Deflection Of Beams

Chapter Objectives 

Determine the deflection and slope at specific points on beams and shafts, using various analytical methods including:  The integration method  The use of discontinuity functions  The method of superposition

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In-class Activities 1. Applications 2. Elastic Curve 3. Integration Method 4. Use of discontinuity functions 5. Method of superposition

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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:

convex

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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

1    y •

Comparing with the Hooke’s Law ε = σ / E and the flexure formula σ = -My/I

1 M 1   or   EI  Ey 6 Copyright © 2011 Pearson Education South Asia Pte Ltd

SLOPE AND DISPLACEMENT BY INTEGRATION • Kinematic relationship between radius of curvature ρ and location x:

1   •

d 2 y dx 2

1   dy dx 

232

Then using the moment curvature equation, we have M 1   EI 

d 2 y dx 2

1   dy dx 

2 3/ 2



d2y dx 2

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SLOPE AND DISPLACEMENT BY INTEGRATION (cont) •

Sign convention:

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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

M   Px

•

Applying Eq. 12–10 and integrating twice yields

EI

d2y dx 2

  Px

(1)

dy Px 2 EI   C1 dx 2 Px 3 EIy    C1x  C2 6

(2) (3) 11

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EXAMPLE 1 (cont) Solutions •

Using the boundary conditions dy/dx = 0 at x = L and y = 0 at x = L, equations 2 and 3 become

PL2 0  C1 2 PL3 0  C1 L  C2 6 PL2 PL3  C1  and C2   2 3 •

Substituting these results, we get

 



P 2  L  x2 2 EI P y  x 3  3L2 x  2 L3 6 EI



(Ans)

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12

EXAMPLE 1 (cont) Solutions •

Maximum slope and displacement occur at for which A(x =0), 2

PL A  2 EI

(4)

PL3 yA   3EI •

(5)

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)

A 

30 5 2 1000 2

  

2 200 84.4 106

yA  

30 5 2 1000  2

 0.0222 rad

  

3 200 84.4 10

6

 74.1 mm

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EXAMPLE 2 The simply supported beam shown in Fig. 12–11a 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,

0 x  L/2 • The equation for the distributed loading is w  •

2 w0 x. L

Hence

  M NA  0;

2

w0 x  x  w0 L  x  0 M    L  3 4 w0 x 2 w0 L M   x 3L 4 15

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EXAMPLE 2 (cont) Solutions •

Integrating twice, we have

d2y

•

w0 3 w0 L EI M  x  x 2 3L 4 dx w0 4 w0 L 2 dy EI  x  x  C1 dx 12 L 8 w0 5 w0 L 3 EIy   x  x  C1x  C2 60 L 24 For boundary condition, y  0, x  0 and dy dx  0, x  L 2 3

5w0 L C1   , C2  0 192 16 Copyright © 2011 Pearson Education South Asia Pte Ltd

EXAMPLE 2 (cont) Solutions •

Hence

w0 5 w0 L 3 5w0 L3 EIy   x  x  x 60 L 24 192 •

For maximum deflection at x = L/2,

w0 L4 ymax   120 EI

(Ans)

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USE OF CONTINUOUS FUNCTIONS •

Macaulay functions

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USE OF CONTINUOUS FUNCTIONS •

Macaulay functions xa

n

for x  a



for x  a



 0

 x  a n

na

•

Integration of Macaulay functions:



n

x  a dx 

xa

n 1

n 1

C

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USE OF CONTINUOUS FUNCTIONS (cont) •

Singularity Functions:

w P xa

1

w  M0 x  a

 0 for x  a   P for x  a

2

for x  a  0   M 0 for x  a 20

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USE OF CONTINUOUS FUNCTIONS (cont) •

Note: Integration of these two singularity functions yields results that are different from those of Macaulay functions. Specifically,

 •

n

x  a dx  x  a

n 1

, n  1,2

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. 12–18a. EI is constant.

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EXAMPLE 3 (cont) Solutions •

The beam deflects as shown in Fig. 12–18a. The boundary conditions require zero displacement at A and B.

•

The loading function for the beam can be written as

w  8 x  0

1

 6 x  10

1

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EXAMPLE 3 (cont) Solutions •

Integrating, we have 0

V  8 x  0  6 x  10 •

0

In a similar manner, 1

M  8 x  0  6 x  10

1

   8 x  6 x  10  kN  m 1

•

Integrating twice yields 2

EI

d y

 8 x  x  10

1

dx 2 dy 2 EI  4 x 2  3 x  10  C1 dx 4 3 3 EIy   x  x  10  C1x  C2 (1) 3 Copyright © 2011 Pearson Education South Asia Pte Ltd

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EXAMPLE 3 (cont) Solutions •

From Eq. 1, the boundary condition y = 0 at x = 10 m and at x = 30 m gives

0  1333  10  10   C1 10   C2 3

0  36000   30  10   C1  30  C2 3

 C1  1333 and C2  12000 •

Thus,

dy 2 2 EI  4 x  3 x  10  1333 (2) dx 4 3 3 EIy   x  x  10  1333 x  12000 (3) 3 25 Copyright © 2011 Pearson Education South Asia Pte Ltd

EXAMPLE 3 (cont) Solutions •

To obtain the displacement of C, set x = 0 in Eq. 3.

12000 yC   kN  m3 (Ans) EI •

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 dy/dx = 0, 2

0   x  3 xD  10  1333 2 D

xD2  60 xD  1633  0 Solving for the positive root, xD  20.3 m 26 Copyright © 2011 Pearson Education South Asia Pte Ltd

EXAMPLE 3 (cont) Solutions •

Hence, from Eq. 3,

4 EIy D    20.3 3   20.3  10 3  1333 20.3  12000 3 5006 yD  kN  m3 EI •

Comparing this value with vC, we see that ymax = yC.

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EXAMPLE 4 Determine the equation of the elastic curve for the cantilevered beam shown in Fig. 12–19a. EI is constant.

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EXAMPLE 4 (cont) Solutions •

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,

w  52 x  0

1

 258 x  0

2

 x0

0

 50 x  5

2

8 x 5

0

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EXAMPLE 4 (cont) Solutions •

Since

dV dx   w x  and dM dx  V 0

V  52 x  0  258 x  0

1

1

 8 x  0  50 x  5

1

8 x5

1

1 0 1 1 2 0 2 M  258 x  0  52 x  0   8 x  0  50 x  5   8 x  5 2 2



0



2

  258  52 x  4 x 2  50 x  5  4 x  5 kN  m

•

Integrating twice, we have

EI

d2y

 258  52 x  4 x 2  50 x  5

0

 4 x5

2

dx 2 dy 4 1 4 3 EI  258 x  26 x 2  x 3  50 x  5  x  5  C1 dx 3 3 26 3 1 4 2 1 4 EIy  129 x 2  x  x  25 x  5  x  5  C1x  C2 3 3 3 Copyright © 2011 Pearson Education South Asia Pte Ltd

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EXAMPLE 4 (cont) Solutions •

Since dy/dx = 0, x = 0, C1 = 0; and y = 0, C2 = 0. Thus

1  2 1 2 26 3 1 4 y x  x  25 x  5  x  5   129 x  EI  3 3 3

4

 m (Ans) 

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STATISTICALLY INDETERMINATE BEAMS AND SHAFT • 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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STATISTICALLY INDETERMINATE BEAMS AND SHAFT (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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EXAMPLE 5 – USE OF THE INTEGRATRION METHOD The beam is subjected to the distributed loading shown in Fig. 12–34a. Determine the reaction at A. EI is constant.

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EXAMPLE 5 (cont) Solutions •

The beam deflects as shown in Fig. 12–34a. Only one coordinate x is needed.

•

The beam is indeterminate to the first degree as indicated from the freebody diagram, Fig. 12–34b 1 x3 M  Ay x  w0 6 L

•

Applying Eq. 12–10, we have d2y

1 x3 EI  Ay x  wo 2 6 L dx dy 1 1 x4 2 EI  Ay x  wo  C1 dx 2 24 L 1 1 x5 3 EIy  Ay x  wo  C1x  C2 6 120 L Copyright © 2011 Pearson Education South Asia Pte Ltd

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EXAMPLE 5 (cont) Solutions •

The 3 unknowns Ay, C1 and C2 are determined from the boundary conditions x = 0 and y = 0; x = L, dv/dx = 0; and x = L, y = 0.

•

Applying these conditions yields,

x  0, y  0;

0  0  0  0  C2

dy x  L,  0; dx

1 1 2 0  Ay L  w0 L3  C1 2 24 1 1 0  Ay L3  w0 L4  C1L  C2 6 120

x  L, y  0; •

Solving,

1 w0 L (Ans) 10 1 C1   w0 L3 C2  0 120 Ay 

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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 statistically determinate and stable.

•

Using the principle of superposition, draw the statistically indeterminate beam and show it equal to a sequence of corresponding statistically determinate beams.

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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. 38 Copyright © 2011 Pearson Education South Asia Pte Ltd

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. 12.2 through 12.5.

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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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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 6 Determine the reactions at the roller support B of the beam shown in Fig. 12–44a, then draw the shear and moment diagrams. EI is constant.

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EXAMPLE 6 (cont) Solutions •

By inspection, the beam is statically indeterminate to the first degree.

•

Taking positive displacement as downward, the compatibility equation at B is

  

•

0  yB  y'B

(1)

Displacements can be obtained from Appendix C.

wL4 5 PL3 83.25 kN  m3 yB     8 EI 48 EI EI 3

PL y'B  3EI

 9 m3 B y   EI

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EXAMPLE 6 (cont) Solutions •

Substituting into Eq. 1 and solving yields

83.25 9 B y 0  EI EI B y  9.25 kN

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EXAMPLE 7 Determine the moment at B for the beam shown in Fig. 12–46a. EI is constant. Neglect the effects of axial load.

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EXAMPLE 7 (cont) Solutions •

Since the axial load on the beam is neglected, there will be a vertical force and moment at A and B.

•

Referring to the displacement and slope at B, we require

 

  

0   B   ' 'B 0  y B  y'B  y' 'B

(1) (2)

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EXAMPLE 7 (cont) Solutions •

Use Appx C to calculate slopes and displacements, wL3 12 kN  m3 B   48 EI EI 7 wL4 42 kN  m3 yB    384 EI EI PL2 8 B y  'B   2 EI EI PL3 21.33B y y'B    3EI EI ML 4 M B  ' 'B   EI EI ML2 8M B y' 'B    2 EI EI 47 Copyright © 2011 Pearson Education South Asia Pte Ltd

EXAMPLE 7 (cont) Solutions •

Substituting these values into Eqs. 1 and 2 and cancelling out the common factor EI, we get

 

   •

0  12  8 B y  4 M B 0  42  21.33B y  8M B

Solving these equations simultaneously gives B y  3.375 kN M B  3.75 kN  m (Ans)

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Example 8 • Determine the reactions at the roller A, B and pin at C. EI is constant. • Draw the shear force and bending moment diagrams.

49

Example 9 Determine the reactions at the supports A and B. EI is constant. Draw the shear force and bending moment diagrams.

50