9_deflection Of Beams.pdf

CHAPTER

9

MECHANICS OF SOLIDS Deflection of Beams

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MECHANICS OF SOLIDS Deflection of Beams Deformation of a Beam Under Transverse Loading

Sample Problem 9.8

Equation of the Elastic Curve Direct Determination of the Elastic Curve From the Load Di...

Statically Indeterminate Beams Sample Problem 9.1 Sample Problem 9.3

Method of Superposition Sample Problem 9.7 Application of Superposition to Statically Indeterminate ...

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MECHANICS OF SOLIDS Deformation of a Beam Under Transverse Loading • Relationship between bending moment and curvature for pure bending remains valid for general transverse loadings. 1





M ( x) EI

• Cantilever beam subjected to concentrated load at the free end, 1





Px EI

• Curvature varies linearly with x 1 • At the free end A, ρ  0, A

• At the support B, GIK Institute of Engineering Sciences and Technology

1

B

ρA  

 0,  B 

EI PL 9-3

MECHANICS OF SOLIDS Deformation of a Beam Under Transverse Loading • Overhanging beam • Reactions at A and C • Bending moment diagram • Curvature is zero at points where the bending moment is zero, i.e., at each end and at E. 1





M ( x) EI

• Beam is concave upwards where the bending moment is positive and concave downwards where it is negative. • Maximum curvature occurs where the moment magnitude is a maximum. • An equation for the beam shape or elastic curve is required to determine maximum deflection and slope. GIK Institute of Engineering Sciences and Technology

9-4

MECHANICS OF SOLIDS Equation of the Elastic Curve • From elementary calculus, simplified for beam parameters, d2y 1





dx2 2 3 2

  dy  1       dx  



d2y dx2

• Substituting and integrating, EI

1



 EI

d2y dx

2

 M x x

dy EI   EI  M  x dx  C1 dx  0

x

x

0

0

EI y   dx  M  x  dx  C1x  C2 GIK Institute of Engineering Sciences and Technology

9-5

MECHANICS OF SOLIDS Equation of the Elastic Curve • Constants are determined from boundary conditions x

x

0

0

EI y   dx  M  x  dx  C1x  C2

• Three cases for statically determinant beams, – Simply supported beam y A  0,

yB  0

– Overhanging beam y A  0,

yB  0

– Cantilever beam y A  0,  A  0

• More complicated loadings require multiple integrals and application of requirement for continuity of displacement and slope. GIK Institute of Engineering Sciences and Technology

9-6

MECHANICS OF SOLIDS Sample Problem 9.1 SOLUTION: • Develop an expression for M(x) and derive differential equation for elastic curve. W 14  68

I  723 in 4

P  50 kips L  15 ft

E  29  106 psi a  4 ft

• Integrate differential equation twice and apply boundary conditions to obtain elastic curve.

For portion AB of the overhanging beam, • Locate point of zero slope or point (a) derive the equation for the elastic curve, of maximum deflection. (b) determine the maximum deflection, • Evaluate corresponding maximum (c) evaluate ymax. deflection.

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9-7

MECHANICS OF SOLIDS Sample Problem 9.1 SOLUTION:

• Develop an expression for M(x) and derive differential equation for elastic curve. - Reactions: RA 

Pa  a  RB  P1    L  L

- From the free-body diagram for section AD, a M  P x L

0  x  L 

- The differential equation for the elastic curve, EI GIK Institute of Engineering Sciences and Technology

d2y

a   P x 2 L dx 9-8

MECHANICS OF SOLIDS Sample Problem 9.1 • Integrate differential equation twice and apply boundary conditions to obtain elastic curve. EI

dy 1 a   P x 2  C1 dx 2 L

1 a EI y   P x3  C1x  C2 6 L 2

EI

d y

a   P x 2 L dx

at x  0, y  0 : C2  0 1 a 1 at x  L, y  0 : 0   P L3  C1L C1  PaL 6 L 6

Substituting, 2 dy 1 a 2 1 dy PaL   x  EI   P x  PaL  1  3   dx 2 L 6 dx 6 EI   L  

1 a 1 EI y   P x3  PaLx 6 L 6 GIK Institute of Engineering Sciences and Technology

PaL2  x  x  y    6 EI  L  L 

3

9-9

 

MECHANICS OF SOLIDS Sample Problem 9.1 • Locate point of zero slope or point of maximum deflection. 2 dy PaL   xm   0 1  3   dx 6 EI   L  

PaL2  x  x  y    6 EI  L  L 

3

 

xm 

L  0.577 L 3

• Evaluate corresponding maximum deflection.



PaL2 ymax  0.577  0.577 3 6 EI



PaL2 ymax  0.0642 6 EI ymax

 50 kips48 in 180 in 2  0.0642





6 29  106 psi 723 in 4



ymax  0.238 in GIK Institute of Engineering Sciences and Technology

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MECHANICS OF SOLIDS

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9 - 11

MECHANICS OF SOLIDS Method of Superposition

Principle of Superposition:

• Deformations of beams subjected to combinations of loadings may be obtained as the linear combination of the deformations from the individual loadings

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• Procedure is facilitated by tables of solutions for common types of loadings and supports.

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MECHANICS OF SOLIDS

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9 - 13

MECHANICS OF SOLIDS Sample Problem 9.7 For the beam and loading shown, determine the slope and deflection at point B.

SOLUTION: Superpose the deformations due to Loading I and Loading II as shown.

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9 - 14

MECHANICS OF SOLIDS Sample Problem 9.7 Loading I wL3  B I   6 EI

wL4  yB I   8EI

Loading II wL3 C II  48 EI

wL4  yC II  128 EI

In beam segment CB, the bending moment is zero and the elastic curve is a straight line. wL3  B II  C II  48 EI wL4 wL3  L  7 wL4  yB II     128 EI 48 EI  2  384 EI GIK Institute of Engineering Sciences and Technology

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MECHANICS OF SOLIDS Sample Problem 9.7

Combine the two solutions, wL3 wL3  B   B I   B II    6 EI 48 EI

7 wL3 B  48 EI

wL4 7 wL4 yB   yB I   yB II    8EI 384 EI

41wL4 yB  384 EI

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MECHANICS OF SOLIDS Application of Superposition to Statically Indeterminate Beams

• Method of superposition may be • Determine the beam deformation applied to determine the reactions at without the redundant support. the supports of statically indeterminate • Treat the redundant reaction as an beams. unknown load which, together with • Designate one of the reactions as the other loads, must produce redundant and eliminate or modify deformations compatible with the the support. original supports.

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MECHANICS OF SOLIDS Sample Problem 9.8 For the uniform beam and loading shown, determine the reaction at each support and the slope at end A.

SOLUTION: • Release the “redundant” support at B, and find deformation. • Apply reaction at B as an unknown load to force zero displacement at B.

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MECHANICS OF SOLIDS Sample Problem 9.8 • Distributed Loading: 4 3  w  2  2  3 2   yB w    L   2 L L   L  L  24 EI  3  3   3 

wL4  0.01132 EI

• Redundant Reaction Loading: 2

2

RB  2   L  RB L3  yB R   L     0.01646 3EIL  3   3  EI

• For compatibility with original supports, yB = 0 wL4 RB L3 0   y B w   y B R  0.01132  0.01646 EI EI

RB  0.688 wL 

• From statics, RA  0.271wL  GIK Institute of Engineering Sciences and Technology

RC  0.0413 wL  9 - 19

MECHANICS OF SOLIDS Sample Problem 9.8

Slope at end A, wL3 wL3  A w    0.04167 24 EI EI 2 0.0688 wL  L   2  L   wL3  A R     L      0.03398 6 EIL  3   EI  3  

wL3 wL3  A   A w   A R  0.04167  0.03398 EI EI GIK Institute of Engineering Sciences and Technology

wL3  A  0.00769 EI 9 - 20

MECHANICS OF SOLIDS

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