Ch5_Derivation for CantileverLast updated at 2014-04-10

What

(Task)

Why

(Purpose)

When

(Flow)

How

(Method)

Where

(Place)

Who

(Person)

Derivation for

Cantilever

Derive for the

equation, i.e.

k=3 EI

L3

1 Define the problem

2 Draw the free body

diagram

2.1 Force equilibrium

2.2 Moment equilibrium

3 By deflection

equation,

3.1 Put Mx=−Fx

3.2 Integrate by indefinite

integral once

3.3 Integrate by indefinite

integral again

4 Consider B.C.,

4.1 By B .C .: y ' L=0 ,

4.2 By B .C .: y L=0 ,

4.3 ¿C1 ,C2 into y ,

4.4 SinceF=ky ,

6. Solve the deflection

equation by define integral

6.1. Graphical method

6.2. Right example

6.3. Wrong example

1. Draw free body diagram

2. Deflection equation, i.e.

d2 yd x2 =

−M x

EI

3. Integration4. Boundary

condition (B.C.)

5. Hooke’s law

Ch5_Derivation for CantileverLast updated at 2014-04-10

1 Symbol meaning

Symbol SI unit Meaning

x m Displacement

y m Deflection at displacement x

L m Total length

F N(kg·m/s2) Force

m kg Mass

a m/s2 Acceleration

V N(kg·m/s2) Shear force

k N/m Stiffness

E Pa Young modulus

I kg·m2 Moment of inertia

2 Define the problem

y (0 )=?

Ch5_Derivation for CantileverLast updated at 2014-04-10

3 Draw the free body diagram

=

3.1 Force equilibrium 3.2 Moment equilibrium

∑ F=ma

F−V=0

V=F

∑M=I α

Vx+M x=0(SinceV=F )

M x=−Fx

4 By deflection equation,

4.1 Put M x=−Fx

d2 yd x2 =

−M x

EI

¿−(−Fx )EI

¿ FxEI

4.2 Integrate by indefinite integral once

dydx

= FEI ( x2

2 )+C1

¿ F2EI

x2+C1

Ch5_Derivation for CantileverLast updated at 2014-04-10

4.3 Integrate by indefinite integral again

y= F2 EI ( x

3

3 )+C1 x+C2

¿ F6 EI

x3+C1 x+C2

5 Consider B.C.,

{y ' (L )=0y (L )=0

5.1 By B .C .: y ' (L )=0 ,

F2EI

L2+C1=0

C1=−F2 EI

L2

5.2 By B .C .: y (L )=0 ,

F6 EI

L3+(−F2 EI

L2)L+C2=0

−F3EI

L3+C2=0

C2=F

3 EIL3

Ch5_Derivation for CantileverLast updated at 2014-04-10

5.3 ¿C1,C2into y ,

y= −F6 EI

x3− F2EI

L2 x+ F3EI

L3

y (0 )= −F6 EI

(0 )3−( F2EI

L2) (0 )+ F3 EI

L3

¿ F3EI

L3

5.4 SinceF=ky ,

y= k y3 EI

L3

k=3 EI

L3

6. Solve the deflection equation by define integral

6.1. Graphical method

y} = {{d} ^ {2} y} over {d {x} ^ {2} ¿ y '=dydx

y

`

Ch5_Derivation for CantileverLast updated at 2014-04-10

6.2. Right example

d2 yd x2 =

FEI

x

∫x

Ld2 yd x2 dx=

FEI

∫x

L

x dx

[ dydx ]x

L

=FEI [ x

2

2 ]x

L

dydx|x=L

−dydx

= F2 EI

[L2−x2 ]

0−dydx

= F2 EI

[L2−x2 ]

dydx

= F2 EI

[ x2−L2 ]

6.3. Wrong example

d2 yd x2 =

FEI

x

∫0

xd2 yd x2 dx=

FEI

∫0

x

x dx

[ dydx ]0

x

= F2EI

x2

dydx

−dydx |x=o

= F2EI

x2

Sincethe graphabove ,dydx|x=o

isunknown ,cannot be solved .