Introduction to 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) $u(x,y) = Ax^2 + By^2 + Cxy + Dx + Ey + F$ ]

Solution

Step 1: Boundary conditions

\begin{align} \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{align}

Step 2: Assume a solution

Let us assume antiplane strain

u_3(x_1,x_2) = Ax_1^2 + Bx_2^2 + Cx_1x_2 + Dx_1 + Ex_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{align} \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{align}

Step 4: Satisfy stress BCs

Thus we have,

\begin{align} 0 & = \mu(Cx_2 + D) \\ 0 & = \mu(Cx_1 + E) \\ S & = \mu(2bB + Cx_1 + E) \end{align}

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

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

\begin{align} 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{align}

Assume that the body force is zero. Then the equilibrium condition is $\nabla^2{u_3} = 0$ . Therefore,

\begin{align} & 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{align}

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{align} & \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) + F x_2 \right]\right|_0^b = 0\\ \text{or,} \quad & \frac{S}{2\mu b} \left(-a^2 b + \frac{b^3}{3}\right) + F b = 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{align}

Therefore,

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

This article is issued from Wikiversity - version of the Monday, February 01, 2016. The text is available under the Creative Commons Attribution/Share Alike but additional terms may apply for the media files.