Parallel Axis Theorem / Bending Stress

[ksukhin]

Homework Statement

find bending stress in x and y dir

Homework Equations

I = bh^3/12 + ad^2
Stress = Mc/I

The Attempt at a Solution

I = bh^3/12 + ad^2
Stress = Mc/I

see attached calculations

My prof gave us a question where we have a motor (20" tall) sitting on a frame with a load of F=3300lb. I have to find the bending stress in x and y direction. I have attached my solution but one part differs from my profs.

When looking at Iy my prof does not use Parallel Axis Theorem but I do, and we get completely different answers. Why? Why are my stresses so low?

Diagram: https://ibb.co/gUs2Hx
Solution: https://ibb.co/fegOPc

 

[PhanthomJay]

There are some errors here. And missing info on loading. Is the 3300 pound applied at the top of the motor along the x-axis in case 1 and then along the y-axis in case 2? Or are the loads applied along both axes at the same time?
Now when applied along the x-axis in case 1, it makes a big difference whether the walls of the frame act independently or as a joined pair. If independently, each wall takes half the load and the parallel axis theorem does not apply. Your Ix about the y-axis will be quite small without the extra term. If however as I suspect the side walls are tied together, then you use the parallel axis theorem and on determining bending stress, use full load not half load, and your Ix must be doubled since there are 2 sided and your 'c ' distance is quite wrong...what should it be?
Now for the other axis Iy about the x-axis is determined without any distance to the parallel x axis, because the centroid of the walls and frame coincide.

 

[ksukhin]

There are some errors here. And missing info on loading. Is the 3300 pound applied at the top of the motor along the x-axis in case 1 and then along the y-axis in case 2? Or are the loads applied along both axes at the same time?

2 separate cases, first along X axis and then along Y axis starting from the centre line.

Now when applied along the x-axis in case 1, it makes a big difference whether the walls of the frame act independently or as a joined pair. If independently, each wall takes half the load and the parallel axis theorem does not apply.

So can i think of this case as a cantilever beam with a load at the end?

If however as I suspect the side walls are tied together, then you use the parallel axis theorem and on determining bending stress, use full load not half load, and your Ix must be doubled since there are 2 sided

So M = (3300)(35)= 115500 lb-in
Ix = 2 [bh³/12 + Ad²] = 2(270.11) = 540.22 in⁴

and your 'c ' distance is quite wrong...what should it be?

from my understanding, C is the distance from the Neutral Axis to the highest stress point. Since the cross section is a rectangle, half would be in tension and half in compression starting at the midpoint. Since the height is 1.25" i took the half of that value as my distance c.

Do I have to take into the 5.75" from the centreline aswell? c would then be 5.75+(1.25/2)=6.375in >>>which is the same as my distance d above
Substituting my new values σ=Mc/I = (115500)(6.375)/540.22 = 1363 psi = 9.4 MPa, still relatively smaller than in other direction.

Now for the other axis Iy about the x-axis is determined without any distance to the parallel x axis, because the centroid of the walls and frame coincide.

Thank you, I had a feeling that was why but wasn't sure.
But in this case, since I took half the load and analyses only 1 side of the frame, there is no need to double it like in the above case?

 

[PhanthomJay]

2 separate cases, first along X axis and then along Y axis starting from the centre line.

ok , good.

So can i think of this case as a cantilever beam with a load at the end?

yes

So M = (3300)(35)= 115500 lb-in
Ix = 2 [bh³/12 + Ad²] = 2(270.11) = 540.22 in⁴

yes

from my understanding, C is the distance from the Neutral Axis to the highest stress point. Since the cross section is a rectangle, half would be in tension and half in compression starting at the midpoint. Since the height is 1.25" i took the half of that value as my distance c.

that would be correct if the walls acted independently, but if acting together as a combined section, that would be incorrect. I think they act together, but perhaps not as well as a boxed section that is all one piece, but close enough it seems.

Do I have to take into the 5.75" from the centreline aswell? c would then be 5.75+(1.25/2)=6.375in >>>which is the same as my distance d above

edit: not quite since 'c' is the distance to the extreme outside edge

Substituting my new values σ=Mc/I = (115500)(6.375)/540.22 = 1363 psi = 9.4 MPa, still relatively smaller than in other direction.

right again! (After correcting c value)

Thank you, I had a feeling that was why but wasn't sure.
But in this case, since I took half the load and analyses only 1 side of the frame, there is no need to double it like in the above case?

well you get the same answer either way, but technically if you are asked about the Iy of the combined section, it's 2(bh^3/12). Good work.

 

[ksukhin]

edit: not quite since 'c' is the distance to the extreme outside edge

So does that mean my C distance would be 5.75+1.25 = 7" - which is up to the furthest point on the wall thickness.

How would my cross section look as far as tension/compression? Initially I thought that half the wall thickness would be in tension and half in compression.
Since I'm looking at the walls as one piece does this midpoint happen at the center line now? Say anything above the center line is tension and below is compression? - I'm having a tough time wrapping my head around this because of the gap in between the walls. No material there to carry the stress.

 

[PhanthomJay]

So does that mean my C distance would be 5.75+1.25 = 7" - which is up to the furthest point on the wall thickness.

That is correct.

How would my cross section look as far as tension/compression? Initially I thought that half the wall thickness would be in tension and half in compression.
Since I'm looking at the walls as one piece does this midpoint happen at the center line now?

yes, the centroid of the box frame, which is the neutral axis, the centerline y-y axis.

Say anything above the center line is tension and below is compression?

yes, where material exists

I'm having a tough time wrapping my head around this because of the gap in between the walls. No material there to carry the stress.

tension stresses exist in one wall and compressive stresses in the other. Max tension or compression stresses occur at 7 inches left or right from the centerline. There are no stresses in the gap.

 

[ksukhin]

Thank you for explaining everything!