Question

Problem 3. Determine the average shear stress at point A of the tube. G=27 GPa. 100 mm mm 750 N-m 4 mm Problem 4. For the figure of problem 3 and L = 2m determine the twisting angle . G = 27 GPa.

Answer 1

3 In case of tubes ie R10 R mean radius to thickras Tman Valid in any Cross-section tubes 20+ around mean line TA. Am ant 750 Nm 2 x 0.005784 x 0.000m 18084490.74 N ma 18.084 mpa meanline Ared = loxha =(0.054*0.096) 0.005184 at A = 18.084 mpa 2 N Thortore average Shear stress 4 Angle of two'st Пь 3 Valid for tubes any 4GA²t Th 8 4 GA² 8 = length of mean line S=2(0.054+0.096 t is different = 750 Nax 2m x 0.3m 0.3m 44 27X10'N 9 X 0.005784 mo* x 0.004 t=6mm an t=4mm But here, thickness is So 0= TL 50% varying TL 4GA² 2x (0.054 +0.096 0.004 TG A² 0.006 0.03049 had 750 Nmx 2m x 59 4x 27x109/2XC0.005184) There tore, twisting angle a = 0.0304 rad