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Unformatted text preview: 08Ch08.qxd 9/18/08 11:06 AM Page 677 SECTION 8.5 Combined Loadings Solution 8.5-2 Gondola on a ski lift b W 677 180 mm sallow 12 kN
2 100 MPa (tension) pd
4 2 50 MPa S pd
S pst 3
4W d 4(6 kN) dmin 32 Wb 2 + 8b pd 3
0 13,089.97 1
m2 1.44 m
d 1.44 Solve numerically: d 4W
pd p(100 Mpa) psallow
4W 13,090 d3 3 MAXIMUM TENSILE STRESS
st tallow pst
8b Find dmin
A SUBSTITUTE NUMERICAL VALUES: 6 kN 0 (d meters) 0.04845 m 48.4 mm ; MAXIMUM SHEAR STRESS
Since tallow is one-half of sallow, the minimum diameter
for shear is the same as for tension.
tmax Problem 8.5-3 The hollow drill pipe for an oil well (see figure) is 6.2 in. in outer
diameter and 0.75 in. in thickness. Just above the bit, the compressive force in the
pipe (due to the weight of the pipe) is 62 k and the torque (due to drilling) is 185 k-in.
Determine the maximum tensile, compressive, and shear stresses in the
drill pipe. ...
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This note was uploaded on 12/22/2011 for the course MEEG 310 taught by Professor Staff during the Fall '11 term at University of Delaware.
- Fall '11