PROBLEM 14.59
A steel penstock has a 750–mm outer diameter, a 12–mm wall
thickness, and connects a reservoir at A with a generating station
at B. Knowing that the density of water is
determine
the maximum normal stress and the maximum shearing stress in
the penstock under static conditions.
SOLUTION
3
3
32
6
63 6
13
11
(750) 12 363 mm 363 10 m
22
12 mm 12 10 m
(1000 kg/m )(9.81 m/s )(300 m)
2.943 10 Pa
(2.943 10 )(363 10 ) 89.0 10 Pa
12 10
r dt
t
p gh
pr
t
r
s
−
−
−
−
= −= − = = ×
= = ×
= =
= ×
××
= = = ×
×
max 1
ss
=
max
89.0 MPa
s
=
min 0p
s
=−≈
max max min
1()
2
τ ss
= −
max
44.5 MPa
τ
=
PROBLEM 14.60
A steel penstock has a 750–mm outer diameter and connects a
reservoir at A with a generating station at B. Knowing that the
density of water is
3
1000 kg/m
and that the allowable normal
stress in the steel is 85 MPa, determine the smallest thickness
that can be used for the penstock.
SOLUTION
32
6
6
1
3
1
6
6
(1000 kg/m )(9.81 m/s )(300 m)
2.943 10 Pa
85 MPa 85 10 Pa
11
(750 10 ) 0.375
22
(2.943 10 )(0.375 )
85 10
p gh
r dt t t
pr
t
t
t
r
s
s
−
= =
= ×
= = ×
= −= × −= −
=
×−
×=
663
(87.943 10 ) 1.103625 10 12.549 10 mtt
−
×= × = ×
12.55 mm
t=
PROBLEM 14.61
The cylindrical portion of the compressed–air tank shown is fabricated of 0.25–in.–
thick plate welded along a helix forming an angle
30
β
= °
with the horizontal.
Knowing that the allowable stress normal to the weld is 10.5 ksi, determine the
largest gage pressure that can be used in the tank.
SOLUTION
12
110 0.25 9.75 in.
2
,2
r dt
pr pr
tt
σσ
= −= − =
= =
ave 1 2
12
ave
13
()
24
1
24
cos 60°
5
8
w
pr
t
pr
Rt
R
pr
t
σ σσ
σσ
σσ
= +=
−
= =
= +
=
8 (8)(10.5)(0.25) 0.43077 ksi
5 (5)(9.75)
wt
pr
σ
= = =
431 psip=
consent of McGraw–Hill Education.
PROBLEM 14.62
For the compressed–air tank of Prob. 14.61, determine the gage pressure that will
cause a shearing stress parallel to the weld of 4 ksi.
PROBLEM 14.61 The cylindrical portion of the compressed–air tank shown is
fabricated of 0.25–in.–thick plate welded along a helix forming an angle
30
β
= °
with
the horizontal. Knowing that the allowable stress normal to the weld is 10.5 ksi,
determine the largest gage pressure that can be used in the tank.
SOLUTION
110 0.25 9.75 in.
2
r dt= −= − =
12
,2
pr pr
tt
σσ
= =
ave 1 2
12
13
()
24
1
24
sin 60
3
8
w
pr
t
pr
Rt
R
pr
t
σ σσ
σσ
t
= +=
−
= =
= °
=
8 8 (4)(0.25) 0.47372 ksi
9.75
33
w
t
pr
t
==⋅=
474 psip=
consent of McGraw–Hill Education.
PROBLEM 14.63
The pressure tank shown has an 8–mm wall thickness and butt–welded
seams forming an angle
20
β
= °
with a transverse plane. For a gage
pressure of 600 kPa, determine (a) the normal stress perpendicular to the
weld, (b) the shearing stress parallel to the weld.
SOLUTION
3
36
13
36
23
6
ave 1 2
6
12
1
1.6 m 8 10 m 0.792 m
2
(600 10 )(0.792) 59.4 10 Pa
8 10
(600 10 )(0.792) 29.7 10 Pa
2(2)(8 10 )
1( ) 44.56 10 Pa
2
1( ) 14.85 10 Pa
2
d t r dt
pr
t
pr
t
R
σ
σ
σ σσ
σσ
−
−
−
= =× = −=
×
= = = ×
×
×
= = = ×
×
= += ×
= −= ×
(a)
6
ave
cos 40° 33.17 10 Pa
w
R
σσ
=−=×
33.2 MPa
w
σ
=
(b)
6
sin 40° 9.55 10 Pa
w
R
τ
= = ×
9.55 MPa
w
τ
=
PROBLEM 14.64
For the tank of Prob. 14.63, determine the largest allowable gage pressure,
knowing that the allowable normal stress perpendicular to the weld is
120 MPa and the allowable shearing stress parallel to the weld is 80 MPa.
PROBLEM 14.64 The pressure tank shown has a 8-mm wall thickness and
butt-welded seams forming an angle
20
β
= °
with a transverse plane. For
a gage pressure of 600 kPa, determine (a) the normal stress perpendicular
to the weld, (b) the shearing stress parallel to the weld.
SOLUTION
3
12
ave 1 2
12
ave
1
1.6 m 8 10 m 0.792 m
2
2
13
()
24
11
()
24
cos 40°
31
cos 40 0.5585
44
w
d t r dt
pr pr
tt
pr
t
pr
Rt
R
pr pr
tt
ss
s ss
ss
ss
−
= =× = −=
= =
= +=
= −=
= −

=− °=


63 6
63 6
(120 10 )(8 10 ) 2.17 10 Pa 2.17 MPa
0.5585 (0.5585)(0.792)
1
sin 40 sin 40 0.1607
4
(80 10 )(8 10 ) 5.03 10 Pa 5.03 MPa
0.1607 (0.1607)(0.792)
w
w
w
t
pr
pr pr
Rtt
t
pr
s
t
t
−
−
××
= = =×=

= °= ° =


××
= = =×=
PROBLEM 14.65
The steel pressure tank shown has a 750–mm inner diameter and a 9–mm
wall thickness. Knowing that the butt–welded seams form an angle
50
β
= °
with the longitudinal axis of the tank and that the gage pressure in the tank is
1.5 MPa, determine (a) the normal stress perpendicular to the weld, (b) the
shearing stress parallel to the weld.
PROBLEM 14.66
The pressurized tank shown was fabricated by welding strips of plate along
a helix forming an angle
β
with a transverse plane. Determine the largest
value of
β
that can be used if the normal stress perpendicular to the weld is
not to be larger than 85 percent of the maximum stress in the tank.
SOLUTION
12
2
pr pr
tt
σσ
= =
ave 1 2
12
ave
13
()
24
1
24
cos 2
w
pr
t
pr
Rt
R
σ σσ
σσ
σσ β
= +=
−
= =
= −
31
0.85 cos 2
44
pr pr
tt
β

= −


3
cos 2 4 0.85 0.4
4
2 113.6
β
β

=− −=−


= °
56.8
β
= °
v
consent of McGraw–Hill Education.
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 14.59
A steel penstock has a 750–mm outer diameter, a 12–mm wall
thickness, and connects a reservoir at A with a generating station
at B. Knowing that the density of water is
determine
the maximum normal stress and the maximum shearing stress in
the penstock under static conditions.
SOLUTION
3
3
32
6
63 6
13
11
(750) 12 363 mm 363 10 m
22
12 mm 12 10 m
(1000 kg/m )(9.81 m/s )(300 m)
2.943 10 Pa
(2.943 10 )(363 10 ) 89.0 10 Pa
12 10
r dt
t
p gh
pr
t
r
s
−
−
−
−
= −= − = = ×
= = ×
= =
= ×
××
= = = ×
×
max 1
ss
=
max
89.0 MPa
s
=
min 0p
s
=−≈
max max min
1()
2
τ ss
= −
max
44.5 MPa
τ
=
PROBLEM 14.60
A steel penstock has a 750–mm outer diameter and connects a
reservoir at A with a generating station at B. Knowing that the
density of water is
3
1000 kg/m
and that the allowable normal
stress in the steel is 85 MPa, determine the smallest thickness
that can be used for the penstock.
SOLUTION
32
6
6
1
3
1
6
6
(1000 kg/m )(9.81 m/s )(300 m)
2.943 10 Pa
85 MPa 85 10 Pa
11
(750 10 ) 0.375
22
(2.943 10 )(0.375 )
85 10
p gh
r dt t t
pr
t
t
t
r
s
s
−
= =
= ×
= = ×
= −= × −= −
=
×−
×=
663
(87.943 10 ) 1.103625 10 12.549 10 mtt
−
×= × = ×
12.55 mm
t=
PROBLEM 14.61
The cylindrical portion of the compressed–air tank shown is fabricated of 0.25–in.–
thick plate welded along a helix forming an angle
30
β
= °
with the horizontal.
Knowing that the allowable stress normal to the weld is 10.5 ksi, determine the
largest gage pressure that can be used in the tank.
SOLUTION
12
110 0.25 9.75 in.
2
,2
r dt
pr pr
tt
σσ
= −= − =
= =
ave 1 2
12
ave
13
()
24
1
24
cos 60°
5
8
w
pr
t
pr
Rt
R
pr
t
σ σσ
σσ
σσ
= +=
−
= =
= +
=
8 (8)(10.5)(0.25) 0.43077 ksi
5 (5)(9.75)
wt
pr
σ
= = =
431 psip=
consent of McGraw–Hill Education.
PROBLEM 14.62
For the compressed–air tank of Prob. 14.61, determine the gage pressure that will
cause a shearing stress parallel to the weld of 4 ksi.
PROBLEM 14.61 The cylindrical portion of the compressed–air tank shown is
fabricated of 0.25–in.–thick plate welded along a helix forming an angle
30
β
= °
with
the horizontal. Knowing that the allowable stress normal to the weld is 10.5 ksi,
determine the largest gage pressure that can be used in the tank.
SOLUTION
110 0.25 9.75 in.
2
r dt= −= − =
12
,2
pr pr
tt
σσ
= =
ave 1 2
12
13
()
24
1
24
sin 60
3
8
w
pr
t
pr
Rt
R
pr
t
σ σσ
σσ
t
= +=
−
= =
= °
=
8 8 (4)(0.25) 0.47372 ksi
9.75
33
w
t
pr
t
==⋅=
474 psip=
consent of McGraw–Hill Education.
PROBLEM 14.63
The pressure tank shown has an 8–mm wall thickness and butt–welded
seams forming an angle
20
β
= °
with a transverse plane. For a gage
pressure of 600 kPa, determine (a) the normal stress perpendicular to the
weld, (b) the shearing stress parallel to the weld.
SOLUTION
3
36
13
36
23
6
ave 1 2
6
12
1
1.6 m 8 10 m 0.792 m
2
(600 10 )(0.792) 59.4 10 Pa
8 10
(600 10 )(0.792) 29.7 10 Pa
2(2)(8 10 )
1( ) 44.56 10 Pa
2
1( ) 14.85 10 Pa
2
d t r dt
pr
t
pr
t
R
σ
σ
σ σσ
σσ
−
−
−
= =× = −=
×
= = = ×
×
×
= = = ×
×
= += ×
= −= ×
(a)
6
ave
cos 40° 33.17 10 Pa
w
R
σσ
=−=×
33.2 MPa
w
σ
=
(b)
6
sin 40° 9.55 10 Pa
w
R
τ
= = ×
9.55 MPa
w
τ
=
PROBLEM 14.64
For the tank of Prob. 14.63, determine the largest allowable gage pressure,
knowing that the allowable normal stress perpendicular to the weld is
120 MPa and the allowable shearing stress parallel to the weld is 80 MPa.
PROBLEM 14.64 The pressure tank shown has a 8-mm wall thickness and
butt-welded seams forming an angle
20
β
= °
with a transverse plane. For
a gage pressure of 600 kPa, determine (a) the normal stress perpendicular
to the weld, (b) the shearing stress parallel to the weld.
SOLUTION
3
12
ave 1 2
12
ave
1
1.6 m 8 10 m 0.792 m
2
2
13
()
24
11
()
24
cos 40°
31
cos 40 0.5585
44
w
d t r dt
pr pr
tt
pr
t
pr
Rt
R
pr pr
tt
ss
s ss
ss
ss
−
= =× = −=
= =
= +=
= −=
= −

=− °=


63 6
63 6
(120 10 )(8 10 ) 2.17 10 Pa 2.17 MPa
0.5585 (0.5585)(0.792)
1
sin 40 sin 40 0.1607
4
(80 10 )(8 10 ) 5.03 10 Pa 5.03 MPa
0.1607 (0.1607)(0.792)
w
w
w
t
pr
pr pr
Rtt
t
pr
s
t
t
−
−
××
= = =×=

= °= ° =


××
= = =×=
PROBLEM 14.65
The steel pressure tank shown has a 750–mm inner diameter and a 9–mm
wall thickness. Knowing that the butt–welded seams form an angle
50
β
= °
with the longitudinal axis of the tank and that the gage pressure in the tank is
1.5 MPa, determine (a) the normal stress perpendicular to the weld, (b) the
shearing stress parallel to the weld.
PROBLEM 14.66
The pressurized tank shown was fabricated by welding strips of plate along
a helix forming an angle
β
with a transverse plane. Determine the largest
value of
β
that can be used if the normal stress perpendicular to the weld is
not to be larger than 85 percent of the maximum stress in the tank.
SOLUTION
12
2
pr pr
tt
σσ
= =
ave 1 2
12
ave
13
()
24
1
24
cos 2
w
pr
t
pr
Rt
R
σ σσ
σσ
σσ β
= +=
−
= =
= −
31
0.85 cos 2
44
pr pr
tt
β

= −


3
cos 2 4 0.85 0.4
4
2 113.6
β
β

=− −=−


= °
56.8
β
= °
v
consent of McGraw–Hill Education.
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