## Q. 9.2

A simple beam supports a concentrated downward force P at a distance a from the left support. The flexural rigidity E I is constant. Find w(x).

Given: Loading conditions, reaction forces, length of beam.
Find: Deflection w(x).
Assume: Hooke’s law applies; long, slender beam.

## Verified Solution

We want to integrate the internal moment to find the deflection w(x). We have simple supports, so we know that at A, w(x = 0) = 0 and M(x = 0) = 0, and at B, w(x = L) = 0 and M(x = L) = 0. We use the method of sections to find M(x):

$M_1(x)=-\frac{P b}{L} x,$ $\\M_2(x)=-\frac{P b}{L} x+P(x-a)=-\frac{P(L-a)}{L} x+P(x-a)=\frac{P a}{L}(x-L) .$

We note that there is a discontinuity at x = a, so we have two distinct M(x) expressions. Although M(x) may be discontinuous in this way, neither the slope nor the deflection is allowed to be discontinuous. We can, therefore, integrate the two distinct M(x) expressions for the deflections of the two portions of the beam, and match the two solutions at x = a.

For 0 ≤ x ≤ a,

$E I \frac{\mathrm{d}^2 w_1}{\mathrm{~d} x^2}=-M_1=\frac{P b}{L} x$,

Integrate: $\frac{\mathrm{d} w_1}{\mathrm{~d} x}=\frac{P b}{2 E I L} x^2+A_1.$
Integrate: $w_1(x)=\frac{P b}{6 E I L} x^3+A_1 x+A_2 .$

For a ≤ x ≤ L,

$E I \frac{\mathrm{d}^2 w_2}{\mathrm{~d} x^2}=-M_2=-\frac{P a}{L}(x-L)=P a-\frac{P a x}{L}$,
Integrate: $\frac{\mathrm{d} w_2}{\mathrm{~d} x}=\frac{P a x}{E I}-\frac{P a x^2}{2 E I L}+B_1 .$
Integrate: $w_2(x)=\frac{P a x^2}{2 E I}-\frac{P a x^3}{6 E I L}+B_1 x+B_2 .$

To find the constants of integration $A_i$ and $B_i$ , we will apply the end BCs as well as the continuity condition: both w and $dw/dx$ must be continuous at x = a, so that $w_1(a)=w_2(a)$ and $\left.\frac{\mathrm{d} w_1}{\mathrm{~d} x}\right|_{x=a}=\left.\frac{\mathrm{d} w_2}{\mathrm{~d} x}\right|_{x=a}$. Beginning with the end conditions, we have

\begin{aligned} w_1(0) & =0=A_2 ,\\ w_2(L) & =0=\frac{P a L^2}{3 E I}+B_1 L+B_2, \\ w_1(a) & =w_2(a) \rightarrow \frac{P a^3 b}{6 E I L}+A_1 a=\frac{P a^3}{2 E I}-\frac{P a^4}{6 E I L}+B_1 a+B_2, \\ \left.\frac{\mathrm{d} w_1}{\mathrm{~d} x}\right|_{x=a} & =\left.\frac{\mathrm{d} w_2}{\mathrm{~d} x}\right|_{x=a} \rightarrow \frac{P a^2 b}{2 E I L}+A_1=\frac{P a^2}{E I}-\frac{P a^3}{2 E I L}+B_1 . \end{aligned}

Here we have three equations for three unknown constants, so we can solve the equations simultaneously and obtain the remaining constants:

$A_1=-\frac{P b}{6 E I L}\left(L^2-b^2\right)$ and $B_1=-\frac{P a}{6 E \ I \ L}\left(2 L^2+a^2\right)$,

so

$A_2$ = 0 and $B_2 \frac{Pa^3}{6 \ E \ I }.$

So the deflection of the beam (after some esthetic rearrangements) is given by

$\begin{array}{ll} 0 \leq x \leq a: & w_1(x)=\frac{P b x}{6 E I L}\left(x^2+b^2-L^2\right) \\ a \leq x \leq L: & w_2(x)=-\frac{P a x^3}{6 E I L}+\frac{P a x^2}{2 E I}-\left[\frac{P a}{6 E I L}\left(2 L^2+a^2\right)\right] x+\frac{P a^3}{6 E I} . \end{array}$

Note: the deflection at the point of application of force P may be determined by substituting x = a into either of the above expressions and is $Pa^2b^2/3E I L$.
An alternative method, which does not require the need to explicitly enforce the continuity conditions at a, is to use discontinuity functions. Writing one expression for M(x) that works for the entire beam can be done with Macaulay brackets. There is no need to use the Macaulay bracket notation for x = 0, but it may be done for consistency of notation:

$M(x)=-\frac{P b}{L}\langle x-0\rangle^1+P\langle x-a\rangle^1 .$

We then carry on with the two integration steps:

$E I \frac{\mathrm{d}^2 w}{\mathrm{~d} x^2}=-M=\frac{P b}{L}\langle x-0\rangle^1-P\langle x-a\rangle^1,$

Integrate: $\frac{\mathrm{d} w}{\mathrm{~d} x}=\frac{P b}{2 E I L}\langle x-0\rangle^2-\frac{P}{2 E I}\langle x-a\rangle^2+C_1 .$
Integrate: $w(x)=\frac{P b}{6 E I L}\langle x-0\rangle^3-\frac{P}{6 E I}\langle x-a\rangle^3+C_1 x+C_2 .$

The $BC_s$ we need are just end conditions:

\begin{aligned} & w(0)=0=C_2, \\ & w(L)=0=\frac{P b}{6 E I L} L^3-\frac{P}{6 E I}(L-a)^3+C_1 L . \\ & \end{aligned}

So

$\begin{gathered} C_1=-\frac{P b L}{6 E I}+\frac{P}{6 E I L}(L-a)^3, \\ w(x)=\frac{P b}{6 E I L}\langle x-0\rangle^3-\frac{P}{6 E I}\langle x-a\rangle^3+\left[-\frac{P b L}{6 E I}+\frac{P}{6 E I L}(L-a)^3\right] x, \end{gathered}$

which looks different but is the same as the result obtained above. Plotting both solutions for arbitrary values of P, E I, a, and b is an easier way to demonstrate this than manipulating the expressions further. Remember that the second Macaulay bracket term does not get included until x > a.