Question

Determine all of the member forces in the truss below( take zero force members into consideration):

Determine all of the member forces in the truss below: Question 1 11 points 5 kN 5 kN 8 m 5 kN A 4 m 4 m 4 m Use the strict tension positive sign convention for internal forces, accuracy 0.5kN Units Number Units FBC Number FAB = Number Units Units Number FCD FAG= Number Units Number Units FAF = FBF FCE Number Number Units Units FRE= FEF Number Units Units Number FDE= Units FFG Number

Answer 1

5 KN 5 KN 8 m E 5 KN H, D rm A C. * * 4m 4 m D 4m Suppor Reachions: the condition of equilibrium, From Sum of all vertical forces O VA 5-5-15 + L VA VD 15 we momem alo o w A Taking - V 2. + 5 x12 + 5* 8 +5 *4 0. > VD lo KN, quation From CD . . V, 15-Lo 5 KN. 7 A /,D

Ana, hor izomtal forces O. .. HA O Angde and Ham( 33.69 2 8 BF 5 333 m. 12 5.333 BF tan tan -- = 53.1284 CC AB prees: Membex A pint all he nchng passing hrough this joint the re is mo A s me mberS will have Zero AG G F A : toint AT AF E V O. sin a = O. V. A A F AB sin 53.1284 5 z O. AF

1 AF sin 53.1284 Fr AF oree is positive, direction As the obtai ned offorce FAF free ody diagram in the above shoun by arrow COTTE ch. the C Note: Arrow towards roinh Compre ssive HAe toint: Tensile Arrow away from CCompressive), F AF 6.25 KN EHO. Cos o t F AR =O AF 6.25 Cos 53.1284 +t F O FAB Tensile) 3.75 KN. Joint F : At EH= O 5 KN x Cosoz O FEF FAF -F x CoSo 25x CoS 53. 1284-F EF F EF * Cos 33.69 20. AF BF

4.50 kN Ccompressive) F EP ΣV=0. O. xSm Sin a-5 + F + F E F BF AF Sm 33.69 0. →6:25 x Srm 53.1284-5 + FRF + 4.50千 KN. 2.5 BF (Ten sile) 2. 5 KN BF Aト D Jint DE E V O 5 KN x sm θ 20. DE FCD x Sm 33.69-5-O F lo DE CCompressive) KN 9. 014 EHEO FeD D. DE - FoD x CoS 33. 69 9. 014 CTensile). 7.5 kv CD

AT FCE Jont c FED Fec EHO しし BC 7.5 KN Ten sile) Fer BC FCE O AH to mt E FEF 5 KN. E H O FR CoSO O 2 COS9 + F x CoS C EF キ「BE DE F BE DE 4.507- 9. O14 BE F 4.507 KN. (Compressive) Fa BE