# Homework Help: Statics Question (Using Modulus of Rigidity)

1. Jan 17, 2012

### papasmurf

1. The problem statement, all variables and given/known data

Find the displacement (mm) in the horizontal direction of point A due to the force, P. P=100kN w1=19mm w2=15mm

2. Relevant equations

$\tau$ = G * $\gamma$
$\tau$ = Shear stress = P / A
$\gamma$ = Shear strain = (pi / 2) - $\alpha$

3. The attempt at a solution

I haven't attempted to work out a solution here yet, but I do have a question regarding the separate G values that are given.

Can I just look at the top layer, the layer where P is acting, and use that G value to determine $\delta$? Or do I need to do something with the other G value as well?

If I were to try something, I would find tau by doing 100[kN] / (100[mm] * 2[mm]). So tau would be equal to 1[kN]/2[mm2] = 0.5[GPa]. Next I would find gamma by dividing tau by G (100[GPa]) giving me $\gamma$ = .005rad. I can use trig to define gamme as $\gamma$=sin-1($\delta$/40). Setting this equal to .005 I would get $\delta$= .20[mm].

Even if I do have to do something with both of the G values, I feel like my method is correct. Any help is appreciated, thanks in advance.

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Last edited: Jan 17, 2012
2. Jan 17, 2012

### Dr.PSMokashi

Hi papasmurf

Last edited: Jan 17, 2012
3. Jan 18, 2012

### pongo38

If delta at A is relative to the fixed base, then all the shear displacements of the various layers must be taken into account.

4. Jan 18, 2012

### papasmurf

Am I correct in assuming the shear force will be the same at all the various layers?

5. Jan 18, 2012

### papasmurf

Hi Dr.PSMokashi

6. Jan 18, 2012

### papasmurf

I'm getting closer to the correct answer. First I set V/A, where V is the internal shear force and A is the area of the cross section where the shear force is acting, equal to G*$\gamma$, where G is the modulus of rigidity and gamma is the shear strain.
I rewrote gamma as pi/2 - θ, where θ=cos-1($\delta$/h), h is the height of the "layer", and put it all together so that my equation looks like this:

V/A = G * ( pi/2 - cos-1($\delta$/h) )

Solving for $\delta$ I come up with
$\delta$ = h * cos( (pi/2) - V/AG)

I used this formula for each "layer" and added up all of the deltas.

However after plugging my numbers in and making sure of correct units, I still am off by fractions of a millimeter.

Last edited: Jan 18, 2012
7. Jan 18, 2012

### papasmurf

Also, should the h value be the height of the layer only or should it go from the base to the top of the layer I am looking at? For example if I am looking at the first layer where G=0.1MPa, would my h be simply w2 [mm] or would it be w2+2 [mm]?

8. Jan 18, 2012

### papasmurf

I keep getting an answer that is off by fractions of a millimeter. I can not figure out what I am doing/not doing that keeps giving me a wrong answer.

9. Jan 21, 2012

### pongo38

How do you know that "an answer that is off by fractions of a millimeter" is "a wrong answer"?