iii. basis of mechanics - afdex · iii. basis of mechanics 3.1 general consideration 3.2 newton’s...
TRANSCRIPT
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