Shear Force and Bending Moment

[Mechaman]

Hi everyone here is the question I'm having trouble with. included is my attempt and the question.

1. The Reactions at A (Assuming the boom weight is negligible)
2. The shear force and bending moment along the boom length and indicate the point of maximum bending moment.
3. The maimum bending stress of the boom
4. What design changes would you propse

I'm stumped as to where the support arm comes into it. It's a pin that's attached to the side of the beam. There seems to be an X and Y reaction but there is no counter X so it remains at 0?

Any help would be much appreciated

 

[SteamKing]

Hi everyone here is the question I'm having trouble with. included is my attempt and the question.

1. The Reactions at A (Assuming the boom weight is negligible)
2. The shear force and bending moment along the boom length and indicate the point of maximum bending moment.
3. The maimum bending stress of the boom
4. What design changes would you propse

I'm stumped as to where the support arm comes into it. It's a pin that's attached to the side of the beam. There seems to be an X and Y reaction but there is no counter X so it remains at 0?

Any help would be much appreciated

Your static equilibrium calculations are a little iffy.

Your moment equation is ∑M_A = -0.9R_By + 6(2.4) = 0

When you moved the moment due to the weight of the engine to the RHS of the equation, you forgot to change the sign.

Any axial force at point B obviously must have an equal and opposite reaction at point A in order for the beam to remain in equilibrium. That's basic statics.

Once you get the correct forces acting on the beam, then you can work out the shear and bending moment.

 

[Mechaman]

Thanks for the reply,

I see the mistake with my moment equation.

To find the X axis at point B I would need to ∑M, but everywhere is running along the same axis? If I take the moment about the bottom left point, say point D, it is still along the same axis?

 

[SteamKing]

Thanks for the reply,

I see the mistake with my moment equation.

To find the X axis at point B I would need to ∑M, but everywhere is running along the same axis? If I take the moment about the bottom left point, say point D, it is still along the same axis?

You don't need to write a moment equation to determine the axial force. You have calculated the vertical component of the force in the support arm. You can use trig to calculate the horizontal component.

 

[Mechaman]

So the force is Cos53(16) = 9.62KN(t)? meaning Rax is 9.62KN(c)?

 

[SteamKing]

So the force is Cos53(16) = 9.62KN(t)? meaning Rax is 9.62KN(c)?

Only if 16 kN is the force in the rod, which it isn't. 16kN is the force opposite the angle of 53 deg. Check your trig again.