ch12

Chapter 1CHAPTER OBJECTIVES
Use various methods to determine the deflection and slope at specific pts on beams and shafts:
Integration method
Discontinuity functions
Moment-area method
Use the various methods to solve for the support reactions on a beam or shaft that is statically indeterminate
12. Deflections of Beams and Shafts
CHAPTER OUTLINE
*Discontinuity Functions
Method of Superposition
Statically Indeterminate Beams and Shafts: Method of Integration
CHAPTER OUTLINE
Statically Indeterminate Beams and Shafts: Method of Superposition
12.1 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
creating the elastic curve.
Use beam convention as
and pin supports at B and
D, displacements at B and
D is zero.
concave upwards.
At pt C, there is an inflection pt where curve changes from concave up to concave down (zero moment).
right, along longitudinal axis
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.
x-sections is d.
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.
Moment-Curvature Relationship
Strain in arc ds, at position y from neutral axis, is
Moment-Curvature Relationship
If material is homogeneous and shows linear-elastic behavior, Hooke’s law applies. Since flexure formula also applies, we combing the equations to get
= radius of curvature at a specific pt on elastic curve (1/ is referred to as the curvature).
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
Eqns 12-2 and 12-3 valid for either small or large radii of curvature.
12.2 SLOPE AND DISPLACEMENT BY INTEGRATION
Let’s represent the curvature in terms of and x.
Substitute into Eqn 12-2
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, Eqn 12-4 rewritten as
Differentiate each side w.r.t. x and substitute
V = dM/dx, we get
Flexural rigidity is constant along beam, thus
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.
Sign convention and coordinates
Boundary and continuity conditions
Boundary and continuity conditions
If a single x coordinate cannot be used to express the eqn for beam’s slope or elastic curve, then continuity conditions must be used to evaluate some of the integration constants.
Procedure for analysis
Draw an exaggerated view of the beam’s elastic curve.
Recall that zero slope and zero displacement occur 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 undeflected beam and can have an origin at any pt along the beam, with +ve direction either to the right or to the left.
Elastic curve
If several discontinuous loads are present, establish x coordinates that are valid for each region of the beam between the discontinuties.
Choose these coordinates so that they will simplify subsequent algrebraic 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 eqn of moment equilibrium to determine M = f(x).
Provided EI is constant, apply either the load eqn
EI d4/dx4 = w(x), which requires four integrations to get = (x), or the moment eqns
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 pts where two functions meet.
Slope and elastic curve
Once constants are evaluated and substituted back into slope and deflection eqns, slope and displacement at specific pts on elastic curve can be determined.
The numerical values obtained is 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 12.1
Cantilevered beam shown is subjected to a vertical load P at its end. Determine the eqn of the elastic curve. EI is constant.
EXAMPLE 12.1 (SOLN)
single x coordinate.
Moment function: From free-body diagram, with M acting in the +ve direction, we have
EXAMPLE 12.1 (SOLN)
EXAMPLE 12.1 (SOLN)
Slope and elastic curve:
Using boundary conditions d/dx = 0 at x = L, and = 0 at x = L, Eqn (2) and (3) becomes
EXAMPLE 12.1 (SOLN)
Thus, C1 = PL2/2 and C2 = PL3/3. Substituting these results into Eqns (2) and (3) with = d/dx, we get
Maximum slope and displacement occur at A (x = 0),
EXAMPLE 12.1 (SOLN)
Positive result for A indicates counterclockwise rotation and negative result for A indicates that A is downward.
Consider beam to have a length of 5 m, support load
P = 30 kN and made of A-36 steel having
Est = 200 GPa.
EXAMPLE 12.1 (SOLN)
Using methods in chapter 11.3, assuming allowable normal stress is equal to yield stress allow = 250 MPa, then a W31039 would be adequate
(I = 84.8(106) mm4).
EXAMPLE 12.1 (SOLN)
EXAMPLE 12.1 (SOLN)
Since 2A = (d/dx)2 = 0.000488 << 1, this justifies the use of Eqn 12-10 than the more exact 12-4.
Also, since it is for a cantilevered beam, we’ve obtained larger values for and than would be obtained otherwise.
EXAMPLE 12.1 (SOLN)
SOLUTION 2
Using Eqn 12-8 to solve the problem. Here w(x) = 0 for 0 x L, so that upon integrating once, we get the form of Eqn 12-19
EXAMPLE 12.1 (SOLN)
Solution II
Shear constant C’1 can be evaluated at x = 0, since
VA = P. Thus, C’1 = P. Integrating again yields the form of Eqn 12-10,
Here, M = 0 at x = 0, so C’2 = 0, and as a result, we obtain Eqn 1 and solution proceeds as before.
EXAMPLE 12.4
Beam is subjected to load P at its end. Determine the displacement at C. EI is a constant.
EXAMPLE 12.4 (SOLN)
Elastic curve
Beam deflects into shape shown. Due to loading, two x coordinates will be considered, 0 x1 2a and
0 x2 a, where x2 is directed to the left from C since internal moment is easy to formulate.
EXAMPLE 12.4 (SOLN)
EXAMPLE 12.4 (SOLN)
EXAMPLE 12.4 (SOLN)
Slope and Elastic curve:
The four constants of integration determined using three boundary conditions, 1 = 0 at x1 = 0, 1 = 0 at x1 = 2a, and 2 =0 at x2 = a and a discontinuity eqn.
Here, continuity of slope at roller requires
d1/dx1 = d2/dx2 at x1 = 2a and x2 = a.
EXAMPLE 12.4 (SOLN)
EXAMPLE 12.4 (SOLN)
Displacement at C is determined by setting x2 = 0,
*12.3 DISCONTINUITY FUNCTIONS
A simplified method for finding the eqn 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
Macaulay functions
x represents the coordinate position of a pt along the beam
a is the location on the beam where a “discontinuity” occurs, or the pt where a distributed loading begins.
Integrating Macaulay functions, we get
The functions describe both uniform load and triangular load.
Singularity functions
Used to describe the pt 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.
Singularity functions
Similarly, couple moment M0, considered +ve counterclockwise, is a limitation as →0 of two distributed loadings. Hence,
Integration of the two functions yields
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 occurs at fixed supports.
Establish the x axis so that it extends to the right and has its origin at the beam’s left end.
Calculate the support reactions and then use the discontinuity functions in Table 12-2 to express either the loading w or the internal moment M as a function of x.
Calculate the support reactions and then use the discontinuity functions in Table 12-2 to express either the loading w or the internal moment M as a function of x.
Make sure to follow the sign convention for each loading as it applies for this equation.
Note that the 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.
Substitute w into EI d4/dx4 = w(x) or M into the moment curvature relation EI d2/dx2 = M, and integrate to obtain the eqns 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 eqns to obtain the final results.
When the slope and deflection eqns are evaluated at any pt on the beam, a +ve slope is counterclockwise, and a +ve displacement is upward.
EXAMPLE 12.5
Determine the eqn of the elastic curve for the cantilevered beam shown. EI is constant.
EXAMPLE 12.5 (SOLN)
Elastic curve
The loads cause the beam to deflect as shown. The boundary conditions require zero slope and displacement at A.
EXAMPLE 12.5 (SOLN)
Loading functions
Support reactions shown on free-body diagram. Since distributed loading does not extend to C as required, use superposition of loadings to represent same effect.
By sign convention, the 50-kNm couple moment, the 52-kN force at A, and
portion of distributed
beam are all –ve.
EXAMPLE 12.5 (SOLN)
Therefore,
The 12-kN load is not included, since x cannot be greater than 9 m. Because dV/dx = w(x), then by integrating, neglect constant of integration since reactions are included in load function, we have
EXAMPLE 12.5 (SOLN)
The same result can be obtained directly from Table 12-2.
EXAMPLE 12.5 (SOLN)
EXAMPLE 12.5 (SOLN)
Since d/dx = 0 at x = 0, C1 = 0; and = 0 at x = 0, so C2 = 0. Thus
*12.4 SLOPE & DISPLACEMENT BY THE MOMENT-AREA METHOD
Theorem 2
The vertical deviation of the tangent at a pt (A) on the elastic curve w.r.t. the tangent extended from another pt (B) equals the moment of the area under the ME/I diagram between these two pts
(A and B).
This moment is computed about pt (A) where the vertical deviation (tA/B) is to be determined.
Theorem 2
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.
Draw an exaggerated view of the beam’s elastic curve.
Recall that pts 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.
Realize that when the beam is subjected to a +ve moment, the beam bends concave up, whereas
-ve moment bends the beam concave down.
Elastic curve
An inflection pt 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 12.7
Determine the slope of the beam shown at pts B and C. EI is constant.
EXAMPLE 12.7 (SOLN)
The force P causes the beam to deflect as shown.
EXAMPLE 12.7 (SOLN)
Elastic curve:
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
EXAMPLE 12.7 (SOLN)
Moment-area theorem:
Applying Theorem 1, B/A is equal to the area under the M/EI diagram between pts A and B, that is,
EXAMPLE 12.7 (SOLN)
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
EXAMPLE 12.8
Determine the displacement of pts B and C of beam shown. EI is constant.
EXAMPLE 12.8 (SOLN)
Elastic curve:
The couple moment at C cause the beam to deflect as shown.
EXAMPLE 12.8 (SOLN)
Elastic curve:
The required displacements can be related directly to deviations between the tangents at B and A and C and A. Specifically, B is equal to deviation of tan A from tan B,
EXAMPLE 12.8 (SOLN)
Applying Theorem 2, tB/A is equal to the moment of the shaded area under the M/EI diagram between A and B computed about pt B, since this is the pt where tangential deviation is to be determined. Hence,
EXAMPLE 12.8 (SOLN)
Likewise, for tC/A we must determine the moment of the area under the entire M/EI diagram from A to C about pt C. We have
Since both answers are –ve, they indicate that pts B and C lie below the tangent at A. This checks with the figure.
12.5 METHOD OF SUPERPOSITION
The differential eqn EI d4/dx4 = 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.
EXAMPLE 12.16
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 unstretched. If the bar is loaded with a force of 3 kN at pt C, determine the vertical displacement of the force. Neglect the weight of the bar and take Est = 200 GPa, I = 4.687510-6 m.
EXAMPLE 12.16 (SOLN)
computed and shown. Each
displacements cause it to move
into positions shown. For this
case, the vertical displacement
We can find the displacement at C caused by the deformation of the bar, by using the table in Appendix C. We have
12.6 STATICALLY INDETERMINATE BEAMS AND SHAFTS
A member of any type is classified as statically indeterminate if the no. of unknown reactions exceeds the available no. of equilibrium eqns.
Additional support reactions on beam that are not needed to keep it in stable equilibrium are called redundants.
No. of these redundants is referred to as the degree of indeterminacy.
12.7 STATICALLY INDETERMINATE BEAMS AND SHAFTS:
METHOD OF INTEGRATION
For a statically indeterminate beam, the internal moment M can be expressed in terms of the unknown redundants.
After integrating this eqn twice, there will be two constants of integration and the redundants to be found.
The unknowns can be found from the boundary and/or continuity conditions for the problem.
EXAMPLE 12.17
Beam is subjected to the distributed loading shown. Determine the reactions at A. EI is a constant.
Elastic curve:
Beam deflects as shown. Only one coordinate x is needed. For convenience, we will take it directed to the right, since internal moment is easy to formulate.
Moment function:
Beam is indeterminate to first degree as indicated from the free-body diagram. We can express the internal moment M in terms of the redundant force at A using segment shown below.
The three unknowns Ay, C1 and C2 are determined from the boundary conditions x = 0, = 0; x = L,
d/dx = 0; and x = L, = 0. Applying these conditions yields
Solving,
Using the result for Ay, the reactions at B can be determined from the equations of equilibrium. Show that Bx = 0. By = 2w0L/5 and MB= w0L2/15
MOMENT-AREA METHOD
Draw the ME/I diagrams such that the redundants are represented as unknowns.
Apply the 2 moment-area theorems to get the relationships between the tangents on elastic curve to meet conditions of displacement and/or slope at supports of beam.
For all cases, no. of compatibility conditions is equivalent to no. of redundants.
MOMENT-AREA METHOD
Moment diagrams constructed by method of superposition
Since moment-area theorems needs calculation of both the area under the ME/I diagram and centroidal location of this area, the method of superposition can be used to combine separate ME/I diagrams for each of the known loads.
This will be relevant if the resultant moment diagram is of a complicated shape.
MOMENT-AREA METHOD
Most loadings on beams are a
combination of the four loadings
as shown.
MOMENT-AREA METHOD
MOMENT-AREA METHOD
EXAMPLE 12.20
Beam is subjected to couple moment at its end C as shown. Determine the reaction at B. EI is constant.
Using superposition,
to the simply supported
established.
Since A = B = C = 0, then tangential deviations shown must be proportional,
Equations of equilibrium:
Reactions at A and C can now be determined from the eqns of equilibrium. Show that Ax = 0, Cy = 5M0/4L, and Ay = M0/4L.
Equations of equilibrium:
From figure shown, this problem can also be worked out in terms of the tangential deviations,
METHOD OF SUPERPOSITION
First, identify the redundant support reactions on the beam.
Remove these reactions from the beam to get a primary beam that is statically determinate and stable and subjected to external load only.
Add to this beam with a series of similarly supported beams, each with a separate redundant, then by principle of superposition, the final loaded beam is obtained.
After computing the redundants, the other reactions on the beam determined from the eqns of equilibrium.
This method of analysis is sometimes called the force method.
Elastic curve
Specify unknown redundant forces or moments that must be removed from the beam in order to make it statically determinate and stable.
Use principle of superposition, draw the statically indeterminate beam and show it to be equal to a sequence of corresponding statically determinate beams.
The first beam (primary) supports the same external loads as the statically indeterminate beam, and each of the other beams “added” to the primary beam shows the beam loaded with a separate single redundant force or moment.
Elastic curve
Sketch the deflection curve for each beam and indicate symbolically the displacement or slope at the pt of each redundant force or moment.
Compatibility equations
Write a compatibility eqn for the displacement or slope at each pt where there is a redundant force or moment.
Determine all the displacements or slopes using an appropriate method explained in chapter 12.212.5.
Substitute the results into the compatibility eqns and solve for the unknown redundants.
Compatibility equations
If a numerical value for a redundant is +ve, it has the same sense of direction as originally assumed.
Similarly, a –ve numerical value indicates the redundant acts opposite to its assumed sense of direction.
Equilibrium equations
Once the redundant forces and/or moments have been determined, the remaining unknown reactions can be found from the eqns of equilibrium applied to the loadings shown on the beam’s free-body diagram.
EXAMPLE 12.22
Determine the reactions on the beam shown. Due to loading and poor construction, the roller support at B settles 12 mm.
Take E = 200 GPa and I = 80(106) mm4.
Principle of superposition
to the first degree. Roller support at
B is chosen as the redundant.
Principle of superposition is shown.
Here, By is assumed to act upwards
on the beam.
Thus Eqn (1) becomes
Expressing E and I in units of kN/m2 and m4, we have
Equilibrium equations:
Applying this result to the beam, we then calculate the reactions at A and C using eqns of equilibrium.
EXAMPLE 12.24
Determine the moment at B for beam shown. EI is constant. Neglect the effects of axial load.
EXAMPLE 12.24
Principle of superposition:
Since axial load if neglected, a there is a vertical force and moment at A and B. Since only two eqns of equilibrium are available, problem is indeterminate to the second degree.
Assume that By and MB are redundant, so that by principle of superposition, beam is represented as a cantilever, loaded separately by distributed load and reactions By and MB, as shown.
EXAMPLE 12.24
Compatibility equations:
Referring to displacement and slope at B, we require
Using table in Appendix C to compute slopes and displacements, we have
EXAMPLE 12.24
EXAMPLE 12.24
Substituting these values into Eqns (1) and (2) and canceling out the common factor EI, we have
Solving simultaneously, we get
CHAPTER REVIEW
The elastic curve represents the centerline deflection of a beam or shaft.
Its shape can be determined using the moment diagram.
Positive moments cause the elastic curve to concave upwards and negative moments cause it to concave downwards.
The radius of curvature at any pt is determined from 1/ = M/EI.
Eqn of elastic curve and its slope can be obtained by first finding the internal moment in the member as a function of x.
CHAPTER REVIEW
If several loadings act on the member, then separate moment functions must be determined between each of the loadings.
Integrating these functions once using
EI(d2/dx2) = M(x) gives the eqn for the slope of the elastic curve, and integrating again gives the eqn for the deflection.
The constants of integration are determined from the boundary conditions at the supports, or in cases where several moment functions are involved, continuity of slope and deflection at pts where these functions join must be satisfied.
CHAPTER REVIEW
Discontinuity functions allow us to express the eqn of elastic curve as a continuous function, regardless of the no. of loadings on the member.
This method eliminates the need to use continuity conditions, since the two constants of integration can be determined solely from the two boundary conditions.
The moment-area method if a semi-graphical technique for finding the slope of tangents or the vertical deviation of tangents at specific pts on the elastic curve.
CHAPTER REVIEW
The moment-area method requires finding area segments under the M/EI diagram, or the moment of these segments about pts on the elastic curve.
The method works well for M/EI diagrams composed of simple shapes, such as those produced by concentrated forces and couple moments.
The deflection or slope at a pt on a member subjected to various types of loadings can be determined by using the principle of superposition. The table in the back of the book can be used for this purpose.
CHAPTER REVIEW
Statically indeterminate beams and shafts have more unknown support reactions than available eqns of equilibrium.
To solve them such problems, we first identify the redundant reactions, and the other unknown reactions are written in terms of these redundants.
The method of integration or moment-area theorems can be used to solve for the unknown redundants.
We can also determine the redundants by using the method of superposition, where we consider the continuity of displacement at the redundant.
CHAPTER REVIEW
The displacement due to the external loading is determined with the redundant removed, and again with the redundant applied and external loading removed.
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