Analysis of Tri-Pod Jack Stand with Welded Connections and Axial Load

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SUMMARY

The discussion centers on analyzing a tri-pod jack stand with welded connections subjected to an axial load of 36,000 pounds. The user calculates vertical reaction forces at each leg, determining that each leg bears a load of 12,000 pounds. The horizontal reaction force due to friction is calculated at 6,000 pounds, assuming a static friction coefficient of 0.5 for concrete and steel. The user raises questions about the implications of rigid connections on moment and shear load transmission, particularly regarding zero force members and potential reactions at connection points.

PREREQUISITES
  • Understanding of static equilibrium equations (∑Fy = 0, ∑Fx = 0, ∑M_x = 0)
  • Knowledge of welded connections in structural analysis
  • Familiarity with friction coefficients in material interactions
  • Basic principles of load distribution in truss structures
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  • Study the effects of rigid versus pinned connections in structural mechanics
  • Learn about zero force members in truss analysis
  • Investigate the calculation of moments in welded structures
  • Explore advanced friction models for different material combinations
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Structural engineers, mechanical engineers, and students studying statics and dynamics who are involved in analyzing load-bearing structures and welded connections.

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Homework Statement


A jack-stand with three equally spaced legs (120 degrees) are connected to a center tube is loaded axially with 36,000 pounds. Each leg has a horizontal cross member 11 inches from the ground. All connections are welded. Set up FBD and find reaction forces.

Homework Equations


∑Fy = 0
∑Fx = 0
∑M_x = 0

The Attempt at a Solution


I guess I'm confused about the connections being rigid and welded. Most structures I have seen in examples are assumed to be pinned and therefore do not transmit moment/shear loads. Solving globally for

∑Fy = 0
36,000 = 3*Na = 0
Na = 12,000 (Vertical reaction at each foot contacting ground)
The horizontal reaction at each foot is the opposition of friction:
Fa = μsNa
Fa = (0.5*12000) = 6000 (I use a μs = 0.5 for concrete/steel)

Summing moments about A:
∑Ma = (15*R_dy) + (15*R_cy) - (28*R_cx) = 0
∑Fy = Na - R_dy - R_cy = 0
∑Fx = R_cx - Fa = 0

R_cx = 6000

Plugging everything in I get R_dy = 0?? So DB is a zero force member then if CB is in compression? I would think the 36,000 would transform some load to the member DB. The other thing I'm confused about is there a reaction at point B between the angled leg and the cross-member? I don't know if each of these points would have a bending moment associated with it either since they are rigid and not pinned.
 

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