This** preview**
has intentionally

**sections.**

*blurred***to view the full version.**

*Sign up*
This** preview**
has intentionally

**sections.**

*blurred***to view the full version.**

*Sign up*
**Unformatted text preview: **Columns Civ E 270 Khattak N Columns 0 Strength — Ability to support a speciﬁed load without experiencing excessive stress 0 Serviceability — Ability of the structure to support a speciﬁed load Without undergoing
unacceptable deformations. 0 Stability — Ability of the structure to support a given load without experiencing a sudden
change in its shape and conﬁguration. Stocky or short columns Slender or long columns P l.
75*”
_+_ Short columns fail by crushing Slender columns fail by buckling Column behaviour is a characteristic of the type of the column , Stocky Columns: Pen: Py = oyA Slender Columns: Pc,_< Py Pen-m1 (Pen) is the maximum load the column can support before it buckles. When P< Per. —> Stable
When P> Per. “F Unstable Columns ' Civ E 270 ' Khattak N We can develop an expression for Pcﬁﬁcal by considering a rigid body model of a column. Rigid Bars Rotational
spring with
spring stiffness L/2 Sin 0 Since this model is composed of rigid bars, therefore under the axial effect of the force P, the
roller support will move downwards. When the stiffness of the rotational spring is exceeded, the
spring will start to open as shown in the above ﬁgure. While the force P wants to disturb the
equilibrium of the system, the rotational spring is acting against the action of the force P and is
trying to restore the equilibrium of the system. Hence, Disturbing Moment = P - gSiné’
Restoring Moment = k - 26 ............................... ..(for a linear spring, P = kA , for a rotational spring, M = k9 )
For equilibrium;
Disturbing Moment = Restoring Moment
:> P.%Sz‘n6 = k-20
For small values of 9 (i.e. small angles), Sin 9 = 0 =>P~£6l = k-26l 2
If P = P” ::> PC, = ill—ic— ...... ..known as the state of neutral equilibrium. This is the maximum
axial load that a column can support when it is on the verge of buckling.
If P < PC, , i.e. P < EL]: ........ ..Stable Equilibrium
If P > PC, , i.e. P > fif- ........ ..Unstable Equilibrium Columns Civ E 270 Khattak N Planes of Buckling The critical buckling load for a column as given by Euler’s
equation is:
2
a=”€
(kL) Assuming that the given column can bend about either the x or
the y axis (as shown on the column’s cross section), the above
equation will have to be provided with the appropriate I, which is
relevant to the bending plane. Buckling about x—x axis 2 E1 3
PC, = ﬂ 2" where I X = bi
(kL) 12
Buckling about y-y axis
7:2EI 3
Pa, = 2y where I 2 519-
(kL) y 12 You need to investigate buckling about both the axes to see which case gives the lower critical
buckling load. This case was obvious because just by looking at the shape of the column we can
tell that x axis is the strong axis of the column and y axis is the weak axis. Hence, the column
will buckle about the y axis at a much lower load. However, if the column is provided with a
lateral brace, it may not be easy to judge, about which axis buckling would occur, until both
buckling loads have been evaluated. The ﬁgure here shows the column to be laterally braced at
midpoint, against buckling about the y axis. Since the braced
point cannot move, the critical buckling load, for buckling
about the y axis will now use the unbraced length Lb . 1
However for buckling about the x axis, there is no lateral brace
and the whole length will be used. Buckling about y-y axis Buckling about x-x axis
7r2EI 231 a= : a=”;
(kLb) (kL) Buckled shape If the lateral brace is not placed at mid length of the column, abOUt the y axis we need to use both the values of the unbraced length, above and below the brace, to see which
one gives the smaller value for PCI . The smallest Pcr coming from all possible bending modes,
will govern. BY COLT ENGINEERING CORPORATION
DATE CCLT ENGINEERING JOB No. CLIENT SUBJ ECT' I 7/ I? if?!” r r r; m
n d
_
w
m
w
.
ﬂ
M ...

View
Full Document

- Fall '14
- Civil Engineering