Question

draw shear force diagram and bending moment diagram

10 kN 5 kN 5 kN 2 kN/m 4 m 3 m--- 2m- 2 m - 3 m--

Answer 1

OKN SKN tantų Sino=420p toso = 3 = 6 WIB E um > -X100 pe 3m't am ram T 3m alva. Va Posted load = 5+2=lok. Hoorzontal load a losmo - Skw s a vertical load = loloso=6kw. o ♡ Now apply Efaco > Hd = 8kN Now apply EM a=0 y Vaxlo = 5x7-10x5-5x3. -losimo)x 4 locosoko 2) Va=12.5kn. Now apply EFG 20 27 Vds 5+10+5+6–121521315kw o lokn = (Total load).

PAGE NO. Date EXPT. No........ & shear force orce diagram! ve vetu (a) Shear force in the member CD. CM Figure a a 790-8. . oli I (SF) =-H, sino - Valoso txa - 8x0:8- 13.5x0.6 +2n. 2-14.5+2x. f linearty varyings.. Teacher's Signature

775 (6 Now a limits are oto 5 (SF),20 = - 14,5kw +(SP) = -4.5kN. Shear force in the member BC Now total upward Porce due to store in CD is -locoso = 13.5-6- 7.5KN 5 km 15km Iloko THÉP SF&r= -7.5+53:-2.5KN.SE (SF) = -25+ los 197.5 kn SPR = 7,5+5=19.5kw (c) Shear force in the member AB.. total downward force atß due to force in Bois do 25kw/215kw 19-03 (SF) = +19.5coso 12.5x0.6 7.5 kW (SF)A = 7,5-715 20 so. Va=125 kn

Ave) 17.560 SE I love) 1.5kn 14.5KN. Shear Forre diagram (ve) 78KN / 5k The - Sogging. gging. Il Bending moment diagram't a Bending moment in member CD From the above Figuren (MA)= (HiSmo tvd Coso) xx .- 8Xxx2/4 = Bx0.8+ 13.5 X016)7 -- 33. = 1454 - (Parabolic variation) (M), -0 (0-0) fm)145x5-25 = 47.5kw.m, (2) Bending moment in member be Sku Joko SKN I am te am 47.5 kv.m 115kN ( Sheartore

A IME = 47.5+ (715-5x2=52.5knim (M = 47,5+.2.5x.44 - lox 2= 37.5 knim (C) Bending moment in member AB do 1125KN Ataska. (linearly varyings B 27.5kn (Muz 375-12:5650)yn. MB = 37.5 kwiem im. Mizo." 37.5% * At the) 52-5kw.m 147156cm Askem parabolic tha lineatly varying . I Teacher's Signature .........