Mechanics in Energy Resources Engineering - Chapter 2. Axially Loaded Members (2)

Week 4, 22 March

Mechanics in Energy Resources Engineering - Chapter 2. Axially Loaded Members (2)

Ki B k Mi PhD Ki-Bok Min, PhD

Assistant Professor

E R E i i

Energy Resources Engineering

Seoul National University

Preview

• Introduction

• Changes in Lengths of Axially Loaded Members (축하중을 받는g g y (

부재의 길이변화)

• Changes in Lengths Under Nonuniform Conditions (균일봉 길이변화)g g ( )

• Statically Indeterminate Structures (부정정 구조물)

Th l Eff t Mi fit d P t i (열효과 어긋남 및 사전변형)

• Thermal Effects, Misfits, and Prestrains (열효과, 어긋남 및 사전변형)

• Stresses on Inclined Sections (경사면에서의 응력)

• Strain Energy (변형율 에너지)

• Stress Concentrations* (응력집중)Stress Concentrations (응력집중)

• Impact Loading* (충격하중)

Thermal Effects, misfits, and prestrains

• Other sources of stresses and strains other than ‘external loads’

– Thermal effects: arises from temperature change – Misfits: results from imperfections in construction – Misfits: results from imperfections in construction – Prestrains: produced by initial deformation

Thermal Effects

• Changes in temperature produce expansion or contraction  thermal strains

• Thermal strain, ε

 

  

– α: coefficient of thermal expansion (1/K or 1/°C). e.g., granitite:

~1 0 ×10^-5 /°C

T

 

  

1.0 ×10 / C

– Heated  Expansion (+), Cooled  contraction (-)

• Displacement by thermal expansion

T TL

 

T T 

 

Thermal Effects

• No restraints  free expansion or contraction

– Thermal strain is NOT followed by thermal stressThermal strain is NOT followed by thermal stress

– Generally, statically determinate structures do not produce thermal stress

stress

• With supports that prevent free expansion and contraction  Thermal stress generated

Thermal stress generated

– How much thermal stress?

Thermal Effects

• Two bars in the left were under

uniform temperature increase of ΔT.

Fixed support R_A1

Fixed support R_A2

– E: Elastic Modulus

α: Coefficient of Thermal Expansion

A

R_A1

A

R_A2

– If E₁=E₂ and α₁> α₂, which bar will generate the higher thermal stress?

1 2

– If α₁= α₂ and E₁>E₂, which bar will generate the higher thermal stress?

g g

– Will R_A and R_B the same?

B

R

B

R Fixed support

R_B1

Fixed support R_B2

Thermal Effects

Calc lation of Thermal stress (E ample 2 7) Calculation of Thermal stress (Example 2-7)

– Equilibrium Eq.

 0





– Compatibility Eq.

AB T R 0

   

– Displacement Relations

 

T T L

   R ^A

R L

  EA – Compat. Eq.  Displ. Rel. ^EA

  ^{A 0}

T R

T L R L

     EA 

– Reactions ^EA

 

A B

R  R  EA T

– Thermal Stress in the bar  T ^{A B}  

R R

E T A A

     

Thermal Effects

C l l ti f Th l t

Calculation of Thermal stress

• Thermal Stress in the bar

 

A B

R R

E T

     

– Stress independent of the length (L) & cross-sectional area (A)

 

T E T

A A

     

– Assumptions: ΔT uniform, homogeneous, linearly elastic material – Lateral strain?

Stresses on inclined sections

• Stresses on inclined sections a more complete picture

– Finding the stresses on section pq.

– Resultant of stresses : still P N, Normal Force

– Normal Force (N) and Shear Force (V)

V, Shear Force

cos

N  P  ^{V  Psin}

– Normal Stress (σ) and shear stress (τ)

  N   V A1

 

A1

 

A: area of cross-section A

A1: area of inclined section ¹ cos A A

 

Stresses on inclined sections

• Based on the sign convention (note minus shear stress),

cos2

N P

       V   P sin cos 

1

A Acos

   

1

sin cos

A A

    

 

2 1

cos  1 cos 2  1

sin cos  sin 2

 

cos2 1 cos 2

  2   sin cos sin 2

   2 

 

2 1

1 2

   i   ^x i 2

– Above equation are independent of material (property and elastic…).

 

cos2 1 cos 2

x 2 x

       sin cos sin 2

2

x

 x

       

– Maximum stresses…why is this important?

When θ = -45°

max x

  _max

2

x

  When θ = 0

Stresses on inclined sections

• Element A:

– maximum normal stress maximum normal stress

– no shear _max _x Maximum normal stress Maximum shear stress

• Element B:

– The stresses at θ = 135°, -45°, and -135° can be obtained from previous equations.

– Maximum shear stresses

– One-half the maximum normal

max 2

x

 

One half the maximum normal stress

Stresses on inclined sections

• Same equations can be used for uniaxial compression

• What will happen if material is much weaker in shear than in

• What will happen if material is much weaker in shear than in compression (or tension)

Sh t f il

– Shear stress may cause failure

Strain Energy (변형율 에너지)

• Static load:

– Load applied slowly without dynamic Load applied slowly without dynamic effects due to motion

• P moves through distance δ and

• P moves through distance δ and

does a certain amount of work

Strain Energy

• The work (W) done by the incremental

loading:

W 



– The work done by the load is equal to the area below the load-displacement curve



• Strain Energy (U):

E b b d b th b d i th – Energy absorbed by the bar during the

loading process  internal work

St i k d b th l d

– Strain energy = work done by the load (when no E subtracted in the form of heat)

1 1

U W 



Strain Energy

• Elastic and Inelastic Strain Energy

Strain Energy

Li l El ti B h i Linearly Elastic Behavior

• Strain Energy for linear elastic bar

U _W _ P U W 2

PL

  EA

2 2

P L EA

Positive for both (+) & ( ) P

2 2

P L EA

U EA L

  

– Positive for both (+) & (-) P

– With unchanged load (P), L↑, U↑

– However, E↑ or A↑ - - > U↓

2 2

k  EA _{2 2}

2 2

P k U k

   L

Strain Energy

N if B

Nonuniform Bars

• Total Strain Energy U of a bar consisting of several segments

n n 2

N L

St i i t li f ti f th l d

2

1 1 2

n n

i i i

i i i i

U U N L

  E A







– Strain energy is not a linear function of the loads, even when the material is linearly elastic

2 2

(P  P ) L P L

L1, E1, A1

1 2 1 2 2

1 1 2 2

( )

2 2

P P L P L

U E A E A

  

L2, E2, A2

 2² 0

( ) 2 ( )

L N x dx

U 



Strain Energy E l 2 12 Example 2-12

• Compare the amounts of strain energy stored in the bars,

assuming linearly elastic behavior.

Strain Energy

E ample 2 14 Example 2-14

• Determine the vertical displacement, δ

, of joint B

Strain Energy

E ample 2 13 Example 2-13

• Determine the strain energy of a prismatic bar suspended from its upper end.

• Two cases;

( ) i ht f th b it lf – (a) weight of the bar itself

– (b) weight + a load P at the end

Strain Energy

Di l t d b Si l L d Displacement caused by a Single Load

• Strain energy stored in the structure

2

U _W _ P 2U

  P

– The displacement of a structure can be determined directly from the strain energy – Conditions:

Structure behave in a Linearly elastic manner

Only one load may act

Strain Energy

St i E D it

Strain-Energy Density

• Strain-Energy Density (u)

– Strain energy per unit volume of materialStrain energy per unit volume of material

2 2

2 2

2 2

U P E

AL u EA L

   

– Strain-energy density in terms of stress and strain

2 2

2 2

u E

E

 

  ^{2 2}

2 2

u E E

 

 

– Geometrical interpretation

σ

ε

Stress Concentration*

• Stress concentration

– Occurs when uniform stress are disturbed Occurs when uniform stress are disturbed by abrupt change in geometry (e.g., hole)

• Saint Venant’s Principle

• Saint Venant s Principle

– Peak stress directly under concentrated load P >> P/bt

load P >> P/bt

– Maximum stress diminish rapidly as we move away from the point of load

move away from the point of load application

At di t f th d f th b – At a distance from the end of the bar equal to the width b of the bar, stress distribution is nearly uniform.

Stress concentration

Stress concentration ca sed b the hole Stress concentration caused by the hole

• Stress concentration factor:

– K =3 for this case

max nom

K 

 

K 3 for this case

Stress concentration

With b >> d

Typical underground operation

Mechanics in Energy Resources Eng

Difference from Ci il/Mechanical Eng Difference from Civil/Mechanical Eng

• Mechanics of addition (덧셈의 역학) – Civil/Architectural Engineering, Machineries

– Ex) building with bricks (load on the column)

• Mechanics of removal (뺄셈의 역학) Underground

• Mechanics of removal (뺄셈의 역학) – Underground structure, underground/surface Mines

Deflected steel beam

– Ex) Drilling a borehole, Excavation of rock

Hendersen Mine, Colorado, USA (Molybdenum)

Civil structural problem:

Mechanics of “Addition”

Mechanics in Energy Resources Eng: Mechanics of “Removal”

Side view

1 2

Mechanics of Addition g

Monitoring points

1

2

Before

drilling/excavation

…

2

…

4

3

Start of

drilling/excavation

stress

1 2

drilling/excavation

t i

3

4 1

4

strain

Further advance of drilling/excavation

Impact Loading*

• Static loads: applied slowly and remain constant with time

• Dynamic loads: applied and removed suddenly or vary with time

vary with time

• Impact of an object falling onto the lower end of a

i ti b M i l ti ?

prismatic bar. Maximum elongation?

– Response is very complicated

– Approximate analysis by using the concept of strain energy

Impact Loading*

• Maximum elongation

– Potential energy lost by the falling mass (M) = Potential energy lost by the falling mass (M) maximum strain energy acquired by the bar

2

( ) ( ) EA max

M (h _max) W h( _max) ^max Mg h W h 2

  L

   

2 1/ 2

max WL WL 2 WL

EA EA h EA

  ^^ ^  ^ ^^

– Elongation due to the weight of the collar under static

EA  EA   EA 

loading condition

st

WL

  EA

1/ 2 max

1 1 2

st

st

  h



   

 

   

   

 

Impact Loading*

• Maximum stress

Emax

_max 

 L

2 1/ 2

max

2

W W WhE

A A AL

  A  ^^ A ^  AL ^

st st

E W Mg

A A L

    

A A L

1/ 2 1/ 2 2

max

2 2

1 1

st st st st

st

hE hE

L L

    



   

   

       

     

• Impact factor

 

max

st

Impact Loading*

dd l li d l di suddenly applied loading

• A load is applied suddenly with no initial velocity.

– What is the difference from the static loading?What is the difference from the static loading?

– h  0 from ^{1/ 2}

max

1 1 2

st

st

  h



   

 

   

   

 ^st 

   

 

max 2 _st

  

– Impact factor is 2

Impact Loading*

E l 2 16 Example 2-16

• Calculate the maximum elongation of the bar due to the

impact? Impact factor?

Summary

• Introduction

• Changes in Lengths of Axially Loaded Members (축하중을 받는g g y (

부재의 길이변화)

• Changes in Lengths Under Nonuniform Conditions (균일봉 길이변화)g g ( )

• Statically Indeterminate Structures (부정정 구조물)

Th l Eff t Mi fit d P t i (열효과 어긋남 및 사전변형)

• Thermal Effects, Misfits, and Prestrains (열효과, 어긋남 및 사전변형)

• Stresses on Inclined Sections (경사면에서의 응력)

• Strain Energy (변형율 에너지)

• Impact Loading (충격하중)Impact Loading (충격하중)

• Stress Concentrations* (응력집중)

Chapter 3 Torsion

• Introduction

• Torsional Deformations of a circular bar

• Circular bars of linearly elastic materials Nonuniform torsion

• Nonuniform torsion

• Stresses and Strains in Pure Shear

• Relationship Between Moduli of Elasticity E and G

• Transmission of Power by Circular ShaftsTransmission of Power by Circular Shafts

• Statically Indeterminate Torsional Members

• Strain Energy in Torsion and Pure Shear