00:01
For part a, we would like to find the minimum required length of the rod.
00:06
So to find this, we want to first note our values.
00:10
We have our shear modulus of elasticity to be 80 gpa.
00:17
And we have d to be 12 millimeters.
00:24
Our angle phi is 30 degrees or 0 .523 radians.
00:30
So we want to use our expression to find our length.
00:35
Our minimum length is going to be equal to g -d -5 over 2 times our allowable shear stress.
00:43
So we plug in our 80 gpa, but that's just giga.
00:54
So we want to multiply that by 10 to the 6, or 10 to the 9.
00:59
So 80 times 10 to the 9 times our d.
01:03
Now our d phi is going to be our 12 millimeters times 0 .523 radiance divided by two times our allowable, which is equal to, and that's given in our problem, 300 mpa.
01:27
And an mpa is just equal to a mega pascal...