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Chapter Objectives
To determine the deflection and slope at specific points on beams and shafts using the integration method, discontinuity functions, and the method of superposition.
To use the method of superposition to solve for the support reactions on a beam or shaft that is statically indeterminate.
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In-class Activities
Reading Quiz
Applications
Elastic Curve
Integration Method
Use of discontinuity functions
Method of superposition
Statically indeterminate beams and shafts
Use of the method of superposition
Concept Quiz
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READING QUIZ
1) The slope angle θ in flexure equations is
Measured in degree
Measured in radian
Exactly equal to dv/dx
None of the above
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READING QUIZ (cont)
2) The load must be limited to a magnitude so as to not change significantly the original geometry of the beam. This is the assumption for:
The method of superposition
The moment area method
The method of integration
All of them
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READING QUIZ (cont)
3) A statically indeterminate structure
is always a stable structure
has more number of unknown reactions than the available number of equilibrium equations
is dynamically determinate
None of the above
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APPLICATIONS
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APPLICATIONS
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ELASTIC CURVE
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, which is characterized by the deflection and slope along the curve
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ELASTIC CURVE (cont)
Moment-curvature relationship:
Sign convention:
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ELASTIC CURVE (cont)
Consider a segment of width dx, the strain in are ds, located at a position y from the neutral axis is ε = (ds’ – ds)/ds. However, ds = dx = ρdθ and ds’ = (ρ-y) dθ, and so ε = [(ρ – y) dθ – ρdθ ] / (ρdθ), or
Comparing with the Hooke’s Law ε = σ / E and the flexure formula σ = -My/I
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11. SLOPE AND DISPLACEMENT BY INTEGRATION
Kinematic relationship between radius of curvature ρ and location x:
Then using the moment curvature equation, we have
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12. SLOPE AND DISPLACEMENT BY INTEGRATION (cont)
Sign convention:
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13. SLOPE AND DISPLACEMENT BY INTEGRATION (cont)
Boundary Conditions:
The integration constants can be determined by imposing the boundary conditions, or
Continuity condition at specific locations
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EXAMPLE 1
The cantilevered beam shown in Fig. 12–10a is subjected to a vertical load P at its end. Determine the equation of the elastic curve. EI is constant.
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EXAMPLE 1 (cont)
Solutions
From the free-body diagram, with M acting in the positive direction, Fig. 12–10b, we have
Applying Eq. 12–10 and integrating twice yields
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EXAMPLE 1 (cont)
Solutions
Using the boundary conditions dv/dx = 0 at x = L and v = 0 at x = L, equations 2 and 3 become
Substituting these results, we get
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EXAMPLE 1 (cont)
Solutions
Maximum slope and displacement occur at for which A(x =0),
If this beam was designed without a factor of safety by assuming the allowable normal stress is equal to the yield stress is 250 MPa; then a W310 x 39 would be found to be adequate (I = 84.4(106)mm4)
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EXAMPLE 2
The simply supported beam shown in the below figure supports the triangular distributed loading. Determine its maximum deflection. EI is constant.
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EXAMPLE 2 (cont)
Solutions
Due to symmetry only one x coordinate is needed for the solution,
The equation for the distributed loading is
Hence
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EXAMPLE 2 (cont)
Solutions
Integrating twice, we have
For boundary condition,
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EXAMPLE 2 (cont)
Solutions
Hence
For maximum deflection at x = L/2,
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22. USE OF CONTINUOUS FUNCTIONS
Macaulay functions
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23. USE OF CONTINUOUS FUNCTIONS
Macaulay functions
Integration of Macaulay functions:
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24. USE OF CONTINUOUS FUNCTIONS (cont)
Singularity Functions:
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25. USE OF CONTINUOUS FUNCTIONS (cont)
Note: Integration of these two singularity functions yields results that are different from those of Macaulay functions. Specifically,
Examples of how to use discontinuity functions to describe the loading or internal moment in a beam: :
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EXAMPLE 3
Determine the maximum deflection of the beam shown in Fig. 16–16a. EI is constant.
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EXAMPLE 3 (cont)
Solutions
The beam deflects as shown in Fig. 16–16a. The boundary conditions require zero displacement at A and B.
The loading function for the beam can be written as
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EXAMPLE 3 (cont)
Solutions
Integrating, we have
In a similar manner,
Integrating twice yields
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EXAMPLE 3 (cont)
Solutions
From Eq. 1, the boundary condition v = 0 at x = 10 m and at x = 30 m gives
Thus,
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EXAMPLE 3 (cont)
Solutions
To obtain the displacement of C, set x = 0 in Eq. 3.
The negative sign indicates that the displacement is downward as shown in Fig. 12–18a
To locate point D, use Eq. 2 with x > 10 and dv/dx = 0,
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EXAMPLE 3 (cont)
Solutions
Hence, from Eq. 3,
Comparing this value with vC, we see that vmax = vC.
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EXAMPLE 4
Determine the equation of the elastic curve for the cantilevered beam shown in Fig a. EI is constant.
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EXAMPLE 4 (cont)
Solution
The boundary conditions require zero slope and displacement at A.
The support diagram reactions at A have been calculated by statics and are shown on the free-body,
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EXAMPLE 4 (cont)
Solution
Since
Integrating twice, we have
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EXAMPLE 4 (cont)
Solution
Since dv/dx = 0, x = 0, C1 = 0; and v = 0, C2 = 0. Thus
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36. METHOD OF SUPERPOSITION
Necessary conditions to be satisfied:
The load w(x) is linearly related to the deflection v(x),
The load is assumed not to change significantly the original geometry of the beam of shaft.
Then, it is possible to find the slope and displacement at a point on a beam subjected to several different loadings by algebraically adding the effects of its various component parts.
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37. STATICALLY INDETERMINATE BEAMS AND SHAFTS
Definition: A member of any type is classified statically indeterminate if the number of unknown reactions exceeds the available number of equilibrium equations, e.g. a continuous beam having 4 supports
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38. STATICALLY INDETERMINATE BEAMS AND SHAFTS (cont)
Strategy:
The additional support reactions on the beam or shaft that are not needed to keep it in stable equilibrium are called redundants. It is first necessary to specify those redundant 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.
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39. USE OF THE METHOD OF SUPERPOSITION
Procedures:
Elastic Curve
Specify the unknown redundant forces or moments that must be removed from the beam in order to make it statically determinate and stable.
Using the principle of superposition, draw the statistically indeterminate beam and show it equal to a sequence of corresponding statically determinate beams.
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40. USE OF THE METHOD OF SUPERPOSITION (cont)
Procedures:
Elastic Curve (cont)
The first of these beams, the primary beam, supports the same external loads as the statistically indeterminate beam, and each of the other beams “added” to the primary beam shows the beam loaded with a separate redundant force or moment.
Sketch the deflection curve for each beam and indicate the symbolically the displacement or slope at the point of each redundant force or moment.
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41. USE OF THE METHOD OF SUPERPOSITION (cont)
Procedures:
Compatibility Equations
Write a compatibility equation for the displacement or slope at each point where there is a redundant force or moment.
Determine all the displacements or slopes using an appropriate method as explained in Secs through 12.5.
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42. USE OF THE METHOD OF SUPERPOSITION (cont)
Procedures:
Compatibility Equations (cont)
Substitute the results into the compatibility equations and solve for the unknown redundant.
If the numerical value for a redundant is positive, it has the same sense of direction as originally assumed. Similarly, a negative numerical value indicates the redundant acts opposite to its assumed sense of direction.
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43. USE OF THE METHOD OF SUPERPOSITION (cont)
Procedures:
Equilibrium Equations
Once the redundant forces and/or moments have been determined, the remaining unknown reactions can be found from the equations of equilibrium applied to the loadings shown on the beam’s free body diagram.
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EXAMPLE 5
Determine the reactions at the roller support B of the beam shown in Fig a, then draw the shear and moment diagrams. EI is constant.
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EXAMPLE 5 (cont)
Solutions
By inspection, the beam is statically indeterminate to the first degree.
Taking positive displacement as downward, the compatibility equation at B is
Displacements can be obtained from Appendix C.
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EXAMPLE 5 (cont)
Solutions
Substituting into Eq. 1 and solving yields
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CONCEPT QUIZ
1) The moment-curvature equation 1/ρ = M/EI is applicable to
Statically determined member only
Beams having uniform cross-sections only
Beams having constant Young’s Modulus E only
Beams having varying moment of inertia I.
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CONCEPT QUIZ
2) The flexure equations imply that
Slope and deflection at a point of a beam are independent
Moment and shear at a point of a beam are independent
Maximum moment occurs at the locations where the shear is zero
Maximum moment occurs at the inflection point.
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