![classmate Date Page 15kn 15 KN Jow B E Question: Ef= 13m CL A С D HA AVA 4m 4m P VE Solna- - step 1 :- finding reactions HARO](//img.homeworklib.com/questions/2677eaa0-91c9-11eb-8145-0194cfb837d6.png?x-oss-process=image/resize,w_560)
![classmate Date Page fee 4 Fap 715 - Joint F smo = 315 Los o= 415 Efy=0, FFE X3 5 FFE = - 25 KN FFD + FFE X Y 5 Frot KN = 15 (](//img.homeworklib.com/questions/27dd52a0-91c9-11eb-b4e8-774e1d28c476.png?x-oss-process=image/resize,w_560)
![Date Page Slep Applying finding unct load at Joint member forces. and + Vr P VA sro VA X 12 = 1x8 VA = 2/3 VF = 1/3 a & HA=0](//img.homeworklib.com/questions/297d4c10-91c9-11eb-9832-858c586cbe8e.png?x-oss-process=image/resize,w_560)
![classmate Date Page Fre Joint F Simo= 315 1 Loso=415 FFE X 3 FFP 11 -1 3. FEE = S 9 (tomp ression ). 122 FFE X Y FED O FED =](//img.homeworklib.com/questions/2b40ca30-91c9-11eb-911b-b5dd78a01231.png?x-oss-process=image/resize,w_560)
![classmate Date Page slep IV deflection at C Member N AE n -25 AB AL BE BD length L 5 4 4 - 1019 819 -419 -519 NNL I AE 138.88](//img.homeworklib.com/questions/2cea6040-91c9-11eb-8f05-b15815a2a602.png?x-oss-process=image/resize,w_560)
classmate Date Page 15kn 15 KN Jow B E Question: Ef= 13m CL A С D HA AVA 4m 4m P VE Solna- - step 1 :- finding reactions HARO VA = Vr = 15 KN Free forces in members Efa= (due to Symmelig finding forces in E fy slep I :- FAB Joint A fo * PAL sino = 3/5 los o= 415 1 15 x 3 Efyzo FAB X 3 -15 5 FAB = - 25W ( compression) Efnco, FAB X Y + FAC 20 5 FAC - 25 xy 20kN (Tension 5
classmate Date Page fee 4 Fap 715 - Joint F smo = 315 Los o= 415 Efy=0, FFE X3 5 FFE = - 25 KN FFD + FFE X Y 5 Frot KN = 15 (compression ) 20 (Tension ) FCB and foo Fco = Joint C Efx=0, 20 KN (Tension) E fy=0, FEB yount FAL -o FOB AFDE + Đ 20 smo = 315 4 15 RO E Fx=0, it 20 20 FDB X Y 5 FOB =D - FEB FDE oor Joint E 25 coso = 315 sim 0= 415 2 25 4 =o + FEB FED = 5 - 20 kn ( compression)
Date Page Slep Applying finding unct load at Joint member forces. and + Vr P VA sro VA X 12 = 1x8 VA = 2/3 VF = 1/3 a & HA=0 Joint A A FAR smo = 315 Loso= 45 #FAC 42/2 en Efy co 2 -2 FAB X 3 5 3 FAB 10 9 KN Compression ) Efaco FAB X Y t Fai D FAL 10 x y Kal (Tension g 5 g
classmate Date Page Fre Joint F Simo= 315 1 Loso=415 FFE X 3 FFP 11 -1 3. FEE = S 9 (tomp ression ). 122 FFE X Y FED O FED = stu » FLD 819 os 1 KN FDE y kal ( Tension) g Joint A FB Efazo, Fo 8 KN ( Tension) Efy=0, FCB F. (Tension) FDB A Joint D Simo= 315, Loso = 4/5 FLD = 8 + 8 -4. g GA FDB' -5 KN (compression) g duboni FDB x 3 + FOE A - O FDB X Y - O کل FDE EN 5 x 3 9 = 3 KM (Tension 3 FEB Joint E oso= 315, smo=4154 on sig. NE 3 Ex 4 5 FEB FEB= 4 KM (comp
classmate Date Page slep IV deflection at C Member N AE n -25 AB AL BE BD length L 5 4 4 - 1019 819 -419 -519 NNL I AE 138.8819 71.11/AE 35:55/AE 20. - 20 ВС o 2o -25 FD FE DE AE AR AE AE AE AE AE AE AE to 3 4 5 3 4.19 - 519 319 35:55 AE 69.44/AR 3 0 20 4 71,11/DE E=421.64 AE Aca EnNL AER こ 421.64 AE cross-sellional area mo dulus A = E = * ASI members modu ws Albeming cross-sectional area of all are equal equal and young? also equal for all members