Question

FORCE METHOOL CURLING STATICALLY INDETERMINATE STRUCTURES EXAMPLE ISS 313 Compute the reactions and draw the moment diagram for the structure shown in Figure 155 Figure 155 Solution. Remove as the most. This will change Alle type of JOL Compute the horizontal deflection at Aby virtual work. The result is 56.600 EI Apply a unit horizontal load at A and determine the horizontal deflection

Answer 1

30K 60k B DSI - 2 KAB o E 104 lost 104 Let us take HB → 20 redundant. 204 VO 20K GOK 57 B c XHA "Tin E Primery Dr. 1 RB MA 16_21TO ILLO EN EMA = 0 30X10 +60X20 TRA R3X 36 = RB - 50 K SV 20 Ratha 30460 » RA = 40K El=0 HA M moment we do parimary structure. ty - A 30% 60k 2 FBD of primary shr C B You E 1 ila 2 Isox c yok A fyok Secono stor # condary EMA=0 с RX30 - 1X20 21 1 Rg = 0.6674 By = 0.667 K m> moment in secondary structure . Elzo HAZ lk НА- к K A m RAT 0.667K

FBD *20K-tt 4 stendeuy st. - 1k & 6 3.1k 0.667k 1 0.667K 0.667k lk с 1k & A (0.667K Membes Az Of Limit </mpele M Memes B EL EI AC A 0-20 26.66.667 0 2c 0 EL co 0-10 40 x ل ФЕ с 0 -10 401107x) -30% 20-0.6674 31106.667 2814.296 EI EL 20—0.66711070 44421-667 1036•074 EI EL 0.667u 111 16.667 148.148 EI El En B 0-10 50x Д. - co 86645.001 EL 6665.185 FI Using comportatility egon -ve means dieo of Ho ABT HBS=0 1967-3K 7 sign Ho u offerite to that 1k haszontal In secondary em. hosszontal force ce → HB = 13k 30K 60 % d FBUD - B 30K C13K 2h, 260K-Ft 60K → E 13K- *3K 1. Vg= 41335 48.667k 48.667K 48 667 41:333K 260K-ft 13 K HA A 'VA EH 20 > HA = 13k 13K → EMA=0 48.667k Kg x3 X30 + 13x20 - 30X10760X20 KB = 41.33K 30*60 = VA = 48.667k Erzo VAHUB 413.4K-St ВМР 22607 Kult memert - Heging 2601 H o E B moment 260k-ft A