Problems 1 and 2 The truss below approximates a radio antenna tower. The blue vertical loads represent the weight of the

Post by **answerhappygod** » Wed Mar 09, 2022 9:13 am

: Problems 1 And 2 The Truss Below Approximates A Radio Antenna Tower The Blue Vertical Loads Represent The Weight Of The 1 (264.52 KiB) Viewed 55 times

: Problems 1 And 2 The Truss Below Approximates A Radio Antenna Tower The Blue Vertical Loads Represent The Weight Of The 2 (208.04 KiB) Viewed 55 times

: Problems 1 And 2 The Truss Below Approximates A Radio Antenna Tower The Blue Vertical Loads Represent The Weight Of The 3 (91.95 KiB) Viewed 55 times

Problems 1 and 2 The truss below approximates a radio antenna tower. The blue vertical loads represent the weight of the structure and the antenna, while the blue horizontal load represents wind loading forces. The red forces are the reaction forces of the pin mount at, and the rolling mount at B (not shown) 5001b Member 1: 251 +100; 1001b Member 2: 30; Member 3: 151-50 13 60010 B A Assuming all of structural elements 1 thn 5 are in tension, a force balance at each of the joints results in the following system of equations:

Summing forces on mounting point A: Summing forces on the upper joint: 4 +T, cos(76°) +T; cos( 70.3°) = 0 4, +T, sin (76°) +T; sin ( 70.3°) = 0 -T, cos(76°)+1cos ( 73.3°) + 200 = 0 -T sin(76°)-1; -T; sin (73.3°) - 500 = 0 T.cos(53.1°)-T, cos (70.3°) = 0 T: -1, sin (53.1°)-T; sin ( 70.3°) - 600=0 -T, cos( 73.3°) -T.cos(53.1°) = 0 ; T; sin ( 73.3°) +T, sin (53.1°) + B, = 0 Summing forces on the interior joint: Summing forces on mounting point B: 0 0 0 0 0 The resulting matrix equation: 1 o cos 76° 0 0 1 sin 76° 0 0 0 0 -cos 76° 0 cos 73.30 -sin 76° -1 -sin 73.30 0 0 0 0 0 0 0 0 1 0 0 0 0 0 -cos 73.30 0 0 0 0 sin 73.30 -200 500 cos 70.3° 074 sin 70.3° 04 0 07: 0 0 | T: -cos 70.3° 0 Tz -sin 70.3° 0 || T. 0 0|| T 0 В. 0 0 cos 53.19 -sin 53.1° -cos 53.1° sin 53.10 . 600 ) 1

Problem 2 Use Gauss Elimination with partial pivoting to calculate all of the tensions and reactions forces. Only perform row operations where necessary. eg in columns 1 and 2, all of the values below the diagonal are already zero, so there are NO row operations required in columns 1 and 2 Similarly, there will be only one set of row operations needed in column 3. There is an example of Gauss Elimination on Canvas, including the back substitution process to solve for [x]