Question

Analyse the frame in Figure 1 uning the Stape-Deflection Method. Draw the shear force, axial force and bending moment diagrams. Assume that the flexural rigidity is a multiple of El and differs for each member as stated in the figure. When evaluating the axial forces in members AB and BC, assume they both have the same cross-sectional res, and that their axial stiffness is EAL Write answers in the space provided below and suburn this, along with a cover page and all workings attached I Shear Force dignum ( A 15 / 3 361 32 kN. 7.5&N 5 4-ET 4m ЗЕТ Figure Axial Force Diagram (M) Bending. Moment Diagram (M

this is only possible solving by mechanical engineering none of this civil Engineering guys was solved this question properly. it is my 5th time uploading same question civil guys pretty much copied and pasted from the already uploaded solution which are quite wrong . please do it in your own instead of copying

Answer 1

A Solution; 1stwimy 3m 32 km SET 27.5kN. B E2 E 3m 5 m 361 4m 4EL. D Degree of Kinetic Padeleminay (DK) =4896, SHB Pc, op} using slope deflection Method D. Calculation of fixed and moments MEAB = 15 x 32 - 11.25 km ME BC = MECB =o. 32x313x5-8) 11.25 kN-M. MEBA MEBD - 10.5tnom MEDB- 32X5 (373-8) 2.5kN-m. 82 Slope deflection equation MBA 2 Hongental Mass 11.25 + 2. (3€1} ( +33) $48=8 - 11.25 + 2(351) ( 206+3 0-18.) + 31989) (88-9) des | MBD = 10.5-4.5 + 34E7) (88). МВс 2 it the end is modified deffectional slope deflectico equation pinned support. equilibrium equation as per (Ok) = At Joint 'B' = MBA & MBD + BC =0. -11.95+ 3EIBB + 1.135 EIS + 10.5-1.25 +1.5E1QB +2.25 E1 08 -0.5695 ETS = 0. -2 + 6.75 ECOB + 0.5625 F18-0. -0

equilibrium equation comesponding to Dk{ SHB}=. 15kN/m 21R8 T MBA VBA 7.5kN VBC MBC VOA F - 7.5 VBC. VBA +705 – VBC = 0. VBA - VBC = -7.5 MABI MBA-67.5 MBC 3 4 - 7.5 π 2.25 EI 03- 0.5625EIS - 7.5+ 22.5 5E1DB + 3.125 EIS 3 4 2 15 1.104 EIDB + 1.182 EIS from equation 0 and ① EI OB =-0.8954, EIS - .13, 46

End moments will bez MAB 86.519 carm MDB = 0 MICB=0 MBA2 1.4163 MBC = -9.428 kaum MBD z 8.01 kum BMD 36.579 12 1.4163 9.428 DZA 8.01 20 Shear force diogram A 3 VABIL 36-379 15100/m Vael - VAB2 = 15*3 = 45-0 EMAFO 36.519 +1.4163 - 454105 VAB2= 32/com 2-9.8549.pn = 35.145KN 8:01 [VABI 1.4163 VAB2 VDB VDB VBC 9.428 NDB = 40 =5kn. 8 VBC= 9.498 =].357 kW 4 VBC 35.145 5KN. 9.8549 allone in (KN), 2.357.

Anial Face dropre 36.519 32 hom K 7.5kw D 3 ways of 5kn. 0 (onjuntwo) A с CD B с