Question

The beam supports the distributed load with wmax=3.6 kN/mwmax=3.6 kN/m as shown. The reactions at the supports AA and BB are vertical.

-- 1.5 m - 3 m — 1.5 m

Determine the resultant internal loadings acting on the cross section at point C.

Nc=?,Vc=?,Mc=?

Answer 1

Wmax = 3.6 kN/m wmáx?owa ANT amaz 3•6kN/m 1.5m t3m ti.sm/ RB For Equilibrium RA+RB= 0 x 6x316=10.8- EMB = 0, 62 RAX5 =%&*9x36)*(323) | RA = 72 kN RB = 10.8-7.2 = 3.6 KN RA = 72 KN TRB=3.6 kN AM my Shear p section at a distance of am from a be sx. In triange ABA' and A oo'B which are similar AA 6 Let Ao = x meter, oo'-.Wx OB = 6 - 2)", AB= 6 m, AA'=3.6 kulm 3.6 G WX (6-2) Ww= 3:6(6-2) - 0.616-X) kN/m-6) So Shear Force at section X-a, Sx = 7.2-0.616-7)*1*(6-1) Sx = 7.2 - 0.3(6-772 So Shear Force at c, 2-e = 1.5m Ve = Sc = 7.2-0316-1,5)2 - 1.125 KN Vc = 1.125 Kw

For equation (1) Normal Force at -e x= 1.5 m Wa=Nc = 0.6 (6-1-5) = 2.7 kN/m Ne=2.7kN/m] Any

Moment at c. Me = 3.6x4.5-624.5x2.7%«4:5) = 16:2-9.1125 Me = 7.0875 kNm] Summary For triangular loading J DT TIL B Total load is acting A at a distance of ta from a deft and 2n distance from Right i-e b'