Question

-6 at 30 ft = 180 ft- 25 k 25k 25L I 20 ft 15 ft 5 ft 45k (g) members CD, DI & DJ (method of sections)

Answer 1

Free-body diagram: 25k 25k 25K - H. - -.-. - - 45k 6x3oft = 180 ft - - Z - direction ex-direction, E Fe=0 i. Ax=0

Taking Moment of all the forces about point A EMASO → 25 X30 +25x60 +25x90 + 45X120 +45 X150 - Gyx 180 = 0 → Gy = 92.5%. {fy=0 Ay+ Gy - 25-25-25-45-45=0 Ay + 92.5 - 75-90=0 Ay = 72.5k. Now, using method of sections : on passing a section through CD to IJ 254 255 25k Ax A B C D E SO Au=725K Ay = 720 it ---- 74 Gy=92.5k 45% 45k

255 25k Ax=0 A soft B soft tc FEDT coft 20. *FIJ Talking Moment of all the forces about point I Ay x 60 – 25x30 – Fest (20 +15) = 0 72.5X60 – 750 – 35 FCD = 0 1. Feb = 102.857 K C Tensile). Stand = 30 35 cm {Fx=0 FeD + Firsino + FI COS< =0 35 102.857 + Fs, Sin 40.601°+ FF cos g.462°30 0.651 Frp +0.986 Fig 2-102.857- ΣF=0. + Fcose + Fisina – 25-25+72.5 70 330ft- FID cos 40.601 - Frysing.462 = -22.5 eura 0.759 FID - 0.164 FIJ = -22.5 — N 22.5 - 2 tanx=5 (2=9.462)

JOB 154 FIT FI I on solving equation and ® , we get FID = -45.669 k ccompressive) Fou = -74.164 k ccompressive) Now, for equilibrium of Joint I A FOJ 130ft tanp = 5/0 1 B = 9.46, E Fx=0 Tx = g. 462 - FI Cosx + FJ K Cosp = 0 - (-74.164 cos 9.462) + FJk Cos 9.462 = 0 - Fyk = - 74.164 k ccompressive) E Fy=0 FJK FDJ + Fij sinx & Fysing zo FDJ – 74.164 sin g. 462 – 74.164 sin 9.462 20 Fog = 24.384 k Tensile. Nature of force Member Force CK) Tensile 102.857 CD DI 45.669 Compressive 24.384 DI Tensile.