Stress Problem Solution for Homework with Simple Equations

  • Thread starter yecko
  • Start date
  • Tags
    Stress
In summary: Both reaction forces are upward. R1+R2-2P=0Eδ/(2L-x)*A+Eδ/x*A-2P=0δAE[1/(2L-x)+1/x]-2P=0δE(πD^2/4)[1/(2L-x)+1/x]=2Pδ=2P/{E(πD^2/4)[1/(2L-x)+1/x]}In summary, the homework statement is that:- δ is the distance between points A and B, and is proportional to the product of the lengths of those
  • #1
yecko
Gold Member
277
15

Homework Statement


螢幕快照 2018-04-07 下午6.09.28.png


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
 

Attachments

  • 螢幕快照 2018-04-07 下午6.09.28.png
    螢幕快照 2018-04-07 下午6.09.28.png
    32.5 KB · Views: 918
Physics news on Phys.org
  • #2
yecko said:
a: (2L-x)/x
That makes σ12 when x<L. Does that seem right?
yecko said:
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?
 
  • #3
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?
 
  • #4
haruspex said:
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
Chestermiller said:
Kinematically, if the downward displacement of point C is δδ\delta, what is the tensile strain in AC?
ε=δ/(2L-x)
Chestermiller said:
From Hooke's law, what is the tensile stress in AC? In BC?
σ=Eε=Eδ/(2L-x)
haruspex said:
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
 
  • #5
yecko said:
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).
yecko said:
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?
 
  • #6
What is the magnitude and direction of the reaction force at A (in terms of ##\delta##)? At B?
 
  • #7
haruspex said:
That sounds like a guess. Can you show some logic to arrive at that?
yecko said:
σ=Eε=Eδ/(2L-x)
From this, the length is inversely proportional to the stress
haruspex said:
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
 
  • #8
Chestermiller said:
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
 
  • #9
yecko said:
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?
 
  • #10
Chestermiller said:
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
 
  • #11
yecko said:
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##?
 
  • #12
Chestermiller said:
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
 
  • #13
yecko said:
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.
 
Last edited:
  • #14
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
 
  • #15
yecko said:
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:
$$\delta=\frac{4Px(2L-x)}{\pi D^2LE}$$
 

What is a simple stress problem?

A simple stress problem is a situation where an individual experiences stress that is caused by a specific event or circumstance. It typically involves a single source of stress and can be resolved with relatively straightforward solutions.

What are some examples of simple stress problems?

Examples of simple stress problems include public speaking, taking a test, meeting a deadline, and dealing with a difficult coworker. These situations can cause temporary stress and can be resolved by addressing the specific issue causing stress.

What causes simple stress problems?

Simple stress problems are usually caused by external factors, such as work, relationships, or finances. They can also be caused by internal factors, such as anxiety or perfectionism. These stressors trigger the body's fight or flight response, leading to feelings of stress.

How can simple stress problems be managed?

Simple stress problems can be managed by identifying the source of stress and developing effective coping strategies. This may include practicing relaxation techniques, setting boundaries, and seeking support from friends or a therapist.

When should I seek professional help for a simple stress problem?

If you are experiencing prolonged or severe stress that is interfering with your daily life, it may be helpful to seek professional help. A therapist or counselor can provide support and guidance in managing your stress and developing healthy coping mechanisms.

Similar threads

  • Introductory Physics Homework Help
Replies
10
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
19K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
7K
  • Introductory Physics Homework Help
Replies
1
Views
649
  • Introductory Physics Homework Help
Replies
3
Views
856
  • Introductory Physics Homework Help
Replies
7
Views
4K
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
4K
Back
Top