lec3 - 2.001 - MECHANICS AND MATERIALS I Lecture #3...

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± ± ± 2.001 - MECHANICS AND MATERIALS I Lecture #3 9/13/2006 Prof. Carol Livermore Recall from last time: FBD: Solve equations of motion. M A =0 F x F y See 9/11/06 Notes. Solution: 1
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± ± ± Draw each component separately. FBD of Bar 1 F x =0 F A x + F C x F y F A y + F C y M A l cos θF C y l sin C x Solve. F A x = F C x F A y = F C y F C y sin θ == t a n θ Forces track with angle. F C x 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
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± ± ± Now look at the FBDs of 1, 2, and C: Solve equations of equilibrium for Pin C. F x =0 P + F 2 sin φ F 1 sin φ F y F 2 cos φ F 1 cos φ F 1 = F 2 M No additional info.( ± Γ=0) P +2 F 2 sin φ Note φ =30 ,so : F 2 = P F 1 = P Now look at FBD of 1, 2, and A: 3
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± Look at equilibrium of Pin A: F X =0 P cos θ + F 3 F 3 = P cos θ Note θ =60 ,so : P F 3 = 2 So: EXAMPLE: 4
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± ± ± Q: What are the reactions at the supports? FBD: F x =0 R A x + R D x F y R A y + R D y P M A 2 dP +
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This note was uploaded on 05/05/2010 for the course MSE 2.001 taught by Professor Carollivermore during the Fall '06 term at MIT.

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lec3 - 2.001 - MECHANICS AND MATERIALS I Lecture #3...

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