5 Engineering Example 2

5.1 Deflection of a uniformly loaded beam

Introduction

A uniformly loaded beam of length $L$ is supported at both ends. The deflection $y\left(x\right)$ is a function of horizontal position $x$ and obeys the differential equation

$\frac{d^{4}y}{d{x}^4}\left(x\right)=\frac{1}{EI}q\left(x\right)$ (1)

where $E$ is Young’s modulus, $I$ is the moment of inertia and $q\left(x\right)$ is the load per unit length at point $x$ . We assume in this problem that $q\left(x\right)=q$ (a constant). The boundary conditions are (i) no deflection at $x=0$ and $x=L$ (ii) no curvature of the beam at $x=0$ and $x=L$ .

Figure 20

Problem in words

In addition to being subject to a uniformly distributed load, a beam is supported so that there is no deflection and no curvature of the beam at its ends. Applying a Laplace Transform to the differential equation (1), find the deflection of the beam as function of horizontal position along the beam.

Mathematical formulation of the problem

Find the equation of the curve $y\left(x\right)$ assumed by the bending beam that solves (1). Use the coordinate system shown in Figure 1 where the origin is at the left extremity of the beam. In this coordinate system, the mathematical formulations of the boundary conditions which require that there is no deflection at $x=0$ and $x=L$ , and that there is no curvature of the beam at $x=0$ and $x=L,$ are

1. $y\left(0\right)=0$
2. $y\left(L\right)=0$
3. $\frac{d^{2}y}{d{x}^2}{\left|\right.}_{x=0}=0$
4. $\frac{d^{2}y}{d{x}^2}{\left|\right.}_{x=L}=0$

Note that $\frac{dy\left(x\right)}{dx}$ and $\frac{d^{2}y\left(x\right)}{d{x}^2}$ are respectively the slope and the radius of curvature of the curve at point $\left(x,y\right).$

Mathematical analysis

The following Laplace transform properties are needed:

$\mathcal{L}\left\{\frac{d^{n}f\left(t\right)}{d{t}^n}\right\}=s^{n}F\left(s\right)-{\sum }_{k=1}^n{s}^{k-1}\frac{d^{n-k}f}{d{x}^{n-k}}{\left|\right.}_{x=0}$ (P1)

$\mathcal{L}\left\{1\right\}=1∕s$ (P2)

$\mathcal{L}\left\{t^n\right\}=n!∕s^{n+1}$ (P3)

${\mathcal{L}}^{-1}\left\{\mathcal{L}\left\{f\left(t\right)\right\}\right\}=f\left(t\right)$ (P4)

To solve a differential equation involving the unknown function $f\left(t\right)$ using Laplace transforms

1. Write the Laplace transform of the differential equation using property (P1)
2. Solve for the function $\mathcal{L}\left\{f\left(t\right)\right\}$ using properties (P2) and (P3)
3. Use the inverse Laplace transform to obtain $f\left(t\right)$ using property (P4)

Using the linearity properties of the Laplace transform, (1) becomes

$\mathcal{L}\left\{\frac{d^{4}y}{d{x}^4}\left(x\right)\right\}-\mathcal{L}\left\{\frac{q}{EI}\right\}=0.$

Using (P1) and (P2)

$s^4\mathcal{L}\left\{y\left(x\right)\right\}-{\sum }_{k=1}^4{s}^{k-1}\frac{d^{4-k}y}{d{x}^{4-k}}{\left|\right.}_{x=0}-\frac{q}{EI}\frac{1}{s}=0.$ (2)

The four terms of the sum are

${\sum }_{k=1}^4{s}^{k-1}\frac{d^{4-k}y}{d{x}^{4-k}}=\frac{d^{3}y}{d{x}^3}{\left|\right.}_{x=0}+d\frac{d^{2}y}{d{x}^2}{\left|\right.}_{x=0}+s^2\frac{dy}{dx}{\left|\right.}_{x=0}+s^{3}y\left(0\right).$

The boundary conditions give $y\left(0\right)=0$ and $\frac{d^{2}y}{d{x}^2}=0.$ So (2) becomes

$s^{4}L\left\{y\left(x\right)\right\}-\frac{d^{3}y}{d{x}^3}{\left|\right.}_{x=0}-s^2\frac{dy}{dx}{\left|\right.}_{x=0}-\frac{q}{EI}\frac{1}{s}=0.$ (3)

Here $\frac{d^{3}y}{d{x}^3}{\left|\right.}_{x=0}$ and $\frac{dy}{dx}{\left|\right.}_{x=0}$ are unknown constants , but they can be determined by using the remaining two boundary conditions $y\left(L\right)=0$ and $\frac{d^{2}y}{d{x}^2}{\left|\right.}_{x=L}=0.$

Solving for $\mathcal{L}\left\{y\left(x\right)\right\}$ , (3) leads to

$\mathcal{L}\left\{y\left(x\right)\right\}=\frac{1}{s^4}\frac{d^{3}y}{d{x}^3}{\left|\right.}_{x=0}+\frac{1}{s^2}\frac{dy}{dx}{\left|\right.}_{x=0}+\frac{q}{EI}\frac{1}{s^5}.$

Using the linearity of the Laplace transform, the inverse Laplace transform of this equation gives

${\mathcal{L}}^{-1}\left\{\mathcal{L}\left\{y\left(x\right)\right\}\right\}=\frac{d^{3}y}{d{x}^3}{\left|\right.}_{x=0}\times {\mathcal{L}}^{-1}\left\{\frac{1}{s^4}\right\}+\frac{dy}{dx}{\left|\right.}_{x=0}\times {\mathcal{L}}^{-1}\left\{\frac{1}{s^2}\right\}+\frac{q}{EI}{\mathcal{L}}^{-1}\left\{\frac{1}{s^5}\right\}.$

Hence

$y\left(x\right)=\frac{d^{3}y}{d{x}^3}{\left|\right.}_{x=0}\times {\mathcal{L}}^{-1}\left\{3!\frac{1}{s^4}\right\}∕3!+\frac{dy}{dx}{\left|\right.}_{x=0}\times {\mathcal{L}}^{-1}\left\{\frac{1}{s^2}\right\}+\frac{q}{EI}{\mathcal{L}}^{-1}\left\{4!\frac{1}{s^5}\right\}∕4!$

So using (P3)

$y\left(x\right)=\frac{d^{3}y}{d{x}^3}{\left|\right.}_{x=0}\times {\mathcal{L}}^{-1}\left\{\mathcal{L}\left\{x^3\right\}\right\}∕6+\frac{dy}{dx}{\left|\right.}_{x=0}\times {\mathcal{L}}^{-1}\left\{\mathcal{L}\left\{x^1\right\}\right\}+\frac{q}{EI}{\mathcal{L}}^{-1}\left\{\mathcal{L}\left\{x^4\right\}\right\}∕24.$

Simplifying by means of (P4)

$y\left(x\right)=\frac{d^{3}y}{d{x}^3}{\left|\right.}_{x=0}\times x^3∕6+\frac{dy}{dx}{\left|\right.}_{x=0}\times x+\frac{q}{EI}{x}^4∕24.$ (4)

To use the boundary condition $\frac{d^{2}y}{d{x}^2}{\left|\right.}_{x=L}=0$ , take the second derivative of (4), to obtain

$\frac{d^{2}y}{d{x}^2}\left(x\right)=\frac{d^{3}y}{d{x}^3}{\left|\right.}_{x=0}\times x+\frac{q}{2EI}{x}^2.$

The boundary condition $\frac{d^{2}y}{d{x}^2}{\left|\right.}_{x=L}=0$ implies

$\frac{d^{3}y}{d{x}^3}{\left|\right.}_{x=0}=-\frac{q}{2EI}L.$ (5)

Using the last boundary condition $y\left(L\right)=0$ with (5) in (4)

$\frac{dy}{dx}{\left|\right.}_{x=0}=\frac{q{L}^3}{24EI}$ (6)

Finally substituting (5) and (6) in (4) gives

$y\left(x\right)=\frac{q}{24EI}{x}^4-\frac{qL}{12EI}{x}^3+\frac{q{L}^3}{24EI}x.$

Interpretation

The predicted deflection is zero at both ends as required.

Note  This problem was solved by an entirely different means (integrating the ODE) in HELM booklet  19.4, page 65.