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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
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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
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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.
Slope and Elastic Curve.
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
Slope and Elastic Curve.
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.
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