Question

QUESTION 2 (TOTAL 25 MARKS) A statically indeterminate frame is given in the figure below. El is constant for all members. Determine the member end moments of the frame using the moment distribution method. 5 kN/m BETTI 2. m mun. 10KN 2 m 8 m Note: The frame can be classified as a frame with sidesway. Stiffness factors 4EI. for a standard member for a member with a pin/roller end A table for fixed-end moments is given at the end of the exam paper

Answer 1

Solution : Given that: (A statically indeterminale frama) (A statically in EI is constant , w=5knm. P= ton , L8 m (horizont - 5kN/m K& 10KN K & K-8mm Note: Stiltres factors K- EL 1 for Standard member Ń = 3ET for member with a pin/rolien end free Body Diagram: 5 kN/m - 8m - zom lotin MA AR

Step 1: Finding Fixed End moments 8 MFAB = - Walter 10-5 kWm MFBA = twee = 1069 35 kw-m My Face - = -56812 = - 26.667 kn-m MFZB = +WE+ 5(88) * = 26.667 kvm Step 2: Calculating Distribution Factors . Toint I member Relative stiffness ki) | Total . stiffness ieki) .. Eki BA MET = Yep 10.727 1.375 EI 0-273

Step 3: Moment Distribution Table. I Joint Member AB BA BCCB DF 6:721 0.273 FEM 5 -26.661 26.667 -26-667 Balanung Com -13.33 Nermomet I 5 -40 Distribution -0.127X -0.273X (5-0) (5-40) = 29.445 4.55 Сом 12.723 Total moment 1.723 KN- m 30.445 k -m) -20.445 KN.no 0 moments : MAB = 7.723 KN-m (Sagging) MB = 30.445 KN-m (hoaging) Me=o Step 4: Finding Reaction torces ( Ax, ny, Cy) At member BC , Emp=0 Cy (8) + 30.445-518) 4 160 - 30.44S C4 = 1275. 16.194 tin Cy= 16.194 IN 1]

& Fe =0 ; Ax= 10 KN EFy=0} Arp Ay+y = 5(8) Ay = 40 - CZ = 40-16.194 - 40 - 16.194 Ay = 23.806 EN he Reaction at joints . Ax= 10 KN The AY = 23.806 KN. Cy = 16.194 KN ↑