SOLUTION: I was given a problem and am having difficulty setting up the problem. We have a square sheared beam and know the temperatures along the four sides of the beam. We must use the ave

Algebra ->  College  -> Linear Algebra -> SOLUTION: I was given a problem and am having difficulty setting up the problem. We have a square sheared beam and know the temperatures along the four sides of the beam. We must use the ave      Log On


   



Question 351940: I was given a problem and am having difficulty setting up the problem. We have a square sheared beam and know the temperatures along the four sides of the beam. We must use the average temperates in order to estimate the internal temperature distribution in that beam using linear algebra. We made a diagram and a grid system showing internal points T1, T2, T3, and T4 where each is and average of the surrounding points.
The diagram looks like a grid of 9 boxes which make up one large square. The four corners of the large square have no data points. Every other intersection has a value assigned to it. The top two intersections have values of 20, meaning that the top surface is 20 degrees, the left intersections have values of 10, the right 40 and the bottom 30. The internal intersections are labeled T1, T2, T3, T4 read left to right from top to bottom.
I have four equations as follows, but then I am stuck.
T1 = (10+20+T2+T3)/4
T2 = (T1+20+40+T4)/4
T3 = (10+T1+T4+30)/4
T4 = (T3+T2+40+30)/4
In some manner, I have to rearrange these equations in order to put them into a matrix and solve.

Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
T1+=+%2810%2B20%2BT2%2BT3%29%2F4
4T1=30%2BT2%2BT3
4T1-T2-T3=30
.
.
.
T2+=+%28T1%2B20%2B40%2BT4%29%2F4
4T2=T1%2BT4%2B60
-T1%2B4T2-T4=60
.
.
.
T3+=+%2810%2BT1%2BT4%2B30%29%2F4
4T3=T1%2BT4%2B40
-T1%2B4T3-T4=40
.
.
.
T4+=+%28T3%2BT2%2B40%2B30%29%2F4
4T4=T2%2BT3%2B70
-T2-T3%2B4T4=70
.
.
.
[A]=
[T]=%28matrix%284%2C1%2CT1%2CT2%2CT3%2CT4%29%29
[b]=%28matrix%284%2C1%2C30%2C60%2C40%2C70%29%29
.
.
.
[A][T]=[b]