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Stress in an accelerating body

  1. Aug 26, 2013 #1
    I don't really understand how to calculate the stress in a deformable body that is unconstrained and acted on by an external force. For example, consider a rod with a circular cross-section lying parallel to the x-axis. If one end is constrained from translating in the x direction and the other end has a pressure applied to it, the normal stress in the x direction is basic to calculate. But what if the situation is identical except the one end is not constrained so that the rod can accelerate? How does one determine the stress in such a situation? Thank you
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  3. Aug 27, 2013 #2


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    If a horizontal inextensible rod on a horizontal frictionless table has a mass m and is subject to a constant horizontal pulling force F applied axially at one end, its acceleration, a, is F/m, per Newton 2. Now if you take a free body diagram of a cut section, you should see that the tension in the rod will vary linearly from 0 at the free end to F at the pulling end.

    Well now check it out : if a vertical rod of mass m is dropped vertically in a gravity field , it will be in free fall (ignoring air resistance) , and you should prove to yourself that the tension in the rod for this case will be 0 throughout. No internal force at all!

    Welcome to PF!
    Last edited: Aug 27, 2013
  4. Aug 27, 2013 #3


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    To turn the previous answer into a general purpose algorithm, if the deformations of the body are small:

    1. Find the acceleration (linear and angular) of the center of mass of the body, assuming the body is rigid.

    2. Solve the problem in an accelerating coordinate frame fixed to the CG, and include the (fictitious) d'Alembert forces distributed over the body as well as the external forces.

    Finite element software may have this as a built-in procedure, called "inertia relief" or something similar.
  5. Jan 6, 2015 #4
    I have the same question.

    If a force is suddently put on a elastic rod resting on a frictionless table, how will the stress distriution along the rod look lika a function of time?

  6. Jan 6, 2015 #5
    Do you have any thoughts on what the stress distribution in this problem will look like? Will all parts of the rod accelerate identically? (Think slinky toy).

  7. Jan 7, 2015 #6
    My gut feeling is that from t=0, the stress will be zero at all other places than at the end where the force is applied. Then will the stress gradually increase down the bar until a linear distribution with zero stress at end after some time.
  8. Jan 7, 2015 #7
    Yes. This is right on target.
    This isn't exactly what will happen. The bar is elastic, so there will initially be a tensile wave that travels backwards along the bar at the speed of sound. Ahead of this wave front, no motion will yet have occurred. To quantify what is happening in detail, one needs to solve the wave equation. Of course, at the trailing end the stress will always be zero, and at the leading end, the stress will always be F/A. These will be boundary conditions that are used in the wave equation solution.

  9. Jan 7, 2015 #8


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    What Chet said makes a lot of sense: I figured out that the the stress would propagate from right to left along the rod (the slinky example is very nice), but I did not realize that this movement would be at the speed of sound in the material, which is what is by definition what it is: the speed at which a mechanical deformation travels through the material.

    Eventually, after some time and if the force is continuously applied, all the rod will be accelerating at the same rate, a=F/m, and there will be no relative movement between any two points of the rod. At any intersection plane along the rod, the stress will be indicated by the mass of the bar behind the section (m'), which has to be accelerated by the force F'=m'·a. If the bar is homogeneous and has a constant section, the stress will be maximum at the tip where the force is applied, and will decrease linearly to zero at the opposite end. I think that is what havsula was referring to.

    As we are talking about pure traction, if S is the area of the section, the stress at any point will be σ=(F·m)/(S·m’)
  10. Jan 7, 2015 #9
    Agreement Witherspoon that. But how to solve it. How will the differential equation look like?
  11. Jan 7, 2015 #10
    The differential equation will look like this:
    [tex]\frac{\partial ^2 ε}{\partial t^2}=c^2\frac{\partial ^2 ε}{\partial x^2}[/tex]
    where ε is the tensile strain and c (the speed of sound) is given by:

    ##c^2=\frac{E}{ρ}##, where E is Young's modulus and ρ is the density of the bar metal.

    The boundary conditions will be

    ##ε=0## at x = 0

    ##ε=\frac{F}{EA}## at x = L

  12. Jan 8, 2015 #11
    First one question.
    In all references I have found the variable in the wave equation is displacment (normally written as u). But in your equation you have written ε, which most often is used for strain. What is ε in your equation:

    Will this be correct:
    Is it possible to look at the constant force as a source which gives the
    ##\frac{\partial ^2 u}{\partial t^2}-c^2\frac{\partial ^2 u}{\partial x^2}=F(x,t)##
    ##F(t)=\frac{F}{Aρ} ## at ##x=L##
    Boundary conditions:
    ##u(0,0) = u(L,0)=0##



    I think this is called the "Inhomogeneous Wave Equation".
    But I am pretty sure that something is wrong in my equations. I do not think that I understood this problem.
    Last edited: Jan 8, 2015
  13. Jan 8, 2015 #12
    As I said in my previous post, ε is the tensile strain. The tensile strain is defined in terms of the displacement u by ##ε=∂u/∂x##.

    You have an F on the right hand side of your equation. Can you please identify where this F came from? If the F weren't there, and you took the partial derivative of your equation with respect to x, you would obtain my equation, and your boundary conditions would then be the same as mine.

  14. Jan 8, 2015 #13
    Yes I can see that it is the same equation.
    The F should be the constant force applied to the end. Do I not need to add it? Is it no possible to look at this as "forced vibration"? or will the boundary condition take care of that constant force?
    I beginning to understand that it is a lot of thing I do not understand here.
  15. Jan 8, 2015 #14
    No, it shouldn't be in the equation. The derivation is on a differential length of bar from x to x + dx, and includes only the adjacent tensile stresses on the differential element.
    If there are oscillations, the equation and bc's will capture that. If F is varying with time, they will capture that also.
  16. Jan 8, 2015 #15
    Will it be correct to do if there is a force F(x,t) which works along the whole length of the beam (also the differential element)?
  17. Jan 8, 2015 #16
    If there were a distributed shear load per unit length along the bar (which is not the case in this problem), then that would have to be included, and it would appear in some way in the differential equation.

  18. Feb 16, 2016 #17
    I have a question similar to the original question, but with the focus more on the deformation. After reading the answers above I could not answer the question I am struggling with:

    If at each end of a free spring (with axial stiffness 'k' and mass 'm') a force is working with both forces working on the axis of the spring but in opposite direction and different magnitude (F1>F2). The axial acceleration of the spring = (F1-F2)/M.
    Deformation vs Acceleration.jpg
    Q1) How can I calculate the axial deformation of the spring? (In case F1=F2 it is easy: delta_x = F1/k; but what if F1 is not equal to F2?)
    Q2) Is the axial deformation determined by the minimum magnitude of the 2 forces? Thus if F1>F2, the acceleration is determined by F1-F2 and the deformation by F2?

    Could you please help me?

  19. Feb 16, 2016 #18
    The first step is to show that, if F1 is the compression force at x = 0 and F2 is the compression force at x = L, the compression force F at an arbitrary location x along the spring is given by:
    Do you think that you can do that?

  20. Feb 17, 2016 #19
    Hello Chet, thank you very much for your response. With this insight I can solve my questions.
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