2. 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.
5. The rotation
0
0
at the joint B is given by
=e/L( Lcosec L-1)+M0/PL( Lcot L-1)----{1}
For beam element BC
0=M0 L1/2EI1
M0=2EI1 0/L1 ----{2}
6. Eliminate M0from {1} and {2}
----(3)
0
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- Lcot L)+1=0
8. For typical case where I=I1
L=L1
Then equation {4} becomes
cot L=1/( L)+( L)/2----{5}
By trial and modification , the lowest root of
transcendental equation is given by
L=3.59
Therefore,
Pcr=(3.59)2EI/L2
Pcr= 2EI/(0.875L)2----{6}
11. 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
12. In this case due to symmetry
MA=-MB= M0
B=-
A= 0
Above equation becomes,
0=
M0 L/6EI [2
= M0 L/6EI
1(
)+
2(
)]
----{2}
13. 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}
14. 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}