Unformatted text preview: Sec_3.3.qxd 9/27/08 1:27 PM Page 265 265 SECTION 3.3 Circular Bars and Tubes Problem 3.3-17 A circular tube of inner radius r1 and outer radius r2 is
subjected to a torque produced by forces P 900 lb (see figure). The
forces have their lines of action at a distance b 5.5 in. from the outside
of the tube. P If the allowable shear stress in the tube is 6300 psi and the inner
radius r1 1.2 in., what is the minimum permissible outer radius r2?
b b 2 r2 Solution 3.3-17 Circular tube in torsion
SOLUTION OF EQUATION
UNITS: Pounds, Inches
Substitute numerical values:
6300 psi 4(900 lb)(5.5 in. + r2)(r2)
2 (1.2 in.)4] or
P 900 lb b 5.5 in. allow r1 6300 psi 2.07360
r2(r2 + 5.5)
or 1.2 in. r4
2 0.181891 0.181891 r2
2 Find minimum permissible radius r2 r2 1.000402 r2 Solve numerically: TORSION FORMULA
T 2P(b IP p4
1 2P(b + r2)r2
4P(b + r2)r2
All terms in this equation are known except r2.
IP 1.3988 in. MINIMUM PERMISSIBLE RADIUS
r2 1.40 in. ; 0 2.07360 0 ...
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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