Akc
1 dO
2 dxi
The non zero stress components are obtained from Hooke's law dO
10 We need to check that this state of stress satisfies equilibrium dTij/dxj identically satisfied, whereas the other two yield
constant
Physically, this means that equilibrium is only satisfied if the increment in angular rotation (twist per unit length) is a constant.
11 We next determine the corresponding surface tractions. On the lateral surface we have a unit normal vector n = a(x2e2 + x3e3), therefore the surface traction on the lateral surface is given by
0 
Ti2 
Tis ' 
\ 0  
T2i 
0 
0 
< x2  
Tsi 
0 
0 
x2 Ti2 0 0 which is in agreement with the fact that the bar is twisted by end moments only, the lateral surface is traction free. 13 On the face xi = L, we have a unit normal n = ei and a surface traction t = Te i = Ï2ie2 + T3ie3 this distribution of surface traction on the end face gives rise to the following resultants We note that J(x2, + x])2dA is the polar moment of inertia of the cross section and is equal to J = na4/2, and we also note that J x2dA = J x3dA = 0 because the area is symmetric with respect to the axes.
14 From the last equation we note that which implies that the shear modulus j can be determined froma simple torsion experiment. 15 Finally, in terms of the twisting couple M, the stress tensor becomes Mx3 J MX2 J

Post a comment