7.10 Numerical Examples
257
∴(3) can be written as
398τ 1 + 254S 4 = 25,322Gθ
−254τ 2 + 2 × 254 × τ 2 + 254τ 5 = 64,516Gθ
−254τ 2 + 2 × 254 × τ 3 + 254τ 6 = 64,516Gθ
(4)
Now, solving (1), (2) and (4) we get
τ 1 = 40.4 N/mm
2
τ 2 = 55.2 N/mm
2
τ 3 = 48.9 N/mm
2
τ 4 = −12.7 N/mm
2
τ 6 = 36.6 N/mm
2
7.11 Exercises
1. Develop expression for the shear stress and torsional rigidity of single cell tube
under torque.
2. The cross-section of hollow prismatic tube is as shown in Fig. 7.17. Evaluate the
angle of twist per unit length produced taking rigidity modulus as 75 GPa. Also
evaluate the maximum shear stress produced.
3. A hollow thin-wall torsion member has two compartments as shown in Fig. 7.18.
The material is aluminium alloy for which G = 26 GPa. Determine the torque
and unit angle of the twist if the maximum shearing stress is 40 MPa.
4. The hollow circular and square thin-wall members are having identical values of
b and t as shown in Fig. 7.19. Determine the ratio of the torques and unit angle of
twists for the two torsion members if shearing stresses are equal in both sections.
Neglect the effect of stress concentration.
5. A hollow thin-wall member has dimensions as shown in Fig. 7.20. It has a total
Fig. 7.17 Hollow prismatic
tube