Question

The cross section of a bar is given by (1+(x^2/100)) where x cm is the distance from one end.Find the extension under a load of 20KN on a length of 10cm and Young's Modulus E=200,000N/mm^2

Answer 1

stress=E x e, the little e is the extension, stress is , stress=F/A, where force is 20 KN, and A is the cross sectional area and can be calculated by x and y where x is calculated from (1+x^2/100) and y is 10cm. this should help you hopefully...

Answer 2

I am using extension=(load x length)/(Area x E) and integrating using the given area but I am getting the answer wrong.So if anyone could either solve it or tell me if there is something wrong in what I'm doing

Answer 3

You're right to integrate since each infinitesimal cross-sectional slice will carry a different stress and therefore react with a different strain.

The equation for change in length is given as\[\Delta l = {F L \over A E}\]

The way to question reads, \((1+(x^2/100))\) is the cross-sectional area. It makes since since the units of this expression will be cm^2.

Substituting yields\[\Delta l = {20,000 \times 10 \over (1+(x^2/100)) \times 20,000,000} \left [ \rm N \times cm \over cm^2 \times (N/cm^2) \right ]\] The square braket shows the units, which reduce properly to cm.

We need to integrate w.r.t. x over the interval 0 to 10 cm. First, it would be helpful to reduce. \[\Delta l = {1 \over \left (1 + \left ( x^2 \over 100 \right) \right) \times 100}\]\[\Delta l = { 1 \over 100 +x^2 }\]
Establishing the integrand\[\Delta l = \int\limits_0^{10} {1 \over 100 + x^2} dx\]
This integral has the identity\[\int\limits\limits {1 \over a^2 + x^2 } = {1 \over a} \arctan \left ({x \over a} \right)\] where \(a = 10\)

Answer 4

Oops. It has the identity of\[\int\limits {dx \over a^2 + x^2}\]

Answer 5

Thanx for the answer
So the answer is 1/10 arctan1=0.08cm right.But the answer is given as 0.008cm.It might be a printing error.

Answer 6

This is interesting to me. Intuitively it should be 0.008 cm, but this integral is giving us a higher value.

Consider that if the bar were to be 1cm^2 along its entire length, the extension would be 0.01 cm. Since the bar's area is increasing along its length, the extension should be less than 0.01 cm.

Upon further inspection, I got the equation wrong.

We shouldn't take L to be the entire length, but instead be the infinitesimally small section, dx.

This yields, \[\Delta l = {F dx \over (1+x^2/100) E}\]

Upon substitution, we get\[\Delta l = {1 \over 1000 + 10x^2}\]
To get into the form \[\int\limits {dx \over a^2 + x^2}\] we need to multiply by (1/10)/(1/10), which yields, \[\Delta l = 0.1 \int\limits_0^{10}{ dx \over 100 + x^2}\]

This should give you the correct answer.

Answer 7

Oh yeah now I noticed that.I was working on this from yesterday not noticing I was putting l instead of dx.This finally makes sense.Thanx a lot!