Statics: four crates, four springs, free body diagram

[Alison A.]

Homework Statement

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 δ_A, δ_B.

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.

Homework Equations

Spring Law: F_s=kδ=k(L-L₀)

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!

 

[billy_joule]

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?

Can you define your variables, it's hard to interpret your attempt. There is δ_a, δ_b, δ_c and δ_d 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:

D 2000lb = 4500 (lb/in) δ

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

 

[Alison A.]

Can you define your variables, it's hard to interpret your attempt. There is δ_a, δ_b, δ_c and δ_d 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)

My bad,

A 1000lb = 4500 (lb/in) (δ_A+δ_B+δ_C+δ_D)
B 2000lb + 1000lb = 4500 (lb/in) (δ_B+δ_C+δ_D)
C 2000lb +2000lb +1000lb = 45000 (lb/in) (δ_C+δ_D)
D 2000lb +2000lb +2000lb +1000lb = 4500 (lb/in) δ_D

Maybe? The weights are confusing me.

 

[billy_joule]

That's closer but not quite there.

A 1000lb = 4500 (lb/in) (δA+δB+δC+δD)

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

EDIT:Fixed incorrect subscripts
so
δ_D =δ_DE
δ_C =δ_CD + δ_DE
δ_B =δ_BC + δ_CD + δ_DE
And so on

Can you see the problem with your attempt now?

 

[Alison A.]

That's closer but not quite there.
δ_A is the sum of the compressions of all four springs, Let's call the compression of just the top spring δ_AB, (that is, how much closer A gets to B) and similarly for the other springs.

so
δ_D =δ_DE
δ_C =δ_DE + δ_CE
δ_B =δ_BC + δ_DE + δ_CE
And so on

Can you see the problem with your attempt now?

Ohh ok, so it would be
A 1000lb = 4500 (lb/in) (δ_AE+δ_BE+δ_CE+δ_DE)
B 2000lb + 1000lb = 4500 (lb/in) (δ_BE+δ_CE+δ_DE)
C 2000lb +2000lb +1000lb = 45000 (lb/in) (δ_CE+δ_DE)
D 2000lb +2000lb +2000lb +1000lb = 4500 (lb/in) δ_DE
?

 

[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 δ_C :

δ_C = δ_DE + δ_CD

F_s=kδ

F_D+C+B+A = kδ_DE
(7,000lb) = (4500 lb/in) δ_DE

and

F_C+B+A = kδ_CD
(5,000lb) = (4500 lb/in) δ_CD

So
δ_C = ( F_D+C+B+A / k ) + (F_C+B+A / k)

And so on for the other crates.

 

[Alison A.]

Can you see the problem with your attempt now?[/QUOTE]

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 δ_C :

δ_C = δ_DE + δ_CD

F_s=kδ

F_D+C+B+A = kδ_DE
(7,000lb) = (4500 lb/in) δ_DE

and

F_C+B+A = kδ_CD
(5,000lb) = (4500 lb/in) δ_CD

So
δ_C = ( F_D+C+B+A / k ) + (F_C+B+A / k)

And so on for the other crates.

ALRIGHT, SO
δ_A=(F_D+C+B+A/k)+(F_C+B+A/k)+(F_B+A/k)+(F_A/k)
δ_B=(F_D+C+B+A/k)+(F_C+B+A/k)+(F_B+A/k)
δ_C=(F_D+C+B+A/k)+(F_C+B+A/k)
δ_D=(F_D+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.