Calculate Average Shear Stress in Glue Bond - 50kN, 80mm

[Ry122]

I need to determine the average shear stress developed in the glue bond.
P=50kN and P2=80mm
http://users.on.net/~rohanlal/Untitled99.jpg

My attempt:
stress=force/area
stress=(50/(200*80))+((50/2)/(200*80))*2
Is this correct?

 

[PhanthomJay]

I need to determine the average shear stress developed in the glue bond.
P=50kN and P2=80mm
http://users.on.net/~rohanlal/Untitled99.jpg

My attempt:
stress=force/area
stress=(50/(200*80))+((50/2)/(200*80))*2
Is this correct?

No, the 50 kN force on the left (sandwiched) plate is balanced by the two forces, on the top and bottom plate. How much shear force does each glued surface (top and bottom) see? Don't forget to use proper units of stress.

 

[Ry122]

So I have to subtract the shearing stress from the two forces?
Is this correct then?
stress=(50/(200*80))-((50/2)/(200*80))*2 Mpa

 

[PhanthomJay]

No, you are not looking correctly at the forces involved. The plates are in equilibrium (sum of forces in x direction =0). The 50kN applied force at the left end, pointing left, is balnced by the two 25kN forces at the right end, pointing right. This is given, and is in accord with Newton's first law. Now draw a free body diagram of the top plate only...isolate it and examine the forces on it. There is a 25kN force acting at the right end of this top plate, pointing right. The top plate is in equilibrium, so you know that all the forces acting on the top plate must add to 0. So what must the force of the glued surface be on the left end of the top plate, to balance the 25kN force? Once you determine that force, the avg shear stress is that force divided by the shear area you have noted. The same applies for the bottom plate.

 

[Ry122]

So is it just
stress=25/(200*80)

 

[PhanthomJay]

So is it just
stress=25/(200*80)

Yes, correct, that is the average shear stress in the glue bond on the top, and it is also the average shear stress in the glue bond on the bottom. But watch your units! Your result is in kN/mm^2, you might want to convert it to Pascals (N/m^2).

 

[Ry122]

is it currently in mega pascals?

 

[PhanthomJay]

is it currently in mega pascals?

Looks like it's giga-pascals. I work in USA units of pounds per square inch, such that I don't get a good feel for the SI units, and there are so many decimals to confuse you. 25kN/16000mm^2 = .00156kN/mm^2, which is .00156 GPa, or 1.56 MPa, if I did the math right.