Propped Cantilever with an end Moment first created
09/11/09 - last modified
09/11/09 Page Author: Ty Harness
A silicon beam 1[mm] long and a square cross section 10um x 10um fixed one end and propped
the other is shown in figure 1. The free end is subjected to moment,M of 1[uNm].
Find the maximum deflection and the slope at the propped end using an analytic method
and compare with any available finite element analysis program.
Work together in groups of 2 or 3 and present your solutions at the next tutorial meet.
Figure 1 - Propped Cantilever
Analytical Soloution
$E*I*(d^2y)/dx^2 = M - F*(L-x)$
$E*I*(dy)/dx = M*x - F*(L*x-x^2/2) + A$
$E*I*y = M*x^2/2 - F*(L*x^2/2-x^3/6) + A*x + B$
$y = 0$ when $x=0$ therefore $B=0$
$dy/dx=0$ when $x = 0$ therefore $A = 0$
$y = 0$ when $x=L$ therefore:
$0 = M*L^2/2 - F*(L*L^2/2-L^3/6)$
$F = (3*M)/(2*L)$
The slope equations is then:
$ E*I*dy/dx = -(M/2) * x + (3*M)/(4*L)*x^2 $
The slope at the propped end x=L:
$ dy/dx = (M*L)/(4*E*I) $
The deflection equation is then:
$ y(x) = (1/(E*I))* (-M/4*x^2 + M/(4*L)*x^3 )$
The max/min deflection will be given when dy/dx = 0
$0 = -M/2 * x + (3*M)/(4*L)*x^2 $
Using the quadratic formula you can obtain the values of x:
$x = 0$ and $x=2/3*L$
Substituting 2/3L into the deflection equation gives us the max deflection:
$y_max = y(2/3*L) = -(M*L^2)/27 $
Applying the analytical solution to the above problem
Use the following material properties:
A typical value of 1.08e11 can be used for the Young's Modulus for silicon[1].
$y_max = -(M*L^2)/27 = (-1*10^-6*(1*10^-3)^2)/27 =4.115*10^-4 [m]$
Finte Element Solution
I've chosen to use Prof. Rieg's Z88 Finite Elements Program because it's free and unlimited and ideal for student use.
The theory behind FEM is beyond the scope of this tutorial but you can learn a great deal from the Z88 manual.
The input files Z88I1.txt and Z88I2.txt are below if you wish to repeat the analysis using the Z88 software.
Z88i1.txt
