Question

C2 Done session.masteringengine AA <HW11 - Attempt 1 Problem 1 1 of 4 > The moment of inertia of each span of the beam is indicated in . Assume the support at B is a roller 800 lb/ft 30 k А LAB = 900 in 1sc = 1200 in 24 ft --8ft ---8-- and A and C are fixed. E = 29(103) ksi. Part A Determine the moment Martina an cnan <

Answer 1

IBC 3ok 800 lb lft 24 ft А Kafk 8fttc By L 16ft I AB 900 in 4 fined OA=0c 2o. since A & Care 2 (FEM) WL -0.8 kl ft X24² AB - 12 12 - 38.4 k-ft =-460.8 kain (FEM) WEZ 38.4 k-ft BA = 12 = +460.8 kin (FEM) BO -WL ~30x16 0. - 8 8 -Gok-ft-720 Kin (FEM) CB WL 으 +60 Kft + 720 k-in.

From slove - deflection method - 26 ае[ + ].6 отбе - 10 ) + Тем), MAB - ge 2.Герб Ast ©р -о) - 38 4 xiz ач Х12 MA (4.15 6.- 460 - 8) К - in : 6.2 x 29 X 10 3g— +60 .8 MAB - 1812 SO 90 +60 - 2 МВР 2x29 Хto 3 ход 220В + о-о) + +60 - а+x12 2) М ВА - 39 Гоо On +4 So , Р M ge де) 1.0 + 0-0 )- кем BC ( 2х21 Xo'x 180o (og to –о) - go 16 x 2 Р) MBC За гоо авсао

MCB - 26/1) (20 + 08-9) + FEM CB - 2x29x109x1200 (200) + Op-1) + 720 (16812) 2 (Мсь 362500 OB +720 From Campatability ean MBAT MBC 20 362500 Op + 460.8) + CO 725000 Op - 720 OB -4 rad +2.383X10 Part- from een = 181250 X (2.383810*) - 460.8 MAB - 417.61.k-in MAB - 34.80 Kft

Part -B From ein (2 MBA 362500 (2.383X10 ++460.8 547.183 k-in MBA = 45.60 k-ft Part ac from ean (3 MBC 725000 (2.383X164) -720 = -547.184 kain MPC a 45.00 Kaftv Part -D From ein MCB W 362500(2.383X101) +720 . 806.384 k-in Mc B 67.20 Kft.

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