Force Applied

1. Jun 2, 2012

raiderUM

1. The problem statement, all variables and given/known data

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

2. Relevant equations

L=610mm
ΔL=.4064mm
E=29,000,000

3. 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

2. Jun 2, 2012

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

3. Jun 2, 2012

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.

4. Jun 2, 2012

Staff: Mentor

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

Chet

5. Jun 2, 2012

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.