Deflection/stress/torque in a horizontal crane

Question

I made a google drive link to the problem (as I cannot paste images on here): https://drive.google.com/file/d/1r9vwwLlm0KNfPZJfgtfgu6KW6wBKw9Ym/view?usp=sharing

How would you set up this problem? If you can provide an explanation for the answers to these questions that would be super helpful!

Answer 1

https://structx.com/Beam_Formulas_026.html

The above is a reference you can use for overhanging beams. I will use the variable names in the above reference. (P for the vertical load, L for the span between supports etc).

Notice that your problem has 2 pins. The only thing this matters for is the axial compression caused by the cord. Since the pins cannot translate horizontally, the axial strain in the beam between the pin supports is zero. Therefore the right pin is responsible for reacting against the entirety of the axial force caused by the cord. Therefore, axial force is P (compression) in region a of the beam, and zero in region L.

Since stress due to axial load is F/A

We can now find the stress due to axial compression:

σ_axial = 0 in region L

σ_axial = P/(bh) in region a

Now for the bending component:

Again, referencing the link provided at the top, we can find the moment at any point in the beam, which we will use to calculate bending stress.

There are 2 spots along the beam where we need to determine the stress due to bending, so that we can determine the maximum tensile stress in the beam: either side of the right pin.

For both of these locations, internal moment in the beam is equal to M_max from the link, so

M = Pa

The equation for maximum stress due to bending is σ_bending = Mc/I, where c is the distance from the neutral axis to the edge, and I is the moment of inertia with respect to the neutral axis.

To the left of the pin:

σ_total = σ_bending

To the right of the pin:

σ_total = σ_bending + σ_axial

Since σ_axial can only be negative, it will always decrease the tensile stress in the cross-section. Therefore, to the left of the pin is where the max tensile stress occurs.

calculating other variables we'll need to find stress due to bending:

I = bh³/12

c = h/2

σ_bending = Mc/I = (Pa)(h/2)/(bh³/12) = 6Pa/(bh²)

and so the maximum tensile stress would be 6Pa/(bh²) (I'll leave it to you to convert back to your figure's variables)

And note that a is taken from the figure of the link I provided, not your diagram. Since the beam is in negative bending here, the location of this max tensile stress is along the top edge of the beam, and as stated earlier, directly to the left of the inner pin where the axial load drops to zero.

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Deflection of the right end of the beam would be Δ_max from my provided link, so

Δ_max = Pa²(L+a)/(3EI)

I'll leave it to you to convert back to your figure's variables and plug in I.

If you aren't allowed to use this deflection table and have to determine deflection by hand, you'll need create a free body diagram to determine the equation for moment along the beam M(x). Then you can integrate this equation twice and use the pins as boundary conditions, and divide by EI to obtain the deflection at the right end, since M(x)/(EI) = d²y/dx². (See https://www.youtube.com/watch?v=EWhL-mixfaI)

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To raise the object, the force in the cord needs to exceed W. Since torque is just force * distance,

T_required > Wd/2