Stress Problem Solution for Homework with Simple Equations

[yecko]

Homework Statement

Homework Equations

δ=PL/AE
σ=Eε

The Attempt at a Solution

a: (2L-x)/x
b: For AC, force acting on it = 2P*x/(2L-x) ##linearly proportional
δ=(2P*x/(2L-x))*(2L-x)/[(πD^2/4)*E]=(8P*x)/[(πD^2)*E]

am i wrong in part b?
thanks

 

[haruspex]

a: (2L-x)/x

That makes σ₁>σ₂ when x<L. Does that seem right?

For AC, force acting on it = 2P*x/(2L-x) ##linearly proportional

How do you arrive at that?
Wouldn't it be more useful to write the force required to extend AC by δ in terms of D, E and δ?

 

[Chestermiller]

Kinematically, if the downward displacement of point C is $\delta$, what is the tensile strain in AC? In BC? From Hooke's law, what is the tensile stress in AC? In BC?

 

[yecko]

Wouldn't it be more useful to write the force required to extend AC by δ in terms of D, E and δ?

as δ is the final answer required, force is just a medium for expressing as part of δ

Kinematically, if the downward displacement of point C is δδ\delta, what is the tensile strain in AC?

ε=δ/(2L-x)

From Hooke's law, what is the tensile stress in AC? In BC?

σ=Eε=Eδ/(2L-x)

That makes σ1>σ2 when x<L. Does that seem right?

so that should it be x/(2L-x) for part a?

what about part b, what should I do?
thanks

 

[haruspex]

as δ is the final answer required, force is just a medium for expressing as part of δ

I don't get your point.
I am suggesting that you can find the force needed to extend the upper part by δ, as a function of δ, D, E etc., and the force required to compress the lower part by δ in the same way. 2P is then the sum of these forces.
You have not explained how you arrived at 2Px/(2L-x).

so that should it be x/(2L-x) for part a?

That sounds like a guess. Can you show some logic to arrive at that?

 

[Chestermiller]

What is the magnitude and direction of the reaction force at A (in terms of $\delta$)? At B?

 

[yecko]

That sounds like a guess. Can you show some logic to arrive at that?

σ=Eε=Eδ/(2L-x)

From this, the length is inversely proportional to the stress

I am suggesting that you can find the force needed to extend the upper part by δ, as a function of δ, D, E etc., and the force required to compress the lower part by δ in the same way. 2P is then the sum of these forces.

I am not sure what do you mean by that... let me try

By δ=PL/AE
F₁*(2L-x)/AE -F₂*x/AE = 0 ---equation 1
F₁+ F₂ = 2P --- equation 2
F₂ = (4PL-2Px)/(1+2L-x)
F₁=2P-(4PL-2Px)/(1+2L-x)

therefore δ=(4PL-2Px)/(1+2L-x)*x/[(πD^2/4)E]
is it correct? thanks

 

[yecko]

What is the magnitude and direction of the reaction force at A (in terms of $\delta$)? At B?

σ=F/A
σ₁=R₁/A
By σ=Eε=Eδ/L
σ₁=Eδ/(2L-x)=R₁/A
R₁=Eδ/(2L-x)*A
similarly R₂=Eδ/x*A

 

[Chestermiller]

σ=F/A
σ₁=R₁/A
By σ=Eε=Eδ/L
σ₁=Eδ/(2L-x)=R₁/A
R₁=Eδ/(2L-x)*A
similarly R₂=Eδ/x*A

In terms of these reaction forces and the force 2P, what is the overall force balance on the rod?

 

[yecko]

In terms of these reaction forces and the force 2P, what is the overall force balance on the rod?

R₁+R₂+2P=0
Eδ/(2L-x)*A+Eδ/x*A+2P=0
δAE[1/(2L-x)+1/x]+2P=0
δE(πD^2/4)[1/(2L-x)+1/x]+2P=0
δ=-2P/{E(πD^2/4)[1/(2L-x)+1/x]}
is it something like this? thanks

 

[Chestermiller]

R₁+R₂+2P=0
Eδ/(2L-x)*A+Eδ/x*A+2P=0
δAE[1/(2L-x)+1/x]+2P=0
δE(πD^2/4)[1/(2L-x)+1/x]+2P=0
δ=-2P/{E(πD^2/4)[1/(2L-x)+1/x]}
is it something like this? thanks

Ya see, this is why I asked you for the directions of the reaction forces. Now, again, of $\delta$ is downward, what are the directions of the reaction forces at A and B? So, what is the sign of $\delta$?

 

[yecko]

Ya see, this is why I asked you for the directions of the reaction forces. Now, again, of $\delta$ is downward, what are the directions of the reaction forces at A and B? So, what is the sign of $\delta$?

R2 is upward and R1 is downward i guess

R1-R2+2P=0
Eδ/(2L-x)*A-Eδ/x*A+2P=0
δAE[1/(2L-x)-1/x]+2P=0
δE(πD^2/4)[1/(2L-x)-1/x]+2P=0
δ=-2P/{E(πD^2/4)[1/(2L-x)-1/x]}
now, is it something like this? thanks

 

[Chestermiller]

R2 is upward and R1 is downward i guess

R1-R2+2P=0
Eδ/(2L-x)*A-Eδ/x*A+2P=0
δAE[1/(2L-x)-1/x]+2P=0
δE(πD^2/4)[1/(2L-x)-1/x]+2P=0
δ=-2P/{E(πD^2/4)[1/(2L-x)-1/x]}
now, is it something like this? thanks

Both reaction forces are upward.

 

[yecko]

R1+R2-2P=0
Eδ/(2L-x)*A+Eδ/x*A-2P=0
δAE[1/(2L-x)+1/x]-2P=0
δE(πD^2/4)[1/(2L-x)+1/x]=2P
δ=2P/{E(πD^2/4)[1/(2L-x)+1/x]}
now should it be something like this? thanks

 

[Chestermiller]

R1+R2-2P=0
Eδ/(2L-x)*A+Eδ/x*A-2P=0
δAE[1/(2L-x)+1/x]-2P=0
δE(πD^2/4)[1/(2L-x)+1/x]=2P
δ=2P/{E(πD^2/4)[1/(2L-x)+1/x]}
now should it be something like this? thanks

Technically, yes. But, I would have reduced the equation to a simpler mathematical form with a least common denominator:
$$\delta=\frac{4Px(2L-x)}{\pi D^2LE}$$