Part B As shown, a truss is loaded by the forces P 1 = 819N and P 2 = 449N and h
ID: 1844019 • Letter: P
Question
Part B
As shown, a truss is loaded by the forces P1 = 819N and P2 = 449N and has the dimension a = 4.30m .
Determine FCG and FGH, the magnitudes of the forces in members CG and GH, respectively, using the method of sections. Assume for your calculations that each member is in tension, and include in your response the sign of each force that you obtain by applying this assumption.
Express your answers numerically in newtons to three significant figures separated by a comma.
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Explanation / Answer
Assume truss members are mass-less
Assume clockwise is the positive moment
Assume to right and up are positive direction
First find the reaction forces at A and E
To find vertical reaction E_y, sum moments about A to zero.
Assume E_y is positive
0 = P_1(a) - P_2(a) - E_y(3a)
E_y = (P_1(a) - P_2(a)) / (3a)
E_y = (P_1 - P_2) / 3
E_y = (819 - 449) / 3
E_y = 123.33 N
To find A_y, sum forces in the vertical direction to zero.
Assume A_y is positive
0 = A_y + E_y + P_1
A_y = - P_1 - E_y
A_y = -(-819) - 123.33
A_y = 695.667 N
To find A_x, sum forces in the horizontal direction to zero.
Assume A_x is positive.
A_x +P_2 = 0
A_x = -P_2
A_x = -(- 449)
A_x = 449 N
Now if we imagine a vertical cut through the three members GH, GC, and BC. we can eliminate the truss members and forces to the right of our cut and replace the cut stubs with equivalent assumed tensors which I will call GH, GC, and BC to represent the forces in each respective member.
the angle ? from vertical of member GC is
tan? = (a/2)/a
tan? = 1/2
? = 26.57