PROBLEM 14.49
A basketball has a 9.5–in. outer diameter and a 0.125–in. wall thickness. Determine the normal stress in the
wall when the basketball is inflated to a 9–psi gage pressure.
PROBLEM 14.50
A spherical gas container made of steel has a 20–ft outer diameter and a wall thickness of
7
16 in.
Knowing that
the internal pressure is 75 psi, determine the maximum normal stress and the maximum shearing stress in the
container.
SOLUTION
20 ft 240 in.= =d
7in. 0.4375 in.
16
1119.56 in.
2
t
r dt
= =
= −=
3
(75)(119.56) 10.25 10 psi
2 (2)(0.4375)
s
= = = ×
pr
t
10.25 ksi
s
=
max
10.25 ksi
s
=
min 0
s
≈
(Neglecting small radial stress)
max max min
1()
2
τ ss
= −
max
5.12 ksi
τ
=
consent of McGraw–Hill Education.
PROBLEM 14.51
The maximum gage pressure is known to be 1150 psi in a spherical steel pressure vessel having a 10–in. outer
diameter and a 0.25–in. wall thickness. Knowing that the ultimate stress in the steel used is
s
U = 60 ksi,
determine the factor of safety with respect to tensile failure.
SOLUTION
12
max
2
10 in. 0.25 in.
2
4.75 in.
2
(1150 psi)(4.75 in.)
2(0.25 in.)
10.925 ksi
60 ksi
F.S. 10.925 ksi
U
d
rt
pr
t
= −
= −
=
= =
=
=
= =
ss
s
s
F.S. 5.49=
PROBLEM 14.52
A spherical gas container having an outer diameter of 5 m and a wall thickness of 22 mm is made of steel for
which
200 GPaE=
and
0.29.v=
Knowing that the gage pressure in the container is increased from zero to
1.7 MPa, determine (a) the maximum normal stress in the container, (b) the corresponding increase in the
diameter of the container.
SOLUTION
33
522 10 2.478 m, 22 10 m
22
d
rt t
−−
= −= − × = = ×
(a)
66
12 3
(1.7 10 Pa)(2.478m) 95.741 10 Pa
22(22 10 m)
pr
t
σσ
−
×
= = = = ×
×
max 95.7 MPa
σ
=
(b)
1 12 1
66
9
11
()
(1 0.29)(95.741 10 Pa) 339.88 10
200 10 Pa
v
v
EE
ε σσ σ
−
−
= −=
−×
= = ×
×
36
1(5 10 mm)(339.88 10 )dd
ε
−
∆= = × ×
1.699 mmd∆=
consent of McGraw–Hill Education.
PROBLEM 14.53
A spherical pressure vessel has an outer diameter of 3 m and a wall thickness of 12 mm. Knowing that for
the steel used
σ
all = 80 MPa, E = 200 GPa, and
ν
= 0.29, determine (a) the allowable gage pressure, (b) the
corresponding increase in the diameter of the vessel.
SOLUTION
3
6
1 2 all
11
(3) 12 10 1.488 m
22
80 10 Pa
σσσ
−
= −= − × =
= = = ×
r dt
(a)
12
36
1
2
2 (2)(12 10 )(80 10 )
1.488
pr
t
t
pr
−
= =
××
= =
σσ
σ
6
1.290 10 Pap= ×
1.290 MPap=
(b)
1 12
66
19
1()
1 1 0.29 (80 10 ) 284 10
200 10
ε σ υσ
υσ
−
= −
−−
= = ×=×
×
E
E
66
1(3)(284 10 ) 852 10 m
ε
−−
∆= = × = ×dd
0.852 mm∆=d
consent of McGraw–Hill Education.
PROBLEM 14.54
A spherical pressure vessel of 750–mm outer diameter is to be fabricated from a steel having an ultimate stress
σ
U = 400 MPa. Knowing that a factor of safety of 4.0 is desired and that the gage pressure can reach 4.2 MPa,
determine the smallest wall thickness that should be used.
SOLUTION
1
2
1(0.750 m)
2
0.375 (m)
= −
= −
=
r dt
t
t
We have
max 1 2
2
σ σσ
= = = pr
t
and
max
F.S.
σ
σ
=
U
Combining these two equations gives
2
F.S.
σ
=U
t
pr
or
2 (F.S.)
U
t pr
=
σ
Substituting for r gives
66
66
3
2(400 10 Pa) (4)(4.2 10 Pa)(0.375 )
816.80 10 6.30 10
7.71 10 m
−
×=× −
×=×
= ×
tt
t
t
7.71mm=t
v
PROBLEM 14.55
Determine the largest internal pressure that can be applied to a cylindrical tank of 5.5–ft outer diameter and
5
8
-in.
wall thickness if the ultimate normal stress of the steel used is 65 ksi and a factor of safety of 5.0
is desired.
SOLUTION
3
1
65 ksi 13 ksi 13 10 psi
. . 5.0
(5.5)(12) 0.625 32.375 in.
22
U
FS
d
rt
s
s
= = = = ×
= −= − =
3
1
1(0.625)(13 10 )
32.375
pr t
p
tr
s
s
×
= = =
251 psip=
v
consent of McGraw–Hill Education.
PROBLEM 14.56
The unpressurized cylindrical storage tank shown has a
3
16
-in.
wall thickness
and is made of steel having a 60-ksi ultimate strength in tension. Determine the
maximum height h to which it can be filled with water if a factor of safety of
4.0 is desired.
3
(Specific weight of water 62.4 lb/ft .)=
SOLUTION
( )
0
3
all
all
3
3
16 2
all
(25)(12) 300 in.
13
150 149.81in.
2 16
60 ksi 15 ksi 15 10 psi
. . 4.0
(15 10 ) 18.77 psi 2703 lb/ft
149.81
U
d
r dt
FS
pr
t
t
pr
s
s
s
s
= =
= −= − =
= = = = ×
=
×
= = = =
But
,ph
γ
=
2
3
2703 lb/ft
62.4 lb/ft
p
h
γ
= =
43.3 fth=
PROBLEM 14.57
For the storage tank of Prob. 14.56, determine the maximum normal stress and
the maximum shearing stress in the cylindrical wall when the tank is filled to
capacity
( 48 ft).h=
PROBLEM 14.56 The unpressurized cylindrical storage tank shown has a
3
16
-in.
wall thickness and is made of steel having a 60–ksi ultimate strength in
tension. Determine the maximum height h to which it can be filled with water
if a factor of safety of 4.0 is desired.
3
(Specific weight of water 62.4 lb/ft .)=
PROBLEM 14.49
A basketball has a 9.5–in. outer diameter and a 0.125–in. wall thickness. Determine the normal stress in the
wall when the basketball is inflated to a 9–psi gage pressure.
PROBLEM 14.50
A spherical gas container made of steel has a 20–ft outer diameter and a wall thickness of
7
16 in.
Knowing that
the internal pressure is 75 psi, determine the maximum normal stress and the maximum shearing stress in the
container.
SOLUTION
20 ft 240 in.= =d
7in. 0.4375 in.
16
1119.56 in.
2
t
r dt
= =
= −=
3
(75)(119.56) 10.25 10 psi
2 (2)(0.4375)
s
= = = ×
pr
t
10.25 ksi
s
=
max
10.25 ksi
s
=
min 0
s
≈
(Neglecting small radial stress)
max max min
1()
2
τ ss
= −
max
5.12 ksi
τ
=
consent of McGraw–Hill Education.
PROBLEM 14.51
The maximum gage pressure is known to be 1150 psi in a spherical steel pressure vessel having a 10–in. outer
diameter and a 0.25–in. wall thickness. Knowing that the ultimate stress in the steel used is
s
U = 60 ksi,
determine the factor of safety with respect to tensile failure.
SOLUTION
12
max
2
10 in. 0.25 in.
2
4.75 in.
2
(1150 psi)(4.75 in.)
2(0.25 in.)
10.925 ksi
60 ksi
F.S. 10.925 ksi
U
d
rt
pr
t
= −
= −
=
= =
=
=
= =
ss
s
s
F.S. 5.49=
PROBLEM 14.52
A spherical gas container having an outer diameter of 5 m and a wall thickness of 22 mm is made of steel for
which
200 GPaE=
and
0.29.v=
Knowing that the gage pressure in the container is increased from zero to
1.7 MPa, determine (a) the maximum normal stress in the container, (b) the corresponding increase in the
diameter of the container.
SOLUTION
33
522 10 2.478 m, 22 10 m
22
d
rt t
−−
= −= − × = = ×
(a)
66
12 3
(1.7 10 Pa)(2.478m) 95.741 10 Pa
22(22 10 m)
pr
t
σσ
−
×
= = = = ×
×
max 95.7 MPa
σ
=
(b)
1 12 1
66
9
11
()
(1 0.29)(95.741 10 Pa) 339.88 10
200 10 Pa
v
v
EE
ε σσ σ
−
−
= −=
−×
= = ×
×
36
1(5 10 mm)(339.88 10 )dd
ε
−
∆= = × ×
1.699 mmd∆=
consent of McGraw–Hill Education.
PROBLEM 14.53
A spherical pressure vessel has an outer diameter of 3 m and a wall thickness of 12 mm. Knowing that for
the steel used
σ
all = 80 MPa, E = 200 GPa, and
ν
= 0.29, determine (a) the allowable gage pressure, (b) the
corresponding increase in the diameter of the vessel.
SOLUTION
3
6
1 2 all
11
(3) 12 10 1.488 m
22
80 10 Pa
σσσ
−
= −= − × =
= = = ×
r dt
(a)
12
36
1
2
2 (2)(12 10 )(80 10 )
1.488
pr
t
t
pr
−
= =
××
= =
σσ
σ
6
1.290 10 Pap= ×
1.290 MPap=
(b)
1 12
66
19
1()
1 1 0.29 (80 10 ) 284 10
200 10
ε σ υσ
υσ
−
= −
−−
= = ×=×
×
E
E
66
1(3)(284 10 ) 852 10 m
ε
−−
∆= = × = ×dd
0.852 mm∆=d
consent of McGraw–Hill Education.
PROBLEM 14.54
A spherical pressure vessel of 750–mm outer diameter is to be fabricated from a steel having an ultimate stress
σ
U = 400 MPa. Knowing that a factor of safety of 4.0 is desired and that the gage pressure can reach 4.2 MPa,
determine the smallest wall thickness that should be used.
SOLUTION
1
2
1(0.750 m)
2
0.375 (m)
= −
= −
=
r dt
t
t
We have
max 1 2
2
σ σσ
= = = pr
t
and
max
F.S.
σ
σ
=
U
Combining these two equations gives
2
F.S.
σ
=U
t
pr
or
2 (F.S.)
U
t pr
=
σ
Substituting for r gives
66
66
3
2(400 10 Pa) (4)(4.2 10 Pa)(0.375 )
816.80 10 6.30 10
7.71 10 m
−
×=× −
×=×
= ×
tt
t
t
7.71mm=t
v
PROBLEM 14.55
Determine the largest internal pressure that can be applied to a cylindrical tank of 5.5–ft outer diameter and
5
8
-in.
wall thickness if the ultimate normal stress of the steel used is 65 ksi and a factor of safety of 5.0
is desired.
SOLUTION
3
1
65 ksi 13 ksi 13 10 psi
. . 5.0
(5.5)(12) 0.625 32.375 in.
22
U
FS
d
rt
s
s
= = = = ×
= −= − =
3
1
1(0.625)(13 10 )
32.375
pr t
p
tr
s
s
×
= = =
251 psip=
v
consent of McGraw–Hill Education.
PROBLEM 14.56
The unpressurized cylindrical storage tank shown has a
3
16
-in.
wall thickness
and is made of steel having a 60-ksi ultimate strength in tension. Determine the
maximum height h to which it can be filled with water if a factor of safety of
4.0 is desired.
3
(Specific weight of water 62.4 lb/ft .)=
SOLUTION
( )
0
3
all
all
3
3
16 2
all
(25)(12) 300 in.
13
150 149.81in.
2 16
60 ksi 15 ksi 15 10 psi
. . 4.0
(15 10 ) 18.77 psi 2703 lb/ft
149.81
U
d
r dt
FS
pr
t
t
pr
s
s
s
s
= =
= −= − =
= = = = ×
=
×
= = = =
But
,ph
γ
=
2
3
2703 lb/ft
62.4 lb/ft
p
h
γ
= =
43.3 fth=
PROBLEM 14.57
For the storage tank of Prob. 14.56, determine the maximum normal stress and
the maximum shearing stress in the cylindrical wall when the tank is filled to
capacity
( 48 ft).h=
PROBLEM 14.56 The unpressurized cylindrical storage tank shown has a
3
16
-in.
wall thickness and is made of steel having a 60–ksi ultimate strength in
tension. Determine the maximum height h to which it can be filled with water
if a factor of safety of 4.0 is desired.
3
(Specific weight of water 62.4 lb/ft .)=