# Introduction to Elasticity/Beam bending example 1

## Example 1

Given:

A long rectangular beam with cross section ${\displaystyle ab}$

Find:

A solution for the displacement and stress fields, using strong boundary conditions on the edges ${\displaystyle x_{2}=0}$ and ${\displaystyle x_{2}=b}$.

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

### Solution

Step 1: Boundary conditions

{\displaystyle {\begin{aligned}{\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{aligned}}}

Step 2: Assume a solution

Let us assume antiplane strain

${\displaystyle u_{3}(x_{1},x_{2})=Ax_{1}^{2}+Bx_{2}^{2}+Cx_{1}x_{2}+Dx_{1}+Ex_{2}+F~;~~u_{1}=u_{2}=0~.}$

Step 3: Calculate the stresses

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

{\displaystyle {\begin{aligned}\sigma _{13}&=\mu u_{3,1}=\mu (2Ax_{1}+Cx_{2}+D)\\\sigma _{23}&=\mu u_{3,2}=\mu (2Bx_{2}+Cx_{1}+E)\end{aligned}}}

Step 4: Satisfy stress BCs

Thus we have,

{\displaystyle {\begin{aligned}0&=\mu (Cx_{2}+D)\\0&=\mu (Cx_{1}+E)\\S&=\mu (2bB+Cx_{1}+E)\end{aligned}}}

Since ${\displaystyle x_{1}}$ and ${\displaystyle x_{2}}$ can be arbitrary, ${\displaystyle C=D=E=0}$.

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

{\displaystyle {\begin{aligned}u_{3}&=Ax_{1}^{2}+{\frac {S}{2\mu b}}x_{2}^{2}+F\\\sigma _{13}&=\mu (2Ax_{1})\\\sigma _{23}&=\mu (2{\frac {S}{2\mu b}}x_{2})\end{aligned}}}

Assume that the body force is zero. Then the equilibrium condition is ${\displaystyle \nabla ^{2}{u_{3}}=0}$. Therefore,

{\displaystyle {\begin{aligned}&u_{3,11}+u_{3,22}=0\\{\text{or,}}\quad &2A+2{\frac {S}{2\mu b}}=0\\{\text{or,}}\quad &A=-{\frac {S}{2\mu b}}\end{aligned}}}

Therefore, the stresses are given by

${\displaystyle {\sigma _{13}=-{\frac {S}{b}}x_{1}~;~~\sigma _{23}={\frac {S}{b}}x_{2}}}$

Step 5: Satisfy displacement BCs

The displacement is given by

${\displaystyle u_{3}=-{\frac {S}{2\mu b}}x_{1}^{2}+{\frac {S}{2\mu b}}x_{2}^{2}+F}$

If we substitute ${\displaystyle x_{1}=a}$, we cannot determine the constant ${\displaystyle F}$ uniquely.

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

{\displaystyle {\begin{aligned}&\int _{0}^{b}u_{3}(a,x_{2})dx_{2}=0\\{\text{or,}}\quad &\int _{0}^{b}\left(-{\frac {S}{2\mu b}}a^{2}+{\frac {S}{2\mu b}}x_{2}^{2}+F\right)dx_{2}=0\\{\text{or,}}\quad &\left.\left[{\frac {S}{2\mu b}}\left(-a^{2}x_{2}+{\frac {x_{2}^{3}}{3}}\right)+Fx_{2}\right]\right|_{0}^{b}=0\\{\text{or,}}\quad &{\frac {S}{2\mu b}}\left(-a^{2}b+{\frac {b^{3}}{3}}\right)+Fb=0\\{\text{or,}}\quad &{\frac {S}{2\mu }}\left(-a^{2}+{\frac {b^{2}}{3}}\right)+Fb=0\\{\text{or,}}\quad &{\frac {S}{2\mu }}\left(-{\frac {a^{2}}{b}}+{\frac {b}{3}}\right)+F=0\\{\text{or,}}\quad &F={\frac {S}{2\mu b}}\left(a^{2}-{\frac {b^{2}}{3}}\right)\end{aligned}}}

Therefore,

${\displaystyle {u_{3}={\frac {S}{2\mu b}}\left(x_{2}^{2}-x_{1}^{2}+a^{2}-{\frac {b^{2}}{3}}\right)}}$