Question

3 kips/ft. 3 kN/m 7 kN 80 lb./ft. 500 lb. B A В LA 200 lb. 300 lb. 6 ft. 10 ft. 6 ft. 172 m 12 m 3 m 5 ft. 5 ft. 5 ft. 5 ft.

Calculate the reaction force and moments then draw a complete shear and moment diagram

Answer 1

3 kipsite BA Af Balock uilibrium. I Halizontal force so Ranzo I leatical force ao RA TRE - (1/2 x 346) – 13x10) - ( + 2 X 3 X 6) =0 RA+RB2 48 kips. Moment a RB X 10 + (1 x 3X6 ( 121 126) (3x10)(5) - (2 x 3X 6) (lot so Roxlo +18 - 150 - 108=0 RB = 24 kips RA= 24 kips

Shead lonce at a to Srl (Just left) = -( 1346) = 9 lips (SP) A = – 9+ 24 = 15 kins (SF) & Just left) = 15- (3810) = -15 kips (S.f) = – 15 + 24 = a kips (SF) b = 9 (2 x 3 x6) = 9-9=0 Bending moment (.m) a = -6 x386) (Š x4) =- (9) (2) = -18 kips oft (B.M) B= - (4 x 346] ( 7 X6 + 10) teadlo - (3410) (5) - - (a) (12) 4240 - 150 = -18 kips & (BM) = (2x 3 X6) (16+16) & RAXIL - (3410) (1) & RB X 6 (X 5 X 6 ) ( 2 / 3XL) - (9) C18) + 384-3304144 -36 (BM) at mid point = (-(1X3x4) (513 XL) +24x5 - (3X5)(2-4) - 634 120 37.5 = 19.5 kipe.ft

Shear force Iskips Diaghram. akin o e curve 9. Bending Moment Draghram 19.5 kinet A clapety '18 kips. ff ! 6ft loft Gft

B arn. Bknim N. s 3kN/m dummy P 12 t /m. RAVY dummy "GCN/m 3m by 11.5m 1. Af equilibrium we = 2 =) Lox= x I Hanigontal toaceio Ran=0 Ileatical force o RAF I (389) + 7 = 46.5+7= 11.5kN RA= 11.5kN Consider a crore rection & ex at a distance & Cin between Ac) forum point B o alto aulaat de Moment at x = – 2 (xxx). ( 13 ) = 7(x- p<) + 3(x+3)(-3) + ( + x3*{->)/) 2. CM -46-48) +2013744(3

B.Mgc fx=3) = (3x3)(1)-9 - -4.5 Cam. =-4.5X10 Nm (40+3 (40--3774046-5) x= 4.0 = - 15.1875 + 3.375 at 1.125 = -10.6875 (BM) =(-+(1«) +30+100€ = -36 - port13.rty.r =- 28.5 kNam.

Dlaghram Shear foace 11.5kal 4.KIN Ара — Grkam I .685 cam! 288 Kalm 1.5m 1Tm 3m

given. қорье 500 lb - RBH rç 10 E 30006 At equilibrium Horizontal forces 0 PB +500 +200 - 800 - 100 to RB= 400 lb soые soublet 4000 E V ANWA con parant 200lb, 1300 14 sober 11 K x -X Spf Of Th St : a from point A B.M at (x-x) at & in between distance a B.

(B.M he = 2004 - (SO XD) ($T +300468-10) -500(x+16) + (80 X (X-10) (x+0) @Me - 2004 - 80 (x-5) +300 X (-20) – 500 (x-15) + 80 x (x-103" 21 (B.MLA (x=0) =0 (BM) (x=5) = 200x5= 1000 lb. He (Bima) o (x=16) = 200x10 - 80 (35 = 2000 - 1000 = 1000 lb.ft (BM) (x = 15) = 200x15 - s0l102 +30045 +80(5) 2- - 3000 - 400D + 1500 +1000 = 1500 lb. ft (BM) B = 200 (20) - 8o (15) +300 (10) (4=20 4000 - 9000 + 3000 - 2000 2 + 4000 = - 500 lb.ff

Shear from Diaghrom. 200lb 100lb - 20016 400 lb Bending moment Diaghrom nebitt 1500 lbff 1000lt 100bt - 2 curve toolbl toolb.t soolb.ft 5H y soo . Sff St Сf St