Question

8. Determine and Draw the quantitative shear, bending moment diagrams and qualitative deflected curve for the following questions by using the force method

86 k FM B C А ID ΟΙΟ. 4 ft 4 ft 8ft

Answer 1

Solution...

free body diagram of beinen beam - RD MA 66k R C sche t H A B C D C RA

0-1 degree of static indeterminacy (DSJ) DSI = 2 thed us consider vertical reaction at point c and D as redundard farlice. - We solve the given problem by using freice method flexibility mateix method. In this method we we compatibility condition. &GK By с D O primary be (after removing redundant farce. Aue=deflection at point D due to given testing loading. Redundant beam of DCL Don < or 0 D . A & etus B с. DCD Doc DDD where A ce = deflection at point ® when will load act at point Dog - deflection at point © when unit load act at point ®

A De=deflection at point when unit load act at ADD = deflection at point 6 when undt load act at @ . Then 3 as per flexibility matix method • Compatibility condition & Vertical deflection at point ® & fs equal to zero De Ap=0 5 We write the following equation .

Der Rc Ac + ADL Asc RD AD Where a DeL= deflection at a due to given loading Deli 86*43 66 Xe + 2 xu 2E] - 27520 661 3 Ei 3 ADL 86x U 3EI 364 U & EI 2 .X 12 60 suu GEI

Daca 1x83 512 BEI 3 EI 4 Дос 11 1883 1882 + 3 EI 2560 6 EI 2 EI . - DDD 1 X 16 BEI 4096 3EI 2560 GEI Then arsao GEI 52 2560 6 EI Rc 0 3E1 + Gosuu GEI 2560 GEI 4096 3 EI Ro 1024 6 EL 2560 GEL Trke Q7520 GEI 1 2560 GEI 8192 6E1 Ro Gosuy GEI solving Both side, we get

Ref - 38.39286 K R = 4.60714K opposite Negative sign Ps. means, direction of reaction Re of as to assume in free body diagram. Then free body of beam . A 86k * 1 38.39286 4.60714 RA • We solve it by wing equilibsium equation a 1 E Hoizontal force = Evertical force =0 - + 38:39286-86-4.60714 = 0 52021428 KT CA] (emoment=0 s) 4060714 X16 – 38039286 x3 RAT a) IRA +86 X U + Me=0 MA = -110.57136 K-fel Negative sign indicate direction is reversed.

We use Calculation foar shear fosice, bending moment diagrama following sign Convention (m-moment) (5= shear force) 110057136 86k 7 1 52.21428 K 38,3928610 4.60/14 * 5241428 52.21428 Shear force diagram €) 4.60714 4.60714 -33.78572 -33.78572 9 828 576 k-ft Moment fol181 Bending Diagoam & -36.85712 kaft - 110.57136 k- et Deflected shape (By red colour) point of inflexim