Shear stress on beam loading

[Mechaman]

1. Determine average shear stress. Question attached2. Shear Stress = F/A , Sum of Forces and Moments = 0, Trig3. Attempt attached.

Having a hard time determining reaction at pins. I worked out that the opposite side is 4.5m. I assume the next step is to determine reaction then apply Shear Stress = F/A to the Y forces on the pins only??

 

[SteamKing]

Where do you get that the acute angle is 37 degrees? The figure with the problem statement says the angle is 30 degrees, as does the sketch attached to your work.

 

[Mechaman]

Where do you get that the acute angle is 37 degrees? The figure with the problem statement says the angle is 30 degrees, as does the sketch attached to your work.

You're right, I don't know why I put 37 down. Even so, any idea of finding the reactions?

 

[SteamKing]

You're right, I don't know why I put 37 down. Even so, any idea of finding the reactions?

Yes. And you should know how, too, but you stopped your calculations for some reason.

At point B, the ΣF = 0. You have calculated R_By = 40 kN. The unknown force in the support rod BC must have a vertical component which is equal to and opposite of R_By. Based on the geometry, you can work out what the tension force in the support rod BC must be.

I noticed that in writing your equations to sum forces in the vertical direction, you included the force at Pin C. You should restrict your FBD to the beam AB and write equilibrium equations only for that member. You can always treat the support rod BC as a separate free body.

 

[Mechaman]

So is the Y force running through point A the same as point C?

 

[SteamKing]

So is the Y force running through point A the same as point C?

Beats me. But the way you had your original equilibrium equations set up, you couldn't solve them. That's why I recommended that you treat the beam AB as a separate free body and the support rod BC as another free body. The equilibrium equations you write for each free body can be solved.

Pins A and C are not connected to each other in any way, so the vertical forces in each will not necessarily be equal either.