Mechanics in Energy Resources Engineering - Chapter 10. Statically Indeterminate Beams

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Week 14, 31 May Week 15, 7 June

Mechanics in Energy Resources Engineering - Chapter 10. Statically Indeterminate Beams

Ki B k Mi PhD Ki-Bok Min, PhD

Assistant Professor

E R E i i

Energy Resources Engineering

Seoul National University

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Deflections of Beams

• Introduction

• Differential Equations of the Deflection Curve (처짐곡선의q (

미분방정식)

• Deflections by Integration of the Bending-Moment Equation y g g q (굽힘모멘트 방정식의 적분에 의한 처짐)

• Deflections by Integration of the Shear-Force and Load Equations y g q (전단력과 하중방정식의 적분에 의한 처짐)

• Method of Superposition (중첩법)

2 2

EI d v EIv M dx  

Method of Superposition (중첩법)

• Moment-Area Method (모멘트-면적법)

N i i B (불균일단면 보)

3 3

EI d v EIv V dx  

• Nonprismatic Beams (불균일단면 보) ⁴

4

EI d v EIv q dx   

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Chapter 10. Statically Indeterminate Beams

• Introduction

• Types of Statically Indeterminate Beams (부정정보의 형태)

• Analysis by the Differential Equations of the Deflection Curve (처짐곡선의 미분방정식에 의한 해석)

(처짐곡선의 미분방정식에 의한 해석)

• Method of Superposition (중첩법) Method of Superposition (중첩법)

propped cantilever beam

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Statically indeterminate structure M h 2010

March 2010

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Introduction

• Statically indeterminate structures:

– Number of reactions > Number of independent Eqs. of Equil.Number of reactions Number of independent Eqs. of Equil.

– Method of Analysis

1) S l t E il E ith Diff ti l E f d fl ti

1) Supplement Equil. Eq. with Differential Eq. of deflection curve

2 2

2) Method of superposition

Supplement Equil. Eq. with Compatability Eq. and Force-Displacement

dx

Supplement Equil. Eq. with Compatability Eq. and Force Displacement (Deflection) Eq.

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Types of Statically Indeterminate Beams

• Most of structures in everyday life are statically indeterminate

– Automobile frames, buildings & underground structuresAutomobile frames, buildings & underground structures – More sophisticated techniques are needed for analysis

• Degree of static indeterminacy (부정정차수):

– Number of unknown reactions - number of independent p equilibrium equation

– Static redundants: excess reactions

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Types of Statically Indeterminate B

Beams

• Degree of static indeterminacy (부정정차수):

Statically indeterminate to the 1^st degree

Statically indeterminate to the 3y ^rd degreeg

Clamped beam

Statically indeterminate to the 1^st degree

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Analysis by the differential equations f th d fl ti

of the deflection curve Example 10-1

– Reactions, shear forces, bending moments, slopes and deflections?

– 3 Unknown reactions: M_A,R R_A, _B

– 2 independent equilibrium Equations

0 F 



A A B

0 L R  R q

v 0 F 



0 M 



A B L 0

R  R q 

2

2 0

A B

M qL R

  L  2

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Analysis by the differential equations f th d fl ti

of the deflection curve Example 10-1

– Bending moment, M at distance x

A B

R  qLR qx2

M  R x M qL² qx²

M  qLx R x  R L

2

Differential equation of the def curve

A A 2

M  R xM 

2 2

B B

M  qLx R x  R L

2

A 2 B

M  qL R L

– Differential equation of the def. curve

2 2

B B

qL qx

EIv  M  qLx R x  R L

– Two successive integrations

x

2 2

B B

EIv M qLx R x  R L

Two successive integrations

2

1

2 2 3

2 2 2 6

B

x

qLx qL x qx

EIv   R   R Lx C

3 2

3 2 2 4

1 2

6 6 4 2 24

Bx BLx

qLx R qL x R q

x C

EIv       xC

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Analysis by the differential equations f th d fl ti

of the deflection curve Example 10-1

– Boundary Conditions

(0) 0

v(0)  0 C₂  0 v

( ) 0 v L 

(0) 0 v 

2 0

C

1 0

C  3

B 8

R  qL



F_v 0 ₅

A

R  qL ²

A 8 M  qL

– Shear Forces and Bending Moments

( ) ^B 8

0 M 



^A ⁸ ^A ⁸

x

5

A 8

V  R qx  qL qx

2 2 2

5

qx qLx qL qx

M R M x

2 8 8 2

A A

q q q q

M  R xM    

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Analysis by the differential equations f th d fl ti

of the deflection curve Example 10-1

– Slopes and Deflections of the beam

 

qx ²

Maximum deflection



⁶ ² ¹⁵ ⁸ ²



48

v qx L Lx x

  EI    ²



³ ² ⁵ ² ²



48

v qx L Lx x

  EI  

– Maximum deflection



² ²



0 6 15 8

v    L  Lx x x 15 33 0, 16

x   L ₁ 15 33

0.5785 x  16 L  L

4 4

L L

– Point of inflection

 

1

4 4

max 39 55 33

65, 536 184.6

x x

qL qL

v EI EI

   _   

Point of inflection

Curvature and bending moment change sign 5qL4

₀ _{/ 4}

x L 2048 v q

   _  EI

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Statically Indeterminate Beams E l 10 2

Example 10-2

– Reactions, shear forces, bending moments, slopes and deflections?

– 3 Unknown reactions: M_A,M R R_B_, _A, _B

– 2 independent equilibrium Equations

,

A B A B

0 F 



^F^v ^ ⁰ ^R ^^R ^P ⁰



0 M 



A B 0

R R  P 2 0

A B B

M L R

P L M

   

2

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Statically Indeterminate Beams E l 10 2

Example 10-2

– The number of Boundary Conditions and other conditions is always sufficient to evaluate constants & redundant reactions

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Statically Indeterminate Beams E l 10 2

Example 10-2

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Summary

Ch t 10 St ti ll I d t i t B

Chapter 10. Statically Indeterminate Beams

• Key Words/Questions:

– Statically Indeterminate BeamsStatically Indeterminate Beams

– Can we use to find reactions for statically indeterminate structure?

2 2

indeterminate structure?

– Point of inflection (변곡점)

– Conditions for Method of Superposition?

= +

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Final Exam

• 14 June 08:30 – 11:00

– If you can solve the home assignment with confidence, you will do If you can solve the home assignment with confidence, you will do a good job.

– More than 50% from the home assignments (with some More than 50% from the home assignments (with some modification).

– ~90% from the examples and the problems from the textbook90% from the examples and the problems from the textbook.

– Level of difficulty will be similar to that of the 1^st & 2^nd exam.

– Scope: Entire Textbook except ch.6 and ch.11

– Partial point will be minimized this time (at most 30%)

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Method of Superposition

• Separate redundants

– RR_B_B as redundantsas redundants oror – M_A as redundants

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Statically Indeterminate Structure

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Method of Superposition R R d d t (여분력) R

_B

as Redundant (여분력)

• Separate redundants

Released structure (이완구조물):

t i t di t d d t remove restraints corresponding to redundants

= +

= ++

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Method of Superposition

• An example of when Method of Superposition does not apply?

– Strain energy

2 2

P L EA

2 2

P L EA

U EA L

  

δ₁

δ δ

P1 P2 P1+P2

1 δ₂ δ₃

2 2 2

1 2 ( 1 2)

P L P L P P L

EA EA EA

  

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Method of Superposition R R d d t (여분력) R

_B

as Redundant (여분력)

• Total defl. = defl. due to uniform load + defl due to redundant

• Equation of compatibility

– Deflection due to uniform load - deflection due to redundant

    ⁰

B    B ₁ B ₂ ⁰

     

 ₁ ⁴

B 8

qL

  EI  ₂ ³

3

B B

R L

  EI

8EI 3EI

3 4

8 3 0

B B

R L qL

EI EI

    3

B 8

R  qL 5

A 8

R  qL ²

A 8 M  qL Equil. Eq.

+

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Method of Superposition R R d d t (여분력) R

_B

as Redundant (여분력)

• Total defl. = defl. due to uniform load + defl due to redundant

2

2 2

(6 4 )

v  qx L Lx x R x^B ² (3 )

v  L x

1 (6 4 )

v 24 L Lx x

  EI   ₂ (3 )

v 6 L x

 EI 

2

2 2

qx ₂ ₂

1 2 (3 5 2 )

48

v v v qx L Lx x

    EI  

+

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Method of Superposition M R d d t (여분력) M

_A

as Redundant (여분력)

= +

+

 ₁ ³

A 24

qL

  EI  ₂

3

A A

M L

  EI

   ₁ ₂ ³ ⁰

24 3

A

A A

qL M L

EI EI

        ²

A 8 M  qL

8 5

A 8 R  qL

Equil. Eq. 3

B 8 R  qL

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Method of Superposition E l 10 3

Example 10-3

– Reactions?

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Method of Superposition E l 10 4

Example 10-4

– Reactions & deflection at point D?

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Method of Superposition E l 10 5

Example 10-5

– Reactions?

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Method of Superposition E l 10 6

Example 10-6

• Magnitude of tensile force T?

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Summary

Ch t 10 St ti ll I d t i t B

Chapter 10. Statically Indeterminate Beams

• Key Words/Questions:

– Statically Indeterminate BeamsStatically Indeterminate Beams

– Can we use to find reactions for statically indeterminate structure?

2 2

indeterminate structure?

– Point of inflection (변곡점)

– Conditions for Method of Superposition?