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PROBLEM 14.68
For the compressed-air tank and loading of Prob. 14.67, determine
the maximum normal stress and the maximum in-plane shearing
stress at point b on the top of the tank.
PROBLEM 14.67 The compressed-air tank AB has an inner
diameter of 450 mm and a uniform wall thickness of 6 mm. Knowing
that the gage pressure inside the tank is 1.2 MPa, determine the
maximum normal stress and the maximum in-plane shearing stress at
point a on the top of the tank.
SOLUTION
PROBLEM 14.69
A pressure vessel of 10-in. inner diameter and 0.25-in. wall thickness
is fabricated from a 4-ft section of spirally-welded pipe AB and is
equipped with two rigid end plates. The gage pressure inside the vessel
is 300 psi and 10-kip centric axial forces P and P′ are applied to the
end plates. Determine (a) the normal stress perpendicular to the weld,
(b) the shearing stress parallel to the weld.
SOLUTION
( )
1
2
0
22 2 2 2
0
3
11
(10) 5 in. 0.25 in.
22
(300)(5) 6000 psi 6 ksi
0.25
(300)(5) 3000 psi 3 ksi
2 (2)(0.25)
5 0.25 5.25 in.
(5.25 5.00 ) 8.0503 in
10 10 1242 psi 1.242 ksi
8.0803
rd t
pr
t
pr
t
r rt
A rr
P
A
= = = =
= = = =
= = = =
=+=+ =
= −= − =
×
=−=− =− =−
s
s
pp
s
Total stresses. Longitudinal:
3 1.242 1.758 ksi
x
s
=−=
Circumferential:
6 ksi
y
s
=
Shear:
0
xy
τ
=
Plotted points for Mohr’s circle:
: (1.758, 0)
: (6, 0)
: (3.879)
X
Y
C
ave
2
2
2
1( ) 3.879 ksi
2
2
(1.758 6) 0 2.121 ksi
2
xy
xy xy
R
s ss
ss τ
= +=
−
= +
−
= +=
PROBLEM 14.70
Solve Prob. 14.69, assuming that the magnitude P of the two forces is
increased to 30 kips.
PROBLEM 14.69 A pressure vessel of 10-in. inner diameter and 0.25-in.
wall thickness is fabricated from a 4-ft section of spirally-welded pipe
AB and is equipped with two rigid end plates. The gage pressure inside
the vessel is 300 psi and 10-kip centric axial forces P and P′ are applied
to the end plates. Determine (a) the normal stress perpendicular to the
weld, (b) the shearing stress parallel to the weld.
SOLUTION
( )
1
2
0
22 22 2
0
3
11
(10) 5 in. 0.25 in.
22
(300)(5) 6000 psi 6 ksi
0.25
(300)(5) 3000 psi 3 ksi
2 (2)(0.25)
5 0.25 5.25 in.
(5.25 5 ) 8.0503 in
30 10 3727 psi 3.727 ksi
8.0503
rd t
pr
t
pr
t
r rt
A rr
P
A
s
s
pp
s
= = = =
= = = =
= = = =
=+=+ =
= −= −=
×
=−=− =− =−
Total stresses. Longitudinal:
3 3.727 0.727 ksi
x
s
=−=−
Circumferential:
6 ksi
y
s
=
Shear:
0
xy
τ
=
Plotted points for Mohr’s circle:
: ( 0.727, 0)
: (6, 0)
: (2.6365, 0)
X
Y
C
−
ave
2
2
2
1( ) 2.6365 ksi
2
2
0.727 6 0 3.3635 ksi
2
xy
xy xy
R
s ss
ss τ
= +=
−
= +
−−
= +=
PROBLEM 14.71
The cylindrical tank AB has an 8-in. diameter and a 0.32-in. wall
thickness. Knowing that the pressure inside the tank is 600 psi, determine
the maximum normal stress and the maximum in-plane shearing stress at
point K.
SOLUTION
1
21
4 in. 4.32 in.
2(600)(4) 7500 psi 7.50 ksi
0.32
13.75 ksi
2
i
i oi
i
d
r r rt
pr
t
s
ss
= = = +=
= = = =
= =
Torsion: No applied torque.
Bending: Point K lies on neutral axis.
Transverse shear: V = 9 kips
2
3
For semicircle:
4
23
2
3
r
A ry
Qr
π
π
= =
=
3 3 33 3
44 44 4
222
(4.32 4 ) 11.081 in
333
( )(0.32) 0.64 in.
( ) (4.32 4 ) 72.481 in
44
(9)(11.081) 2.15 ksi
(72.481)(0.64)
oi o i
oi
QQ Q r r
tt
I rr
VQ
It
ππ
t
=−= − = − =
= =
= −= −=
= = =
PROBLEM 14.71 (Continued)
Summary of stresses: Longitudinal:
13.75 ksi
x
ss
= =
Circumferential:
2
7.50 ksi
y
ss
= =
Shear:
xy =2.15 ksi
t
2
2
2
1( ) 5.625 ksi
2
R= 2
=2.853 ksi
8.48 ksi
2.77 ksi
=0
ave x y
xy
xy
a ave
b ave
R
R
s ss
ss t
ss
ss
s
= +=
−
+
= +=
= −=
max 8.48 ksi
s
=
min 0
s
=
max (in-plane) 2.85 ksiR
t
= =
consent of McGraw-Hill Education.
PROBLEM 14.72
Solve Prob. 14.71, assuming that the 9-kip force applied at point D is
directed vertically downward.
Prob. 14.71 The cylindrical tank AB has an 8-in. diameter and a 0.32-in.
wall thickness. Knowing that the pressure inside the tank is 600 psi,
determine the maximum normal stress and the maximum in-plane
shearing stress at point K.
SOLUTION
1
21
4 in. 4.32 in.
2(600)(4) 7500 psi 7.50 ksi
0.32
13.75 ksi
2
i
i oi
i
d
r r rt
pr
t
s
ss
= = = +=
= = = =
= =
Torsion:
44 4
( ) 144.96 in 4.32 in.
2
=(9)(10) 90 kip in
(90)(4.32)
= 2.68 ksi
144.96
oi o
J r r cr
T
Tc
J
p
τ
= −= =
=
= ⋅
= =
Bending:
4
172.48 in 4.32 in.
2(135)(4.32)
(9)(15) 135 kip in = 8.05 ksi
72.48
o
m
I J cr
Mc
MI
s
= = = =
==⋅==
Transverse shear: At point K,
0
VQ
It =
Summary of stresses:
1
y2
xy
Longitudinal: 3.75 8.05 11.80 ksi
Circumferential: = 7.50 ksi
Shear: =2.68 ksi
x
ss
ss
τ
==+=
=
SOLUTION (Continued)
2
2
2
1(11.80 7.60) 9.65 ksi
2
11.80 7.50
R= (2.68) =3.44 ksi
2
13.09 ksi
6.21 ksi
=0
ave
a ave
b ave
R
R
s
ss
ss
s
= +=
+
+
= +=
= −=
max
13.09 ksi
s
=
min 0
s
=
max (in-plane) 3.44 ksiR
τ
= =
PROBLEM 14.73
Two members of uniform cross section
50 80 mm×
are glued together along plane
a-a that forms an angle of 25° with the horizontal. Knowing that the allowable
stresses for the glued joint are
800 kPa
σ
=
and
600 kPa,
τ
=
determine the largest
centric load P that can be applied.
SOLUTION
For plane a-a,
65 .
θ
= °
22 2
33 3 3
22
22
3
0, 0,
cos sin 2 sin cos 0 sin 65 0
(50 10 )(80 10 )(800 10 ) 3.90 10 N
sin 65 sin 65
( )sin cos (cos sin ) sin65 cos65 0
(50 10 )(8
sin65 cos65
x xy y
x y xy
x y xy
P
AP
A
A
P
P
A
A
P
στ σ
σσ θσ θ τ θ θ
σ
τ σ σ θ θτ θ θ
τ
−−
−
= = =
= + + = + °+
×× ×
= = = ×
°°
=− − + − = ° °+
×
= =
°°
33 3
0 10 )(600 10 ) 6.27 10 N
sin65 cos65
−
××
= ×
°°
Allowable value of P is the smaller one.
3.90 kNP=
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written
consent of McGraw-Hill Education.
PROBLEM 14.68
For the compressed-air tank and loading of Prob. 14.67, determine
the maximum normal stress and the maximum in-plane shearing
stress at point b on the top of the tank.
PROBLEM 14.67 The compressed-air tank AB has an inner
diameter of 450 mm and a uniform wall thickness of 6 mm. Knowing
that the gage pressure inside the tank is 1.2 MPa, determine the
maximum normal stress and the maximum in-plane shearing stress at
point a on the top of the tank.
SOLUTION
PROBLEM 14.69
A pressure vessel of 10-in. inner diameter and 0.25-in. wall thickness
is fabricated from a 4-ft section of spirally-welded pipe AB and is
equipped with two rigid end plates. The gage pressure inside the vessel
is 300 psi and 10-kip centric axial forces P and P′ are applied to the
end plates. Determine (a) the normal stress perpendicular to the weld,
(b) the shearing stress parallel to the weld.
SOLUTION
( )
1
2
0
22 2 2 2
0
3
11
(10) 5 in. 0.25 in.
22
(300)(5) 6000 psi 6 ksi
0.25
(300)(5) 3000 psi 3 ksi
2 (2)(0.25)
5 0.25 5.25 in.
(5.25 5.00 ) 8.0503 in
10 10 1242 psi 1.242 ksi
8.0803
rd t
pr
t
pr
t
r rt
A rr
P
A
= = = =
= = = =
= = = =
=+=+ =
= −= − =
×
=−=− =− =−
s
s
pp
s
Total stresses. Longitudinal:
3 1.242 1.758 ksi
x
s
=−=
Circumferential:
6 ksi
y
s
=
Shear:
0
xy
τ
=
Plotted points for Mohr’s circle:
: (1.758, 0)
: (6, 0)
: (3.879)
X
Y
C
ave
2
2
2
1( ) 3.879 ksi
2
2
(1.758 6) 0 2.121 ksi
2
xy
xy xy
R
s ss
ss τ
= +=
−
= +
−
= +=
PROBLEM 14.70
Solve Prob. 14.69, assuming that the magnitude P of the two forces is
increased to 30 kips.
PROBLEM 14.69 A pressure vessel of 10-in. inner diameter and 0.25-in.
wall thickness is fabricated from a 4-ft section of spirally-welded pipe
AB and is equipped with two rigid end plates. The gage pressure inside
the vessel is 300 psi and 10-kip centric axial forces P and P′ are applied
to the end plates. Determine (a) the normal stress perpendicular to the
weld, (b) the shearing stress parallel to the weld.
SOLUTION
( )
1
2
0
22 22 2
0
3
11
(10) 5 in. 0.25 in.
22
(300)(5) 6000 psi 6 ksi
0.25
(300)(5) 3000 psi 3 ksi
2 (2)(0.25)
5 0.25 5.25 in.
(5.25 5 ) 8.0503 in
30 10 3727 psi 3.727 ksi
8.0503
rd t
pr
t
pr
t
r rt
A rr
P
A
s
s
pp
s
= = = =
= = = =
= = = =
=+=+ =
= −= −=
×
=−=− =− =−
Total stresses. Longitudinal:
3 3.727 0.727 ksi
x
s
=−=−
Circumferential:
6 ksi
y
s
=
Shear:
0
xy
τ
=
Plotted points for Mohr’s circle:
: ( 0.727, 0)
: (6, 0)
: (2.6365, 0)
X
Y
C
−
ave
2
2
2
1( ) 2.6365 ksi
2
2
0.727 6 0 3.3635 ksi
2
xy
xy xy
R
s ss
ss τ
= +=
−
= +
−−
= +=
PROBLEM 14.71
The cylindrical tank AB has an 8-in. diameter and a 0.32-in. wall
thickness. Knowing that the pressure inside the tank is 600 psi, determine
the maximum normal stress and the maximum in-plane shearing stress at
point K.
SOLUTION
1
21
4 in. 4.32 in.
2(600)(4) 7500 psi 7.50 ksi
0.32
13.75 ksi
2
i
i oi
i
d
r r rt
pr
t
s
ss
= = = +=
= = = =
= =
Torsion: No applied torque.
Bending: Point K lies on neutral axis.
Transverse shear: V = 9 kips
2
3
For semicircle:
4
23
2
3
r
A ry
Qr
π
π
= =
=
3 3 33 3
44 44 4
222
(4.32 4 ) 11.081 in
333
( )(0.32) 0.64 in.
( ) (4.32 4 ) 72.481 in
44
(9)(11.081) 2.15 ksi
(72.481)(0.64)
oi o i
oi
QQ Q r r
tt
I rr
VQ
It
ππ
t
=−= − = − =
= =
= −= −=
= = =
PROBLEM 14.71 (Continued)
Summary of stresses: Longitudinal:
13.75 ksi
x
ss
= =
Circumferential:
2
7.50 ksi
y
ss
= =
Shear:
xy =2.15 ksi
t
2
2
2
1( ) 5.625 ksi
2
R= 2
=2.853 ksi
8.48 ksi
2.77 ksi
=0
ave x y
xy
xy
a ave
b ave
R
R
s ss
ss t
ss
ss
s
= +=
−
+
= +=
= −=
max 8.48 ksi
s
=
min 0
s
=
max (in-plane) 2.85 ksiR
t
= =
consent of McGraw-Hill Education.
PROBLEM 14.72
Solve Prob. 14.71, assuming that the 9-kip force applied at point D is
directed vertically downward.
Prob. 14.71 The cylindrical tank AB has an 8-in. diameter and a 0.32-in.
wall thickness. Knowing that the pressure inside the tank is 600 psi,
determine the maximum normal stress and the maximum in-plane
shearing stress at point K.
SOLUTION
1
21
4 in. 4.32 in.
2(600)(4) 7500 psi 7.50 ksi
0.32
13.75 ksi
2
i
i oi
i
d
r r rt
pr
t
s
ss
= = = +=
= = = =
= =
Torsion:
44 4
( ) 144.96 in 4.32 in.
2
=(9)(10) 90 kip in
(90)(4.32)
= 2.68 ksi
144.96
oi o
J r r cr
T
Tc
J
p
τ
= −= =
=
= ⋅
= =
Bending:
4
172.48 in 4.32 in.
2(135)(4.32)
(9)(15) 135 kip in = 8.05 ksi
72.48
o
m
I J cr
Mc
MI
s
= = = =
==⋅==
Transverse shear: At point K,
0
VQ
It =
Summary of stresses:
1
y2
xy
Longitudinal: 3.75 8.05 11.80 ksi
Circumferential: = 7.50 ksi
Shear: =2.68 ksi
x
ss
ss
τ
==+=
=
SOLUTION (Continued)
2
2
2
1(11.80 7.60) 9.65 ksi
2
11.80 7.50
R= (2.68) =3.44 ksi
2
13.09 ksi
6.21 ksi
=0
ave
a ave
b ave
R
R
s
ss
ss
s
= +=
+
+
= +=
= −=
max
13.09 ksi
s
=
min 0
s
=
max (in-plane) 3.44 ksiR
τ
= =
PROBLEM 14.73
Two members of uniform cross section
50 80 mm×
are glued together along plane
a-a that forms an angle of 25° with the horizontal. Knowing that the allowable
stresses for the glued joint are
800 kPa
σ
=
and
600 kPa,
τ
=
determine the largest
centric load P that can be applied.
SOLUTION
For plane a-a,
65 .
θ
= °
22 2
33 3 3
22
22
3
0, 0,
cos sin 2 sin cos 0 sin 65 0
(50 10 )(80 10 )(800 10 ) 3.90 10 N
sin 65 sin 65
( )sin cos (cos sin ) sin65 cos65 0
(50 10 )(8
sin65 cos65
x xy y
x y xy
x y xy
P
AP
A
A
P
P
A
A
P
στ σ
σσ θσ θ τ θ θ
σ
τ σ σ θ θτ θ θ
τ
−−
−
= = =
= + + = + °+
×× ×
= = = ×
°°
=− − + − = ° °+
×
= =
°°
33 3
0 10 )(600 10 ) 6.27 10 N
sin65 cos65
−
××
= ×
°°
Allowable value of P is the smaller one.
3.90 kNP=
Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written
consent of McGraw-Hill Education.
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