00:01
So the deflection delta is given by the formula delta, the pl divided by ae.
00:23
So we're dealing with the stress over to the strain, but we have a composite rod.
00:29
So we need to consider the deflection in each portion separately.
00:35
So let's denote the lengths of portions, a, c, and b, c as lac, and l -c -b respectively.
00:51
So the portion ac, this becomes the delta ac, is equal to p, l -a -c, divided it into a -e.
01:19
And so for portion c -b, we have delta c -b, is equal to p, l -c -b, divided into a .e.
01:53
So if we get the components to find first, cross -sectional area of each portion, we'll call this a -1750 squared millimeters.
02:18
Then we have the youngs modulus e for both portions, and this is the 200 gigapascals.
02:29
Then the yield stress, proportion ac, this becomes the sigma ac, this is 250 megapascals.
03:01
Then the yield stress, proportion cb, that again is the sigma cb, that's 3, 4, 5 megapascals.
03:19
Then our maximum deflection called this delta m is 0 .3 millimeters.
03:44
Okay, so how to solve it, the problem here, and then we have the specifications as stated.
03:59
Second, so the total deflection, delta m is the sum of the deflections of both portions, so that's delta m is equal to delta ac plus delta.
04:10
So we'll call that delta ac plus delta c b...