III. Basis of Mechanics

3.1 General Consideration

3.2 Newton’s Law of Motion and Statics

3.3 Unit and Dimension

3.4 Material

3.5 Solid Mechanics of Slender Members

3.6 Uniaxial Loading

3.7 Torsion

3.8 Beam Theory

3.9 Buckling

3.10 Special phenomenon

3.1 General Consideration

<Continuum>

Basis of mechanics

Solid

Three major factors

<Grain and boundary><Grain>

Rigid-body : No change in distance between particles occurs under external force.

Elastic deformation: Deformation due to external force disappearing when it removes.

Plastic deformation: Deformation due to external force remaining when it removes.

Fluid including gas

Material and Continuum, Deformation of material

Force Material Displacement

Rigid-body motion

Deformation

Force

Rigid-body motion

Deformation

Displacement or motion

Body force Gravity, Magnetic force

ForceInternal

External

Action and reaction force

Traction Exerted load, Reaction force

Material: Set of particles continuously distributed Continuum

Rigid-body: No relationship between force and displacement is needed.

Elastically deformable body: Linear elastic(Hooke’s law) and Non-elastic,

Isotropic(Common material) and Anisotropic (Composite material)

Plastic body: Yield criterion, Isotropic hardening, Kinematic hardening

Solid

Fluid including gas

Some details of the 3 major factors

Classification of solid mechanics (SM)

Displacement Rigid-body Statics

Rigid-body

dynamicsGeneral SM

Narrow SM

Flexible

dynamics

Vibration

Plasticity

Elasticity

Fracture

mechanics

Deformation

3.2 Newton’s Law of Motion

and Statics

F R 0i ij

j

F

F

F

F

F

R

R

R

RR

RR

RR

R

RRR

RR

R

R R

RR

F

r

r

A

Newton’s law of motion

R 0r ij

i

i j

Newton’s law

2nd

law

Sum of all the forces exerting on a particle ( f ) is

equal to the acceleration ( a ) multiplied by the mass

( m ), i.e., f = ma.

3rd

law

Action and reaction law: Two internal forces exerting

between two particles have the same magnitude and

line of action and the opposite direction.

•

•

•

•

:

:

i

ij

i

j

i

A material is a set of infinite number of particles

F Sum of external force exerting

on particle

R Internal force exering on particle

from particle

All forces are bounding vectors.

.

, 1,2, ,i j

Requirement

on equilibrium

• The above requirement of equilibrium should satisfy

for all subsystem as well as the whole system.

Leading to differential equations, for example,

equations of equilibrium.

or

or

23 32

2 3r R r R o

( )F R 0 F R 0i ij i ij

i j i i j

R Rij ij R 0

ij

i j

F 0i

r F R 0 r F r R 0i ij i ij

i i i

i j i i j

r F 0i

i

i

F 0i F 0

M 0A r F 0i

i

i

5F

2 0 (O)5 0

3 0 (X)

xx y

y

F

F

F

F

F

R

R

R

RR

RR

RR

R

RRR

RR

R

R R

RR

F

r

r

A

1 12 13 14 15 16

2 21 23 24 25 26

3 31 32 34 35 36

4 41 42 43 45 46

5 51 52 53 54 56

6 61 62 63 64 65

F + R + R + R + R + R = 0

F + R + R + R + R + R = 0

F + R + R + R + R + R = 0

F + R + R + R + R + R = 0

F + R + R + R + R + R = 0

F + R + R + R + R + R = 0

F = 0i ij ji

R = -R

1 12 13 14 15 16

1 1 1 1 1 1

2 21 23 24 25 26

2 2 2 2 2 2

3 31 32 34 35 36

3 3 3 3 3 3

4 41 42 43 45 46

4 4 4 4 4 4

5 51 52 53 5

5 5 5 5 5

r F + r R + r R + r R + r R + r R = 0

r F + r R + r R + r R + r R + r R = 0

r F + r R + r R + r R + r R + r R = 0

r F + r R + r R + r R + r R + r R = 0

r F + r R + r R + r R + r R

4 56

5

6 61 62 63 64 65

6 6 6 6 6 6

i

i

+ r R = 0

r F + r R + r R + r R + r R + r R = 0

r F =0 ij ij

i jr R = -r R

※Actual number of particles is infinite.

Derivation of requirement on equilibrium

using a six-particle body

Subsystem

※ The requirement of equilibrium should satisfy for any subsystem.

This statement is the same with the following statement: The

requirement of equilibrium should satisfy for arbitrary infinitesimal area

in 2D or volume in 3D, leading to differential equation. Note that we

should define stress, i.e., force per unit area to define the force

exerting on boundary of the infinitesimal area in 2D or volume in 3D.

2

yF2

xF

1

xF

1

yF

4

yF4

xF

3

xF

3

xF

y

x

0

0 0

0

x

y

z

F

F F

F

Requirement on equilibrium in 2D and 3D

,

,

,

0

0 0

0

A x

A A y

A z

M

M M

M

2D plane 3D space

00

0

x

y

FF

F

,0 0 A zAM M

3 equations 6 equations

Examples

Tip

x,y, and z can be

replaced by any

three independent

directions

D

B

E

F

C2L2L

2L

P

A

AR

BR

W

x

y

45 45

BDf

DDFf

ADf

'x

'y

A AB B

i ir r r

0

0

i

A

A i i

i

F

M r F? BM

Caution in applying requirements on equilibrium

( )

0

0, 0 0

0, 0 0

B

B i i

i

A AB

i i

i

A AB

i i i

i i

A AB

i

A B

A B

M r F

r r F

r F r F

M r F

F M M

M M F

andA B

i ir r

AB

r

are dependent

on the position of point i,

while is fixed.

A B

AB

r

B

irA

ir

i iF

1F1

A

r1

B

r

nF

2F

(If )ABr F

Statics and solid mechanics with three factors of mechanics

Ground and support

Subsystem

Entire system

Separation of subsystems

Essential separation of subsystems from support or ground

Division of subsystems if needed

-When number of equations should be greater than that of new knowns

-When internal resultant forces is to be seen

Newtonian mechanics Analytical mechanics

based on vector quantities based on energy

Factor Statics Solid mechanics

Force Requirement on equilibrium Equation of equilibrium or motion

Deform-

ation

Rigid-body Displacement-strain relation, Geometric

compatibility

Material Rigid-bodyElastic body: Hooke’s law

Plastic body: Plastic flow rule

BC Geometric(essential) BC Geometric(essential) BC, mechanical(natural)

BC

ground

ground

Subsystem 1

Subsystem 1 Subsystem 2

Statics

Simply supported beam

Free-body diagram and application of RE on it

0 ; 0( / )

0 ; 0 0( / )

0; 0

y A B

A

x

B

A B

F R R PR b L P

FR a L P

M L R a P

F.B.D. No. 1

F.B.D. No. 2

0 ; 0 0

0 ; 0 ( / )

( / )0; 0

y A B A

x A A

BA B

F V V P H

F H V b L P

V a L PM LV a p

F.B.D. No. 3

F.B.D. No. 5

0; 0( / )

0 ; 0 0( / )

0; 0

A B

A

x

B

B A

M L R a PR b L P

FR a L P

M L R b P

Requirement on equilibrium

P

L

ba

F.B.D. No. 6

F.B.D. No. 4

P

BVAR

a

P

BRAR

a b

P

BRAR

a b

b

P

BYAR

a b

P

( )

A

A

H

X ( )B BV Y( )A AV Y

a b

( )

A

A

H

X

P

( )B BV Y( )A AV Y( )

B

B

H

X

a b

No. 1

No. 2

No. 3

0 ; 0 0

0 ; 0

0 ; 2 ( / 2) 0

0 ; ( / 2) 0

0 ; 2 (3/ 2) 0

x

y A B C

A B C

B A C

C A B

F

F R R R P

M L R L R L P

M LR LR L P

M LR LR L P

Statically indeterminate system

①

③

④

⑤

Eqs. ① ~ ⑤ have only two independent equations.

Infinite number of equations can be made applying

moment balance requirement at arbitrary point.

Number of unknowns > number of equations

⇒ Statically indeterminate system

3 points supporting beam

F.B.D.

Requirement on equilibrium

P

LA B/ 2L/ 2L

P

AR BR CR

②

Reactions in a statically indeterminate system can be determined

not by statics but by solid mechanics.

Division of total system into subsystems

xR

yR

xR

yR

M

Division of total system into subsystems

Division of total system into subsystems

20 ; 2 0

2

20 ; 2 0

2

,

x BD

y CD

BD CD

F f P

F F P

f P f P

20 ; 0

2

20 ; 0

2

2 ,

x CE DE

y DE

DE CE

F f f

F P f

f P f P

20 ; 0

2

20 ; 0

2

2 , 2

x AC BC

y BC

BC AC

F P f f

F P f

f P f P

. 2 ( )

2 ( ),

AC

BC

Ans f P in tension

f P in compression

A C

B D

E

CEf

DEf

ACf

BCf

PP

P

P

At D

At E

At C

BDf

CDf2P

Forces in members of a truss structure

0 ; sin30 0

0 ; cos30 300 0

0; 8cos30 4 cos30 300 0

x B

y A B

B A

F T R

F R R

M R

2D static problem

F.B.D.

Tension of cable, T ?

Applying RE on FBD

30

T

AR

300kg

BR

3 6 , 5.5 , 3 4.5CA CD CBr i k r i r i k

20; 046.2

2.52 046.2

2 046.2

0; ( )

(2 )

( ) 0

x x

y

z z

CA x y zC

CD

CB x z

TF A B

TA

TA B

M r A i A j A k

r j

r B i B k

.

Cable tension T ?

F.B.D. Requirement on equilibrium

( 2.5 ) ( 3 6 ) 2 2.5 6

(2 2.5 6 ) / 46.2

CE i j i k i j k

CE ce i j kCE

Vector expression of line segments CA and CE

xA yA

zA

A

D

C

B

xB

zB

T

2kN

x y

z

E

E

D

3D static problem

3.3 Unit and Dimension

System of Units

Unit and dimension

Science unit: Basis units are length, mass, time, current, temperature, etc.

Conventional unit: Basis units are length, weight, time, current, temperature, etc.

Dimension

[Length] ≡ [L], [Mass] ≡ [M], [Time] ≡ [T], [Weight] ≡ [F]

※ Three basis units characterize the system of units.

Basis units of system of science units: [L], [M], [T]

Basis units of system of conventional units : [L], [F], [T]

British unit Basis unit: ft, lb, sec

SI unit (The International System of Units)

British unit belongs to conventional unit, i.e, lb means basically lbf(pound force).

Unit for mass : slug = lb · sec2/ft

Basis unit: m, kg, s

In SI unit, kg means basically kgm(kilogram mass).

Entangled or mixed use of conventional and science units

Unit for force: N = kg · m/s2

The other countries are using SI unit. However, most people are using the SI unit like British unit, i.e,

they are using kg (i.e., kg force) instead of N (Newton).

Sometimes mass is expressed in kg · s2/m, where kg means kgf.

British unit and SI unit

Many countries belonging to the Commonwealth of Nations are using British unit.

T(1012), G(109), M(106), k(103), c(10-2), m(10-3), (10-6), n(10-9), p(10-12)

Basis unit

Length : m, ft

Assembled unit

rad

Complement unit

Velocity : m/s, ft/sec 1N = 1kg · m/s2 , lb, kg

Pressure : 1Pa = 1N/m2, psi = lb/in2, psf = lb/ft2, kg/m2, kg/mm2

Mass : kg, slug Time : s, sec

Current : A Temperature : K

Work : 1J = 1N · m, lb · ft, kg · m Power : 1W = 1J/s, lb · ft/sec, kg · m/s, PS, hp

Moment : N · m, lb · ft, kg · m

Basis unit, complement unit, assembled unit

Factors for unit

Relationship between the two systems of unit

1ft = 0.3048 m, 1 in = 25.4 mm, 1lb = 0.4536 kg

Voltage : V

All units should be expressed in Roman characters while variables in Italic or

bold Roman characters.

Example 1

, 1kg/mm, 1kgn

kk m

m

4

f m m

2

m m m

1kg /mm 1kg 1kg 10 mm

1kg 1kg mm 1kg mm s

1100

s

n

g

m

k

Mistake due to gravitational acceleration.

1 50(Hz)

2n

k

m

2m/s27kg/cm

2cm

Calculation of blow energy per stroke in hammer forging

E = (m × g + p × A)H

E = (56500kg×9.8+7kg/㎠×17671.5㎠)0.75m

=(553700kg +123700kg)0.75

E = 508,050kg.m → 4,982,250 N-m

From a lecture note floating in internet

Physical quantities

and dimensions

SI U.S customary

mass [M] kg slug

length [L] m ft

time [T] s sec

force [F] N lb

22

2

sec/.806.9),sec/()()(

sec/

mgmgkgmweightW

mkgNmaF

2

2 2

N kg m /s

( ) (kg) (m /s ), 9.806m /s

m

W weight m g g

F a

Original

Modified

3.4 Material

Tensile test

Engineering strain (mm/mm)

En

gin

ee

rin

gstr

ess

(MP

a)

0 0.1 0.2 0.3 0.40

200

400

600

800

1000

1200

Experiment (SCM435)

Analysis (SCM435)

Experiment (ESW95)

Analysis (ESW95)

Experiment (ESW105)

Analysis (ESW105)

Tensile test

0 0

0

0

0

0 0

,

ln ln 1

1

e e

f

t e

t

f

e e

P

L A

L

L

AP P

A A A

LP

A L

Predictions of tensile test

Hooke's law in uniaxial loading

,t tE E

0L

0

LLf

P

Current

area A

P

Initial

area A0

,x x xx xxE E

: engineering

: true

e

t

Engineering strain, Engineering stress

(Engineering = Conventional=Nominal)Before necking occurs

Tensile test

Engineering stress-engineering strain curve

Definition of and ee

: Yield strength

: Tensile strength

: Elongation

Y

U

max 100(%)

lat t

long a

Poisson’s ratio

0

xx

y z x

xy yz zx

E

v

0

y

y

x z y

xy yz zx

E

v

0

zz

x y z

xy yz zx

E

v

1

1

1

1 1 1, ,

x x y z

y y x z

z z x y

xy xy yx yz zx zx

T

T

T

vE

vE

vE

G G G

Small deformation

Principle of superposition

Coefficient of

thermal

expansion

x

y

z

xy

z x

y

z

isotropic

0i

Hooke’s law

for an

isotropic

material

xx xx

xx1

Poisson’s ratio1

1 1 ( )xx

yy zz xx xxE

Generalized Hooke's law for an isotropic material

2(1 )

EG

Thermal load and shrink fit

779MPa -18MPa

CL

3.5 Solid Mechanics

of Slender Members

Mechanics of slender members

Truss member or rod – Uniaxial loading

Circular shaft - Torsion

Stresses in beam

Axial force

Normal stress:

Twisting moment

Shear stress:

Beam

Lateral force – Bending

moment and shear force

Normal and shear stress,

I-beam, H-beam

z

Tr

J

, ,bx xy xz

zz zz zz

M y VQ VQ

I bI tI

Compressive axial force

Buckling

Column

x xx

F

A

PP

ColumnBuckled

t

b

x

Flange

Web

x

( )z x

:

, :

:

, :

xx

xz xy

xx

xz xy

M twisting moment

M M bending moment

F axial force

F F shear force

Definition of axes Definition of cross-section

2D

First subscript: Direction of face

Second subscript: Direction of force

x

y

z

Axis of symmetry

Neutral axis

Negative x-face

x

y

z

Positive x-face

1x

Force and moment exerting cross-section y

x

V

F

bM

V

F

bM

xxMxzM

xyM

y

xz

xxF

xyF

xzF

, , xx xy b xzF F V F M M

3D

Internal force

3.6 Uniaxial Loading

Deformation and stress in uniaxial loading

Spring) Rod, truss member

P kx

1x Pk

2 21 1 12 2

U kx Pk

, ,AE PP E EL A L

L PAE

2 21 12 2eq

LU k PAE

eqAEkL

Force-deformation

relationship

Hooke’s law

Displacement

Strain energy

P

x1

k

/P A

/ L

1

E

,P xP ,P P

: spring constantk : modulus of elasticityE

Stress

Strain

Displacement, strain and stress in rod

x

( )u x

P

0L 0A

x

( )x x

x

( )x x

x

0/ L

0/P A

( )xF x

x

P

P 2P

( )u x

x

( )x x

x

( )xF x

x

0/ L02 / L

P2P

( )x x

x

0/P A02 /P A

2P P

( )u x

x

( )x x

( )xF x

x

3P

x0/ L

02 / L0/ L

P

P

2PP

( )x x

x0/P A

02 /P A0/P A

P

( )u x

x/ 2

( )x x

x

( )xF x

x

0/ L

P

( )x x

x

0/P A

P P

( )u x

x

( )x x

x

/ 4

( )xF x

x

0/ L

P

( )x x

x

0/P A

( )F x ( )F x dx

( )u x

x

( )x x

x

du

dx

( )x x

x

x x x

duE

dx

( )xF x

x

Rod

(X) (X) (X) (O)P/n n 개

UP

Down UP

Down

P

P

P

P

P

P

P

P

P

P P

P

P

P

Saint-Venant’s principle

=

=

●●●

Principle of

symmetry

⊙ Two statically equivalent loads

have nearly the same influence on

the material except the region near

to the load exerting area.

End Effect

End-Effect

Non-end Effect

0

0

3

2

0

2

2 0

1

2

Internal force, stress vector in tensile test

θ 60θ 45

θ 30θ 90

θθ

θ 0

θ=0

n

nnn

n

0

0A

0

θo 0

sin

AA

cos sini j n

0 0P A i

θ

( )

0 sinP

iA

n

(n)T t

θ

=

o o o oo

o o0 sin

0 cos sin nt

2

0 sin nn

x

y

0

3

4

0

1

4

i

j

0 0P A

nt

Stress vector

( ) 2

0 0sin cos sinn t nt

0 0P A

θ

0

0

3

4n

0

1

2n

0

1

4n

Internal force and stress components

θ 60θ 45

θ 30θ 90

θ

θ

θ

θ = 0

n

nnn

n

0

0A

0

o

o o o oo

θ

0

3

4nt

0

1

2nt

0

3

4nt

= sinN P

= cosT P

0

sin

AA

2 2

0

0

sin sinnn

N P

A A

0

0

cos sin cos sinnt

T P

A A

t

t

t

P P P P

θ

n

o

t

NT

P

0

Normal stress

Shear stress

0 0P A

( ) 2

0 0sin cos sin nn ntn t n t nt

1 2 2 3

0

( ) 0

y B C D

B C D

F P R R R

M l P l R l l R

32 2 3

2

: ( ): (1 )C D D C

ll l l

l

5 6,C DC D

C C D D

R l R l

A E A E

※ x-directional displacement is neglected under

small deformation.2 2 3 4: ( ):C El l l l

④

⑤

③

①

②

<F. B. D.>

Force equilibriumⅠ

Geometric compatibilityⅡ

Force-deformation relationshipⅢ

Reaction and displacement of point E

③④→⑤:

, ,B C DR R R

5 3

6 2

(1 )D CD C

C D

A R l lR R

A R l l

①,②,⑥ →

⑥

P

A B C D E

1l 2l 3l 4l

5l 6l

PBR

CRDR

B

BED

C

2l 3l 4l

Example

1 1 1 1

2 2 2 2

(- / )

(- / )

BA

CB

L P EA T L

L P EA T L

1 21 2

1 2

( ) ( )

CA BA CB

L LP T L L

EA EA

Geometric compatibility

0CA

1 2 1 2 1 2

1 2 1 2 2 1

1 2

( ) ( )

( )

E T L L A A E T L LP

L L L A L A

A A

①

② From Eqs. ① and ②

< Method Ⅱ>

1 2

1 2

1 2

( )

( ) ( )

CA

CA CA

CA

dueto T T L L

T PPL PLdueto P

A E A E

<F. B. D.>

PP

ⅡForce-deformation relationshipⅢ

1 1 1, ,A L E

1 1,

2 2 2, ,A L E

2 2,

A B C

T

Statically indeterminate system – thermal load

< Method I>

0 ;xF

< Method Ⅰ>

02 sin 0TF rpb d

< Method Ⅱ>

2 2 0T TF p b r F pbr

( )2 ( ) ( )2 2

( )T R

t tpbr r pr r

bt E tE

2 ( ) 2 2R Rr r 2

TR

<F. B. D.> Force equilibriumⅠ

Force-deformation relationshipⅡ

d

sFsF

0sF TFTF

TF TFp

Deformation of cylindrical pressure vessel

( ) ( ) , sin , cos 12 2 2

dTT T

d

F N

0 ; sin( / 2) ( )sin( / 2) 0

0 ; cos( / 2) ( )cos( / 2) 0

rF N T T

F N T T

Applying RE on the FBD and governing equation

Law of Coulomb friction

Magnitude:

Direction: Preventing relative motion

: Coefficient of Coulomb friction,

2

02

0

dNT

N T T dTdT

dN T dN T

d

0

ln

dTd

T

T C

T Ce

Solving

0

0

(0)T T

T T e

Boundary condition

Brake band N

vF

N

N

NμΔN

Tension in belt

, , ,P AEE E PA L L

,PL L PAE AE

2

21 1 1/2 2 2 2

12

eqP L P PL PU k L AL VAE A AE A L

UuV

eqAEkL

Force-deformation

relationship

Hooke’s law

Strain energy

2

2 2

2

1 12 2

1 12 2

V V V

L A L

U udV dV dVE

P PdAdx dxAEA E

/P A

/ xduL

dx

1

E

,P P

: modulus of elasticityE

Stress

Strain,P xP

2 2

21 1 12 2 2

x x

V L A L

du duU E dV E dAdx EA dx

dx dx

strainenergydensity(function)

eqk

Strain energy in rod

Generalization of uniaxial loading

0 0

0 0

0 0

x

x

x

v

v

0

' ' ( ) ( )lim xx

A B AB u x x u x du

AB x dx

Strain in rod under uniaxial loading

:Load per unit lengthq x

x x

( )u x ( )u x x

x x

x x

A BUndeformed

( )q x x( )F x (( ) )

dFF xF x x x

dx

2( ) ( ) 0( )x x

xq x dt q x x x

Deformed

( ) ( ) ( ) 0

( )

xF F x q x x F x x

dFq x

dx

E

( ) ( )d du

AE q xdx dx

Governing equation

Ⅲ Stress-strain relation

( )du

F x A AE AEdx

Hooke’s law

( )dF

q xdx

Ⅰ Force equilibrium

3.7 Torsion

Torsion of cylindrical shaft, problem definition

Cross-section: Circular shaft (CS)

Role of CS: Power transmission, spring, etc.

Assumption to apply rule of symmetry

End-effects are negligible (Saint-Venant Principle) Uniform cross-section

Geometry and material are axisymmetric

Symmetric expansion and contraction are neglected

Lengthening and shortening are neglected

Shaft

tM

,t

M T

,t

M T

upsidedown

Cavity

tM

tM

tM

tM

Summary Diametrical straight line remains straight line

Plane section, perpendicular to the central line, remains plane

Rule of symmetry

Twisting moment, torque

z

x

y

z

r

z

r

( , , )x y z•

y

x

( , , )r z

cos

sin

x r

y r

z z

Reference coordinate systemLocal coordinate system

Strain components

Relation of strain and angle of twist

0rx

r z plane

r plane

0rr zz

0r r 0rz zr

z

z

z z

r z

rz

dr

dz

xx xy xz rr r rz

yx yy yz r z

zr z zzzx zy zz

: Angle of twist

: Rate of twistd

dz

r

( )z z

( )z

z

z

z

r rz

z Shear strain and angle of twist

Normal strain:

From assumption

Shear strain: Shear strain:

z

r

z

z

: Angle of twist

: Rate of twistd

dz

Stress components

z planez

rz r plane

r plane

r face

( ) face

zz

zr

z

rz

rr

r

z

r

z

r

z face

yface

xface

zz

zyzxyz

yyyxxy

xx

xz

z

x

y

z face

y z plane

z

xy

z x plane

x y plane

x-y-z coordinate

system Cylindrical coordinate

system

rr r rz

r z

zr z zz

xx xy yz

xy yy yz

zx zy zz

Lz z

dG Gr Gr

dz L

Torsion test

Hooke’s law

G

1

1

E

2(1 )

EG

E

Lxx xx

uduE E E

dx L

Tensile test

G

1E

1E

1E

2 1 1E

2 1 1E

2 1 1E

rr rr zz

zz rr

zz zz rr

r r r

z z z

zr zr zr

G

G

G

0 0 0

0 0 /

0 / 0

rr r rz

r z

zr z zz

rd dz

rd dz

0 0 0

0 0 /

0 / 0

rr r rz

r z

zr z zz

Grd dz

Grd dz

Homogeneous shaft

Requirement on equilibrium

r

dr

2dA r dr

dF dA

dAA

r

2

2

tA A A

tA

t tz z

dM rdF r dA Gr dA

dz

d dM G r dA GJ

dz dz

M M rd

dz GJ J

Stress

1 1A G2 2A G

rr

GJ

1 2

2

2 2

1 2 1 1 2 2

1 1 2 2 1 1 2 2

( )

,

tA A A

tA A

t i t

dM rdF r dA Gr dA

dz

d dM G r dA G r dA G J G J

dz dz

M G rMd

dz G J G J G J G J

02 3 4 4

02 ( )2i

R

iA R

J r dA r dr R R

1 1

2 2

i

G if r AG

G if r A

Composite shaft

Polar second moment of inertia

at the cross-section

: Torsional

rigidity

dF

1 2

max

1[( ) ]

( )( / 2)

CA B C C

B C

T T L T LGJ

T T d

J

< F.B.D. >

Example 1 – Statically determinate system

AT

Moment balance

0;

0

x

A B C A B C

M

T T T T T T

,

,

t AB B C

t BC C

M T T

M T

Maximum angle of twist and stress

A B C1L 2L

B CT T

BT CT

Example 2 – Statically indeterminate system

< F.B.D. >

,

,

0;

0

x

A C C A

t AB A

t BC C A

M

T T T T T T

M T

M T T T

0CA

1 2

2

1 2

1[ ( ) ] 0CA BA CB A A

A

T L T T LGJ

L TT

L L

A B C

AT TCT

Moment balance Geometric compatibility:

, 1 1 2 1 2

1 2 1 2

maxmax max

1

( )

, max( , )

t AB ABA

A C

M T L L L T L L Tdx

GJ GJ GJ L L GJ L L

T rT T T

J

Angle of twist at Point B and maximum shear stress

Example 3 – Design of circular shaft

Ship

1hp 76kg m / s

1 176 lb ft / sec

0.453 0.3048

260 hp, 3800 rpm, 30,000 psiaP n

6

3

33

max 3 24

260 hp 260 6600 in lb / s 1.716 10 in lb / sec

2 rad rad3800 rpm 3800 398

60 sec sec

4.31 10 in lb

( / 2) 16 lb 16( 30 10 )

in

32

a

a

P

P T T

T d T Td

dd

①

②

① + ②

Given value

Determination of diameter of the shaft

316

0.90 ina

Td

Summary of torsional problem of circular shaft

z

r

Strain Stress

Hooke’s law Force

equilibriumrr r rz

r z

zr z zz

0 0 0

0 0

0 0

dGr

dz

dGr

dz

tMd

dz GJ

•

•

•

•

•

•

1

1

1

rr rr zz

zz rr

zz zz rr

rr

zz

zrzr

E

E

E

G

G

G

For composite shaft

1 1 2 2

...tMd

dz G J G J

tz z

M r

J

21

2

t

L L

t

L

Mddz dz

dz GJ

MU dz

GJ

•Rule of symmetry

•Assumption:

•Geometric compatibility:

0rr zz

0

0

z z

r r

rz zr

dr

dz

r

( )z z

( )z

z

z

z

3.8 Beam Theory

Beam and external load

( )q x( )q x

( )q x ( )q x

Saint-Venant’s principle

Resultant force

Statically the same concentrated force with

a set of distributed force = Vector sum

passing through centroid of loading diagram

Concentrated force

Concentrated moment

Statically

Statically

( )q x = Load intensity function = Load/length

y

xBA

( )q xR

X x

x

y

x

Loading diagram

( )

( )

L A

L A

xq x dx xdAX x

q x dx dA

:

, :

:

, :

xx

xz xy

xx

xz xy

M twisting moment

M M bending moment

F axial force

F F shear force

Definition of axes Definition of cross-section

2D

First subscript: Direction of face

Second subscript: Direction of force

x

y

z

Axis of symmetry

Neutral axis

Negative x-face

x

y

z

Positive x-face

1x

Force and moment exerting cross-section y

x

V

F

bM

V

F

bM

xxMxzM

xyM

y

xz

xxF

xyF

xzF

, , xx xy b xzF F V F M M

3D

Internal force (Shear force and bending moment)

Pure bending

a a

Pa

x

M

L

( ) 0, ( ) bV x M x constant

Pure bending

Derivation of relationship between and

1 d d dx

ds dx ds

2

( )

1 ( )

d v x

dx v x

2 2

2

2

2

1

1

1

1

1 ( )

s x v

vx

x

x

s v

x

dx

ds v x

3

2 2

1 ( )( )

1 ( )

d d dx v xv x

ds dx dsv x

( ) tanv x 2

2

1cos

1 ( )v x

?

2( ) secv x

?dx

ds

sv

xs

( )y v x

x

s Slope

0

( )v x

x

( )v x

Curvature

200

30 ; ( )

2 8C bM M x x Lx

① 02

Lx

00 ; ( )8

C b

LM M x L x

②2

Lx L

0

8

L

C

bMV

( )L xx

03

8

L VbM

C

x

0 x

0

2

LR

38

x L

0

8

L4L

o 2L

L( )V x

L0

( )bM x

L0

Reactions

0 0

30 ; ( )

8yF V x x L

00 ; ( )8

y

LF V x

8

L

3

8

L

29128

L

V

bM

03

8

L

Example – Find SFD and BMD by cut method

0

0; ( ) ( ) ( ) ( ) 02

( ) ( ) ( )lim ( ) ( )

( )( ), ( ) ( ) 0

( ) ( )( )

( ), ( ) ( ) 0

C b b

b b b

x

bb

b

xM M x x M x V x x q x x

M x x M x dM xV x V x

x dx

dM xV x M x V x

dxM x q x

dV xq x V x q x

dx

C( )bM x

( )V x

( )V x x

( )bM x x

( )q x

x x

( )q x x

0; ( ) ( ) ( ) 0 yF V x V x x q x x

0

( ) ( )lim ( )

x

V x x V xq x

x

( )( )

dV xq x

dx

( )V x

L0

( )bM x

L0

Reactions

0 0

1

3( )

8 2

oo o

L Lq x x x x

Load intensity function

Shear force and bending moment diagram

o2L

L

0 1 13( )

8 2

oo o

L LV x x x x

1 2 23 1 1( )

8 2 2 2

ob o o

L LM x x x x

V

bM

0

2

LR

03

8

L 0

8

L4L

Example – Find SFD and BMD using DE

Shear force and shear stress

Distribution of internal force

Bending moment and bending stress

=

= bM

x

xy

xz

Strain

( )( )xx x

P P y yyv x

P

Q Q

Q

Pure bending – deflection curve-strain relation

y z x

vyv

Anticlastic curvature

0, 0, 0xy yz zx

x

yx

z

y

z

0xy 0zx 0yz

yQP

R S x

A B C

D E FNeutral axis

QP

RS

y

y

RS PQ

bM

bM

Plane of symmetry

0 0

0 0 0

0 0 0

x

0 0

0 0

0 0

x

x

x

v

v

( / )zz z

y

,x x yy zz xx

yE E ν

•

•

•

•

•

•

1

1

1

1 12 2

1 12 2

1 12 2

xx xx yy zz

yy yy zz xx

zz zz xx yy

xy

xy xy xy

yz

yz yz yz

zxzx zx zx

E

E

E

E G

E G

E G

Hooke’s law of an isotropic material

Assumption: 0, 0y z

0, 0, 0( 0, 0, 0)xy yz zx xy yz zx

Stress

Application of Hooke’s law

x xy xz

yx y yz

zx zy z

ij ji

xface

zface

yy

yxyzxy

xxxz

zxzz

zy

y

z

x

yface

0y

0z

Flange

Webxx x

zz z yy y

xx x

yy y

zz zz

Pure bending - stress-strain relation

0 ; 0x xA

F dA 1

0A

EydA

( )A 0

A

ydA

If E constant

21b

AEy dA M

zzb

EIM

2

zzA

I y dA

,x

Ey

yM

b

0 ;z x bA

M y dA M

0 ; 0y xA A

EM z dA yzdA

( )A

If E constant

,x

Ey

Automatically satisfied

for symmetric beam2nd moment of inertia

of the cross-section

zzEI : Flexural rigidity

Application of requirement of force equilibrium

Pure bending – Force equilibrium

bx

zz

M y

I

Summary of pure bending beam theory

( ) ( ) ( ) b

q x v x M x Curvature-deflection curve :1

( ) v x

Curvature-strain :

xx

y

Constitutive law :

xx xx

yE E

Axial force : 0 0 xxA AdA ydA

Bending moment :1

( )

b

xx bA

MydA M v x

EI

( ) ( ) b

EIv x M x

( ( )) ( ) b

EIv x M q x

Boundary conditions

• Geometric :

• Mechanical :

( ) 0,v L ( ) 0 v L

(0) ,V P (0) ,b

M M ( ) ,V L P

(0) (0) (0) b

MM EIv M v

EI

( ) (0) ( (0)) (0) b

PV M x V EIv v

EI

( )( )

dV xq x

dx

( )( ) b

dM xV x

dx

(0) 0 v(0) 0,v

2

2

( )( )b

d M xq x

dx

b

xx

zz

M y

I

Engineering beam theory

Assumption of engineering beam theory

The following relationships obtained by the pure bending beam theory are valid even

though shear force does not vanish.

,bx

zz

M y

I

1( ) b

zz

Mv x

EI

Purpose of engineering beam theory

When shear force exists,

-bending moment varies from position to position.

-shear stress as well as bending stress exist.

Purpose of engineering beam theory is to calculate

the shear stress

xy

xz

xz

x

( )bM x

( )bM x x

x

x

Results of engineering beam theory

yx xy

zz

V

bI

Q

1AydA Q

1

( ) ( ) 0x x x x x x yxA

F dA F

Engineering beam theory – Shear stress

( ) ( ) yxb b

zz

FM x x M x

I x x

Q

( ) ( )( ) ( )

b b

x x x x x

zz

M x x M x y

I

yx b

zz zz

dF dM Vf

dx dx I I

Q Q

1AydA Q

yx xy

zz

V

bI

Qyx

yx yx yx

dFF b x b

dx

Assume: is uniformly distributed over width yxF b

Shear flow

Take a cut fraction having -y face in the cut surface at y = y1

: Cross-sectional area

defined by y = y11A

Assumption: uniform cross-section

Example 1 – Shear and bending stresses

)max max,4 2

b

LP PM V

Given values

1000psi, 6 , 1000lba b h P

Calculated maximum shear force and bending moment

Allowable maximum beam length maxL

)max

)max 3

)max2

3/ 2 / 2 / 4 3 4

3/12 4 2

2

xy

xy

x

PVQ P bh h P hbh

LPbI b bh bh L

bh

Ratio of maximum shear stress to bending stress

)max 3 2

/ 2

2 3 3

)max max 2

/ 4 / 2 3

/12 2

2 2 6 in 1000lb144in

3 3000lb in

bx

z y h

ax a

LP hM y LP

I bh bh

bhL

P

0

L

( )V x

0

2

L

0

2

L

x

L0

( )bM x2

0

8

L

xL0

0

1 1( )2 2

o o

o

L Lq x x x x L

0( )q x

0R L0

2

L 0

2

L

0 1 0

1 2 1

( )2 2

1( )

2 2 2

o oo

o ob o

L LV x x x x L

L LM x x x x L

V

bM

22

01( )2 2 2 2 4 8

ob o

L LL L LM

max

0

0)

3

32 2 41 4

12

xy

L h hb

L

bhb bh

max

2

02

0) 2

3

38 2

1 4

12

xy

L h

L

bhbh

max

max

)

)

xy

xy

h

L

b

h

Example 2 – Ratio of shear to bending stresses

001 0

(4) 001 0

0 100

1 20 0

( )2

( ) ( )2

( )2

( )2 2

A

A B

Lq x x x

LEIv x q x x x

LEIv x x x C

LEIv x x x C x C

2 30 01

3 40 01 2

( )4 6

( )12 24

LEIv x x x C

LEIv x x x C x C

23

4 40 0 01 1

(0) 0 0

( ) 0 012 24 24

v C

Lv L L L C L C

33 40 0 0

3 4 30

1( )

12 24 24

224

L Lv x x x x

EI

Lx x L xEI

•

•

•

•

•

•

•

•

Problem

Reactions

Boundary conditions

(0) (0) 0

( ) ( ) 0

b

b

M EIv

M L EIv L

(0) 0

( ) 0

v

v L

?( )v x

Determination of 1 2,C C0

2

L 0

2

LL

0

Example 3 – Beam deflection

P

P P

3 3

2

3 1 1 7( )

4 3 6 8 16

5( )

8

PL PLv L

EI EI

PLL

EI

2 1 1

1 1

3( ) 2

2 2

3( ) 2

2 2

b

Lq x PL x P x P x

LM x PL P x P x

2 2

1 2

2

1 2

1 5 7

3 48 16

5

8

PL PLv v v

EI EI

PL

EI

•

•

•

•

•

•

•

2 2

1

2 3 3

2

( ) ( )

3( ) ( (0) 0)

2 2 2

3( ) ( (0) 0)

4 3 6 2

bEIv x M x

P LEIv x PLx Px x C v

P P LEIv x PLx x x C v

Deflection and angle due to each force

Principle of superposition

P

2L

2P

32PL

P

2L

3

1

2

1

5

48

8

PL

EI

PL

EI

2

2

2

2

3

2

PL

EI

PL

EI

Problem description

Reactions

( )bM x

( ), ( ), ( ) ( )v x v L L v L

Application of principle of superposition

Boundary conditions

( ) 0, ( ) 0v L v L

(0) 0, (0) 0v v

3.9 Buckling

Displacement

in the longitudinal

direction is

neglected

Pv

x,EI L

P

( ) ( )V x x V x V x

( ) ( )b b bM x x M x M x

( ) ( )v x x v x v x

P

( )bM x

( )v x

( )V x

xx

C

Governing equation of buckling of column

0bdM dvV P

dx dx

Relationship of , , , and

bM V

P

Similarity and difference between beam and column

Similarity: Bending moment and shear force are exerted and pure bending theory is applied.

In beam theory, axial force ( ) is neglected in calculating bending moment.

However, in column theory, axial force should be considered in calculating bending moment.

( )q x

2

0 ; ( ) ( ) ( )

( )( ( ) ( )) 0

2

C b bM M x x M x V x x

q xx P v x x v x

0 ; ( ) 0 ( ) 0 ( )y

dV dV dVF q x x x q x q x

dx dx dx

2

2( ) ( )bd M d dvP q x

dx dx dx

2

2( ) 0bd M dV d dvP

dx dx dx dx

( )q x

( )v x

P

2

2( ) ( ) ( )b

d d dvM P q x

dx dx dx

Assume: Pure bending theory is applied.

(4) 2 2( ) ( ) 0,P

v x v xEI

, =

( ) 0

EI P

q x

일정 일정

1( )v x

( ) ( )

1

0

b

b

A

x

x

EIv x M x

M

EI

ydA

yE

y

Pure bending

Governing equations

2 2

2 2( ) ( ) ( )

d d v d dvEI P q x

dx dx dx dx

( ) 0dV

q xdx

0bdM dvV P

dx dx

4 2 2

1 2 3 4

0 0( )

( ) sin cos

s s s s i

v x C C x C x C x

중근 ,

( ) sxv x Ce

1 2 3 4, , ,C C C C are calculated by BCs.

Governing equation of buckling of column

Example – buckling loads and modes

2 2

3 4

1 4

1 2 3 4

2

4

2 2

3 4

( ) sin cos

(0) 0

( ) sin cos 0

(0) 0

( ) sin cos 0

v x C x C x

v C C

v L C C L C L C L

v C

v L C L C L

2

2

crcr

P EIL P

EI L

①

②

③

④

③①

P

,EI L

(4) 2( ) ( ) 0,P P

v x v xEI EI

(0) 0, ( ) 0

(0) 0 (0) 0

( ) 0 ( ) 0

b

b

v v L

M v

M L v L

sin 0 / ( 1,2, , )L L n n L n

GE and homegeneous solution

BCs

Solving

3( ) sinv x C x

2 2

2( ) sinn

n n

P EI n xL n P n v x C

EI L L

Buckling load

Problem

P

EI

1 2 3 4( ) sin cosv x C C x C x C x

1( ) sin

xv x C

L

3

3( ) sin

xv x C

L

2

2( ) sin

xv x C

L

Because , from Eq. .

from Eq. . In order to obtain non-trivial

solution, .

0 4 0C 1 0C

sin 0L

Buckling mode 1

(n=1)Buckling mode 2

(n=2)

Mode of buckling

Non-buckled

3.10 Special phenomenon

Stress concentration(SC) and SC factor K