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Stability analysis of rigid frames
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STABILITY ANALYSIS OF RIGID FRAMES
Omkar M. Salunkhe. M.Tech-Structures-I
(132040010) VJTI,Mumbai
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1. The buckling of rigid jointed frame implies the
buckling of its compression members.
2. Rigid frame is separated in to isolated frame
members.
3. In analysis of rigid frame , the results for the
beam-column derived can be directly applied to
determine the critical load for isolated frame
member.
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Rigid frame subjected to end moment and axial thrust
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Isolated beam-column subjected to axial thrust & moment at one end only.
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The rotation 0 at the joint B is given by
0 =e/L(LcosecL-1)+M0/PL(LcotL-1)----{1}
For beam element BC
0=M0 L1/2EI1
M0=2EI10/L1 ----{2}
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Eliminate M0from {1} and {2}
0 ----(3)
At the buckling of the frame rotation becomes very large i.e. It tends to infinity. This occurs when the denominator vanishes i.e.
(1/PL)( 2EI1/L1)(1-LcotL)+1=0
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(LcotL-1)=PLL1/2EI1
=1/2(P/EI)[(I/I1)(L1/L)]L2
=1/2(L2)[(I/I1)(L1/L)]
LcotL=1+1/2(L2)[(I/I1)(L1/L)]----{4}
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For typical case where I=I1
L=L1
Then equation {4} becomes cotL=1/(L)+(L)/2----{5} By trial and modification , the lowest root of transcendental equation is given by L=3.59Therefore, Pcr=(3.59)2EI/L2
Pcr=2EI/(0.875L)2----{6}
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Closed rigid frame subjected to end moment and axial thrust
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Isolated beam-column subjected to axial thrust & moment at both end.
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The rotation at the end of columns are given by
A =[MAL/3EI]1()+[MBL/6EL]2()
B =[MAL/6EI]1()+[MBL/3EL]2()---{1}
Where
1()=
2()=
2=L= , Pe =2EI/L2
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In this case due to symmetry MA=-MB= M0
B=- A=0
Above equation becomes, 0 = M0 L/6EI [21()+2()]
= M0 L/6EI ----{2}
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For compatability , the rotation 0 of the column
must be same as that of horizantal member is given by 0=-M0 L1/2EI1----{3}
Equate {2} and {3}, we get = ----{4}
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For typical case I=I1 L=L1
then equation {4} becomes = or tan = -
The lowest root of this transcendental equation is given by =(L)/2=2.02916 Therefore, Pcr=(4.0583)2EI/L2
Pcr=2EI/(0.774L)2----{5}
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THANK
YOU.....