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Chapter 10: Deflection of beam (Ch. 12 in Shames)

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445.204

Introduction to Mechanics of Materials (재료역학개론)

Chapter 10: Deflection of beam (Ch. 12 in Shames)

Myoung-Gyu Lee, 이명규

Tel. 880-1711; Email: myounglee@snu.ac.kr

TA: Chanmi Moon, 문찬미

Lab: Materials Mechanics lab.(Office: 30-521) Email: chanmi0705@snu.ac.kr

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Contents

- Differential equations for deflection of symmetric beams - Statically indeterminate beams

- Superposition methods - Energy method

2

(3)

Differential equations

• Recall pure bending

• For general loading, the above equation gives local radius of curvature of the neutral surface

• Normal stress (or normal strain) results in deflection of the beam

• Radius of curvature as a function of deflection, from analytic geometry

zz z

R EI

= M

( )

2

2 2 3/ 2

1

1

d v dx

R dv

dx κ

= =    +   

2 2

1 d v R = dx

Simplification/small deformation

(4)

Differential equations

• Basic differential equations

• From

• Therefore,

zz z

R EI

= M

2

2

1 d v R = dx

&

2 2

z zz

M d v

dx = EI

2

zz 2 z

EI d v M dx =

or

z

y

dM V

dx =

and y y

dV w

dx = −

2 2

z

y

d M w

dx = −

2

2 2

2 2 2

z

zz y

d M

d d v

EI w

dx dx dx

 

= = −

 

 

4

zz 4 y

EI d v w dx = −

or

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Differential equations: example 12.1

2 2

z zz

M d v

dx = EI

• Boundary conditions

dv 0

v = dx = at x = L

(6)

Differential equations: example 12.2

( ) ( )

1

1

n n

n

P x a

P x a dx C

n

+

− = +

∫ +

(7)

Differential equations: example 12.2

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Differential equations: example 12.2

Boundary conditions

(9)

Differential equations: example 12.2

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Differential equations: example 12.4

Pin connected beam Q: Find max. deflection

Solution step

1. Apply moment equilibrium at the position of pin And determine the reaction force at A

2. Obtain bending moment along the beam axis Note: divide the region at the pin position

3. Apply boundary conditions

BCs. 0 deflection at x=0 & x=9, zero slope at x=9 Apply patch condition.

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Differential equations: example 12.4

Pin connected beam Q: Find max. deflection

Solution step

1. Apply moment equilibrium at the position of pin And determine the reaction force at A

2. Obtain bending moment along the beam axis Note: divide the region at the pin position

3. Apply boundary conditions

BCs. 0 deflection at x=0 & x=9, zero slope at x=9 Apply patch condition.

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Differential equations: example 12.4

Pin connected beam

Q: Find max. deflection

(13)

Statically indeterminate problem: example 12.5

Solution step

1. Calculate bending moment as a function of x with unknown R1 (considered as redundant constant) 2. Obtain deflection and slope for each

region between 0~5, and 5~10

3. Apply proper boundary condition at x=0, x=10, and patch conditions at x=5 4. Obtain additional unknowns for R2 and

M2 at the wall (or x=10)

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Statically indeterminate problem: example 12.5

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Superposition method

Homework!!!!

Derive the left equations!

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Superposition method

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Energy method

• Strain energy related 1) normal stress, 2) transverse shear stress

• For normal stress

• For shear stress resulting from Vy

z xx

zz

M y σ = − I

2 2 2 2

* 2

0 2 0 2

2 2 2

L L

xx z z

V A zz zz A

M y M

U dv dA dx y dA dx

E EI EI

σ

= ∫ = ∫ ∫   = ∫  ∫ 

   

2

*

0

2

L z

zz

U M dx

EI

 

= ∫  

 

2

*

2

xy

U

V

dv

G

= ∫ τ

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Energy method

• Shear stress may vary across the cross section as a function of both y and z

• Therefore, instead of determining the exact distribution of shear stress, use “tabulated shape factors”

• The shape factors correlate the energy computed using actual shear stress distribution with that of a cross section with equal area but with a constant distribution of shear stress

2 2

*

2 2

xy actual xy simplified

xy xy

V y V

U dv dv

G G

τ τ

τ τ

α

−> −>

= ∫ = ∫

2 2

*

0 2

2

0

2

L y L y

y y

A

V V

U dA dx dx

GA GA

α α

= ∫ ∫  = ∫

 

 

xy simplified

V

y

τ −>

= A

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Energy method – Example 12.11

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Energy method

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Energy method- Example 12.12

(22)

Homework by yourself

Solve examples 12.3, 12.6, 12.8, 12.13, 12.14

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