Structural Stability Fall 2006
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Prof. Kang,YoungJong
Chap 5. Continuous Beams
Rigid Frames
§5.1 Continuous Beams.
The slope deflection equations from §4.3 are
(5.1-1)
(5.1-2)
where
Modification of S.D.E for a far end hinged condition ( b = hinge)
by eliminating from (5.1-2) (5.1-2).
(5.1-3)
where
a limiting value for when 3.
Apply L’Hopital rule. ( to set 3 ).
Example
FOR AB.
For BC ,
Recall Sign Convention. ( §4.3 )
Stability Criterion
Where are function of .
again are functions of P. hence the smallest value of P which satisfy this equation is .
H.W. If
Example
side sway
Sol.)
2 unknowns ;
①
1 additional equation is needed.
----- ⓐ
-----ⓑ
＠ Joint B.
--- ②
Sub ⓐ ＆ ⓑinto ②
gives the general form of
Set det = 0
§Buckling Modes of Frames.
40 ~ 20
30
(where, side sway lower limit is
Prevented.) H-Fix in frame.
(side sway Permitted).
(Strong Beam) (Weak Beam).
§5.3 Critical Load of a frame by using Slope-Deflection Equations.
Review of chap. 2 D.E. Method.
The Equil. of moment of Vertical member (Fig. c)
Gives
or
B.C. ＠
＠
Thus,
ⓐ
Denoting the horizontal displ. ＠ be ,
Then, ⓑ
Moment Equil. (Fig. b) dictates that
ⓒ
No Shear.
Sub eq. ⓒinto eq ⓑ, gives.
------①
The equation for horizontal member.
ⓓ
(※No axial force ! ordinary
slope-deflection Equation.)
Since ( Positive )
ⓔ
※The compatibility condition at joint B requires that of Eq. ⓔmust be equal to the slope of eq. ⓐ ＠ .
Hence
or
------②
Ordinarily, a frame with n numbers would require n such ( Eq. ① ＆ ②) eqns. However in this case, the two vertical members are identical, two eqns suffice.
We set the det for ＆ equal to zero.
stability condition equation.
The critical load is the smallest root of this equation.
For example, then
which is as expected.
Similarly for Portal frame where side sway prevented.
Same Example ) by slope-deflection Equation.
a) Portal Frame, Side-Sway prevented.
Horizontal
force is
needed.
Mode 1(symm) Mode 2(antisymm)
Applying slope-deflection Eqn.
,
Since there is no axial force in member BC.
and
If and .
then , or .
or in matrix form.
Noting
we have
Set the det = 0 .
which gives.
Using graphical methods, or by computer program.
( Second mode ).
Example) Portal Frame Side Sway Allowed.
Axial Force Shearing Small bending
Force in Column.
Axial shortening in BC is negligible same.
①
for member AB.
( externally applied load )
Hence ②
Express in terms of , Likewise.
Only two independent kinematic DOF – since the are equal ＆ the rotation of , (The dependent on each other as so are ＆ ).
For
Solving DET = 0 For ＆ gives
§5.5 Effect of Primary Bending and Plasticity on Frame Behavior.
No Primary Bending Primary Bending.
Note : a) Primary Bending does not significantly lower the critical load of a frame as long as stresses remain elastic. ( small displ. theory ) , Hence, Primary Bending is neglected in the determination of the critical load of framed structures. However, they should be reflected in design. Must be treated as beam – column.
For Example, AISC 8th ed. demands that when , amplification must be considered.
b)Inelastic Buckling.
i) If instability were the only factor leading to collapse, failure should occur at the critical load.
ii)If collapse were only due to plasticity effect, the frame will fail when it becomes a mechanism due to formation of plastic hinge.
iii) In the actual case, both instability and plasticity might be present and collapse occurs due to a interaction of these two at a load that is lower than either the critical load or the mechanism load.
To Predict this failure load ;
(Usually yields conservative load yet reasonably accurate.
where, : Failure load.
: Elastic Buckling load.
: Plastic Mechanism load.
◎Strength Requirement.
◎Stiffness Requirement.
( serviceability ? – deflection).
§5.6. Design of Framed columns.
①No side sway.
②Side sway.
1)No side sway.
Add column loads from each story.
( A simple beam approx. will do ).
Approx. Methods.
(1) Assume an inflection point at the mid-height.
h
Portal Frame
＠ inflection point. No Moment.
then Determinate Structure
So Can solve it.
(2)Cantilever Method.
(behave as cantilever beam)
For tall building gives wild results.
(3)K-coeff method.
Conduct a stability analysis for the entire frame. Practically impossible to carry out without compute.
Improved Approx. method.
Side sway Prevented Case.
Assume Rigid frame.
0
likewise
Assume ＆ are distributed between cols AB and the one above AB in the ratio of .
( relative stiffness, assuming homogeneous material ).
Joint Equilibrium require that
--- ⓐ
Similarly
--- ⓑ
In of ⓐ ＆ ⓑ two unknowns ＆ .
hence,
Set the det = 0 and expanding, yields.
--- ⓒ
where, P 193
After Some manipulations eq.ⓒmay be rewritten as
where, ; effective col. length factor.
Corresponding to .
above Equation on K is the basis for Jackson, Mooreland alignment chart for effective column length.
Braced frame
(side sway prevented )
for unbraced frame ( Side sway Permitted).
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