26357891 8 Beam Deflection

8/8/2019 26357891 8 Beam Deflection

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DMV 4343JAN ~ JUN `07

DEPARTMENT MANUFACTURING / PRODUCT DESIGN /MOULD / TOOL AND DIE

SEMESTER 4

COURSE MECHANICS OF MATERIALS DURATION 8 hrsCOURSE CODE DMV 4343 / DMV 5343 REF. NO.

VTOS NAME MISS AFZAN BINTI ROZALIMR RIDHWAN BIN RAMELI

PAGE 17

TOPICBEAM DEFLECTION

SUB TOPIC8.1 Elastic Curve Equations8.2 Slope and Deflection Equations by Integration8.3 Sign Convention8.4 Singularity Function Macaulays Method8.5 Statically Indeterminate Beams8.6 Extended and Compensation Loads

Often limits must be placed on the amount of deflection a beam or shaft may undergo whenit is subjected to a load, and so in this chapter we will discuss various methods fordetermining the deflection and slope at specific points on beams and shafts. The analyticalmethods include the integration method, the use of discontinuity functions, and the methodof superposition. Also, a semigraphical technique, called the moment-area method, will bepresented. At the end of the chapter, we will use these methods to solve for the supportreactions on a beam or shaft that is statically indeterminate.

Chapter 8 DEAM DEFLECTION p1

INFORMATION SHEET

REF NO. :PAGE :


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8.1 Elastic Curve Equations

Before the slope or the displacement at a point on a beam (or shaft) is determined, it is often

helpful to sketch the deflected shape of the beam when it is loaded, in order to "visualize"

any computed results and thereby partially check these results. The deflection diagram of

the longitudinal axis that passes through the centroid of each cross-sectional area of the

beam is called the elastic curve. For most beams the elastic curve can be sketched without

much difficulty. When doing so, however, it is necessary to know how the slope or

displacement is restricted at various types of supports. In general, supports that resist a

force, such as a pin, restrict displacement, and those that resist a moment, such as a fixed

wall, restrict rotation orslope. With this in mind, two typical examples of the elastic curves

for loaded beams (or shafts), sketched to a greatly exaggerated scale, are shown in Figure

8.1.

FIGURE 8.1 Elastic curve due to

transverse load

If the elastic curve for a beam seems difficult to establish, it is suggested that the moment

diagram for the beam be drawn first. A positive internal moment tends to bend the beam

concave upward, Figure 8.1a. Likewise, a negative moment tends to bend the beam

concave downward, Figure 8.1b. Therefore, if the moment diagram is known, it will be easy

to construct the elastic curve.

Moment-Curvature Relationship.

We will now develop an important relationship between the internal moment in the beam

and the radius of curvaturep (rho) of the elastic curve at a point.

To derive the relationship between the internal moment and p, we will limit the analysis to

the most common case of an initially straight beam that is elastically deformed by loads

applied perpendicular to the team's xaxis and lying in the x-vplane of symmetry for the

beam's cross-sectional area. Due to the loading, the deformation of the beam is caused by

both the internal shear force and bending moment. If the beam has a length that is much

greater than its depth, the greatest deformation will be caused by bending, and therefore we

will direct our attention to its effects.



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In previous section 5.2, we have prove that

1=

M EI

And from Bernoulli Euler Law

M = d

2

v Bernoulli Euler LawEI dx2

Moment can be found by manipulating Bernoulli Euler Law where we get

M = EId2vdx2

By integrating or differentiating the law we gain the following sequence of expressions

1 Deflection = v

2 Slope = =dv

= vdx

3 Moment = M = EId2v

= EIv'dx2

4 Shear = V = -dM

= -(EIv')dx

5 Load = w =dV

= (EIv')dx2

We assume flexural rigidity, EI is constant.

Thus

EI v = MEI v = -vEI v = w

Deflection v can thus be found by successively integrating any one of above equations.



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8.2 Slope and Deflection Equations by Integration

One method to determine deflection v is by direct integration method

Load ; EIv = w

Shear ; EIv = w dx + C1

Moment ; EIv = dx w dx + C1x + C2

Slope ; EIv = dx dx w dx +C1x2 + C2x + C32

Deflection ; EIv =dx dx dx w

dx+

C1x3 +C2x2 + C3x + C46 2

The limits and unknown from the direct integration methods, C 1, C2, C3, and C4 can be

determine using boundary condition

Thus for a beam of uniform (constant) flexural rigidity EI,

Where the constant C1 and C2 are found from the geometric boundary conditions

Boundary condition

TABLE 8.1 Boundary condition by type of support

Fixed or clamped endv = 0, x=0 = 0, x=0

Simply supported endv = 0, x=0M = 0, x=0

Free end (roller)v = 0, x=0M = 0, x=0


EI v = MEI v = Mdx + C1EI v = dx Mdx + C1x + C2


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8.3 Sign Convention

FIGURE 8.2 Positive loads and internal force

Figure 8.2 above showing positive deflection of a beam. Thus;

Upward deflection has positive sign; +v for deflection

Angle of deflection is considered positive in counterclockwise direction

Where for small deflection, = dv/dx

The following procedure provides a method for determine the slope and deflection of a

beam (or shaft) using the method of integration

Elastic Curve.

Draw an exaggerated view of the beam's elastic curve. Recall that all points of zeroslope and zero displacement occur at a fixed support, and zero displacement occurs at all

pin and roller supports.

Establish thexand ycoordinate axes. Thexaxis must be parallel to the undeflected

beam and can have an origin at any point along the beam, with a positive 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.

In all cases, the associated positive vaxis should be directed upward.

Load or Moment Function.

For each region in which there is an xcoordinate, express the loading w or the

internal moment M as a function ofx. In particular, always assume that M acts in the

positive direction when applying the equation of moment equilibrium to determine M = f(x).



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Slope and Elastic Curve.

Provided Elis constant, apply either the load equation El d4v/dx4 = - w(x), which

requires four integrations to get v = v(x), or the moment equation El d2vldx2 = M(x), which

requires only two integrations. For each integration it is important to include a constant of

integration.

The constants are evaluated using the boundary conditions for the supports

(Table 8.2) and the continuity conditions that apply to slope and displacement at points

where two functions meet. Once the constants are evaluated and substituted back into the

slope and deflection equations, the slope and displacement at specific points on the elastic

curve can then be determined.

The numerical values obtained can be checked graphically by comparing them

with the sketch of the elastic curve. Realize that positive values for slope arecounterclockwise if the x axis extends positive to the right, and clockwise if the x axis

extendspositive to the left. In either of these cases,positive displacement is upward.



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8.4 Singularity Function Macaulays Method

The method of integration, used to find the equation of the elastic curve for a beam or shaft,

is convenient if the load or internal moment can be expressed as a continuous function

throughout the beam's entire length. If several different loadings act on the beam, however,

the method becomes more tedious to apply, because separate loading or moment functions

must be written for each region of the beam. Furthermore, integration of these functions

requires the evaluation of integration constants using boundary conditions and/or continuity

conditions.

In this section we will discuss a method for finding the equation of the elastic curve for a

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

beam, w = w(x), or the beam's internal moment, M = M(x). If the expression for w is

substituted into El d4v/dx4 = -w(x) and integrated four times, or if the expression for M is

substituted into El d2v/dx2 = M(x), and integrated twice, the constants of integration will be

determined only from the boundary conditions. Since the continuity equations will not be

involved, the analysis will be greatly simplified.

Discontinuity Functions.

In order to express the load on the beam or the internal moment within it using a single

expression, we will use two types of mathematical operators known as discontinuity

functions.

TABLE 8.1 Common loading and corresponding bending moments



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Macaulay Functions.

For purposes of beam or shaft deflection, Macaulay functions, named after the

mathematician W. H. Macaulay, can be used to describe distributed loadings. They can be

written in general form as

Herexrepresents the coordinate position of a point along the beam, and a is the location on

the beam where a "discontinuity" occurs, namely the point where a distributed loading

begins. Note that the Macaulay function n is written with angle brackets to distinguish

it from the ordinary function (x - a)n, written with parentheses. As stated by the equation,

only when x a is n = n, otherwise it is zero. Furthermore, these functions are

valid only for exponential values n 0. Integration of Macaulay functions follows the same

rules as for ordinary functions, i.e.,

Note how the Macaulay functions describe both the uniform load wo (n = 0) and triangular

load (n = 1), shown in Table 8.2, items 3 and 4. This type of description can, of course, be

extended to distributed loadings having other forms. Also, it is possible to use superposition

with the uniform and triangular loadings to create the Macaulay function for a trapezoidal

loading. Using integration, the Macaulay functions for shear, V = - w(x) dx, and moment, M

= V dx, are also shown in the table.



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Singularity Functions.

These functions are used only to describe the point location of concentrated forces or

couple moments acting on a beam or shaft. Specifically, a concentrated force P can be

considered as a special case of a distributed loading, where the intensity of the loading is w

= P/such that its width is where 0, Figure 8.4.

Figure 8.4

The area under this loading diagram is equivalent to P, positive downward, and so we will

use the singularity function to describe the force P.

Note that here n = -1 so that the units for w are force per length, as it should be.

Furthermore, the function takes on the value of P only at the point x = a where the load

occurs, otherwise it is zero.

In a similar manner, a couple moment M0, considered positive counterclockwise, is alimitation as 0 of two distributed loadings as shown in Figure 8.5.



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FIGURE 8.5

Here the following function describes its value

The exponent n = -2, in order to ensure that the units of w, force per length, are

maintained.

Integration of the above two singularity functions follow the rules of operational calculus

and yields results that are differentfrom those of Macaulay functions. Specifically,

Here, only the exponent n increases by one, and no constant of integration will be

associated with this operation. Using this formula, notice how M0 and P, described in

Table 8.2, items I and 2, are integrated once, then twice, to obtain the internal shear and

moment in the beam.

Application of previous equations provides a rather direct means for expressing the loadingor the internal moment in a beam as a function of x. When doing so, close attention must be

paid to the signs of the external loadings.



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As an example ofhow to apply discontinuity functions to describe the loading or internal

moment in a beam, we will consider the beam loaded as shown in Figure 8. 6a. Here the

reactive force P created by the pin, Figure 8.6b, is negative since it acts upward, and M 0 is

negative since it acts clockwise.

(a) (b)FIGURE 8.6

Using Table 8.2, the loading at any pointxon the beam is therefore,

w = -R1 -1 + P-1 M0 -2 + w00

The reactive force at the roller is not included here since x is never greater than L, and

furthermore, this value is of no consequence in computing slope or deflection. Note that

when x = a, w = P, all other terms being zero. Also, whenx > c, w = w0, etc.

Integrating this equation twice yields the expression that describes the internal moment in

the beam. The constants of integration will be ignored here since the boundary conditions,

or the end shear and moment, have been calculated (V = R1and M = 0) and these valuesare incorporated into the beam loading w. One can also obtain this result directly from Table

8.2. In either case,

M = R1 - P + M0(x - b)0 w02 (12-16)

The validity of this expression may be checked by using the method of sections, say, within

the region b < x< c, Figure 8.7.

(a) (b)FIGURE 8.7



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Moment equilibrium requires that

M = R1 x - P(x - a) + M0

This result agrees with that obtained from the discontinuity functions.

As a second example, consider the beam in Figure 8.7a.The support reaction atA has been

computed in Figure 8.7b, and the trapezoidal loading has been separated into triangular and

uniform loadings. From Table 8.2, the loading is therefore

w= -2.75 kN-1 - 1.5 kN m


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The following procedure provides a method for using discontinuity functions to

determine a beam's elastic curve. This method is particularly advantageous for solving

problems involving beams or shafts subjected to several loadings, since the constants of

integration can be evaluated by using only the boundary conditions, while the compatibility

conditions are automatically satisfied.

Elastic Curve.

Sketch the beam's elastic curve and identify the boundary conditions at the supports.

Zero displacement occurs at all pin and roller supports, and zero slope and zero

displacement occur at fixed supports.

Establish the x axis so that it extends to the right and has its origin at the beam's left

end.

Load or Moment Function.

Calculate the support reactions and then use the discontinuity functions in Table 8.2

to express either the loading w or the internal moment Mas a function ofx. Make sure to

follow the sign convention for each loading as it applies for this equation.

Note that me distributed loadings must extend all the way to the beam's right end to

be valid. If this does not occur, use the method of superposition, which is illustrated in

Example 8-5.


Substitute w into El d4v/dx4 = -w(x) or M into the moment-curvature relation

El d2vldx2 = M, and integrate to obtain the equations for the beam's slope and deflection.

Evaluate the constants of integration using the boundary conditions, and substitute

these constants into the slope and deflection equations to obtain the final results.

When the slope and deflection equations are evaluated at any point on the beam, a

positive slope is counterclockwise, and apositive displacement is upward



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8.5 Statically Indeterminate Beams

In this section we will illustrate a general method for determining the reactions on statically

indeterminate beams and shafts. Specifically, a member of any type is classified as statically

indeterminate if the number of unknown reactions exceeds the available number of

equilibrium equations.

The additional support reactions on the beam or shaft that are not needed to keep it in

stable equilibrium are called redundants. The number of these redundants is referred to as

the degree of indeterminacy. For example, consider the beam shown in Figure 8.6 a. If the

free-body diagram is drawn. Figure 8.6b, there will be four unknown support reactions, and

since three equilibrium equations are available for solution, the beam is classified as being

indeterminate to the first degree.

FIGURE 8.6

Either Ay, By, or MA can be classified as the redundant, for if any one of these reactions is

removed, the beam remains stable and in equilibrium (Ax cannot be classified as the

redundant, for if it were removed, Fy = 0 would not be satisfied.) In a similar manner, the

continuous beam in Figure 8.7a is indeterminate to the second degree, since there are five

unknown reactions and only three available equilibrium equations, Figure 8.7b. Here the two

redundant support reactions can be chosen among Ay, By, Cy, and Dy.

FIGURE 8.7



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To determine the reactions on a beam (or shaft) that is statically indeterminate, it is first

necessary to specify the redundant reactions. We can determine these redundants from

conditions of geometry known as compatibility conditions. Once determined, the redundants

are then applied to the beam, and the remaining reactions are determined from the

equations of equilibrium.

STATICALLY INDETERMINATE BEAMS AND SHAFTSMETHOD OF

INTEGRATION

The method of integration, discussed in Section 8.2, requires two integrations of the

differential equation d2v/dx2 = M/EI once the internal moment M in the beam is expressed as

a function of position x. If the beam is statically indeterminate, however, M can also be

expressed in terms of the unknown redundants. After integrating this equation twice, there

will be two constants of integration and the redundants to be determined. Although this is

the case, these unknowns can always be found from the boundary and/or continuity

conditions for the problem. For example, the beam in Figure 8.8a has one redundant. It can

either be Ay, MA, or By, Figure 8.8b.

FIGURE 8.8

Once it is chosen the internal moment M can be written in terms of the redundant, and

integrating the moment-displacement relationship, we can then determine the two constants

of integration and the redundant from the three boundary conditions v = 0 at x = 0, dv/dx = 0

at x = 0, and v = 0 at x = L.

The following example problems illustrate specific applications of this method using the

procedure for analysis outlined in Section 8.2.



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EXAMPLE 8.1

The beam shown is fixed supported at both ends and is subjected to the uniform loading

shown. Determine the reactions at the supports. Neglect the effect of axial load.

SOLUTION

Elastic Curve.

As in the previous problem, only one x coordinate is necessary for the solution since the

loading is continuous across the span.Moment Function.

From the free-body diagram

the respective shear and moment reactions at A and B must be equal, since there is

symmetry of both loading and geometry. Because of this, the equation of equilibrium, Fy = 0, requires

VA = VB =wL2

The beam is indeterminate to the first degree, where M' is redundant. Using the beam

segment shown below, the internal moment M can be expressed in terms ofM'as follows:

M =wL

x -w

x2 - M2 2


EId2v

=wL

x -w

x2 - Mdx2 2 2

EIdv

=wL

x2 -w

x3 - Mx + C1dx 4 6

EIv =wL

x3 -w

x4 -Mx2

+ C1x + C212 24 2

The three unknowns, M', C1, and C2, can be determined from the boundary conditions v= 0

at x = 0, which yields C2 = 0; dv/dx= 0 at x = 0, which yields C1 = 0; and v= 0 at x = L,which yields



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M =wL2

12

Using these results, notice that because of symmetry the remaining boundary condition

dvldx= 0 atx = L is automatically satisfied.

It should be realized that this method of solution is generally suitable when only one x

coordinate is needed to describe the elastic curve. If several x coordinates are needed,

equations of continuity must be written, thus complicating the solution process



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8.6 Extended and Compensation Loads

The above loading will cause the beam to deflect as follows

The above boundary condition will evolve following

It is noted that the maximum deflection not occur at the applied load.

26357891 8 Beam Deflection

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