Question

Chapter 4, Reserve Problem 4/019 Determine the force in each member of the loaded truss. Forces are positive if in tension, negative if in compression. am 1.7 m 1.5 m 1.7 m 1.3 m 560 kg 2.6 m Answers: Click if you would like to Show Work for this question: Open Show Work

Answer 1

1.97 7 11.7m My 1.7m B 107m .sm GF lsm El-3 Tang= 1.97 [B= 49-16° B 1.97m/ A ANA TAMTH Length of BG is hat given, assume Lap=room. Length of AB = √(26) 2 (1.772 = 1.97m 1.97m. Length of C = √ (2.6) 2 - (1.7) 2 = ) 1997 m. 1097m. s s & 0 ton bell moment about 4 will be o. Let Rxn at A be VAkta & at a be ved Hea. from Toom equilibrium equations equilibrio an Ema=0 an sign convent VAT Va = 5600 - VA*1.7+HA* 1.97 + 860*(1.711.7*1-97) & fx=0 1.7VA +1197 HA =3807. 2 0 Hat HA = 0 Hu=-HA 6 moment about A will be zero Tana = 1:7 1.97 Vox1.7 that 1.97 +560*(1.7*1.97) =0 1.7 Va t 1.97 ha & 2055. 2 = 0 EMA=0 ron L = 40.80

From FB.D a point D FED Toe cast & fyzo - For sina - 560=0 FDE sin (40.8°) = -560 FDE = - 857-16 kg Esina FDE asboky compressione Efnd - FED - FoE Cosa = 0 Fed=-FoE casos) fco =-(-857-16) C08 (408) FED= 648.866 kg Tensile. member Joins " at a are points callinear then It 3 If two members in force in thril member will be FEF= FOE = – 857.164 TFF = -857.1 kry compressine. en/ & # e from f.B.D of point c & Fx=0 - Fes sinus & Fo=0 Fossinus! - FeB = to sinus = 648.866 sinus IFR Cosus FCF FOR FOR = 917.636 by Tensiee.

Efyzo - Feb casus - FF =0 FCF =- FCB casus =-917.636 Confuso) For = - 648.866 kg 1 (compressive)

* 4497. gada ohit # From F.B:D of point F { Fx=0 - FFBT FFA CUSB & FFE Causa to 4 FF Fresina FFE Fresca Frecas2 FFB flaging FFB + 0.654 FFA = - 857. 16 (408) FFA I FFB + 0.654 FFA = - 648.865 PARA o {fyzo For A Fresina - FFA Sinß=0 6 - 648-866 – 8 57.16 sin (40-8) - Fasih (49.16) = 0 FFA = ~1598.0 kg compressive. put the value of FFA in eah FFB +0.654«(-1598-0) = - 648.865 FFB - 1045:0g = – 648.865 [FFB=396.23 kg / Etensile) Forsinus # prom FB:D of point Bi Efy=0 For Sinus - FBA TO FOA = 917-636 sin uso Foa= 648 867 kg / Etensile) for b For Foc casuso

Efx=0 - Foat For & FBc Casus = 0 FB4 = 396.23 + 917 636 Cosuso [Fou= 1045-1 kg (Tensiee) Hence Take g = 10 m/82 FAB= 648.867 kg = 6.49KN (Tensiec) FAF = - 1598.0 kg = - 15.98 KN ( compressive) FBc 2 917-636kg = 9.18 KN (Tensile) FBF = 396.23 kg = 3.96 KN (Tensiee) FBG= 1045.1 kg = 10.451 KN (Tensive) FeD = 648.866 kg = 6.49 KN (Tensiee) FCE = 0 FCF = - 648.866 kg = - 6:49AN (Compressive) FOE = - 857.16 kg = - 8.57 KN (Compressine) FEF = - 857. 16ky = - 8.57 kN (compressive)