Question 9.6: Use Castigliano’s second theorem to find the deflection of t......

Castigliano’s second theorem lets us find the deflection at B in a straightforward manner. We find the strain energy in the beam, take the partial derivative of it with respect to force P, and the result is the deflection at B, the point of application of force P. It pays to think ahead to the other deflection that is requested before beginning, however. There is no point load at C, so how can we find the deflection there? We can cleverly add a “dummy” force Q at C, and then the partial derivative of strain energy with respect to Q is the deflection at C. We then set the value of Q to its real value, zero.
To use Castigliano’s second theorem, we need the function M(x) which we can write using Macaulay bracket notation. Starting with the load q(x) and including the reaction force R_1 and moment M_1 at the left end,

q(x)=-M_A\langle x-0\rangle^{-2}+R_A\langle x-0\rangle^{-1}-P\left\langle x-\frac{1}{2} L\right\rangle^{-1}-Q\langle x-L\rangle^{-1} .

The final term is always zero and is dropped in the following expressions. Integrating twice and remembering the negative sign in dV/dx = −q, we obtain

\begin{aligned} & V(x)=M_A\langle x-0\rangle^{-1}-R_A\langle x-0\rangle^0+P\left\langle x-\frac{1}{2} L\right\rangle^0 ,\\ & M(x)=M_A\langle x-0\rangle^0-R_A\langle x-0\rangle^1+P\left\langle x-\frac{1}{2} L\right\rangle^1. \end{aligned}

Using equilibrium, we can eliminate M_A = P L/2 + QL and R_A = P + Q. And we recognize that the Macaulay brackets are not needed on the first two terms:

M(x)=\frac{1}{2} P L+Q L-P x-Q x+P\left\langle x-\frac{1}{2} L\right\rangle^1 .

Now the strain energy can be broken up into two integrals:
\begin{aligned} U & =\frac{1}{2} \int_0^L \frac{[M(x)]^2}{E I} \mathrm{~d} x \\ & =\frac{1}{2 E I} \int_0^{\frac{1}{2} L}\left[\frac{1}{2} P L+Q L-P x-Q x\right]^2 \mathrm{~d} x+\frac{1}{2 E I} \int_{\frac{1}{2} L}^L[Q L-Q x]^2 \mathrm{~d} x \\ & =\frac{L^3}{48 E I}\left(P^2+5 P Q+7 Q^2\right)+\frac{L^3}{48 E I} Q^2=\frac{L^3}{48 E I}\left(P^2+5 P Q+8 Q^2\right) \end{aligned}
Implementing Castigliano’s second theorem,

\begin{aligned} & \Delta_B=\frac{\partial U}{\partial P}=\frac{L^3}{48 E I}(2 P+5 Q), \\ & \Delta_C=\frac{\partial U}{\partial Q}=\frac{L^3}{48 E I}(5 P+16 Q). \end{aligned}

These deflections are both downward, the direction of P and Q. Now setting Q to its real value, zero, we get the final result:

\Delta_B=\frac{P L^3}{24 E I} and \Delta_C=\frac{5 P L^3}{48 E I} .