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Statics Question (Using Modulus of Rigidity)

by papasmurf
Tags: modulus, rigidity, statics
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papasmurf
#1
Jan17-12, 12:07 PM
P: 22
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

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

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 [itex]\delta[/itex]? 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 [itex]\gamma[/itex] = .005rad. I can use trig to define gamme as [itex]\gamma[/itex]=sin-1([itex]\delta[/itex]/40). Setting this equal to .005 I would get [itex]\delta[/itex]= .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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Dr.PSMokashi
#2
Jan17-12, 06:06 PM
P: 1
Hi papasmurf
pongo38
#3
Jan18-12, 11:26 AM
P: 696
If delta at A is relative to the fixed base, then all the shear displacements of the various layers must be taken into account.

papasmurf
#4
Jan18-12, 12:56 PM
P: 22
Statics Question (Using Modulus of Rigidity)

Am I correct in assuming the shear force will be the same at all the various layers?
papasmurf
#5
Jan18-12, 12:56 PM
P: 22
Hi Dr.PSMokashi
papasmurf
#6
Jan18-12, 02:13 PM
P: 22
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*[itex]\gamma[/itex], where G is the modulus of rigidity and gamma is the shear strain.
I rewrote gamma as pi/2 - θ, where θ=cos-1([itex]\delta[/itex]/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([itex]\delta[/itex]/h) )

Solving for [itex]\delta[/itex] I come up with
[itex]\delta[/itex] = 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.
papasmurf
#7
Jan18-12, 02:37 PM
P: 22
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]?
papasmurf
#8
Jan18-12, 03:34 PM
P: 22
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.
pongo38
#9
Jan21-12, 06:11 AM
P: 696
How do you know that "an answer that is off by fractions of a millimeter" is "a wrong answer"?


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