Question 3.11: A prismatic bar AB, fixed at one end and free at the other, ......
A prismatic bar AB, fixed at one end and free at the other, is loaded by a distributed torque of constant intensity t per unit distance along the axis of the bar (Fig. 3-42).
(a) Derive a formula for the strain energy of the bar.
(b) Evaluate the strain energy of a hollow shaft used for drilling into the earth if the data are t= 2100 N·m/m, L= 3.7 m , G= 80 GPa, and I_p=7.15 × 10^{-6} m^4
Use a four-step problem-solving approach.
1,2. Conceptualize, Categorize:
Part (a): Strain energy of the bar.
The first step in the solution is to determine the internal torque T(x) acting at distance x from the free end of the bar (Fig. 3-42). This internal torque is equal to the total torque acting on the part of the bar between x = 0 and x = x . This latter torque is equal to the intensity t of torque times the distance x over which it acts:
T( x)= tx (a)
Substitute into Eq. (3-58)
U=\int_{0}^{L}{\frac{[T(x)]^2dx}{2GI_p(x)} } (3-58)
to obtain
U=\int_{0}^{L}{\frac{[T(x)]^2dx}{2GI_p} }=\frac{1}{2GI_p}\int_{0}^{L}(tx)^2dx=\frac{t^2L^3}{6GI_p} \hookleftarrow (3-64)
This expression gives the total strain energy stored in the bar.
3. Analyze:
Part (b): Numerical results.
To evaluate the strain energy of the hollow shaft, substitute the given data into Eq. (3-64):
U=\frac{t^2L^3}{6GI_p} =\frac{(2100 \ N\cdot m/m)^2(3.7\ m)^3}{6(80\ GPa)(7.15\times 10^{-6} m^4)} =65.1\ N\cdot m \hookleftarrow
4. Finalize: This example illustrates the use of integration to evaluate the strain energy of a bar subjected to a distributed torque.