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Force exerted by a rod on supports

  • Thread starter dvep
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  • #1
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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.
 

Answers and Replies

  • #2
nvn
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You are on the right track. Your last equation looks correct.
 
  • #3
dvep please can you answer the complete question please..
 
  • #4
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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.
 
  • #5
nvn
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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.
 
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  • #6
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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?
 
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  • #7
nvn
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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.
 
  • #8
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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
 
  • #9
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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 dont 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?
 
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  • #10
nvn
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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
nvn
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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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