
#1
Jan1712, 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[mm^{2}] = 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. 



#2
Jan1712, 06:06 PM

P: 1

Hi papasmurf




#3
Jan1812, 11:26 AM

P: 692

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
Jan1812, 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?




#5
Jan1812, 12:56 PM

P: 22

Hi Dr.PSMokashi




#6
Jan1812, 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. 



#7
Jan1812, 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 w_{2} [mm] or would it be w_{2}+2 [mm]?




#8
Jan1812, 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.




#9
Jan2112, 06:11 AM

P: 692

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



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