Tj

per metre bd3 db3 bd ,2 l2 Now J =/„ + /„ = — + — = — (¿2 + ¿>2)

0.0476 rad/m

'2.73 - 2.77' Percentage error = (--) 100 = —1.44%

Example 52

Compare the torsional stiffness of the following cross-sections which can be assumed to be of unit length. Compare also the maximum shear stresses set up in each case:

(a) a hollow tube 40 mm mean diameter and 2 mm wall thickness;

(b) the same tube with a 2 mm wide saw-cut along its length;

(c) a rectangular solid bar, side ratio 4 to 1, having the same cross-sectional area as that enclosed by the mean diameter of the hollow tube;

(d) an equal-leg angle section having the same perimeter and thickness as the tube;

(e) a square box section having the same perimeter and thickness as the tube.

Solution

(a) In the case of the closed hollow tube we can apply the standard torsion equation

T _ Gd _ x 1 ~~L ~~r together with the simplified formula for the polar moment of area / of thin tubes,

T GJ 27T x (20 x 10"3)3 x 2 x 10~3G torsional stiffness = — = — =-

J 27T x (20 x 10"3)3 x 2 x 10-3 = 0.198 X 106r

angle of twist/unit length = - = -——r— = ——--y—

Maximum shear stress

0 0

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