Mechanics 1I

Mechanics 1I - 2.001 - MECHANICS AND MATERIALS I Lecture #6...

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2.001 - MECHANICS AND MATERIALS I Lecture #6 9/27/2006 Prof. Carol Livermore Recall: Sign Convention Positive internal forces and moments shown. Distributed Loads 1
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± ± q ( x )= q 0 ² L ³ L F Total Loading = q ( x ) dx = q 0 x = q 0 L 0 0 ² L 2 ³ L M A Distributed Load = q ( x ) xdx = q 0 2 x = q 0 2 L 2 0 0 Lump: (Equivalent Forces) FBD 2
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± ± ± F x =0 R A x F y R A y + R B y q 0 L M A L q 0 L + R B y L 2 q 0 L R B y = 2 Substitute: R A y + q 0 L q 0 L 2 q o L R A y = 2 Cut beam. 3
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± ± ²³ ± Lump: F x =0 N F y q 0 LL q 0 x V y V y q 0 x 22 M q 0 L x + q 0 x x + M z Lx 2 M z = q 0 x Plot
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± ± ²³ Sanity Check Pinned and Pinned on rollers at end Cannot support moment M at ends Can support x and y loads OK F y =0 V y ( V y + δV y )+ qδx y = Limit as δx 0 qdV y →⇒ dx M 0 M z + M z + δM z V y ( V y + y )= 0 22 z 2 V y y z = V y y 2 z y V y =+ 2 Take limit 0 dM z V y = dx dV y q = dx But what if load is not uniformly distributed?
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± ± ± EXAMPLE: A more interesting structure Q: Find all internal forces and moments in all members. Plot.
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Mechanics 1I - 2.001 - MECHANICS AND MATERIALS I Lecture #6...

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