russell c. hibbeler chapter 1: stresses - justanswer · 2014/4/17 · chapter 1: stress mechanics...

Russell C. Hibbeler

Chapter 1: Stresses

© 2008 Pearson Education South Asia Pte Ltd

Chapter 1: Stress

Mechanics of Material 7th Edition

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.


Chapter 1: Stress


Stress

Distribution of internal loading is important in

mechanics of materials.

We will consider the material to be continuous.

This intensity of internal force at a point is called

stress.


Chapter 1: Stress


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

Az

0lim

A

F

A

F

y

Azy

x

Azx

0

0

lim

lim


Chapter 1: Stress


Average Normal Stress in an Axially Loaded Bar

When a cross-sectional area bar is subjected to

axial force through the centroid, it is only subjected

to normal stress.

Stress is assumed to be averaged over the area.


Chapter 1: Stress


Average Normal Stress in an Axially Loaded Bar

Average Normal Stress Distribution

When a bar is subjected to a

constant deformation,

Equilibrium

2 normal stress components

that are equal in magnitude

but opposite in direction.

A

P

AP

dAdFA

σ = average normal stress

P = resultant normal force

A = cross sectional area of bar


Chapter 1: Stress


Example 1.6 The bar has a constant width of 35 mm and a thickness of 10 mm. Determine the

maximum average normal stress in the bar when it is subjected to the loading

shown.

Solution: By inspection, different sections have different internal forces.


Chapter 1: Stress


Graphically, the normal force diagram is as shown.

Solution:

By inspection, the largest loading is in region BC,

kN 30BCP

Since the cross-sectional area of the bar is constant,

the largest average normal stress is

(Ans) MPa 7.8501.0035.0

1030 3

A

PBCBC


Chapter 1: Stress


3kN/m 80st

Example 1.8 The casting is made of steel that has a specific weight of

. Determine the average compressive stress

acting at points A and B.

Solution: By drawing a free-body diagram of the top segment,

the internal axial force P at the section is

kN 042.8

02.08.080

0 ;0

2

P

P

WPF stz

The average compressive stress becomes

(Ans) kN/m 0.642.0

042.8 2

2A

P


Chapter 1: Stress


Average Shear Stress

The average shear stress distributed over each

sectioned area that develops a shear force.

2 different types of shear:

A

Vavg

τ = average shear stress

P = internal resultant shear force

A = area at that section

a) Single Shear b) Double Shear


Chapter 1: Stress


Example 1.12 The inclined member is subjected to a compressive force of 3000 N. Determine

the average compressive stress along the smooth areas of contact defined by AB

and BC, and the average shear stress along the horizontal plane defined by

EDB.

Solution: The compressive forces acting on the areas of contact are

N 240003000 ;0

N 180003000 ;0

54

53

BCBCy

ABABx

FFF

FFF


Chapter 1: Stress


The shear force acting on the sectioned horizontal plane EDB is

Solution:

N 1800 ;0 VFx

Average compressive stresses along the AB and BC planes are

(Ans) N/mm 20.14050

2400

(Ans) N/mm 80.14025

1800

2

2

BC

AB

(Ans) N/mm 60.04075

1800 2

avg

Average shear stress acting on the BD plane is


Chapter 1: Stress


Allowable Stress

Many unknown factors that influence the actual

stress in a member.

A factor of safety is needed to obtained allowable

load.

The factor of safety (F.S.) is a ratio of the failure

load divided by the allowable load

allow

fail

allow

fail

allow

fail

SF

SF

F

FSF

.

.

.

Russell C. Hibbeler

Chapter 2: Strain


Chapter 1: Stress


Deformation

When a force is applied to a body, it will change the

body’s shape and size.

These changes are deformation.

Note the before and after positions of 3 line

segments where the material is subjected

to tension.


Chapter 1: Stress


Strain

Normal Strain

The elongation / contraction of a line segment per

unit of length is referred to as normal strain.

Average normal strain is defined as

If the normal strain is known, then the approximate

final length is

s

ssavg

'

ss 1'

+ε line elongate

-ε line contracts


Chapter 1: Stress


Strain

Units

Normal strain is a dimensionless quantity since it is a

ratio of two lengths.

Shear Strain

Change in angle between 2 line segments that were

perpendicular to one another refers to shear strain.

'lim2

along along tACnAB

nt

θ<90 +shear strain

θ>90 -shear strain


Chapter 1: Stress


Example 2.1 The slender rod creates a normal strain in the rod of where z is in

meters. Determine (a) displacement of end B due to the temperature increase,

and (b) the average normal strain in the rod.

Solution: Part (a)

Since the normal strain is reported at each point along the rod, it has

a deformed length of

The sum along the axis yields the deformed length of the rod is

The displacement of the end of the rod is therefore

2/131040 zz

dzzdz 2/1310401'

m 20239.010401'

2.0

0

2/13 dzzz

(Ans) mm39.2m00239.02.020239.0B


Chapter 1: Stress


Part (b)

Assumes the rod has an original length of 200 mm and a change in length of 2.39

mm. Hence,

Solution:

(Ans) mm/mm 0119.0200

39.2'

s

ssavg

Example 2.3 The plate is deformed into the dashed shape. If, in this deformed shape, horizontal

lines on the plate remain horizontal and do not change their length, determine (a)

the average normal strain along the side AB, and (b) the average shear strain in the

plate relative to the x and y axes.


Chapter 1: Stress


Part (a)

Line AB, coincident with the y axis, becomes line after deformation, thus the

length of this line is

The average normal strain for AB is therefore

The negative sign indicates the strain causes a contraction of AB.

Solution:

mm 018.24832250' 22AB

(Ans) mm/mm 1093.7240

250018.248' 3

AB

ABABavgAB


Chapter 1: Stress


Part (b)

As noted, the once 90° angle BAC between the sides of the plate, referenced

from the x, y axes, changes to θ’ due to the displacement of B to B’.

Since then is the angle shown in the figure. Thus,

Solution:

'2xy

xy

(Ans) rad 121.02250

3tan 1

xy

Russell C. Hibbeler

Chapter 3: Mechanical Properties

of Materials


Chapter 1: Stress


The Tension and Compression Test

The strength of a material depends on its ability to

sustain a load.

This property is to perform under the tension or

compression test.

The following machine is designed to read the load

required to maintain specimen stretching.


Chapter 1: Stress


The Stress–Strain Diagram

Conventional Stress–Strain Diagram

Nominal or engineering stress is obtained by

dividing the applied load P by the specimen’s original

cross-sectional area.

Nominal or engineering strain is obtained by

dividing the change in the specimen’s gauge length

by the specimen’s original gauge length.

0A

P

0L


Chapter 1: Stress




Stress-Strain Diagram

Elastic Behaviour Stress is proportional to the strain.

Material is said to be

linearly elastic.

Yielding Increase in stress above

elastic limit will cause material

to deform permanently.


Chapter 1: Stress




Stress-Strain Diagram

Strain Hardening. After yielding a further load will

reaches a ultimate stress.

Necking At ultimate stress, cross-sectional

area begins to decrease in a

localized region of the specimen.

Specimen breaks at the

fracture stress.


Chapter 1: Stress



True Stress–Strain Diagram

The values of stress and strain computed from these

measurements are called true stress and true strain.

Use this diagram since most engineering design is

done within the elastic range.


Chapter 1: Stress


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.


Chapter 1: Stress


Hooke’s Law

Hooke’s Law defines the linear relationship between

stress and strain within the elastic region.

E can be used only if a material has linear–elastic

behaviour.

Eσ = stress

E = modulus of elasticity or Young’s modulus

ε = strain


Chapter 1: Stress


Hooke’s Law

Strain Hardening

When ductile material is loaded into the plastic

region and then unloaded, elastic strain is recovered.

The plastic strain remains and material is subjected

to a permanent set.


Chapter 1: Stress


Strain Energy

When material is deformed by external loading, it will

store energy internally throughout its volume.

Energy is related to the strains called strain energy.

Modulus of Resilience

When stress reaches the proportional limit, the

strain-energy density is the modulus of resilience,

ur.

Eu

pl

plplr

2

2

1

2

1


Chapter 1: Stress


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.


Chapter 1: Stress


Example 3.2 The stress–strain diagram for an aluminum alloy that is used for making aircraft

parts is shown. When material is stressed to 600 MPa, find the permanent strain

that remains in the specimen when load is released. Also, compute the modulus of

resilience both before and after the load application.

Solution: When the specimen is subjected to the load,

the strain is approximately 0.023 mm/mm.

The slope of line OA is the modulus of elasticity,

From triangle CBD,

mm/mm 008.0100.7510600 9

6

CDCDCD

BDE

GPa 0.75006.0

450E


Chapter 1: Stress


Solution: This strain represents the amount of recovered elastic strain.

The permanent strain is

(Ans) MJ/m 40.2008.06002

1

2

1

(Ans) MJ/m 35.1006.04502

1

2

1

3

3

plplfinalr

plplinitialr

u

u

(Ans) mm/mm 0150.0008.0023.0OC

Computing the modulus of resilience,

Note that the SI system of units is measured in joules, where 1 J = 1 N • m.


Chapter 1: Stress


Poisson’s Ratio

Poisson’s ratio, v (nu), states that in the elastic

range, the ratio of these strains is a constant since

the deformations are proportional.

Negative sign since longitudinal elongation (positive

strain) causes lateral contraction (negative strain),

and vice versa.

long

latv Poisson’s ratio is dimensionless.

Typical values are 1/3 or 1/4.


Chapter 1: Stress


Example 3.4 A bar made of A-36 steel has the dimensions shown. If an axial force of is applied to

the bar, determine the change in its length and the change in the dimensions of its

cross section after applying the load. The material behaves elastically.

Solution: The normal stress in the bar is

mm/mm 108010200

100.16 6

6

6

st

zz

E

Pa 100.1605.01.0

1080 63

A

Pz

From the table for A-36 steel, Est = 200 GPa


Chapter 1: Stress


The axial elongation of the bar is therefore

Solution:

The contraction strains in both the x and y directions are

m/m 6.25108032.0 6

zstyx v

(Ans) m1205.11080 6

z zz L

The changes in the dimensions of the cross section are

(Ans) m28.105.0106.25

(Ans) m56.21.0106.25

6

6

yyy

xxx

L

L


Chapter 1: Stress


*Failure of Materials Due to Creep and Fatigue

Creep

When material support a load for long period of time,

it will deform until a sudden fracture occurs.

This time-dependent permanent deformation is

known as creep.

Both stress and/or temperature play a significant role

in the rate of creep.

Creep strength will decrease

for higher temperatures or

higher applied stresses.


Chapter 1: Stress


*Failure of Materials Due to Creep and Fatigue

Fatigue

When metal subjected to repeated cycles of stress

or strain, it will ultimately leads to fracture.

This behaviour is called fatigue.

Endurance or fatigue limit is a limit which no failure

can be detected after applying a load for a specified

number of cycles.

This limit can be

determined in S-N diagram.

russell c. hibbeler chapter 1: stresses - justanswer · 2014/4/17 · chapter 1: stress mechanics...

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