6 Engineering Example 4

6.1 Deflection of a uniformly loaded beam

Introduction

A uniformly loaded beam of length $L$ is supported at both ends as shown in Figure 9. The deflection $y\left(x\right)$ is a function of horizontal position $x$ and obeys the ordinary differential equation (ODE)

$\frac{d^{4}y}{d{x}^4}\left(x\right)=\frac{1}{EI}q\left(x\right)$ (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 9 :

Problem in words

Find the deflection of a beam, supported so that that there is no deflection and no curvature of the beam at its ends, subject to a uniformly distributed load, as a function of position along the beam.

Mathematical statement of problem

Find the equation of the curve $y\left(x\right)$ assumed by the bending beam that satisfies the ODE (1). Use the coordinate system shown in Figure 9 where the origin is at the left extremity of the beam. In this coordinate system, 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. $\frac{d^{2}y}{d{x}^2}{\left|\right.}_{x=L}=0$
5. $\frac{d^{2}y}{d{x}^2}{\left|\right.}_{x=L}=0$

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

Integrating Equation (1) leads to:

$EI\frac{d^{3}y}{d{x}^3}\left(x\right)=qx+A$ (2)

Integrating a second time:

$EI\frac{d^{2}y}{d{x}^2}\left(x\right)=q{x}^2∕2+Ax+B$ (3)

Integrating a third time:

$EI\frac{dy}{dx}\left(x\right)=q{x}^3∕6+A{x}^2∕2+Bx+C$ (4)

Integrating a fourth time:

$EIy\left(x\right)=q{x}^4∕24+A{x}^3∕6+B{x}^2∕2+Cx+D.$ (5)

The boundary conditions 1. and 2. enable determination of the constants of integration $A,B,C,D.$ Indeed, the boundary condition 1., $y\left(0\right)=0,$ and Equation (5) give

$EIy\left(0\right)=q\times {\left(0\right)}^4∕24+A\times {\left(0\right)}^3∕6+B\times {\left(0\right)}^2∕2+C\times \left(0\right)+D=0$

which yields $D=0$ .

The boundary condition 2., $y\left(L\right)=0,$ and Equation (5) give

$EIy\left(L\right)=q{L}^4∕24+A{L}^3∕6+B{L}^2∕2+CL+D.$

Using the newly found value for $D$ one writes

$q{L}^4∕24+A{L}^3∕6+B{L}^2∕2+CL=0$ (6)

The boundary condition 3. obtained from the definition of the radius of curvature, $\frac{d^{2}y}{d{x}^2}\left(0\right)=0,$ and Equation (3) give

$I\frac{d^{2}y}{d{x}^2}\left(0\right)=q\times {\left(0\right)}^2∕2+A\times \left(0\right)+B$

which yields $B=0$ . The boundary condition 4., $\frac{d^{2}y}{d{x}^2}\left(L\right)=0,$ and Equation (3) give

$EI\frac{d^{2}y}{d{x}^2}\left(L\right)=q{L}^2∕2+AL=0$

which yields $A=-qL∕2$ . The expressions for $A,B,D$ are introduced in Equation (6) to find the last unknown constant $C.$ This leads to $q{L}^4∕24-q{L}^4∕12+CL=0$ or $C=q{L}^3∕24.$ Finally, Equation (5) and the values of constants lead to the solution

$y\left(x\right)=\left[q{x}^4∕24-qL{x}^3∕12+q{L}^{3}x∕24\right]∕EI.$ (7)

Interpretation

The predicted deflection is zero at both ends as required, and you may check that it is symmetrical about the centre of the beam by switching to the coordinate system $\left(X,Y\right)$ with $L∕2-x=X$ and $y=Y$ and verifying that the deflection $Y\left(X\right)$ is symmetrical about the vertical axis, i.e. $Y\left(X\right)=Y\left(-X\right).$

Exercises

1. In an RC circuit (a resistor and a capacitor in series) the applied emf is a constant $E$ . Given that $\frac{dq}{dt}=i$ where $q$ is the charge in the capacitor, $i$ the current in the circuit, $R$ the resistance

   and $C$ the capacitance the equation for the circuit is

   $Ri+\frac{q}{C}=E.$

   If the initial charge is zero find the charge subsequently.

2. If the voltage in the RC circuit is $E=E_{0}cos\omega t$ find the charge and the current at time $t$ .
3. An object is projected from the Earth’s surface. What is the least velocity (the escape velocity) of projection in order to escape the gravitational field, ignoring air resistance.

   The equation of motion is

   $mv\frac{dv}{dx}=-mg\frac{R^2}{x^2}$

   where the mass of the object is $m$ , its distance from the centre of the Earth is $x$ and the radius of the Earth is $R$ .

4. The radial stress $p$ at distance $r$ from the axis of a thick cylinder subjected to internal pressure is given by $p+r\frac{dp}{dr}=A-p$ where $A$ is a constant. If $p=p_0$ at the inner wall ( $r=r_1$ ) and is negligible $\left(p=0\right)$ at the outer wall ( $r=r_2$ ) find an expression for $p$ .
5. The equation for an LCR circuit with applied voltage $E$ is

   $L\frac{di}{dt}+Ri+\frac{1}{C}q=E.$

   By differentiating this equation find the solution for $q\left(t\right)$ and $i\left(t\right)$ if $L=1$ , $R=100$ , $C=10^{-4}$ and $E=1000$ given that $q=0$ and $i=0$ at $t=0$ .

6. Consider the free vibration problem in Section 19.4 subsection 2 (page 57) when $m=1,n=1$ and $k=2$ (critical damping).

   Find the solution for $x\left(t\right)$ .

7. Repeat Exercise 6 for the case    $m=1,n=1$ and $k=1.5$  (light damping)
8. Consider the forced vibration problem in Section 19.4 subsection 2 with $m=1,n=25,k=8,E=sin3t$ , $x_0=0$ with an initial velocity of 3.
9. This refers to the Task on page 55 concerning modelling the dissolving of a pill in the stomach.

   An alternative model supposes that the pill is very rapidly permeated by stomach acids and the small granules contained in the capsule dissolve individually. In this case, the rate of change of volume is assumed to be directly proportional to the volume. Using the experimental data given in the Task, estimate the time for 95% of the pill to dissolve, based on this alternative model, and compare results.

Answer