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Unformatted text preview: Statically Indeterminate Flexural Members Determine the bending moment diagrammes of the following statically indeterminate beams. Example 1: Reactions A and B are rollers and C is a fixed end (no rotation or translation) Equivalent Load on Beam with Possible Deflected Shape Unknowns are: R A , R B , R C and M C . If we use the integration of the M/EI function we will find a further 2 unknowns, namely C 1 and C 2 the integration constants. We require 6 equations to solve the unknowns. Possible equations, , , x = 0 v = 0, x = 4 v = 0, x = 10 v = 0, x = 10 dv/dx = 0. ∑ = Y ∑ = M Use McCaulay notation and set up the equation for the bending moment. 7 10 2 4 5 4 2 5 2 2 2 2 − ⋅ − − ⋅ + − ⋅ + ⋅ − ⋅ = = x x x R x x R dx v d EI M B A 1 2 3 2 3 2 2 7 10 6 4 5 2 4 6 5 2 C x x x R x x R dx dv EI B A + − ⋅ − − ⋅ + − ⋅ + ⋅ − ⋅ = 2 1 3 4 3 4 3 6 7 10 24 4 5 6 4 24 5 6 C x C x x x R x x R EIv B A + ⋅ + − ⋅ − − ⋅ + − ⋅ + ⋅ − ⋅ = x = 0 v = 0 C 2 = 0 (1) x = 4 v = 0 10,6667 R A –53,333 + 4C 1 = 0 (2) x = 10 v = 0 166,667 R A – 2083,333 + 36 R B + 270 – 45 + 10 C 1 = 0 166,667 R A + 36 R B + 10 C 1 – 1858,333 = 0 (3) x = 10 dv/dx = 0 50 R A – 833,333 + 18 R B + 180 – 45 + C 1 = 0 50 R A + 18 R B + C 1 – 698,333 = 0 (4) Solve for R A , R B and C 1 R A = 7,8301 kN R B = 17,463 kN C 1 =  7,549 Stat Indeterminate structures Page 1 of 20 7/23/2003 ∑ = Y R A + R B + R C – 5 x 4 – 10 = 0 R C = 4,7069 kN ∑ = C M R A x 10 – 5 x 4 x 8 + R B x 6 – 10 x 3 + M C = 0 M C = 6,921 kN.m Draw the bending moment diagramme. Example 2: Reactions A and B are rollers, D is fixed and there is a hinge at C. Loaded Beam with possible deflected shape. Unknowns are: R A , R B , R D and M D and θ C . If we use the integration of the M/EI function we will find a further 2 unknowns, namely C 1 and C 2 the integration constants. We require 7 equations to solve the unknowns. Possible equations, , , , x = 0 v = 0, x = 5 v = 0, x = 13 v = 0, x = 13 dv/dx = 0. ∑ = Y ∑ = C M ∑ = D M Use McCaulay notation and set up the equation for the bending moment. 5 2 10 2 2 2 − ⋅ + ⋅ − ⋅ = = x R x x R dx v d EI M B A 1 2 3 2 9 2 5 6 10 2 C x EI x R x x R dx dv EI C B A + − + − ⋅ + ⋅ − ⋅ = θ 2 1 1 3 4 3 9 6 5 24 10 6 C x C x EI x R x x R EIv C B A + ⋅ + − + − ⋅ + ⋅ − ⋅ = θ Stat Indeterminate structures Page 2 of 20 7/23/2003 x = 0 v = 0 C 2 = 0 (1) x = 5 v = 0 20,8333 R A – 260,4167 + 5 C 1 = 0 (2) x = 13 v = 0 366,1667 R A – 11900,4167 + 85,333 R B + 4 θ C EI + 13 C 1 = 0 366,1667 R A + 85,3333 R B + 4 θ C EI + 13 C 1 – 11900,4167 = 0 (3) x = 13 dv/dx = 0 84,5 R A –3661,667 + 32 R B + θ C EI + C 1 = 0 84,5 R A + 32 R B + θ C EI + C 1 –3661,667 = 0 (4) ∑ = C M 9 R A – 405 + 4 R B = 0 (5) Solve for R A , R B , θ c and C 1 R A = 12,6725 kN R B = 72,737 kN θ C = 363,97/EI C 1 = 0,71875 ∑ = Y R 13 10 = ⋅ − + + D B A R R R D = 44,591 kN ∑ = D M 2 13 10...
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This note was uploaded on 12/16/2011 for the course STRUCTURAL 112 taught by Professor Colin during the Spring '11 term at Université Robert Schuman.
 Spring '11
 Colin

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