00:01
Nodes 1 and 2 are fixed, only node 3 can move so we solve the 2x u3y.
00:08
E is 200 times 10 to the 9th, 8 is 5 times 10 to the negative 4th the coordinates are 0 .0 .2 and 0 .50 from the figure and the load at 3 is 0 negative 5 for a bar from fixed node to node 3, k is equal to ae over l, c is delta x over l, s is delta y over l.
00:46
K3 plus k, well, k3 plus it was going to be equal to k times c squared, cs, cs, s squared.
01:09
Now for number 1, l1 is 0 and cs is 1 .0.
01:15
So k1 is ae over l1, that's 5 times 10 to the negative 4 times 200 times 10 to the 9th over 0 .5, which is 2 .0 times 10 to the 8th.
01:26
Number 2, delta x is 0 .5 and delta y is negative 2.
01:30
L2 is a square root of 0 .5 squared plus 0 .2 squared, which is 0 .3385.
01:35
So k2 is a, e over l2, which is 1 .857 times 10 to the 8.
01:53
C is 0 .5 over 0 .5385, which is 0 .9285...