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Chapter10

DEFLECTION OF BEAMS:10.1 Introduction

10.2 The Elastic Curve

10.3 Boundary Conditions

10.4 Method of integration

10.5 Use of Discontinuity Functions

10.6 Method of Superposition

10.7 Moment-Area Method

10.1 Introduction: Chapter 9 dealt with an introduction to failure theories using yield criterion for

ductile materials under static loading.

In this chapter deflection due to loads determined by various methods is discussed.

Equation for the elastic curve or deflection curve to determine the deflections at

specific points along the beam axis will be presented.

10.2 The Elastic Curve: It is useful to sketch the deflected

shape of the loaded beam, to

“visualize” computed results and

partially check the results.

The deflection diagram of the

longitudinal axis that passes through

the centroid of each x-sectional area

of the beam is called the elastic curve.

Draw the moment diagram

for the beam first before

creating the elastic curve.

Use beam convention as

shown and established

in chapter 6.

For example, due to roller and pin supports at B and D, displacements and

D is zero.

For region of - MAC, elastic curve concave downwards.

Within region of + MCD, elastic curve concave upwards.

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At point C, there is an inflection

point where curve changes from

concave up to concave down (zero

moment).

Moment-Curvature Relationship:

x axis extends +ve to the right, along

longitudinal axis of beam.

A differential element of undeformed

width dx is located.

axis extends +ve upwards from x

axis. It measures the

displacement of the centroid on

x-sectional area of element.

A “localized” y coordinate is

specified for the position of a

fiber in the element.

It is measured +ve upward from

the neutral axis.

Limit analysis to the case of initially straight

beam elastically deformed by loads applied

perpendicular to beam’s x axis and lying in the x-

plane of symmetry for beam’s x-sectional area.

Internal moment M deforms element such that

angle between x-sections is d.

Arc dx is a part of the elastic curve that intersects

the neutral axis for each x-section.

Radius of curvature for this arc defined as the

distance , measured from center of curvature O’

to dx.

Strain in arc ds, at position y from neutral axis, is

If material is homogeneous and shows linear-elastic behavior, Hooke’s law applies.

Since flexure formula also applies, we combing the equations to get:

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κ=1ρ= M

EI(10 -2 )

= radius of curvature at a specific point on elastic curve (1/ is referred to as

the curvature, κ ).

M = internal moment in beam at pt where is to be determined.

E = material’s modulus of elasticity.

I = beam’s moment of inertia computed about neutral axis.

EI is the flexural rigidity and is

always positive.

Sign for depends on the direction

of the moment.

As shown, when M is + ve ,

extends above the beam. When M is

–ve, extends below the beam.

Using flexure formula, = My/I, curvature is also

κ= 1ρ=− σ

Ey(10−3 )

Eqs. 10-2 and 10-3 valid for either small or large radii of curvature.

SLOPE AND DISPLACEMENT BY INTEGRATION:

Let’s represent the curvature in terms of and x.

κ=1ρ=

d2 υdx2

[1+(dυdx)

2]3

2

. .. .. .. . .. .(a )

Substitute into Eq. 10-2:

d2υdx2

[1+(dυdx)

2]3

2

= MEI

(10−4 )

Most engineering codes specify limitations on deflections for tolerance or aesthetic

purposes.

Slope of elastic curve determined from d/dx is very small and its square will be

negligible compared with unity.

Therefore, by approximation 1/ = d2 /dx2, Eq. 10-4 rewritten as:

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d2υdx2 =

MEI

(10−5 )

Differentiate each side w.r.t. x and substitute V = dM/dx, we get:

ddx (EI d2 υ

dx 2 )=V ( x ) (10−6 )

Differentiating again, using w = dV/dx yields:

d2

dx2 (EI d2 υdx 2 )=−w ( x ) (10−7 )

Flexural rigidity is constant along beam, thus:

EI d 4 υdx 4 =−w ( x )

EI d3υdx 3 =V ( x ) .. .. . .. .. . .. .. . .. .. . ..(10−8)

EI d2υdx 2 =M ( x )

Sign Convention:

The deflection v is measured from the x axis to the elastic curve. An upward deflection is

therefore positive. The angle of rotation, θ (measured in radians) is the angle between the

x axis and the tangent to the curve at a point. The sense of the θ vector, also of the M

vector, follows the right-hand rule. That is, θ is positive when counterclockwise, as

portrayed in the figure. For small displacements, the angle of rotation is approximately

equal to the slope (dv /dx) of the elastic curve.

Deflection=v

Slope=θ=EI dθdx

=v ' .

Momement=M=EI dθdx

=EIv '' .. .. . .. .. . .. .. . .. .. .. (10−9 )

Shear=V =dMdx

=(EIv '' )'

Load=w=dVdx

=(EIv'' )''

10.3 Boundary conditions:

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Possible boundary conditions are shown here:

10.4 Method of Integration:• Generally, it is easier to determine the internal

moment M as a function of x, integrate twice,

and evaluate only two integration constants.

• For convenience in writing each moment

expression, the origin for each x coordinate

can be selected arbitrarily.

• Use the proper signs for M, V and w.

• If a single x coordinate cannot be used to express the Eq. for beam’s slope or elastic

curve, then continuity conditions must be used to evaluate some of the integration

constants.

Procedure for analysis

Elastic curve:

• Draw an exaggerated view of the beam’s elastic curve.

• Recall that zero slope and zero displacement occurs at all fixed supports, and zero

displacement occurs at all pin and roller supports.

• Establish the x and coordinate axes.

• The x axis must be parallel to the un-deflected beam and can have an origin at any

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point along the beam, with +ve direction either to the right or to the left.

• If several discontinuous loads are present, establish x coordinates that are valid for

each region of the beam between the discontinuities.

• Choose these coordinates so that they will simplify subsequent algebraic work.

Load or moment function:

• For each region in which there is an x coordinate, express that loading w or the

internal moment M as a function of x.

• In particular, always assume that M acts in the +ve direction when applying the Eq.

of moment equilibrium to determine, M=f (x).

• Slope and elastic curve

• Provided EI is constant, apply either the load Eq., EI d4 /dx4=w(x ), which requires

four integrations to get ¿(x ), or the moment Eqs.,EI d2/dx2=M (x ), which requires

only two integrations. For each integration we include a constant of integration.

• Constants are evaluated using boundary conditions for the supports and the

continuity conditions that apply to slope and displacement at points where two

functions meet.

• Once constants are evaluated and substituted back into slope and deflection Eqs.,

slope and displacement at specific points on elastic curve can be determined.

• The numerical values obtained are checked graphically by comparing them with

sketch of the elastic curve.

• Realize that +ve values for slope are counterclockwise if the x axis extends +ve to

the right, and clockwise if the x axis extends +ve to the left. For both cases, +ve

displacement is upwards.

Example (Deflection of a Simple Beam with a Uniform Load):

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Given: A simply supported beam AB carries a uniform

load w per unit length, as shown in Fig. 10.3a.

Find:

a) Determine the equation of the elastic curve

using the double-integration approach.

b) Derive the equation of the elastic curve using the

multiple-integration approach.

c) Obtain the maximum deflection and the slopes.

Solution:

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Example of multi-interval method:

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Example of Cantilevered Beam:

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Cantilevered beam shown is subjected to a vertical load P at its end. Determine the Eq. of

the elastic curve. EI is constant.

10.5 Use of Discontinuity Functions:The integration procedures sometimes become quite cumbersome for cases

in which the beam carries concentrated forces and moments or

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discontinuous distributed loads, and thus several intervals and several sets

of continuity conditions exist.

The use of discontinuity functions may simplify the computations involved

in solving problems of this kind. These functions can be efficiently applied

to obtain the deflections of statically determinate and indeterminate beams

with constant flexural rigidity EI.

A simplified method for finding the Eq. of the elastic curve for a multiply

loaded beam using a single expression, formulated from the loading on the

beam, w=w(x), or the beam’s internal moment, M=M (x ) is discussed

below.

Macaulay functions:

Such functions can be used to describe distributed loadings, written generally as:

⟨ x−a ⟩n={ 0 for x<a( x−a )n for x≥ a

…………..(10-10)

x represents the coordinate position of a point along the beam a is the location on

the beam where a “discontinuity” occurs, or the point where a distributed loading

begins.

Integrating Macaulay functions, we get:

∫⟨ x−a ⟩n dx=⟨ x−a ⟩n+1

n+1 +C (10−11)

The functions describe both uniform load and triangular load.

Singularity functions

Used to describe the point location of concentrated forces or couple

moments acting on a beam.

A concentrated force P can be considered as a special case of distributed

loading, where w=P/e such that its width is , → 0.

w=P ⟨ x−a ⟩−1={0 x≠ a

p x=a(10−12)

Similarly, couple moment Mo, considered +ve counterclockwise, is a

limitation as → 0 of two distributed loadings. Hence,

w=M ⟨ x−a ⟩−2={ 0 x ≠ a

M o x=a(10−13)

Integration of the two functions yields:

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∫⟨ x−a ⟩n dx=⟨x−a⟩n+1 , n=−1 ,−2 (10−14 )

Example(PROBLEM (9.79)):Determine the Eq. of the elastic curve for the cantilevered beam shown. EI is constant.

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10.6 Method of Superposition:• The differential Eq. EI d4 /d x4=w (x), satisfies the two necessary requirements for

applying the principle of superposition

• The load w(x) is linearly related to the deflection (x)

• The load is assumed not to change significantly the original geometry of the beam

or shaft.

• Works well for beams with small deflections

that follow linear Hooke’s law.

• Consider, for example, the beam illustrated in

next Figure, the total deflection vc at point C is:

vC=(vC )w+ (vC )P• For special case in which a=b=L/2, from case

7 and case 10 of table B.14, we have:

vC=5w L4

384 EI+ P L3

48 EI↓

Example:

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Steel bar shown is supported by two springs at its ends A and B. Each spring has a stiffness

k = 45 kN/m and is originally un-stretched. If

the bar is loaded with a force of 3 kN at point C,

determine the vertical displacement of the force.

Neglect the weight of the bar and take Est = 200

GPa, I = 4.687510-6 m.

Solution:

End reactions at A and B are computed

and shown. Each spring deflects by an

amount:

If bar is considered rigid, these

displacements cause it to move into

positions shown. For this case, the

vertical displacement

at C is:

We can find the displacement at C caused by the deformation of the bar, by

using the table in Appendix B14-8. We have:

Adding the two displacement components, we get:

(+ )↓υC=0 .0370 m+0 .001422 m¿0 . 0384 m=38 . 4 mm

(υC )2=Pab6 EIL

(L2−b2−a2 )

=(3 kN ) (1 m ) (2 m ) [ (3 m )2−(2 m )2−(1 m )2 ]6 (200 ) (106 ) kN/m2 (4 . 6875 ) (10−6 ) m4 (3 m )

¿1 .422 mm

(υC )1=(υB )1+2 m3 m [ (υA )1−(υB )1 ]

=0 . 0222 m+23

[0 .0444 m−0.0282 m ]

=0 . 0370 m

(υ A )1=2 kN45 kN/m

=0 .0444 m

(υ B)1=1 kN45 kN/m

=0.0222 m

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10.7 MOMENT-AREA METHOD:

Moment-area method is a semi-graphical

technique that can be used to determine the

deformation of uniform beams, stepped shafts,

etc.

Assumptions:

• beam is initially straight,

• is elastically deformed by the loads,

such that the slope and deflection of

the elastic curve are very small, and

• deformations are caused by bending.

Theorem 1

The angle between the tangents at any two

points on the elastic curve equals the area

under the M/EI diagram between these two

points.

θ BA

=area under the MEI

diagram between points A∧B

Theorem 2

The vertical deviation of the tangent at

a point (A) on the elastic curve w.r.t.

the tangent extended from another point

(B) equals the moment of the area

under the ME/I diagram between these

two points (A and B).

This moment is computed about point

(A) where the vertical deviation (tA/B) is

to be determined.

θB/ A=∫A

B MEI

dx (10−16 )

t A /B=∫A

Bx1

MEI

dx (10−17 )

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t AB

=(area under the MEI

diagrambetween points A∧B)× x1

Procedure for analysis

M/EI Diagram:

Determine the support reactions and

draw the beam’s M/EI diagram.

If the beam is loaded with concentrated

forces, the M/EI diagram will consist

of a series of straight line segments,

and the areas and their moments

required for the moment-area theorems

will be relatively easy

to compute.

If the loading consists

of a series of

distributed loads, the

M/EI diagram will

consist of parabolic

or perhaps higher-

order curves, and we

use the table on the

inside front cover to

locate the area and

centroid under each

curve.

Elastic curve:

Draw an exaggerated

view of the beam’s

elastic curve.

Recall that points of zero slope and zero displacement always occur at a fixed

support, and zero displacement occurs at all pin and roller supports.

If it is difficult to draw the general shape of the elastic curve, use the moment

(M/EI) diagram.

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Realize that when the beam is subjected to a +ve moment, the beam bends concave

up, whereas -ve moment bends the beam concave down.

An inflection point or change in curvature occurs when the moment if the beam (or

M/EI) is zero.

The unknown displacement and slope to be determined should be indicated on the

curve.

Since moment-area theorems apply only between two tangents, attention should be

given as to which tangents should be constructed so that the angles or deviations

between them will lead to the solution of the problem.

The tangents at the supports should be considered, since the beam usually has zero

displacement and/or zero slope at the supports.

Moment-area theorems

Apply Theorem 1 to determine the angle between any two tangents on the elastic

curve and Theorem 2 to determine the tangential deviation.

The algebraic sign of the answer can be checked from the angle or deviation

indicated on the elastic curve.

A positive B/A represents a counterclockwise rotation of the tangent at B w.r.t.

tangent at A, and a +ve tB/A indicates that pt B on the elastic curve lies above the

extended tangent from pt A.

Example:

Determine the slope of the beam shown at pts

B and C. EI is constant.

Solution:

Elastic curve: The force P causes the beam to

deflect as shown.

The tangents at B and C are indicated since we

are required to find B and C. Also, the tangent

at the support (A) is shown. This tangent has a

known zero slope. By construction, the angle

between tan A and tan B, ❑B /A, is equivalent

to ❑B, or

Moment-area theorem:θB=θB/ A and θC=θC / A

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Applying Theorem 1, ❑B /A is equal to the area under the M /EI diagram between pts A and

B, that is,

The negative sign indicates that angle measured from tangent at A to tangent at B is

clockwise. This checks, since beam slopes downward at B. Similarly, area under the M /EI

diagram between pts A and C equals ❑C/ A. We have

θC=θC / A=12 (−PL

EI )L¿−PL2

2 EI

θB=θB/ A=(−PL2EI )(L2 )+1

2 (−PL2 EI )(L

2 )¿−3 PL2

8 EI

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