STABILITY ANALYSIS OF RIGID FRAMES

Omkar M. Salunkhe. M.Tech-Structures-I

(132040010) VJTI,Mumbai

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.

Rigid frame subjected to end moment and axial thrust

Isolated beam-column subjected to axial thrust & moment at one end only.

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}

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

(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}

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}

Closed rigid frame subjected to end moment and axial thrust

Isolated beam-column subjected to axial thrust & moment at both end.

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

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}

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}

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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