2.1 Stresses
2.1.1 Equilibrium of Deformable bodies and Internal Forces
F2I
F3I
(I)
F2II
Q
Fi
F2I
I
F2II
PI
(II)
(I)
Q
Q
(II)
P II
F1I
F1II
F1I
F1II
F3I
F2I
F2II
F2I
P III
(IV)
(III)
F1I
(III)
Q
F1II
Q
F1I
* equilibrium
- body in equilibrium:  F  0 , F1I  F2I  F3I  F1II  F2II  0
- equilibrium of isolated free bodies: PI, PII = resultant of internal forces on the section (reactions)
isolated body (I) : F1I  F2I  F3I  P I  0
isolated body (II) : F1II  F2II  P II  0
 P I   P II (a pair of forces with equal magnitudes but opposite directions)

Internal force is a tensor.
* method of section to find internal forces
- Isolate (or, cut out) a part of body through the section of interest: to find the internal forces on the section.
- Free body diagram of the isolated body; external forces and internal forces(unknown vectors)
- Static equilibrium equations to solve for the internal forces.
* Even for the same points, different section results in different internal forces: cf. PI and PIII.
* Internal forces are point-dependent as well as section-dependent(direction-dependent): 2nd-order tensor
2.1.2 Definitions and Classification of Stresses
s
P
Ps
A
Pn
n = direction vector normal (vertical) to the section
s = direction vector tangential (parallel) to the section
 = direction angle of P w.r.t n

n
A = differential area of the section
section
* components of internal force : normal and tangential to the section
- normal component:
Pn  P cos
- tangential component: Pn  P sin 
* stress = magnitude of internal force per unit area
Pn
A0 A
P
  lim s
A0 A
- normal stress (tensile/compressive stress):   lim
(normal component per unit area)
- tangential stress (shearing stress):
(tangential component per unit area)
* examples of stresses
- normal stress: cable tension, column compression, truss members, beam bending
- shearing stress: transmission shaft, key in shaft and wheel, connections(pin, rivet, bolt), punch, scissors
- bearing stress (contact stress): contact surfaces, gear surface, rivet hole, bearing plate, footing
 rivet connection of tension/compression plates
- normal (tensile/compressive) stress in plate section:  
- shearing stress in rivet section:  
P
d 2 / 4
- bearing stress on rivet hole surface:  b 
P
td
P
bt
2.1.3 Components and Sign Convention of Stresses
* right-hand Cartesian coordinate system
- orthogonal coordinates: x-y-z system, positive and negative direction
- positive face, negative face: normal direction of surface = coordinate direction
* sign convention (mathematical sign convention)
- positive(+): +ve coordinate direction on +ve face, or ve coordinate direction on ve surface
- negative(): ve coordinate direction on +ve face, or +ve coordinate direction on ve surface
- positive normal stress = tensile stress
 xy
 xx
 xz
 yy
D
C
 yz
Q
Y
A
y
 xy
B
 zy
Z
 zz
 yx
 zx
X
 xz
H
x
z
E
F
 yy
 yx
A
B
 xy
y
 xx
 xx
x
dy
 xy
E
dx
 yx
F
 yy
* stresses at a point
 xx  xy  xz 


- space stresses (3D xyz-spatial components) :  yx  yy  yz 
  zx  zy  zz 
 xx  xy 
- plane stresses (2D xy-planar components) : 

 yx  yy 
  = main diagonal = normal stress;
 = off-diagonal = shearing stress
 The first subscript means the ‘surface’, and the second ‘direction’:  face,direction .
 xx
G
2.1.4 Units of Stresses
* definition: [ stress ] 
[ force]
[area ]
 dimension: FL2
* units
- SI units: Pa (pascal)  N/m2  SI prefixes: k (kilo, 103), M(mega, 106), G(giga, 109),
- metric units: kg/cm2 , t/m2
- British units: psi (pound per square inch)  lb/in2, ksi  kip/in2, psf  lb/ft2,
* conversion of units
1 Pa = 1.0210-5 kg/cm2 = 14.510-5 psi ;
9.80104 Pa = 1 kg/cm2 = 14.2 psi
2.1.5 Examples
* Example 2.1.1 [tension/compression of prismatic bar]
- stress at Q
p
m
P
d
Q
n

q
solution
(i) stress at the point Q on mn-section
- FBD of isolated part (SEE) resultant of internal force on mn-section = P
- The mn-section is far enough from the concentrated load P. (St Venant’s Principle) uniform stress
P
- Pn  P , Ps  0  Pn  A, Ps  A   0  ,  0  0
A
 non-uniform stress distribution: Pn    dA
A
m
Q
P
p
P
Q
P

=0
n

n
Ps
x
P
A
q A  cos 
A  A0  14 d 2
(ii) stress at the point Q on pq-section
- Pn  P cos , Ps  P sin 
1P
P
P cos 
P
1  cos 2  ,
 cos 2  
   n 
2A
A A / cos  A
Pn
 
Ps
P sin 
P
1P

 cos  sin  
sin 2
A A / cos  A
2A
Discussions
(1) The stresses on mn-section should be the same as the ones which are obtained by substituting  = 0 in the
stresses on pq-section.
P
   0    0 ,    0  0   0
A
(2) As can be seen in   and   , the stresses at a point Q are the function of .
Or, the stresses are dependent on the direction of section.
In general, stresses are different at the different locations(points) even on the same section.
For the transformation of stresses, the direction vector (here, cos or sin) are twice multiplied, which is
a general property of the second-order tensor.
(3) No material property is involved in the stress formulae.
Dimension of material (i.e., cross-sectional area for this example) is necessary for the stress distribution.
(4) The necessary condition for the normal stress   to be extremum (maximum or minimum) is
The solutions of the equation are  = 0, 90; or  max   0 
d 
0.
d
P
and  min   90  0 , where it is easily
A
found    0 ; or,  0  0 and  90  0 .
In general, the shearing stress is zero on the section in which the normal stress becomes extrema.
(5) The maximum value of shearing stress is 12  0 which occurs at  = 45 ( d/d = 0).
In general, the direction of maximum shearing stress is the one which bisects the directions of maximum
and minimum normal stresses.
 max   45  12  0
* Example 2.1.2 [punch]
* Example 2.1.3 [classification of internal forces]
- axial force: compressive or tensile force in truss members
- shear force: connecting bolt of lap joint or butt joint
- bearing force: contact surface
- bending moment: beams, rigid frames
- torque: twisting moment in a shaft
Figure E2.1.3 Internal forces of structural elements
2.2 Strains
2.2.1 Normal Strain
* deformation = change of dimension
 x-direction : l  l'
 y-direction : w  w'
 z-direction : t  t'
l
y
w
x
w'
Px
Px
l'
* normal strain =
* amount of deformation (elongation/shortening)
= [deformed length] – [original length]
 x-direction : x = l' – l
 y-direction : y = w' – w
 z-direction : z = t' – t
amount of deformatio n
original length
l  l  x

l
l
w  w  y
 y-direction :  y 

w
w
t  t  z

 z-direction :  z 
t
t
 x-direction :  x 
* longitudinal strain --- direction of load action :  x
lateral strain --- perpendicular to the load direction :  y and  z
* Strain is a dimensionless quantity. [sign: elongation(+), shortening(–)]
2.2.2 Shearing Strain
Ps

* amount of deformation (shearing deformation)
= sliding length in the direction of shearing force : 
or, angle change form right angle (/2 rad = 90) :  rad
l
* shearing strain

Ps
 sliding length :  s 

l
λ
0
 angle change   tan γ 

tan   
l

:
s  
2.2.3 Volumetric Strain
* volume change [dilatation(+); contraction(–)]
- V  V   V  (a)3  a 3
- normal strain  : a   a (1   )
p
a
a'
p
a'
a
p
a'
a
* volumetric strain
V a 3  a 3 [a (1   )]3  a 3
v 


 3  3 2   3
V
a3
a3
  v  3 (   0 )
2.2.4 Poisson’s Ratio
* definition: The ratio of lateral strain and longitudinal strain is constant for a material.
*Poisson’s ratio (isotropic material):  
1 y


 z ;
m x
x
m = Poisson’s number
* Poisson’s ratio for isotropic materials: Table 2.2.1
* Poisson’s ratio for an anisotropic material:  xy 

1
 y ,
mxy  x
 xz 
1

 z
mxz  x
* range of Poisson’s ratio: 0    12
- volume change due to the tensile load in x-direction:  x   ;  y     z
V  l wt  l (1   ) w(1   ) t (1   )


 1 +  - 2  0
V
lwt
lwt
V
 1 .  1 +  - 2  1 , or  (1  2 )  0    12 ( > 0, for a tensile load)
For any tensile load,
V
- volume change due to the compressive load in x-direction:  x   ;  y     z
V
 1 .   (1  2 )  0    12 ( < 0, for a compressive load)
For any compressive load,
V
- definition of the Poisson’s ratio,   0.
2.3 Hooke’s Law and Modulus of Elasticity
* basic requirements for structural design
- strength of materials (hardness, toughness, ductility, etc)
- stiffness of structures and members (stability, too)
* test of materials
- fundamental: simple tension test, simple torsion test
- other tests for mechanical properties: compression, shear, bending, impact, fatigue, creep, hardness
2.3.1 Simple Tension Test and Stress-Strain Diagram
* test specimen for tension test [KS B0801]
- specimen #4: L = 50 mm, D = 14 mm
* load-elongation diagram (P- diagram)
- nominal stress (P/A) vs. actual stress
- stress-strain diagram (- diagram):  = P/A,
 = /L
* typical stress-strain diagram for mild steel: Fig. 2.3.1
stress
 = P/A
H
u
y
E
p
G
C
B
A
nominal
stress-strain
diagram
D
E
E
1
1
O
actual
stress-strain
diagram
K
F
F'
 = /L
strain
A = proportional limit (limit point of linear elasticity: OA)
- The slope of OA is the modulus of elasticity(E):   E [Hooke’ law]
- unloading at any point on the straight line OA  return to point O.
 method of superposition in linear elastic region
B = elastic limit (limit point of non-linear elasticity: AB)
- It is not easy to determine the point B.
- unloading on curve AB  return to point O: No residual strain, however, the Hooke’s law is not valid.
- AB is not a straight line but a curved shape.
* unloading at any point beyond the point B  No return to point O:
- plastic deformation: residual strain, or permanent set
[ ex, unloading at the point G and reloading along the straight line GF, which is parallel to OA]
(total strain at G  OF') = (residual strain OF) + (elastic strain  FF')
C, D = upper and lower yield point (slip)
- determination of yield point (i.e., cast iron)  0.2-offset method (the point of permanent strain 0.002)
* actual stress-strain curve vs. nominal stress-strain curve
DH = strain-hardening, work-hardening
H = ultimate strength
K = fracture
2.3.2 Young’s Modulus
* Hooke’s law for axial tension/compression in linear elastic range (Robert Hooke, UK, 1635-1703)
- assumption: prismatic specimen, homogeneous material, axial load and constant stress in a section
 specimen: length (L), cross-sectional area (A), axial load (P), elongation/shortening ()

P
L
E
A
P
- Hooke in 1678, Ut tensio sic vis (As strain, so force):  
PL
A
- In 1807, Thomas Young(UK, 1773-1829) introduced a constant E :  
PL
EA
 E = Young’s modulus, or modulus of elasticity; same unit as stress
* considerations on Eq.(2.3.3)
 EA 
(1) P  
  F  kd for a linear spring with a spring constant k
 L 
EA

= equivalent spring constant of tension/compression rod (k = EA/L)
L
= axial force needed to induce the unit deformation ( = 1  P = EA/L)
= stiffness of axial element
 L 
(2)   
P
 EA 
L

= inverse of equivalent spring constant
EA
= amount of deformation due to unit load (P = 1   = L/EA)
= flexibility of axial element
(2.3.3)
P/ A 

 /L 
 E = modulus of elasticity
= tangent of the - curve  constant in linear elastic range
= material-dependent constant
(3) E 
(4) A 
LP
: design formula
E
2.3.3 Modulus of Rigidity
* linear elastic range:   G
 G = modulus of rigidity
P

 s G
As
l
2.3.4 Bulk Modulus
* Constant normal pressure() on all surfaces of elastic body induce the volumetric strain (v):   K v
PV
 volume change V due to stress  = P/A: V 
AK
Table 2.3.1 Mechanical properties of materials
density
ultimate strength(tension, compression, shear)
yield stress(tension, compression)
modulus of elasticity(Young’s modulus, modulus of rigidity)
thermal expansion coefficient
- steel, aluminum, alloy, wood, concrete, others
2.4 Stress-Strain Relations of Materials
* nonlinear stress-strain diagram  no linear elastic range
- polynomial approximation of - relation
- piece-wise linear idealization of - relation

mild steel
leather
cast iron,
concrete
rubber

* modulus of nonlinear material
- tangent modulus, Et
- secant modulus, Es
 For concrete,  Q  0.5 28

- curve
tan-1Et
Q
Q
tan-1Es
Q

2.5 Relations between Elastic Moduli
E = Young’s modulus, modulus of elasticity
G = modulus of rigidity
K = bulk modulus
 = Poisson’s ratio
2.5.1 Tri-axial State and E-K- Relations
* tri-axial stress state : only normal stresses on all three faces, no shearing stress
y
1+x
1+y
x
z
1+z
* strain due to stress components
x
- strain due to x :
 x-strain =
- strain due to y :
 x-strain =  
E
y
E

 x-strain =   z
E
- strain due to z :
 y-strain =  
 y-strain =
x
E
y
E

 y-strain =   z
E
 z-strain =  
 z-strain =  
 z-strain =
x
E
y
E
z
E
* stress-strain relations for tri-axial stress state (or, generalized Hooke’s law): superposition of above
1
1
1
 x  [ x   ( y   z )] ,
 y  [ y   ( z   x )] ,
 z  [ z   ( x   y )]
(2.5.1)
E
E
E
* volumetric strain :
v   x   y  z
V = 111,
V   1   x 1   y 1   z  ; neglecting higher-order terms…
(v = dilatation of the material)
(2.5.2)
(1  2 )
 x   y   z 
E
- special case of constant stress:  x   y   z     x   y   z     v  3
- From (2.5.1) & (2.5.2),
v 
 definition of volumetric strain :
* E-K- relation :
K
E
3(1  2 )

K
  v  3 
3(1  2 )

E
(2.5.5)
2.5.2 Pure Shear and E-G- Relations
* bi-axial stress state
- special case of plane stress state:  xy   yx  0
- stress-strain relations for bi-axial stress state  special case of tri-axial stress state
1
1
 x  [ x   y ] ,
 y  [ y   x ]
E
E
(2.5.6)
* special case of bi-axial stress state:  x   y  
- stress-strain relations:  x 
 (1   )
  y
E
- element of 45-rotation (rhombus abcd)  pure-shear deformation [discussed in Ex. 4.4.1]
y = -
y
y
 = x = y
=
45
x
b
y
x
x
b'
c'
x
c
a
O
a'
x= 
d'
y
d
(a) bi-axial stress state
(a)' special case: x = -y = 
(b) pure shear stress state
* pure shear stress state: Normal stresses are zero, and    .
- deformation of Oa and Ob
  (1   ) 
  (1   ) 
Oa   Oa (1   x )  Oa 1 
Ob  Ob (1   y )  Ob 1 
,

E 
E 


- shearing strain of square abcd (see Oa'b' )
    Ob 
tan Oa b  tan    
 4 2  Oa 
tan

 tan

1

  
4
2 
2
 tan    
2 2
 4 2  1  tan  tan  1  
4
2
2
  (1   ) 

1
1

Ob 
E 
  
2


Oa  Ob  tan    
 4 2  1   Oa  1   (1   ) 

2
E 
tan



* E-G- relation : G 
E
2(1   )


E

 2(1   )
(2.5.7)
2.5.3 Additional Relations
* reformation of Eqs. (2.5.5) & (2.5.7)
E  2G (1   )  3K (1  2 ) 
9 KG
G  3K
G
E
3K (1  2 )
3KE


2(1   )
2(1   )
3K  E
K
E
2G(1   )
GE


3(1  2 ) 3(1  2 ) 9G  3E

E  2G 3K  E 3K  2G


2G
6K
6 K  2G
2.6 Allowable Stress and Safety Factor
* working stress (w) = stress occurred during the use of structures or the operation of machine elements
allowable stress (a) = maximum working stress in a material which is safe
* purpose of practical design: w = a  p (proportional limit)
* factor of safety (N): Yield stress or ultimate strength is used instead of the proportional limit.
- with respect to yield stress:  a 
y
Ny
- with respect to ultimate strength:  a 
u
Nu
* Why the factor of safety in design?  ‘Uncertainties’
- characteristics of load
- properties of materials
- mechanical theories
- design formula
- characteristics of design data
* considerations on determining the factor of safety
- material properties used
- applied loads
- mechanical theories and computation scheme: reasonable, accurate
- environmental condition
- importance of the project
- manufacturing and construction: accurate, workmanship
* In practice, N  4
2.7 Examples
* Example 2.7.1 [specimen of simple tension test]
- specimen #4 of KS B0801: L = 50 mm, D = 14 mm
- axial tensile load: P = 100 kN
- deformation measurement: L' = 50.433 mm, D' = 13.970 mm
(1) maximum normal stress
(2) modulus of elasticity of the material
(3) Poisson’s ratio
(4) modulus of rigidity
(5) maximum shearing stress
(6) maximum shearing strain
(7) factor of safety for the ultimate strength, u = 1,440 MPa
(8) bulk modulus
Remarks: