 dx =
L0
1
1500 43
x
4
L
dA =
4(12)
P(x) dx
1
3
1
1500 4
1500
b a b(48)3
=
x3 dx = a
AE
4
(3)(35)(108)(4) 7
(3)(35)(106) L0
L0
dA = 0.0128 in.
Ans.
124
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4–7. The load of 800 lb is supported by the four 304 stainless
steel wires that are connected to the rigid members AB and
DC. Determine the vertical displacement of the load if the
members were horizontal before the load was applied. Each
wire has a cross-sectional area of 0.05 in2.
E
F
4 ft
H
D
C
2 ft
Referring to the FBD of member AB, Fig. a
5 ft
4.5 ft
a + ©MA = 0;
FBC (5) - 800(1) = 0
FBC = 160 lb
a + ©MB = 0;
800(4) - FAH (5) = 0
FAH = 640 lb
800 lb
A
B
1 ft
Using the results of FBC and FAH, and referring to the FBD of member DC, Fig. b
a + ©MD = 0;
FCF (7) - 160(7) - 640(2) = 0
FCF = 342.86 lb
a + ©MC = 0;
640(5) - FDE(7) = 0
FDE = 457.14 lb
Since E and F are fixed,
dD =
457.14(4)(2)
FDE LDE
=
= 0.01567 in T
A Est
0.05 C 28.0 (106) D
dC =
342.86 (4)(12)
FCF LCF
=
= 0.01176 in T
A Est
0.05 C 28.0 (106) D
From the geometry shown in Fig. c,
dH = 0.01176 +
5
(0.01567 - 0.01176) = 0.01455 in T
7
Subsequently,
dA>H =
640(4.5)(12)
FAH LAH
=
= 0.02469 in T
A Est
0.05 C 28.0(106) D
dB>C =
160(4.5)(12)
FBC LBC
=
= 0.006171 in T
A Est
0.05 C 28.0(106) D
Thus,
A + T B dA = dH + dA>H = 0.01455 + 0.02469 = 0.03924 in T
A + T B dB = dC + dB>C = 0.01176 + 0.006171 = 0.01793 in T
From the geometry shown in Fig. d,
dP = 0.01793 +
4
(0.03924 - 0.01793) = 0.0350 in T
5
125
Ans.
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*4–8. The load of 800 lb is supported by the four
304 stainless steel wires that are connected to the rigid
members AB and DC. Determine the angle of tilt of each
member after the load is applied.The members were originally
horizontal, and each wire has a cross-sectional area of 0.05 in2.
E
F
4 ft
H
D
C
2 ft
Referring to the FBD of member AB, Fig. a,
5 ft
4.5 ft
a + ©MA = 0;
FBC (5) - 800(1) = 0
FBC = 160 lb
a + ©MB = 0;
800(4) - FAH (5) = 0
FAH = 640 lb
800 lb
A
B
1 ft
Using the results of FBC and FAH and referring to the FBD of member DC, Fig. b,
a + ©MD = 0;
FCF (7) - 160(7) - 640(2) = 0
FCF = 342.86 lb
a + ©MC = 0;
640(5) - FDE (7) = 0
FDE = 457.14 lb
Since E and F are fixed,
dD =
457.14 (4)(12)
FDE LDE
=
= 0.01567 in T
A Est
0.05 C 28.0(106) D
dC =
342.86 (4)(12)
FCF LCF
=
= 0.01176 in T
A Est
0.05 C 28.0(106) D
From the geometry shown in Fig. c
dH = 0.01176 +
u =
5
(0.01567 - 0.01176) = 0.01455 in T
7
0.01567 - 0.01176
= 46.6(10 - 6) rad
7(12)
Ans.
Subsequently,
dA>H =
640 (4.5)(12)
FAH LAH
=
= 0.02469 in T
A Est
0.05 C 28.0(106) D
dB>C =
160 (4.5)(12)
FBC LBC
=
= 0.006171 in T
A Est
0.05 C 28.0(106) D
Thus,
A + T B dA = dH + dA>H = 0.01455 + 0.02469 = 0.03924 in T
A + T B dB = dC + dB>C = 0.01176 + 0.006171 = 0.01793 in T
From the geometry shown in Fig. d
f =
0.03924 - 0.01793
= 0.355(10 - 3) rad
5(12)
Ans.
126
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4–8. Continued
127
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•4–9.
The assembly consists of three titanium (Ti-6A1-4V)
rods and a rigid bar AC. The cross-sectional area of each rod
is given in the figure. If a force of 6 kip is applied to the ring
F, determine the horizontal displacement of point F.
D
4 ft
C
ACD ⫽ 1 in2
2 ft
E
Internal Force in the Rods:
a + ©MA = 0;
FCD(3) - 6(1) = 0
FCD = 2.00 kip
+ ©F = 0;
:
x
6 - 2.00 - FAB = 0
FAB = 4.00 kip
AAB ⫽ 1.5 in2
6 ft
B
1 ft F
6 kip
2
1 ft AEF ⫽ 2 in
A
Displacement:
dC =
2.00(4)(12)
FCD LCD
= 0.0055172 in.
=
ACD E
(1)(17.4)(103)
dA =
4.00(6)(12)
FAB LAB
= 0.0110344 in.
=
AAB E
(1.5)(17.4)(103)
dF>E =
6.00(1)(12)
FEF LEF
= 0.0020690 in.
=
AEF E
(2)(17.4)(103)
œ
dE
0.0055172
=
;
2
3
œ
dE
= 0.0036782 in.
œ
dE = dC + dE
= 0.0055172 + 0.0036782 = 0.0091954 in.
dF = dE + dF>E
= 0.0091954 + 0.0020690 = 0.0113 in.
Ans.
4–10. The assembly consists of three titanium (Ti-6A1-4V)
rods and a rigid bar AC. The cross-sectional area of each rod
is given in the figure. If a force of 6 kip is applied to the ring
F, determine the angle of tilt of bar AC.
D
C
4 ft
ACD ⫽ 1 in
2
2 ft
E
Internal Force in the Rods:
a + ©MA = 0;
FCD(3) - 6(1) = 0
FCD = 2.00 kip
+ ©F = 0;
:
x
6 - 2.00 - FAB = 0
FAB = 4.00 kip
AAB ⫽ 1.5 in2
B
Displacement:
dC =
2.00(4)(12)
FCD LCD
= 0.0055172 in.
=
ACD E
(1)(17.4)(103)
dA =
4.00(6)(12)
FAB LAB
= 0.0110344 in.
=
AAB E
(1.5)(17.4)(103)
u = tan - 1
dA - dC
0.0110344 - 0.0055172
= tan - 1
3(12)
3(12)
= 0.00878°
Ans.
128
6 ft
A
1 ft F
6 kip
2
1 ft AEF ⫽ 2 in
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4–11. The load is supported by the four 304 stainless steel
wires that are connected to the rigid members AB and DC.
Determine the vertical displacement of the 500-lb load if
the members were originally horizontal when the load was
applied. Each wire has a cross-sectional area of 0.025 in2.
E
F
G
3 ft
5 ft
H
D
C
1 ft
2 ft
1.8 ft
I
Internal Forces in the wires:
A
FBD (b)
FBC(4) - 500(3) = 0
+ c ©Fy = 0;
FAH + 375.0 - 500 = 0
FAH = 125.0 lb
a + ©MD = 0;
FCF(3) - 125.0(1) = 0
FCF = 41.67 lb
+ c ©Fy = 0;
FDE + 41.67 - 125.0 = 0
FDE = 83.33 lb
FBC = 375.0 lb
FBD (a)
Displacement:
dD =
83.33(3)(12)
FDELDE
= 0.0042857 in.
=
ADEE
0.025(28.0)(106)
dC =
41.67(3)(12)
FCFLCF
= 0.0021429 in.
=
ACFE
0.025(28.0)(106)
œ
dH
= 0.0014286 in.
dH = 0.0014286 + 0.0021429 = 0.0035714 in.
dA>H =
125.0(1.8)(12)
FAHLAH
= 0.0038571 in.
=
AAHE
0.025(28.0)(106)
dA = dH + dA>H = 0.0035714 + 0.0038571 = 0.0074286 in.
dB =
375.0(5)(12)
FBGLBG
= 0.0321428 in.
=
ABGE
0.025(28.0)(106)
dlœ
0.0247143
=
;
3
4
1 ft
500 lb
a + ©MA = 0;
œ
dH
0.0021429
=
;
2
3
B
3 ft
dlœ = 0.0185357 in.
dl = 0.0074286 + 0.0185357 = 0.0260 in.
Ans.
129
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*4–12. The load is supported by the four 304 stainless
steel wires that are connected to the rigid members AB and
DC. Determine the angle of tilt of each member after the
500-lb load is applied. The members were originally
horizontal, and each wire has a cross-sectional area of
0.025 in2.
E
F
G
3 ft
5 ft
H
D
C
1 ft
2 ft
1.8 ft
I
Internal Forces in the wires:
A
FBD (b)
FBG(4) - 500(3) = 0
+ c ©Fy = 0;
FAH + 375.0 - 500 = 0
FAH = 125.0 lb
a + ©MD = 0;
FCF(3) - 125.0(1) = 0
FCF = 41.67 lb
+ c ©Fy = 0;
FDE + 41.67 - 125.0 = 0
FDE = 83.33 lb
FBG = 375.0 lb
FBD (a)
Displacement:
dD =
83.33(3)(12)
FDELDE
= 0.0042857 in.
=
ADEE
0.025(28.0)(106)
dC =
41.67(3)(12)
FCFLCF
= 0.0021429 in.
=
ACFE
0.025(28.0)(106)
œ
dH
0.0021429
=
;
2
3
œ
dH
= 0.0014286 in.
œ
+ dC = 0.0014286 + 0.0021429 = 0.0035714 in.
dH = dH
tan a =
0.0021429
;
36
dA>H =
125.0(1.8)(12)
FAHLAH
= 0.0038571 in.
=
AAHE
0.025(28.0)(106)
a = 0.00341°
Ans.
dA = dH + dA>H = 0.0035714 + 0.0038571 = 0.0074286 in.
375.0(5)(12)
FBGLBG
= 0.0321428 in.
=
ABGE
0.025(28.0)(106)
tan b =
1 ft
500 lb
a + ©MA = 0;
dB =
B
3 ft
0.0247143
;
48
b = 0.0295°
Ans.
130
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•4–13. The bar has a length L and cross-sectional area A.
Determine its elongation due to the force P and its own
weight.The material has a specific weight g (weight>volume)
and a modulus of elasticity E.
d =
=
L
P(x) dx
1
=
(gAx + P) dx
AE L0
L A(x) E
L
gAL2
gL2
1
PL
a
+ PLb =
+
AE
2
2E
AE
Ans.
P
4–14. The post is made of Douglas fir and has a diameter
of 60 mm. If it is subjected to the load of 20 kN and the soil
provides a frictional resistance that is uniformly distributed
along its sides of w = 4 kN>m, determine the force F at its
bottom needed for equilibrium.Also, what is the displacement
of the top of the post A with respect to its bottom B?
Neglect the weight of the post.
20 kN
A
y
w
2m
Equation of Equilibrium: For entire post [FBD (a)]
+ c ©Fy = 0;
F + 8.00 - 20 = 0
B
F = 12.0 kN
Ans.
Internal Force: FBD (b)
+ c ©Fy = 0;
-F(y) + 4y - 20 = 0
F(y) = {4y - 20} kN
Displacement:
L
dA>B =
2m
F(y)dy
1
=
(4y - 20)dy
AE L0
L0 A(y)E
=
2m
1
A 2y2 - 20y B 冷0
AE
= -
32.0 kN # m
AE
32.0(103)
= -p
2
9
4 (0.06 ) 13.1 (10 )
= - 0.8639 A 10 - 3 B m
Ans.
= - 0.864 mm
Negative sign indicates that end A moves toward end B.
131
F
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4–15. The post is made of Douglas fir and has a diameter
of 60 mm. If it is subjected to the load of 20 kN and the soil
provides a frictional resistance that is distributed along its
length and varies linearly from w = 0 at y = 0 to w =
3 kN>m at y = 2 m, determine the force F at its bottom
needed for equilibrium. Also, what is the displacement of
the top of the post A with respect to its bottom B? Neglect
the weight of the post.
20 kN
A
y
w
2m
B
F
Equation of Equilibrium: For entire post [FBD (a)]
+ c ©Fy = 0;
F + 3.00 - 20 = 0
F = 17.0 kN
Ans.
Internal Force: FBD (b)
+ c ©Fy = 0;
-F(y) +
1 3y
a b y - 20 = 0
2 2
3
F(y) = e y2 - 20 f kN
4
Displacement:
L
dA>B =
2m
F(y) dy
1
3
=
a y2 - 20b dy
AE L0
4
L0 A(y)E
=
2m
y3
1
a
- 20y b 2
AE 4
0
= -
38.0 kN # m
AE
38.0(103)
= -p
2
9
4 (0.06 ) 13.1 (10 )
= -1.026 A 10 - 3 B m
Ans.
= -1.03 mm
Negative sign indicates that end A moves toward end B.
132
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*4–16. The linkage is made of two pin-connected A-36
steel members, each having a cross-sectional area of 1.5 in2.
If a vertical force of P = 50 kip is applied to point A,
determine its vertical displacement at A.
P
A
2 ft
B
C
1.5 ft
Analysing the equilibrium of Joint A by referring to its FBD, Fig. a,
+ ©F = 0 ;
:
x
3
3
FAC a b - FAB a b = 0
5
5
+ c ©Fy = 0
4
-2Fa b - 50 = 0
5
The
initial
FAC = FAB = F
F = -31.25 kip
length
of
members
AB
and
AC
is
12
in
b = 30 in. The axial deformation of members
L = 21.52 + 22 = (2.50 ft)a
1 ft
AB and AC is
d =
(-31.25)(30)
FL
=
= -0.02155 in.
AE
(1.5) C 29.0(103) D
The negative sign indicates that end A moves toward B and C. From the geometry
1.5
shown in Fig. b, u = tan - 1 a
b = 36.87°. Thus,
2
A dA B g =
d
0.02155
=
= 0.0269 in. T
cos u
cos 36.87°
Ans.
133
1.5 ft
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•4–17.
The linkage is made of two pin-connected A-36
steel members, each having a cross-sectional area of 1.5 in2.
Determine the magnitude of the force P needed to displace
point A 0.025 in. downward.
P
A
2 ft
Analysing the equilibrium of joint A by referring to its FBD, Fig. a
+ ©F = 0;
:
x
3
3
FAC a b - FAB a b = 0
5
5
+ c ©Fy = 0;
4
-2Fa b - P = 0
5
The
initial
B
1.5 ft
F = -0.625 P
of
members
AB
and
AC
are
12 in
b = 30 in. The axial deformation of members
L = 21.5 + 2 = (2.50 ft)a
1 ft
AB and AC is
2
length
2
d =
C
FAC = FAB = F
-0.625P(30)
FL
=
= -0.4310(10 - 3) P
AE
(1.5) C 29.0(103) D
The negative sign indicates that end A moves toward B and C. From the geometry
1.5
shown in Fig. b, we obtain u = tan - 1 a
b = 36.87°. Thus
2
(dA)g =
d
cos u
0.025 =
0.4310(10 - 3) P
cos 36.87°
P = 46.4 kips
Ans.
134
1.5 ft
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4–18. The assembly consists of two A-36 steel rods and a
rigid bar BD. Each rod has a diameter of 0.75 in. If a force
of 10 kip is applied to the bar as shown, determine the
vertical displacement of the load.
C
A
3 ft
2 ft
B
Here, FEF = 10 kip. Referring to the FBD shown in Fig. a,
1.25 ft
a + ©MB = 0;
FCD (2) - 10(1.25) = 0
FCD = 6.25 kip
a + ©MD = 0;
10(0.75) - FAB(2) = 0
FAB = 3.75 kip
The cross-sectional area of the rods is A =
A and C are fixed,
3.75 (2)(12)
FAB LAB
=
= 0.007025 in. T
A Est
0.140625p C 29.0(103) D
dD =
6.25(3)(12)
FCD LCD
=
= 0.01756 in T
A Est
0.140625p C 29.0(103) D
From the geometry shown in Fig. b
dE = 0.007025 +
1.25
(0.01756 - 0.00725) = 0.01361 in. T
2
Here,
dF>E =
10 (1) (12)
FEF LEF
=
= 0.009366 in T
A Est
0.140625p C 29.0(103) D
Thus,
A + T B dF = dE + dF>E = 0.01361 + 0.009366 = 0.0230 in T
Ans.
135
D
0.75 ft
F
p
(0.752) = 0.140625p in2. Since points
4
dB =
E
10 kip
1 ft
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4–19. The assembly consists of two A-36 steel rods and a
rigid bar BD. Each rod has a diameter of 0.75 in. If a force
of 10 kip is applied to the bar, determine the angle of tilt of
the bar.
C
A
Here, FEF = 10 kip. Referring to the FBD shown in Fig. a,
3 ft
a + ©MB = 0;
FCD (2) - 10(1.25) = 0
FCD = 6.25 kip
a + ©MD = 0;
10(0.75) - FAB (2) = 0
FAB = 3.75 kip
2 ft
B
p
The cross-sectional area of the rods is A = (0.752) = 0.140625p in2. Since points
4
A and C are fixed then,
dB =
3.75 (2)(12)
FAB LAB
=
= 0.007025 in T
A Est
0.140625p C 29.0(103) D
dD =
6.25 (3)(12)
FCD LCD
=
= 0.01756 in T
A Est
0.140625p C 29.0(103) D
0.01756 - 0.007025
= 0.439(10 - 3) rad
2(12)
Ans.
136
D
0.75 ft
F
10 kip
From the geometry shown in Fig. b,
u =
1.25 ft
E
1 ft
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*4–20. The rigid bar is supported by the pin-connected
rod CB that has a cross-sectional area of 500 mm2 and is
made of A-36 steel. Determine the vertical displacement of
the bar at B when the load is applied.
C
3m
45 kN/m
Force In The Rod. Referring to the FBD of member AB, Fig. a
a + ©MA = 0;
3
1
1
FBC a b (4) - (45)(4) c (4) d = 0
5
2
3
4m
Displacement. The initial length of rod BC is LBC = 232 + 42 = 5 m. The axial
deformation of this rod is
dBC =
50.0(103)(5)
FBC LBC
=
= 2.50 (10 - 3) m
ABC Est
0.5(10 - 3) C 200(109) D
3
From the geometry shown in Fig. b, u = tan - 1 a b = 36.87°. Thus,
4
(dB)g =
2.50(10 - 3)
dBC
=
= 4.167 (10 - 3) m = 4.17 mm
sin u
sin 36.87°
137
B
A
FBC = 50.0 kN
Ans.
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•4–21.
A spring-supported pipe hanger consists of two
springs which are originally unstretched and have a stiffness
of k = 60 kN>m, three 304 stainless steel rods, AB and CD,
which have a diameter of 5 mm, and EF, which has a
diameter of 12 mm, and a rigid beam GH. If the pipe and
the fluid it carries have a total weight of 4 kN, determine the
displacement of the pipe when it is attached to the support.
F
B
D
k
G
0.75 m
0.75 m
k
H
E
A
C
Internal Force in the Rods:
0.25 m 0.25 m
FBD (a)
a + ©MA = 0;
FCD (0.5) - 4(0.25) = 0
+ c ©Fy = 0;
FAB + 2.00 - 4 = 0
FCD = 2.00 kN
FAB = 2.00 kN
FBD (b)
FEF - 2.00 - 2.00 = 0
+ c ©Fy = 0;
FEF = 4.00 kN
Displacement:
dD = dE =
4.00(103)(750)
FEFLEF
= 0.1374 mm
= p
2
9
AEFE
4 (0.012) (193)(10 )
dA>B = dC>D =
2(103)(750)
PCDLCD
= 0.3958 mm
= p
2
9
ACDE
4 (0.005) (193)(10 )
dC = dD + dC>D = 0.1374 + 0.3958 = 0.5332 mm
Displacement of the spring
dsp =
Fsp
k
=
2.00
= 0.0333333 m = 33.3333 mm
60
dlat = dC + dsp
= 0.5332 + 33.3333 = 33.9 mm
Ans.
138
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4–22. A spring-supported pipe hanger consists of two
springs, which are originally unstretched and have a
stiffness of k = 60 kN>m, three 304 stainless steel rods, AB
and CD, which have a diameter of 5 mm, and EF, which has
a diameter of 12 mm, and a rigid beam GH. If the pipe is
displaced 82 mm when it is filled with fluid, determine the
weight of the fluid.
F
B
D
k
G
0.75 m
0.75 m
k
H
E
Internal Force in the Rods:
A
C
FBD (a)
a + ©MA = 0;
FCD(0.5) - W(0.25) = 0
FCD =
+ c ©Fy = 0;
FAB +
W
- W = 0
2
W
2
FAB =
0.25 m 0.25 m
W
2
FBD (b)
FEF -
+ c ©Fy = 0;
W
W
= 0
2
2
FEF = W
Displacement:
dD = dE =
W(750)
FEFLEF
= p
2
9
AEFE
4 (0.012) (193)(10 )
= 34.35988(10 - 6) W
dA>B = dC>D =
W
FCDLCD
2 (750)
= p
2
9
ACDE
4 (0.005) (193)(10 )
= 98.95644(10 - 6) W
dC = dD + dC>D
= 34.35988(10 - 6) W + 98.95644(10 - 6) W
= 0.133316(10 - 3) W
Displacement of the spring
dsp =
W
2
Fsp
k
=
60(103)
(1000) = 0.008333 W
dlat = dC + dsp
82 = 0.133316(10 - 3) W + 0.008333W
W = 9685 N = 9.69 kN
Ans.
139
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4–23. The rod has a slight taper and length L. It is
suspended from the ceiling and supports a load P at its end.
Show that the displacement of its end due to this load is
d = PL>1pEr2r12. Neglect the weight of the material. The
modulus of elasticity is E.
r(x) = r1 +
A(x) =
r2
r1L + (r2 - r1)x
r2 - r1
x =
L
L
L
p
(r1L + (r2 - r1)x)2
L2
r1
L
PL2
Pdx
dx
d =
=
2
A(x)E
pE
[r
L
+
(r
L0
L
1
2 - r1)x]
= -
L
1
PL2
c
dƒ
p E (r2 - r2)(r1L + (r2 - r1)x) 0
= -
=
= -
P
1
1
PL2
c
d
p E(r2 - r1) r1L + (r2 - r1)L
r1L
r1 - r2
PL2
1
1
PL2
c
d = c
d
p E(r2 - r1) r2L
r1L
p E(r2 - r1) r2r1L
r2 - r1
PL2
PL
c
d =
p E(r2 - r1) r2r1L
p E r2r1
QED
*4–24. Determine the relative displacement of one end of
the tapered plate with respect to the other end when it is
subjected to an axial load P.
P
d2
t
w = d1 +
d1 h + (d2 - d1)x
d2 - d1
x =
h
h
h
h
P(x) dx
P
dx
=
d =
[d1h + ( d 2 - d1 )x ] t
A(x)E
E
L0
L
h
h
=
Ph
dx
E t L0 d1 h + (d2 - d1)x
d1
P
h
dx
Ph
=
E t d1 h L0 1 + d2 - d1
d1 h
h
d1 h
d2 - d1
Ph
=
a
b cln a1 +
xb d ƒ
E t d1 h d2 - d1
d1 h
0
x
d2 - d1
d1 + d2 - d1
Ph
Ph
cln a1 +
bd =
cln a
bd
=
E t(d2 - d1)
d1
E t(d2 - d1)
d1
=
d2
Ph
cln d
E t(d2 - d1)
d1
Ans.
140
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4–25. Determine the elongation of the A-36 steel member
when it is subjected to an axial force of 30 kN. The member
is 10 mm thick. Use the result of Prob. 4–24.
20 mm
30 kN
30 kN
75 mm
0.5 m
Using the result of prob. 4-24 by substituting d1 = 0.02 m, d2 = 0.075 m t = 0.01 m
and L = 0.5 m.
d = 2c
= 2c
d2
PL
ln d
Est t(d2 - d1) d1
30(103) (0.5)
9
200(10 )(0.01)(0.075 - 0.02)
ln a
0.075
bd
0.02
= 0.360(10 - 3) m = 0.360 mm
Ans.
4–26. The casting is made of a material that has a specific
weight g and modulus of elasticity E. If it is formed into a
pyramid having the dimensions shown, determine how far
its end is displaced due to gravity when it is suspended in
the vertical position.
b0
b0
L
Internal Forces:
+ c ©Fz = 0;
P(z) -
1
gAz = 0
3
P(z) =
1
gAz
3
Displacement:
L
d =
P(z) dz
L0 A(z) E
L1
3 gAz
=
L0 A E
dz
=
L
g
z dz
3E L0
=
gL2
6E
Ans.
141
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4–27. The circular bar has a variable radius of r = r0eax
and is made of a material having a modulus of elasticity
of E. Determine the displacement of end A when it is
subjected to the axial force P.
L
Displacements: The cross-sectional area of the bar as a function of x is
A(x) = pr2 = pr0 2e2ax. We have
x
B
L
d =
L
P(x)dx
P
dx
=
2
A(x)E
pr0 E L0 e2ax
L0
r0
r ⫽ r0 eax
L
1
P
2
c
d
=
pr0 2E 2ae2ax 0
= -
A
P
a1 - e - 2aL b
2apr0 2E
P
Ans.
*4–28. The pedestal is made in a shape that has a radius
defined by the function r = 2>12 + y1>22 ft, where y is
in feet. If the modulus of elasticity for the material is
E = 1411032 psi, determine the displacement of its top
when it supports the 500-lb load.
y
500 lb
0.5 ft
r⫽
2
(2 ⫹ y 1/ 2)
4 ft
d =
=
P(y) dy
L A(y) E
y
4
dy
500
3
14(10 )(144) L0 p(2 + y2-1
2
)
2
1 ft
4
-3
= 0.01974(10 )
L0
r
1
2
(4 + 4y + y) dy
4
2 3
1
= 0.01974(10 - 3)c4y + 4 a y2 b + y2 d
3
2
0
= 0.01974(10 - 3)(45.33)
= 0.8947(10 - 3) ft = 0.0107 in.
Ans.
142
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•4–29.
The support is made by cutting off the two
opposite sides of a sphere that has a radius r0 . If the original
height of the support is r0>2, determine how far it shortens
when it supports a load P. The modulus of elasticity is E.
P
r0
Geometry:
A = p r2 = p(r0 cos u)2 = p r20 cos2 u
y = r0 sin u;
dy = r0 cos u du
Displacement:
L
P(y) dy
L0 A(y) E
d =
= 2B
When y =
u
u
r0 cos u du
P
P
du
=
2
R
B
R
E L0 p r20 cos2 u
p r0 E L0 cos u
=
u
2P
[ln (sec u + tan u)] 2
p r0 E
0
=
2P
[ln (sec u + tan u)]
p r0 E
r0
;
4
u = 14.48°
d =
=
2P
[ln (sec 14.48° + tan 14.48°)]
p r0 E
0.511P
p r0 E
Ans.
Also,
Geometry:
A (y) = p x2 = p (r20 - y2)
Displacement:
L
d =
P(y) dy
L0 A(y) E
0
0
r0 + y p
2P 1
2P p dy
=
ln
=
B
R 2
2
2
pE L0 r0 - y
p E 2r0 r0 - y 0
=
P
[ln 1.667 - ln 1]
p r0 E
=
0.511 P
p r0 E
Ans.
143
r0
2
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4–30. The weight of the kentledge exerts an axial force of
P ⫽ 1500 kN on the 300-mm diameter high strength
concrete bore pile. If the distribution of the resisting skin
friction developed from the interaction between the soil
and the surface of the pile is approximated as shown, and
the resisting bearing force F is required to be zero,
determine the maximum intensity p0 kN>m for equilibrium.
Also, find the corresponding elastic shortening of the pile.
Neglect the weight of the pile.
P
p0
12 m
Internal Loading: By considering the equilibrium of the pile with reference to its
entire free-body diagram shown in Fig. a. We have
1
p (12) - 1500 = 0
2 0
+ c ©Fy = 0;
p0 = 250 kN>m
Ans.
Thus,
p(y) =
250
y = 20.83y kN>m
12
The normal force developed in the pile as a function of y can be determined by
considering the equilibrium of a section of the pile shown in Fig. b.
1
(20.83y)y - P(y) = 0
2
+ c ©Fy = 0;
P(y) = 10.42y2 kN
Displacement: The cross-sectional area of the pile is A =
p
(0.32) = 0.0225p m2.
4
We have
L
d =
12 m
P(y)dy
10.42(103)y2dy
=
0.0225p(29.0)(109)
L0
L0 A(y)E
12 m
=
L0
5.0816(10 - 6)y2dy
= 1.6939(10 - 6)y3 冷 0
12 m
= 2.9270(10 - 3)m = 2.93 mm
Ans.
144
F
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4–31. The column is constructed from high-strength
concrete and six A-36 steel reinforcing rods. If it is subjected
to an axial force of 30 kip, determine the average normal
stress in the concrete and in each rod. Each rod has a
diameter of 0.75 in.
4 in.
30 kip
Equations of Equilibrium:
6Pst + Pcon - 30 = 0
+ c ©Fy = 0;
3 ft
[1]
Compatibility:
dst = dcon
Pcon(3)(12)
Pst(3)(12)
p
2
(0.75
)(29.0)(103)
4
=
[p4 (82) - 6(p4 )(0.75)2](4.20)(103)
Pst = 0.064065 Pcon
[2]
Solving Eqs. [1] and [2] yields:
Pst = 1.388 kip
Pcon = 21.670 kip
Average Normal Stress:
Pst
1.388
= 3.14 ksi
= p
2
Ast
(0.75
)
4
Ans.
Pcon
21.670
=
= 0.455 ksi
p 2
Acon
(8 ) - 6 A p B (0.752)
Ans.
sst =
scon =
4
4
*4–32. The column is constructed from high-strength
concrete and six A-36 steel reinforcing rods. If it is subjected
to an axial force of 30 kip, determine the required diameter
of each rod so that one-fourth of the load is carried by the
concrete and three-fourths by the steel.
4 in.
30 kip
Equilibrium: The force of 30 kip is required to distribute in such a manner that 3/4
of the force is carried by steel and 1/4 of the force is carried by concrete. Hence
Pst =
3
(30) = 22.5 kip
4
Pcon =
1
(30) = 7.50 kip
4
3 ft
Compatibility:
dst = dcon
PstL
Pcon L
=
AstEst
Acon Econ
Ast =
22.5Acon Econ
7.50 Est
3 C 4 (8 ) - 6 A 4 B d D (4.20)(10 )
p
6 a bd2 =
4
29.0(103)
p
2
p
2
3
d = 1.80 in.
Ans.
145
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•4–33.
The steel pipe is filled with concrete and subjected
to a compressive force of 80 kN. Determine the average
normal stress in the concrete and the steel due to this
loading. The pipe has an outer diameter of 80 mm and an
inner diameter of 70 mm. Est = 200 GPa, Ec = 24 GPa.
80 kN
Pst + Pcon - 80 = 0
+ c ©Fy = 0;
(1)
500 mm
dst = dcon
Pcon L
Pst L
p
2
2
(0.08
0.07
) (200) (109)
4
= p
2
9
4 (0.07 ) (24) (10 )
Pst = 2.5510 Pcon
(2)
Solving Eqs. (1) and (2) yields
Pst = 57.47 kN
sst =
Pcon = 22.53 kN
57.47 (103)
Pst
= 48.8 MPa
= p
2
2
Ast
4 (0.08 - 0.07 )
Ans.
22.53 (103)
Pcon
= 5.85 MPa
= p
2
Acon
4 (0.07 )
Ans.
scon =
4–34. The 304 stainless steel post A has a diameter of
d = 2 in. and is surrounded by a red brass C83400 tube B.
Both rest on the rigid surface. If a force of 5 kip is applied
to the rigid cap, determine the average normal stress
developed in the post and the tube.
5 kip
B
B
A
8 in.
Equations of Equilibrium:
+ c ©Fy = 0;
3 in.
Pst + Pbr - 5 = 0[1]
Compatibility:
d
dst = dbr
Pst(8)
p 2
3
4 (2 )(28.0)(10 )
Pbr(8)
= p
2
4 (6
- 52)(14.6)(103)
Pst = 0.69738 Pbr
[2]
Solving Eqs. [1] and [2] yields:
Pbr = 2.9457 kip
Pst = 2.0543 kip
Average Normal Stress:
sbr =
Pbr
2.9457
= 0.341 ksi
= p 2
Abr
(6
- 52)
4
Ans.
sst =
Pst
2.0543
= p 2 = 0.654 ksi
Ast
4 (2 )
Ans.
146
0.5 in.
A
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4–35. The 304 stainless steel post A is surrounded by a red
brass C83400 tube B. Both rest on the rigid surface. If a
force of 5 kip is applied to the rigid cap, determine the
required diameter d of the steel post so that the load is
shared equally between the post and tube.
5 kip
B
B
A
A
8 in.
Equilibrium: The force of 5 kip is shared equally by the brass and steel. Hence
3 in.
Pst = Pbr = P = 2.50 kip
Compatibility:
d
0.5 in.
dst = dbr
PL
PL
=
AstEst
AbrEbr
Ast =
AbrEbr
Est
p
2
2
3
p
4 (6 - 5 )(14.6)(10 )
a b d2 =
4
28.0(103)
d = 2.39 in.
Ans.
*4–36. The composite bar consists of a 20-mm-diameter
A-36 steel segment AB and 50-mm-diameter red brass
C83400 end segments DA and CB. Determine the average
normal stress in each segment due to the applied load.
+ ©F = 0;
;
x
(1)
0 = ¢ D - dD
50(0.25)
150(0.5)
0 = p
2
9
4 (0.02) (200)(10 )
- p
2
9
4 (0.05 )(101)(10 )
FD(0.5)
FD(0.5)
- p
2
9
4 (0.05 )(101)(10 )
500 mm
50 mm
20 mm
75 kN 100 kN
D
FC - FD + 75 + 75 - 100 - 100 = 0
FC - FD - 50 = 0
+
;
250 mm
- p
2
9
4 (0.02 )(200)(10 )
FD = 107.89 kN
From Eq. (1), FC = 157.89 kN
sAD =
107.89(103)
PAD
= 55.0 MPa
= p
2
AAD
4 (0.05 )
Ans.
sAB =
42.11(103)
PAB
= 134 MPa
= p
2
AAB
4 (0.02 )
Ans.
sBC =
157.89(103)
PBC
= 80.4 MPa
= p
2
ABC
4 (0.05 )
Ans.
147
A
75 kN
100 kN B
250 mm
C
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•4–37.
The composite bar consists of a 20-mm-diameter
A-36 steel segment AB and 50-mm-diameter red brass
C83400 end segments DA and CB. Determine
the displacement of A with respect to B due to the
applied load.
250 mm
500 mm
50 mm
20 mm
75 kN 100 kN
D
+
;
A
75 kN
250 mm
100 kN B
C
0 = ¢ D - dD
150(103)(500)
0 = p
50(103)(250)
2
9
4 (0.02 )(200)(10 )
- p
2
9
4 (0.05 )(101)(10 )
FD(500)
FD(500)
- p
2
9
4 (0.05 )(101)(10 )
- p
2
9
4 (0.02) (200)(10 )
FD = 107.89 kN
Displacement:
dA>B =
42.11(103)(500)
PABLAB
= p
2
9
AABEst
4 (0.02 )200(10 )
= 0.335 mm
Ans.
4–38. The A-36 steel column, having a cross-sectional area
of 18 in2, is encased in high-strength concrete as shown. If
an axial force of 60 kip is applied to the column, determine
the average compressive stress in the concrete and in the
steel. How far does the column shorten? It has an original
length of 8 ft.
16 in.
Pst + Pcon - 60 = 0
+ c ©Fy = 0;
dst = dcon ;
60 kip
Pst(8)(12)
18(29)(103)
(1)
Pcon(8)(12)
=
[(9)(16) - 18](4.20)(103)
Pst = 0.98639 Pcon
(2)
Solving Eqs. (1) and (2) yields
Pst = 29.795 kip;
sst =
Pcon = 30.205 kip
Pst
29.795
=
= 1.66 ksi
Ast
18
Ans.
Pcon
30.205
=
= 0.240 ksi
Acon
9(16) - 18
Ans.
scon =
Either the concrete or steel can be used for the deflection calculation.
d =
29.795(8)(12)
PstL
= 0.0055 in.
=
AstE
18(29)(103)
Ans.
148
9 in.
8 ft
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4–39. The A-36 steel column is encased in high-strength
concrete as shown. If an axial force of 60 kip is applied to
the column, determine the required area of the steel so that
the force is shared equally between the steel and concrete.
How far does the column shorten? It has an original length
of 8 ft.
60 kip
16 in.
9 in.
8 ft
The force of 60 kip is shared equally by the concrete and steel. Hence
Pst = Pcon = P = 30 kip
dcon = dst;
Ast =
PL
PL
=
Acon Econ
Ast Est
[9(16) - Ast] 4.20(103)
AconEcon
=
Est
29(103)
= 18.2 in2
d =
Ans.
30(8)(12)
PstL
= 0.00545 in.
=
AstEst
18.2(29)(103)
Ans.
*4–40. The rigid member is held in the position shown by
three A-36 steel tie rods. Each rod has an unstretched length
of 0.75 m and a cross-sectional area of 125 mm2. Determine
the forces in the rods if a turnbuckle on rod EF undergoes
one full turn. The lead of the screw is 1.5 mm. Neglect the
size of the turnbuckle and assume that it is rigid. Note: The
lead would cause the rod, when unloaded, to shorten 1.5 mm
when the turnbuckle is rotated one revolution.
B
D
0.75 m
E
A
0.5 m
0.5 m
C
0.75 m
a + ©ME = 0;
-TAB(0.5) + TCD(0.5) = 0
F
TAB = TCD = T
+ T ©Fy = 0;
(1)
TEF - 2T = 0
TEF = 2T
(2)
Rod EF shortens 1.5mm causing AB (and DC) to elongate. Thus:
0.0015 = dA>B + dE>F
0.0015 =
T(0.75)
-6
2T(0.75)
9
(125)(10 )(200)(10 )
+
(125)(10 - 6)(200)(109)
2.25T = 37500
T = 16666.67 N
TAB = TCD = 16.7 kN
Ans.
TEF = 33.3 kN
Ans.
149
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•4–41.
The concrete post is reinforced using six steel
reinforcing rods, each having a diameter of 20 mm.
Determine the stress in the concrete and the steel if the post
is subjected to an axial load of 900 kN. Est = 200 GPa,
Ec = 25 GPa.
900 kN
250 mm
375 mm
Referring to the FBD of the upper portion of the cut concrete post shown in Fig. a
+ c ©Fy = 0;
Pcon + 6Pst - 900 = 0
(1)
Since the steel rods and the concrete are firmly bonded, their deformation must be
the same. Thus
0 con = dst
Pcon L
Pst L
=
Acon Econ
Ast Est
C
Pcon L
0.25(0.375) - 6(p4 )(0.022)
D C 25(10 ) D
Pst L
=
9
C
(p4 )(0.022) 200(109)
D
Pcon = 36.552 Pst
(2)
Solving Eqs (1) and (2) yields
Pst = 21.15 kN
Pcon = 773.10 kN
Thus,
scon =
773.10(103)
Pcon
= 8.42 MPa
=
Acon
0.15(0.375) - 6(p4 )(0.022)
Ans.
sst =
21.15(103)
Pst
= 67.3 MPa
= p
2
Ast
4 (0.02 )
Ans.
150
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4–42. The post is constructed from concrete and six A-36
steel reinforcing rods. If it is subjected to an axial force of
900 kN, determine the required diameter of each rod so that
one-fifth of the load is carried by the steel and four-fifths by
the concrete. Est = 200 GPa, Ec = 25 GPa.
900 kN
250 mm
375 mm
The normal force in each steel rod is
Pst =
1
5 (900)
6
= 30 kN
The normal force in concrete is
Pcon =
4
(900) = 720 kN
5
Since the steel rods and the concrete are firmly bonded, their deformation must be
the same. Thus
dcon = dst
Pcon L
Pst L
=
Acon Econ
Ast Est
720(103) L
30(103)L
C 0.25(0.375) - 6(p4 d2) D C 25(109) D
=
p 2
4 d
49.5p d2 = 0.09375
C 200(109) D
d = 0.02455 m = 24.6 mm
Ans.
4–43. The assembly consists of two red brass C83400
copper alloy rods AB and CD of diameter 30 mm, a stainless
304 steel alloy rod EF of diameter 40 mm, and a rigid cap G.
If the supports at A, C and F are rigid, determine the
average normal stress developed in rods AB, CD and EF.
300 mm
450 mm
40 kN
A
B
E
30 mm
F
40 mm
C
Equation of Equilibrium: Due to symmetry, FAB = FCD = F. Referring to the freebody diagram of the assembly shown in Fig. a,
2F + FEF - 2 C 40(103) D = 0
+ ©F = 0;
:
x
(1)
Compatibility Equation: Using the method of superposition, Fig. b,
+ B 0 = -d + d
A:
P
EF
0 = -p
40(103)(300)
2
9
4 (0.03 )(101)(10 )
+ cp
FEF (450)
2
9
4 (0.04 )(193)(10 )
+ p
A FEF>2 B (300)
2
9
4 (0.03 )(101)(10 )
FEF = 42 483.23 N
Substituting this result into Eq. (1),
F = 18 758.38 N
151
d
30 mm
40 kN
D
G
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4–43.
Continued
Normal Stress: We have,
sAB = sCD =
sEF =
F
18 758.38
= 26.5 MPa
= p
2
ACD
4 (0.03 )
Ans.
FEF
42 483.23
= 33.8 MPa
= p
2
AEF
4 (0.04 )
Ans.
152
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*4–44. The two pipes are made of the same material and
are connected as shown. If the cross-sectional area of BC is
A and that of CD is 2A, determine the reactions at B and D
when a force P is applied at the junction C.
B
L
–
2
Equations of Equilibrium:
+ ©F = 0;
;
x
FB + FD - P = 0
[1]
Compatibility:
+ B
A:
0 = dP - dB
0 =
0 =
P A L2 B
2AE
- C
FB A L2 B
AE
+
FB A L2 B
2AE
S
3FBL
PL
4AE
4AE
FB =
P
3
Ans.
From Eq. [1]
FD =
C
2
P
3
Ans.
153
D
P
L
–
2
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•4–45.
The bolt has a diameter of 20 mm and passes
through a tube that has an inner diameter of 50 mm and an
outer diameter of 60 mm. If the bolt and tube are made of
A-36 steel, determine the normal stress in the tube and bolt
when a force of 40 kN is applied to the bolt. Assume the end
caps are rigid.
160 mm
40 kN
Referring to the FBD of left portion of the cut assembly, Fig. a
+ ©F = 0;
:
x
40(103) - Fb - Ft = 0
(1)
Here, it is required that the bolt and the tube have the same deformation. Thus
dt = db
Ft(150)
C
p
2
2
9
4 (0.06 - 0.05 ) 200(10 )
D
Fb(160)
=
C
p
2
9
4 (0.02 ) 200(10 )
D
Ft = 2.9333 Fb
(2)
Solving Eqs (1) and (2) yields
Fb = 10.17 (103) N
40 kN
150 mm
Ft = 29.83 (103) N
Thus,
sb =
10.17(103)
Fb
= 32.4 MPa
= p
2
Ab
4 (0.02 )
Ans.
st =
29.83 (103)
Ft
= 34.5 MPa
= p
2
2
At
4 (0.06 - 0.05 )
Ans.
154
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4–46. If the gap between C and the rigid wall at D is
initially 0.15 mm, determine the support reactions at A and
D when the force P = 200 kN is applied. The assembly
is made of A36 steel.
50 mm
Equation of Equilibrium: Referring to the free-body diagram of the assembly
shown in Fig. a,
200(103) - FD - FA = 0
(1)
Compatibility Equation: Using the method of superposition, Fig. b,
+ B
A:
d = dP - dFD
0.15 = p
200(103)(600)
2
9
4 (0.05 )(200)(10 )
- Cp
FD (600)
2
9
4 (0.05 )(200)(10 )
0.15 mm
P
A
+ ©F = 0;
:
x
600 mm
600 mm
FD (600)
+ p
2
9
4 (0.025 )(200)(10 )
FD = 20 365.05 N = 20.4 kN
Ans.
Substituting this result into Eq. (1),
FA = 179 634.95 N = 180 kN
Ans.
155
S
D
B
25 mm
C
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4–47. Two A-36 steel wires are used to support the 650-lb
engine. Originally, AB is 32 in. long and A¿B¿ is 32.008 in.
long. Determine the force supported by each wire when the
engine is suspended from them. Each wire has a crosssectional area of 0.01 in2.
B¿ B
A¿ A
TA¿B¿ + TAB - 650 = 0
+ c ©Fy = 0;
(1)
dAB = dA¿B¿ + 0.008
TA¿B¿ (32.008)
TAB (32)
(0.01)(29)(106)
=
(0.01)(29)(106)
+ 0.008
32TAB - 32.008TA¿B¿ = 2320
TAB = 361 lb
Ans.
TA¿B¿ = 289 lb
Ans.
*4–48. Rod AB has a diameter d and fits snugly between
the rigid supports at A and B when it is unloaded. The
modulus of elasticity is E. Determine the support reactions
at A and B if the rod is subjected to the linearly distributed
axial load.
p⫽
A
1
p L - FA - FB = 0
2 0
+ ©F = 0;
:
x
(1)
Compatibility Equation: Using the method of superposition, Fig. b,
+ B
A:
0 = dP - dFA
L
0 =
L0
FA (L)
P(x)dx
AE
AE
L
0 =
L0
B
x
Equation of Equilibrium: Referring to the free-body diagram of rod AB shown in
Fig. a,
P(x)dx - FAL
156
p0
p0
x
L
L
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4–48.
Continued
Here, P(x) =
p0 2
1 p0
a xb x =
x . Thus,
2 L
2L
0 =
L
p0
x2 dx - FAL
2L L0
0 =
p0 x3 L
¢ ≤ ` - FAL
2L 3 0
FA =
p0L
6
Ans.
Substituting this result into Eq. (1),
FB =
p0L
3
Ans.
157
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•4–49.
The tapered member is fixed connected at its ends
A and B and is subjected to a load P = 7 kip at x = 30 in.
Determine the reactions at the supports. The material is
2 in. thick and is made from 2014-T6 aluminum.
A
B
3 in.
P
6 in.
x
60 in.
y
1.5
=
120 - x
60
y = 3 - 0.025 x
+ ©F = 0;
:
x
FA + FB - 7 = 0
(1)
dA>B = 0
30
-
L0
60
FA dx
FBdx
+
= 0
2(3 - 0.025 x)(2)(E)
L30 2(3 - 0.025 x)(2)(E)
30
-FA
L0
60
dx
dx
+ FB
= 0
(3 - 0.025 x)
L30 (3 - 0.025x)
60
40 FA ln(3 - 0.025 x)|30
0 - 40 FB ln(3 - 0.025x)|30 = 0
-FA(0.2876) + 0.40547 FB = 0
FA = 1.40942 FB
Thus, from Eq. (1).
FA = 4.09 kip
Ans.
FB = 2.91 kip
Ans.
158
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4–50. The tapered member is fixed connected at its ends
A and B and is subjected to a load P. Determine the location
x of the load and its greatest magnitude so that the average
normal stress in the bar does not exceed sallow = 4 ksi. The
member is 2 in. thick.
A
B
3 in.
P
6 in.
x
60 in.
y
1.5
=
120 - x
60
y = 3 - 0.025 x
+ ©F = 0;
:
x
FA + FB - P = 0
dA>B = 0
x
-
60
FA dx
FBdx
+
= 0
Lx 2(3 - 0.025 x)(2)(E)
L0 2(3 - 0.025 x)(2)(E)
x
-FA
60
dx
dx
+ FB
= 0
L0 (3 - 0.025 x)
Lx (3 - 0.025 x)
FA(40) ln (3 - 0.025 x)|x0 - FB(40) ln (3 - 0.025x)|60
x = 0
FA ln (1 -
0.025 x
0.025x
) = -FB ln (2 )
3
1.5
For greatest magnitude of P require,
4 =
FA
;
2(3 - 0.025 x)(2)
4 =
FB
;
2(3)
FA = 48 - 0.4 x
FB = 24 kip
Thus,
(48 - 0.4 x) ln a 1 -
0.025 x
0.025 x
b = -24 ln a2 b
3
1.5
Solving by trial and error,
x = 28.9 in.
Ans.
Therefore,
FA = 36.4 kip
P = 60.4 kip
Ans.
159
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4–51. The rigid bar supports the uniform distributed load
of 6 kip>ft. Determine the force in each cable if each cable
has a cross-sectional area of 0.05 in2, and E = 3111032 ksi.
C
6 ft
6 kip/ft
A
D
B
3 ft
a + ©MA = 0;
TCB a
u = tan - 1
6
= 45°
6
2
25
b(3) - 54(4.5) + TCD a
2
25
b9 = 0
(1)
L2B¿C¿ = (3)2 + (8.4853)2 - 2(3)(8.4853) cos u¿
Also,
L2D¿C¿ = (9)2 + (8.4853)2 - 2(9)(8.4853) cos u¿
(2)
Thus, eliminating cos u¿ .
-L2B¿C¿(0.019642) + 1.5910 = -L2D¿C¿(0.0065473) + 1.001735
L2B¿C¿(0.019642) = 0.0065473 L2D¿C¿ + 0.589256
L2B¿C¿ = 0.333 L2D¿C¿ + 30
But,
LB¿C = 245 + dBC¿ ,
LD¿C = 245 + dDC¿
Neglect squares or d¿ B since small strain occurs.
L2D¿C = (245 + dBC)2 = 45 + 2 245 dBC
L2D¿C = (245 + dDC)2 = 45 + 2 245 dDC
45 + 2245 dBC = 0.333(45 + 2245 dDC) + 30
2 245 dBC = 0.333(2245 dDC)
dDC = 3dBC
Thus,
TCD 245
TCB 245
= 3
AE
AE
TCD = 3 TCB
From Eq. (1).
TCD = 27.1682 kip = 27.2 kip
Ans.
TCB = 9.06 kip
Ans.
160
3 ft
3 ft
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*4–52. The rigid bar is originally horizontal and is
supported by two cables each having a cross-sectional area
of 0.05 in2, and E = 3111032 ksi. Determine the slight
rotation of the bar when the uniform load is applied.
C
See solution of Prob. 4-51.
6 ft
TCD = 27.1682 kip
dDC =
TCD 245
0.05(31)(103)
27.1682245
= 0.1175806 ft
0.05(31)(103)
=
6 kip/ft
A
D
B
Using Eq. (2) of Prob. 4-51,
3 ft
3 ft
3 ft
(245 + 0.1175806)2 = (9)2 + (8.4852)2 - 2(9)(8.4852) cos u¿
u¿ = 45.838°
Thus,
¢u = 45.838° - 45° = 0.838°
Ans.
•4–53.
The press consists of two rigid heads that are held
together by the two A-36 steel 12-in.-diameter rods. A 6061T6-solid-aluminum cylinder is placed in the press and the
screw is adjusted so that it just presses up against the
cylinder. If it is then tightened one-half turn, determine
the average normal stress in the rods and in the cylinder.
The single-threaded screw on the bolt has a lead of 0.01 in.
Note: The lead represents the distance the screw advances
along its axis for one complete turn of the screw.
12 in.
2 in.
10 in.
+ ©F = 0;
:
x
2Fst - Fal = 0
dst = 0.005 - dal
Fst(12)
p
( 4 )(0.5)2(29)(103)
= 0.005 -
Fal(10)
p(1)2(10)(103)
Solving,
Fst = 1.822 kip
Fal = 3.644 kip
srod =
Fst
1.822
= p
= 9.28 ksi
Ast
( 4 )(0.5)2
Ans.
scyl =
Fal
3.644
=
= 1.16 ksi
Aal
p(1)2
Ans.
161
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4–54. The press consists of two rigid heads that are held
together by the two A-36 steel 12-in.-diameter rods. A 6061T6-solid-aluminum cylinder is placed in the press and the
screw is adjusted so that it just presses up against the
cylinder. Determine the angle through which the screw can
be turned before the rods or the specimen begin to yield.
The single-threaded screw on the bolt has a lead of 0.01 in.
Note: The lead represents the distance the screw advances
along its axis for one complete turn of the screw.
12 in.
2 in.
10 in.
+ ©F = 0;
:
x
2Fst - Fal = 0
dst = d - dal
Fst(12)
(p4 )(0.5)2(29)(103)
= d-
Fal(10)
(1)
p(1)2(10)(103)
Assume steel yields first,
sY = 36 =
Fst
(p4 )(0.5)2
;
Fst = 7.068 kip
Then Fal = 14.137 kip;
sal =
14.137
= 4.50 ksi
p(1)2
4.50 ksi 6 37 ksi steel yields first as assumed. From Eq. (1),
d = 0.01940 in.
Thus,
0.01940
u
=
360°
0.01
u = 698°
Ans.
162
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4–55. The three suspender bars are made of A-36 steel
and have equal cross-sectional areas of 450 mm2.
Determine the average normal stress in each bar if the rigid
beam is subjected to the loading shown.
A
2m
1m
+ c ©Fy = 0;
FAD + FBE + FCF - 50(103) - 80(103) = 0
(1)
a + ©MD = 0;
FBE(2) + FCF(4) - 50(103)(1) - 80(103)(3) = 0
(2)
Referring to the geometry shown in Fig. b,
dBE = dAD + a
dBE =
dCF - dAD
b(2)
4
1
A d + dCF B
2 AD
FBE L
FCF L
1 FADL
= a
+
b
AE
2
AE
AE
FAD + FCF = 2 FBE
(3)
Solving Eqs. (1), (2) and (3) yields
FBE = 43.33(103) N
FAD = 35.83(103) N
FCF = 50.83(103) N
Thus,
sBE =
43.33(103)
FBE
= 96.3 MPa
=
A
0.45(10 - 3)
Ans.
sAD =
35.83(103)
FAD
= 79.6 MPa
=
A
0.45(10 - 3)
Ans.
sCF = 113 MPa
Ans.
163
80 kN
50 kN
E
D
Referring to the FBD of the rigid beam, Fig. a,
C
B
1m
1m
F
1m
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*4–56. The rigid bar supports the 800-lb load. Determine
the normal stress in each A-36 steel cable if each cable has a
cross-sectional area of 0.04 in2.
C
12 ft
800 lb
B
A
5 ft
Referring to the FBD of the rigid bar, Fig. a,
FBC a
a + ©MA = 0;
12
3
b(5) + FCD a b (16) - 800(10) = 0
13
5
(1)
The unstretched length of wires BC and CD are LBC = 2122 + 52 = 13 ft and
LCD = 2122 + 162 = 20 ft. The stretches of wires BC and CD are
dBC =
FBC (13)
FBC LBC
=
AE
AE
dCD =
FCD(20)
FCD LCD
=
AE
AE
Referring to the geometry shown in Fig. b, the vertical displacement of the points on
d
12
3
the rigid bar is dg =
. For points B and D, cos uB =
and cos uD = . Thus,
cos u
13
5
the vertical displacement of points B and D are
A dB B g =
FBC (13)>AE
dBC
169 FBC
=
=
cos uB
12>13
12AE
A dD B g =
FCD (20)>AE
dCD
100 FCD
=
=
cos uD
3>5
3 AE
The similar triangles shown in Fig. c give
A dB B g
5
=
A dD B g
16
1 169 FBC
1 100 FCD
b =
b
a
a
5 12 AE
16
3AE
FBC =
125
F
169 CD
(2)
Solving Eqs. (1) and (2), yields
FCD = 614.73 lb
FBC = 454.69 lb
Thus,
sCD =
FCD
614.73
= 15.37(103) psi = 15.4 ksi
=
ACD
0.04
Ans.
sBC =
FBC
454.69
=
= 11.37(103) psi = 11.4 ksi
ABC
0.04
Ans.
164
D
5 ft
6 ft
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4–56.
Continued
165
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•4–57.
The rigid bar is originally horizontal and is
supported by two A-36 steel cables each having a crosssectional area of 0.04 in2. Determine the rotation of the bar
when the 800-lb load is applied.
C
12 ft
800 lb
B
A
Referring to the FBD of the rigid bar Fig. a,
a + ©MA = 0;
FBC a
12
3
b(5) + FCD a b (16) - 800(10) = 0
13
5
5 ft
(1)
The unstretched length of wires BC and CD are LBC = 2122 + 52 = 13 ft and
LCD = 2122 + 162 = 20 ft. The stretch of wires BC and CD are
dBC =
FBC (13)
FBC LBC
=
AE
AE
dCD =
FCD(20)
FCD LCD
=
AE
AE
Referring to the geometry shown in Fig. b, the vertical displacement of the points on
d
12
3
the rigid bar is dg =
. For points B and D, cos uB =
and cos uD = . Thus,
cos u
13
5
the vertical displacement of points B and D are
A dB B g =
FBC (13)>AE
dBC
169 FBC
=
=
cos uB
12>13
12AE
A dD B g =
FCD (20)>AE
dCD
100 FCD
=
=
cos uD
3>5
3 AE
The similar triangles shown in Fig. c gives
A dB B g
5
=
A dD B g
16
1 169 FBC
1 100 FCD
a
b =
a
b
5 12 AE
16
3 AE
FBC =
125
F
169 CD
(2)
Solving Eqs (1) and (2), yields
FCD = 614.73 lb
FBC = 454.69 lb
Thus,
A dD B g =
100(614.73)
3(0.04) C 29.0 (106) D
= 0.01766 ft
Then
u = a
0.01766 ft 180°
ba
b = 0.0633°
p
16 ft
Ans.
166
D
5 ft
6 ft
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4–58. The horizontal beam is assumed to be rigid and
supports the distributed load shown. Determine the vertical
reactions at the supports. Each support consists of a wooden
post having a diameter of 120 mm and an unloaded
(original) length of 1.40 m. Take Ew = 12 GPa.
18 kN/m
A
B
C
1.40 m
2m
a + ©MB = 0;
FC(1) - FA(2) = 0
(1)
+ c ©Fy = 0;
FA + FB + FC - 27 = 0
(2)
dB - dA
dC - dA
=
;
2
3
1m
3dB - dA = 2dC
3FBL
FAL
2FCL
=
;
AE
AE
AE
3FB - FA = 2FC
(3)
Solving Eqs. (1)–(3) yields :
FA = 5.79 kN
Ans.
FB = 9.64 kN
Ans.
FC = 11.6 kN
Ans.
4–59. The horizontal beam is assumed to be rigid and
supports the distributed load shown. Determine the angle
of tilt of the beam after the load is applied. Each support
consists of a wooden post having a diameter of 120 mm and
an unloaded (original) length of 1.40 m. Take Ew = 12 GPa.
18 kN/m
A
a + ©MB = 0;
FC(1) - FA(2) = 0
(1)
c + ©Fy = 0;
FA + FB + FC - 27 = 0
(2)
dB - dA
dC - dA
=
;
2
3
3dB - dA = 2dC
3FBL
FAL
2FCL
=
;
AE
AE
AE
3FB - FA = 2FC
(3)
FB = 9.6428 kN;
FC = 11.5714 kN
3
dA =
5.7857(10 )(1.40)
FAL
= 0.0597(10 - 3) m
= p
2
9
AE
(0.12
)12(10
)
4
dC =
11.5714(103)(1.40)
FCL
= 0.1194(10 - 3) m
= p
2
9
AE
(0.12
)12(10
)
4
tan u =
0.1194 - 0.0597
(10 - 3)
3
u = 0.0199(10 - 3) rad = 1.14(10 - 3)°
Ans.
167
C
1.40 m
2m
Solving Eqs. (1)–(3) yields :
FA = 5.7857 kN;
B
1m
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*4–60. The assembly consists of two posts AD and CF
made of A-36 steel and having a cross-sectional area of
1000 mm2, and a 2014-T6 aluminum post BE having a crosssectional area of 1500 mm2. If a central load of 400 kN is
applied to the rigid cap, determine the normal stress in each
post. There is a small gap of 0.1 mm between the post BE
and the rigid member ABC.
400 kN
0.5 m
A
Equation of Equilibrium. Due to symmetry, FAD = FCF = F. Referring to the
FBD of the rigid cap, Fig. a,
FBE + 2F - 400(103) = 0
(1)
Compatibility Equation. Referring to the initial and final position of rods AD (CF)
and BE, Fig. b,
d = 0.1 + dBE
F(400)
1(10 ) C 200(10 ) D
-3
9
= 0.1 +
FBE (399.9)
1.5(10 - 3) C 73.1(109) D
F = 1.8235 FBE + 50(103)
(2)
Solving Eqs (1) and (2) yield
FBE = 64.56(103) N
F = 167.72(103) N
Normal Stress.
sAD = sCF =
sBE =
B
C
0.4 m
D
+ c ©Fy = 0;
0.5 m
167.72(103)
F
= 168 MPa
=
Ast
1(10 - 3)
Ans.
64.56(103)
FBE
= 43.0 MPa
=
Aal
1.5(10 - 3)
Ans.
168
E
F
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•4–61.
The distributed loading is supported by the three
suspender bars. AB and EF are made of aluminum and CD
is made of steel. If each bar has a cross-sectional area of
450 mm2, determine the maximum intensity w of the
distributed loading so that an allowable stress of 1sallow2st =
180 MPa in the steel and 1sallow2al = 94 MPa in the
aluminum is not exceeded. Est = 200 GPa, Eal = 70 GPa.
Assume ACE is rigid.
1.5 m
1.5 m
B
al
D
st
A
F
al
C
2m
E
w
a + ©MC = 0;
FEF(1.5) - FAB(1.5) = 0
FEF = FAB = F
+ c ©Fy = 0;
2F + FCD - 3w = 0
(1)
Compatibility condition :
dA = dC
FCDL
FL
=
;
9
A(70)(10 )
A(200)(109)
F = 0.35 FCD
(2)
Assume failure of AB and EF:
F = (sallow)al A
= 94(106)(450)(10 - 6)
= 42300 N
From Eq. (2) FCD = 120857.14 N
From Eq. (1) w = 68.5 kN>m
Assume failure of CD:
FCD = (sallow)st A
= 180(106)(450)(10 - 6)
= 81000 N
From Eq. (2) F = 28350 N
From Eq. (1) w = 45.9 kN>m
(controls)
Ans.
169
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4–62. The rigid link is supported by a pin at A, a steel wire
BC having an unstretched length of 200 mm and crosssectional area of 22.5 mm2, and a short aluminum block
having an unloaded length of 50 mm and cross-sectional
area of 40 mm2. If the link is subjected to the vertical load
shown, determine the average normal stress in the wire and
the block. Est = 200 GPa, Eal = 70 GPa.
C
200 mm
B
100 mm
D
450(250) - FBC(150) - FD(150) = 0
50 mm
750 - FBC - FD = 0
[1]
Compatibility:
dBC = dD
FD(50)
FBC(200)
22.5(10 - 6)200(109)
=
40(10 - 6)70(109)
FBC = 0.40179 FD
[2]
Solving Eqs. [1] and [2] yields:
FD = 535.03 N
150 mm
450 N
Equations of Equilibrium:
a + ©MA = 0;
A
150 mm
FBC = 214.97 N
Average Normal Stress:
sD =
FD
535.03
= 13.4 MPa
=
AD
40(10 - 6)
Ans.
sBC =
FBC
214.97
= 9.55 MPa
=
ABC
22.5(10 - 6)
Ans.
170
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4–63. The rigid link is supported by a pin at A, a steel wire
BC having an unstretched length of 200 mm and crosssectional area of 22.5 mm2, and a short aluminum block
having an unloaded length of 50 mm and cross-sectional area
of 40 mm2. If the link is subjected to the vertical load shown,
determine the rotation of the link about the pin A. Report
the answer in radians. Est = 200 GPa, Eal = 70 GPa.
C
200 mm
B
100 mm
D
450(250) - FBC(150) - FD(150) = 0
50 mm
750 - FBC - FD = 0
[1]
Compatibility:
dBC = dD
FD(50)
FBC(200)
-6
9
22.5(10 )200(10 )
=
40(10 - 6)70(109)
FBC = 0.40179 FD
[2]
Solving Eqs. [1] and [2] yields :
FD = 535.03 N
FBC = 214.97 N
Displacement:
dD =
535.03(50)
FDLD
= 0.009554 mm
=
ADEal
40(10 - 6)(70)(109)
tan u =
150 mm
450 N
Equations of Equilibrium:
a + ©MA = 0;
A
150 mm
dD
0.009554
=
150
150
u = 63.7(10 - 6) rad = 0.00365°
Ans.
171
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*4–64. The center post B of the assembly has an original
length of 124.7 mm, whereas posts A and C have a length of
125 mm. If the caps on the top and bottom can be
considered rigid, determine the average normal stress in
each post. The posts are made of aluminum and have a
cross-sectional area of 400 mm2. Eal = 70 GPa.
800 kN/m
A
a + ©MB = 0;
100 mm
B
C
-FA(100) + FC(100) = 0
(1)
2F + FB - 160 = 0
(2)
dA = dB + 0.0003
F (0.125)
FB (0.1247)
400 (10 - 6)(70)(106)
=
400 (10 - 6)(70)(106)
+ 0.0003
0.125 F - 0.1247FB = 8.4
(3)
Solving Eqs. (2) and (3)
F = 75.762 kN
FB = 8.547 kN
sA = sC =
sB =
75.726 (103)
400(10 - 6)
8.547 (103)
400 (10 - 6)
= 189 MPa
Ans.
= 21.4 MPa
Ans.
•4–65.
The assembly consists of an A-36 steel bolt and a
C83400 red brass tube. If the nut is drawn up snug against
the tube so that L = 75 mm, then turned an additional
amount so that it advances 0.02 mm on the bolt, determine
the force in the bolt and the tube. The bolt has a diameter of
7 mm and the tube has a cross-sectional area of 100 mm2.
L
Equilibrium: Since no external load is applied, the force acting on the tube and the
bolt is the same.
Compatibility:
0.02 = dt + db
0.02 =
P(75)
P(75)
-6
9
100(10 )(101)(10 )
125 mm
800 kN/m
FA = FC = F
+ c ©Fy = 0;
100 mm
+ p
2
9
4 (0.007 )(200)(10 )
P = 1164.83 N = 1.16 kN
Ans.
172
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4–66. The assembly consists of an A-36 steel bolt and a
C83400 red brass tube. The nut is drawn up snug against
the tube so that L = 75 mm. Determine the maximum
additional amount of advance of the nut on the bolt so that
none of the material will yield. The bolt has a diameter of
7 mm and the tube has a cross-sectional area of 100 mm2.
L
Allowable Normal Stress:
(sg)st = 250 A 106 B = p
Pst
2
4 (0.007)
Pst = 9.621 kN
(sg)br = 70.0 A 106 B =
Pbr
100(10 - 6)
Pbr = 7.00 kN
Since Pst 7 Pbr, by comparison he brass will yield first.
Compatibility:
a = dt + db
7.00(103)(75)
=
100(10 - 6)(101)(109)
7.00(103)(75)
+ p
2
9
4 (0.007) (200)(10 )
= 0.120 mm
Ans.
173
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4–67. The three suspender bars are made of the same
material and have equal cross-sectional areas A. Determine
the average normal stress in each bar if the rigid beam ACE
is subjected to the force P.
a + ©MA = 0;
D
d
FCD(d) + FEF(2d) - Pa b = 0
2
FCD + 2FEF =
+ c ©Fy = 0;
B
F
L
P
P
2
(1)
FAB + FCD + FEF - P = 0
A
C
d
2
(2)
d
2
E
d
dC - dE
dA - dE
=
d
2d
2dC = dA + dE
2FCDL
FABL
FEFL
=
+
AE
AE
AE
2FCD - FAB - FEF = 0
(3)
Solving Eqs. (1), (2) and (3) yields
P
3
P
12
FAB =
7P
12
sAB =
7P
12A
Ans.
sCD =
P
3A
Ans.
sEF =
P
12A
Ans.
FCD =
FEF =
*4–68. A steel surveyor’s tape is to be used to measure the
length of a line. The tape has a rectangular cross section of
0.05 in. by 0.2 in. and a length of 100 ft when T1 = 60°F and
the tension or pull on the tape is 20 lb. Determine the
true length of the line if the tape shows the reading to
be 463.25 ft when used with a pull of 35 lb at T2 = 90°F. The
ground on which it is placed is flat. ast = 9.60110-62>°F,
Est = 2911032 ksi.
P
P
0.2 in.
0.05 in.
dT = a¢TL = 9.6(10 - 6)(90 - 60)(463.25) = 0.133416 ft
d =
(35 - 20)(463.25)
PL
= 0.023961 ft
=
AE
(0.2)(0.05)(29)(106)
L = 463.25 + 0.133416 + 0.023961 = 463.41 ft
Ans.
174
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•4–69.
Three bars each made of different materials are
connected together and placed between two walls when the
temperature is T1 = 12°C. Determine the force exerted
on the (rigid) supports when the temperature becomes
T2 = 18°C. The material properties and cross-sectional
area of each bar are given in the figure.
+ )
(;
Copper
Steel
Brass
Est ⫽ 200 GPa
Ebr ⫽ 100 GPa
Ecu ⫽ 120 GPa
ast ⫽ 12(10⫺6)/⬚C abr ⫽ 21(10⫺6)/°C acu ⫽ 17(10⫺6)/⬚C
Acu ⫽ 515 mm2
Ast ⫽ 200 mm2
Abr ⫽ 450 mm2
300 mm
200 mm
100 mm
0 = ¢T - d
0 = 12(10 - 6)(6)(0.3) + 21 (10 - 6)(6)(0.2) + 17 (10 - 6)(6)(0.1)
F(0.3)
-
-6
F(0.2)
9
200(10 )(200)(10 )
-
-6
F(0.1)
9
-
450(10 )(100)(10 )
515(10 - 6)(120)(109)
F = 4203 N = 4.20 kN
Ans.
k ⫽ 1000 lb/in.
4–70. The rod is made of A-36 steel and has a diameter of
0.25 in. If the rod is 4 ft long when the springs are compressed
0.5 in. and the temperature of the rod is T = 40°F, determine
the force in the rod when its temperature is T = 160°F.
k ⫽ 1000 lb/ in.
4 ft
Compatibility:
+ B
A:
x = dT - dF
x = 6.60(10 - 6)(160 - 40)(2)(12) - p
1.00(0.5)(2)(12)
2
3
4 (0.25 )(29.0)(10 )
x = 0.01869 in.
F = 1.00(0.01869 + 0.5) = 0.519 kip
Ans.
4–71. A 6-ft-long steam pipe is made of A-36 steel with
sY = 40 ksi. It is connected directly to two turbines A
and B as shown. The pipe has an outer diameter of 4 in.
and a wall thickness of 0.25 in. The connection was made
at T1 = 70°F. If the turbines’ points of attachment are
assumed rigid, determine the force the pipe exerts on
the turbines when the steam and thus the pipe reach a
temperature of T2 = 275°F.
6 ft
A
Compatibility:
+ B
A:
0 = dT - dF
0 = 6.60(10 - 6)(275 - 70)(6)(12) - p
F(6)(12)
2
4 (4
- 3.52)(29.0)(103)
F = 116 kip
Ans.
175
B
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*4–72. A 6-ft-long steam pipe is made of A-36 steel with
sY = 40 ksi. It is connected directly to two turbines A and
B as shown. The pipe has an outer diameter of 4 in. and a
wall thickness of 0.25 in. The connection was made at
T1 = 70°F. If the turbines’ points of attachment are
assumed to have a stiffness of k = 8011032 kip>in., determine
the force the pipe exerts on the turbines when the steam
and thus the pipe reach a temperature of T2 = 275°F.
6 ft
A
B
Compatibility:
x = dT - dF
80(103)(x)(3)(12)
x = 6.60(10 - 6)(275 - 70)(3)(12) - p
2
2
3
4 (4 - 3.5 )(29.0)(10 )
x = 0.001403 in.
F = k x = 80(103)(0.001403) = 112 kip
Ans.
•4–73.
The pipe is made of A-36 steel and is connected to
the collars at A and B. When the temperature is 60° F, there
is no axial load in the pipe. If hot gas traveling through the
pipe causes its temperature to rise by ¢T = 140 + 15x2°F,
where x is in feet, determine the average normal stress in the
pipe. The inner diameter is 2 in., the wall thickness is 0.15 in.
A
Compatibility:
L
0 = dT - dF
Where
0 = 6.60 A 10 - 6 B
8ft
L0
dT =
L0
(40 + 15 x) dx -
0 = 6.60 A 10 - 6 B B 40(8) +
a ¢T dx
F(8)
A(29.0)(103)
15(8)2
F(8)
R 2
A(29.0)(103)
F = 19.14 A
Average Normal Stress:
s =
B
8 ft
19.14 A
= 19.1 ksi
A
Ans.
176
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4–74. The bronze C86100 pipe has an inner radius of
0.5 in. and a wall thickness of 0.2 in. If the gas flowing
through it changes the temperature of the pipe uniformly
from TA = 200°F at A to TB = 60°F at B, determine the
axial force it exerts on the walls. The pipe was fitted between
the walls when T = 60°F.
A
B
8 ft
Temperature Gradient:
T(x) = 60 + a
8 - x
b140 = 200 - 17.5x
8
Compatibility:
0 = dT - dF
Where
0 = 9.60 A 10 - 6 B
2ft
0 = 9.60 A 10 - 6 B
L0
dT = 1 a¢Tdx
2
2
3
4 (1.4 - 1 )15.0(10 )
2ft
L0
F(8)
[(200 - 17.5x) - 60] dx - p
(140 - 17.5x) dx - p
F(8)
2
2
3
4 (1.4 - 1 ) 15.0(10 )
F = 7.60 kip
Ans.
4–75. The 40-ft-long A-36 steel rails on a train track are laid
with a small gap between them to allow for thermal
expansion. Determine the required gap d so that the rails just
touch one another when the temperature is increased from
T1 = -20°F to T2 = 90°F. Using this gap, what would be the
axial force in the rails if the temperature were to rise to
T3 = 110°F? The cross-sectional area of each rail is 5.10 in2.
d
40 ft
Thermal Expansion: Note that since adjacent rails expand, each rail will be
d
required to expand on each end, or d for the entine rail.
2
d = a¢TL = 6.60(10 - 6)[90 - (-20)](40)(12)
Ans.
= 0.34848 in. = 0.348 in.
Compatibility:
+ B
A:
0.34848 = dT - dF
0.34848 = 6.60(10 - 6)[110 - (-20)](40)(12) -
d
F(40)(12)
5.10(29.0)(103)
F = 19.5 kip
Ans.
177
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*4–76. The device is used to measure a change in
temperature. Bars AB and CD are made of A-36 steel and
2014-T6 aluminum alloy respectively. When the temperature
is at 75°F, ACE is in the horizontal position. Determine the
vertical displacement of the pointer at E when the
temperature rises to 150°F.
0.25 in.
A
3 in.
C
E
1.5 in.
Thermal Expansion:
A dT B CD = aal ¢TLCD = 12.8(10 - 6)(150 - 75)(1.5) = 1.44(10 - 3) in.
B
D
A dT B AB = ast ¢TLAB = 6.60(10 - 6)(150 - 75)(1.5) = 0.7425(10 - 3) in.
From the geometry of the deflected bar AE shown Fig. b,
dE = A dT B AB + C
= 0.7425(10 - 3) + B
A dT B CD - A dT B AB
0.25
S(3.25)
1.44(10 - 3) - 0.7425(10 - 3)
R (3.25)
0.25
= 0.00981 in.
Ans.
•4–77. The bar has a cross-sectional area A, length L,
modulus of elasticity E, and coefficient of thermal
expansion a. The temperature of the bar changes uniformly
along its length from TA at A to TB at B so that at any point
x along the bar T = TA + x1TB - TA2>L. Determine the
force the bar exerts on the rigid walls. Initially no axial force
is in the bar and the bar has a temperature of TA.
+
:
x
A
TA
0 = ¢ T - dF
(1)
However,
d¢ T = a¢ T dx = a(TA +
TB - TA
x - TA)dx
L
L
¢T = a
= ac
L
TB - TA
TB - TA 2
x dx = ac
x d冷
L
2L
L0
0
TB - TA
aL
Ld =
(TB - TA)
2
2
From Eq.(1).
0 =
FL
aL
(TB - TA) 2
AE
F =
a AE
(TB - TA)
2
B
Ans.
178
TB
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4–78. The A-36 steel rod has a diameter of 50 mm and is
lightly attached to the rigid supports at A and B when
T1 = 80°C. If the temperature becomes T2 = 20°C and an
axial force of P = 200 kN is applied to its center, determine
the reactions at A and B.
0.5 m
FB - FA + 200(103) = 0
(1)
When the rod is unconstrained at B, it has a free contraction of
dT = ast ¢ TL = 12(10 - 6)(80 - 20)(1000) = 0.72 mm. Also, under force P and FB
with unconstrained at B, the deformation of the rod are
dP =
200(103)(500)
PLAC
= 0.2546 mm
=
p
AE
(0.052) C 200(109) D
4
dFB =
FB (1000)
FB LAB
= 2.5465(10 - 6) FB
=
p
AE
(0.052) C 200(109) D
4
Using the method of super position, Fig. b,
+ B
A:
B
P
Referring to the FBD of the rod, Fig. a
+ ©F = 0;
:
x
C
A
0 = -dT + dP + dFB
0 = -0.72 + 0.2546 + 2.5465(10 - 6) FB
FB = 182.74(103) N = 183 kN
Ans.
Substitute the result of FB into Eq (1),
FA = 382.74(103) N = 383 kN
Ans.
179
0.5 m
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4–79. The A-36 steel rod has a diameter of 50 mm and is
lightly attached to the rigid supports at A and B when
T1 = 50°C. Determine the force P that must be applied to
the collar at its midpoint so that, when T2 = 30°C, the
reaction at B is zero.
C
A
B
P
0.5 m
0.5 m
When the rod is unconstrained at B, it has a free contraction of
dT = ast ¢TL = 12(10 - 6)(50 - 30)(1000) = 0.24 mm. Also, under force P with
unconstrained at B, the deformation of the rod is
dP =
P(500)
PLAC
= 1.2732(10 - 6) P
=
p
AE
(0.052) C 200(109) D
4
Since FB is required to be zero, the method of superposition, Fig. b, gives
+ B
A:
0 = -dT + dP
0 = -0.24 + 1.2732(10 - 6)P
P = 188.50(103) N = 188 kN
Ans.
*4–80. The rigid block has a weight of 80 kip and is to be
supported by posts A and B, which are made of A-36 steel,
and the post C, which is made of C83400 red brass. If all the
posts have the same original length before they are loaded,
determine the average normal stress developed in each post
when post C is heated so that its temperature is increased
by 20°F. Each post has a cross-sectional area of 8 in2.
A
Equations of Equilibrium:
a + ©MC = 0;
FB(3) - FA(3) = 0
+ c ©Fy = 0;
2F + FC - 80 = 0
FA = FB = F
[1]
Compatibility:
(dC)F - (dC)T = dF
(+ T)
FCL
8(14.6)(103)
- 9.80 A 10 - 5 B (20)L =
FL
8(29.0)(103)
8.5616 FC - 4.3103 F = 196
[2]
Solving Eqs. [1] and [2] yields:
F = 22.81 kip
FC = 34.38 kip
average Normal Sress:
sA = sB =
sC =
F
22.81
=
= 2.85 ksi
A
8
Ans.
FC
34.38
=
= 4.30 ksi
A
8
Ans.
180
C
B
3 ft
3 ft
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•4–81.
The three bars are made of A-36 steel and form
a pin-connected truss. If the truss is constructed when
T1 = 50°F, determine the force in each bar when
T2 = 110°F. Each bar has a cross-sectional area of 2 in2.
t
5f
5f
t
A
4 ft
B
D
3 ft
(dT ¿)AB - (dF ¿)AB = (dT)AD + (dF)AD
(1)
œ
cos u;
However, dAB = dAB
œ
dAB
=
dAB
5
= dAB
cos u
4
Substitute into Eq. (1)
5
5
(dT)AB - (dF)AB = (dT)AD + (dF)AD
4
4
FAB(5)(12)
5
d
c6.60(10 - 6)(110° - 50°)(5)(12) 4
2(29)(103)
= 6.60(10 - 6)(110° - 50°)(4)(12) +
FAD(4)(12)
2(29)(103)
620.136 = 75FAB + 48FAD
+ ©F = 0;
:
x
3
3
F
- FAB = 0;
5 AC
5
+ c ©Fy = 0;
4
FAD - 2a FAB b = 0
5
(2)
FAC = FAB
(3)
Solving Eqs. (2) and (3) yields :
FAD = 6.54 kip
Ans.
FAC = FAB = 4.09 kip
Ans.
181
C
3 ft
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4–82. The three bars are made of A-36 steel and form a pinconnected truss. If the truss is constructed when T1 = 50°F,
determine the vertical displacement of joint A when
T2 = 150°F. Each bar has a cross-sectional area of 2 in2.
t
5f
5f
t
A
4 ft
(dT ¿)AB - (dF ¿)AB = (dT)AD + (dF)AD
(1)
œ
However, dAB = dAB
cos u;
œ
dAB
=
B
dAB
5
= dAB
cos u
4
3 ft
Substitute into Eq. (1)
5
5
(d )
- (dT)AB = (dT)AD + (dF)AD
4 T AB
4
FAB(5)(12)
5
d
c6.60(10 - 6)(150° - 50°)(5)(12) 4
2(29)(103)
= 6.60(10 - 6)(150° - 50°)(4)(12) +
FAD(4)(12)
2(29)(103)
239.25 - 6.25FAB = 153.12 + 4 FAD
4 FAD + 6.25FAB = 86.13
+ © F = 0;
:
x
3
3
F
- FAB = 0;
5 AC
5
+ c © Fy = 0;
4
FAD - 2 a FAB b = 0;
5
(2)
FAC = FAB
FAD = 1.6FAB
(3)
Solving Eqs. (2) and (3) yields:
FAB = 6.8086 kip:
FAD = 10.8939 kip
(dA)r = (dT)AD + (dT)AD
= 6.60(10 - 6)(150° - 50°)(4)(12) +
D
10.8939(4)(12)
2(29)(103)
= 0.0407 in. c
Ans.
182
C
3 ft
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4–83. The wires AB and AC are made of steel, and wire
AD is made of copper. Before the 150-lb force is applied,
AB and AC are each 60 in. long and AD is 40 in. long. If
the temperature is increased by 80°F, determine the
force in each wire needed to support the load. Take
Est = 29(103) ksi, Ecu = 17(103) ksi, ast = 8(10-6)>°F,
acu = 9.60(10-6)>°F. Each wire has a cross-sectional area of
0.0123 in2.
40 in.
60 in.
45⬚
45⬚
A
150 lb
Equations of Equilibrium:
+ ©F = 0;
:
x
FAC cos 45° - FAB cos 45° = 0
FAC = FAB = F
2F sin 45° + FAD - 150 = 0
+ c ©Fy = 0;
[1]
Compatibility:
(dAC)T = 8.0 A 10 - 6 B (80)(60) = 0.03840 in.
(dAC)Tr =
(dAC)T
0.03840
=
= 0.05431 in.
cos 45°
cos 45°
(dAD)T = 9.60 A 10 - 6 B (80)(40) = 0.03072 in.
d0 = (dAC)Tr - (dAD)T = 0.05431 - 0.03072 = 0.02359 in.
(dAD)F = (dAC)Fr + d0
F(60)
FAD (40)
6
0.0123(17.0)(10 )
=
0.0123(29.0)(106) cos 45°
C
D
B
+ 0.02359
0.1913FAD - 0.2379F = 23.5858
[2]
Solving Eq. [1] and [2] yields:
FAC = FAB = F = 10.0 lb
Ans.
FAD = 136 lb
Ans.
183
60 in.
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*4–84. The AM1004-T61 magnesium alloy tube AB is
capped with a rigid plate E. The gap between E and end C of
the 6061-T6 aluminum alloy solid circular rod CD is 0.2 mm
when the temperature is at 30° C. Determine the normal
stress developed in the tube and the rod if the temperature
rises to 80° C. Neglect the thickness of the rigid cap.
25 mm
a
Section a-a
E
B
A
20 mm
C
25 mm
a
0.2 mm
300 mm
Compatibility Equation: If tube AB and rod CD are unconstrained, they will have a
free expansion of A dT B AB = amg ¢TLAB = 26(10 - 6)(80 - 30)(300) = 0.39 mm and
A dT)CD = aal ¢TLCD = 24(10 - 6)(80 - 30)(450) = 0.54 mm. Referring
deformation diagram of the tube and the rod shown in Fig. a,
to
the
d = C A dT B AB - A dF B AB D + C A dT B CD - A dF B CD D
0.2 = C 0.39 -
F(300)
p A 0.025 - 0.02 B (44.7)(10 )
2
2
9
S + C 0.54 -
A
F(450)
B
p
2
9
4 0.025 (68.9)(10 )
S
F = 32 017.60 N
Normal Stress:
sAB =
F
32 017.60
=
= 45.3 MPa
AAB
p A 0.0252 - 0.022 B
sCD =
F
32 017.60
=
= 65.2 MPa
p
ACD
A 0.0252 B
Ans.
Ans.
4
F = 107 442.47 N
184
450 mm
D
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•4–85. The AM1004-T61 magnesium alloy tube AB is
capped with a rigid plate. The gap between E and end C of
the 6061-T6 aluminum alloy solid circular rod CD is 0.2 mm
when the temperature is at 30° C. Determine the highest
temperature to which it can be raised without causing
yielding either in the tube or the rod. Neglect the thickness
of the rigid cap.
25 mm
a
Section a-a
E
B
A
20 mm
C
25 mm
a
0.2 mm
300 mm
Then
sCD =
F
107 442.47
=
= 218.88MPa 6 (sY)al
p
ACD
A 0.0252 B
(O.K.!)
4
Compatibility Equation: If tube AB and rod CD are unconstrained, they will have
a free expansion of A dT B AB = amg ¢TLAB = 26(10 - 6)(T - 30)(300) = 7.8(10 - 6)
(T - 30) and
A dT B CD = aal ¢TLCD = 24(10 - 6)(T - 30)(450) = 0.0108(T - 30).
Referring to the deformation diagram of the tube and the rod shown in Fig. a,
d = C A dT B AB - A dF B AB D + C A dT B CD - A dF B CD D
0.2 = C 7.8(10 - 3)(T - 30) -
+ C 0.0108(T - 30) -
107 442.47(300)
p A 0.0252 - 0.022 B (44.7)(109)
107 442.47(450)
p
4
A 0.0252 B (68.9)(109)
S
S
T = 172° C
Ans.
185
450 mm
D
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4–86. The steel bolt has a diameter of 7 mm and fits
through an aluminum sleeve as shown. The sleeve has an
inner diameter of 8 mm and an outer diameter of 10 mm.
The nut at A is adjusted so that it just presses up against
the sleeve. If the assembly is originally at a temperature
of T1 = 20°C and then is heated to a temperature of
T2 = 100°C, determine the average normal stress in the
bolt and the sleeve. Est = 200 GPa, Eal = 70 GPa, ast =
14(10-6)>°C, aal = 23(10-6)>°C.
A
Compatibility:
(ds)T - (db)T = (ds)F + (db)F
23(10 - 6)(100 - 20)L - 14(10 - 6)(100 - 20)L
FL
FL
+ p
2
2
9
2
(0.01
0.008
)70(10
)
(0.007
)200(109)
4
4
= p
F = 1133.54 N
Average Normal Stress:
ss =
F
1133.54
= 40.1 MPa
= p
2
As
(0.01
- 0.0082)
4
Ans.
sb =
F
1133.54
= 29.5 MPa
= p
2
Ab
4 (0.007 )
Ans.
4–87. Determine the maximum normal stress developed
in the bar when it is subjected to a tension of P = 8 kN.
5 mm
40 mm
20 mm
P
P
For the fillet:
r ⫽ 10 mm
20 mm
r
10
=
= 0.5
h
20
w
40
=
= 2
h
20
From Fig. 10-24. K = 1.4
smax = Ksavg
= 1.4 a
8 (103)
b
0.02 (0.005)
= 112 MPa
For the hole:
r
10
=
= 0.25
w
40
From Fig. 4-25. K = 2.375
smax = Ksavg
= 2.375 a
8 (103)
b
(0.04 - 0.02)(0.005)
= 190 MPa
Ans.
186
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*4–88. If the allowable normal stress for the bar is
sallow = 120 MPa, determine the maximum axial force P
that can be applied to the bar.
5 mm
40 mm
20 mm
P
P
Assume failure of the fillet.
r ⫽ 10 mm
20 mm
r
10
=
= 0.5
h
20
w
40
=
= 2;
h
20
From Fig. 4-24. K = 1.4
sallow = smax = Ksavg
120 (106) = 1.4 a
P
b
0.02 (0.005)
P = 8.57 kN
Assume failure of the hole.
r
10
=
= 0.25
w
20
From Fig. 4-25. K = 2.375
sallow = smax = Ksavg
120 (104) = 2.375 a
P
b
(0.04 - 0.02) (0.005)
P = 5.05 kN (controls)
Ans.
•4–89.
The member is to be made from a steel plate that is
0.25 in. thick. If a 1-in. hole is drilled through its center,
determine the approximate width w of the plate so that it
can support an axial force of 3350 lb. The allowable stress is
sallow = 22 ksi.
0.25 in.
w
3350 lb
sallow = smax = Ksavg
1 in.
3.35
d
22 = K c
(w - 1)(0.25)
w =
3.35K + 5.5
5.5
By trial and error, from Fig. 4-25, choose
w =
r
= 0.2;
w
K = 2.45
3.35(2.45) + 5.5
= 2.49 in.
5.5
Since
r
0.5
=
= 0.2
w
2.49
3350 lb
Ans.
OK
187
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4–90. The A-36 steel plate has a thickness of 12 mm. If
there are shoulder fillets at B and C, and sallow = 150 MPa,
determine the maximum axial load P that it can support.
Calculate its elongation, neglecting the effect of the fillets.
r = 30 mm
120 mm
r = 30 mm
60 mm
P
A
w
120
=
= 2
h
60
and
60 mm
P
D
B
Maximum Normal Stress at fillet:
r
30
=
= 0.5
h
60
C
800 mm
200 mm
200 mm
From the text, K = 1.4
smax = sallow = Ksavg
150(106) = 1.4 B
P
R
0.06(0.012)
P = 77142.86 N = 77.1 kN
Ans.
Displacement:
d = ©
PL
AE
77142.86(800)
77142.86(400)
=
9
+
(0.06)(0.012)(200)(10 )
(0.12)(0.012)(200)(109)
= 0.429 mm
Ans.
4–91. Determine the maximum axial force P that can be
applied to the bar. The bar is made from steel and has an
allowable stress of sallow = 21 ksi.
0.125 in.
1.25 in.
1.875 in.
P
Assume failure of the fillet.
r
0.25
=
= 0.2
h
1.25
P
w
1.875
=
= 1.5
h
1.25
0.75 in.
From Fig. 4-24, K = 1.73
sallow = smax = Ksavg
21 = 1.73 a
P
b
1.25 (0.125)
P = 1.897 kip
Assume failure of the hole.
r
0.375
=
= 0.20
w
1.875
From Fig. 4-25, K = 2.45
sallow = smax = Ksavg
21 = 2.45 a
P
b
(1.875 - 0.75)(0.125)
P = 1.21 kip (controls)
Ans.
188
r ⫽ 0.25 in.
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*4–92. Determine the maximum normal stress developed
in the bar when it is subjected to a tension of P = 2 kip.
0.125 in.
1.25 in.
1.875 in.
At fillet:
P
r
0.25
=
= 0.2
h
1.25
P
w
1.875
=
= 1.5
h
1.25
0.75 in.
From Fig. 4-24, K = 1.73
smax = Ka
r ⫽ 0.25 in.
P
2
d = 22.1 ksi
b = 1.73 c
A
1.25(0.125)
At hole:
r
0.375
=
= 0.20
w
1.875
From Fig. 4-25, K = 2.45
smax = 2.45 c
2
d = 34.8 ksi
(1.875 - 0.75)(0.125)
(Controls)
Ans.
•4–93.
Determine the maximum normal stress developed
in the bar when it is subjected to a tension of P = 8 kN.
5 mm
60 mm
P
Maximum Normal Stress at fillet:
r
15
=
= 0.5
h
30
P
ht
= 1.4 B
8(103)
R = 74.7 MPa
(0.03)(0.005)
Maximum Normal Stress at the hole:
r
6
=
= 0.1
w
60
From the text, K = 2.65
smax = K savg = K
P
(w - 2r) t
= 2.65 B
8(103)
R
(0.06 - 0.012)(0.005)
= 88.3 MPa
P
r = 15 mm
12 mm
w
60
=
= 2
h
30
and
From the text, K = 1.4
smax = Ksavg = K
30 mm
(Controls)
Ans.
189
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4–94. The resulting stress distribution along section AB
for the bar is shown. From this distribution, determine the
approximate resultant axial force P applied to the bar. Also,
what is the stress-concentration factor for this geometry?
0.5 in.
A
P
4 in.
1 in.
B
12 ksi
P =
L
3 ksi
sdA = Volume under curve
Number of squares = 10
P = 10(3)(1)(0.5) = 15 kip
savg =
K =
Ans.
15 kip
P
=
= 7.5 ksi
A
(4 in.)(0.5 in.)
smax
12 ksi
=
= 1.60
savg
7.5 ksi
Ans.
4–95. The resulting stress distribution along section AB
for the bar is shown. From this distribution, determine the
approximate resultant axial force P applied to the bar. Also,
what is the stress-concentration factor for this geometry?
0.5 in.
A
0.6 in.
0.8 in.
Number of squares = 28
P = 28(6)(0.2)(0.5) = 16.8 kip
0.6 in.
Ans.
B
P
16.8
=
= 28 ksi
savg =
A
2(0.6)(0.5)
K =
6 ksi
36 ksi
smax
36
=
= 1.29
savg
28
Ans.
*4–96. The resulting stress distribution along section AB
for the bar is shown. From this distribution, determine the
approximate resultant axial force P applied to the bar. Also,
what is the stress-concentration factor for this geometry?
10 mm
A
20 mm
80 mm
B
5 MPa
Number of squares = 19
30 MPa
6
P = 19(5)(10 )(0.02)(0.01) = 19 kN
savg =
K =
P
0.2 in.
Ans.
19(103)
P
=
= 23.75 MPa
A
0.08(0.01)
smax
30 MPa
=
= 1.26
savg
23.75 MPa
Ans.
190
P
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•4–97.
The 300-kip weight is slowly set on the top of a post
made of 2014-T6 aluminum with an A-36 steel core. If both
materials can be considered elastic perfectly plastic,
determine the stress in each material.
Aluminum
1 in.
2 in.
Steel
Equations of Equilibrium:
+ c ©Fy = 0;
Pst + Pal - 300 = 0
[1]
Elastic Analysis: Assume both materials still behave elastically under the load.
dst = dal
Pst L
p
2
(2)
(29)(103)
4
Pal L
= p
2
2
3
4 (4 - 2 )(10.6)(10 )
Pst = 0.9119 Pal
Solving Eqs. [1] and [2] yields:
Pal = 156.91 kip
Pst = 143.09 kip
Average Normal Stress:
sal =
Pal
156.91
= p 2
Aal
(4
- 22)
4
(OK!)
= 16.65 ksi 6 (sg)al = 60.0 ksi
sst =
Pst
143.09
= p 2
Ast
4 (2 )
= 45.55 ksi 7 (sg)st = 36.0 ksi
Therefore, the steel core yields and so the elastic analysis is invalid. The stress in the
steel is
sst = (sy)st = 36.0 ksi
Ans.
p
Pst = (sg)stAst = 36.0a b A 22 B = 113.10 kip
4
From Eq. [1] Pal = 186.90 kip
sal =
Pal
186.90
= 19.83 ksi 6 (sg)al = 60.0 ksi
= p 2
2
Aal
4 (4 - 2 )
Then sal = 19.8 ksi
Ans.
191
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4–98. The bar has a cross-sectional area of 0.5 in2 and is
made of a material that has a stress–strain diagram that
can be approximated by the two line segments shown.
Determine the elongation of the bar due to the applied
loading.
A
B
5 ft
8 kip
C 5 kip
2 ft
s(ksi)
40
20
Average Normal Stress and Strain: For segment BC
sBC =
0.001
PBC
5
=
= 10.0 ksi
ABC
0.5
10.0
20
=
;
eBC
0.001
eBC =
0.001
(10.0) = 0.00050 in.>in.
20
Average Normal Stress and Strain: For segment AB
sAB =
PAB
13
=
= 26.0 ksi
AAB
0.5
40 - 20
26.0 - 20
=
eAB - 0.001
0.021 - 0.001
eAB = 0.0070 in.>in.
Elongation:
dBC = eBCLBC = 0.00050(2)(12) = 0.0120 in.
dAB = eAB LAB = 0.0070(5)(12) = 0.420 in.
dTot = dBC + dAB = 0.0120 + 0.420 = 0.432 in.
Ans.
192
0.021
P (in./in.)
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4–99. The rigid bar is supported by a pin at A and two
steel wires, each having a diameter of 4 mm. If the yield
stress for the wires is sY = 530 MPa, and Est = 200 GPa,
determine the intensity of the distributed load w that can
be placed on the beam and will just cause wire EB to
yield. What is the displacement of point G for this case?
For the calculation, assume that the steel is elastic
perfectly plastic.
E
D
800 mm
A
B
C
G
w
400 mm
Equations of Equilibrium:
a + ©MA = 0;
FBE(0.4) + FCD(0.65) - 0.8w (0.4) = 0
0.4 FBE + 0.65FCD = 0.32w
[1]
Plastic Analysis: Wire CD will yield first followed by wire BE. When both wires yield
FBE = FCD = (sg)A
p
= 530 A 106 B a b A 0.0042 B = 6.660 kN
4
Substituting the results into Eq. [1] yields:
w = 21.9 kN>m
Ans.
Displacement: When wire BE achieves yield stress, the corresponding yield strain is
eg =
sg
E
530(106)
=
200(109)
= 0.002650 mm>mm
dBE = eg LBE = 0.002650(800) = 2.120 mm
From the geometry
dBE
dG
=
0.8
0.4
dG = 2dBE = 2(2.120) = 4.24 mm
Ans.
193
250 mm
150 mm
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*4–100. The rigid bar is supported by a pin at A and two
steel wires, each having a diameter of 4 mm. If the yield
stress for the wires is sY = 530 MPa, and Est = 200 GPa,
determine (a) the intensity of the distributed load w that
can be placed on the beam that will cause only one of the
wires to start to yield and (b) the smallest intensity of
the distributed load that will cause both wires to yield. For
the calculation, assume that the steel is elastic perfectly
plastic.
E
D
800 mm
A
B
C
G
w
400 mm
Equations of Equilibrium:
a + ©MA = 0;
FBE(0.4) + FCD(0.65) - 0.8w (0.4) = 0
0.4 FBE + 0.65 FCD = 0.32w
[1]
(a) By observation, wire CD will yield first.
p
Then FCD = sg A = 530 A 106 B a b A 0.0042 B = 6.660 kN.
4
From the geometry
dCD
dBE
=
;
0.4
0.65
dCD = 1.625dBE
FBEL
FCDL
= 1.625
AE
AE
FCD = 1.625 FBE
[2]
Using FCD = 6.660 kN and solving Eqs. [1] and [2] yields:
FBE = 4.099 kN
w = 18.7 kN>m
Ans.
(b) When both wires yield
FBE = FCD = (sg)A
p
= 530 A 106 B a b A 0.0042 B = 6.660 kN
4
Substituting the results into Eq. [1] yields:
w = 21.9 kN>m
Ans.
194
250 mm
150 mm
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•4–101.
The rigid lever arm is supported by two A-36 steel
wires having the same diameter of 4 mm. If a force of
P = 3 kN is applied to the handle, determine the force
developed in both wires and their corresponding elongations.
Consider A-36 steel as an elastic-perfectly plastic material.
P
450 mm
150 mm
150 mm
30⬚
A
C
300 mm
B
Equation of Equilibrium. Refering to the free-body diagram of the lever shown
in Fig. a,
FAB (300) + FCD (150) - 3 A 103 B (450) = 0
a + ©ME = 0;
2FAB + FCD = 9 A 103 B
(1)
Elastic Analysis. Assuming that both wires AB and CD behave as linearly elastic,
the compatibility equation can be written by referring to the geometry of Fig. b.
dAB = a
300
bd
150 CD
dAB = 2dCD
(2)
FAB L
FCD L
= 2a
b
AE
AE
FAB = 2FCD
(3)
Solving Eqs. (1) and (3),
FCD = 1800 N
FAB = 3600 N
Normal Stress.
sCD =
FCD
1800
=
= 143.24 MPa 6 (sY)st
p
ACD
A 0.0042 B
(O.K.)
FAB
3600
=
= 286.48 MPa 7 (sY)st
p
2
AAB
0.004
A
B
(N.G.)
4
sAB =
4
Since wire AB yields, the elastic analysis is not valid. The solution must be reworked
using
FAB = (sY)st AAB = 250 A 106 B c
p
A 0.0042 B d
4
Ans.
= 3141.59 N = 3.14 kN
Substituting this result into Eq. (1),
FCD = 2716.81 N = 2.72 kN
sCD =
Ans.
FCD
2716.81
=
= 216.20 MPa 6 (sY)st
p
ACD
A 0.0042 B
(O.K.)
4
195
D
E
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4–101.
Continued
Since wire CD is linearly elastic, its elongation can be determined by
dCD =
2716.81(300)
FCD LCD
=
p
ACD Est
A 0.0042 B (200) A 109 B
4
Ans.
= 0.3243 mm = 0.324 mm
From Eq. (2),
dAB = 2dCD = 2(0.3243) = 0.649 mm
Ans.
196
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4–102. The rigid lever arm is supported by two A-36 steel
wires having the same diameter of 4 mm. Determine the
smallest force P that will cause (a) only one of the wires to
yield; (b) both wires to yield. Consider A-36 steel as an
elastic-perfectly plastic material.
P
450 mm
150 mm
150 mm
30⬚
A
C
300 mm
B
Equation of Equilibrium. Refering to the free-body diagram of the lever arm shown
in Fig. a,
a + ©ME = 0;
FAB (300) + FCD (150) - P(450) = 0
2FAB + FCD = 3P
(1)
Elastic Analysis. The compatibility equation can be written by referring to the
geometry of Fig. b.
dAB = a
300
bd
150 CD
dAB = 2dCD
FAB L
FCD L
= 2a
b
AE
AE
FCD =
1
F
2 AB
(2)
Assuming that wire AB is about to yield first,
FAB = (sY)st AAB = 250 A 106 B c
p
A 0.0042 B d = 3141.59 N
4
From Eq. (2),
FCD =
1
(3141.59) = 1570.80 N
2
Substituting the result of FAB and FCD into Eq. (1),
P = 2618.00 N = 2.62 kN
Ans.
Plastic Analysis. Since both wires AB and CD are required to yield,
FAB = FCD = (sY)st A = 250 A 106 B c
p
A 0.0042 B d = 3141.59 N
4
Substituting this result into Eq. (1),
197
D
E
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4–103. The three bars are pinned together and subjected
to the load P. If each bar has a cross-sectional area A, length
L, and is made from an elastic perfectly plastic material, for
which the yield stress is sY, determine the largest load
(ultimate load) that can be supported by the bars, i.e., the
load P that causes all the bars to yield. Also, what is the
horizontal displacement of point A when the load reaches
its ultimate value? The modulus of elasticity is E.
B
L
u
C
L
P
u
L
D
P = 3141.59 N = 3.14 kN
A
Ans.
When all bars yield, the force in each bar is, FY = sYA
+ ©F = 0;
:
x
P - 2sYA cos u - sYA = 0
P = sYA(2 cos u + 1)
Ans.
Bar AC will yield first followed by bars AB and AD.
dAB = dAD =
dA =
FY(L)
sYAL
sYL
=
=
AE
AE
E
dAB
sYL
=
cos u
E cos u
Ans.
*4–104. The rigid beam is supported by three 25-mm
diameter A-36 steel rods. If the beam supports the force of
P = 230 kN, determine the force developed in each rod.
Consider the steel to be an elastic perfectly-plastic material.
D
F
E
600 mm
P
Equation of Equilibrium. Referring to the free-body diagram of the beam shown in
Fig. a,
+ c ©Fy = 0;
FAD + FBE + FCF - 230 A 103 B = 0
(1)
FBE + 3FCF = 460 A 103 B
(2)
FBE(400) + FCF(1200) - 230 A 103 B (800) = 0
a + ©MA = 0;
Elastic Analysis. Referring to the deflection diagram of the beam shown in Fig. b,
the compatibility equation can be written as
dBE = dAD + a
dBE =
dCF - dAD
b(400)
1200
2
1
d
+ dCF
3 AD
3
FBEL
2 FCDL
1 FCF L
= a
b + a
b
AE
3 AE
3 AE
FBE =
2
1
F
+ FCF
3 AD
3
(3)
Solving Eqs. (1), (2), and (3)
FCF = 131 428.57 N
FBE = 65 714.29 N FAD = 32 857.14 N
198
A
B
400 mm
400 mm
C
400 mm
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4–104.
Continued
Normal Stress.
sCF =
FCF
131428.57
=
= 267.74 MPa 7 (sY)st
p
ACF
A 0.0252 B
(N.G.)
FBE
65714.29
=
= 133.87 MPa 6 (sY)st
p
ABE
A 0.0252 B
(O.K.)
FAD
32857.14
=
= 66.94 MPa 6 (sY)st
p
AAD
A 0.0252 B
(O.K.)
4
sBE =
4
sAD =
4
Since rod CF yields, the elastic analysis is not valid. The solution must be
reworked using
FCF = (sY)st ACF = 250 A 106 B c
p
A 0.0252 B d = 122 718.46 N = 123 kN
4
Ans.
Substituting this result into Eq. (2),
FBE = 91844.61 N = 91.8 kN
Ans.
Substituting the result for FCF and FBE into Eq. (1),
FAD = 15436.93 N = 15.4 kN
sBE =
Ans.
FBE
91844.61
=
= 187.10 MPa 6 (sY)st
p
ABE
A 0.0252 B
(O.K.)
FAD
15436.93
=
= 31.45 MPa 6 (sY)st
p
AAD
A 0.0252 B
(O.K.)
4
sAD =
4
199
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•4–105.
The rigid beam is supported by three 25-mm
diameter A-36 steel rods. If the force of P = 230 kN is
applied on the beam and removed, determine the residual
stresses in each rod. Consider the steel to be an elastic
perfectly-plastic material.
D
600 mm
P
Equation of Equilibrium. Referring to the free-body diagram of the beam shown in
Fig. a,
+ c ©Fy = 0;
FAD + FBE + FCF - 230 A 103 B = 0
(1)
FBE + 3FCF = 460 A 103 B
(2)
FBE(400) + FCF(1200) - 230 A 103 B (800) = 0
a + ©MA = 0;
Elastic Analysis. Referring to the deflection diagram of the beam shown in Fig. b,
the compatibility equation can be written as
dBE = dAD + a
dBE =
dCF - dAD
b(400)
1200
2
1
d
+ dCF
3 AD
3
(3)
FBE L
2 FCD L
1 FCF L
= a
b + a
b
AE
3 AE
3 AE
FBE =
2
1
F
+ FCF
3 AD
3
(4)
Solving Eqs. (1), (2), and (4)
FCF = 131428.57 N
FBE = 65714.29 N
FAD = 32857.14 N
Normal Stress.
sCF =
FCF
131428.57
=
= 267.74 MPa (T) 7 (sY)st
p
ACF
A 0.0252 B
(N.G.)
FBE
65714.29
=
= 133.87 MPa (T) 6 (sY)st
p
ABE
A 0.0252 B
(O.K.)
FAD
32857.14
=
= 66.94 MPa (T) 6 (sY)st
p
AAD
A 0.0252 B
(O.K.)
4
sBE =
4
sAD =
4
Since rod CF yields, the elastic analysis is not valid. The solution must be
reworked using
sCF = (sY)st = 250 MPa (T)
FCF = sCF ACF = 250 A 106 B c
F
E
p
A 0.0252 B d = 122718.46 N
4
200
A
B
400 mm
400 mm
C
400 mm
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4–105.
Continued
Substituting this result into Eq. (2),
FBE = 91844.61 N
Substituting the result for FCF and FBE into Eq. (1),
FAD = 15436.93N
sBE =
FBE
91844.61
=
= 187.10 MPa (T) 6 (sY)st
p
ABE
A 0.0252 B
(O.K.)
FAD
15436.93
=
= 31.45 MPa (T) 6 (sY)st
p
AAD
A 0.0252 B
(O.K.)
4
sAD =
4
Residual Stresses. The process of removing P can be represented by applying the
force P¿ , which has a magnitude equal to that of P but is opposite in sense, Fig. c.
Since the process occurs in a linear manner, the corresponding normal stress must
have the same magnitude but opposite sense to that obtained from the elastic
analysis. Thus,
œ
sCF
= 267.74 MPa (C)
œ
sBE
= 133.87 MPa (C)
œ
sAD
= 66.94 MPa (C)
Considering the tensile stress as positive and the compressive stress as negative,
œ
= 250 + (-267.74) = -17.7 MPa = 17.7 MPa (C)
(sCF)r = sCF + sCF
Ans.
œ
= 187.10 + (-133.87) = 53.2 MPa (T)
(sBE)r = sBE + sBE
Ans.
œ
(sAD)r = sAD + sAD
= 31.45 + (-66.94) = -35.5 MPa = 35.5 MPa (C)
Ans.
201
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4–106. The distributed loading is applied to the rigid
beam, which is supported by the three bars. Each bar has a
cross-sectional area of 1.25 in2 and is made from a
s(ksi)
material having a stress–strain diagram that can be
approximated by the two line segments shown. If a load of
60
w = 25 kip>ft is applied to the beam, determine the stress
in each bar and the vertical displacement of the beam.
4 ft
A
B
5 ft
36
A
0.0012
a + ©MB = 0;
0.2
FC(4) - FA(4) = 0;
FA = FC = F
+ c ©Fy = 0;
2F + FB - 200 = 0
(1)
Since the loading and geometry are symmetrical, the bar will remain horizontal.
Therefore, the displacement of the bars is the same and hence, the force in each bar
is the same. From Eq. (1).
F = FB = 66.67 kip
Thus,
sA = sB = sC =
66.67
= 53.33 ksi
1.25
Ans.
From the stress-strain diagram:
60 - 36
53.33 - 36
=
:
e - 0.0012
0.2 - 0.0012
e = 0.14477 in.>in.
d = eL = 0.14477(5)(12) = 8.69 in.
Ans.
202
4 ft
∋ (in./in.)
B
C
w
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4–107. The distributed loading is applied to the
rigid beam, which is supported by the three bars.
Each bar has a cross-sectional area of 0.75 in2 and is
s(ksi)
made from a material having a stress–strain diagram
that can be approximated by the two line segments
60
shown. Determine the intensity of the distributed
loading w needed to cause the beam to be displaced
36
downward 1.5 in.
0.0012
a + ©MB = 0;
FC(4) - FA(4) = 0;
+ c ©Fy = 0;
2F + FB - 8 w = 0
4 ft
A
A
0.2
(1)
Since the system and the loading are symmetrical, the bar will remain horizontal.
Hence the displacement of the bars is the same and the force supported by each bar
is the same.
From Eq. (1),
(2)
From the stress-strain diagram:
e =
1.5
= 0.025 in.>in.
5 (12)
60 - 36
s - 36
=
;
0.025 - 0.0012
0.2 - 0.0012
s = 38.87 ksi
Hence F = sA = 38.87 (0.75) = 29.15 kip
From Eq. (2), w = 10.9 kip>ft
Ans.
203
B
5 ft
FA = FC = F
FB = F = 2.6667 w
4 ft
∋ (in./in.)
B
C
w
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*4–108. The rigid beam is supported by the three posts A,
B, and C of equal length. Posts A and C have a diameter of
75 mm and are made of aluminum, for which Eal = 70 GPa
and 1sY2al = 20 MPa. Post B has a diameter of 20 mm and
is made of brass, for which Ebr = 100 GPa and
1sY2br = 590 MPa. Determine the smallest magnitude of P
so that (a) only rods A and C yield and (b) all the posts yield.
P
A
B
FA = FC = Fal
Fat + 2Fat - 2P = 0
+ c ©Fy = 0;
(1)
(a) Post A and C will yield,
Fal = (st)alA
= 20(104)(pa )(0.075)2
= 88.36 kN
(Eal)r =
(sr)al
20(104)
= 0.0002857
=
Eal
70(104)
Compatibility condition:
dbr = dal
= 0.0002857(L)
Fbr (L)
p
2
(0.02)
(100)(104)
4
= 0.0002857 L
Fbr = 8.976 kN
sbr =
8.976(103)
p
3
4 (0.02 )
= 28.6 MPa 6 sr
OK.
From Eq. (1),
8.976 + 2(88.36) - 2P = 0
P = 92.8 kN
Ans.
(b) All the posts yield:
Fbr = (sr)brA
= (590)(104)(p4 )(0.022)
= 185.35 kN
Fal = 88.36 kN
From Eq. (1); 185.35 + 2(88.36) - 2P = 0
P = 181 kN
Ans.
204
C
br
al
2m
©MB = 0;
P
2m
2m
al
2m
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•4–109.
The rigid beam is supported by the three posts
A, B, and C. Posts A and C have a diameter of 60 mm
and are made of aluminum, for which Eal = 70 GPa and
1sY2al = 20 MPa. Post B is made of brass, for which
Ebr = 100 GPa and 1sY2br = 590 MPa. If P = 130 kN,
determine the largest diameter of post B so that all the
posts yield at the same time.
P
A
B
2(Fg)al + Fbr - 260 = 0
(1)
(Fal)g = (sg)al A
= 20(106)(p4 )(0.06)2 = 56.55 kN
From Eq. (1),
2(56.55) + Fbr - 260 = 0
Fbr = 146.9 kN
(sg)br = 590(106) =
146.9(103)
p
3
4 (dB)
dB = 0.01779 m = 17.8 mm
Ans.
205
C
br
al
2m
+ c ©Fy = 0;
P
2m
2m
al
2m
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4–110. The wire BC has a diameter of 0.125 in. and the
material has the stress–strain characteristics shown in the
figure. Determine the vertical displacement of the handle at
D if the pull at the grip is slowly increased and reaches a
magnitude of (a) P = 450 lb, (b) P = 600 lb.
C
40 in.
A
D
B
50 in.
30 in.
P
s (ksi)
Equations of Equilibrium:
a + ©MA = 0;
FBC(50) - P(80) = 0
(a) From Eq. [1] when P = 450 lb,
[1]
80
70
FBC = 720 lb
Average Normal Stress and Strain:
sBC =
FBC
720
= 58.67 ksi
= p
2
ABC
4 (0.125 )
P (in./in.)
0.007
From the Stress–Strain diagram
58.67
70
=
;
eBC
0.007
eBC = 0.005867 in.>in.
Displacement:
dBC = eBCLBC = 0.005867(40) = 0.2347 in.
dBC
dD
=
;
80
50
dD =
8
(0.2347) = 0.375 in.
5
(b) From Eq. [1] when P = 600 lb,
Ans.
FBC = 960 lb
Average Normal Stress and Strain:
sBC =
FBC
960
= p
= 78.23 ksi
2
ABC
4 (0.125)
From Stress–Strain diagram
78.23 - 70
80 - 70
=
eBC - 0.007
0.12 - 0.007
eBC = 0.09997 in.>in.
Displacement:
dBC = eBCLBC = 0.09997(40) = 3.9990 in.
dD
dBC
=
;
80
50
dD =
8
(3.9990) = 6.40 in.
5
Ans.
206
0.12
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4–111. The bar having a diameter of 2 in. is fixed
connected at its ends and supports the axial load P. If the
material is elastic perfectly plastic as shown by the
stress–strain diagram, determine the smallest load P needed
to cause segment CB to yield. If this load is released,
determine the permanent displacement of point C.
P
A
C
2 ft
B
3 ft
s (ksi)
20
0.001
When P is increased, region AC will become plastic first, then CB will become
plastic. Thus,
FA = FB = sA = 20(p)(1)2 = 62.832 kip
+ ©F = 0;
:
x
FA + FB - P = 0
(1)
P = 2(62.832) = 125.66 kip
P = 126 kip
Ans.
The deflection of point C is,
dC = eL = (0.001)(3)(12) = 0.036 in. ;
Consider the reverse of P on the bar.
FB ¿(3)
FA ¿(2)
=
AE
AE
FA ¿ = 1.5 FB ¿
So that from Eq. (1)
FB ¿ = 0.4P
FA ¿ = 0.6P
dC ¿ =
0.4(P)(3)(12)
0.4(125.66)(3)(12)
FB ¿L
= 0.02880 in. :
=
=
AE
AE
p(1)2(20>0.001)
¢d = 0.036 - 0.0288 = 0.00720 in. ;
Ans.
207
P (in./in.)
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*4–112. Determine the elongation of the bar in
Prob. 4–111 when both the load P and the supports are
removed.
P
A
C
2 ft
B
3 ft
s (ksi)
20
0.001
When P is increased, region AC will become plastic first, then CB will become
plastic. Thus,
FA = FB = sA = 20(p)(1)2 = 62.832 kip
+ ©F = 0;
:
x
FA + FB - P = 0
(1)
P = 2(62.832) = 125.66 kip
P = 126 kip
Ans.
The deflection of point C is,
dC = eL = (0.001)(3)(12) = 0.036 in. ;
Consider the reverse of P on the bar.
FB ¿(3)
FA ¿(2)
=
AE
AE
FA ¿ = 1.5 FB ¿
So that from Eq. (1)
FB ¿ = 0.4P
FA ¿ = 0.6P
The resultant reactions are
FA ¿¿ = FB ¿¿ = -62.832 + 0.6(125.66) = 62.832 - 0.4(125.66) = 12.568 kip
When the supports are removed the elongation will be,
d =
12.568(5)(12)
PL
= 0.0120 in.
=
AE
p(1)2(20>0.001)
Ans.
208
P (in./in.)
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s
•4–113.
A material has a stress–strain diagram that can be
described by the curve s = cP1>2. Determine the deflection
d of the end of a rod made from this material if it has a
length L, cross-sectional area A, and a specific weight g.
L
1
2
A
s2 = c2 e
s = ce ;
s2(x) = c2e(x)
P
(1) d
P(x)
;
A
However s(x) =
e(x) =
dd
dx
From Eq. (1),
P2(x)
= c2
A2
P2(x)
dd
=
dx
A2c2
dd
;
dx
L
d =
1
1
P2(x) dx = 2 2
(gAx)2 dx
2 2
Ac L
A c L0
g2
=
d =
L
c2 L0
x2 dx =
g2 x3 L
冷
c2 3 0
g3L3
Ans.
3c2
4–114. The 2014-T6 aluminum rod has a diameter of 0.5 in.
and is lightly attached to the rigid supports at A and B when
T1 = 70°F. If the temperature becomes T2 = -10°F, and
an axial force of P = 16 lb is applied to the rigid collar as
shown, determine the reactions at A and B.
A
B
P/2
P/2
5 in.
8 in.
+ 0 = ¢ - ¢ + d
:
B
T
B
0 = p
0.016(5)
2
3
4 (0.5 )(10.6)(10 )
- 12.8(10 - 6)[70° - (-10°)](13) + p
FB(13)
2
3
4 (0.5 )(10.6)(10 )
FB = 2.1251 kip = 2.13 kip
+ ©F = 0;
:
x
Ans.
2(0.008) + 2.1251 - FA = 0
FA = 2.14 kip
Ans.
4–115. The 2014-T6 aluminum rod has a diameter of
0.5 in. and is lightly attached to the rigid supports at A
and B when T1 = 70°F. Determine the force P that must
be applied to the collar so that, when T = 0°F, the
reaction at B is zero.
+
:
A
P/2
5 in.
0 = ¢ B - ¢ T + dB
0 = p
P(5)
2
3
4 (0.5 )(10.6)(10 )
B
P/2
- 12.8(10 - 6)[(70)(13)] + 0
P = 4.85 kip
Ans.
209
8 in.
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*4–116. The rods each have the same 25-mm diameter
and 600-mm length. If they are made of A-36 steel,
determine the forces developed in each rod when the
temperature increases to 50° C.
C
600 mm
60⬚
B
A
60⬚
600 mm
Equation of Equilibrium: Referring to the free-body diagram of joint A shown in
Fig. a,
+ c ©Fx = 0;
FAD sin 60° - FAC sin 60° = 0
+ ©F = 0;
:
x
FAB - 2F cos 60° = 0
FAC = FAD = F
FAB = F
(1)
Compatibility Equation: If AB and AC are unconstrained, they will have
a free expansion of A dT B AB = A dT B AC = ast ¢TL = 12(10 - 6)(50)(600) = 0.36 mm.
Referring to the initial and final position of joint A,
dFAB - A dT B AB = a dT ¿ b
AC
- dFAC ¿
Due to symmetry, joint A will displace horizontally, and dAC ¿ =
a dT ¿ b
AC
dAC
= 2dAC. Thus,
cos 60°
= 2(dT)AC and dFAC ¿ = 2dFAC. Thus, this equation becomes
dFAB - A dT B AB = 2 A dT B AC - 2dAC
A
FAB (600)
B
p
2
9
4 0.025 (200)(10 )
- 0.36 = 2(0.36) - 2 C
A
F(600)
B
p
2
9
4 0.025 (200)(10 )
FAB + 2F = 176 714.59
S
(2)
Solving Eqs. (1) and (2),
FAB = FAC = FAD = 58 904.86 N = 58.9 kN
Ans.
210
D
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•4–117.
Two A-36 steel pipes, each having a crosssectional area of 0.32 in2, are screwed together using a
union at B as shown. Originally the assembly is adjusted so
that no load is on the pipe. If the union is then tightened
so that its screw, having a lead of 0.15 in., undergoes two full
turns, determine the average normal stress developed in the
pipe. Assume that the union at B and couplings at A and C
are rigid. Neglect the size of the union. Note: The lead would
cause the pipe, when unloaded, to shorten 0.15 in. when the
union is rotated one revolution.
B
A
3 ft
C
2 ft
The loads acting on both segments AB and BC are the same since no external load
acts on the system.
0.3 = dB>A + dB>C
0.3 =
P(2)(12)
P(3)(12)
3
0.32(29)(10 )
+
0.32(29)(103)
P = 46.4 kip
P
46.4
=
= 145 ksi
A
0.32
sAB = sBC =
Ans.
4–118. The brass plug is force-fitted into the rigid casting.
The uniform normal bearing pressure on the plug is
estimated to be 15 MPa. If the coefficient of static friction
between the plug and casting is ms = 0.3, determine the
axial force P needed to pull the plug out. Also, calculate the
displacement of end B relative to end A just before the plug
starts to slip out. Ebr = 98 GPa.
100 mm
B
15 MPa
P - 4.50(106)(2)(p)(0.02)(0.1) = 0
P = 56.549 kN = 56.5 kN
Ans.
Displacement:
dB>A = a
PL
AE
0.1 m
56.549(103)(0.15)
=
2
9
p(0.02 )(98)(10 )
+
L0
P
A
Equations of Equilibrium:
+ ©F = 0;
:
x
150 mm
0.56549(106) x dx
p(0.022)(98)(109)
= 0.00009184 m = 0.0918 mm
Ans.
211
20 mm
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4–119. The assembly consists of two bars AB and CD of
the same material having a modulus of elasticity E1 and
coefficient of thermal expansion a1, and a bar EF having a
modulus of elasticity E2 and coefficient of thermal
expansion a2. All the bars have the same length L and
cross-sectional area A. If the rigid beam is originally
horizontal at temperature T1, determine the angle it makes
with the horizontal when the temperature is increased
to T2.
D
B
L
A
C
d
Equations of Equilibrium:
a + ©MC = 0;
FAB = FEF = F
+ c ©Fy = 0;
FCD - 2F = 0
[1]
Compatibility:
dAB = (dAB)T - (dAB)F
dCD = (dCD)T + (dCD)F
dEF = (dEF)T - (dEF)F
From the geometry
dCD - dAB
dEF - dAB
=
d
2d
2dCD = dEF + dAB
2 C (dCD)T + (dCD)F D = (dEF)T - (dEF)F + (dAB)T - (dAB)F
2 B a1 (T2 - T1) L +
FCD (L)
R
AE1
= a2 (T2 - T1) L -
F(L)
F(L)
+ a1 (T2 - T1) L AE2
AE1
[2]
Substitute Eq. [1] into [2].
2a1 (T2 - T1) L +
4FL
FL
FL
= a2 (T2 - T1)L + a1 (T2 - T1)L AE1
AE2
AE1
F
5F
+
= a2 (T2 - T1) - a1 (T2 - T1)
AE1
AE2
F¢
5E2 + E1
b = (T2 - T1)(a2 - a1) ;
AE1E2
F =
AE1E2 (T2 - T1)(a2 - a1)
5E2 + E1
(dEF)T = a2 (T2 - T1) L
(dEF)F =
AE1E2 (T2 - T1)(a2 - a1)(L)
E1 (T2 - T1)(a2 - a1)(L)
=
AE2 (5E2 + E1)
5E2 + E1
dEF = (dEF)T - (dEF)F =
a2 L(T2 - T1)(5E2 - E1) - E1L(T2 - T1)(a2 - a1)
5E2 + E1
(dAB)T = a1 (T2 - T1) L
(dAB)F =
AE1E2 (T2 - T1)(a2 - a1)(L)
E2 (T2 - T1)(a2 - a1)(L)
=
AE1 (5E2 + E1)
5E2 + E1
dAB = (dAB)T - (dAB)F =
F
a1 L(5E2 + E1)(T2 - T1) - E2 L(T2 - T1)(a2 - a1)
5E2 + E1
212
E
d
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4–119.
Continued
dEF - dAB =
L(T2 - T1)
[a2 (5E2 + E1) - E1 (a2 - a1) - a1 (5E2 + E1)
5E2 + E1
+ E2 (a2 - a1)]
=
L(T2 - T1)
C (5E2 + E1)(a2 - a1) + (a2 - a1)(E2 - E1) D
5E2 + E1
=
L(T2 - T1)(a2 - a1)
(5E2 + E1 + E2 - E1)
5E2 + E1
=
L(T2 - T1)(a2 - a1)(6E2)
5E2 + E1
u =
3E2L(T2 - T1)(a2 - a1)
dEF - dAB
=
2d
d(5E2 + E1)
Ans.
*4–120. The rigid link is supported by a pin at A and two
A-36 steel wires, each having an unstretched length of 12 in.
and cross-sectional area of 0.0125 in2. Determine the force
developed in the wires when the link supports the vertical
load of 350 lb.
12 in.
C
5 in.
B
Equations of Equilibrium:
a + ©MA = 0;
4 in.
A
-FC(9) - FB (4) + 350(6) = 0
[1]
Compatibility:
6 in.
dC
dB
=
4
9
350 lb
FC(L)
FB (L)
=
4AE
9AE
9FB - 4FC = 0‚
[2]
FB = 86.6 lb
Ans.
FC = 195 lb
Ans.
Solving Eqs. [1] and [2] yields:
213