Question

CE 3115 - Structural Analysis EXAMPLE #1 - Internal Forces at a Point Find F, V, and M at point B. zok с D. с. — А 1+ 3kAft у т 5' s т 5' 5' 5' t Stel 2 mare cu ter ☺ Get RxN.S Ax 544 A569 Fax. 51 57

Answer 1

20K B OB+5K 3yet В 5k 5- 5 571 5 e 5 - Compact 5 4 5 Prow Free tedy dogteam 20K Waal C 5K RATK 5K 5k 57 5 Here Wedi = 3W/HXO = 30k beam By equating the forces on the I RAx = 5k Pre Rayt RC = bludl. +20 Rayt RC = 30+20 Raya Re 250k) Taking moments about point A. ... Rexto = {20x15) +C 30x 5) Rexion 450 Shot on realme 2 RC =45k

Rayt Rc = 50k -- Ray = 50 -45 Ray = 5k] Bkler AR 53k s 5k k sok 45K X A a) Take a section x-x' at a distance ex foom Righteed Bending Moments kx Shear force ! (SF axx = 0 SE@ x=0 so (BM)@xx 20 BM@X-0 =0 SF@2=5 = 20 BM@2= =0 figuin consider another section **2 @ a distance Shear force Bending Moment? (SF)@ x,= 20K (BM@xa =-(20(425) BM@ 72=5 = 0 (SF)@x2=5=20$ BM@X=10 = -20X5 (SP) @ 22:10 = 20$ =-100K-PC Riant

Consider another Section X-X3 from a distance na from Right end zkler & 20K Shear force. (Sfax = 20+ ( 38(09-10)) – 45 ($#@z=10 = 20+ (344*6=10)) -45 = 20-45 --25% (Spa x= 15 = 20+(3x (15–10)) -45 = 20 + 15-45 = -10k (@x=20. = 20 Right - 20+ (39(20 –108) - 45 2 20+ (3x1) mus - 5 (SP@z=20161 = 20+ (3x (20-10)) -45-5 = 20 +30-45-5 .-0

Bending moment : (omare :f20x (5+ (13-0) 46153 (1)-(609 (oder: 10 = -2024634 cm) 650yfras) becifras - - 2045 -100 K-PL (BM)& 2 = 15 =(-20%C5+(15-0) + 641515-10) (3 (15-1)(1510 (-20X10) +(45 * 5) – 63x545 ) - -200 + 225 - 37.5 = - 1215 kaft (BM)2 x = 90 = (-20%(5+ (20-0))+(45 (20-10)+(2120-1) 23:17 = (-1016) + (45xio) – (3x10x1)

-- SF!- Bending Moments (SP)ED 50 (SP) E = 206 - 20K CBM) @ D 0 (BM)@ E =0 (BM) QC - -100k-ft (St) ( Right = 20$ (Sp@c left = -25K - 21 (BM) @B = -12.5K-ft (BM) QA = 0 (SP) @ B = -10K SF @ A Right = 5.k SF @ Aleft = 0

By lova Shot on realme 2 klet - 56 w 20K the 4 5 To Jant to take k 5 K 45K 5 K 5 SFD -106 ATTITUTIIT72 BMD -12.5k-ft -100K-Ft At point B iF=0 V=10K M = -12.5 kaft