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Analyze the frame as shown in fig by flexibility method. Draw BMD.

Subject:- Structural Analysis II

Title:- Frames

Difficulty:- Hard

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1 Answer
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Step 1:-

Degree of btatic indeterminacy(DSE)

(DSE) = r-3 = 5-3 = 2

Step 2:-

Select redundant as

VC=R1

HC=R2

Step 3:- Draw M and M1andM2 diagram remove the redundant part only.

Released Structure (M)

To draw M1andM2 remove all the external load and apply 1 KN on removed load.

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Step 4:-

Past Origin Limit M m1|m2
CB C 0-4.47 0 0.45x 0.84x
BA B 0-4 10x22=5x2 1(x+2) 1*4

^R1=l0M.M1EIdx 1EI4.470(0)(0.45)dx+40(5x2)(x+2)dx [^R1=533EI] ^R2=l0M.M1EIdx = 1EI4.470(0)(0.89x)dx+40(5x2)(4)dx [^R2=426.67EI] *Step 5:-* ***Flexibility Coefficient:-*** f11=l0(m1)2EIdx = 4.470(0.45x)2EIdx+40(0.89)2EIdx f11=75.36EI f22=l0(m2)2EIdx = 4.470(0.89x)2EIdx+40(42EIdx f22=85.53EI f12=f21=l0M1.M2EIdx f12=f21=75.92EI *Step 6:-* **Apply flexibility equation** ^I = {^R} + [F]{R} 0 = -533.33 + 75.36VC+75.92HC 0 = -426.67 + 75.92 VC+87.57HC 75.36 VC+75.92HC = 533.33 ----(1) 75.92 VC+87.53HC = 426.67 -----(2) *Solving (1) and (2) we get, R1=VC=17.29KN R2=HC=10.12KN *Step 7:-* Draw BMD ![enter image description here][5] Free Bonding moment (FBM) =$\frac{we^2}{8} = \frac{104^2}{8}=20KN.mBMCalculation:B.M_C = 0B.M_B=17.92210.124=5.9KN.mB.M_A = 17.296 - 10.124 - 104*\frac{4}{2}$ = -16.74 kN.m

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