Force Method—Part III
6.3 Indeterminate Frame Analysis
Example 6.5
Analyze the frame shown and draw the moment and deflection diagrams.
EI is constant for all members.
L
2L P
A rigid frame with one degree of redundancy.
Solution
We choose the moment at midspan of the beam member as the redundant force: Mc.
P c
P c
θc Mcθcc
Mc
+
=
θc = 0
Principle of superposition and compatibility equation.
The compatibility equation is
θ = θ +c c Mc ccθ =0
To find θc, we use the unit load method. It turns out that θ =c 0 because the contribution of the column members cancels out each other and the contribution from the beam member is zero due to the antisymmetry of M and symmetry of m. Consequently, there is no need to find θcc and Mc is identically zero.
c θc
1 P
L
L L
P/2
P/2 P/2
P/2
1/L 1/L
PL/2
M –1 m
Unit load method to find relative angle of rotation at C.
The moment diagram shown above is the correct moment diagram for the frame and the deflection diagram is shown next.
∆
Deflection diagram of the frame.
Example 6.6
Analyze the frame shown and draw the moment and deflection diagrams. EI is constant for the two members.
Mo
c
L
L
A frame example with one degree of redundancy.
Solution
We choose the horizontal reaction at C as the redundant force: Rch.
Mo
Principle of superposition and compatibility equation.
The compatibility equation is
c= c+Rch ccδ =0
We use the unit load method to compute Δc ′ and δcc.
Load diagrams for applied load and unit load.
Moment diagrams for applied load and unit load.
∆c = Σ ∫m EIMdx =
Load, moment, and deflection diagrams.
Example 6.7
Outline the formulation of the compatibility equation of the rigid frame shown. EI is constant for all members.
L
L P
A rigid frame with three degrees of redundancy.
Solution
We choose three internal moments as the redundant forces. The resulting primary structure is one with three hinges as shown in the following figure (the circles at 1 and 3 are meant to represent hinges). At each of the three hinges, the cumulative effect on the relative rotation must be zero. That is the compatibility condition, which can be put in a matrix form.
P
Primary structure and the relative rotation at each hinge.
.
The matrix on the left-hand side is symmetric because of the Maxwell’s reciprocal law.
Beam deflection formulas. For. statically. determinate. beam. configurations,.
simple. deflection. formulas. can. be. easily. derived.. They. are. useful. for. the.
solution. of. indeterminate. beam. problems. using. the. method. of. consistent.
deformations..Some.of.the.formulas.are.given.in.the.upcoming.table.
Approximate methods for statically indeterminate frames..As.we.can.see.from.
the.previous.examples,.the.force.method.of.analysis.for.frames.is.practical.
for. hand. calculation. only. for. cases. of. one. to. two. degrees. of. redundancy..
Although. we. can. computerize. the. process. for. high. redundancy. cases,. an.
easier.way.for.computerization.is.through.the.displacement.method,.which.is.
covered.next.in.Chapter 8..In.the.meantime,.for.practical.applications,.we.can.
use.approximate.methods.for.preliminary.design.purposes..The.approximate.
methods.described.herein.give.good.approximation.to.the.correct.solutions.
Beam Deflection Formulas
Beam Configuration Formulas for Any Point Formulas for Special Points θ = –
The.basic.concept.of.the.approximate.methods.is.to.assume.the.location.of.
zero.internal.moment..At.the.point.of.zero.moment,.the.conditions.of.con-struction.apply,.that.is,.additional.equations.are.available..For.rigid.frames.of.
regular.geometry,.we.can.guess.at.the.location.of.zero.moment.fairly.accu-rately.from.experience..When.enough.conditions.of.construction.are.added,.
the.original.problem.becomes.statically.determinate..We.shall.deal.with.two.
classes.of.problems.separately.according.to.loading.conditions.
Vertical loads..For.regular-shaped.rigid.frames.loaded.with.vertical.floor.
loads.such.as.shown.in.the.following.figure,.the.deflection.of.the.beams.are.
such.that.zero.moment.exists.at.a.location.approximately.one-tenth.of.the.
span.from.each.end.
L
0.1L 0.1L
Vertically.loaded.frame.and.approximate.location.of.zero.internal.moment.
Once.we.put.a.pair.of.hinge-and-roller.at.the.location.of.zero.moment.in.
the.beams,.the.resulting.frame.is.statically.determinate.and.can.be.analyzed.
easily..The.following.figure.illustrates.the.solution.process.
Roller Hinge
Beams.and.columns.as.statically.determinate.components.
This. approach. neglects. any. shear. force. in. the. columns. and. axial. force.
in. the. beams,. which. is. a. fairly. good. assumption. for. preliminary. design.
purposes.
Horizontal loads. Depending. on. the. configuration. of. the. frame,. we. can.
apply.either.the.portal.method.or.the.cantilever.method..The.portal method.is.
generally.applicable.to.low-rise.building.frames.of.no.more.than.five.stories.
high..The.assumptions.are:
. 1..Every.midpoint.of.a.beam.or.a.column.is.a.point.of.zero.moment.
. 2..Interior.columns.carry.twice.the.shear.as.that.of.exterior.columns.
2V V
V P
0.5P 0.25P
P
0.25P
Assumptions.of.the.portal.method.
The.shear.forces.in.the.columns.are.computed.first.from.the.free-body.diagram.
(FBD).in.the.preceding.figure.using.the.horizontal.equilibrium.condition..The.
rest.of.the.unknowns.are.computed.from.the.FBDs.in.the.sequence.shown.in.the.
following.figure.one.at.a.time..Each.FBD.contains.no.more.than.three.unknowns..
The.curved.arrows.link.dashed.circles.containing.internal.force.pairs.
0.5P 0.25P
0.25P
1 3
2
2
4 4 0.25P
P
0.25P
FBDs.to.compute.internal.forces.in.the.sequence.indicated.
The.assumptions.of.the.portal.method.are.based.on.the.observation.that.the.
deflection.pattern.of.low-rise.building.frames.is.similar.to.that.of.the.shear.
deformation.of.a.deep.beam..This.similarity.is.illustrated.next.
Deflections.of.a.low-rise.building.frame.and.a.deep.beam.
On. the. other. hand,. the. cantilever method. is. generally. applicable. to. high-rise.building.frames,.whose.configurations.are.similar.to.those.of.vertical.
cantilevers..We.can.then.borrow.the.pattern.of.normal.stress.distribution.in.
a.cantilever.and.apply.it.to.the.high-rise.building.frame.
+ – + –
Normal.stress.distribution.in.a.cantilever.and.axial.force.distribution.in.a.frame.
The.assumptions.of.the.cantilever.method.are:
. 1..The. axial. forces. in. columns. are. proportional. to. the. column’s. dis-tance.to.the.center.line.of.the.frame.
. 2..The.midpoints.of.beams.and.columns.are.points.of.zero.moment.
The.solution.process.is.slightly.different.from.that.of.the.portal.method..It.
starts.from.the.FBD.of.the.upper.story.to.find.the.axial.forces..Then,.it.pro-ceeds.to.find.the.column.shears.and.axial.forces.in.beams.one.FBD.at.a.time..
This.solution.process.is.illustrated.in.the.following.figure..Note.that.the.FBD.
of.the.upper.story.cuts.through.midheight,.not.the.base,.of.the.story.
10 kN
3 m
3 m 3 m
3 m 4 m
10 kN
3 m
3 m 3 m 2 m 3 m
10 kN
1.5 m
S 2.5S
2.5S a
S
a 2.5S 10 kN
S 2 m
Cantilever.method.and.the.FBDs.
In.the.figure,.the.external.columns.have.an.axial.force.2.5.times.that.of.the.
interior.columns.because.their.distance.to.the.center.line.is.2.5.times.that.of.
the.interior.columns..The.solution.for.the.axial.force,.S,.is.obtained.by.taking.
moment.about.any.point.on.the.midheight.line:
Σ Ma = (1.5) (10) – (2.5S) (10) – S (7 – 3) = 0 S = 0.52 kN
The.rest.of.the.computation.goes.from.one.FBD.to.another,.each.with.no.more.
than.three.unknowns.and.each.takes.advantage.of.the.results.from.the.previ-ous.one..Readers.are.encouraged.to.complete.the.solution.of.all.internal.forces.
PROBLEM 6.1
Find.all.the.reaction.forces.and.moments.at.a.and.b..EI.is.constant.and.
the.beam.length.is.L.
a
b Mb
Problem.6.1 PROBLEM 6.2
Find.all.the.reaction.forces.and.moments.at.a.and.b,.taking.advantage.of.
the.symmetry.of.the.problem..EI.is.constant.
a b
L/2 L/2
P
Problem.6.2
PROBLEM 6.3
Find.the.horizontal.reaction.force.at.d.
L
2L
PL
b c
d 2EI
EI EI a
Problem.6.3
PROBLEM 6.4
Find.the.internal.moment.at.b.
L
2L
PL
a
b c
d 2EI
EI EI
Problem.6.4
165