The aluminum test specimen shown is subjected to two equal...

The aluminum test specimen shown is subjected to two equal and opposite centric axial forces of magnitude $P$ (a) Knowing that $E=70 \mathrm{GPa}$ and $\sigma_{\mathrm{all}}=200 \mathrm{MPa}$, determine the maximum allowable value of $P$ and the corresponding total clongation of the specimen. (b) Solve part $a$, assuming that the specimen has been replaced by an aluminum bar of the same length and a uniform $60 \times 15$ -mm rectangular cross section.

Transcript

00:01 Hello students, there is an aluminum test specimen.

00:05 There are two equal and opposite central axial forces acting here.

00:12 So this is force, this is again force.

00:15 So it's an elongation happening.

00:18 So we are given that the young's modulus is equal to 70 gigapascal and the allowed stress is equal to 200 megapascal.

00:28 Here we need to determine the pmax which is the maximum allowable pressure for total elongation and we need to find out the delta l which is the change in length that is the elongation of this thing due to the pressure.

00:46 So we can use the equation for maximum allowable pressure pmax will be equal to sigma allowed or sigma allowed divided by a square not a square sigma allowed times a by 2.

01:09 That is the equation 8 by 2.

01:12 So this is the pmax equation and the elongation can be found out delta l will be equal to pl by ae where p is the pressure here pmax, l is the length, a is the area and e is the elastic the coefficient of modulus of elasticity.

01:35 So we are given the values.

01:38 So let's find substitute and find the values here.

01:41 So pmax will be equal to sigma is equal to 200 times 10 to the power 6 times area.

01:49 We are given that the dimension is 60 times 15.

01:53 So 60 times 15 millimeter in rectangular cross section.

02:01 So 16 times 15 millimeter.

02:04 So that is equal to again 10 to the power 2 times you have to convert this millimeter into meter...

Question

Please give Ace some feedback