## Q. 9.3

For the beam with the given loading, with a maximum load intensity of $q_0$ , find (a) the reaction at A, (b) the equation of the elastic curve w(x), and (c) the slope at A.

Find: Reactions, deflection w(x), slope of neutral axis at A.
Assume: Hooke’s law applies; long, slender beam. ## Verified Solution

\begin{aligned} & \sum F_y=0=R_A+R_B-\frac{1}{2} q_0 L, \\ \curvearrowleft & \sum M_{\text {about } A}=0=-M_B+R_B L-\left(\frac{1}{2} q_0 L\right)\left(\frac{2 L}{3}\right) \text {, } \\ \curvearrowleft & \sum M_{\text {about } B}=0=-M_B-R_A L+\left(\frac{1}{2} q_0 L\right)\left(\frac{L}{3}\right) . \\ & \end{aligned}

We have three unknowns ($R_A, \ R_B$, and $M_B$) and only two relevant equilibrium equations, so this problem is statically indeterminate. We will proceed with the solution for w(x), leaving the reactions as unknowns, and hope that our boundary conditions for V, M, dw/dx, and w may help us out. First, we will make an imaginary cut at some x to determine the form of M(x).

We require moment equilibrium about our point x, that is, $-M(x)+\left(\frac{1}{2} q_0 x^2 / L\right)(x / 3)- R_A x=0$. Thus, $M(x)=-R_A x+q_0 x^3 / 6 L$. Having this expression for internal bending moment as a function of x allows us to integrate the second-order equation for deflection w(x):

$E I \frac{d^2 w(x)}{d x^2}=-M(x)=R_A x-\frac{q_0 x^3}{6 L} .$

Integrate: $E I \frac{d w(x)}{d x}=\frac{1}{2} R_A x^2-\frac{q_0 x^4}{24 L}+C_1 .$
Integrate: $\operatorname{EI} w(x)=\frac{1}{6} R_A x^3-\frac{q_0 x^5}{120 L}+C_1 x+C_2 .$

Note that the numbering scheme for our constants of integration is not tied to the numbered $C_i$ cited in Section 9.3. Although this scheme was followed in Example 9.1, there is no need to stick to it. In working problems, we will most often be integrating the second-order equation and so will have only two constants to find, so they may be named in any manner the problem solver deems appropriate.
We now need some boundary conditions to find the constants $C_1$ and $C_2$ above. At A, where x = 0, we have a pin support, at which we are sure both moment and deflection are zero. Then one of these that helps us is w(x = 0) = 0. At B, or x = L, we have a fixed support, where deflection and slope must both be zero. Applying these three BCs gets us

\begin{aligned} w(x=0)=0 \quad \rightarrow \quad C_2=0, \\ w(x=L)=0 \quad \rightarrow \quad \frac{1}{6} R_A L^3-\frac{q_0 L^4}{120}+C_1 L=0, \\ \left.\frac{\mathrm{d} w}{\mathrm{~d} x}\right|_{x=L}=0 \quad \rightarrow \quad \frac{1}{2} R_A L^2-\frac{q_0 L^3}{24}+C_1=0 . \end{aligned}

At last, we have two equations and two unknowns, a soluble system. We choose arbitrarily to solve for $R_A$ first, and do this by multiplying the slope boundary condition by L and then subtracting the deflection condition:

$\left(\frac{1}{2} R_A L^3-\frac{q_0 L^4}{24}+C_1 L\right)-\left(\frac{1}{6} R_A L^3-\frac{q_0 L^4}{120}+C_1 L\right)=0 .$

So,

$\frac{1}{3} R_A L^3-\frac{q_0 L^4}{30}=0$

This allows us to solve for $R_A=\frac{1}{10} q_0 L$, which is an upward force as assumed in the FBD, and which we note is independent of E I. By substituting this $R_A$ into either condition at x = L, we are able to find that the constant $C_1=-\frac{1}{120} q_0 L^3$. Putting both of these into our expression for the deflection of the neutral axis, we have

$E I w(x)=\frac{1}{6}\left(\frac{1}{10} q_0 L\right) x^3-\frac{q_0 x^5}{120 L}+\left(-\frac{1}{120} q_0 L^3\right) x ,$

or

$w(x)=\frac{q_0 / L}{120 E I}\left(-x^5+2 L^2 x^3-L^4 x\right)$

We could then find a general expression for the slope dw/dx of the neutral axis along the beam, and find the slope at A as requested in part (c):

$\frac{\mathrm{d} w}{\mathrm{~d} x}(x=0)=-\frac{q_0 L^3}{120 E I}$

Note: We could also have solved this complex problem by recognizing the loading on the beam as the superposition of two more straightforward conditions:

The superposition of w(x) for both these loading conditions is exactly the result achieved above. Superposition is quite a useful technique for finding the deflections of beams.   