Question

For the beam shown, use the Conjugate Beam Method to determine a) The deflection at F, b) The slopes on both sides of the hinge at C. 90 KN 160 kN.m E B с D F 2 m 2 m 2 m E 2 m EI = Constant

Answer 1

- 1B YOKN G E 2M Atz 2 m 2m 21 2M 160km Calculate support reations 160knm DIMA 190KN 2 Gx C 2 E 2 F 2 B cacy 1 Cx AXTA cy Ey. Ay for Right segment EFx = 6 - Ce=0 40KN E Fy L. Cy= > Ey = 40KN = Cyt Ey=0 EM-O → 4 Ey - 160 so For left segment MA LookN AR A 2 M i Horn 2 m 40KN Ay Ax=0 E Fx = 0 MA= Efy =o Ay +40=90 => / Ay = 50kn EMA=0 = MA-(4x40) + 90x2 =0 ZOKNM A Giso qokN zoknm #160kNm Trokni с 50KN — 50 quel SFD. 40 "80KN 86 20

2 ✓ qokN с IGOKNM E 2 TORN B 2 2 2 40KN sokwa UBMD) 20KNMO 8INY EL MEI diagn 20%ECH 80 EI 80/ET Corgugate bean E 'சிப்பாய f 2M 20EL 80/EI 80/EI A 0.4m latt, B 20 L For stople at we calculate shear force at & of conjugak bean- t 2M 2m Oc = -1/2 x 0.4x 20 ЕТ :) +(**3 x 80 EI kret so se 4. 2 + 144.0 El = + Oc - 140 EI EI 2-ze re 8% 26 4x = 2-2 ga - 2 for deflection at 't we calculate moment at 'F' in 2 / 4 = 0.Am Conjugate beam. 816L Ey ME 0,4 ↑ Ex D. 2M 2m N + 2 m А 1.6 B 2m Roy j का Bler 8%

For lest portion the cale. E Fe=0 Erezo & Fy -0 20 4x 80 )#3x36% Cy -Ey, (1/4x0. -&* - (2x)0 1604 EI EI + 144 EI Cy - Ey - 80 EI 160 E1 100 EI Σ ΜΕ Cy - Ey Cy - Ey = Cyx + - 140x2) - (0 (260))+ *5:33) + [121.6 x (4:53)]-(eve 7-38) 5 (Cyx4) - SO -160 EL 212.8 EI + 4-964 EI 418:11 -314.4 EI EI 157:31 4 =-39.3 उप EI from egr ) get 1100 + lere 39.3 EI -1393 EI for Right portion - ME 2m Fx Ex=0 ley- 139.3 EI & Fy = o » Ey - Fy Fy = 139.3 E Fx =0 = Fezo > M = 0 ME =-139 3 X2 EI -2786 EI = ME Now deflection at 'f' is the B.M at F in conjugate Om C sigu, endiote deffection is down word be che 1-278-6 EI