written 6.6 years ago by | • modified 3.2 years ago |
Subject:- Structural Analysis II
Title:- Frames
Difficulty:- Hard
written 6.6 years ago by | • modified 3.2 years ago |
Subject:- Structural Analysis II
Title:- Frames
Difficulty:- Hard
written 6.6 years ago by |
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.
To draw M1andM2 remove all the external load and apply 1 KN on removed load.
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 1EI∫4.470(0)(0.45)dx+∫40(−5x2)(x+2)dx [^R1=533EI] ^R2=∫l0M.M1EIdx = 1EI∫4.470(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) =$\frac{we^2}{8} = \frac{104^2}{8}=20KN.m∗∗∗BMCalculation:−∗∗B.M_C = 0B.M_B=17.92∗2−10.12∗4=−5.9KN.mB.M_A = 17.296 - 10.124 - 104*\frac{4}{2}$ = -16.74 kN.m