Question

Q2. In the 3-span beam shown in the Figure 2, the support at C settles by 15 mm. Using the method of moment distribution method, a) Calculate joint moments (10 marks); b) Draw the bending moment diagram (10 marks); c) Calculate the supports' vertical reactions (10 marks); and d) Draw the shear force diagram (5 marks). Hint: For the calculation of Fixed End Moments, you can use the principle of superposition to add the settlement-related and the load-related FEMs together You need not show the absolute maximum values of the bending moments in the spans. Assume E = 200x10' N/m² and I = 200x106mm for all members. X7 X8kN/m X4m X5m Xom A B с D Figure 2

x7=12KN , x8=19kN/m,  x4=4m, x5=39m, x6=5m,

Answer 1

2 EN д. N D 19 kNlm ш ДЕ itsm 39 од Д no Lue 4 t 1150 th И Tv V $ с ои fixed End Moments 2 N Арь с. 0 МБА _c No 2 -we 8 Ge2 Д 2 - Э! 6 х 200 х 2-oo HSx lo - 1993 R 392- о 2:3e (s-85-) Сч- Nй 98 «Ф9- с rmit Gen A t® Mce: Q" 8 25 b et Ses) о + 5% 5 а: 26 n yl 56:14 KN-'m d

MAID' alwe ? + GEZA 12 12 = (–19215)? + 6 * 2004 200415 416 12 52 = (-39.58) + 144 E lou.42 kNam Mpya 13 twee 12 GEZA L? 68200 12001 istio + (+19)(5) - 12 52 183.58 KN~m 고 (39158) +144 a

Moment distribution Moble Member BA Be CB CD DC 4 Il4 9) 1 Ilga 15 39 1 0.079 0,121 011136 0.886 0 FEM O O +60186 5614 lou.42183.58 Balance 53:49 7.36 -18.23 142.25 COM -9.115 368 o -71.129 Bolanel 7.87 1.10 -0.418 -3.26 0 Com H.63 o -0.20910:55 Balone + 0.188100251-0062-048 COM o o -010311010125 -024 Oordo Bolonel is o 161.575|-61.75 M16A2-41.57 110.46 M End Moments are MAB =o e) Meg=-ulieta a) MBA = 61.75 b) f) MAC = +110.46 N 8c 5- ده MCB tul.GA2 d)

110,40 kn mnt5:10 2 Degree UICAIN MBA e) tum Support Reactions EMRO MBA CO (VA) *4 + (VA) 4 vel MBA Vi -61.52 --1539 KN (VA) 4 consider member co Med 19 kN lm Zme unu ch5m + ad -cuaxs)+(varsmini) quel Tuan +(More) +(Meo) = 0 (5v.) = (1945 m)+ (+10:46) – (41,57) n VD 61.228 KN

Now comiden systen equibrio = 12 + 1985 - 1ozen VA + VGA ve avo Emezo (UA) (43) +(VB)(39) -12439 + (19x545) 21:278)(5) +110.46 =0 (Vg)(34) = (234)- -237.5t 806.39 -110.46 + G61.77 VB) = 21.90 kn Also Ver (107) - 615.39) -21.90-6128 Ne = 39.21 kW Reactions NA ~ 15:39 KN VB = 21.90 KN Loka Ne 39.21 KN No G1 278 KN

SHEAR FORCE DIAGRAM 33.72 B,51 + + п 5.49 15:39 IS. 61.278kW