Calculate Force to Elongate Steel Bar

[raiderUM]

Homework Statement

What Force must be applied to a steel bar, 1in [25.4mm] square and 2ft [610mm] long, to produce an elongation of .016in. [.4064mm]?

Homework Equations

L=610mm
ΔL=.4064mm
E=29,000,000

The Attempt at a Solution

What I know is:

E=Stress/Strain

Strain=.4064/610 = 6.66*10^-4

Stress=F/A
Stress=19329.65

 

[raiderUM]

I am not getting the correct answer for FORCE. F=stress(Area) Am I messing up the Area some how?? The answer is suppose to be 86KN of Force or 19,333lb

 

[Sleepy_time]

Hi raiderUM. What are the units for Young's modulus that you are using. From your numerical answers you're using SI units. But, E=29x10⁶ is small for steel, it should be about 10³ to 10⁴ times as large as that number. This would give you the same approximately the same error from what you had.

 

[Chestermiller]

Hi raiderUM. What are the units for Young's modulus that you are using. From your numerical answers you're using SI units. But, E=29x10⁶ is small for steel, it should be about 10³ to 10⁴ times as large as that number. This would give you the same approximately the same error from what you had.

The units of the Young's modulus that the OP used were psi. He already has the right answer (aside from roundoff).

Chet

 

[Sleepy_time]

Ok, that's the reason. You need to convert it into Pa or Nm^-2. On wikipedia it says that for steel $E=200\times10^9 Nm^{-2}$. This will give you the answer that you need.