Force exerted by a rod on supports

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Discussion Overview

The discussion revolves around calculating the force exerted by a steel rod on its supports under varying temperature conditions and constraints. Participants explore the effects of thermal expansion and contraction, as well as the implications of yielding supports, within the context of a homework problem.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants propose using the equation F = α * ΔT * A * E to calculate the force, where E is the modulus of elasticity, α is the expansion coefficient, and ΔT is the temperature change.
  • Others argue about the sign of the force, suggesting that a negative temperature change should lead to a positive tensile force when the rod is constrained.
  • A participant questions the method of calculating the change in length and whether it should be treated as negative due to contraction.
  • There is a discussion about the difference between compressive and tensile forces, with some participants stating that compressive forces can be expressed as positive values in certain contexts.
  • One participant expresses confusion regarding the calculations for part (b) of the problem, arriving at a different numerical answer than others, prompting requests for clarification on the methodology used.

Areas of Agreement / Disagreement

Participants generally agree on the use of certain equations but disagree on the interpretation of signs and the resulting forces. There is no consensus on the correct numerical answer for part (b) of the problem, as different participants arrive at different results.

Contextual Notes

Participants note the importance of correctly applying the concepts of thermal expansion, yielding, and the definitions of tensile and compressive forces. Some express uncertainty about the implications of negative signs in their calculations and the interpretation of results.

Who May Find This Useful

This discussion may be useful for students studying mechanics of materials, thermal expansion, or those working on similar homework problems involving forces in constrained systems.

dvep
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Homework Statement



A steel rod having a cross-sectional area of 0.005 m^2and length 3 m is attached to supports at both ends. Find (a) the force exerted by the rod on the supports if the temperature falls 60ºC and the supports are unyielding, and (b) the force exerted by the rod on the supports if they yield 0.0002 m while the temperature raises 50ºC. Take E = 210 GPa, coefficient of thermal expansion =10.5 x 10^(-6) °C^(-1)

Homework Equations



dl = length x temperature change x expansion coefficient
E = stress/strain
F = α x ΔT x A x E


The Attempt at a Solution



a)
rod unrestrained : dl = original length x temperature change x expansion coefficient

= 3 x 60 x 10.5 x 10^(-6)
= 1.89 x10^-3 m

Tensile stress due to supports : E = stress/strain = (Force*cross-sect area)/(dl*orig length)

at this point I have no force, do I just neglect it in this equation?

Or do I just use the equation:
F = α x ΔT x A x E

Where E is the modulus of elasticity
α is the expansion coefficient

Am I on the right track? Would be grateful for any help of guidance.

Thanks.
 
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You are on the right track. Your last equation looks correct.
 
dvep please can you answer the complete question please..
 
nvn said:
You are on the right track. Your last equation looks correct.

a) F = α * ΔT * A * E
= -661.500 kN

b) F= (E x Change in length x Area)/length
= 481.250 MN

Is the correct method? I have derived equations to give me that.

thanks.
 
dvep: I forgot to mention, in this particular scenario, put a negative sign on your last equation in post 1. Regarding post 4: (a) Your answer is correct, except change negative answer to positive. Rod and supports are in tension. (b) Your answer is correct, except change positive answer to negative. Rod and supports are in compression. Also, the units should be kN, not MN. Also, see the last paragraph of post https://www.physicsforums.com/showthread.php?t=463768#post3087641", regarding rounding your final answer.
 
Last edited by a moderator:
nvn said:
dvep: I forgot to mention, in this particular scenario, put a negative sign on your last equation in post 1. Regarding post 4: (a) Your answer is correct, except change negative answer to positive. Rod and supports are in tension. (b) Your answer is correct, except change positive answer to negative. Rod and supports are in compression. Also, the units should be kN, not MN. Also, see the last paragraph of post https://www.physicsforums.com/showthread.php?t=463768#post3087641", regarding rounding your final answer.

Thanks for your reply nvn

Can I ask why we would put the negative sign in the equation? When putting ∆T in the equation, seeing that it goes down by 60C, I input it as -60C right?
 
Last edited by a moderator:
Yes, put a negative sign on the equation. And the temperature change is -60 C; therefore, input -60 C. This gives a positive (tensile) force.
 
nvn said:
Yes, put a negative sign on the equation. And the temperature change is -60 C; therefore, input -60 C. This gives a positive (tensile) force.

Thank you
 
nvn said:
dvep: I forgot to mention, in this particular scenario, put a negative sign on your last equation in post 1. Regarding post 4: (a) Your answer is correct, except change negative answer to positive. Rod and supports are in tension. (b) Your answer is correct, except change positive answer to negative. Rod and supports are in compression. Also, the units should be kN, not MN. Also, see the last paragraph of post https://www.physicsforums.com/showthread.php?t=463768#post3087641", regarding rounding your final answer.

Can I just ask for a) wouldn't the length change be a negative number as it contracts?
and for b) i don't understand where the negative comes from in the equations.
I get the theory, but I can't see it in the maths.
Would it not be dt =yield - unrestrained change in length
= 0.0002-0.001575
= -0.001375 m

Then apply this to the equation which would give us a negative number, which then would mean it is compressive?

Also,I have seen a fairly similar problem in a book and it gives the compression force as a positive number, why do you think this is?
 
Last edited by a moderator:
  • #10
dvep: (a) Yes, a decrease in length is a negative length change (contraction) and a negative strain. If you constrain this contraction, then it causes a positive (tensile) force and a positive (tensile) stress. If unconstrained, epsilon = alpha*deltaT, where epsilon = strain = dL/L. If constrained, epsilon = -alpha*deltaT. Finally, F = E*epsilon*A.

(b) For part b in post 9, because the rod is constrained, yes, we have epsilon = dLyield/L - dLthermal/L = dLyield/L - alpha*deltaT = (0.2 mm)/(3000 mm) - (10.5e-6/C)(50 C) = 0.000 066 667 - 0.000 525 000 = -0.000 458 333. Or, we might say, dL = epsilon*L = dLyield - dLthermal = dLyield - alpha*deltaT*L = 0.2 mm - (10.5e-6/C)(50 C)(3000 mm) = -1.375 mm. And epsilon = dL/L = (-1.375 mm)/(3000 mm) = -0.000 458 333. Finally, F = E*epsilon*A.

Sometimes compressive force is stated as a positive number, followed (or preceded) by the word "compression."
 
  • #11
Hi guys,

For some reason I am getting different answer for (b). I am getting 551.250 KN.
Can someone explain to me how dvnp got 481.50 MN. Please

Thanks
 
  • #12
ankhkeansoo: No, you must list relevant equations yourself, and show your work; and then someone might check your math. Also, you misspelled kN.
 

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