To be updated Announcement

Announcement

@Copyright Prof. Juhyuk Moon (문주혁), CEE, SNU

To be updated

Chap. 9 Deflections of Beams

9.1 Introduction

9.2 Differential equations of the deflection curve

9.3 Deflection by integration of the bending-moment equation

9.4 Deflections by integration of the shear-force and load equations 9.5 Method of superposition

9.7 Nonprismatic beams

Differential equations of the deflection curve

Geometry of deflected beam

𝜅 = 1

𝜌 = 𝑑𝜃 𝑑𝑠

𝑑𝑣

𝑑𝑥 = tan𝜃 𝜃 = arctan 𝑑𝑣

𝑑𝑥

Differential equations of the deflection curve

Beams with small angles of rotation

𝜅 = 1

𝜌 = 𝑑𝜃 𝑑𝑠

𝜅 = 1

𝜌 = 𝑑

𝑣 𝑑𝑥

𝑑

𝑣

𝑑𝑥

= 𝑀

𝐸𝐼

Differential equations of the deflection curve

Beams with small angles of rotation

𝑑

𝑣

𝑑𝑥

= 𝑀

𝐸𝐼

Differential equations of the deflection curve

Beams with small angles of rotation

𝑑

𝑣

𝑑𝑥

= 𝑀 𝐸𝐼

𝐸𝐼 𝑑²𝑣

𝑑𝑥² = 𝑀 𝐸𝐼 𝑑³𝑣

𝑑𝑥³ = 𝑉 𝐸𝐼 𝑑⁴𝑣

𝑑𝑥⁴ = −𝑞 𝐸𝐼𝑣^′′ = 𝑀 𝐸𝐼𝑣^′′′ = 𝑉 𝐸𝐼𝑣^′′′′ = −𝑞

Differential equations of the deflection curve

Beams with large angles of rotation (Exact expression for curvature)

Deflections by integration of the bending-moment equation

𝑑²𝑣

𝑑𝑥² = 𝑀 𝐸𝐼

𝑑³𝑣

𝑑𝑥³ = 𝑉 𝐸𝐼

𝑑⁴𝑣

𝑑𝑥⁴ = −𝑞 𝐸𝐼 𝑑𝑣

𝑑𝑥 = 𝜃 𝑣

𝜅 = 1

𝜌 = 𝑑²𝑣 𝑑𝑥²

Deflections by integration of the bending-moment equation

𝑑²𝑣

𝑑𝑥² = 𝑀 𝐸𝐼

𝑑³𝑣

𝑑𝑥³ = 𝑉 𝐸𝐼

𝑑⁴𝑣

𝑑𝑥⁴ = −𝑞 𝐸𝐼 𝑑𝑣

𝑑𝑥 = 𝜃 𝑣

𝜃_𝐵 = −𝑣′(𝐿) = 𝑞𝐿³ 6𝐸𝐼

𝜃_𝐴 = −𝑣′(0)

𝜃_𝐵 = 𝑣′(𝐿) 𝛿_𝐵 = −𝑣(𝐿) = 𝑞𝐿⁴

8𝐸𝐼 𝛿_𝐶 = −𝑣 𝐿

2

Deflections by integration of the bending-moment equation

𝑀 = −𝑞𝐿²

2 + 𝑞𝐿𝑥 − 𝑞𝑥² 2

𝐸𝐼𝑣′′ = −𝑞𝐿²

2 + 𝑞𝐿𝑥 − 𝑞𝑥² 2

𝐸𝐼𝑣′ = −𝑞𝐿²

2 𝑥 + 𝑞𝐿𝑥²

2 − 𝑞𝑥³

6 + 𝐶₁

𝐸𝐼𝑣′ = −𝑞𝐿²

2 𝑥 + 𝑞𝐿𝑥²

2 − 𝑞𝑥³ 6

𝑣′ = − 𝑞𝑥

6𝐸𝐼 3𝐿² − 3𝐿𝑥 + 𝑥²

𝑣′(0) = 0

Deflections by integration of the bending-moment equation

𝑣(0) = 0 𝑣′ = − 𝑞𝑥

6𝐸𝐼 3𝐿² − 3𝐿𝑥 + 𝑥²

𝐸𝐼𝑣 = −𝑞𝐿²𝑥²

4 + 𝑞𝐿𝑥³

6 − 𝑞𝑥⁴

24 + 𝐶₂

𝑣 = − 𝑞𝑥²

24𝐸𝐼 6𝐿² − 4𝐿𝑥 + 𝑥²

𝜃_𝐵 = −𝑣′(𝐿) = 𝑞𝐿³ 6𝐸𝐼

𝛿_𝐵 = −𝑣(𝐿) = 𝑞𝐿⁴ 8𝐸𝐼

Deflections by integration of the bending-moment equation

𝑀 = 𝑃𝑏𝑥 𝐿 𝑀 = 𝑃𝑏𝑥

𝐿 − 𝑃 𝑥 − 𝑎

0 ≤ 𝑥 ≤ 𝑎 𝑎 ≤ 𝑥 ≤ 𝐿

𝐸𝐼𝑣′′ = 𝑃𝑏𝑥 𝐿 𝐸𝐼𝑣′′ = 𝑃𝑏𝑥

𝐿 − 𝑃 𝑥 − 𝑎

𝐸𝐼𝑣′ = 𝑃𝑏𝑥²

2𝐿 + 𝐶₁ 𝐸𝐼𝑣′ = 𝑃𝑏𝑥²

2𝐿 −𝑃 𝑥 − 𝑎 ²

2 + 𝐶₂

𝐸𝐼𝑣 = 𝑃𝑏𝑥³

6𝐿 −𝑃 𝑥 − 𝑎 ³

6 + 𝐶₂𝑥 + 𝐶₄ 𝐸𝐼𝑣 = 𝑃𝑏𝑥³

6𝐿 + 𝐶₁𝑥 + 𝐶₃

0 ≤ 𝑥 ≤ 𝑎 𝑎 ≤ 𝑥 ≤ 𝐿

0 ≤ 𝑥 ≤ 𝑎 𝑎 ≤ 𝑥 ≤ 𝐿

0 ≤ 𝑥 ≤ 𝑎 𝑎 ≤ 𝑥 ≤ 𝐿

Deflections by integration of the bending-moment equation

𝐸𝐼𝑣 = 𝑃𝑏𝑥³

6𝐿 − 𝑃 𝑥 − 𝑎 ³

6 + 𝐶₂𝑥 + 𝐶₄ 𝐸𝐼𝑣 = 𝑃𝑏𝑥³

6𝐿 + 𝐶₁𝑥 + 𝐶₃ 0 ≤ 𝑥 ≤ 𝑎 𝑎 ≤ 𝑥 ≤ 𝐿

Boundary condition (BC)

1. At x = a, the slopes v’ for the two parts of the beam are the same.

2. At x = a, the deflections v for the two parts of the beam are the same.

3. At x = 0, the deflection v is zero.

4. At x = L, the deflection v is zero.

Deflection equations

𝐸𝐼𝑣′ = 𝑃𝑏𝑥²

2𝐿 + 𝐶₁ 𝐸𝐼𝑣′ = 𝑃𝑏𝑥²

2𝐿 −𝑃 𝑥 − 𝑎 ²

2 + 𝐶₂

0 ≤ 𝑥 ≤ 𝑎 𝑎 ≤ 𝑥 ≤ 𝐿 Angle equations

Deflections by integration of the bending-moment equation

Boundary condition (BC)

1. At x = a, the slopes v’ for the two parts of the beam are the same.

2. At x = a, the deflections v for the two parts of the beam are the same.

3. At x = 0, the deflection v is zero.

4. At x = L, the deflection v is zero.

𝐸𝐼𝑣 = 𝑃𝑏𝑥³

6𝐿 − 𝑃 𝑥 − 𝑎 ³

6 + 𝐶₂𝑥 + 𝐶₄ 𝐸𝐼𝑣 = 𝑃𝑏𝑥³

6𝐿 + 𝐶₁𝑥 + 𝐶₃ 0 ≤ 𝑥 ≤ 𝑎 𝑎 ≤ 𝑥 ≤ 𝐿 Deflection equations

𝐸𝐼𝑣′ = 𝑃𝑏𝑥²

2𝐿 + 𝐶₁ 𝐸𝐼𝑣′ = 𝑃𝑏𝑥²

2𝐿 − 𝑃 𝑥 − 𝑎 ²

2 + 𝐶₂

0 ≤ 𝑥 ≤ 𝑎 𝑎 ≤ 𝑥 ≤ 𝐿 Angle equations

𝑃𝑏𝑎²

2𝐿 + 𝐶₁ = 𝑃𝑏𝑎²

2𝐿 + 𝐶₂ 𝐶₁ = 𝐶₂

𝑃𝑏𝑎³

6𝐿 + 𝐶₁𝑎 + 𝐶₃ = 𝑃𝑏𝑎³

6𝐿 + 𝐶₂𝑎 + 𝐶₄ 𝐶₃ = 𝐶₄

𝐶₃ = 𝐶₄ = 0

𝑃𝑏𝐿³

6𝐿 − 𝑃𝑏³

6 + 𝐶₂𝐿 = 0 𝐶₁ = 𝐶₂ = −𝑃𝑏 𝐿² − 𝑏² 6𝐿

or

or

or

Deflections by integration of the bending-moment equation

Finally..

0 ≤ 𝑥 ≤ 𝑎 𝑎 ≤ 𝑥 ≤ 𝐿 𝑣 = − 𝑃𝑏𝑥

6𝐿𝐸𝐼 𝐿² − 𝑏² − 𝑥² 𝑣 = − 𝑃𝑏𝑥

6𝐿𝐸𝐼 𝐿² − 𝑏² − 𝑥² −𝑃 𝑥 − 𝑎 ³ 6𝐿𝐸𝐼

𝑣′ = − 𝑃𝑏

6𝐿𝐸𝐼 𝐿² − 𝑏² − 3𝑥² 𝑣′ = − 𝑃𝑏

6𝐿𝐸𝐼 𝐿² − 𝑏² − 3𝑥² − 𝑃 𝑥 − 𝑎 ² 2𝐸𝐼

0 ≤ 𝑥 ≤ 𝑎 𝑎 ≤ 𝑥 ≤ 𝐿

𝜃_𝐴 = −𝑣′(0) = 𝑃𝑏 𝐿² − 𝑏²

6𝐿𝐸𝐼 = 𝑃𝑎𝑏 𝐿 + 𝑏 6𝐿𝐸𝐼 𝜃_𝐵 = 𝑣′(𝐿) = 𝑃𝑏 2𝐿² − 3𝑏𝐿 + 𝑏²

6𝐿𝐸𝐼 = 𝑃𝑎𝑏 𝐿 + 𝑎 6𝐿𝐸𝐼 𝛿_𝐶 = 𝑣( 𝐿 2) = − 𝑃𝑏𝑥

6𝐿𝐸𝐼 𝐿² − 𝑏² − 𝐿⁴ 4

Deflections by integration of the bending-moment equation

Maximum angle of rotation

Maximum deflection of the beam 𝜃_𝐴 = −𝑣′(0) = 𝑃𝑏 𝐿² − 𝑏²

6𝐿𝐸𝐼 𝑑𝜃

𝑑𝑏 = 0 ; 𝑏 = 𝐿 3 𝜃_{𝐴 max} = 𝑃𝐿² 3 27𝐸𝐼

𝛿_max = − 𝑣 _𝑥=𝑥1 = 𝑃𝑏 𝐿² − 𝑏^{2 3 2} 9 3𝐿𝐸𝐼

; 𝑥₁= 𝐿² − 𝑏²

𝑣′ = − ^𝑃𝑏 3

6𝐿𝐸𝐼 𝐿² − 𝑏² − 3𝑥² =0

0 ≤ 𝑥 ≤ 𝑎 Condition:

0 ≤ 𝑥 ≤ 𝑎 𝑣 = − 𝑃𝑏𝑥

6𝐿𝐸𝐼 𝐿² − 𝑏² − 𝑥² Condition:

Deflections by integration of the shear-force and load equations

𝑉 = −𝑃 2

𝑉 = 𝑃 𝐿 < 𝑥 < 3𝐿 2 0 < 𝑥 < 𝐿

𝐸𝐼𝑣′′′ = −𝑃 2 𝐸𝐼𝑣′′′ = 𝑃

𝑀 = 𝐸𝐼𝑣′′ = 𝑃𝑥 + 𝐶₂ 𝑀 = 𝐸𝐼𝑣′′ = −𝑃𝑥

2 + 𝐶₁

𝐿 < 𝑥 < 3𝐿 2 0 < 𝑥 < 𝐿

𝐿 < 𝑥 < 3𝐿 2 0 < 𝑥 < 𝐿

Deflections by integration of the shear-force and load equations

𝑀 = 𝐸𝐼𝑣′′ = 𝑃𝑥 + 𝐶₂ 𝑀 = 𝐸𝐼𝑣′′ = −𝑃𝑥

2 + 𝐶₁

𝐿 < 𝑥 < 3𝐿 2 0 < 𝑥 < 𝐿

𝑣′′ 0 = 0

𝑣′′ 3𝐿

2 = 0

𝐶₁ = 0

𝐶₂ = −3𝑃𝐿 2

𝑀 = 𝐸𝐼𝑣′′ = 𝑃𝑥 − 3𝑃𝐿 2 𝑀 = 𝐸𝐼𝑣′′ = −𝑃𝑥

2

𝐿 < 𝑥 < 3𝐿 2 0 < 𝑥 < 𝐿

Deflections by integration of the shear-force and load equations

𝑀 = 𝐸𝐼𝑣′′ = 𝑃𝑥 − 3𝑃𝐿 2 𝑀 = 𝐸𝐼𝑣′′ = −𝑃𝑥

2

𝐿 < 𝑥 < 3𝐿 2 0 < 𝑥 < 𝐿

𝐸𝐼𝑣′ = −𝑃𝑥²

4 + 𝐶₃ 𝐸𝐼𝑣′ = −𝑃𝑥 3𝐿 − 𝑥

2 + 𝐶₄ 𝐿 < 𝑥 < 3𝐿 2 0 < 𝑥 < 𝐿

𝐸𝐼𝑣 = −𝑃𝑥² 9𝐿 − 2𝑥

12 + 𝐶₄𝑥 + 𝐶₆ 𝐸𝐼𝑣 = −𝑃𝑥³

12 + 𝐶₃𝑥 + 𝐶₅

𝐿 < 𝑥 < 3𝐿 2 0 < 𝑥 < 𝐿

𝑣′ 𝐿 = 𝑣′ 𝐿

𝑣 0 = 0

𝑣 𝐿 = 0

Deflections by integration of the shear-force and load equations

𝐸𝐼𝑣 = −𝑃𝑥² 9𝐿 − 2𝑥

12 + 𝐶₄𝑥 + 𝐶₆ 𝐸𝐼𝑣 = −𝑃𝑥³

12 + 𝐶₃𝑥 + 𝐶₅

𝐿 < 𝑥 < 3𝐿 2 0 < 𝑥 < 𝐿

𝐶₅ = 0; 𝐶₃ = 𝑃𝐿²

12 ; 𝐶₄= 5𝑃𝐿²

6 ; 𝐶₆ = −𝑃𝐿³ 4

𝐸𝐼𝑣 = −𝑃𝑥² 9𝐿 − 2𝑥

12 + 5𝑃𝐿²

6 𝑥 − 𝑃𝐿³ 4 𝐸𝐼𝑣 = −𝑃𝑥³

12 + 𝑃𝐿² 12 𝑥

𝐿 < 𝑥 < 3𝐿 2 0 < 𝑥 < 𝐿

𝛿_𝐶 = −𝑣 3𝐿

2 = 𝑃𝐿³ 8𝐸𝐼

= +

Method of superposition

𝑣′ = − 𝑞

24𝐸𝐼 𝐿³ − 6𝐿𝑥² + 4𝑥³ 𝑣 = − 𝑞𝑥

24𝐸𝐼 𝐿³ − 2𝐿𝑥² + 𝑥³ 𝑣 = − 𝑃𝑥

48𝐸𝐼 3𝐿² − 4𝑥² 𝑣′ = − 𝑃

16𝐸𝐼 𝐿² − 4𝑥²

= +

= +

Method of superposition

𝑣′ = − 𝑞

24𝐸𝐼 𝐿³ − 6𝐿𝑥² + 4𝑥³ 𝑣 = − 𝑞𝑥

24𝐸𝐼 𝐿³ − 2𝐿𝑥² + 𝑥³ 𝑣 = − 𝑃𝑥

48𝐸𝐼 3𝐿² − 4𝑥² 𝑣′ = − 𝑃

16𝐸𝐼 𝐿² − 4𝑥²

Method of superposition

Appendix G

Method of superposition

Appendix G

Method of superposition

Distributed loads

Angle of rotation Midpoint deflection

Use Table G-2, Appendix G

Method of superposition

Distributed loads Midpoint deflection Angle of rotation

Nonprismatic beams

𝑀 = 𝑃𝑥

2 0 ≤ 𝑥 ≤ 𝐿

2 𝐸𝐼𝑣′′ = 𝑃𝑥

2 𝐸 2𝐼 𝑣′′ = 𝑃𝑥

2

0 ≤ 𝑥 ≤ 𝐿 4 𝐿

4 ≤ 𝑥 ≤ 𝐿 2 Equation for Moment

Conditions

1. At support A (x=0), deflection is zero (v=0) 2. At point C (x=L/2), slope is zero (v’=0)

3. At point B (x=L/4), slope should be continuous 4. At point B (x=L/4), deflection should be continuous

Nonprismatic beams

0 ≤ 𝑥 ≤ 𝐿 4 𝐿

4 ≤ 𝑥 ≤ 𝐿 2

𝑣′ = 𝑃𝑥²

4𝐸𝐼 + 𝐶₁

𝑣′ = 𝑃𝑥²

8𝐸𝐼 + 𝐶₂

2. At point C (x=L/2), slope is zero (v’=0)

𝐶₂ = − 𝑃𝐿² 32𝐸𝐼

3. At point B (x=L/4), slope should be continuous

𝑃 4𝐸𝐼

𝐿 4

2

+ 𝐶₁ = − 3𝑃𝐿²

128𝐸𝐼 𝐶₁ = − 5𝑃𝐿² 128𝐸𝐼

Nonprismatic beams

0 ≤ 𝑥 ≤ 𝐿 4 𝐿

4 ≤ 𝑥 ≤ 𝐿 2 𝑣′ = 𝑃𝑥²

4𝐸𝐼 − 5𝑃𝐿² 128𝐸𝐼 𝑣′ = 𝑃𝑥²

8𝐸𝐼 − 𝑃𝐿² 32𝐸𝐼

𝑣 = − 𝑃

128𝐸𝐼 5𝐿²𝑥 −32𝑥³

3 + 𝐶₃

𝑣 = − 𝑃

32𝐸𝐼 𝐿²𝑥 − 4𝑥³

3 + 𝐶₄

0 ≤ 𝑥 ≤ 𝐿 4 𝐿

4 ≤ 𝑥 ≤ 𝐿 2

1. At support A (x=0), deflection is zero (v=0) 𝐶₃ = 0

4. At point B (x=L/4), deflection should be continuous

− 𝑃

32𝐸𝐼 𝐿² 𝐿

4 −4 3

𝐿 4

3

+ 𝐶₄ = 𝑣 𝐿

4 = − 13𝑃𝐿³

1536𝐸𝐼 𝐶₄ = − 𝑃𝐿³ 768𝐸𝐼