Upcoming SlideShare
×

Lec10 MECH ENG STRucture

348 views

Published on

0 Likes
Statistics
Notes
• Full Name
Comment goes here.

Are you sure you want to Yes No
• Be the first to comment

• Be the first to like this

Views
Total views
348
On SlideShare
0
From Embeds
0
Number of Embeds
6
Actions
Shares
0
2
0
Likes
0
Embeds 0
No embeds

No notes for slide

Lec10 MECH ENG STRucture

1. 1. 2.001 - MECHANICS AND MATERIALS I Lecture #10 10/16/2006 Prof. Carol LivermoreFrom last time:1. Equilibrium FBC = −P/ cos θ FAB = P tan θ2. Force deformation relationships (”P = kδ”) P L tan θ sin θ δ AB = AE −P L δ BC = AE cos θ3. Compatibility a. uB x 1
2. 2. δ1 = uB AB x δ1 = uB sin θ BC x b. uB y AB δ2 = 0 BC B δ2 = ux cos θSum: δ AB AB = δ1 + δ2AB = ux + 0 = uB = δ AB B x δ BC BC = δ1 + δ2 BC = ux sin θ + 0 = uy cos θ = δ BC B B P L tan θ sin θ uB = x AE δ BC uB sin θ B uy = − x cos θ cos θ −P L P L tan θ sin2 θ uB = y − AE cos2 θ AE cos θ 2
3. 3. 2, 3, 4 are statically indeterminate! (more bars than are needed to support the structure)A, E for all barsForces? Deformations? Displacement of point D?Unconstrained D.O.F 1. uD x 2. uD y Equilibrium FDx = 0 P + FDC sin θ − FAD sin θ = 0 FDy = 0 −FDC cos θ − FAD cos θ − FBD = 0 Force-Deformation Relationships 3
4. 4. AE AD FAD = kAD δ AD = δ L AE CD FCD = kCD δ CD = δ L AE BD FBD = kBD δ BD = δ L cos θCompatibilityuD x BD δ1 = 0 δ1 = uD sin θ AD x 4
5. 5. δ1 = uD sin θ CD xuD y δ2 = uD BD y δ2 = uD cos θ AD y δ2 = uD cos θ CD ySum: δ AD = uD sin θ + uD cos θ x y δ BD = 0 + uD = uD y y δ CD = −uD sin θ + uD cos θ x ySubstitute δ into F AE D FAD = (ux sin θ + uD cos θ) y L 5
6. 6. AE D FBD = u L cos θ y AE FCD = (−uD sin θ + uD cos θ) x y LSubstitue F into equilibrium P + FCD sin θ − FAD sin θ = 0 AE AE D P+ (−uD sin θ + uD cos θ) sin θ − x y (ux sin θ + uD cos θ) sin θ = 0 y L L D uy cos θ → 0 2AE D 2 P− u sin θ = 0 L x PL uD = x 2AE sin2 θ FDC cos θ + FAD cos θ + FBD = 0AE D AE D AE D (uy cos θ − uD sin θ) cos θ + x (ux sin θ + uD cos θ) cos θ + y u =0 L L L cos θ y AE uD y 2uD cos2 θ + y =0 L cos θ AE D 1 uy 2 cos2 θ + =0 L cos θ uD = 0 y P FAD = 2 sin θ FBD = 0 −P FCD = 2 sin θ 6