1 Learning outcomes Equilibrium body & Free Body Diagram - - PowerPoint PPT Presentation

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Learning outcomes

• Equilibrium body & Free Body Diagram
• Elastic & Plastic Deformation
• Engineering Stress & Strain
• This subject also involves the deformations and

stability of a body when subjected to external forces.

Introduction

• Mechanics of materials is a study of the

relationship between the external loads on a body and the intensity of the internal loads within the body.

• This subject also involves the deformations and

stability of a body when subjected to external forces.

Equilibrium of a Deformable Body

External Forces

1.Surface/ Linear/ Concentrated

Forces

• caused by direct contact of
• ther body’s surface

2.Body Forces

• other body exerts a force

without contact

Equilibrium of a Deformable Body

Internal Resultant Loadings

 Objective of free body diagram (FBD) is to determine the

resultant force and moment acting within a body.

 In general, there are 4 different types of resultant

loadings: a) Normal force, N b) Shear force, V c) Torsional moment or torque, T d) Bending moment, M

Free-body diagrams = diagrams used to show the relative magnitude and

direction of all forces acting upon an object in a given situation. A free-body diagram is a special example of the vector diagrams.

Equilibrium of a Deformable Body

Equations of Equilibrium

 Equilibrium of a body requires a balance of forces

and a balance of moments

 For a body with x, y, z coordinate system with origin

O,

 Best way to account for these forces is to draw

the body’s free-body diagram (FBD). M F  

 

O

, , , ,      

     

z y x z y x

M M M F F F

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Elastic Deformation

• 1. Initial
• 2. Small load
• 3. Unload

F d

bonds stretch return to initial

Elastic means reversible!

F d

Linear- elastic Non-Linear- elastic

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Plastic Deformation (Metals)

Plastic means permanent!

F d

linear elastic linear elastic

dplastic

• 1. Initial
• 2. Larger load
• 3. Unload

p lanes still sheared

F

d elastic + plastic bonds stretch & planes shear d plastic

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Engineering Stress

 Stress has units: N/m2 or lbf/in2

• Shear stress, t:

Area, A

Ft Ft Fs F F Fs t = Fs Ao

• Normal/Tensile stress, s:
• riginal area

before loading Area, A

Ft Ft s = Ft Ao

2 f 2

m N

• r

in lb =

Stress

Normal Stress σ

 Force per unit area acting normal to ΔA

Shear Stress τ

 Force per unit area acting tangent to ΔA

A Fz

A z

  

 

lim s A F A F

y A zy x A zx

     

   

lim lim t t

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Common States of Stress

• Simple tension: cable

Note: t = M/AcR here. Ao = cross sectional area (when unloaded)

F F

• s  F

A

• t  Fs

A s s M M Ao

2R

Fs Ac

• Torsion (a form of shear): drive shaft

Ski lift (photo courtesy

P.M. Anderson)

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OTHER COMMON STRESS STATES

(photo courtesy P.M. Anderson)

Canyon Bridge, Los Alamos, NM

• s F

A

• Simple compression:

Note: compressive structure member (s < 0 here).

(photo courtesy P.M. Anderson)

Ao

Balanced Rock, Arches National Park

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OTHER COMMON STRESS STATES

• Bi-axial tension:
• Hydrostatic compression:

Pressurized tank

s < 0

h

(photo courtesy P.M. Anderson) (photo courtesy P.M. Anderson)

Fish under water

s z > 0 s q > 0

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Engineering Strain

• Tensile strain:
• Lateral strain:
• Shear strain:

Strain is always dimensionless.

q

90º 90º - q

y x

q g = x/y = tan e  d Lo

• d

eL 

L

w o

d /2 d L/2

Lo wo

(terikan)

 Strain-energy density is strain energy per unit volume

• f material

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u = ∆U ∆V σe 2 =

If material behavior is linear elastic, Hooke’s law applies,

Strain Energy

Strain u = σ 2 σ2 2E = σ e

( )

E

Strain Energy

 When material is deformed by external loading,

energy is stored internally throughout its volume

 Internal energy is also referred to as strain energy  Stress develops a force,

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F = σ A = σ (x y)

STRAIN ENERGY

 Hooke’s law for shear

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τ = Gγ G is shear modulus of elasticity or modulus

• f rigidity

G can be measured as slope of line on τ-γ diagram, G = τpl/ γpl The three material constants E, ν, and G is related by G = E 2(1 + ν)

Shear

REVIEW

Tensile Strength (TS), M Fracture point, f Stress-Strain Graph Proportional Limit

Shear Stress-strain

 Use thin-tube specimens and subject it to torsional

loading

 Record measurements of applied torque and

resulting angle of twist

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Shear Stress-strain

 Material will exhibit linear-elastic behavior till its

proportional limit, τpl

 Strain-hardening continues till it reaches ultimate

shear stress, τu

 Material loses shear strength till it fractures, at stress

• f τf

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Stress-Strain Testing

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• Typical tensile test

machine specimen extensometer

• Typical tensile

specimen gauge length

Stress–Strain Behavior

Table 1 - Room-Temperature Elastic and Shear Modulus, and Poisson’s Ratio for Various Metal Alloys SE

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• Modulus of Elasticity, E:

(also known as Young's modulus)

• Hooke's Law:

s = E e s

Linear- elastic

E e F F

simple tension test

STRESS–STRAIN BEHAVIOR

Stress-Strain Behavior

Schematic stress–strain diagram showing non-linear elastic behavior, and how secant and tangent moduli are determined.

There are some material such as gray cast iron, concrete and many polymers which this elastic portion of the stress-strain curve is not linear Tangent modulus = E (Modulus of elastic)

Stress-Strain Behavior

 Slope of stress strain plot (which is proportional to the elastic modulus)

depends on bond strength of metal

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Force versus inter-atomic separation for weakly and strongly bonded

• atoms. The magnitude of the

modulus of elasticity is proportional to the slope of each curve at the equilibrium inter-atomic separation ro.

EXERCISE 1 A piece of copper originally 305 mm long is pulled in tension with a stress of 276MPa. If the deformation is entirely elastic, what will be the resultant elongation?

EXERCISE 2 A piece of aluminum originally long is half of 600mm pulled in tension with a stress of

• 275MPa. If the deformation is entirely elastic,

what will be the resultant elongation? Ans: 1.195x10-3

Poisson’s Ratio

 When body subjected to axial tensile force, it

elongates and contracts laterally

 Similarly, it will contract and its sides expand

laterally when subjected to an axial compressive force

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Poisson’s Ratio

 Strains of the bar are:

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Early 1800s, S.D. Poisson realized that within elastic range, ration of the two strains is a constant value, since both are proportional.

• ν is unique for homogenous and isotropic material
• Why negative sign? Longitudinal elongation cause lateral contraction (-ve strain) and

vice versa

• Lateral strain is the same in all lateral (radial) directions
• Poisson’s ratio is dimensionless, 0 ≤ ν ≤ 0.5

El Elas asti tic c Proper erties ties Of Mat ater erials als

Poisson’s ratio

From the tensile stress-strain behavior for the brass specimen shown in the figure, determine the following:

a) The modulus of elasticity b) The yield strength at a strain offset of 0.002 c) The maximum load that can be sustained by a cylindrical specimen having an original diameter of 12.8 mm d) The change in length of a specimen

• riginally 250 mm long that is subjected to a

tensile stress of 345MPa

Exercise III

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Yield Strength, y

• Stress at which noticeable plastic deformation has occurred.

when e = 0.002 @ 0.2%

sy = yield strength

Note: for 2 inch sample e = 0.002 = z/z  z = 0.004 in

tensile stress, s engineering strain, e

sy

e = 0.002

Plastic Elastic 0.002 Strain Offset Method

P (yielding point)

s

A typical s-e behavior for metal showing elastic/plastic deformation, proportional limit (P) & sY determined using 0.002 strain offset value.

At 0.2% strain, extrapolate line (dashed) parallel to OA till it intersects stress-strain curve at A’

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σYS = 469 MPa

EXAMPLE 3.1 (SOLN)

Yield Strength

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Yield Strength : Comparison

Room T values

Graphite/ Ceramics/ Semicond Metals/ Alloys Composites/ fibers Polymers

Yield strength, sy (MPa)

PVC

Hard to measure

, since in tension, fracture usually occurs before yield. Nylon 6,6 LDPE

70 20 40 60 50 100 10 30 200 300 400 500 600 700 10 00 2000

Tin (pure) Al (6061) a Al (6061) ag Cu (71500) hr Ta (pure) Ti (pure) a Steel (1020) hr Steel (1020) cd Steel (4140) a Steel (4140) qt Ti (5Al-2.5Sn) a W(pure) Mo (pure) Cu (71500) cw

Hard to measure,

in ceramic matrix and epoxy matrix composites, since in tension, fracture usually occurs before yield. HDPE PP

humid dry

PC PET ¨

Tensile Strength, TS

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• Metals: occurs when noticeable necking starts.
• Polymers: occurs when polymer backbone chains are

aligned and about to break.

sy

strain

Typical response of a metal

F = fracture or ultimate strength Neck – acts as stress concentrator

engineering TS stress engineering strain

• Maximum stress on engineering stress-strain curve.

Tensile Strength : Comparison

38 Si crystal

<100>

Graphite/ Ceramics/ Semicond Metals/ Alloys Composites/ fibers Polymers

Tensile strength, TS (MPa)

PVC Nylon 6,6

10 100 200 300 1000

Al (6061) a Al (6061) ag Cu (71500) hr Ta (pure) Ti (pure) a Steel (1020) Steel (4140) a Steel (4140) qt Ti (5Al-2.5Sn) a W(pure) Cu (71500) cw LDPE PP PC PET

20 30 40 2000 3000 5000

Graphite Al oxide Concrete Diamond Glass-soda Si nitride HDPE wood ( fiber) wood(|| fiber)

1

GFRE (|| fiber) GFRE ( fiber) CFRE (|| fiber) CFRE ( fiber) AFRE (|| fiber) AFRE( fiber) E-glass fib C fibers Aramid fib

Room Temp. values

Stress–Strain Behavior of Ductile and Brittle Materials

Ductile Materials

 Material that can subjected to large strains before it

ruptures is called a ductile material. Brittle Materials

 Materials that exhibit little or no yielding before failure are

referred to as brittle materials.

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Ductility

• Plastic tensile strain at failure:
• Another ductility measure:

100 x A A A RA %

• f
• -

=

x 100 L L L EL %

• f -



Engineering tensile strain, e Engineering tensile stress, s smaller %EL (elongation) larger %EL

Lf Ao Af Lo

• A measure of the degree of plastic deformation that has been sustained at

fracture.

• can express as %EL (elongation) or %RA (reduction in area)

Lf = Fracture Length

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DUCTILITY -Toughness

• Toughness = Energy to break a unit volume of material
• Approximate by the area under the stress-strain curve.

Brittle fracture : elastic energy Ductile fracture : elastic + plastic energy

very small toughness (unreinforced polymers)

Engineering tensile strain, e Engineering tensile stress, s

small toughness (ceramics) (Brittle) large toughness (metals) (Ductile)

Adapted from Fig. 6.13, Callister 7e.

Strain Energy

Modulus of Toughness

 Modulus of toughness, ut, represents the entire

area under the stress–strain diagram.

 It indicates the strain-energy density of the material

just before it fractures.

Elastic Strain Recovery

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Schematic tensile stress-strain diagram showing the phenomenon of elastic strain recovery and strain hardening. syo = the initial yield strength syi = yield strength after releasing the load at point D and then upon reloading

Hardness

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• Def  a measure of a material’s resistance to localized plastic deformation

eg small dent or scratch.

• Large hardness means:
• -high resistance to plastic deformation or cracking in compression.

e.g., 10 mm sphere apply known force measure size

• f indent after

removing load

d

D

Smaller indents mean larger hardness.

most plastics brasses Al alloys easy to machine steels file hard cutting tools nitrided steels diamond

increasing hardness

Hardness: Measurement

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Table 6.5

Hardness: Measurement

 Rockwell

 Combination of various indenters (spherical hardened steel balls dia 1/16, 1/8,

¼ and ½ in & conical diamond).

 Scale = HR (Rockwell Hardness no)  Hardness no = diff in depth of penetration with

various load applied.

 Minor load = 10 kg  Major load = 60 (A), 100 (B) & 150 (C) kg



A = diamond, B = 1/16 in. ball, C = diamond  Brinell

• Spherical indenter (harden steel dia 10mm) forced into surface of



specimen

• Load range between 500 to 3000 kg
• Scale = HB (Brinell Hardness no)
• Hardness Conversion :

 TS (psia) = 500 x HB  TS (MPa) = 3.45 x HB 46

Hardening

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• Curve fit to the stress-strain response:

sT  K eT

( )

n “true” stress (F/A) “true” strain: ln(L/Lo) hardening exponent: n = 0.15 (some steels) to n = 0.5 (some coppers)

• An increase in sy due to plastic deformation.

s e

large hardening small hardening

sy sy

1

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Design or Safety Factors

• Design uncertainties for actual stress in-service applications
• Need to consider Safety Factor, N

N

y working

s  s

Often N is between 1.2 and 4

• Example: Calculate a diameter, d, to ensure that yield does not occur in the 1045

carbon steel rod below. Use a factor of safety of 5.

( )

4 000 220

2 /

d N , 

5

N

y working

s  s

1045 plain carbon steel: sy = 310 MPa TS = 565 MPa F = 220,000N d L o d = 0.067 m = 6.7 cm

EXERCISE



A tensile testing is to be constructed that must withstand a maximum load of 110 kN. The design calls for two cylindrical support posts, each which is to half of the maximum load. Furthermore, plain-carbon steel ground and polished shafting rounds are to be used: the minimum yield of this alloy are 310 MPa respectively. Specify a suitable diameter for these support posts. Use a suitable safety factor.

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Summary

• Stress and strain: These are size-independent measures
• f load and displacement, respectively.
• Elastic behavior: This reversible behavior often shows a

linear relation between stress and strain. To minimize deformation, select a material with a large elastic modulus (E or G).

• Toughness: The energy needed to break a unit volume
• f material.
• Ductility : The plastic strain at failure.
• Plastic behavior: This permanent deformation behavior
• ccurs when the tensile (or compressive) uniaxial stress

reaches sy.

Chapter Review



Tension test is the most important test for determining material strengths. Results

• f

normal stress and normal strain can then be plotted.



Many engineering materials behave in a linear- elastic manner, where stress is proportional to strain, defined by Hooke’s law, σ = Ee. E is the modulus of elasticity, and is measured from slope of a stress-strain diagram



When material stressed beyond yield point, permanent deformation will occur.

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Chapter Review



Strain hardening causes further yielding of material with increasing stress



At ultimate stress, localized region on specimen begin to constrict, and starts “necking”. Fracture

• ccurs.



Ductile materials exhibit both plastic and elastic behavior. Ductility specified by permanent elongation to failure or by the permanent reduction in cross-sectional area



Brittle materials exhibit little or no yielding before failure

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

Yield point for material can be increased by strain hardening, by applying load great enough to cause increase in stress causing yielding, then releasing the load. The larger stress produced becomes the new yield point for the material



Deformations of material under load causes strain energy to be stored. Strain energy per unit volume/strain energy density is equivalent to area under stress-strain curve.

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

The area up to the yield point of stress-strain diagram is referred to as the modulus

• f

resilience



The entire area under the stress-strain diagram is referred to as the modulus of toughness



Poisson’s ratio (ν), a dimensionless property that measures the lateral strain to the longitudinal strain [0 ≤ ν ≤ 0.5]



For shear stress vs. strain diagram: within elastic region, τ = Gγ, where G is the shearing modulus, found from the slope of the line within elastic region



G can also be obtained from the relationship of G = E/[2(1+ ν)]

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s = E e

Tensile stress, s:

s = F A

100 x A A A RA %

• f
• -

=

x 100 L L L EL %

• f
• -



N

y working

s  s

e  ΔL Lo e  ΔD D

Thank you…