2.001 - MECHANICS AND MATERIALS I                                 Lecture #3                                 9/13/2006    ...
Draw each component separately.   FBD of Bar 1                                           Fx = 0                           ...
Now look at the FBDs of 1, 2, and C:    Solve equations of equilibrium for Pin C.                                         ...
Look at equilibrium of Pin A:                                FX = 0                          P cos θ + F3 = 0             ...
Q: What are the reactions at the supports?FBD:                                     Fx = 0                                R...
Solve for Pin D.                                      Fx = 0                            2P + FBD cos θ = 0                ...
FBD                                      Fx = 0                                    R Ax = 0                               ...
Upcoming SlideShare
Loading in...5
×

Lec3

224

Published on

0 Comments
0 Likes
Statistics
Notes
  • Be the first to comment

  • Be the first to like this

No Downloads
Views
Total Views
224
On Slideshare
0
From Embeds
0
Number of Embeds
0
Actions
Shares
0
Downloads
2
Comments
0
Likes
0
Embeds 0
No embeds

No notes for slide

Lec3

  1. 1. 2.001 - MECHANICS AND MATERIALS I Lecture #3 9/13/2006 Prof. Carol LivermoreRecall from last time:FBD:Solve equations of motion. Fx = 0 Fy = 0 MA = 0See 9/11/06 Notes.Solution: 1
  2. 2. Draw each component separately. FBD of Bar 1 Fx = 0 FAx + FCx = 0 Fy = 0 FAy + FCy = 0 MA = 0 l cos θFCy − l sin θFCx = 0 Solve. FAx = FCx FAy = −FCy FCy sin θ = = tan θ Forces track with angle. FCx cos θ Two Force Member: If forces are only applied at 2 points, then: 1. Forces are equal and opposite. 2. Forces are aligned (colinear) with the vector tha connects the two points of application of forces. So the FBD could be rewritten as: 2
  3. 3. Now look at the FBDs of 1, 2, and C: Solve equations of equilibrium for Pin C. Fx = 0 P + F2 sin φ − F1 sin φ = 0 Fy = 0 −F2 cos φ − F1 cos φ = 0 ⇒ F1 = −F2 M = 0 ⇒ No additional info.(Γ = 0) P + 2F2 sin φ = 0 Note φ = 30◦ , so: F2 = −P F1 = P Now look at FBD of 1, 2, and A: 3
  4. 4. Look at equilibrium of Pin A: FX = 0 P cos θ + F3 = 0 F3 = −P cos θ Note θ = 60◦ , so: P F3 = − 2 So: EXAMPLE: 4
  5. 5. Q: What are the reactions at the supports?FBD: Fx = 0 RAx + RDx = 0 Fy = 0 R A y + R Dy − P = 0 MA = 0 −2dP + dRDx = 0Solve. RDx = 2P RAx = −2P RAy + RDy = PNote: BD is a 2-Force member.Draw FBD of BD and Pin D. 5
  6. 6. Solve for Pin D. Fx = 0 2P + FBD cos θ = 0 Fy = 0 RDy + FBD sin θ = 0Solve. 2P FBD = − cos θ 2P R Dy − sin θ = 0 cos θ RDy = 2P tan θNote: θ = 45◦ , so: RDy = 2PSubstituting: RAy + 2P − P = 0 RAy = −PCheck:Statically Determinate: A situation in which the equations of equilibrium de-termine the forces and moments that support the structure.EXAMPLE: 6
  7. 7. FBD Fx = 0 R Ax = 0 Fy = 0 R Ay + R By − P = 0 MB = 0 −aRAy + MA − (l − a)P = 0Note: 4 unknowns and 3 equations.This is statically indeterminate. 7

×