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Chapter 4
Inelastic Material Behavior
Introduction
Load – stress and load – deflection
relation requires stress – strain
relations
Strain Tensor → Stress Tensor
(Material Behavior)
First Law of thermodynamics
Law : The net external work and heat into the system is equal
to increase in potential and kinetic energies
W H U K
   
  
Variation of work δW of the external forces that act on the volume
V with surface S is equal to δWs and δWB
     
S B
xx xy xz yx yy yz zx zy zz x y z
V
W W W
u v w u v w u v w B u B v B w dV
x y z
  
                    
  
 
  
          
 
  
 

First Law of thermodynamics
 
2 2 2
xx xx yy yy zz zz xy xy yz yz zx zx
V
W dV
            
     

which simplifies to
𝑈 =
𝑉
𝑈𝑜𝑑𝑉
Internal Energy per unit
volume
𝛿𝑈 =
𝑉
𝛿𝑈𝑜𝑑𝑉
Variation of internal
energy
First Law of thermodynamics
.
Relation between strain components and strain energy Uo
0 2 2 2
xx xx yy yy zz zz xy xy xz xz yz yz
U
            
     
Strain energy density function Uo depends on strain components,
the coordinates and the temperature
 
0 0 , , , , , , , , ,
xx yy zz xy xz yz
U U x y z T
     

Variations in strain components and the function Uo are related by
0 0 0 0 0 0
0 xx yy zz xy xz yz
xx yy zz xy xz yz
U U U U U U
U
      
     
     
     
     
Hooke’s law: Isotropic Elasticity
Isotropy : Directional independence
Homogeneity : Spatial independence
A material can be anisotropic and homogeneous or inhomogeneous and
anisotropic
Isotropic material - Two independent constants are required to describe
the material
1
( )
1
( )
1
( )
1 1
2
1 1
2
xx xx yy zz
yy yy xx zz
zz zz xx yy
xy xy xy
yz yz yz
E
E
E
G E
G E
      
      
      
 
    
 
    
Strain-stress relations are given
as :-
Hooke’s law: Isotropic Elasticity contd.
[(1 ) ( )]
(1 )(1 2 )
[(1 ) ( )]
(1 )(1 2 )
[(1 ) ( )]
(1 )(1 2 )
, ,
1 1 1
xx xx yy zz
yy yy xx zz
zz zz xx yy
xy xy xz xz yz yz
E
E
E
E E E
         
   
         
   
         
   
        
     
Alternatively the stress- strain relation can also be
written as :
Hooke’s law: Isotropic Elasticity contd.
In the case of plane stress,
0
zz xz yz
  
  
2
2
( )
1
( )
1
1
xx xx yy
yy xx yy
xy xy
E
E
E
  

  

 

 

 



In the case of plane strain,
0
zz xz yz
  
  
[(1 ) ]
(1 )(1 2 )
[ (1 ) ]
(1 )(1 2 )
( )
(1 )(1 2 )
, 0
(1 )
xx xx yy
yy xx yy
zz xx yy
xy xy xz yz
E
E
E
E
   
 
   
 

  
 
   

  
 
  
 
 
 
  

Hooke’s law: Isotropic Elasticity
contd.
Strain Energy Density
𝑈𝑜
=
1
2𝐸
𝜎2
𝑥𝑥 + 𝜎2
𝑦𝑦 + 𝜎2
𝑧𝑧
− 2𝜐 𝜎𝑥𝑥𝜎𝑦𝑦 + 𝜎𝑥𝑥𝜎𝑧𝑧 + 𝜎𝑧𝑧𝜎𝑦𝑦
Objectives
Nonlinear material behavior
Yield criteria
Yielding in ductile materials
Sections
4.1 Limitations of Uniaxial Stress- Strain data
4.2 Nonlinear Material Response
4.3 Yield Criteria : General Concepts
4.4 Yielding of Ductile Materials
4.6 General Yielding
Introduction
When a material is elastic, it returns to the same state (at
macroscopic, microscopic and atomistic levels) upon removal
of all external load
Any material is not elastic can be assumed to be inelastic
E.g.. Viscoelastic, Viscoplastic, and plastic
 To use the measured quantities like yield strength etc. we
need some criteria
The criteria are mathematical concepts motivated by strong
experimental observations
E.g. Ductile materials fail by shear stress on planes of
maximum shear stress
Brittle materials by direct tensile loading without much
yielding
 Other factors affecting material behavior
- Temperature
- Rate of loading
- Loading/ Unloading cycles
Types of Loading
Nonlinear Material Response
Models for Uniaxial stress-strain
All constitutive equations are models that are supposed to
represent the physical behavior as described by experimental
stress-strain response
Experimental Stress strain curvesIdealized stress strain curves
Elastic- perfectly plastic
response
Models for Uniaxial stress-strain contd.
. Linear elastic response Elastic strain hardening response
Models for Uniaxial stress-strain contd.
.
Rigid models
Rigid- perfectly plastic
response
Rigid- strain hardening plastic
response
Ideal Stress Strain Curves
The Yield Criteria : General concepts
General Theory of Plasticity defines
Yield criteria : predicts material yield under multi-axial state of
stress
Flow rule : relation between plastic strain increment and stress
increment
Hardening rule: Evolution of yield surface with strain
Yield Criterion is a mathematical postulate and is defined by a
yield function  ,
( )
ij
f f Y


where Y is the yield strength in uniaxial load, and is correlated
with the history of stress state.
Maximum Principal Stress Criterion:- used for brittle materials
Maximum Principal Strain Criterion:- sometimes used for brittle materials
Strain energy density criterion:- ellipse in the principal stress plane
Maximum shear stress criterion (a.k.a Tresca):- popularly used for ductile materials
Von Mises or Distortional energy criterion:- most popular for ductile materials
Some Yield criteria developed over the years are:
Maximum Principal Stress Criterion
Originally proposed by Rankine
 
1 2 3
max , ,
f Y
  
 
1 Y
  
2 Y
  
3 Y
  
Yield surface is:
Maximum Principal Strain
This was originally proposed by St. Venant
1 1 2 3 0
f Y
  
     1 2 3 Y
  
   
2 1 2 3 0
f Y
  
     2 1 3 Y
  
   
3 3 1 2 0
f Y
  
     3 1 2 Y
  
   
or
or
or
Hence the effective stress may be defined as
max
e i j k
i j k
   
 
  
The yield function may be defined as
e
f Y

 
Strain Energy Density Criterion
.
This was originally proposed by Beltrami
Strain energy density is found as
 
2 2 2
0 1 1 1 1 2 1 3 2 3
1
2 0
2
U
E
         
 
      
 
Strain energy density at yield in uniaxial tension test
2
0
2
Y
Y
U
E

Yield surface is given by
 
2 2 2 2
1 1 1 1 2 1 3 2 3
2 0
Y
         
      
2 2
e
f Y

 
 
2 2 2
1 1 1 1 2 1 3 2 3
2
e
          
     
Maximum Shear stress (Tresca) Criterion
.
This was originally proposed by Tresca
2
e
Y
f 
 
Yield function is defined as
where the effective stress is
max
e
 

2 3
1
3 1
2
1 2
3
2
2
2
 

 

 







Magnitude of the extreme values of the stresses
are
Conditions in which yielding
can occur in a
multi-axial stress state
2 3
3 1
1 2
Y
Y
Y
 
 
 
  
  
  
Distortional Energy Density (von Mises)
Criterion
Originally proposed by von Mises & is the most popular for ductile materials
Total strain energy density = SED due to volumetric change +SED due to distortion
       
2 2 2 2
1 2 3 1 2 2 3 3 1
0
18 12
U
G
        
      
 
     
2 2 2
1 2 2 3 3 1
12
D
U
G
     
    

The yield surface is given by
2
2
1
3
J Y

     
2 2 2 2
1 2 2 3 3 1
1 1
6 3
f Y
     
 
      
 
Distortional Energy Density (von Mises)
Criterion contd.
Alternate form of the yield function
2 2
e
f Y

 
where the effective stress is
     
2 2 2
1 2 2 3 3 1 2
1
3
2
e J
      
 
      
 
       
2 2 2 2 2 2
1
3
2
e xx yy yy zz zz xx xy yz xz
         
 
        
 
 
J2 and the octahedral shear stress are related by
2
2
3
2
oct
J 
 
Hence the von Mises yield criterion can be written as
2
3
oct
f Y

 
Problems
4.7: The design loads of a member made of a brittle
material produce the following nonzero stress
components at the critical section in the member
𝜎𝑥𝑥 = −60𝑀𝑃𝑎, 𝜎𝑦𝑦 = 80𝑀𝑃𝑎, 𝑎𝑛𝑑 𝜎𝑥𝑦 = 70𝑀𝑃𝑎. The
ultimate strength of the material is 460MPa.
Determine the factor of safety used in the design.
a. Apply the maximum principal stress criterion
b. Apply the maximum principal strain criterion
and use 𝜐 = 0.20
Problems
4.10: A member made of steel (𝐸 =
200𝐺𝑃𝑎 𝑎𝑛𝑑 𝜐 = 0.29) is subjected to state of
plane strain when the design loads are applied. At
the critical point in the member, three state of the
stress components are 𝜎𝑥𝑥 = 60𝑀𝑃𝑎, 𝜎𝑦𝑦 =
240𝑀𝑃𝑎, 𝑎𝑛𝑑 𝜎𝑥𝑦 = −80𝑀𝑃𝑎. The material has a
yield stress 𝑌 = 490. Based on the maximum shear-
stress criterion determine the factor of safety.
4.11: Solve problem 4.10 using octahedral shear-
stress criterion.
General Yielding
The failure of a material is when the structure cannot support
the intended function
For some special cases, the loading will continue to increase
even beyond the initial load
At this point, part of the member will still be in elastic range.
When the entire member reaches the inelastic range, then the
general yielding occurs
2
,
6
Y Y
bh
P Ybh M Y
 
P Y
P Ybh P
 
2
1.5
4
P Y
bh
M Y M
 
Elastic Plastic Bending
Consider a beam made up of elastic-perfectly plastic material
subjected to bending. We want to find the maximum bending
moment the beam can sustain
1 ( )
,
( )
zz Y
Y
k a
where
Y
b
E
  

 

( )
2
Y
h
y c
k

0 ( )
Z zz
F dA d

  

/2
0
/2
0
2 2 0
2 2 ( )
Y
Y
Y
Y
y h
x ZZ y
y
y h
EP zz
y
M M ydA Y dA
or
M M ydA Y ydA e


    
  
 
 
Elastic Plastic Bending contd.
2
2 2
3 1 3 1
(4.43)
6 2 2 2 2
EP Y
Ybh
M M
k k
   
   
   
   
3
2
EP Y P
M M M
 
2
, /6
Y
where M Ybh

as k becomes large
Fully Plastic Bending
Definition: Bending required to
cause yielding either in tension
or compression over the entire
cross section
0
z zz
F dA

 
 
Equilibrium condition
Fully plastic moment is
2
P
t b
M Ybt

 
  
 
Comparison of failure yield criteria
For a tensile specimen
of ductile steel the
following six quantities
attain their critical
values at the same load
PY
1. Maximum principal stress reaches the yield strength Y
2. Maximum principal strain reaches the value
3. Strain energy Uo absorbed by the material per unit volume reaches
the value
4. The maximum shear stress reaches the
tresca shear strength
5. The distortional energy density UD reaches
6. The octahedral shear stress
max
( / )
Y
P A
 
max max
( / )
E
 
 /
Y Y E
 
2
0 / 2
Y
U Y E

max
( / 2 )
Y
P A
 
( / 2)
Y Y
 
2
/ 6
DY
U Y G

2 /3 0.471
oct Y Y
  
Failure criteria for general yielding
Interpretation of failure criteria for general yielding
Combined Bending and Loading
2 2 2
2
According to Maximum shear stress criteria, yielding starts when
or 4 1
2 2 2
Y
Y
  

     
   
     
     
2 2
2 2
According to the octahedral shear-stress criterion, yielding starts when
2 6 2
or 3 1
3 3
Y
Y Y
   
    
  
   
   
Interpretation of failure criteria for general yielding
Comparison of von Mises and Tresca criteria
Suggested Problems
4.7, 4.8, 4.9, 4.12, 4.13,
4.14, 4.17, 4.25, 4.26, 4.28,
4.29, 4.33, 4.34, 4.35, 4.36,
4.37, 4.38, 4.40, 4.41

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Chapter - 4.pptx

  • 2. Introduction Load – stress and load – deflection relation requires stress – strain relations Strain Tensor → Stress Tensor (Material Behavior)
  • 3. First Law of thermodynamics Law : The net external work and heat into the system is equal to increase in potential and kinetic energies W H U K        Variation of work δW of the external forces that act on the volume V with surface S is equal to δWs and δWB       S B xx xy xz yx yy yz zx zy zz x y z V W W W u v w u v w u v w B u B v B w dV x y z                                                   
  • 4. First Law of thermodynamics   2 2 2 xx xx yy yy zz zz xy xy yz yz zx zx V W dV                     which simplifies to 𝑈 = 𝑉 𝑈𝑜𝑑𝑉 Internal Energy per unit volume 𝛿𝑈 = 𝑉 𝛿𝑈𝑜𝑑𝑉 Variation of internal energy
  • 5. First Law of thermodynamics . Relation between strain components and strain energy Uo 0 2 2 2 xx xx yy yy zz zz xy xy xz xz yz yz U                    Strain energy density function Uo depends on strain components, the coordinates and the temperature   0 0 , , , , , , , , , xx yy zz xy xz yz U U x y z T        Variations in strain components and the function Uo are related by 0 0 0 0 0 0 0 xx yy zz xy xz yz xx yy zz xy xz yz U U U U U U U                               
  • 6. Hooke’s law: Isotropic Elasticity Isotropy : Directional independence Homogeneity : Spatial independence A material can be anisotropic and homogeneous or inhomogeneous and anisotropic Isotropic material - Two independent constants are required to describe the material 1 ( ) 1 ( ) 1 ( ) 1 1 2 1 1 2 xx xx yy zz yy yy xx zz zz zz xx yy xy xy xy yz yz yz E E E G E G E                                    Strain-stress relations are given as :-
  • 7. Hooke’s law: Isotropic Elasticity contd. [(1 ) ( )] (1 )(1 2 ) [(1 ) ( )] (1 )(1 2 ) [(1 ) ( )] (1 )(1 2 ) , , 1 1 1 xx xx yy zz yy yy xx zz zz zz xx yy xy xy xz xz yz yz E E E E E E                                                          Alternatively the stress- strain relation can also be written as :
  • 8. Hooke’s law: Isotropic Elasticity contd. In the case of plane stress, 0 zz xz yz       2 2 ( ) 1 ( ) 1 1 xx xx yy yy xx yy xy xy E E E                    In the case of plane strain, 0 zz xz yz       [(1 ) ] (1 )(1 2 ) [ (1 ) ] (1 )(1 2 ) ( ) (1 )(1 2 ) , 0 (1 ) xx xx yy yy xx yy zz xx yy xy xy xz yz E E E E                                         
  • 9. Hooke’s law: Isotropic Elasticity contd. Strain Energy Density 𝑈𝑜 = 1 2𝐸 𝜎2 𝑥𝑥 + 𝜎2 𝑦𝑦 + 𝜎2 𝑧𝑧 − 2𝜐 𝜎𝑥𝑥𝜎𝑦𝑦 + 𝜎𝑥𝑥𝜎𝑧𝑧 + 𝜎𝑧𝑧𝜎𝑦𝑦
  • 10. Objectives Nonlinear material behavior Yield criteria Yielding in ductile materials Sections 4.1 Limitations of Uniaxial Stress- Strain data 4.2 Nonlinear Material Response 4.3 Yield Criteria : General Concepts 4.4 Yielding of Ductile Materials 4.6 General Yielding
  • 11. Introduction When a material is elastic, it returns to the same state (at macroscopic, microscopic and atomistic levels) upon removal of all external load Any material is not elastic can be assumed to be inelastic E.g.. Viscoelastic, Viscoplastic, and plastic  To use the measured quantities like yield strength etc. we need some criteria The criteria are mathematical concepts motivated by strong experimental observations E.g. Ductile materials fail by shear stress on planes of maximum shear stress Brittle materials by direct tensile loading without much yielding  Other factors affecting material behavior - Temperature - Rate of loading - Loading/ Unloading cycles
  • 12.
  • 13.
  • 16. Models for Uniaxial stress-strain All constitutive equations are models that are supposed to represent the physical behavior as described by experimental stress-strain response Experimental Stress strain curvesIdealized stress strain curves Elastic- perfectly plastic response
  • 17. Models for Uniaxial stress-strain contd. . Linear elastic response Elastic strain hardening response
  • 18. Models for Uniaxial stress-strain contd. . Rigid models Rigid- perfectly plastic response Rigid- strain hardening plastic response
  • 20. The Yield Criteria : General concepts General Theory of Plasticity defines Yield criteria : predicts material yield under multi-axial state of stress Flow rule : relation between plastic strain increment and stress increment Hardening rule: Evolution of yield surface with strain Yield Criterion is a mathematical postulate and is defined by a yield function  , ( ) ij f f Y   where Y is the yield strength in uniaxial load, and is correlated with the history of stress state. Maximum Principal Stress Criterion:- used for brittle materials Maximum Principal Strain Criterion:- sometimes used for brittle materials Strain energy density criterion:- ellipse in the principal stress plane Maximum shear stress criterion (a.k.a Tresca):- popularly used for ductile materials Von Mises or Distortional energy criterion:- most popular for ductile materials Some Yield criteria developed over the years are:
  • 21. Maximum Principal Stress Criterion Originally proposed by Rankine   1 2 3 max , , f Y      1 Y    2 Y    3 Y    Yield surface is:
  • 22. Maximum Principal Strain This was originally proposed by St. Venant 1 1 2 3 0 f Y         1 2 3 Y        2 1 2 3 0 f Y         2 1 3 Y        3 3 1 2 0 f Y         3 1 2 Y        or or or Hence the effective stress may be defined as max e i j k i j k          The yield function may be defined as e f Y   
  • 23. Strain Energy Density Criterion . This was originally proposed by Beltrami Strain energy density is found as   2 2 2 0 1 1 1 1 2 1 3 2 3 1 2 0 2 U E                      Strain energy density at yield in uniaxial tension test 2 0 2 Y Y U E  Yield surface is given by   2 2 2 2 1 1 1 1 2 1 3 2 3 2 0 Y                  2 2 e f Y      2 2 2 1 1 1 1 2 1 3 2 3 2 e                 
  • 24. Maximum Shear stress (Tresca) Criterion . This was originally proposed by Tresca 2 e Y f    Yield function is defined as where the effective stress is max e    2 3 1 3 1 2 1 2 3 2 2 2                Magnitude of the extreme values of the stresses are Conditions in which yielding can occur in a multi-axial stress state 2 3 3 1 1 2 Y Y Y               
  • 25. Distortional Energy Density (von Mises) Criterion Originally proposed by von Mises & is the most popular for ductile materials Total strain energy density = SED due to volumetric change +SED due to distortion         2 2 2 2 1 2 3 1 2 2 3 3 1 0 18 12 U G                         2 2 2 1 2 2 3 3 1 12 D U G             The yield surface is given by 2 2 1 3 J Y        2 2 2 2 1 2 2 3 3 1 1 1 6 3 f Y                 
  • 26. Distortional Energy Density (von Mises) Criterion contd. Alternate form of the yield function 2 2 e f Y    where the effective stress is       2 2 2 1 2 2 3 3 1 2 1 3 2 e J                           2 2 2 2 2 2 1 3 2 e xx yy yy zz zz xx xy yz xz                          J2 and the octahedral shear stress are related by 2 2 3 2 oct J    Hence the von Mises yield criterion can be written as 2 3 oct f Y   
  • 27. Problems 4.7: The design loads of a member made of a brittle material produce the following nonzero stress components at the critical section in the member 𝜎𝑥𝑥 = −60𝑀𝑃𝑎, 𝜎𝑦𝑦 = 80𝑀𝑃𝑎, 𝑎𝑛𝑑 𝜎𝑥𝑦 = 70𝑀𝑃𝑎. The ultimate strength of the material is 460MPa. Determine the factor of safety used in the design. a. Apply the maximum principal stress criterion b. Apply the maximum principal strain criterion and use 𝜐 = 0.20
  • 28. Problems 4.10: A member made of steel (𝐸 = 200𝐺𝑃𝑎 𝑎𝑛𝑑 𝜐 = 0.29) is subjected to state of plane strain when the design loads are applied. At the critical point in the member, three state of the stress components are 𝜎𝑥𝑥 = 60𝑀𝑃𝑎, 𝜎𝑦𝑦 = 240𝑀𝑃𝑎, 𝑎𝑛𝑑 𝜎𝑥𝑦 = −80𝑀𝑃𝑎. The material has a yield stress 𝑌 = 490. Based on the maximum shear- stress criterion determine the factor of safety. 4.11: Solve problem 4.10 using octahedral shear- stress criterion.
  • 29. General Yielding The failure of a material is when the structure cannot support the intended function For some special cases, the loading will continue to increase even beyond the initial load At this point, part of the member will still be in elastic range. When the entire member reaches the inelastic range, then the general yielding occurs 2 , 6 Y Y bh P Ybh M Y   P Y P Ybh P   2 1.5 4 P Y bh M Y M  
  • 30. Elastic Plastic Bending Consider a beam made up of elastic-perfectly plastic material subjected to bending. We want to find the maximum bending moment the beam can sustain 1 ( ) , ( ) zz Y Y k a where Y b E        ( ) 2 Y h y c k  0 ( ) Z zz F dA d      /2 0 /2 0 2 2 0 2 2 ( ) Y Y Y Y y h x ZZ y y y h EP zz y M M ydA Y dA or M M ydA Y ydA e              
  • 31. Elastic Plastic Bending contd. 2 2 2 3 1 3 1 (4.43) 6 2 2 2 2 EP Y Ybh M M k k                 3 2 EP Y P M M M   2 , /6 Y where M Ybh  as k becomes large
  • 32. Fully Plastic Bending Definition: Bending required to cause yielding either in tension or compression over the entire cross section 0 z zz F dA      Equilibrium condition Fully plastic moment is 2 P t b M Ybt        
  • 33. Comparison of failure yield criteria For a tensile specimen of ductile steel the following six quantities attain their critical values at the same load PY 1. Maximum principal stress reaches the yield strength Y 2. Maximum principal strain reaches the value 3. Strain energy Uo absorbed by the material per unit volume reaches the value 4. The maximum shear stress reaches the tresca shear strength 5. The distortional energy density UD reaches 6. The octahedral shear stress max ( / ) Y P A   max max ( / ) E    / Y Y E   2 0 / 2 Y U Y E  max ( / 2 ) Y P A   ( / 2) Y Y   2 / 6 DY U Y G  2 /3 0.471 oct Y Y   
  • 34. Failure criteria for general yielding
  • 35. Interpretation of failure criteria for general yielding
  • 36. Combined Bending and Loading 2 2 2 2 According to Maximum shear stress criteria, yielding starts when or 4 1 2 2 2 Y Y                           2 2 2 2 According to the octahedral shear-stress criterion, yielding starts when 2 6 2 or 3 1 3 3 Y Y Y                    
  • 37. Interpretation of failure criteria for general yielding Comparison of von Mises and Tresca criteria
  • 38.
  • 39.
  • 40. Suggested Problems 4.7, 4.8, 4.9, 4.12, 4.13, 4.14, 4.17, 4.25, 4.26, 4.28, 4.29, 4.33, 4.34, 4.35, 4.36, 4.37, 4.38, 4.40, 4.41

Editor's Notes

  1. The practical application of mathematics and science to create, design, test, improve, and develop knowledge, research, money, business, economics, and technology. This is why engineering is such a challenging and demanding field of study: It involves areas of expertise that continue to evolve independently yet are required to perform together as part of the engineering process. Thus, an engineer must be expert in many areas, must know how to communicate knowledge between those areas, and must apply that knowledge to create, design, study, research, and invent all kinds of things. It is not uncommon for engineers to begin their careers as mathematicians, applied scientists, or even economists
  2. The practical application of mathematics and science to create, design, test, improve, and develop knowledge, research, money, business, economics, and technology. This is why engineering is such a challenging and demanding field of study: It involves areas of expertise that continue to evolve independently yet are required to perform together as part of the engineering process. Thus, an engineer must be expert in many areas, must know how to communicate knowledge between those areas, and must apply that knowledge to create, design, study, research, and invent all kinds of things. It is not uncommon for engineers to begin their careers as mathematicians, applied scientists, or even economists
  3. The practical application of mathematics and science to create, design, test, improve, and develop knowledge, research, money, business, economics, and technology. This is why engineering is such a challenging and demanding field of study: It involves areas of expertise that continue to evolve independently yet are required to perform together as part of the engineering process. Thus, an engineer must be expert in many areas, must know how to communicate knowledge between those areas, and must apply that knowledge to create, design, study, research, and invent all kinds of things. It is not uncommon for engineers to begin their careers as mathematicians, applied scientists, or even economists
  4. The practical application of mathematics and science to create, design, test, improve, and develop knowledge, research, money, business, economics, and technology. This is why engineering is such a challenging and demanding field of study: It involves areas of expertise that continue to evolve independently yet are required to perform together as part of the engineering process. Thus, an engineer must be expert in many areas, must know how to communicate knowledge between those areas, and must apply that knowledge to create, design, study, research, and invent all kinds of things. It is not uncommon for engineers to begin their careers as mathematicians, applied scientists, or even economists
  5. The practical application of mathematics and science to create, design, test, improve, and develop knowledge, research, money, business, economics, and technology. This is why engineering is such a challenging and demanding field of study: It involves areas of expertise that continue to evolve independently yet are required to perform together as part of the engineering process. Thus, an engineer must be expert in many areas, must know how to communicate knowledge between those areas, and must apply that knowledge to create, design, study, research, and invent all kinds of things. It is not uncommon for engineers to begin their careers as mathematicians, applied scientists, or even economists
  6. The practical application of mathematics and science to create, design, test, improve, and develop knowledge, research, money, business, economics, and technology. This is why engineering is such a challenging and demanding field of study: It involves areas of expertise that continue to evolve independently yet are required to perform together as part of the engineering process. Thus, an engineer must be expert in many areas, must know how to communicate knowledge between those areas, and must apply that knowledge to create, design, study, research, and invent all kinds of things. It is not uncommon for engineers to begin their careers as mathematicians, applied scientists, or even economists
  7. The practical application of mathematics and science to create, design, test, improve, and develop knowledge, research, money, business, economics, and technology. This is why engineering is such a challenging and demanding field of study: It involves areas of expertise that continue to evolve independently yet are required to perform together as part of the engineering process. Thus, an engineer must be expert in many areas, must know how to communicate knowledge between those areas, and must apply that knowledge to create, design, study, research, and invent all kinds of things. It is not uncommon for engineers to begin their careers as mathematicians, applied scientists, or even economists
  8. The practical application of mathematics and science to create, design, test, improve, and develop knowledge, research, money, business, economics, and technology. This is why engineering is such a challenging and demanding field of study: It involves areas of expertise that continue to evolve independently yet are required to perform together as part of the engineering process. Thus, an engineer must be expert in many areas, must know how to communicate knowledge between those areas, and must apply that knowledge to create, design, study, research, and invent all kinds of things. It is not uncommon for engineers to begin their careers as mathematicians, applied scientists, or even economists
  9. The practical application of mathematics and science to create, design, test, improve, and develop knowledge, research, money, business, economics, and technology. This is why engineering is such a challenging and demanding field of study: It involves areas of expertise that continue to evolve independently yet are required to perform together as part of the engineering process. Thus, an engineer must be expert in many areas, must know how to communicate knowledge between those areas, and must apply that knowledge to create, design, study, research, and invent all kinds of things. It is not uncommon for engineers to begin their careers as mathematicians, applied scientists, or even economists
  10. The practical application of mathematics and science to create, design, test, improve, and develop knowledge, research, money, business, economics, and technology. This is why engineering is such a challenging and demanding field of study: It involves areas of expertise that continue to evolve independently yet are required to perform together as part of the engineering process. Thus, an engineer must be expert in many areas, must know how to communicate knowledge between those areas, and must apply that knowledge to create, design, study, research, and invent all kinds of things. It is not uncommon for engineers to begin their careers as mathematicians, applied scientists, or even economists
  11. The practical application of mathematics and science to create, design, test, improve, and develop knowledge, research, money, business, economics, and technology. This is why engineering is such a challenging and demanding field of study: It involves areas of expertise that continue to evolve independently yet are required to perform together as part of the engineering process. Thus, an engineer must be expert in many areas, must know how to communicate knowledge between those areas, and must apply that knowledge to create, design, study, research, and invent all kinds of things. It is not uncommon for engineers to begin their careers as mathematicians, applied scientists, or even economists
  12. The practical application of mathematics and science to create, design, test, improve, and develop knowledge, research, money, business, economics, and technology. This is why engineering is such a challenging and demanding field of study: It involves areas of expertise that continue to evolve independently yet are required to perform together as part of the engineering process. Thus, an engineer must be expert in many areas, must know how to communicate knowledge between those areas, and must apply that knowledge to create, design, study, research, and invent all kinds of things. It is not uncommon for engineers to begin their careers as mathematicians, applied scientists, or even economists
  13. The practical application of mathematics and science to create, design, test, improve, and develop knowledge, research, money, business, economics, and technology. This is why engineering is such a challenging and demanding field of study: It involves areas of expertise that continue to evolve independently yet are required to perform together as part of the engineering process. Thus, an engineer must be expert in many areas, must know how to communicate knowledge between those areas, and must apply that knowledge to create, design, study, research, and invent all kinds of things. It is not uncommon for engineers to begin their careers as mathematicians, applied scientists, or even economists