Finding axial extension in steel bar

[cd19]

Homework Statement

A cylindrical steel bar, 20-mm in diameter and 5.0-m in length, is subjected to an axial tensile load of 40-kN applied at the ends of the steel bar. Determine the axial extension, in millimeters, of the steel bar under this loading condition?

Homework Equations

E=$$\sigma$$/e

$$\sigma$$ = F/A

e= change in L / Lo

The Attempt at a Solution

To find the extension we need to find (change in Lo)

$$\sigma$$ = 40x10³/((.1²x pi)/4) = 50265482.46

Even though not given in the question I am presuming E for steel = 210Gpa

therefore, e = 50265482.46/ 210x10⁹ = 2.393594403x10-4

as e = (change in L/Lo), the change in length = (2.393594403x10-4 x 5) = 1.2x10-3 m and 1.2 mm
This is a multiple choice question and this is not one of the answers, can anyone see where I am going wrong?

ANSWERS: 5x10-5 , 0.05, 3.2x10-3, 5.0, 3.2

 

[radou]

First of all, the area equals D^2 x PI / 4, where D is the diameter. You calculated as if it were the radius.

 

[cd19]

Thanks for the quick reply

I did new calculations and still am getting a wrong answer,

I found the $$\sigma$$ = 1273239.545

the strain = 6.063045452x10-6

and the extension I found to be .0303 mm

I reckon it has to do with the value I am using for E but it wasn't given in the question so I am not sure, any thoughts?

 

[radou]

Make sure your units are consistent - use milimeters for length and Newtons for force (i.e. megapascals for stress), and convert the assumed steel "Gpa" to "Mpa".

 

[DeeNos]

Your approach to finding the axial extension is correct. However, your calculation for the change in length may have a small error. The correct value for the change in length should be 2.393594403x10-4 x 5 = 1.1967972015x10-3 m, or approximately 1.197 mm. This would make the closest answer choice to be 5x10-5, which is the same as 0.05 mm. So the correct answer would be 0.05 mm.