106
CHAPTER 2
Axially Loaded Members
Problem 2.5-3 A rigid bar of weight W 750 lb hangs from three
equally spaced wires, two of steel and one of aluminum (see figure).
The diameter of the wires is 18 in. Before they were loaded, all three
wires had the same length.
What temperature increase T in all three wires will result in the
entire load being carried by the steel wires? (Assume Es 30 106 psi,
s 6.5 106/F, and a 12 106/F.)
W = 750 lb
Solution 2.5-3
Bar supported by three wires
2 increase in length of a steel wire due to load
W/2
S
WL
2Es As
3 increase in length of aluminum wire due to
temperature increase T
S steel
a(T)L
Rigid
Bar
For no load in the aluminum wire:
1 2 3
A aluminum
W 750 lb
d
As
s (T)L
1
in.
8
WL
a (T )L
2Es As
or
d2
0.012272 in.2
4
T
Es 30 106 psi
W
2Es As (a s )
Substitute numerical values:
EsAs 368,155 lb
T
s 6.5 106/F
185F
a 12 106/F
NOTE: If the temperature increase is larger than T,
the aluminum wire would be in compression, which
is not possible. Therefore, the steel wires continue to
carry all of the load. If the temperature increase is
less than T, the aluminum wire will be in tension
and carry part of the load.
L Initial length of wires
S
S
1
3
W
2
750 lb
(2)(368,155 lb)(5.5 10 6F)
2
W
2
1 increase in length of a steel wire due to
temperature increase T
s (T)L
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SECTION 2.5
Problem 2.5-4 A steel rod of diameter 15 mm is held snugly (but
without any initial stresses) between rigid walls by the arrangement
shown in the figure.
Calculate the temperature drop T (degrees Celsius) at which
the average shear stress in the 12-mm diameter bolt becomes 45 MPa.
(For the steel rod, use 12 106/C and E 200 GPa.)
Solution 2.5-4
B
107
Thermal Effects
12 mm diameter bolt
15 mm
Steel rod with bolted connection
Solve for T: T
15 mm
AB
d2B
where dB diameter of bolt
4
12 mm diameter bolt
R rod
B bolt
P tensile force in steel rod due to temperature drop
T
AR cross-sectional area of steel rod
From Eq. (2-17) of Example 2-7: P EAR(T)
Bolt is in double shear.
V shear force acting over one cross section of the
bolt
2tAB
EAR
AR
d2R
where dR diameter of steel rod
4
T
2td2B
Ed2R
SUBSTITUTE NUMERICAL VALUES:
45 MPa
dB 12 mm
12 106/C
T
dR 15 mm
E 200 GPa
2(45 MPa)(12 mm) 2
(200 GPa)(12 10 6C)(15 mm) 2
T 24C
1
V P2 EAR(T)
2
average shear stress on cross section of the bolt
AB cross-sectional area of bolt
t
EAR (T)
V
AB
2AB
Problem 2.5-5 A bar AB of length L is held between rigid supports and
heated nonuniformly in such a manner that the temperature increase T at
distance x from end A is given by the expression T TBx3/L3, where
TB is the increase in temperature at end B of the bar (see figure).
Derive a formula for the compressive stress c in the bar. (Assume
that the material has modulus of elasticity E and coefficient of thermal
expansion .)
0
A
B
x
L
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TB
108
CHAPTER 2
Axially Loaded Members
Solution 2.5-5
Bar with nonuniform temperature change
d Elongation of element dx
TB
d (T )dx (TB )
0
A
x3
dx
L3
elongation of bar
B
x
d
At distance x:
(T )
B
x3
3
L
dx
1
(TB )L
4
COMPRESSIVE FORCE P REQUIRED TO SHORTEN THE BAR
x3
T TB 3
L
BY THE AMOUNT
P
REMOVE THE SUPPORT AT END B OF THE BAR:
A
COMPRESSIVE STRESS IN THE BAR
dx
EA 1
EA(TB )
L
4
sc
P E(TB )
A
4
Consider an element dx at a distance x from end A.
Problem 2.5-6 A plastic bar ACB having two different solid circular cross
sections is held between rigid supports as shown in the figure. The diameters
in the left- and right-hand parts are 50 mm and 75 mm, respectively. The
corresponding lengths are 225 mm and 300 mm. Also, the modulus of elasticity
E is 6.0 GPa, and the coefficient of thermal expansion is 100 106/C. The
bar is subjected to a uniform temperature increase of 30C.
Calculate the following quantities: (a) the compressive force P in the bar;
(b) the maximum compressive stress c; and (c) the displacement C of point C.
Solution 2.5-6
75 mm
50 mm C
225 mm
300 mm
Bar with rigid supports
75 mm
50 mm
225 mm
LEFT-HAND PART:
d2 75 mm
2
d (75 mm) 2 4417.9 mm2
4 2 4
(a) COMPRESSIVE FORCE P
Remove the support at end B.
d1 50 mm
A1 d21 (50 mm) 2
4
4
1963.5 mm2
L2 300 mm
A2
300 mm
100 106/C
E 6.0 GPa
L1 225 mm
RIGHT-HAND PART:
B
T 30C
A
L1
A1
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L2
A2
SECTION 2.5
T elongation due to temperature
109
(b) MAXIMUM COMPRESSIVE STRESS
P (T)(L1L2)
sc
1.5750 mm
P
51.78 kN
26.4 MPa
A1 1963.5 mm2
(c) DISPLACEMENT OF POINT C
P shortening due to P
Thermal Effects
C Shortening of AC
PL1 PL2
EA1 EA2
P(19.0986 109 m/N11.3177 109 m/N)
C
PL1
(T )L1
EA1
0.9890 mm 0.6750 mm
(30.4163 109 m/N)P
C 0.314 mm
(P newtons)
Compatibility: T P
(Positive means AC shortens and point C displaces to
the left.)
1.5750 103 m (30.4163 109 m/N)P
P 51,781 NorP 51.8 kN
d1
Problem 2.5-7 A circular steel rod AB (diameter d1 1.0 in., length
A
L1 3.0 ft) has a bronze sleeve (outer diameter d2 1.25 in., length
L2 1.0 ft) shrunk onto it so that the two parts are securely bonded
(see figure).
Calculate the total elongation of the steel bar due to a temperature rise
T 500F. (Material properties are as follows: for steel, Es 30 106 psi
and s 6.5 106/F; for bronze, Eb 15 106 psi and b 11 106/F.)
Solution 2.5-7
d2
d1
L2
L1
SUBSTITUTE NUMERICAL VALUES:
s 6.5 106/F
b 11 106/F
L2
Es 30 106 psi
Eb 15 106 psi
L1
d1 1.0 in.
L2 12 in.
As
ELONGATION OF THE TWO OUTER PARTS OF THE BAR
1 s(T)(L1 L2)
2
d 0.78540 in.2
4 1
d2 1.25 in.
2
(d d12) 0.44179 in.2
4 2
(6.5 106/F)(500F)(36 in. 12 in.)
Ab
0.07800 in.
T 500F
ELONGATION OF THE MIDDLE PART OF THE BAR
The steel rod and bronze sleeve lengthen the same
amount, so they are in the same condition as the bolt
and sleeve of Example 2-8. Thus, we can calculate
the elongation from Eq. (2-21):
2
Steel rod with bronze sleeve
L1 36 in.
d2
L2 12.0 in.
2 0.04493 in.
TOTAL ELONGATION
1 2 0.123 in.
(s Es As b Eb Ab )(T)L2
Es As Eb Ab
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110
CHAPTER 2
Axially Loaded Members
Problem 2.5-8 A brass sleeve S is fitted over a steel bolt B (see figure),
and the nut is tightened until it is just snug. The bolt has a diameter
dB 25 mm, and the sleeve has inside and outside diameters
d1 26 mm and d2 36 mm, respectively.
Calculate the temperature rise T that is required to produce a
compressive stress of 25 MPa in the sleeve. (Use material properties
as follows: for the sleeve, S 21 106/C and ES 100 GPa;
for the bolt, B 10 106/C and EB 200 GPa.)
(Suggestion: Use the results of Example 2-8.)
Solution 2.5-8
dB
Sleeve (S)
Bolt (B)
SUBSTITUTE NUMERICAL VALUES:
S 25 MPa
d2 36 mm
Steel Bolt
Brass Sleeve
Subscript S means sleeve.
Subscript B means bolt.
Use the results of Example 2-8.
S compressive force in sleeve
(S B )(T)ES EB AB
(Compression)
ES AS EB AB
SOLVE FOR T:
sS (ES AS EB AB )
(S B )ES EB AB
d1 26 mm
EB 200 GPa
S 21 106/C
B 10 106/C
AS
2
(d2 d21 ) (620 mm2 )
4
4
AB
(dB ) 2 (625 mm2 )
4
4
ES AS
1.496
EB AB
T
25 MPa (1.496)
(100 GPa)(11 10 6C)
T 34C
(Increase in temperature)
or
T
dB 25 mm
ES 100 GPa
1
EQUATION (2-20a):
T
d1
Brass sleeve fitted over a Steel bolt
sS
d2
sS
ES AS
1
ES (S B )
EB AB
Problem 2.5-9 Rectangular bars of copper and aluminum are held
by pins at their ends, as shown in the figure. Thin spacers provide a
separation between the bars. The copper bars have cross-sectional
dimensions 0.5 in. 2.0 in., and the aluminum bar has dimensions
1.0 in. 2.0 in.
Determine the shear stress in the 7/16 in. diameter pins if the
temperature is raised by 100F. (For copper, Ec 18,000 ksi and
c 9.5 106/F; for aluminum, Ea 10,000 ksi and
a 13 106/F.) Suggestion: Use the results of Example 2-8.
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Copper bar
Aluminum bar
Copper bar
SECTION 2.5
Solution 2.5-9
111
Thermal Effects
Rectangular bars held by pins
C
0.5 in. 2.0 in.
1.0 in. 2.0 in.
0.5 in. 2.0 in.
A
C
Pin
Diameter of pin: dP
Area of pin: AP
Copper
Aluminum
7
in. 0.4375 in.
16
2
d 0.15033 in.2
4 P
SUBSTITUTE NUMERICAL VALUES:
(3.5 10 6F)(100F)(18,000 ksi)(2 in.2 )
Pa Pc
18 2.0
1
10 2.0
4,500 lb
Area of two copper bars: Ac 2.0 in.2
Area of aluminum bar: Aa 2.0 in.2
FREE-BODY DIAGRAM OF PIN AT THE LEFT END
T 100F
Copper: Ec 18,000 ksi
c 9.5 106/F
Aluminum: Ea 10,000 ksi
a 13 106/F
Pc
2
Pa
Pc
2
Use the results of Example 2-8.
Find the forces Pa and Pc in the aluminum bar and
copper bar, respectively, from Eq. (2-19).
Replace the subscript S in that equation by a
(for aluminum) and replace the subscript B
by c (for copper), because for aluminum
is larger than for copper.
(a c )(T)Ea Aa Ec Ac
Pa Pc
Ea Aa Ec Ac
Note that Pa is the compressive force in the
aluminum bar and Pc is the combined tensile
force in the two copper bars.
Pa Pc
V shear force in pin
Pc /2
2,250 lb
average shear stress on cross section of pin
t
V
2,250 lb
AP 0.15033 in.2
t 15.0 ksi
(a c )(T)Ec Ac
Ec Ac
1
Ea Aa
Problem 2.5-10 A rigid bar ABCD is pinned at end A and supported by
two cables at points B and C (see figure). The cable at B has nominal
diameter dB 12 mm and the cable at C has nominal diameter dC 20 mm.
A load P acts at end D of the bar.
What is the allowable load P if the temperature rises by 60C and
each cable is required to have a factor of safety of at least 5 against its
ultimate load?
(Note: The cables have effective modulus of elasticity E 140 GPa
and coefficient of thermal expansion 12 106/C. Other properties
of the cables can be found in Table 2-1, Section 2.2.)
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dC
dB
B
2b
2b
D
b
P
112
CHAPTER 2
Solution 2.5-10
Axially Loaded Members
Rigid bar supported by two cables
FREE-BODY DIAGRAM OF BAR ABCD
TB
TC
RAH
SUBSTITUTE EQS. (3) AND (4) INTO EQ. (2):
2b
2b
RAV
TB force in cable B
dB 12 mm
TC force in cable C
E 140 GPa
EQUATION OF EQUILIBRIUM
SUBSTITUTE NUMERICAL VALUES INTO EQ. (5):
(Eq. 6)
2TB 4TC 5P
(Eq. 1)
DISPLACEMENT DIAGRAM
B
TB 0.2494 P 3,480
(Eq. 7)
TC 1.1253 P 1,740
(Eq. 8)
in which P has units of newtons.
MA 0 TB (2b) TC (4b) P(5b) 0
2b
(Eq. 5)
SOLVE SIMULTANEOUSLY EQS. (1) AND (6):
AC 173 mm2
12 106/C
2TB AC TC AB E(T)AB AC
in which TB and TC have units of newtons.
mm2
T 60C
or
or
TB(346) TC(76.7) 1,338,000
dC 20 mm
From Table 2-1:
AB 76.7
TC L
2TB L
(T)L
2(T)L
EAC
EAB
2b
SOLVE EQS. (7) AND (8) FOR THE LOAD P:
PB 4.0096 TB 13,953
(Eq. 9)
PC 0.8887 TC 1,546
(Eq. 10)
ALLOWABLE LOADS
From Table 2-1:
B
COMPATIBILITY:
(TB)ULT 102,000 N
(TC)ULT 231,000 N
Factor of safety 5
C
C 2B
(Eq. 2)
(TB)allow 20,400 N
(TC)allow 46,200 N
From Eq. (9): PB (4.0096)(20,400 N) 13,953 N
FORCE-DISPLACEMENT AND TEMPERATURE-
95,700 N
DISPLACEMENT RELATIONS
From Eq. (10): PC (0.8887)(46,200 N) 1546 N
B
TB L
(T )L
EAB
(Eq. 3)
C
TC L
(T )L
EAC
(Eq. 4)
39,500 N
Cable C governs.
Pallow 39.5 kN
Problem 2.5-11 A rigid triangular frame is pivoted at C and held by two identical horizontal wires at points A and B (see figure). Each wire has axial rigidity
EA 120 k and coefficient of thermal expansion 12.5 106/F.
(a) If a vertical load P 500 lb acts at point D, what are the tensile forces
TA and TB in the wires at A and B, respectively?
(b) If, while the load P is acting, both wires have their temperatures raised
by 180F, what are the forces TA and TB?
(c) What further increase in temperature will cause the wire at B to become
slack?
A
b
B
b
D
P
2b
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SECTION 2.5
Thermal Effects
Solution 2.5-11 Triangular frame held by two wires
FREE-BODY DIAGRAM OF FRAME
TA
(b) LOAD P AND TEMPERATURE INCREASE T
Force-displacement and temperaturedisplacement relations:
b
TB
B
b
D
A
TAL
(T)L
EA
(Eq. 8)
B
TBL
(T)L
EA
(Eq. 9)
2b
Substitute (8) and (9) into Eq. (2):
EQUATION OF EQUILIBRIUM
TAL
2TBL
(T)L
2(T)L
EA
EA
MC 0
P(2b) TA(2b) TB(b) 0
or
2TA TB 2P (Eq. 1)
A
A
b
B
B
b
A 2B
1
TA [4P EA(T ) ]
5
(Eq. 11)
2
TB [P EA(T ) ]
5
(Eq. 12)
Substitute numerical values:
P 500 lb
T 180F
EQUATION OF COMPATIBILITY
(Eq. 2)
EA 120,000 lb
12.5 106/F
1
TA (2000 lb 270 lb) 454 lb
5
(a) LOAD P ONLY
Force-displacement relations:
TA L
TB L
B
EA
EA
(Eq. 3, 4)
2
TB (500 lb 270 lb) 92 lb
5
(L length of wires at A and B.)
(c) WIRE B BECOMES SLACK
Substitute (3) and (4) into Eq. (2):
Set TB 0 in Eq. (12):
P EA(T)
TA L 2TB L
EA
EA
or
or
TA 2TB
(Eq. 5)
Solve simultaneously Eqs. (1) and (5):
TA
4P
2P
TB
5
5
Numerical values:
(Eqs. 6, 7)
T
P
500 lb
EA (120,000 lb)(12.5 10 6F)
333.3F
Further increase in temperature:
T 333.3F 180F
P 500 lb
TA 400 lb
(Eq. 10)
Solve simultaneously Eqs. (1) and (10):
DISPLACEMENT DIAGRAM
A
TA 2TB EA(T)
or
153F
TB 200 lb
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113
114
CHAPTER 2
Axially Loaded Members
Misfits and Prestrains
Problem 2.5-12 A steel wire AB is stretched between rigid supports
(see figure). The initial prestress in the wire is 42 MPa when the
temperature is 20C.
(a) What is the stress in the wire when the temperature drops
to 0C?
(b) At what temperature T will the stress in the wire become zero?
(Assume 14 106/C and E 200 GPa.)
Solution 2.5-12
Steel wire
Steel wire with initial prestress
From Eq. (2-18): 2 E(T)
s s1 s2 s1 E(T )
42 MPa (200 GPa)(14 10 6C)(20C)
Initial prestress: 1 42 MPa
42 MPa 56 MPa 98 MPa
Initial temperature: T1 20C
E 200 GPa
14
(b) TEMPERATURE WHEN STRESS EQUALS ZERO
106/C
1 2 0
(a) STRESS WHEN TEMPERATURE DROPS TO 0C
T2 0C
T 20C
Note: Positive T means a decrease in temperature
and an increase in the stress in the wire.
Negative T means an increase in temperature and a
decrease in the stress.
T
1 E(T) 0
s1
E
(Negative means increase in temp.)
T
42 MPa
15C
(200 GPa)(14 10 6C)
T 20C 15C 35C
Stress equals the initial stress 1 plus the
additional stress 2 due to the temperature drop.
Problem 2.5-13 A copper bar AB of length 25 in. is placed in position
at room temperature with a gap of 0.008 in. between end A and a rigid
restraint (see figure).
Calculate the axial compressive stress c in the bar if the temperature
rises 50F. (For copper, use 9.6 106/F and E 16 106 psi.)
0.008 in.
A
25 in.
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SECTION 2.5
Solution 2.5-13
S
115
Misfits and Prestrains
Bar with a gap
c stress in the bar
L 25 in.
EeC
S 0.008 in.
L
T 50F (increase)
9.6
EC E
[(T)L S]
L
L
Note: This result is valid only if (T)L S.
(Otherwise, the gap is not closed).
106/F
E 16 106 psi
Substitute numerical values:
elongation of the bar if it is free to expand
sc
(T)L
16 106 psi
[ (9.6 10 6F)(50F)(25 in.)
25 in.
0.008 in.] 2,560 psi
C elongation that is prevented by the support
(T)L S
eC strain in the bar due to the restraint
C /L
Problem 2.5-14 A bar AB having length L and axial rigidity EA is fixed
at end A (see figure). At the other end a small gap of dimension s exists
between the end of the bar and a rigid surface. A load P acts on the bar at
point C, which is two-thirds of the length from the fixed end.
If the support reactions produced by the load P are to be equal in
magnitude, what should be the size s of the gap?
Solution 2.5-14
2L
B
P
Bar with a gap (load P)
2L
COMPATIBILITY EQUATION
1 2 S
or
2PL RBL
S
3EA
EA
A
L length of bar
EQUILIBRIUM EQUATION
S size of gap
RA
EA axial rigidity
RB
RA reaction at end A (to the left)
Reactions must be equal; find S.
RB reaction at end B (to the left)
FORCE-DISPLACEMENT RELATIONS
2L
(Eq. 1)
P RA RB
P
1
RB
2
1
P( 2L
3)
EA
Reactions must be equal.
2
RBL
EA
Substitute for RB in Eq. (1):
2PL
PL
PL
SorS
3EA 2EA
6EA
NOTE: The gap closes when the load reaches the
value P/4. When the load reaches the value P, equal
to 6EAs/L, the reactions are equal (RA RB P/2).
When the load is between P/4 and P, RA is greater
than RB. If the load exceeds P, RB is greater than RA.
RA RB
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P 2RB
RB
P
2
116
CHAPTER 2
Axially Loaded Members
Problem 2.5-15 Wires B and C are attached to a support at the left-hand
end and to a pin-supported rigid bar at the right-hand end (see figure).
Each wire has cross-sectional area A 0.03 in.2 and modulus of elasticity
E 30 106 psi. When the bar is in a vertical position, the length of
each wire is L 80 in. However, before being attached to the bar, the
length of wire B was 79.98 in. and of wire C was 79.95 in.
Find the tensile forces TB and TC in the wires under the action of
a force P 700 lb acting at the upper end of the bar.
700 lb
B
b
b
80 in.
Solution 2.5-15
Wires B and C attached to a bar
P = 700 lb
Elongation of wires:
B SB 2
(Eq. 2)
C SC
(Eq. 3)
FORCE-DISPLACEMENT RELATIONS
L = 80 in.
B
P 700 lb
TC L
TBL
C
EA
EA
A 0.03 in.2
SOLUTION OF EQUATIONS
E 30106 psi
Combine Eqs. (2) and (4):
TBL
SB 2
EA
LB 79.98 in.
LC 79.95 in.
TCL
SC
EA
P = 700 lb
TC
(Eq. 6)
Combine Eqs. (3) and (5):
EQUILIBRIUM EQUATION
TB
(Eqs. 4, 5)
Mpin 0
Eliminate between Eqs. (6) and (7):
TC(b) TB(2b) P(3b)
TB 2TC
b
Pin
2TB TC 3P
(Eq. 1)
SC 80 in. LC 0.05 in.
(Eq. 8)
Solve simultaneously Eqs. (1) and (8):
TB
6P EASB 2EASC
5
5L
5L
TC
3P 2EASB 4EASC
5
5L
5L
DISPLACEMENT DIAGRAM
SB 80 in. LB 0.02 in.
EASB 2EASC
L
L
(Eq. 7)
SUBSTITUTE NUMERICAL VALUES:
B
SB
2
SC
L = 80 in.
EA
2250 lbin.
5L
TB 840 lb 45 lb 225 lb 660 lb
TC 420 lb 90 lb 450 lb 780 lb
(Both forces are positive, which means tension, as
required for wires.)
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SECTION 2.5
Problem 2.5-16 A rigid steel plate is supported by three posts of
high-strength concrete each having an effective cross-sectional area
A 40,000 mm2 and length L 2 m (see figure). Before the load
P is applied, the middle post is shorter than the others by an amount
s 1.0 mm.
Determine the maximum allowable load Pallow if the allowable
compressive stress in the concrete is allow 20 MPa. (Use E 30 GPa
for concrete.)
Solution 2.5-16
Misfits and Prestrains
S
s
Plate supported by three posts
P
EQUILIBRIUM EQUATION
Steel plate
2P1 P2 P
P1
P2
(Eq. 1)
P1
COMPATIBILITY EQUATION
1 shortening of outer posts
2 shortening of inner post
s size of gap 1.0 mm
1 2 s
L length of posts 2.0 m
(Eq. 2)
FORCE-DISPLACEMENT RELATIONS
A 40,000 mm2
1
allow 20 MPa
E 30 GPa
P1L
P2L
2
EA
EA
(Eqs. 3, 4)
SOLUTION OF EQUATIONS
C concrete post
Substitute (3) and (4) into Eq. (2):
DOES THE GAP CLOSE?
Stress in the two outer posts when the gap is just
closed:
s
1.0 mm
s Ee E (30 GPa)
L
2.0 m
15 MPa
Since this stress is less than the allowable stress, the
allowable force P will close the gap.
P1L P2L
EAs
sorP1 P2
EA
EA
L
(Eq. 5)
Solve simultaneously Eqs. (1) and (5):
P 3P1
EAs
L
By inspection, we know that P1 is larger than P2.
Therefore, P1 will control and will be equal to
allow A.
Pallow 3sallow A
EAs
L
2400 kN 600 kN 1800 kN
1.8 MN
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117
118
CHAPTER 2
Axially Loaded Members
Problem 2.5-17 A copper tube is fitted around a steel bolt and the nut
is turned until it is just snug (see figure). What stresses s and c will be
produced in the steel and copper, respectively, if the bolt is now tightened
by a quarter turn of the nut?
The copper tube has length L 16 in. and cross-sectional area
Ac 0.6 in.2, and the steel bolt has cross-sectional area As 0.2 in.2 The
pitch of the threads of the bolt is p 52 mils (a mil is one-thousandth
of an inch). Also, the moduli of elasticity of the steel and copper are
Es 30 106 psi and Ec 16 106 psi, respectively.
Note: The pitch of the threads is the distance advanced by the nut
in one complete turn (see Eq. 2-22).
Solution 2.5-17
Copper tube
Steel bolt
Steel bolt and copper tube
Copper tube
FORCE-DISPLACEMENT RELATIONS
c
Ps L
Pc L
s
Ec Ac
Es As
(Eq. 3, Eq. 4)
Steel bolt
SOLUTION OF EQUATIONS
L 16 in.
Substitute (3) and (4) into Eq. (2):
p 52 mils 0.052 in.
PsL
PcL
np
EcAc EsAs
1
n (See Eq. 2-22)
4
Solve simultaneously Eqs. (1) and (5):
Steel bolt: As 0.2 in.2
npEs As Ec Ac
L(Es As Ec Ac )
Es 30 106 psi
Ps Pc
Copper tube: Ac 0.6 in.2
Substitute numerical values:
Ec 16 106 psi
Ps Pc 3,000 lb
EQUILIBRIUM EQUATION
Pc
STRESSES
Steel bolt:
Ps
ss
Ps tensile force in steel bolt
Pc Ps
Ps 3,000 lb
15 ksi (tension)
As
0.2 in.2
Copper tube:
Pc compressive force in copper tube
(Eq. 1)
sc
Pc 3,000 lb
Ac
0.6 in.2
5 ksi (compression)
COMPATIBILITY EQUATION
Ps Pc
np
c shortening of copper tube
s elongation of steel bolt
c s np
(Eq. 5)
(Eq. 2)
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(Eq. 6)
SECTION 2.5
Problem 2.5-18 A plastic cylinder is held snugly between a rigid plate
and a foundation by two steel bolts (see figure).
Determine the compressive stress p in the plastic when the nuts on
the steel bolts are tightened by one complete turn.
Data for the assembly are as follows: length L 200 mm, pitch of
the bolt threads p 1.0 mm, modulus of elasticity for steel Es 200 GPa,
modulus of elasticity for the plastic Ep 7.5 GPa, cross-sectional area of
one bolt As 36.0 mm2, and cross-sectional area of the plastic cylinder
Ap 960 mm2.
Solution 2.5-18
119
Misfits and Prestrains
Steel
bolt
Probs. 2.5-18 and 2.5-19
Plastic cylinder and two steel bolts
FORCE-DISPLACEMENT RELATIONS
s
L 200 mm
SOLUTION OF EQUATIONS
P 1.0 mm
Substitute (3) and (4) into Eq. (2):
Es 200 GPa
As 36.0
mm2
Pp L
Ps L
p
Es As
Ep Ap
Pp L
Ps L
np
Es As Ep Ap
(for one bolt)
Ep 7.5 GPa
Solve simultaneously Eqs. (1) and (5):
Ap 960 mm2
Pp
n 1 (See Eq. 2-22)
EQUILIBRIUM EQUATION
2npEs As Ep Ap
L(Ep Ap 2Es As )
STRESS IN THE PLASTIC CYLINDER
Ps
Ps
sp
Pp
Ap
2 np Es As Ep
L(Ep Ap 2Es As )
SUBSTITUTE NUMERICAL VALUES:
N Es As Ep 54.0 1015 N2/m2
Pp
D Ep Ap 2Es As 21.6 106 N
Ps tensile force in one steel bolt
Pp compressive force in plastic cylinder
Pp 2Ps
sp
(Eq. 1)
2np N
2(1)(1.0 mm) N
L D
200 mm
D
25.0 MPa
COMPATIBILITY EQUATION
Ps
Pp
Ps
np
s elongation of steel bolt
p shortening of plastic cylinder
s p np
(Eq. 2)
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(Eq. 3, Eq. 4)
(Eq. 5)
120
CHAPTER 2
Axially Loaded Members
Problem 2.5-19 Solve the preceding problem if the data for the
assembly are as follows: length L 10 in., pitch of the bolt threads
p 0.058 in., modulus of elasticity for steel Es 30 106 psi,
modulus of elasticity for the plastic Ep 500 ksi, cross-sectional
area of one bolt As 0.06 in.2, and cross-sectional area of the
plastic cylinder Ap 1.5 in.2
Solution 2.5-19
Plastic cylinder and two steel bolts
FORCE-DISPLACEMENT RELATIONS
s
L 10 in.
SOLUTION OF EQUATIONS
p 0.058 in.
Substitute (3) and (4) into Eq. (2):
Es 30 106 psi
As 0.06
in.2
(for one bolt)
Ep 500 ksi
Ap 1.5
Pp L
Ps L
p
Es As
Ep Ap
Pp L
Ps L
np
Es As Ep Ap
Solve simultaneously Eqs. (1) and (5):
in.2
Pp
n 1 (see Eq. 2-22)
2 np Es As Ep Ap
L(Ep Ap 2Es As )
EQUILIBRIUM EQUATION
STRESS IN THE PLASTIC CYLINDER
Ps tensile force in one steel bolt
sp
Pp compressive force in plastic cylinder
Pp 2Ps
(Eq. 1)
Ps
Pp
Ap
2 np Es As Ep
L(Ep Ap 2Es As )
SUBSTITUTE NUMERICAL VALUES:
N Es As Ep 900 109 lb2/in.2
Ps
D Ep Ap 2Es As 4350 103 lb
2np N
2(1)(0.058 in.) N
L D
10 in.
D
2400 psi
sP
Pp
COMPATIBILITY EQUATION
s elongation of steel bolt
p shortening of plastic cylinder
s p np
Ps
(Eq. 2)
Pp
Ps
np
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(Eq. 3, Eq. 4)
(Eq. 5)
SECTION 2.5
Problem 2.5-20 Prestressed concrete beams are sometimes
manufactured in the following manner. High-strength steel wires are
stretched by a jacking mechanism that applies a force Q, as represented
schematically in part (a) of the figure. Concrete is then poured around
the wires to form a beam, as shown in part (b).
After the concrete sets properly, the jacks are released and the
force Q is removed [see part (c) of the figure]. Thus, the beam is
left in a prestressed condition, with the wires in tension and the
concrete in compression.
Let us assume that the prestressing force Q produces in the steel
wires an initial stress 0 620 MPa. If the moduli of elasticity of the
steel and concrete are in the ratio 12:1 and the cross-sectional areas are in
the ratio 1:50, what are the final stresses s and c in the two materials?
Misfits and Prestrains
Steel wires
Q
(a)
Concrete
Q
(b)
(c)
Solution 2.5-20
Prestressed concrete beam
L length
Steel wires
0 initial stress in wires
Q
Q
620 MPa
As
As total area of steel wires
Concrete
Ac area of concrete
50 As
Ps
Es 12 Ec
Pc
Ps final tensile force in steel wires
Pc final compressive force in concrete
EQUILIBRIUM EQUATION
Ps Pc
(Eq. 1)
COMPATIBILITY EQUATION AND
FORCE-DISPLACEMENT RELATIONS
STRESSES
1 initial elongation of steel wires
s0L
QL
EsAs
Es
2 final elongation of steel wires
ss
sc
Ps L
Es As
3 shortening of concrete
Ps
As
s0
Es As
1
Ec Ac
s0
Pc
Ac Ac Es
As Ec
SUBSTITUTE NUMERICAL VALUES:
Pc L
Ec Ac
s0 L Ps L
Pc L
1 2 3or
(Eq. 2, Eq. 3)
Es
Es As Ec Ac
Solve simultaneously Eqs. (1) and (3):
s0 As
Ps Pc
Es As
1
Ec Ac
121
Es
As
1
s0 620 MPa 12
Ec
Ac 50
ss
620 MPa
500 MPa (Tension)
12
1
50
sc
620 MPa
10 MPa (Compression)
50 12
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