Question

03: Given the statically indeterminate frame as shown in Fig. Q3 below, determine the member end moments and the support reactions of the frame using moment distribution method. Moment distribution to be done for 3 iterations of locking and unlocking cycle. A point load of 100KN is acting on span BC at 1/4 point of the span as shown. Support A and D are fixed supports. There is a roller support at joint C to prevent the frame from going side- sway horizontally. Er is constant for the frame. (25 marks) 100 KN - d7 16m TITT 2 m CS Scanned withk 246m. CamScanner

Answer 1

100 KN 2m-> 6m Fixed End Moments Moca - 16OXON -112,5 kstrom 2 = + 100 X 6X92 + 37.5 kn.m CT 82 MAR=MBA = M = Moc=0 | Joint Member Stiffness Total Relative Stiffness factor BA عبد العال - الحل - Hla عالم

moment distribution is carried out in the following tobler 0 -112.5 0 +37.5 0 - 16.071|-21-429- +24.107 +32-143 -10.714 - 10.332-13.775 +1.792 +2.2964 -6.888 +64.286 +48-214- 1-8.0364 +4.592 +3.444 |-5.1664 +2.952 +2.214 - --0-3694 +0.211 +0.158 +72.041 -72.041 -0.492 +1.476 +0.105 +36-02 -0.738 -0.984 +1.107 -0.474-0.633 +36-821 -36-82 -0.316 1. A = MAB+MBA +36.02+72.04) = +18.01 KN ) Dx= Moc+ McD = -18.41–36.821 = -9.205 kN (*) & Fx=0 Ax + Dx+Cx=0 18.01 – 9.205 + Cx=0 Cx=-8.805 D = Moc+Men+ 100x2 _ -72.041+36.821 +200 - 20.6 KN (1) Ay = 100 – 20.6 = 79.4KN (1)