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

Contents

- Differential equations for deflection of symmetric beams- Statically indeterminate beams- Superposition methods- Energy method

2

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

EIRM

=

( )

22

3/22

1

1

d vdx

R dvdx

κ= = +

2

2

1 d vR dx=

Simplification/small deformation

Differential equations

• Basic differential equations

• From

• Therefore,

zz

z

EIRM

=2

2

1 d vR dx=&

2

2z

zz

Md vdx EI

=2

2zz zd vEI Mdx

=or

zy

dM Vdx

= and yy

dVw

dx= −

2

2z

yd M wdx

= −

22 2

2 2 2z

zz yd Md d vEI w

dx dx dx

= = −

4

4zz yd vEI wdx

= −or

Differential equations: example 12.1

2

2z

zz

Md vdx EI

=

• Boundary conditions

0dvvdx

= = at x L=

Differential equations: example 12.2

( ) ( ) 1

1

nn

n

P x aP x a dx C

n

+−− = +∫

+

Differential equations: example 12.2

Differential equations: example 12.2

Boundary conditions

Differential equations: example 12.2

Differential equations: example 12.4

Pin connected beam

Q: Find max. deflection

Solution step

1. Apply moment equilibrium at the position of pinAnd determine the reaction force at A2. Obtain bending moment along the beam axisNote: divide the region at the pin position3. Apply boundary conditionsBCs. 0 deflection at x=0 & x=9, zero slope at x=9

Apply patch condition.

Differential equations: example 12.4

Pin connected beam

Q: Find max. deflection

Solution step

1. Apply moment equilibrium at the position of pinAnd determine the reaction force at A2. Obtain bending moment along the beam axisNote: divide the region at the pin position3. Apply boundary conditionsBCs. 0 deflection at x=0 & x=9, zero slope at x=9

Apply patch condition.

Differential equations: example 12.4

Pin connected beam

Q: Find max. deflection

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)

Statically indeterminate problem: example 12.5

Superposition method

Homework!!!!Derive the left equations!

Superposition method

Energy method

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

• For shear stress resulting from Vy

zxx

zz

M yI

σ = −

2 2 2 2* 2

0 02 22 2 2L Lxx z z

V A Azz zz

M y MU dv dA dx y dA dxE EI EI

σ = = =∫ ∫ ∫ ∫ ∫

2*

0 2L z

zz

MU dxEI

= ∫

2*

2xy

VU dvGτ

= ∫

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 2xy actual xy simplified

xy xyyV VU dv dv

G Gτ τ

τ τα

−> −>

= =∫ ∫

2 2*

0 202 2

LL y y

y yA

V VU dA dx dx

GA GAα α

= =∫ ∫ ∫

xy simplified

yVAτ −>

=

Energy method – Example 12.11

Energy method

Energy method- Example 12.12

Homework by yourself

Solve examples 12.3, 12.6, 12.8, 12.13, 12.14

22