Fig. 5.13.

5.8 (B/C). Show that for the symmetrical section shown in Fig. 5.13 there is no stress in the central web. Show also that the shear stress in the remainder of the section has a value of T/4tb2.

5.9 (C). A washing machine agitator of the cross-section shown in Fig. 5.14 acts as a torsional member subjected to a torque T. The central tube is 100 mm internal diameter and 12 mm thick; the rectangular bars are 50 mm x 18 mm section. Assuming that the total torque carried by the member is given by

T = 7"tllbe + 47"bar determine the maximum value of t which the shaft can carry if the maximum stress is limited to 80 MN/ra2. (Hint: equate angles of twist of tube and bar.) [19.1 kNm.]

5.10 (C). The cross-section of an aeroplane elevator is shown in Fig. 5.15. If the elevator is 2 m long and constructed from aluminium alloy with G = 30 GN/m2, calculate the total angle of twist of the section and the magnitude of the shear stress in each part for an applied torque of 40 Nm.

Fig. 5.15.

5.11 (B/C). Develop a relationship between torque and angle of twist for a closed uniform tube of thin-walled non-circular section and use this to derive the twist per unit length for a strip of thin rectangular cross-section.

Use the above relationship to show that, for the same torque, the ratio of angular twist per until length for a closed square-section tube to that for the same section but opened by a longitudinal slit and free to warp is approximately At2/3b2, where t, the material thickness, is much less than the mean width b of the cross-section.

5.12 (C). A torsional member used for stirring a chemical process is made of a circular tube to which is welded four rectangular strips as shown in Fig. 5.16. The tube has inner and outer diameters of 94 mm and 100 mm respectively, each strip is 50 mm x 18 mm, and the stirrer is 3 m in length.

If the maximum shearing stress in any part of the cross-section is limited to 56 MN/m2, neglecting any stress concentration, calculate the maximum torque which can be carried by the stirrer and the resulting angle of twist over the full length.

For torsion of rectangular sections the torque T is related to the maximum shearing stress, rmax> and angle of twist, 6, in radians per unit length, as follows:

where b is the longer and d the shorter side of the rectangle and in this case, k\ = 0.264, ki =0.258 and G = 83 GN/m2. [C.E.I.] [2.83 kNm, 2.4°.]

5.13 (C). A long tube is subjected to a torque of 200 Nm. The tube has the double-cell, thin-walled, effective cross-section illustrated in Fig. 5.17. Assuming .that no buckling occurs and that the twist per unit length of the tube is constant, determine the maximum shear stresses in each wall of the tube.

5 mm

I 5mm


4mm i 5mm

100 mm



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