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.

M_o x

A cantilever beam loaded by a moment at the tip.

Solution

The moment diagram is easily obtained as shown next.

M_o M

Moment diagram.

Clearly,

M x( )=Mo

Integrate once: EIv = M_o EIv = M_o x + C₁ Condition 1: v = 0 at x = L C₁ = –M_o L

EIv = M_o (x–L) Integrate again: EIv = M_o (x – L) EIv = M_o (

2

x²– Lx) + C₂

Condition 2: v = 0 at x = L C₂ = M_o 2 L² EIv = M_o (

2 x² + L² – Lx)

Rotation: θ = v = EI

M_o (x – L) at x = 0, v= – EI M_oL

Deflection: v = EI Mo(

2

x² + L²– Lx) at x = 0, v = MEIo

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.

a.deflection.solution..On.the.other.hand,.we.notice.the.process.is.nothing.but.

that.of.integration,.similar.to.what.we.have.used.for.the.shear.and.moment.

diagram.solutions..Can.we.devise.a.way.of.drawing.rotation.and.deflection.

diagrams.in.much.the.same.way.as.drawing.shear.and.moment.diagrams?.The.

answer.is.yes.and.the.method.is.called.the.conjugate.beam.method.

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

∫

V= −q dx M

⁼ ∫∫

⁻q dxdx

Clearly,.the.operations.in.Equation.5.6.and.Equation.5.7.are.parallel.to.those.

in.the.previous.equations..If.we.define.–M/EI.as.“elastic.load”.in.parallel.to.

q.as.the.real.load,.then.the.two.processes.of.finding.shear.and.moment.dia- grams.and.rotation.and.deflection.diagrams.are.identical.

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.

zero.and.the.shear.and.moment.are.not..At.the.same.location.of.the.conjugate.

beam,.to.preserve.the.parallel,.the.shear.and.moment.should.be.zero..But,.

that.is.the.condition.of.a.free.end..Thus,.the.conjugate.beam.should.have.a.

free.end.at.where.the.original.beam.has.a.fixed.end..The.other.conditions.are.

derived.in.a.similar.way.

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.

to.remember.is.a.fixed.support.turns.into.a.free.support.and.vice.versa.while.

an.internal.connection.turns.into.an.internal.support.and.vice.versa.

θ = 0 θ ≠ 0

v = 0 v ≠ 0 θ = 0

v = 0 v ≠ 0 V ≠ 0

M ≠ 0

V ≠ 0

M = 0 V ≠ 0 M = 0

V = 0 M = 0 V = 0

M = 0 V ≠ 0

M ≠ 0 V ≠ 0

M ≠ 0 V ≠ 0

M = 0 V ≠ 0 M = 0 Real

Conjugate θ ≠ 0

v = 0 θ ≠ 0

v = 0 θ ≠ 0 V = 0

M = 0 V ≠ 0

M ≠ 0 V = 0 M = 0

Conversion.from.a.real.beam.to.a.conjugate.beam.

We.can.now.summarize.the.process.of.constructing.the.conjugate.beam.

and.drawing.the.rotation.and.deflection.diagrams:

. 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.

M_o 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

M_o/EI

Conjugate beam and elastic load.

3. Analyze the conjugate beam to find all reactions.

x

M_o/EI M_oL/EI

M_oL²/2EI

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).

M_oL²/2EI

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.

M_o

2a a

Find the tip rotation and deflection.

Solution

The solution is presented next in a series of diagrams.

Mo

M_o

2a a

Mo/2a

M_o/2a

2a a

Mo/EI

M_o/EI

2aM_o/3EI aM_o/3EI

M_o/EI

5aM_o/3EI

7a²M_o/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 = 5aM_o/3EI θ = 5aM_o/3EI Moment = 7a²M_o/6EI v = 7a²M_o/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)

(5)

L/2 L/2

1 kN-m

L/2 L/2

1 kN

L/2 L/2

M_o = 1 kN-m

L/2 L/2

1 kN

2a a a

2Pa

Problem.5.1