FBD for tri-pod jack stand?

  • Thread starter gomerpyle
  • Start date
  • #1
46
0

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.
 

Attachments

Last edited by a moderator:

Answers and Replies

  • #2
CWatters
Science Advisor
Homework Helper
Gold Member
10,532
2,298
See previous thread.
 

Related Threads on FBD for tri-pod jack stand?

  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
0
Views
1K
Replies
2
Views
3K
  • Last Post
Replies
1
Views
2K
Replies
0
Views
2K
  • Last Post
Replies
1
Views
4K
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
0
Views
5K
Replies
2
Views
696
Top