Moment Deflection Relationship Problem

[simonsid]

Homework Statement

An electrical engineer in your semiconductor company is working on a design of an electrical contact with a Si wafer. The electrical contact is in the form of a thin Cu “finger” touching the back side of a wafer as shown below. The electrical engineer claims that he is getting poor signals in his electrical work because the contact between the “finger” and the wafer is a “point contact”. He has asked you to find an optimum angle at which the “finger” should approach the wafer so that the contact will be more like an “area contact” without causing permanent deformation of the finger. The electrical contact “finger” is sufficiently stiff and assume that it can be modeled as a cantilever beam. Note that the angle of the support relative to the wafer dictates the angle at which the finger approaches the wafer. Can you find such an “optimum” angle? If not, show rigorously that it is not possible to make this into an “area contact” as the electrical engineer claims.

Homework Equations

I considered the geometry of the beams
Taking the Length of the si wafter (diameter)=300mm
Thickness=1mm
Length of the CU finger=10mm
Thickness of the Cu finger=0.3mm

Slope (θ)=∫ (M/EI)dx
Deflection(v)=∫∫(M/EI)dx*dx

The Attempt at a Solution

Slope (θ)=∫ (M/EI)dx
Deflection(v)=∫∫(M/EI)dx*dx

I(si wafer)=(300*1^3)/12=25mm
I(cu finger)=(10*0.3^3)/12=0.0225mm

E (si wafer)=163Gpa
E(cu finger)=117Gpa

Iam really confused how to solve this problem,there is no load mentioned.

Theoretically slope= (-P*L^2)/(2*E*I) for an end point load cantilever beam,can anyone suggest the right approach to this problem?

 

[jlefevre76]

So, the angle at which the beam begins to deform permanently is when the maximum stress in the beam approaches the yield strength of copper. That is dependent on this equation:

$\sigma=\frac{My_{max}}{I}$
$y_{max}=thickness/2$
$\sigma=$stress in beam, should be set equal to yielding strength of copper

So, once you've found M from the above equation, plug that into your integral for the slope angle, θ

$Slope(\theta)=\int\frac{M}{EI}dx$