Torque Problem Involving elastic modulus?

Question

A 20 cm long animal tendon was found to stretch 3.5 mm by a force of 13.9 N. The tendon was approximately round with an average diameter of 7.0 mm. Calculate the elastic modulus of this tendon.

I calculated both answers and submitted them. Each were wrong. I calculated again making sure I did not make any stupid mistakes. My calculator still yielded the same numbers. Still need help.

I calculated 460340.7349 as well as 459874.107, and 361184.279. None work. What am I doing wrong?

Answer 1

E = stress/strain = ((13.9N/(7²π/4)) / (3.5/200) = 20.639 N/mm²

Answer 2

This problem does not involve torque. It is a straightforward stress/strain situation. The elastic modulus is

E = stress/ strain.

In this case, strain is 3.5*10^-3 / 20*10^-2

The stress is force divided by area. The force is 13.9N, the area is (π/4)*d^2

Stress is 13.9 /[(π/4)*(7.0*10^-3)^2]

You can do the math. Note all units are converted to meters.

Answer 3

I can't see anything wrong the the above answers, 20.639 N/mm^2 looks right.

The only things I would suggest is check your rounding and your units to make sure they are correct when entering the answer.
N/mm^2 is equal to MPa

Is there any other information available? The fact you used torque in the questions makes me wonder if there is something missing, as this is not a torque problem.

Answer 4

properly, the equation is: (F/A) = Y (?L/L) the place (F/A is the stress) and ( Y is the youngs modulus) so which you dont even would desire to organize the equation. the main element to realize is that the upward thrust in length (?L) is 0.10% , with the objective to make issues much less perplexing, think of of the unique length to be 100m meaning the rigidity, (?L/L) is 0.a million/a hundred it is 10^-3 as a result, purely exchange for Y and (?L/L) you get F/A = (2 x 10^11) x 10^-3 which provides 2x10^8