How can stress develop on a moving uniform elastic plank

  • #1
208
0

Homework Statement


A uniform elastic plank moves due to a constant force F distributed uniformly over the end face. The area of the ned face is S and Young's modulus = E . What is the average strain produced in the direction of force ?


Homework Equations


E*strain=stress....



The Attempt at a Solution



I am not getting any clue.....how can stress or strain develop unless equal and opposite forces are acting? How ro approach this sum?
 

Answers and Replies

  • #2
208
0


It would be great if you help me out.......I have used my head .....and i guess thats why i have posted it here......
 
  • #3
PhanthomJay
Science Advisor
Homework Helper
Gold Member
7,167
507


I am unsure what is meant by the word 'moved'. It could be that the beam is supported at one end and just moves while deforming (stretches or compresseses) without moving like a toy car being pushed along a floor. In which case the reaction force is equal and opposite to the applied force, and the strain is per your equation (after first indicating the stress a function of F and S). If the beam is actually moving like a train under that force, there is no strain at the front (free) end and max strain at the force end, but I don't quite understand the wording.
 
  • #4
208
0


Firstly PhanthomJay.....thanks for the reply......i got one ater a long time....which is uncommon in Physicsforums(i usually get replies in one or two days....)

And ya I posted the exact question......and i dont think its atatched to any end......i guess it will be the case of a toy car being pushed along a floor......
Hey i can do one thing.....in my book i have another question almost same as this but has a diagram......i will somehow make the diagram and post that question in my next post.........
 
  • #5
208
0


If the beam is actually moving like a train under that force, there is no strain at the front (free) end and max strain at the force end, but I don't quite understand the wording.

Can you elaborate on this......i did not exactly get how you come to these conclusions.....Maybe this is the concept the book is hinting at....as they are asking for the average strain.....
 
  • #6
208
0


Here goes the other question.....which i think is similar....

A constant force F is applied on a uniform elastic string placed placed over a smooth horizontal surface as shown in figure. Young's Modulus of string is Y and area of cross section is S. Find the strain produced in the string in the direction of Force.....

PS: Refer to diagram in attachment......It is same as the one given in the book but i have no clue how that thing is a string....it rather looks like a rectangular block....anyway i think we can asume it as a string and go ahead....
 

Attachments

  • strain.bmp
    238.1 KB · Views: 544
  • #7
PhanthomJay
Science Advisor
Homework Helper
Gold Member
7,167
507


Here goes the other question.....which i think is similar....

A constant force F is applied on a uniform elastic string placed placed over a smooth horizontal surface as shown in figure. Young's Modulus of string is Y and area of cross section is S. Find the strain produced in the string in the direction of Force.....

PS: Refer to diagram in attachment......It is same as the one given in the book but i have no clue how that thing is a string....it rather looks like a rectangular block....anyway i think we can asume it as a string and go ahead....
Since the applied force is constant, the internal stress, strain, and force varies linearly from 0 at the free end to a maximum at the applied force end; the average normal strain can then be computed by looking at the internal force at the midpoint of the beam, at x = L/2. What is the axial force at the midpoint of the beam?
 
  • #8
208
0


Since the applied force is constant, the internal stress, strain, and force varies linearly from 0 at the free end to a maximum at the applied force end; the average normal strain can then be computed by looking at the internal force at the midpoint of the beam, at x = L/2. What is the axial force at the midpoint of the beam?

Ok....but how do you say this.....that it will be linear....

I will tell you how i think you got the fact that stress is maximum at the force end and mimimum at the free end......correct me if i am wrong.....

At the free end if a take the extreme last element of mass(cross section A) then there is no opposing force from the other end so stress is zero......
And at the force end.....dm*a (here a is acceleration) almost = 0 because dm is very small.....so forces must be equal on it.....so Force F on it from right equal to internal force on it ......So stress = F/A.....Is this all correct ....
and even so how do I go about with the other parts of the rod.....how to find forces on them?
 
  • #9
PhanthomJay
Science Advisor
Homework Helper
Gold Member
7,167
507


Ok....but how do you say this.....that it will be linear....

I will tell you how i think you got the fact that stress is maximum at the force end and mimimum at the free end......correct me if i am wrong.....

At the free end if a take the extreme last element of mass(cross section A) then there is no opposing force from the other end so stress is zero......
And at the force end.....dm*a (here a is acceleration) almost = 0 because dm is very small.....so forces must be equal on it.....so Force F on it from right equal to internal force on it ......So stress = F/A.....Is this all correct ....
and even so how do I go about with the other parts of the rod.....how to find forces on them?
Yes, that's right....engineers don't like to do calculus :smile:...the average stress is [(F/A) + 0]/2 = (F)/(2A). If you look at the entire beam, F = Ma, so a= F/M; now cut a section through the midpoint (a Free body diagram of half the beam).... There is an unknown axial force there , call it P. Using Newton 2 for the free end to mid point part of the beam, where the mass of the beam section is a half of M, then P = (1/2)(M)a, or P = F/2, and stress = F/(2A).
 
  • #10
208
0


Ok.....i got the concept.....and ya strain at the mid point comes out to be F/2SE....which is correct....Thanks.....

But still i am wondering how you get the average at the mid point.....
Just tell me what is meant by average strain is it total elongation by length of bar?
You dont have to do the calculus( i am relieving you of that....) but just help me out.....
I am not getting the meaning of average strain....

Going by your way i got the force at a distance x from the force end as F(1-x/L)...here L is the length of the bar.....now what to do next.....Will my force expression help me in the calculus used to find average strain?
 
  • #11
PhanthomJay
Science Advisor
Homework Helper
Gold Member
7,167
507


Ok.....i got the concept.....and ya strain at the mid point comes out to be F/2SE....which is correct....Thanks.....

But still i am wondering how you get the average at the mid point.....
Just tell me what is meant by average strain is it total elongation by length of bar?
yes, indeed, it is total elongation divided by length of bar
You dont have to do the calculus( i am relieving you of that....) but just help me out.....
I am not getting the meaning of average strain....
strain is difficult to visualize...if I asked you what is the average force in the bar, would you agree it is F/2?
Going by your way i got the force at a distance x from the force end as F(1-x/L)...here L is the length of the bar.....now what to do next.....Will my force expression help me in the calculus used to find average strain?
Oh, sure. But start from the free end for simplicity, P = Fx/L. P linearly varies with x. Then you get the average force by using the definition of average, which is 1/L times the integral of Fx/L (integrating from 0 to L) , which, from my rusting calculus skills, since F/L is a constant, is (F/L)(L2/2L = F/2, which is precisely the same result you get by adding the forces at the extreme ends and dividing by 2, which works for linearly varying equations only.

Q.E.D. (Bonus, what does Q.E.D. stand for...if you're younger than 50, you probably won't know):wink:
 
  • #12
208
0


Hi PhanthomJay...sorry i came to the site after a long time......Any way i think i got what you mean to say .....average strain is total extension upon total Length ie..initial length....

So going by that i proceeded like this......As you say P = Fx/L

now the above equation is the force acting on an element dx at a distance of x from the free end.....So i want to calculate the extension of each element dx and divide that by L....

Here Young's modulus = E and area = S so using extension = Force*Length/Area*Youngs modulus we have equation for extension of the element dx = Fxdx/LSE integrating this from 0 to L we get net extension =FL/2SE and diving this by the initial length L we get our answer F/2SE.....I guess this is correct.....

And hey thanks all along......
 
  • #13
PhanthomJay
Science Advisor
Homework Helper
Gold Member
7,167
507


Q. E. D.
 
  • #14
208
0


Ok....great.....
 

Related Threads on How can stress develop on a moving uniform elastic plank

  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
3
Views
3K
Replies
25
Views
8K
  • Last Post
Replies
9
Views
14K
Replies
15
Views
6K
Replies
3
Views
1K
  • Last Post
Replies
7
Views
5K
Replies
22
Views
883
Replies
3
Views
2K
Top