Force Method—Part II
5.2 Integration Methods
Bending.induced.shortening.is.negligible.
We.shall.be.concerned.with.only.transverse.deflection.henceforth..The.
shear.deformation.effect.on.transverse.deflection,.however,.is.also.negli-gible.if.the.length-to-depth.ratio.of.a.member.is.greater.than.10,.as.a.rule.of.
thumb..Consequently,.the.only.effect.to.be.included.in.the.analysis.of.beam.
and.frame.deflection.is.that.of.the.flexural.deformation.caused.by.bending.
moments..As.such,.there.is.no.need.to.distinguish.frames.from.beams..We.
shall.now.introduce.the.applicable.theory.for.the.transverse.deflection.of.
beams.
5.2 Integration Methods
Linear flexural beam theory—classical beam theory. The.classical.beam.theory.is.
based.on.the.following.assumptions:
. 1..Shear.deformation.effect.is.negligible.
. 2..Transverse.deflection.is.small.(<<.depth.of.beam).
Consequently:
. 1..The.normal.to.a.transverse.section.remains.normal.after.deformation.
. 2..The.arc.length.of.a.deformed.beam.element.is.equal.to.the.length.of.
the.beam.element.before.deformation.
n.a.
ds dx
dθ
ρ
ds dθ
Beam.element.deformation.and.the.resulting.curvature.of.the.neutral.axis.(n.a.).
From.the.preceding.figure,.it.is.clear.that.the.rotation.of.a.section.is.equal.
to.the.rotation.of.the.neutral.axis..The.rate.of.change.of.angle.of.the.neutral.
axis.is.defined.as.the.curvature..The.reciprocal.of.the.curvature.is.called.the.
radius.of.curvature,.denoted.by.ρ.
.
θ= =
θ= ρ d ds d ds
rate of change of angle curvature
1 . (5.1)
For.a.beam.made.of.linearly.elastic.materials,.the.curvature.of.an.element,.
represented.by.the.curvature.of.its.neutral.axis,.is.proportional.to.the.bend-ing.moment.acting.on.the.element..The.proportional.constant,.as.derived.in.
textbooks.on.strength.of.materials.or.mechanics.of.deformable.bodies,.is.the.
product.of.the.Young’s.modulus,.E,.and.the.moment.of.inertia.of.the.cross-section.with.respect.to.the.horizontal.neutral.surface.line.of.the.section,.I..
Collectively,.EI.is.called.the.sectional.flexural.rigidity.
M x k ( ) 1x
= ( ) ρ where.k.=.EI.
Rearranging.the.previous.equation,.we.obtain.the.following.moment-cur-vature.formula:
. =
ρ M EI
1. (5.2)
Equation.5.2.is.applicable.to.all.beams.made.of.linearly.elastic.materials.and.
is. independent. of. any. coordinate. system.. In. order. to. compute. any. beam.
deflection,.measured.by.the.deflection.of.its.neutral.axis,.however,.we.need.
to.define.a.coordinate.system.as.shown.next..Henceforth,.it.is.understood.
that.the.line.or.curve.shown.for.a.beam.represents.that.of.the.neutral.axis.of.
the.beam.
y, v
x, u
v(x)
Deflection.curve.and.the.coordinate.system.
In.the.preceding.figure,.u.and.v.are.the.displacements.of.a.point.of.the.neu-tral.axis.in.the.x-.and.y-direction,.respectively..As.explained.earlier,.the.axial.
displacement,. u,. is. separately. considered. and. we. shall. concentrate. on. the.
transverse.displacement,.v..At.a.typical.location,.x,.the.arc.length,.ds,.and.its.
relation.with.its.x-.and.y-components.are.depicted.in.the.next.figure.
x, u
y, v dv
dxθ ds
Arc.length,.its.x-.and.y-components,.and.the.angle.of.rotation.
The.small.deflection.assumption.of.the.classical.beam.theory.allows.us.
to.write
Tan dv
dx v andds dx
θ = θ = = = . (5.3)
where.we.have.replaced.the.differential.operator.d/dx.by.the.simpler.symbol,.
prime.(′).
A.direct.substitution.of.the.previous.formulas.into.Equation.5.1.leads.to
.
θ=
ρ= θ= θ = d
ds
d
dx v
1 . (5.4)
which.in.turn.leads.to,.from.Equation.5.2,
. =
ρ= M
EI 1 v . (5.5)
This.last.equation,.Equation.5.5,.is.the.basis.for.the.solution.of.the.deflection.
curve,.represented.by.v(x)..We.can.solve.for.v′.and.v.from.Equation.5.4.and.
Equation.5.5.by.direct.integration.
Direct integration. If.we.express.M.as.a.function.of.x.from.the.moment.dia-gram,.then.we.can.integrate.Equation.5.5.once.to.obtain.the.rotation
θ =v = ∫M
EIdx. (5.6)
Integrate.again.to.obtain.the.deflection
v= ∫∫M
EIdx dx. (5.7)
We.shall.now.illustrate.the.solution.process.by.the.following.example.
Example 5.1
The following beam has a constant EI and a length L, find the rotation and deflection formulas.
Mo x
A cantilever beam loaded by a moment at the tip.
Solution
The moment diagram is easily obtained as shown next.
Mo M
Moment diagram.
Clearly,
M x( )=Mo
Integrate once: EIv = Mo EIv = Mo x + C1
Deflection: v = EI Mo(
2
x2 + L2– Lx) at x = 0, v = EIMo
L22
The rotation and deflection at x = 0 are commonly referred to as the tip rotation and the tip deflection, respectively.
Example.5.1.demonstrates.the.lengthy.process.one.has.to.go.through.to.obtain.
Conjugate beam method..In.drawing.the.shear.and.moment.diagrams,.the.
basic. equations. we. rely. on. are. Equation. 4.1. and. Equation. 4.3,. which. are.
reproduced.next,.respectively,.in.equivalent.forms
Shear and moment diagrams: q V M
Rotation and deflection diagrams: – EI
M θ v
We.can.now.define.a.“conjugate.beam,”.on.which.an.elastic.load.of.mag-nitude.–M/EI.is.applied..We.can.draw.the.shear.and.moment.diagrams.of.
this.conjugate.beam.and.the.results.are.actually.the.rotation.and.deflection.
diagrams.of.the.original.beam..Before.we.can.do.that,.however,.we.have.to.
find.out.what.kind.of.support.or.connection.conditions.we.need.to.specify.for.
the.conjugate.beam..This.can.be.easily.achieved.by.following.the.reasoning.
in.the.following.table.from.left.to.right,.noting.that.M.and.V.of.the.conjugate.
beam.corresponds.to.deflection.and.rotation.of.the.real.beam,.respectively.
Support and Connection Conditions of a Conjugated Beam
Original Beam Conjugate Beam
Support/Connection v v= θ M V M V Support/Connection
Fixed 0 0 ≠0 ≠0 0 0 Free
Free ≠0 ≠0 0 0 ≠0 ≠0 Fixed
Hinge/Roller End 0 ≠0 0 ≠0 0 ≠0 Hinge/Roller End Internal Support 0 ≠0 ≠0 ≠0 0 ≠0 Internal Connection Internal Connection ≠0 ≠0 0 ≠0 ≠0 ≠0 Internal Support
At.a.fixed.end.of.the.original.beam,.the.rotation.and.deflection.should.be. Note.that.the.support.and.connection.conversion.summarized.in.the.pre-vious. table. can. be. summarized. in. the. following. figure,. which. is. easy. to.
memorize..The.various.quantities.are.also.attached.but.the.important.thing.
. 1..Construct. a. conjugate. beam. of. the. same. dimension. as. the. original.
beam.
. 2..Replace. the. supports. and. connections. in. the. original. beam. with.
another. set. of. supports. and. connections. on. the. conjugate. beam.
according. to. the. previous. table,. that. is,. fixed. becomes. free,. free.
becomes.fixed,.internal.hinge.becomes.internal.support,.and.so.forth.
. 3..Place.the.M/EI.diagram.of.the.original.beam.onto.the.conjugate.beam.
as.a.distributed.load,.turning.positive.moment.into.upward.load.
. 4..Draw.the.shear.diagram.of.the.conjugate.beam;.positive.shear.indi-cates.counterclockwise.rotation.of.the.original.beam.
. 5..Draw.the.moment.diagram.of.the.conjugate.beam;.positive.moment.
indicates.upward.deflection.
Example 5.2
The following beam has a constant EI and a length L. Draw the rotation and deflection diagrams.
Mo x
A cantilever beam load by a moment at the tip.
Solution
1. Draw the moment diagram of the original beam.
Mo
M
Moment diagram.
2. Construct the conjugate beam and apply the elastic load.
x
Mo/EI
Conjugate beam and elastic load.
3. Analyze the conjugate beam to find all reactions.
x
Mo/EI MoL/EI
MoL2/2EI
Conjugate beam, elastic load, and reactions.
4. Draw the rotation diagram (the shear diagram of the conjugate beam).
–MoL/EI
Shear (rotation) diagram indicating clockwise rotation.
5. Draw the deflection diagram (the moment diagram of the conju-gate beam).
MoL2/2EI
Moment (deflection) diagram indicating upward deflection.
Example 5.3
Find the rotation and deflection at the tip of the loaded beam shown. EI is constant.
Mo
2a a
Find the tip rotation and deflection.
Solution
The solution is presented next in a series of diagrams.
Mo
Mo
2a a
Mo/2a
Mo/2a
2a a
Mo/EI
Mo/EI
2aMo/3EI aMo/3EI
Mo/EI
5aMo/3EI
7a2Mo/6EI aMo/EI
Reactions
Moment Diagram
Conjugate Beam
Reactions
Solution process to find tip rotation and deflection.
At the right end (tip of the real beam):
Shear = 5aMo/3EI θ = 5aMo/3EI Moment = 7a2Mo/6EI v = 7a2Mo/6EI
Example 5.4
Draw the rotation and deflection diagrams of the loaded beam shown. EI is constant.
2a a a
P
Beam example on rotation and deflection diagrams.
Solution
The solution is presented next in a series of diagrams. Readers are encour-aged to verify all numerical results.
2a a a
P Reactions Pa
Pa/2
–Pa Moment Diagram
2a a a
Conjugate Beam
Pa/2EI Pa/EI
2a a a
Reactions
Pa/2EI Pa/EI
11Pa2/12EI 5Pa2/12EI
Shear (Rotation) Diagram (Unit: Pa2/EI)
–1 –1/12
1/6 5/12
Moment (Deflection) Diagram (Unit: Pa3/EI)
a/6
–2701/7776 = –0.35 –1/3 = –0.33
P/2 P/2
Solution process for rotation and deflection diagrams.
PROBLEM 5.1
Draw.the.rotation.and.deflection.diagrams.of.the.loaded.beams.shown..
EI.is.constant.in.all.cases.
(1)
(2)
(3)
(4)