Question

90 kN/m 165 kN 120 KN.m 1m 2m { 3m 1mA 2m - find the relative stiffness of the beam between hinge and roller support. - moment at the hinge support. - moment at the fixed support. - Distribution factor of the beam between hinge and fixed considering the modified stiffness. -reaction at Fix support FEM is not given.

fem is not given

Answer 1

Given 280 kN/m 165kN 120kN-m t a B q you D Kim .3m *2m kamkam k om By Moment Distribution Method Step I : calculation of fixed ena moments ecause no external load (FEM)2 = (EEM)BA =0 [Beca is present on this span] 6=fvw -bl3a-e),MO 90x3 + 4(3X2-6) (FEM). EB xizo 30 30 ez 62 -27 ks-m (FEM)BC = (+62-a (36-4)xmo = +90x3²+2(3x4–6)x120 = (40:5+40) 12 62 = 480.5KN-m Step II : calculation of Distribution factors (DF);- Joint member Relative Stiffness (k) Ek DF=Klek BA I 2 -0.51 005 It 6•075I 0.SI -0.86 lo SESI B = 6·575I BC 2 x 5 Z 3T 24 0.0751 oostsi-0014 -0.0FSI

McD = -165KN-m e. The relative stiffness between hinge and roller support = 0.51 Distribution factor between hinge and roller support =0.86 Step Ill: calculation of ena moments by Moment aistribution Table or Joint A B Member AB BA Вс CB CD DE 0.86 0.14 FEM -165 O 1-27 Balance +27 69.23 - 11:27 A A +13.5 COM -165 +82.73 Total -34.61 -69-23 Balance 11061 - 1.89 COM -80.84 +80084 -165 Total -40.41 ". The ena moments are s- MAB = - 40.41 KN-m, M BA=-80:84 KN-m MBC=+80°84 KN-m , McB=0 +80.5 - 34.61 -5-80 4 :.] he moment at hinge support =-80-84 KN-m The moment at fixea support = -40.41 kN-m

Step I - Calculation of reaction at tix Support - For span AB :- 980084 kp-m We know that (Blan RA Fy=0 RB RotRA TO Jaking moment about (B), we get 80084 -2RA=0 RA= - 40142 KN , R6 = 40.42 KN 3. Reaction at fix support = 40°42 KN (4) Thus, we get ů Relative stiffness of the beam between hinge and roller support 20: 5I (19) Moment at the hinge = 80.84 KN-m N-m (3) Support (ii) moment at fixed support=40041 kN-m (iv) Distribution factor between hinge ana fixea support 50.86 v Reaction at tix support 340042 KN (W) Answer