Statics: four crates, four springs, free body diagram

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Discussion Overview

The discussion revolves around a statics problem involving four identical crates stacked on top of each other, each supported by springs. Participants are tasked with determining the deflections at various points in the system based on given weights and spring constants. The focus is on the application of free body diagrams (FBDs) and the relationships between the forces and deflections in the springs.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant outlines the problem and provides initial equations for the deflections at points A, B, C, and D based on the weights of the crates and the spring constant.
  • Another participant requests clarification on the variables used in the equations, noting that the cumulative forces acting on the bottom crate have not been properly accounted for.
  • Subsequent posts refine the equations, with participants suggesting that the deflections should consider the contributions from all crates above each spring.
  • There are discussions about the correct way to express the relationships between the deflections and the forces, with suggestions to isolate each spring's deflection and sum them appropriately.
  • One participant acknowledges errors in their previous attempts and makes corrections to the subscripts in their equations, indicating a progression in understanding the problem.
  • Final posts express satisfaction with the derived equations for deflections, although the accuracy of the final expressions remains unverified by others.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the final equations for deflections, as there are multiple iterations and corrections proposed throughout the discussion. Some participants express confusion about the weights and how they affect the deflections, indicating ongoing uncertainty.

Contextual Notes

Limitations include the lack of specific distances provided in the problem statement, which may affect the ability to solve for deflections directly. The discussion also reflects varying interpretations of how to apply the spring force equations and the cumulative effects of the weights on the deflections.

Alison A.
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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 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.

Homework Equations


Spring Law: Fs=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!
 
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Alison A. said:

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 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:

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)
 
billy_joule said:
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)

My bad,

A 1000lb = 4500 (lb/in) (δA+δB+δC+δD)
B 2000lb + 1000lb = 4500 (lb/in) (δB+δC+δD)
C 2000lb +2000lb +1000lb = 45000 (lb/in) (δC+δD)
D 2000lb +2000lb +2000lb +1000lb = 4500 (lb/in) δD

Maybe? The weights are confusing me.
 
That's closer but not quite there.

Alison A. said:
A 1000lb = 4500 (lb/in) (δA+δB+δC+δD)

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
δD =δDE
δC =δCD + δDE
δB =δBC + δCD + δDE
And so on

Can you see the problem with your attempt now?
 
Last edited:
billy_joule said:
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.

so
δD =δDE
δC =δDE + δCE
δB =δBC + δDE + δCE
And so on

Can you see the problem with your attempt now?
Ohh ok, so it would be
A 1000lb = 4500 (lb/in) (δAE+δBE+δCE+δDE)
B 2000lb + 1000lb = 4500 (lb/in) (δBE+δCE+δDE)
C 2000lb +2000lb +1000lb = 45000 (lb/in) (δCE+δDE)
D 2000lb +2000lb +2000lb +1000lb = 4500 (lb/in) δDE
?
 
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.
 
Can you see the problem with your attempt now?[/QUOTE]
billy_joule said:
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.

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.
 
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