Civil Engineering Reference
In-Depth Information
and the maximum moment obtained by substituting equation 7.74 into equation
7.73 is
+
2 -
.
0.5
N
N cr , y + cot π
N
N cr , y
, 1 +
M max = M
β m cosec π
.
(7.3)
When β m > cos π ( N / N cr , y ) , x max is negative, in which case the maximum
moment in the beam-column is the end moment M at x = 0, whence
M max = M ,
(7.4)
For beam-columns which are bent in single curvature by equal and opposite
endmoments m =− 1 ) , equation7.1forthedeflectedshapesimplifies, andthe
central deflection δ max can be expressed non-dimensionally as
8 2
N / N cr , y
δ max
δ
sec π
2
N
N cr , y 1
=
(7.5)
inwhich δ = ML 2 / 8 EI y isthevalueof δ max when N = 0.Themaximummoment
occurs at the centre of the beam-column, and is given by
M max = M sec π
2
N
N cr , y .
(7.6)
Equations 7.5 and 7.6 are plotted in Figure 7.4.
7.5.2 Beam-columns with initial curvature
If the beam-column shown in Figure 7.3 is not straight, but has an initial
crookedness given by
w 0 = δ 0 sin π x / L ,
then the differential equation of bending becomes
EI y d 2 w
d x 2 = M M ( 1 + β m ) x
L + N ( w + w 0 ) .
(7.75)
The solution of this equation which satisfies the support conditions of ( w ) 0 =
( w ) L = 0is
w = ( M / N ) [ cos µ x m cosec µ L + cot µ L ) sin µ x 1
+ ( 1 + β m ) x / L ]+[ L /π) 2 / { 1 L /π) 2 }] δ 0 sin π x / L ,
(7.76)
 
Search WWH ::




Custom Search