Akc

Figure 10.1: Torsion of a Circular Bar

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.

Ri =

J TiidA

R2 =

J T2idA

R3 =

jTsidA

Mi =

J (x2T3i

M2 =

M3 = 0

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

Mx3

Mx2

J

J

0

0

0

0 0

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