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Lec2    MECH ENG  STRucture
Lec2    MECH ENG  STRucture
Lec2    MECH ENG  STRucture
Lec2    MECH ENG  STRucture
Lec2    MECH ENG  STRucture
Lec2    MECH ENG  STRucture
Lec2    MECH ENG  STRucture
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Lec2 MECH ENG STRucture

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  • 1. 2.001 - MECHANICS AND MATERIALS I Lecture #2 9/11/2006 Prof. Carol LivermoreTOPIC: LOADING AND SUPPORT CONDITIONSSTRUCTURAL ANALYSISTools we need:1. Recall loading conditions (last time)a. Forces - (”Point Loads ”) Point of Application Magnitude Directionb. Moments From forces applied at a distance from a point.2. Recall equation of static equilibrium (last time) F = 0 ⇒ Fx = 0, Fy = 0, Fz = 0 M0 = 0 ⇒ Mx = 0, My = 0, Mz = 03. Distributed Loads - Force per unit length4. Concentrated Moment - Moment applied at a point (EX: Screwdriver) 1
  • 2. FREE BODY DIAGRAMS Draw like a mechanical engineerFixed Support Cannot translate horizontally or vertically. Cannot rotate.Pinned Support - Cannot translate. - Free to rotate.Pinned on Rollers - Cannot translate in y. - Can translate in x. - Free to rotate.REACTIONS: (Forces applied by a support) 2
  • 3. No ⇒ can apply Rx . ↑↓⇒ can apply Ry . No rotate ⇒ can apply M .So:Need Rx , Ry , MNeed Rx , Ry 3
  • 4. Need RyOTHERS:CABLE:SMOOTH SUPPORT:EXAMPLE: DIVING BOARD Find reactions from the supports for this structure. 4
  • 5. 1. How to draw a Free Body Diagram (FBD) i. Remove all external supports. ii. Draw all relevant dimensions. iii. Depict all external forces and moments. iv. Coordinate system.2. Apply the equations of equilibrium. Fx : RAx = 0 Fy : −P + RAy = 0 MAz : MA − LP + (RAx )(0) + (RAy )(0) = 03. Solve. RAx = 0 RAy = P MA = LP4. Check⇒ looks good!EXAMPLE: 5
  • 6. Draw FBD.Apply equilibrium. Fx : RAx = 0 L Fy : RAy + RBy − 0 (−q0 )dx =0 L q0 L2 MA : RBy L + 0 (−q0 )xdx = 0 = RBy L − 2Solve. RAx = 0 2 RBy = q0 L = q0 L ⇒ RBy = q0 L 2L 2 2 RAy + q0 L − q0 L = 0 ⇒ RAy = q0 L 2 2Check.⇒ looks good!EXAMPLE: PLANAR TRUSS 6
  • 7. Q: What are the forces on each member? Approach: Look at whole structure. Look at each piece individually. FBD of Whole Structure RB = RBxˆ + RBy ˆ i j Equations of Equilibrium: Fx = 0 R Bx + P = 0 Fy = 0 R Ay + R B y = 0 MA = 0 RBy l + (P )(− sin θ)l = 0 OR rAC × P + rAB × RB = 0 Solve. RBx = −P RBy = P sin θ RAy = −P sin θCheck. 7

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