# Homework Help: Statics: four crates, four springs, free body diagram

Tags:
1. Sep 20, 2015

### Alison A.

1. The problem statement, all variables and given/known data
In figure (a), four identical crates weighing 2,000 lb each are stacked one on top of another, and in figure (b) a simple model for determining the deformation of the stack of crates is shown. In this model, each spring has the same stiffness k = 4,500 lb/in. and the forces are determined using the following idealization. Half the weight of the top crate is applied to point A and half to point B. Similarly, the half the weight of the next crate is applied to point B and half to point C, and so on. Determine the deflections δA, δB.

http://imgur.com/XhWr2YE

Hint given: Begin by drawing FBDs of joints A, B, C and D. Starting from the top, you should find that each spring carries progressively larger weight. Note that the problem statement requests deflections, which are not necessarily the same as stretches. For the bottom spring, anchored at one end, its deflection is identical to its stretch. For all the spring above it, however, the deflection is the cumulative sum of the stretches below and including that spring.

2. Relevant equations
Spring Law: Fs=kδ=k(L-L0)

3. The attempt at a solution
A 1000lb = 4500 (lb/in) (δ+B+C+D)
B 2000lb = 4500 (lb/in) (δ+C+D)
C 2000lb = 45000 (lb/in) (δ+D)
D 2000lb = 4500 (lb/in) δ

I don't know where I'm going wrong here. No actually distances are given so I'm assuming you just have to solve for δ, but is my approach correct?

Thank you for any help!

2. Sep 20, 2015

### billy_joule

Can you define your variables, it's hard to interpret your attempt. There is δa, δb, δc and δd but you only use δ.
It looks like you've ignored the accumulative forces, that is, the bottom crate D has the force from three crates on top of it. not just:

(Note that that 2000lb comes from the top half of crate D and the bottom half of crate C)

3. Sep 20, 2015

### Alison A.

A 1000lb = 4500 (lb/in) (δABCD)
B 2000lb + 1000lb = 4500 (lb/in) (δBCD)
C 2000lb +2000lb +1000lb = 45000 (lb/in) (δCD)
D 2000lb +2000lb +2000lb +1000lb = 4500 (lb/in) δD

Maybe? The weights are confusing me.

4. Sep 20, 2015

### billy_joule

That's closer but not quite there.

δA is the sum of the compressions of all four springs, Let's call the compression of just the top spring δAB, (that is, how much closer A gets to B) and similarly for the other springs.

EDIT:Fixed incorrect subscripts
so
δDDE
δCCD + δDE
δBBC + δCD + δDE
And so on

Can you see the problem with your attempt now?

Last edited: Sep 20, 2015
5. Sep 20, 2015

### Alison A.

Ohh ok, so it would be
A 1000lb = 4500 (lb/in) (δAEBECEDE)
B 2000lb + 1000lb = 4500 (lb/in) (δBECEDE)
C 2000lb +2000lb +1000lb = 45000 (lb/in) (δCEDE)
D 2000lb +2000lb +2000lb +1000lb = 4500 (lb/in) δDE
?

6. Sep 20, 2015

### billy_joule

Sorry, There were errors in my subscripts in post #4. I've edited in corrections.

The problem is how you're calculating the deflection of each spring, you need to consider each in isolation and then sum them, that is, the parentheses of all you attempts is not right.

eg for δC :

δC = δDE + δCD

Fs=kδ

FD+C+B+A = kδDE
(7,000lb) = (4500 lb/in) δDE

and

FC+B+A = kδCD
(5,000lb) = (4500 lb/in) δCD

So
δC = ( FD+C+B+A / k ) + (FC+B+A / k)

And so on for the other crates.

7. Sep 20, 2015

### Alison A.

Can you see the problem with your attempt now?[/QUOTE]
ALRIGHT, SO
δA=(FD+C+B+A/k)+(FC+B+A/k)+(FB+A/k)+(FA/k)
δB=(FD+C+B+A/k)+(FC+B+A/k)+(FB+A/k)
δC=(FD+C+B+A/k)+(FC+B+A/k)
δD=(FD+C+B+A/k)

I got them correct :)
Thank you so much!

If you have any time do you think you could take a look at my other posts from today? I'd really appreciate it.