Question

The continuous beam ABCD shown in Figure Q2(a) has a flexural rigidity EI = 1000 kNm2. The beam is subjected to a concentrated load at point B and a uniform load from points C to D.

(b) The beam ABCD shown in Figure Q2(b) is identical to that in Figure Q2(a), except that the roller support at point is replaced by a linear spring of stiffness K = 500 kN/m and point D settles (downwards) by 6 cm, calculate the reaction at point C using the force method (flexibility method). (5 Marks) 12 kN 9 kN/m ŞC D k 2m 2m >k 2m > Figure Q2(b)

Answer 1

12 KN 2m 2m 2m Remove elastic and redraw. N/m 1 1 MA=0 tikN JAKN Apply unit lead 12x2 + 9x2x5=86RD RD - 19 KN RAT Ro= 12 + 18 RATIN EMA=0 ht TIKN 1x 4 + 62 = 0 Pon-4 & Ro=-1 | RA , - 1+ - HN Member I Origin limit DC ID 0-2 0-2 192-98² 19x - a - 2 x VIN С В 0-2 198+2)- 9x2(1+2 + x 119 (x+4) I-3 (x+4) - 9*2 (3+x) + (x+2) - 12 x 4B IB 0-2 wdro **** - - $(x+2) + x) + -3 x+4)= (x+2) = 3.55x103 27 TEL

flexibility matrix due to elastic constant (k = sookN/m) = 3.55*10-3+ the = 5:55*10 {P,} = (fJ-'{f} As we know , D settles down by ban = 6x 10-2m and Pi is san at pointe {P, f = 5155 410-3* 6 x 10-2 {pl}z 10.81 KN - Aus