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Engineering Mechanics - Equilibrium of a Rigid Body - Discussion

Discussion Forum : Equilibrium of a Rigid Body - General Questions (Q.No. 4)
Equilibrium of a Rigid Body
4.

The space truss is supported by a ball-and-socket joint at A and short links, two at C and one at D. Determine the x, y, z components of reaction at A and the force in each link.

Ax = -1.050 kN, Ay = 1.050 kN, Az = 0.800 kN, Cy = -1.050 kN, Cz = 0.600 kN, Dx = 1.050 kN
Ax = -1.400 kN, Ay = 1.400 kN, Az = 0.800 kN, Cy = -1.400 kN, Cz = 0.600 kN, Dx = 1.400 kN
Ax = -2.49 kN, Ay = 1.867 kN, Az = 0.800 kN, Cy = -2.49 kN, Cz = 0.600 kN, Dx = 2.49 kN
Ax = -1.867 kN, Ay = 1.867 kN, Az = 0.800 kN, Cy = -1.867 kN, Cz = 0.600 kN, Dx = 1.867 kN
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
2 comments Page 1 of 1.
Shaun said:   2 years ago
Since we know the body is in equilibrium, the sum of all forces and moments must be equal to zero.
1. Moment about x-axis: -.8(4)-.6(4)-Cy(3) = 0, Cy = -1.867kN.
2. Moment about y-axis: -.8(4)-Az(4) = 0, Az = .8kN.
3. Moment about z-axis: Ay(4)-Dx(4) = 0, Ay = Dx.
4. Force in x-direction: Ax+Dx = 0, Ax+Ay=0 (from 3), Ax=-Ay.
5. Force in y-direction: Ay+Cy = 0, Ay = -Cy, Ay = 1.867kN (from 1), Ax = -1.867kN (from 4), Dx = 1.867kN (from 3).
6. Force in z-direction: Az+Cz-.8-.6=0, Cz = 1.4-Az, Cz = .6kN (from 2).
Mark G. Wheaton said:   6 years ago
Please explain the answer.
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