00:01
So we are given a plane truss which is shown here in this figure having a square shape that is equals to 15 millimeter cross 15 millimeter cross section and a modulus of the elasticity is given that is equals to 59 gpa.
00:15
So in the first part of the question we have to assemble the global stiffness matrix for the given part.
00:22
So here basically we are given this figure this is the figure from here a force of 3 kilo newton is acting upward and a force of 5 kilo newton is acting towards the right side this is first and this is point second from here this is point fourth and this is point third from here this distance from here is 1 .5 meter.
00:51
So from here this is a given value so area is equals to 0 .015 multiplied by the 0 .015 that become equals to 225 multiplied by the 10 raised to the power minus 6 meter square and the young's modulus for e that is equals to 69 gigapascal that become equals to 69 multiplied by the 10 raised to the power 9 newton per meter square or we can also write it as 6 .9 multiplied by the 10 raised to the power 3 meter per millimeter square.
01:21
Now we have to constitute for the nodular area so at the node one this point is 0 0 if we are considering about the node 2 this point is 1 .5 and 0 if we are considering about the node 3 this point is at 1 .5 and 1 .5 if we are considering about the node 4 this is at 0 and 1 .5.
01:42
So in the first part we have to draw the stiff matrix so we can say that the stiffness matrix for element 1 is given that become equals to delta 1 e 1 which is divided by l 1 multiplied by the c square c s minus c square minus c s c s c s square minus c s minus s square minus c square minus c s c square c s and this from here is minus c s minus s square c s and s square so this is the stiffness matrix for the first element.
02:28
So we are considering about the element 1 where the value of 9 is at 0 degree 6 c z cos of phi theta 1 which is equals to cos of theta that is equals to 1 and s is equals to sin of theta 1 that is equals to 0.
02:45
So the value of k 1 in this case after solving this matrix so k 1 from here become equals to k 1 matrix become equals to 10 raised to the power 6 which is multiplied by the 10 .35 0 minus 10 .35 0 0 0 0 0 minus 10 .35 0 10 .35 0 and 0 0 0 0.
03:11
These points are q 1 q 2 q 3 q 4 and this from here is phi 1 phi 2 phi 3 and phi 4.
03:21
So this is the value for the element 1 now we are considering about the element 2 for element 2 theta 2 is equals to omega l 2 become equals to under root of 1 .5 raised to the power 2 plus 1 .5 raised to the power 2.
03:37
So solving the term this become equals to 2 .121 meter and c become equals to 0 .707 and s that is sin of theta 2 become equals to 0 .707.
03:49
So the value of k 2 in this case become equals to 10 raised to the power 6 and the matrix that is 3 .655 3 .655 minus 3 .655 minus 3 .655.
04:03
Again all the elements in plus this again is in plus this element is in minus this element is in minus minus same element minus same element plus same element plus same element minus same element minus same element plus same and plus same.
04:24
So this is for k 2 now we are considering about the k 3 so for element 3 we are having the value of theta 3 q 3 that become equals to 0.
04:36
So k 3 become equals to 10 raised to the power 6 to the matrix 10 .35 0 minus 10 .35 0 0 0 0 0 minus 10 .35 0 10 .35 0 0 0 0 0...