DAY 5

STRESS
• Stress is a measure of force per unit area
within a body.
• It is a body's internal distribution of force per
area that reacts to external applied loads.

P
STRESS 
A
 ONE DIMENSIONAL STRESS
• Engineering stress / Nominal stress
– The simplest definition of stress, σ = F/A,
where A is the initial cross-sectional area prior
to the application of the load
• True stress
– True stress is an alternative definition in which
the initial area is replaced by the current area

• Relation between Engineering & Nominal stress

 true  (1   e ) e
 TYPES OF STRESSES

TENSILE COMPRESSIVE

BENDING SHEAR

TORSION
 SHEAR STRESS
dx

2 1 xdxdy
B z z A

zdzdy
zdzdy
2 1
dz D
TORSION xdxdy
C

Taking moment about CD, We get

 z dzdy dx    x dxdy dz 
z  x
This implies that if there is a shear in one plane then there will be a shear in
the plane perpendicular to that
 TWO DIMENSIONAL STRESS
• Plane stress y
 yx
 xy
x x
 xy
 yx
y
• Principal stress

x  y  x  y  2
 1, 2       xy
2  2 
 THREE DIMENSIONAL STRESS
• Cauchy stress
– Force per unit area in the deformed geometry
 xx  xy  xz 
 
 ij   yx  yy  yz 
   
 zx zy zz 

• Second Piola Kirchoff stress

– Relates forces in the reference configuration to
area in the reference configuration

S  JX τ X X – Deformation gradient
IJ I, j ij J, i
 3D PRINCIPAL STRESS
• Stress invariants of the Cauchy stress
I1   x   y   z
I 2   x y   y z   z x   xy   yz   zx
2 2 2

I 3   x y z  2 xy yz zx         
2
x yz
2
y zx
2
z xy

• Characteristic equation of 3D principal stress is

  I1  I 2  I 3  0
3 2

• Invariants in terms of principal stress

I1   1   2   3
I 2   1 2   2 3   3 1
I 3   1 2 3
 VON-MISES STRESS
• Based on distortional energy

v 
 1   2  2
  2   3    3   1 
2 2

v 
1
 x   y    y   z    z   x   6 xy2   yz2   zx2 
2 2 2

2
 STRAIN
• Strain is the geometrical expression of deformation
caused by the action of stress on a physical body.
L
Strain 
L
• Strain – displacement relations
Normal Strain
u v w
 x  y  z
x y z

Shear strain (The angular change at any point

between two lines crossing this point in a body can
be measured as a shear (or shape) strain)
u v v w w u
 xy   yz   zx 
y x z y x z
 VOLUMETRIC STRAIN
• Volumetric strain
V  V0

V0

  x  y  z
TWO DIMENSIONAL STRAIN
• Plane strain y
 yx  xy
 xy
x x
 yx
y
• Principal strain

 x y   x y    xy 
 1, 2      
2  2   2 
 3D STRAIN
Strain tensor   xy  xz 
 xx 2 2 

  yx  yz 
 ij   yy
2 2 
 
  zx  zy
 2
 zz 
2 
Green Lagrangian Strain tensor E  1  F F   
ij 2 ki kj ij

1  ui u j uk uk 

   
2  x j xi xi x j 

Almansi Strain tensor

1

E   ij  Fki Fkj
ij 2
-1 1

STRESS-STRAIN CURVE

Mild steel Copper

Thermoplastic
 BEAM
• A STRUCTURAL MEMBER WHOSE THIRD DIMENSION
IS LARGE COMPARED TO THE OTHER TWO
DIMENSIONS AND SUBJECTED TO TRANSVERSE
LOAD

• A BEAM IS A STRUCTURAL MEMBER THAT CARRIES

LOAD PRIMARILY IN BENDING

• A BEAM IS A BAR CAPABLE OF CARRYING LOADS IN

BENDING. THE LOADS ARE APPLIED IN THE
TRANSVERSE DIRECTION TO ITS LONGEST
DIMENSION
 TERMINOLOGY
• SHEAR FORCE
– A shear force in structural mechanics is an example
of an internal force that is induced in a restrained
structural element when external forces are applied

• BENDING MOMENT
– A bending moment in structural mechanics is an
example of an internal moment that is induced in a
restrained structural element when external forces
are applied

• CONTRAFLEXURE
– Location, where no bending takes place in a beam
 TYPES OF BEAMS

• CANTILEVER BEAM
• SIMPLY SUPPORTED BEAM
• FIXED-FIXED BEAM
• OVER HANGING BEAM
• CONTINUOUS BEAM
 BEAMS (Contd…)
• STATICALLY DETERMINATE
• STATICALLY INDETERMINATE

B
A

C D
 BEAM

•TYPES OF BENDING
Hogging
Sagging
 SHEAR FORCE & BENDING
MOMENT
BEAM P
P
PL/8 PL/8
L
P/2 P/2
SHEAR FORCE FREE BODY DIAGRAM
P/2
P/2

BENDING MOMENT

PL/8 PL/8
 SHEAR FORCE & BENDING
MOMENT
BEAM P
P
3PL/8
L
11P/16 5P/16
SHEAR FORCE FREE BODY DIAGRAM
11P/16
5P/16

BENDING MOMENT

3PL/8
 RELATION BETWEEN
BM, SHEAR & LOAD
w
V V+dV
M O M+dM
dx
Taking moments about O
M O 0
 dx 
M  ( M  dM )  Vdx  wdx   0
 2
dM
V
dx

Force equilibrium gives

V  V  dV   w * dx  0
dV
w
dx
 BEAM THEORY
• ASSUMPTIONS
– MATERIAL IS HOMOGENOUS
– MATERIAL IS ISOTROPIC
– THE BEAM IS SYMMETRICAL
– THE TRANSVERSE PLANE SECTION
REMAIN PLANE AND NORMAL TO THE
LONGITUDIONAL FIBRES AFTER BENDING
(NEUTRAL PLANE REMAINS SAME AFTER
BENDING)
 BENDING STRESS
o

M M
R
c d
ef
a b
Change in length ef Ey
strain ( )   ... (1) dF  stress * area  dA ... (6)
Original length cd R
From similar triangles edf & cod  Ey 
ef de dM  y * dF  y dA ... (7)
 ... (2)  R
cd co
ef y E  EI 
strain ( )   ...(3) M     y 2dA    ... (8)
cd R R  R
Hooks law M E
f (or )   ... (9)
strain ( )  ...(4) I R
E
From (3) & (4) From (5) & (9)
f E M E f
 ...(5)   ... (9)
y R I R y
 FINITE ELEMENTS
• TRUSS / BAR / LINK ELEMENT
• BEAM ELEMENT
3D BEAM ELEMENT
 3D BEAM ELEMENT
q
t q
s q
t
x(r , s, t )  hk l xk   ak hk lVtxk   bk hk lVsxk
k 1 2 qk 1 2 qk 1
q
t s
t
y (r , s, t )  hk yk   ak hk Vty   bk hk lVsyk ……… (1)
l l k

k 1 2 kq1 2 kq1
q
t s
t
z (r , s, t )  hk z k   ak hk Vtz   bk hk lVszk
l l k

k 1 2 k 1 2 k 1

l
x, y, z  Cartesian coordinate of any point in the element
l l

l
xk , y k , z k  Cartesian coordinate of any nodal point k
l l

a k , bk  Cross sectional dimensions of the beam at nodal point k

l
Vtxk , l Vtyk , l Vtzk  Components
l
V s
k l
t
k
of unit vector V in direction
V l k

t at nodal point k
s

l
Vsxk , l Vsyk , l Vszk  Components of unit vector V in direction s at nodal point k
l
s
k

l k
We call Vt and lVsk the normal vectors or director vectors at
nodal point k
 3D BEAM ELEMENT
The displacement components are
x ( r , s , t ) 1 x  0 x
y ( r , s , t ) 1 y  0 y ……… (2)
z ( r , s , t ) 1 z  0 z
From (1) & (2) we get
q
t q s q
x(r , s, t )   hk xk   ak hkVtx   bk hkVsxk
k

k 1 2 qk 1 2 qk 1
q
t s
y (r , s, t )   hk yk   ak hkVty   bk hkVsyk
k
……… (3)
k 1 2 kq1 2 kq1
q
t s
z (r , s, t )   hk z k   ak hkVtzk   bk hkVszk
k 1 2 k 1 2 k 1
Vt  Vt  Vt
k 1 k 0 k

Vs  Vs  Vs
k 1 k 0 k
 3D BEAM ELEMENT
Strain displacement relation
   q
     Bk uˆ k where uˆ k  u k vk wk  xk  yk  zk 
  k 1
  
and the matrices Bk,k=1,…..,q, together constitute the matrix B,

B   B1 . . . . Bq 
 hr k  

  u
 1 (g)1i (g)2i (g)3i   u 
k k

 r  r  
k

   
   k 
 u 
 
k  
q

   k   h 1 (gˆ ) k
(gˆ ) k
(gˆ )
x

 s k 1 
1i 2i 3i
  k 
   y
 
k  
 u  
h 1 (g ) k
( g ) k
( g ) 
3i   k 
 t   k 1i 2i
z
   
 3D BEAM ELEMENT
 0 - 0 Vszk 0Vsyk 
 
bk  0 k
 gˆ  k
 Vsz 0 - 0Vsxk 
2 
- 0 V k 0V k 0 
 sy sx 
 0 - 0Vtzk 0Vtyk 
 
ak  0 k
g k
 Vtz 0 - 0 Vtxk 
2  
- 0V k 0V k 0 
 ty tx 

 g k
ij  s gˆ  ij  t  g  ij
k k

 1 
Jacobian Transformation J
x 
 3D BEAM ELEMENT
Strain displacement relation
 u   1 hk 
 x   J 11 r (G1 )ik1 (G 2 )ik1 (G3 )ik1  u k 
 
    k 
 u  q  h   x 
     J 211 k (G1 )i 2 (G 2 )i 2 (G3 )i 2   
k k k

 y  k 1  r   yk 
    
 u   J 311 hk (G1 )ik3 (G 2 )ik3 (G3 )ik3   k 
 z   r   z 

Where  Gm  k   J 1  g k   hk   J 1  gˆ k   J 1  g k   h
r
in n1 mi n2 mi n3 mi k

1
l
Stiffness K   B T DB 2 d
1 2
1
l2
Load f   B N d  f KT
T

1 2
 STIFFNESS MATRIX
 AE AE 
 L 0 0 0 0 0  0 0 0 0 0 
L
 12 EI 6 EI 12 EI 6 EI 
 0 0 0  0 0 0 
 L3 L2 L3 L2 
 0 12 EI 6 EI 12 EI 6 EI 
0 0 0 0  0 0
 L3 L2 L3 L2 
 GJ GJ 
 0 0 0 0 0 0 0 0  0 0 
 L L 
 0 6 EI 4 EI 6 EI 2 EI
0 0 0 0  0 0 0 
 L2 L L2 L 
 6 EI 4 EI 6 EI 2 EI 
 0 0
L2
0 0
L
0 0
L2
0 0
L 
Ke   
AE AE
 0 0 0 0 0 0 0 0 0 0 
 L L 
 0 12 EI 6 EI 12 EI 6 EI
- 0 0 - 0 0 0 0 - 0 
 L3 L2 L3 L2 
 12 EI 6 EI 12 EI 6 EI 
 0 0 - 0 0 - 0 0 0 0 
 L3 L2 L3 L2 
 0 GJ GJ
0 0 - 0 0 0 0 0 0 0 
 L L 
 6 EI 2 EI 6 EI 4 EI 
 0 0 0 0 0  0 0 0 
L2 L L2 L
 6 EI 2 EI 6 EI 4 EI 
 0 0 0 0 0 0 0 0 
 L2 L L2 L 
THREE MOMENT EQUATION
 THREE MOMENT EQUATION
(Developed by clapeyron)
Continuity condition  L tan C  R tan C

LL LR

Using second moment-area theorem

1  2 1 1 1 
 L tan C   L L
x A  LL
M L
C L
 LL
M L L
L
ELL  3 2 3 2 

1  2 1 1 1 
 R tan C   xR AR  LR M C LR  LR M R LR 
ELR  3 2 3 2 

Equating the above equations

LL  L L  L 6x A 6x A
M L  2 L  R  M C  R M R   L L  R R
EI L  EI L EI R  EI R LL EI L LR EI R
 THREE MOMENT THEOREM

 A1 x1 A2 x2 
M A L1  2M B ( L1  L2 )  M C L2  6  
 L1 L2 

 L1   L1 L2   L2   A1 x1 A2 x2 
M A    2 M B     M C    6  
 I1   I1 I 2   I2   L1 I1 L2 I 2 