08 Solutions 46060 5/28/10 8:34 AM Page 532
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–1. A spherical gas tank has an inner radius of r = 1.5 m.
If it is subjected to an internal pressure of p = 300 kPa,
determine its required thickness if the maximum normal
stress is not to exceed 12 MPa.
pr 300(103)(1.5)
sallow = ; 12(106) =
2t 2t
t = 0.0188 m = 18.8 mm Ans.
8–2. A pressurized spherical tank is to be made of
0.5-in.-thick steel. If it is subjected to an internal pressure
of p = 200 psi, determine its outer radius if the maximum
normal stress is not to exceed 15 ksi.
pr 200 ri
sallow = ; 15(103) =
2t 2(0.5)
ri = 75 in.
ro = 75 in. + 0.5 in. = 75.5 in. Ans.
8–3. The thin-walled cylinder can be supported in one of
two ways as shown. Determine the state of stress in the wall
of the cylinder for both cases if the piston P causes the P P
internal pressure to be 65 psi. The wall has a thickness of
0.25 in. and the inner diameter of the cylinder is 8 in. 8 in. 8 in.
(a) (b)
Case (a):
pr 65(4)
s1 = ; s1 = = 1.04 ksi Ans.
t 0.25
s2 = 0 Ans.
Case (b):
pr 65(4)
s1 = ; s1 = = 1.04 ksi Ans.
t 0.25
pr 65(4)
s2 = ; s2 = = 520 psi Ans.
2t 2(0.25)
532
,08 Solutions 46060 5/28/10 8:34 AM Page 533
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentl
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*8–4. The tank of the air compressor is subjected to an
internal pressure of 90 psi. If the internal diameter of
the tank is 22 in., and the wall thickness is 0.25 in.,
determine the stress components acting at point A. Draw a
volume element of the material at this point, and show the
A
results on the element.
r 11
Hoop Stress for Cylindrical Vessels: Since = = 44 7 10, then thin wall
t 0.25
analysis can be used. Applying Eq. 8–1
pr 90(11)
s1 = = = 3960 psi = 3.96 ksi Ans.
t 0.25
Longitudinal Stress for Cylindrical Vessels: Applying Eq. 8–2
pr 90(11)
s2 = = = 1980 psi = 1.98 ksi Ans.
2t 2(0.25)
•8–5. The spherical gas tank is fabricated by bolting together
two hemispherical thin shells of thickness 30 mm. If the gas
contained in the tank is under a gauge pressure of 2 MPa,
determine the normal stress developed in the wall of the tank
and in each of the bolts.The tank has an inner diameter of 8 m
and is sealed with 900 bolts each 25 mm in diameter.
r 4
Normal Stress: Since = = 133.33 7 10, thin-wall analysis is valid. For the
t 0.03
spherical tank’s wall,
pr 2(4)
s = = = 133 MPa Ans.
2t 2(0.03)
Referring to the free-body diagram shown in Fig. a,
P = pA = 2 A 10 B c A 8 B d = 32p A 10 B N. Thus,
p 2
6 6
4
+ c ©Fy = 0; 32p A 106 B - 450Pb - 450Pb = 0
Pb = 35.56 A 103 B p N
The normal stress developed in each bolt is then
Pb 35.56 A 103 B p
sb = = 228 MPa Ans.
A 0.0252 B
=
Ab p
4
533
, 08 Solutions 46060 5/28/10 8:34 AM Page 534
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–6. The spherical gas tank is fabricated by bolting
together two hemispherical thin shells. If the 8-m inner
diameter tank is to be designed to withstand a gauge pressure
of 2 MPa, determine the minimum wall thickness of the
tank and the minimum number of 25-mm diameter bolts
that must be used to seal it. The tank and the bolts are made
from material having an allowable normal stress of 150 MPa
and 250 MPa, respectively.
Normal Stress: For the spherical tank’s wall,
pr
sallow =
2t
2 A 106 B (4)
150 A 106 B =
2t
t = 0.02667 m = 26.7 mm Ans.
r 4
Since = = 150 7 10, thin-wall analysis is valid.
t 0.02667
Referring to the free-body diagram shown in Fig. a,
P = pA = 2 A 106 B c A 82 B d = 32p A 106 B N. Thus,
p
4
32p A 106 B -
n n
+ c ©Fy = 0; (P ) - (Pb)allow = 0
2 b allow 2
32p A 106 B
n = (1)
(Pb)allow
The allowable tensile force for each bolt is
(Pb)allow = sallowAb = 250 A 106 B c A 0.0252 B d = 39.0625 A 103 B pN
p
4
Substituting this result into Eq. (1),
32p A 106 B
39.0625p A 103 B
n = = 819.2 = 820 Ans.
534
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–1. A spherical gas tank has an inner radius of r = 1.5 m.
If it is subjected to an internal pressure of p = 300 kPa,
determine its required thickness if the maximum normal
stress is not to exceed 12 MPa.
pr 300(103)(1.5)
sallow = ; 12(106) =
2t 2t
t = 0.0188 m = 18.8 mm Ans.
8–2. A pressurized spherical tank is to be made of
0.5-in.-thick steel. If it is subjected to an internal pressure
of p = 200 psi, determine its outer radius if the maximum
normal stress is not to exceed 15 ksi.
pr 200 ri
sallow = ; 15(103) =
2t 2(0.5)
ri = 75 in.
ro = 75 in. + 0.5 in. = 75.5 in. Ans.
8–3. The thin-walled cylinder can be supported in one of
two ways as shown. Determine the state of stress in the wall
of the cylinder for both cases if the piston P causes the P P
internal pressure to be 65 psi. The wall has a thickness of
0.25 in. and the inner diameter of the cylinder is 8 in. 8 in. 8 in.
(a) (b)
Case (a):
pr 65(4)
s1 = ; s1 = = 1.04 ksi Ans.
t 0.25
s2 = 0 Ans.
Case (b):
pr 65(4)
s1 = ; s1 = = 1.04 ksi Ans.
t 0.25
pr 65(4)
s2 = ; s2 = = 520 psi Ans.
2t 2(0.25)
532
,08 Solutions 46060 5/28/10 8:34 AM Page 533
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentl
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*8–4. The tank of the air compressor is subjected to an
internal pressure of 90 psi. If the internal diameter of
the tank is 22 in., and the wall thickness is 0.25 in.,
determine the stress components acting at point A. Draw a
volume element of the material at this point, and show the
A
results on the element.
r 11
Hoop Stress for Cylindrical Vessels: Since = = 44 7 10, then thin wall
t 0.25
analysis can be used. Applying Eq. 8–1
pr 90(11)
s1 = = = 3960 psi = 3.96 ksi Ans.
t 0.25
Longitudinal Stress for Cylindrical Vessels: Applying Eq. 8–2
pr 90(11)
s2 = = = 1980 psi = 1.98 ksi Ans.
2t 2(0.25)
•8–5. The spherical gas tank is fabricated by bolting together
two hemispherical thin shells of thickness 30 mm. If the gas
contained in the tank is under a gauge pressure of 2 MPa,
determine the normal stress developed in the wall of the tank
and in each of the bolts.The tank has an inner diameter of 8 m
and is sealed with 900 bolts each 25 mm in diameter.
r 4
Normal Stress: Since = = 133.33 7 10, thin-wall analysis is valid. For the
t 0.03
spherical tank’s wall,
pr 2(4)
s = = = 133 MPa Ans.
2t 2(0.03)
Referring to the free-body diagram shown in Fig. a,
P = pA = 2 A 10 B c A 8 B d = 32p A 10 B N. Thus,
p 2
6 6
4
+ c ©Fy = 0; 32p A 106 B - 450Pb - 450Pb = 0
Pb = 35.56 A 103 B p N
The normal stress developed in each bolt is then
Pb 35.56 A 103 B p
sb = = 228 MPa Ans.
A 0.0252 B
=
Ab p
4
533
, 08 Solutions 46060 5/28/10 8:34 AM Page 534
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
8–6. The spherical gas tank is fabricated by bolting
together two hemispherical thin shells. If the 8-m inner
diameter tank is to be designed to withstand a gauge pressure
of 2 MPa, determine the minimum wall thickness of the
tank and the minimum number of 25-mm diameter bolts
that must be used to seal it. The tank and the bolts are made
from material having an allowable normal stress of 150 MPa
and 250 MPa, respectively.
Normal Stress: For the spherical tank’s wall,
pr
sallow =
2t
2 A 106 B (4)
150 A 106 B =
2t
t = 0.02667 m = 26.7 mm Ans.
r 4
Since = = 150 7 10, thin-wall analysis is valid.
t 0.02667
Referring to the free-body diagram shown in Fig. a,
P = pA = 2 A 106 B c A 82 B d = 32p A 106 B N. Thus,
p
4
32p A 106 B -
n n
+ c ©Fy = 0; (P ) - (Pb)allow = 0
2 b allow 2
32p A 106 B
n = (1)
(Pb)allow
The allowable tensile force for each bolt is
(Pb)allow = sallowAb = 250 A 106 B c A 0.0252 B d = 39.0625 A 103 B pN
p
4
Substituting this result into Eq. (1),
32p A 106 B
39.0625p A 103 B
n = = 819.2 = 820 Ans.
534
