Question

3 kN/m 60° 4 m 30° 8 m

1) Find the normal force, shear force and bending moment at point C, along section c-c, using lower segment.

2) Redo part (1) using upper segment.

Answer 1

(1) Assumption: Neglecting weight of beam. (9) Prawing the FBD of the beam: y y LP Bx ö Rely 2XL wo 3knim - t Crom. A 24m. 24/2=8m. . 12rm. Ах L= The reference axes are chosen as shown: x: along beam length g: perpendicular to beam lengts. (b) Now; Moment equilibrium about B would givere EMB=0 ce rive ) Ayx ! - RX220 | Ay=% XR= 3x ? xwxL = § 23X12 = 12 KN By force equilibrium: Eryco (gestive ) Ay tBy_R=0 12 +By -4X3X12 =0 to 7By = 6kN]

Also Axe Bx 20 (No force in se- direction) Assuming negligible weight of beam for all above casesi a SO Resultant reaction at A&B: RA= Jay?t Ag? = 12 kW Calong y). RB = V By² + Bap2 = 6kN (alongy)

force (v) Bending me (C) Normal force (F), Shea atc cibout C-C . 7 Let us consider first acul section at Ċ whichis a-a 10 to AB. o we first calculate the F, vem about this section f then resolve it to get F,V&M about e-c section. with cut section ara, consider equilibrium of lower section containing Rfcis. wa (W+W.DAL! R VC- PTT A Lot (FC lama to find wei By symmetry of triangle in original loadi w = we ad B = We 12-3 WC = |kw|m Resultant of trapezoidal loading in AC : R'S Ý (wtwe) XL'= (3+1) x8 = 16 leopold Location of R' from A = location of CG of trapezoid. r erg of trapezoid i Wetu Loya Ř=(2 we tw les e 2-- (2x) *3 ) x Tez 10 m]

Nowi consider force equilibrium: {1370 Casey 14+)). Ay+ (velace-R!=0. | (Ve Jaa = 16-12- 4KNI A 150 EFxz0 C Ax-Flo so X +)) (Fc bq.c=0 i A 180 EMA = 0 (tive')) - R'x = 4 và x + Mg = 0 -16 xlo + 4x8+ M670 me = 21.33kw.m () Nowi Resolving (Felaa, (uclaa , along of perpendicum to c-c section to geb (Fc lcc, (Melco F Mc will . be same for both cases. (Fc laq cos (30) cay Faq Six (30) ve Jace sin(60) Velga (Vc)ag pose6s)

Resolving (Fclara f... soi Normal force along C-6:= LC Fchec] . (Velaa for to cc. Prive) (Fedec = CV Jag sin (60) - (FC) as sin (30) (Ecleca 4 sin (60) - T(Elec: 3.464 KN (9), & Shear force along C-C!= ( (Vc) cc)) Resolving (Fclauf (Velga parallel to G. C.. (c tive) (velec: (Vc)aa cos (63) + (Fc la acos(30) (Velcc = 4xş to | (Velee =2 kN (+) Sul TResultant Internal loading at C, about c-c: Normal force-l Felcc = 3464KW (1) shear force = (Nclcc = 2KN (9) moment: Mc = 21:33 Knım (9)

Tera) Reaction sat 'e' using upper segment. Calong cae). let the narro consider the cut section Loc, force, Shear force & Bending moment at this section be c (Falece, (Ne leve & Mc). BF is (Veloma C uram. ( "C) cc k whe welkN/m casin Co.)). XIX4:2KN., RSWexllo axes y' fol: consider face equilibrium about E fy'=6 (+jvc) (FC) cc -&'Sin (60') + @y sin coo')- efesin (20) (Felcc = ex sin (60) - 6x sin (60) Er lee : 3.464 KN or 3.464 KN (*) 150 EFt - 0 + e'tive ) , (veest Ricos (60) - By cos Go Balcos(301= 0 (Veccs Gros (60) - 2 Cus (60) (VICC=2KN (+)1 By moment equilibrium about c: Emeioco tive) MC-R'x㢠By X L'-D Me=- 6x4 +2X4 Me:-21.33 KN- TOY 2133 kN. MU) f
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