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Unformatted text preview: Pressure Tanks: 1. The diameter of a ping pong ball is 40 111111 and the wall thickness is 0.5
mm. The ball is made of Nylon with an ultimate compressive strength of 95 MPa. Find the
pressure required to crush the ball. Solution:
0 _ po
4!
_ 4t_0
DI"
6 N
I: 0.0005 m 0' = 95 x10 —2 D”I = 0.0395
m _ 4><0.0005x95X106 = 4.810 MPa
0.0395 — Note: The density of sea water is approximately 1030 kg‘m’ so its weight is: 1030 kg x9.81£ = 10,1001} 3 m kg m
The depth. d. to which the ball would have to be submerged in order to crush it is then:
Pressure = weight density x depth 0 2
4.8]0xIO lleim =476m
10,100 N; m' — d: This is a very large value for the depth required to crush a ping pong ball. It is possible that
failure would be caused ﬁrst by some other mechanism like wall buckling. Pressure Tanks: 2. A cylindrical pressure tank with hemispherical end caps is to be fabricated
from mild steel 3/ 8 in thick. The length of the tank is to be three times the diameter. Find the
dimensions of the tank if o.=20 ksi and internal pressure is 500 psi. Gaumbk = 20,000 psi p = 500 psi
1/2 Dolﬂi 2Do Ail/2 Do
Solution:
Hoop stress is higher than longitudinal stress and thus is the one that limits the design.
a = pD’" = w
""°” 2! 2t
PDo pt
0 we or o a” W 2! 2’
call + p = pDo
2 21
2!
[call + = D0
D0 = “W [20,000 +%] = 30.375 in Overall length = 3Do = 91.125 in What is a higher grade of steel were used so that om = 35 ksi? D0 = 2x3/8[ 500 35,000 + — = 52.875 in resulting length = 158.625 in
500 2 — Pressure Tanks: 3 A cylindrical pressure vessel of inner diameter d=4ft (48 inches) and wall
thickness t=1/2 inch is fabricated from a welded pipe with helix angle rp=38°. Two spherical
caps, also of wall thickness t are welded to the cylindrical part. If the pressure inside the vessel is p=200psi, ﬁnd the normal stress O'w perpendicular to the weld. Solution a, = ﬂ= 200'“ =9,600psi
t 0.5 0'2 =ﬂ= 20024 =4,800psi
2t 20.5 Considering the xaxis to be in the vertical (along the cylinder direction,
at the weld, 9 = tp = 38°, 0, = 4800, O", = 9600, 1"}, = 0 . ox+o 0' a ax,=o = r I ’cos29+r sin29=
w 2 2 0
480039600 + 4800—9600 cos(_2 _ 38°) = 6619.4pSi Pressure Tanks: 4 For problem 3, find the shear stress 1'” parallel to the weld. Solution _ _ O'x—O'y , _
rm ¢'w ——Tsm29+rxy c0529—
_4800—9600 sin(—2  38°) = —2328.7 1 psi Pressure Tanks: 5 For the vessel in problem 3, ﬁnd the maximum shear stress anywhere in the
vessel, and in any possible plane. Solution
The maximum shear stress is the maximum semidifference of the principal stresses. In this case,
the maximum semidifference occurs in the cylindrical part and it is at the inside of the vessel, where ol=ﬂ=2°°'24=9,6oopsi
t 0.5
2: 205 2 2 a, = p = 200psi Pressure Tanks: 6 A pipe is formed by bolting together parts as the one shown.
The internal radius of the pipe is r=20cm and the wall thickness is (=1 .2cn1.
While the pipe is subjected to internal pressure ﬁ'om a ﬂuid of p=2,000Pa, ﬁnd
the force on each of the 12 bolts that connect the parts. Solution The pipe wall is subjected to a longitudinal stress (along the direction of the bolts) 0'2 = The total force resulting from this stress along a transverse cross section is F=O’2A =0227trt=g27trt =7t'pr2 =3.l4'20000.22 = 251.2N
force=£=ﬂ=2094 IV
12 12 For each bolt: ...
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This note was uploaded on 04/28/2010 for the course CE 215 taught by Professor Frantziskonis during the Spring '08 term at University of Arizona Tucson.
 Spring '08
 FRANTZISKONIS

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