Deformation under axial loading 8= PL. AE Generalized Hooke's Law Ex Ey = y E E E -V E E E Txy Vxy -V E 5151 Torsion T=Tp/J = TL E7 = Y yz Bending -=0 My/I Shear T=VQ/(It) 4r y = 3 for a semi circle Stress Transformation Tx'y' 2 2 y + cos 20 + Txy T... sin 20 2 - sin 20+ cos 20 , max - x 2 y 2 y +1 max,min 2 2 = 5050501 G= E 2(1+v) Deflection El y" = M(x) Columns Pcr = nEI/Le (b) Figure 3(b) shows a shaft ABCDEF with diameter of 60 mm, which is made of a mild steel (G=80 GPa) and it is fixed at A. In the shaft, a coaxial hole of 30-mm diameter is drilled from B to D. The torque Tc = 6kNm is applied at position C and the torque TE = 1 kNm is applied at position E. The directions of the two torques are opposite to each other. Determine the angle of twist at the end F. Tc = 6 kNm TE = 1 kNm E 0.3 mi 0.3 mi 0.3 mi 0.3 m; 0.3 mi Figure 3(b) x (10 marks)