KIC000119

# KIC000119 - CE 332 Moment Distribution Moment Distribution...

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CE 332 ._ Moment Distribution for frames (Part 4) Moment Distribution for Braced Frames Moment Distribution for braced frames is exactly the same as Moment Distribution for beams. You will just have to add extra members at some joints. Example: 8k 2 k/ft. F. n 12' r: '\\\ I ( ( I I i i i 15' 12' 8' Step 1: Set up the Moment Distribution Table (see below) Step 2: Calculate member stiffness factors (k's) and DF's. We'll use modified stiffness for member AB since A is an end pin. kBD = kDB = 4EI/12 kBc = kcs = 4EI/12 DF cB =0 DFDB = DFAB = 1

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4E/ DF BC = 3E/ 4 1 ];/ 4EI = 0.3846 -+--+-~ 15 12 12 3E/ DF BA = 3E/ 4~/ 4E/ = 0.231 -+--+-- 15 12 12 FEMBA = wL 2 /8 = 56.25 } FEMAB = 0 Note: here we're recognizing A as a pin and using the appropriate figure from the back cover of Hibbeler. DF BD = 1 - DF BA - DF BC = 0.3846 Step 3: Lock appropriate joints and calculate FEM's FEMBD = -wL 2 /12 = -24 FEMDB = wL 2 /12 = 24 FEMBc = FEMCB = 0 FEMDE = -8(8) = -64
Joint A B C D Member AB BA BC BD CB DB DE DF 0 0.231 0.385 0.385 0 1 FEM 0 56.25 0 -24 0 24 -64 Unbalanced Moments I -32.25 I I 40 I Distribute -7.450 -12.403 -12.403 1/40000 Carry Over 20.000 i'" -6.202 ....-6.202 Unbalanced Moments I -20 I I 6.202 I Distribute -4.620 -7.692 . ..

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## This note was uploaded on 10/13/2011 for the course CE 377 taught by Professor Hosteng during the Fall '10 term at Iowa State.

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KIC000119 - CE 332 Moment Distribution Moment Distribution...

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