A brass alloy is known to have a yield strength of 275 mpa...

A brass alloy is known to have a yield strength of 275 MPa (40,000 psi), a tensile strength of 380 MPa (55,000 psi), and an elastic modulus of 103 GPa (15.0×10^6 psi). A cylindrical specimen of this alloy 8.9 mm (0.35 in.) in diameter and 241 mm (9.50 in.) long is stressed in tension and found to elongate 6.2 mm (0.24 in.). On the basis of the information given, is it possible to compute the magnitude of the load that is necessary to produce this change in length? If not, explain why. What is the yield strain? What is the test strain? b) If it is possible to calculate the load, what is the load value in N? If this computation is not possible enter a zero (0)

Transcript

00:01 Strain and force and to do this we will apply the relationship between strain stress modulus of elasticity relation between strain change in length and original length all right so we have these uh this uh this uh brass alloy i'm subjected to intense side force.

00:47 It's called it b.

00:54 We're given the diameter of 0 .0.

01:02 So 8 .9 mm millimeters, 8 .9 millimeters.

01:18 So we can just convert this straight to meters.

01:20 It gives us 0 .009.

01:33 And the length is given as 241.

01:46 1 mm plus of 2 to 4 1 mm.

01:53 If you can just convert to the same thing as equal to 0 .24 1 meters all right so what else given there? so we're given other details we're given the 10 size strength so that's called the 10 size strength that's such a t is equal to 275, as a pasta, 275 per square as 5, 1st2.

02:44 I just convert this to newton per meter square.

02:51 So it gives us 275 times 10, literally power of, basically power of 6, okay? newton per meter square, or you're just saying pasta, it's the same chain.

03:15 Alright, so we have a e also, which is a modulus of elasticity.

03:29 10 plus the power of 3 here pasta which we can just come back to newton per meter square based in as 10 plus a power of 3 or we can just say as 30 times 10 ratio 5 9 or simply 1 times 10 result 12 newton per meter square all right so we've got in a lot of detail then we are given the 10 size stress is two this is the use stress sorry this is yield and stress at which the material we yield okay then the 10 side strength or 10 size stress is given as you got it here as equal to 300 and there which is higher around 80 me faster using as 380 times 10 % of 6 15 per meter square.

05:05 All right so we can just continue now.

05:07 We are told to find, we are told to find the u strain.

05:17 So the u strain, the u strain is, so let's call this the u strain.

05:33 It's equal to the ustress of the modulus of elasticity.

05:49 So we plug in things there now.

05:50 We have this one with two around 75.

05:53 5 times 10 to the power of 6 over 10 times 10 to the power of 9.

06:16 So this gives us, so this is the e -string 0 .00, then we are asked again, is it possible to calculate.

06:43 Okay, we have to find the the test string.

06:53 Test, strain.

07:01 So, we assume that the test was carried out using the brass allerer as a specimen.

07:08 So if we consider it that way, we have all it is, test strain.