Elasticity/Beam bending example 1

Example 1

Given:

A long rectangular beam with cross section $ab$

Find:

A solution for the displacement and stress fields, using strong boundary conditions on the edges $x_2=0$ and $x_2=b$.

[Hint : Assume that the displacement can be expressed as a second degree polynomial (using the Pascal's triangle to determine the terms) $u(x,y)=A{x}^2+B{y}^2+Cxy+Dx+Ey+F$]

Solution

Step 1: Boundary conditions

$\begin{array}{cc}\text{at}{x}_1& =0;{\sigma }_{13}=0\\ \text{at}{x}_1& =a;u_3=0\\ \text{at}{x}_2& =0;{\sigma }_{23}=0\\ \text{at}{x}_2& =b;{\sigma }_{23}=S\end{array}$

Step 2: Assume a solution

Let us assume antiplane strain

$u_3(x_1,x_2)=A{x}_1^2+B{x}_2^2+C{x}_1{x}_2+D{x}_1+E{x}_2+F;u_1=u_2=0.$

Step 3: Calculate the stresses

The stresses are given by ${\sigma }_{\alpha 3}=\mu u_{3,\alpha }$, and ${\sigma }_{11}={\sigma }_{22}={\sigma }_{33}={\sigma }_{12}=0$. Therefore,

$\begin{array}{cc}{\sigma }_{13}& =\mu u_{3,1}=\mu (2A{x}_1+C{x}_2+D)\\ {\sigma }_{23}& =\mu u_{3,2}=\mu (2B{x}_2+C{x}_1+E)\end{array}$

Step 4: Satisfy stress BCs

Thus we have,

$\begin{array}{cc}0& =\mu (C{x}_2+D)\\ 0& =\mu (C{x}_1+E)\\ S& =\mu (2bB+C{x}_1+E)\end{array}$

Since $x_1$ and $x_2$ can be arbitrary, $C=D=E=0$.

Hence, $B=S/2\mu b$ which gives us

$\begin{array}{cc}{u}_3& =A{x}_1^2+\frac{S}{2\mu b}{x}_2^2+F\\ {\sigma }_{13}& =\mu (2A{x}_1)\\ {\sigma }_{23}& =\mu (2\frac{S}{2\mu b}{x}_2)\end{array}$

Assume that the body force is zero. Then the equilibrium condition is ${\mathrm{\nabla }}^2{u}_3=0$. Therefore,

$\begin{array}{cc}& u_{3,11}+u_{3,22}=0\\ \text{or,}& 2A+2\frac{S}{2\mu b}=0\\ \text{or,}& A=-\frac{S}{2\mu b}\end{array}$

Therefore, the stresses are given by

${\sigma }_{13}=-\frac{S}{b}{x}_1;{\sigma }_{23}=\frac{S}{b}{x}_2$

Step 5: Satisfy displacement BCs

The displacement is given by

$u_3=-\frac{S}{2\mu b}{x}_1^2+\frac{S}{2\mu b}{x}_2^2+F$

If we substitute $x_1=a$, we cannot determine the constant $F$ uniquely.

Hence the displacement boundary conditions have to be applied in a weak sense,

$\begin{array}{cc}& \underset{0}{\overset{b}{\int }}{u}_3(a,x_2)d{x}_2=0\\ \text{or,}& \underset{0}{\overset{b}{\int }}(-\frac{S}{2\mu b}{a}^2+\frac{S}{2\mu b}{x}_2^2+F)d{x}_2=0\\ \text{or,}& {[\frac{S}{2\mu b}(-a^2{x}_2+\frac{x_2^3}{3})+F{x}_2]|}_0^b=0\\ \text{or,}& \frac{S}{2\mu b}(-a^{2}b+\frac{b^3}{3})+Fb=0\\ \text{or,}& \frac{S}{2\mu }(-a^2+\frac{b^2}{3})+Fb=0\\ \text{or,}& \frac{S}{2\mu }(-\frac{a^2}{b}+\frac{b}{3})+F=0\\ \text{or,}& F=\frac{S}{2\mu b}(a^2-\frac{b^2}{3})\end{array}$

Therefore,

$u_3=\frac{S}{2\mu b}(x_2^2-x_1^2+a^2-\frac{b^2}{3})$