Question

30° w F M B •E A d ki* L L = 5 m d = 4 m F = 4.6 kN w = 3.9 kN/m M = 9.6 kN.m Determine the internal normal force, shear force and bending moment at point E, located in the middle of BC.

Answer 1

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*fBD:- F=4.6kN M=9.6kNom W=3.9kN/m ! AXA Ay By FĘ 2.5m 4m 5m 5m 5m na j rott & By Reaction at Bin y → Since we have Pinned support at A and roller support at B Hence there will only be reaction in a direction at A and will not be there at Batat Ax Reaction at A in a direction at Ay> > Reaction at a y direction. direction. Since udl acts on span fo ... Resultant force due to udl span FC 3.9 kN/mxum = 15.6 KN = Rw eating at at the centre the Spans FC of lie at am cleft from c. Resolving f in x and y direction tost x component = fsin 30 = = h. Gsin 30 = 2.3 KN Fcos 30 y component = 4.6 cos 30 - 3.98 KN. - (56) 2017 Loreto arofins a بار از امروز ایران امروز 130 الا و ارز در بازار 2012 Dopusti

* FBD (After resolution) E Rex Rw = 15.6 kN 3.98 KN 3-98 Mkat W=3.9KN/m - Ax 9.6 kNom 모 B e كع 2.3KN 2.5m am C IAY le' < 4m 5m 5m از By 5m Considering equilibrium of Beam: > in Y direction.; Efy=0 Ay +By = 15.6+3.98 = 19.58 KN sin x direction Efx=0 Ax= 2.3 kN. → Moment about A=0 (CCW stue 1 cwn-ve) 9.6 + By (io) - RW (6+5+1 +412) - 3.98(15) = 0. 9.6 + By CTO) – 15.6 (13) - 3.98(15) = 6 By= 25.29 KN It is not required, Since Ay+By = 19.58. But I have found it for your ease of understanding Ay = 19.58-25.29 Ay = -5.71 KN Since force An ats along the axis of the beam, and normal to the cross-section of the beam, as a reaction to fsin 30 along the beam. Normal force Axe = 2.3kN) rol Den

moment for finding Shear force and Bending Consider the beam /span on one side Eie EC E 12.5/2m 2-512m of the given point & ie IRS 3.98 KN (3.9 kN/m it 2.3 KN 2.5m f Rs Resultanca force due to udl on on Spania Eco Rs= (3.9 kN/m) x 2.5 9.75 Acting at the centre of the span ie 2.5/2 = 1.25m from C. Shear Force = t3.98) + Rs 3.98 + 9.75 SF:- Sign conventions for =13.73 KN Shear Force at E = 13.73 KN Bending moment about E:- BM (3.98342:5) + (RS) (2:5/2) Sign conventions -9.95 - 9.75x1.25 for BM. Bry -22:1375 KN-m F. Bending Moment about f=22:1375 vitve →tve -ve 1 ☺-ve. w.edu.in Knom