Sway

Moment, MOriginal structure (sway)Restraining R applied to prevent horizontal movement Opposite direction of R, R applied

By moment distribution,get M1(FEM due to external loads)Value of R by static of equilibrium FX = 0Assume R causing a (FEM due to )Moment distribution, get M2Value of R by static of equilibrium FX = 0M2 = R/R (M2), R/R is a correction factorRTherefore, Final Moment M = M1 + R/R (M2)Moment, M1Moment, M2(Non-sway)(sway)Artificial joint

ECS 468 Indeterminate StructuresExample 5:12-6 RC HIBBELER 6TH EDITIONDetermine the moments at each joint of the frame shown. EI is constantQuestion : Is this sway or non-sway frame? Why

Step1: Non-sway analysisFEMCompute the stiffness factor and DF accordingly

Moment distribution table (M1):2.72 kNm1.32 kNmDCDx = 0.81 kN5mTo find R,FX = 0R = 1.73 - 0.81 = 0.92 kN How do we get the reactions?

Step 2: Sway analysisAssume R is applied at C, causing the frame to deflect .Since both B and C happen to be displaced the same , AB and DC have the same E, I and L and both A and D are fixed support, FEMAB = FEMDC

Due to , FEMAB = - 6EI /52 = -0.24 EIFEMBA = - 0.24 EI

FEMCD = - 0.24 EIFEMDC = - 0.24 EIAssume FEMAB = - 0.24 EI = -100 kNmThus, FEMBA = - 100 kNm

FEMCD = - 100 kNmFEMDC = - 100 kNm

FEM

Moment distribution table (M2):To find R,Ax = 28 kNFX = 0R = 28 + 28 = 56 kN

ECS 468 Indeterminate StructuresM = M1 + R/R (M2 )Final moment,R/R = 0.92/56 = 0.0164

ABBABCCBCDDCM12.885.78-5.782.72-2.72-1.32M2= R/R*M2-1.31-0.990.990.99-0.99-1.31M (kNm)1.574.79-4.793.71-3.71-2.63

ExerciseNow lets perform an exercise from your 7th Edition Book for sway frames.

Thanks

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