Question

In the truss shown in the figure: 1. Determine the support reactions at supports A and F. (10 points) 2. Determine the force in members BC, IC, and IH of the truss using method of joints. Explain and justify if the members are in tension or compression. (20 points) 3. Draw FBD for each step of 1 and 2. (10 points) 60KN 40 kN -2 m -2 m -2 m E D 1.5 m F с O -30 kN H 1.5 m B 40 kN 1.5 m A

Answer 1

YOKN FBP 60 KN А м 2m 2 м k E 30KN 67 H 1.5m 40 KN I Bl 1.5m A Ax tan o = 2 Fig-1 'Ay 1.5 EMA = 0 → Ne O = 53.13 40x1.5 + 30 x3 +40.X2 – F *4=O F = 57.5 KN IN EA An- 30-40 = o ² Ax = 70KN Efy =o Ay + 57.5 – 40-60 Ay = 42.5 KN FAI (2) Joint a FAB 153.13 E E Fn = Zokn 70+ FAB sin 53.13 & 87.5 KN FAB E Ty = 0 42.5KN Fig-2 ca) 42.5 + Foz + 6 275 Cap 53.13) CS scanned With Ca Ascanner Fas = -lokn

Joint .B Foc Efy = o Чо EN K 2 of Foc = - 87.5 sin 36.87 36.379 For Foc = -52, S KN -87.5KN Foc N 52.5 KN Fig-2(b) Compressive (c) E Fn = 0 ) Fax +40-87.5 Cop 36.87° = 0 For = 30 KN En Joint- = FEC 1.5 tanß= B 2 30 KN IN ß = 36.87° Fig-2 (c) EF=0 -OEN 30+ El Cos 36.87 = 0 Ffc =-37.5 KN Ffc = 37.5 KN Compressive (e) W Efy F - 37.5 bin 36.87 =-10 ĘEM = 12.5 KN Tensile (I) IH 3 FBD of step-1 is shown in spre Fig-1 and FBD of Step-2 is shown in Fij-2(a), Fij-2 (b), Fig-2 (c) CSScanned with CamScanner