I did the Lindeburg PE sample exam (which I found very hard). and on problem 92, afternoon session asks for the minimum diameter of a shaft subject to both a bending moment and a Torque. I arrived to equivalent normal stress=r/J*(M^2+T^2)^0.5. and the solution arrived to stress=r/J*(4M^2+3T^2)^0.5. I have hard time understanding where the "3" came from. because for me bending stress=Mr/I and Shear stress =Tr/J and the equivalent normal stress = ( (bending stress/2)^2+(shear stress)^2)^0.5. Any ideas?
Thanks!
Thanks!