Question

Question #1: A two-section bar is shown in following figure. Left end is fixed. The deflection at the right end is 0.02 mm. The information for this bar is:

E=210000 N/mm^2
L_1=100mm
L_2=50mm
EA_1/L_1=3000 N/mm
EA_2/L_2=1000 N/mm

Divide the bar into two1D elements with total 3 nodes. Use the FEM method to calculate strain, stress and reaction forces.

u-0.02 A2,E.L2

Answer 1

$k_1=\frac{EA_1}{L_1} \begin{bmatrix} 1 &-1 \\ -1& 1 \end{bmatrix}$

$k_2=\frac{EA_2}{L_2} \begin{bmatrix} 1 &-1 \\ -1& 1 \end{bmatrix}$

$\Rightarrow k_1=3k_2$

Using superposition concept ,we can obtain the global stifness matrix as

$K=\frac{E_2A_2}{L_2}\begin{bmatrix} 3 &-3 &0 \\ -3&3+1 & -1\\ 0& -1 & 1 \end{bmatrix}=\frac{E_2A_2}{L_2}\begin{bmatrix} 3 &-3 &0 \\ -3&4 & -1\\ 0& -1 & 1 \end{bmatrix}$

Equilibrium equation for the whole system

$\frac{E_2A_2}{L_2}\begin{bmatrix} 3 &-3 &0 \\ -3&4 & -1\\ 0& -1 & 1 \end{bmatrix}\begin{bmatrix} u_1\\ u_2\\ u_3 \end{bmatrix}=\begin{bmatrix} F_1\\ F_2\\ F_3 \end{bmatrix}$

we have $u_1=0\: u_3=0.02mm$ and $\frac{E_2A_2}{L_2}=1000N/mm$

$\Rightarrow 1000\begin{bmatrix} 3 &-3 &0 \\ -3&4 & -1\\ 0& -1 & 1 \end{bmatrix}\begin{bmatrix} 0\\ u_2\\ 0.02 \end{bmatrix}=\begin{bmatrix} F_1\\ F_2\\ F_3 \end{bmatrix}$

$\Rightarrow -3000u_2=F_1$

$\Rightarrow 4000u_2-20=F_2$

$\Rightarrow -1000u_2+20=F_3$

Also $F_1=-F_3$

$\Rightarrow -1000u_2+20=-F_1$

$\Rightarrow -1000u_2+20=3000u_2$

$\Rightarrow u_2=0.005mm$
$\Rightarrow F_1=-15N$
$\Rightarrow F_3=15N$

Strain $\delta_1 =\frac{0.005}{100}=0.00005$
     $\delta_2 =\frac{0.015}{100}=0.00015$

stress $\sigma _1=E_1\delta _1=10.5N/mm^2$

  $\sigma _2=E_2\delta _2=31.5N/mm^2$