Question

Problem 2 (40 points). For the given beam with constant El answer the questions. You MUST show your FBD & BMD for the real/virtual systems in your solution: (a) (20 points) Determine the slope at B. 08, and its direction using Unit-Load Method. (b) [20 points) Determine the displacement at C, Ac, and its direction using Castigliano's Theorem. 8 KN A B С 6 m 3 m

Answer 1

a) SKN A B 6m 3m Ay 'By By equilibrium Conditions, EMA=0 >> Byx6 = 849 By = = 12 KN (4) {fy=0 Ay+By = 8 - Ay= Ay = 4 KN () The given structure is determinate hence we take this Ay = -4KN primary Structure We Love, FBD: X2 8 KN 1x1 B A 16m 36 C IXA 12 KN "AKN lxa BMD:7 # For A Bit Mx, = -4X, 30,5 41,56m #For CB.,: Male =-80₂ ; 0,962 s 3m 6m B 3m C * sign Convention to sagging BM=> tve ها Lo Agging BM=>-ve -24 KNim Ics

1 A 3m Now, remove the externally applied load & apply unit moment at 'B'. The secondar structure is. X 1X2 Ą tom -X, 16 146 # For AB, Mn, == = -;OSA, 56om. # For CB , o ; ordan 3 on. 'X1 16 l 6 'Ma Mada EI * from Unit lood method, 0g = SM [ Séan)(-44) de + Og = EI (-802) (0) dla 10o 48 EI (2) Hence, Slope at 'B' is 48 in clockwise direction, EI 6 let, P = 8kN X2 P AXI 45 k da A Tom 3m -X, 0.5P A xi # For AB, Mx, ==o.spri ; osa,56m # For CB, Mr = - Pha iOS7,63m 1X2 11.5P B -3P knom ICS Scanned with CamScanner

ау ap ; By Castigliano's 2nd theorem, Here, 'U' is the strain energy of A = 20 the structure due to bending only ie U= M² dx Д. du Mehmen ap U = sme da da EI 6 6 A = 1 1 1 0.25p (x da, + P A = ti[ [tospu) cosas) dhe JCP) yes prieten De to [02538:42 + 349] 216 Aca EI (+) Hence, the displacement at ć' is 246 in downwand direction. CS Scanned with CamScannes