Determine the strain energy release rate
Gfor the double torsion specimen shown in the Figure using strength of materials analysis and energy methods. Each arm of the specimen may be considered as a rectangular bar built-in at its end of length
a, and cross-section dimensions
Wand
B. The rate of twist for each arm due to torsion is given by
\theta _(t)=(M_(t))/(k\mu B^(3)W),and thus, the total angle of rotation
\alpha for each arm is given by
\alpha =\theta _(t)a. In the above formula,
kdepends on the ratio
(W)/(B). Values for
kare given in the table below. Evaulate the problem under both load control and displacement control conditions. Comment on the nature of the solutions you obtain in terms of the stability conditions and also comment on why the double torsion specimen is a popular specimen for measuring critical strain energy release rate in ceramic materials? How would changing the direction of the the torque on one side change the result?
