Max Stress + Shear Stress for 10,000N Axial Load

[steve2510]

Homework Statement

A draw bar between a tractor and trailer is made from a length of steel bar 0.08m by 0.012m rectangular cross-section. The load is transmitted by means of a 0.015 diameter pin at each end. Determine the maximum stresses in the bar and pin if the axial load is 10,000N

Homework Equations

σ=F/A
ε=ΔL/L
τ=F_s/A

3. The attempt at the solution
Area of steel cross-section = 0.08x0.012=9.6x10^-4
Stress is bar = F/A
= 10000/9.6x10^-4
=10.42x10⁶
Shear in Bolt = F_s/A A = ∏d²/4
=1.767x10^-4m²
Shear in bolt = 10000/1.767x10^-4
=56.588 x 10⁶
But that's the shear in one bolt so surely the max shear is 2x that which would be 113.2MPA

Both values aren't the same as the ones in the back of the book so I'm not sure where I've gone wrong

 

[PhanthomJay]

Yor calc for bar stress looks good, but don't forget units! Why do say shear stress is doubled? Each pin sees the same force. If the pin is in double shear (need connection detail), shear stress is halved.

 

[steve2510]

I think i got confused with the term "maximum stress" is there any difference between asking: Determine the stress in the bar? And Determine the maximum stress in the bar

 

[PhanthomJay]

I think i got confused with the term "maximum stress" is there any difference between asking: Determine the stress in the bar? And Determine the maximum stress in the bar

Yes, generally there is a difference. In your example, the tensile stress in the bar is maximum at the pin holes, due to the reduction in cross section area of the bar at those locations. But I am not sure if the problem is asking for that...does it give the hole size and connection detail?

 

[steve2510]

For the area in the bar it's done (0.08-0.015)x0.012=a but I don't see why.

 

[PhanthomJay]

For the area in the bar it's done (0.08-0.015)x0.012=a but I don't see why.

Oh, OK, the cross section area must be reduced by the area of the hole, and that's what that calculation shows: the net cross section area at the hole. Another way of looking at it is this. The cross section without the hole is .08 X .012, and the area of the hole is .015 X .012. So the net area is (.08 X .012) - (.015 X .012) = (0.08-0.015) X 0.012, same result.

 

[steve2510]

Oh, OK, the cross section area must be reduced by the area of the hole, and that's what that calculation shows: the net cross section area at the hole. Another way of looking at it is this. The cross section without the hole is .08 X .012, and the area of the hole is .015 X .012. So the net area is (.08 X .012) - (.015 X .012) = (0.08-0.015) X 0.012, same result.

Ah i see thanks very much !