Question

For the following beams sketch the shear Force and bending moment diagrams. Also determine the position and magnetude of the maximum bending moment and where applicable the inflection point.

Answer 1

problem! Hoknim 55 m 2.5 m Free body Diagram"- HoX 8 KN = 880KN. 812=4m . m ادماه او TRB * 9.5m To find the reactions at supports, we need to replace the distributed load. with point load. The magnitude of point load = Area of - the distributed point load = lokal load shape. = 80KN Applying Evuilibrium Equations. EFx=0 i Raa= 0 EFy = 0 Assuming (1+ve) TRAIRB = 880 kN) 0

Taking moment EMA=0 about Assuming A. #moment tue. load acts The at point centriod load of of distributed Rectangle load. - RBX55 + 860 X 4 - 0 - Ro*55 =-880x4 RB = 860 5.5. | RB = 640kn substitute RB value in ean o RA tSto = 88 FRA = 240KN - shear Force Diagram. 1. cat the Beam at the point where you I want to find shear force. you con consider left side of section oo) Right side of section to the calculate shear force. for heft side for right side Here I am taking Find the resultant of section by ? tue atue ? -ve. hett side, so force to the left appointing the summing the side Forces

shear Force at A. At A There is only RA present. So ( ) SFA - 24. OKN Shear Force at B. HONI -55m Ях PX - RA-110 X +640 = 240- 110X55 7640 from sign (S.F)B= +295 kN] Just Before point B the shear Force is (ove) point A to B. the shear Force changes from the to ave. We knew at that 0 point (&fx= 240-110xx = 0 We that at don't knew point. Assume distence a 240 - loxx a=2181m ها (۶۴) shear Force at c. llokalm 5.5m (SF) Rate Pp. 110%* (sf) - 2407640 - M10x& (sfe=ol

Shear Force diagram Ti oknim TRA= 840kn TRS= Backa 275 kn 240kn F 2. - 365kn If the to the you take shear Force Reaction value. a section is ave. force & Just before paint & At point o due suddenly raises to The loed given , so lood the is shear uniformly force distributed changes with length

Bending moment Diagram. as sheer procedure same Force Diegrem from het to Right In this if we go we consider ato from Right to left moment possitive we consider & tue Bending moment at A heft side to section A, no force present, so (B.MA =0 Bending moment Between AkB. OK Tko Here T(Bm A - B = RAxa - 1oxxxx | we take section of between AIB have to consider the UAL upto a only to To find the Bending moment we need of the location where it acts we know it acts at controid according to force moment this evtlation e out that mammum where the location Bending is

Bending moment at x =2018ım Bm) men 24.0% 2461 - Hlox (2 161) 16m).men = 261 81 kw.m (8:n ]AB140*- 40x4 Absorbing the above evection we can easily Sey Bending moment varying parabolically a In parabolic curve, we have two types of carves, whether it is concare deplor) - Concave down. By can double differentiating the enuation we find concauity of the curve. sebeb 8.9 la = 240*- 110 m de on 240 - thox = -110 / -ve . carve so is the concave down Bending moment at B. at n=55m. BmlB = 24045-5- 10x (55) 18925. = -343-75 11000lm moment at Bending c. HTRE BR : RAX 5+ RBX 2.5 - (lo X8X13) (BM 240x8+ 640*2.5 - 110x &x4 O

loknlm. TRA = 240kn RB = 640KN the (o mlmcn . where point - sign changee 126181 Inflection >1 12-181m - 343.15 دمع 1 4.363m moment curve Bendina where Inflection point is changes it's sish (omao) (BM)*8= 240 1 - 100x o 240x - llox x? 240x8 = 160mm 110 .480 UPO Tio I inflection point = 4.36 Bm