Question

we need to draw the shear and draw the moment diagram for the following beam

30 kN.m 30 kN.m 27 kN.m B Α. С 1.5 m + 1.5 m 1.5 m

Answer 1

Solution - 30 kNm 30 kN.m 27 kNm bo -1.5m- iy 27 kNm с 0.75m TRE RA 45kn k 3m 3.75m 45KN (30x1.5) • Find support reaction - EMB=0 take clockwise the 2 +- 4 5 (315)- 45 (015) + 3 RA = 0 RA= 58.5 kN E.Fy=o, RA+RB - 90 = 0 RA 31.5 KN

Shear force Diagram (SED) 30kNm 30 kNm 27 kNm 5 ТВ ism .1.5 RA=58.5 31.5 2 for section and - 31.5 + 30 (6) 18 at x=0 こ V= - 3105 KN at x=105 V² – 31.5 + 30 (1.5) 45.0 31.5 13.5 kW for section 2-2 - 31-5+45 = 13.5kN for 3-3- x= 31-5 +45 of 30 (2-3)*58.5 13.5+ 30(x-3) - 58.5 at &-53, Y = 18.5 kA at x= V = .45 KN at x=4.5, V = 136 5 + 30 (1.5) -5805 at x=3, Vs - OKN V (KN) 13.5 B A x - 31.5 -45

Bending moment Diagram - for section 1-1- My = - 31.5(x) + 30 (2). 7/2 da at x = 0, M = 0 2 = -31.5 x + 15x2, d Mx - -31-5 + 30 x = 0, x= 1.05 max. value of BM) at x = 1.5 M = - 31.5 (1.5) + 15 (105) at x=1.05 M = ~31.5 (1.05) +15 (1.0592 -- 13.5 kNm = - 16.0125 kNm for section 2-2 Mx = - 31.5x + 45 (x-0.75) at x = 1.5 M = – 31.5 (1.5) +45 (0.75) I – 13.5kNm at x=3 M = – 31.5 (3) +45 (3-0.75) = 6.75 kNm for section 3-3- at x =3 Mx = - 31.5 x +45 (2-0.75) -58.5 (x - 3) + 30 (2-3) (x-3)/2 M= – 31.5 (3) +45(3-0.75 -0 = 6.75 kNm M= -31-5(4.5) +45(4.5-0.75 - 58.5 (4.5-3) +15 (4.5-32 - 27 kNm = Maximum value at x=4.5, of Bendung Bending (kNm) moment 6.75 C o A B. -13.5 - 16.0125 x=3.45 -27