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PROBLEM 9.2
Two gage marks are placed exactly 250 mm apart on a 12-mm-diameter aluminum rod with E = 73 GPa and
an ultimate strength of 140 MPa. Knowing that the distance between the gage marks is 250.28 mm after a load
is applied, determine (a) the stress in the rod, (b) the factor of safety.
SOLUTION
(a)
0
LL
δ
= −
250.28 mm 250 mm= −
0.28 mm=
0
δ
ε
=L
0.28 mm
250 mm
=
4
1.11643 10
−
= ×
σε
=E
94
(73 10 Pa)(1.11643 10 )
−
=××
7
8.1760 10 Pa= ×
81.8 MPa
σ
=
(b)
F.S. u
σ
σ
=
140 MPa
81.760 MPa
=
1.71233=
F.S. 1.712=
consent of McGraw-Hill Education.
PROBLEM 9.3
A nylon thread is subjected to a 8.5-N tension force. Knowing that
3.3 GPa=E
and that the length of the
thread increases by 1.1%, determine (a) the diameter of the thread, (b) the stress in the thread.
SOLUTION
(a) Strain:
1.1 0.011
100L
δ
ε
= = =
Stress:
96
(3.3 10 )(0.011) 36.3 10 Pa
σε
==×=×E
P
A
σ
=
Area:
92
6
8.5 234.16 10 m
36.3 10
P
A
σ
−
= = = ×
×
Diameter:
96
4 (4)(234.16 10 ) 546 10 m
ππ
−−
×
= = = ×
A
d
0.546 mmd=
(b) Stress:
36.3 MPa
σ
=
consent of McGraw-Hill Education.
PROBLEM 9.4
An 18-m-long steel wire of 5-mm diameter is to be used in the manufacture of a prestressed concrete beam.
It is observed that the wire stretches 45 mm when a tensile force P is applied. Knowing that
200 GPa,E=
determine (a) the magnitude of the force P, (b) the corresponding normal stress in the wire.
SOLUTION
(a)
, or
PL AE
P
AE L
δ
δ
= =
with
2 2 62
11
(0.005) 19.6350 10 m
44
ππ
−
= = = ×Ad
62 9 2
(0.045 m)(19.6350 10 m )(200 10 N/m ) 9817.5 N
18 m
P
−
××
= =
9.82 kNP=
(b)
6
62
9817.5 N 500 10 Pa
19.6350 10 m
σ
−
= = = ×
×
P
A
500 MPa
σ
=
consent of McGraw-Hill Education.
PROBLEM 9.5
A steel control rod is 5.5 ft long and must not stretch more than 0.04 in. when a 2-kip tensile load is applied to it.
Knowing that
6
29 10 psi,E= ×
determine (a) the smallest diameter rod that should be used, (b) the
corresponding normal stress caused by the load.
SOLUTION
(a)
6
(2000 lb)(5.5 12 in.)
0.04 in. (29 10 psi)
:
PL
AE A
δ
×
= = ×
22
10.113793 in
4
Ad
π
= =
0.38063 in.d=
0.381 in.d=
(b)
2
2000 lb 17575.8 psi
0.113793 in
P
A
s
= = =
17.58 ksi
s
=
consent of McGraw-Hill Education.
PROBLEM 9.6
A control rod made of yellow brass must not stretch more than 3 mm when the tension in the wire is 4 kN.
Knowing that E = 105 GPa and that the maximum allowable normal stress is 180 MPa, determine (a) the
smallest diameter rod that should be used, (b) the corresponding maximum length of the rod.
SOLUTION
(a)
2
;4
= =
Pd
A
A
π
σ
Substituting, we have
2
3
6
3
4
4
4(4 10 N)
(180 10 Pa)
5.3192 10 m
PP
d
d
d
d
σσπ
π
π
−
= ⇒=
×
=×
= ×
5.32 mmd=
(b)
;
δ
σε ε
= =EL
Substituting, we have
E
EL
L
δ
σσ
∂
= ⇒=
93
6
(105 10 Pa)(3 10 m)
(180 10 Pa)
−
××
=×
L
1.750mL=
consent of McGraw-Hill Education.
PROBLEM 9.7
An aluminum pipe must not stretch more than 0.05 in. when it is subjected to a tensile load. Knowing that
6
10.1 10E= ×
psi and that the maximum allowable normal stress is 14 ksi, determine (a) the maximum
allowable length of the pipe, (b) the required area of the pipe if the tensile load is 127.5 kips.
SOLUTION
(a)
PL
AE
δ
=
Thus,
6
3
(10.1 10 )(0.05)
14 10
EA E
LP
δδ
σ
×
= = = ×
36.1 in.=L
(b)
P
A
σ
=
Thus,
3
3
127.5 10
14 10
P
A
σ
×
= = ×
2
9.11 in=A
consent of McGraw-Hill Education.
PROBLEM 9.8
A cast-iron tube is used to support a compressive load. Knowing that
6
10 10= ×E
psi and that the maximum
allowable change in length is 0.025%, determine (a) the maximum normal stress in the tube, (b) the minimum
wall thickness for a load of 1600 lb if the outside diameter of the tube is 2.0 in.
SOLUTION
(a)
0.00025
100L
δδ
= =
;
δ
σε ε
= =EL
δ
σ
∴=
EL
6
(10 10 psi)(0.00025)
σ
= ×
3
2.50 10 psi
σ
= ×
2.50 ksi
σ
=
(b)
2
3
1600 lb
; 0.64 in
2.50 10 psi
PP
A
A
σσ
= ∴== =
×
( )
22
4
π
= −
oi
A dd
22
4
π
= −
io
A
dd
2
22 2
4(0.64 in )
(2.0 in.) 3.1851 in
i
d
π
=−=
1.78469 in.
i
d∴=
11
( ) (2.0 in. 1.78469 in.)
22
= −= −
oi
t dd
0.107655 in.t=
0.1077=t
consent of McGraw-Hill Education.
PROBLEM 9.9
A block of 10-in. length and 1.8 × 1.6-in. cross section is to support a centric compressive load P. The
material to be used is a bronze for which E = 14 × 106 psi. Determine the largest load that can be applied,
knowing that the normal stress must not exceed 18 ksi and that the decrease in length of the block should be at
most 0.12% of its original length.
SOLUTION
Considering allowable stress,
3
18 ksi or 18 10 psi= ×
s
Cross-sectional area:
2
(1.8 in.)(1.6 in.) 2.880 inA= =
32
4
(18 10 psi)(2.880 in )
5.1840 10 lb
or 51.840 kips
= ⇒=
= ×
= ×
PPA
A
ss
Considering allowable deformation,
0.12% or 0.0012 in.=
L
δ
26
4
(2.880 in )(14 10 psi)(0.0012 in.)
4.8384 10 lb
or 48.384 kips
PL P AE
AE L
P
δ
δ
= ⇒=
= ×
= ×
48.4 kips
consent of McGraw-Hill Education.
PROBLEM 9.10
A 4-m-long steel rod must not stretch more than 3 mm and the normal stress must not exceed 150 MPa when
the rod is subjected to a 10-kN axial load. Knowing that
200 GPa,E=
determine the required diameter of
the rod.
SOLUTION
4mL=
36
93
3 10 m, 150 10 Pa
200 10 Pa, 10 10 NEP
δσ
−
=×=×
=×=×
Stress:
362 2
6
10 10 N 66.667 10 m 66.667 mm
150 10 Pa
−
=
×
== =×=
×
P
A
P
A
σ
σ
Deformation:
362 2
93
(10 10 )(4) 66.667 10 m 66.667 mm
(200 10 )(3 10 )
δ
δ
−
−
=
×
== =×=
××
PL
AE
PL
AE
The larger value of A governs:
2
66.667 mmA=
2
4 4(66.667)
4
A
A dd
π
ππ
= = =
9.21 mmd=
consent of McGraw-Hill Education.
PROBLEM 9.2
Two gage marks are placed exactly 250 mm apart on a 12-mm-diameter aluminum rod with E = 73 GPa and
an ultimate strength of 140 MPa. Knowing that the distance between the gage marks is 250.28 mm after a load
is applied, determine (a) the stress in the rod, (b) the factor of safety.
SOLUTION
(a)
0
LL
δ
= −
250.28 mm 250 mm= −
0.28 mm=
0
δ
ε
=L
0.28 mm
250 mm
=
4
1.11643 10
−
= ×
σε
=E
94
(73 10 Pa)(1.11643 10 )
−
=××
7
8.1760 10 Pa= ×
81.8 MPa
σ
=
(b)
F.S. u
σ
σ
=
140 MPa
81.760 MPa
=
1.71233=
F.S. 1.712=
consent of McGraw-Hill Education.
PROBLEM 9.3
A nylon thread is subjected to a 8.5-N tension force. Knowing that
3.3 GPa=E
and that the length of the
thread increases by 1.1%, determine (a) the diameter of the thread, (b) the stress in the thread.
SOLUTION
(a) Strain:
1.1 0.011
100L
δ
ε
= = =
Stress:
96
(3.3 10 )(0.011) 36.3 10 Pa
σε
==×=×E
P
A
σ
=
Area:
92
6
8.5 234.16 10 m
36.3 10
P
A
σ
−
= = = ×
×
Diameter:
96
4 (4)(234.16 10 ) 546 10 m
ππ
−−
×
= = = ×
A
d
0.546 mmd=
(b) Stress:
36.3 MPa
σ
=
consent of McGraw-Hill Education.
PROBLEM 9.4
An 18-m-long steel wire of 5-mm diameter is to be used in the manufacture of a prestressed concrete beam.
It is observed that the wire stretches 45 mm when a tensile force P is applied. Knowing that
200 GPa,E=
determine (a) the magnitude of the force P, (b) the corresponding normal stress in the wire.
SOLUTION
(a)
, or
PL AE
P
AE L
δ
δ
= =
with
2 2 62
11
(0.005) 19.6350 10 m
44
ππ
−
= = = ×Ad
62 9 2
(0.045 m)(19.6350 10 m )(200 10 N/m ) 9817.5 N
18 m
P
−
××
= =
9.82 kNP=
(b)
6
62
9817.5 N 500 10 Pa
19.6350 10 m
σ
−
= = = ×
×
P
A
500 MPa
σ
=
consent of McGraw-Hill Education.
PROBLEM 9.5
A steel control rod is 5.5 ft long and must not stretch more than 0.04 in. when a 2-kip tensile load is applied to it.
Knowing that
6
29 10 psi,E= ×
determine (a) the smallest diameter rod that should be used, (b) the
corresponding normal stress caused by the load.
SOLUTION
(a)
6
(2000 lb)(5.5 12 in.)
0.04 in. (29 10 psi)
:
PL
AE A
δ
×
= = ×
22
10.113793 in
4
Ad
π
= =
0.38063 in.d=
0.381 in.d=
(b)
2
2000 lb 17575.8 psi
0.113793 in
P
A
s
= = =
17.58 ksi
s
=
consent of McGraw-Hill Education.
PROBLEM 9.6
A control rod made of yellow brass must not stretch more than 3 mm when the tension in the wire is 4 kN.
Knowing that E = 105 GPa and that the maximum allowable normal stress is 180 MPa, determine (a) the
smallest diameter rod that should be used, (b) the corresponding maximum length of the rod.
SOLUTION
(a)
2
;4
= =
Pd
A
A
π
σ
Substituting, we have
2
3
6
3
4
4
4(4 10 N)
(180 10 Pa)
5.3192 10 m
PP
d
d
d
d
σσπ
π
π
−
= ⇒=
×
=×
= ×
5.32 mmd=
(b)
;
δ
σε ε
= =EL
Substituting, we have
E
EL
L
δ
σσ
∂
= ⇒=
93
6
(105 10 Pa)(3 10 m)
(180 10 Pa)
−
××
=×
L
1.750mL=
consent of McGraw-Hill Education.
PROBLEM 9.7
An aluminum pipe must not stretch more than 0.05 in. when it is subjected to a tensile load. Knowing that
6
10.1 10E= ×
psi and that the maximum allowable normal stress is 14 ksi, determine (a) the maximum
allowable length of the pipe, (b) the required area of the pipe if the tensile load is 127.5 kips.
SOLUTION
(a)
PL
AE
δ
=
Thus,
6
3
(10.1 10 )(0.05)
14 10
EA E
LP
δδ
σ
×
= = = ×
36.1 in.=L
(b)
P
A
σ
=
Thus,
3
3
127.5 10
14 10
P
A
σ
×
= = ×
2
9.11 in=A
consent of McGraw-Hill Education.
PROBLEM 9.8
A cast-iron tube is used to support a compressive load. Knowing that
6
10 10= ×E
psi and that the maximum
allowable change in length is 0.025%, determine (a) the maximum normal stress in the tube, (b) the minimum
wall thickness for a load of 1600 lb if the outside diameter of the tube is 2.0 in.
SOLUTION
(a)
0.00025
100L
δδ
= =
;
δ
σε ε
= =EL
δ
σ
∴=
EL
6
(10 10 psi)(0.00025)
σ
= ×
3
2.50 10 psi
σ
= ×
2.50 ksi
σ
=
(b)
2
3
1600 lb
; 0.64 in
2.50 10 psi
PP
A
A
σσ
= ∴== =
×
( )
22
4
π
= −
oi
A dd
22
4
π
= −
io
A
dd
2
22 2
4(0.64 in )
(2.0 in.) 3.1851 in
i
d
π
=−=
1.78469 in.
i
d∴=
11
( ) (2.0 in. 1.78469 in.)
22
= −= −
oi
t dd
0.107655 in.t=
0.1077=t
consent of McGraw-Hill Education.
PROBLEM 9.9
A block of 10-in. length and 1.8 × 1.6-in. cross section is to support a centric compressive load P. The
material to be used is a bronze for which E = 14 × 106 psi. Determine the largest load that can be applied,
knowing that the normal stress must not exceed 18 ksi and that the decrease in length of the block should be at
most 0.12% of its original length.
SOLUTION
Considering allowable stress,
3
18 ksi or 18 10 psi= ×
s
Cross-sectional area:
2
(1.8 in.)(1.6 in.) 2.880 inA= =
32
4
(18 10 psi)(2.880 in )
5.1840 10 lb
or 51.840 kips
= ⇒=
= ×
= ×
PPA
A
ss
Considering allowable deformation,
0.12% or 0.0012 in.=
L
δ
26
4
(2.880 in )(14 10 psi)(0.0012 in.)
4.8384 10 lb
or 48.384 kips
PL P AE
AE L
P
δ
δ
= ⇒=
= ×
= ×
48.4 kips
consent of McGraw-Hill Education.
PROBLEM 9.10
A 4-m-long steel rod must not stretch more than 3 mm and the normal stress must not exceed 150 MPa when
the rod is subjected to a 10-kN axial load. Knowing that
200 GPa,E=
determine the required diameter of
the rod.
SOLUTION
4mL=
36
93
3 10 m, 150 10 Pa
200 10 Pa, 10 10 NEP
δσ
−
=×=×
=×=×
Stress:
362 2
6
10 10 N 66.667 10 m 66.667 mm
150 10 Pa
−
=
×
== =×=
×
P
A
P
A
σ
σ
Deformation:
362 2
93
(10 10 )(4) 66.667 10 m 66.667 mm
(200 10 )(3 10 )
δ
δ
−
−
=
×
== =×=
××
PL
AE
PL
AE
The larger value of A governs:
2
66.667 mmA=
2
4 4(66.667)
4
A
A dd
π
ππ
= = =
9.21 mmd=
consent of McGraw-Hill Education.
