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EM 319 - 10th Quiz

# EM 319 - 10th Quiz - − = ⇒ = ∑ = ⇒ = ∑ x M M C M...

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Name: Session: 2 1. Derive the equation of the deflection curve for a beam ( EI = constant) subjected to uniformly distributed load (see figure). Also, determine the deflection at the midpoint of the beam. Assume that the spring constant is 3 48 L EI k = . Solution: FBD1 EQM ( ) 2 0 2 2 0 qL R qL R R F qL R L qL L R M A B A y B B A = = + = = = = FBD2 qLx qx x M qLx qx x M M C 2 1 2 1 ) ( 2 1 2 1 ) ( 0 2 2 + = = + = qLx qx x M v EI 2 1 2 1 ) ( 2 + = = 1 2 3 4 1 6 1 C qLx qx v EI + + = 2 1 3 4 12 1 24 1 C x C qLx qx EIv + + + = BCs q L A B k q A B R A R B q A qL / 2 x V ( x ) M ( x ) C

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96 96 ) 0 ( 4 2 4 qL C EI qL EIk R v A = = = EI qL C L v 32 0 ) ( 3 1 = = + + = 1 3 8 4 96 ) ( 3 4 4 L x L x L x EI qL x v EI qL L v 384 7 2 4 = (negative sign downward)
Name: Session: 2 2. A cantilever beam ABC is subjected to a moment M 0 in the midpoint B of the beam (see figure). Derive the equations of the deflection curve for the beam. Also, determine the deflection at point C . Solution: FBD1 EQM 0 0 0 0 = = = = A y A A R F M M M FBD2 FBD3 2 0 L x L x L 2 EQM EQM

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Unformatted text preview: − = ⇒ = ∑ ) ( = ⇒ = ∑ x M M C M v EI L − = ′ ′ = ′ ′ R v EI 1 C x M v EI L + − = ′ 1 D v EI R = ′ Boundary Condition ) ( 1 = ⇒ = ′ C v L Continuity Condition ( ) ( ) 2 2 2 1 L M D L v L v R L − = ⇒ ′ = ′ 2 2 2 1 C x M EIv L + − = 2 2 1 D Lx M EIv R + − = L /2 M L /2 A B C M M A R A M V ( x ) M ( x ) M V ( x ) M ( x ) M C C Boundary Condition ) ( 2 = ⇒ = C v L Continuity Condition ( ) ( ) 2 1 8 1 2 2 L M D L v L v R L = ⇒ = EI x M v L 2 2 1 − = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + − = 2 8 1 2 1 1 L M Lx M EI v R Deflection at point C EI L M L v R 2 8 3 ) ( − = (negative sign ⇒ downward)...
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EM 319 - 10th Quiz - − = ⇒ = ∑ = ⇒ = ∑ x M M C M...

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