Question

The truss shown below is subjected to three Q kN vertical forces and one F kN force. Calculate: 1.5 m 900 45% Q IN

Q = 170 kN F = 80kN

The x component of the reaction at E,Ex (in kN)

The y component of the reaction at E, Ey (in kN)

The y component of the reaction at A, Ay (in kN)

The force in member ED (in kN)

The force in member IB (in kN)

The force in member IH (in kN)

Question 2

W = 115 lb P = 80 lb

The x component of the reaction at H,Hx (in lbs)

The y component of the reaction at H, Hy (in lbs)

The y component of the reaction at A, Ay (in lbs)

The force in member AB (in lbs)

The force in member AM (in lbs)

The force in member KO (in lbs)

Answer 1

Date Page to kn a 5m BOKN Ito KN A Im * 2m f 2m bt im Ev=o Ay tEy = 566.56 KN مجمع Ey X 6 - 170x1 - 170x3 -170 X5 20 a255 KN 311.54 KN 24 zo - Ex + 80 cos(45°)=0 La Ex= 56.56 KN Jointe Sa .56 KM _45. Ev=0 255-FED Sin(45) 20 255KW la FED = 360.62 KN stanned with CamScanner

Date Page Joint A Sokn 950 $45 Evzo - for sin 659) +311.54- so sin 6450) = 0 . 311.54 KN |:: FAB < 520.58 KN T©). El=0 80 cos (450) -- 520.58 cos(450) - FAEZO La fase 31.53 KN (I) | Joint I Test Ev=o fif zo THI 11.53 SH=0 StorFHp2311.53 KN D) $canned with CamScanner

Date Page 115 15 115 lb 15 16 15lb Á Ó 15lb Ev=O Ayt Hy = 115 x 5 = 575 lb EMA=0 HyX 9 - 115 XG-115 X 6 -115 x 4.5 - 115x3 - 80x3 + 80x3 = 0 os Hy = 287.5 lb e Ay = 287.5 16 - - ZH2O -cth +80-8030 Hxe7 $canned with CamScanner

Date Page Joint A 45 lb FAB Qatant/3_ (1.5 o a 63.43 FAM Ev=0 12.87.516 _287.50 – 115 - FAB sin 63.43) 20 FAB = 192.86.26 69 EH=0 LEAM - 192.86 C05(63.43) co. for FAM = 86.26 lb (1) consider the FBD of the downward part i of section - -EVEO - 115 - Fko =0 fro = 115 lb (1) LK 115lb $canned with CamScanner