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Russell C. Hibbeler

Chapter 1: Stress

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2002 The McGraw-Hill Companies, Inc. All rights reserved.

MECHANICS OF MATERIALSThird

Edition

Beer Johnston DeWolf

1 - 2

Contents

Concept of Stress

Review of Statics

Structure Free-Body Diagram

Component Free-Body Diagram

Method of Joints

Stress Analysis

Design

Axial Loading: Normal Stress

Centric & Eccentric Loading

Shearing Stress

Shearing Stress Examples

Bearing Stress in Connections

Stress Analysis & Design Example

Rod & Boom Normal Stresses

Pin Shearing Stresses

Pin Bearing Stresses

Stress in Two Force Members

Stress on an Oblique Plane

Maximum Stresses

Stress Under General Loadings

State of Stress

Factor of Safety
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Edition


1 - 3

Concept of Stress

The main objective of the study of mechanics of materials is to

provide the future engineer with the means of analyzing and

designing various machines and load bearing structures.

Both the analysis and design of a given structure involve thedetermination ofstressesand deformations. This chapter is

devoted to the concept of stress.
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2008 Pearson Education South Asia Pte Ltd

Chapter 1: Stress

Mechanics of Material 7thEdition

Introduction

Mechanics of materials is a study of the relationshipbetween the external loads on a body and the

intensity of the internal loads within the body.

This subject also involves the deformationsand

stability of a body when subjected to external forces.

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Chapter 1: Stress


Equilibrium of a Deformable Body

External Forces1.Surface Forces

- caused by direct contact

of other bodys surface

2.Body Forces

- other body exerts a force

without contact

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Chapter 1: Stress



Reactions Surface forces developed at the supports/points of

contact between bodies.

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7/49 2008 Pearson Education South Asia Pte Ltd

Chapter 1: Stress



Equations of Equilibrium Equilibrium of a body requires a balance of forces

and a balance of moments

For a body withx, y, z coordinate system with origin

O,

Best way to account for these forces is to draw

the bodys free-bo dy diagram (FBD).

0M0F O

0,0,0

0,0,0

zyx

zyx

MMM

FFF

MECHANICS OF MATERIALSTE

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8/49 2002 The McGraw-Hill Companies, Inc. All rights reserved.


Edition


1 - 8

Review of Statics

The structure is designed tosupport a 30 kN load

Perform a static analysis to

determine the internal force in

each structural member and the

reaction forces at the supports

The structure consists of a

boom and rod joined by pins

(zero moment connections) at

the junctions and supports


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Edition


1 - 9

Structure Free-Body Diagram

Structure is detached from supports and

the loads and reaction forces are indicated

Ayand Cycan not be determined from

these equations

kN30

0kN300

kN40

0

kN40

m8.0kN30m6.00

yy

yyy

xx

xxx

x

xC

CA

CAF

AC

CAF

A

AM

Conditions for static equilibrium:

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Edition


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Component Free-Body Diagram

In addition to the complete structure, each

component must satisfy the conditions forstatic equilibrium

Results:

kN30kN40kN40 yx CCA

Reaction forces are directed along boom

and rod

0

m8.00

y

yB

A

AM

Consider a free-body diagram for the boom:

kN30yC

substitute into the structure equilibrium

equation


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Edition


1 - 11

Method of Joints

The boom and rod are 2-force members, i.e.,

the members are subjected to only two forceswhich are applied at member ends

kN50kN40

3

kN30

54

0

BCAB

BCAB

B

FF

FF

F

Joints must satisfy the conditions for static

equilibrium which may be expressed in the

form of a force triangle:

For equilibrium, the forces must be parallel to

to an axis between the force application points,

equal in magnitude, and in opposite directions


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Edition


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Stress Analysis

Conclusion: the strength of memberBCisadequate

MPa165all

From the material properties for steel, theallowable stress is

Can the structure safely support the 30 kN

load?

MPa159m10314

N105026-

3

A

PBC

At any section through member BC, theinternal force is 50 kN with a force intensity

or stress of

dBC= 20 mm

From a statics analysis

FAB= 40 kN (compression)

FBC= 50 kN (tension)


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Edition


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Design

Design of new structures requires selection of

appropriate materials and component dimensions

to meet performance requirements

For reasons based on cost, weight, availability,

etc., the choice is made to construct the rod from

aluminum all= 100 MPa). What is an

appropriate choice for the rod diameter?

mm2.25m1052.2m10500444

m10500Pa10100

N1050

226

2

26

6

3

Ad

dA

PA

A

P

allall

An aluminum rod 26 mm or more in diameter is

adequate

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Chapter 1: Stress



Internal Resultant Loadings Objective of 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, Td) Bending moment, M

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Chapter 1: Stress


Example 1.1Determine the resultant internal loadings acting on the cross section at C of the

beam.

Solution:Free body Diagram

mN1809

270

6 w

w

Distributed loading at C is found by proportion,

Magnitude of the resultant of the distributed load, N5406180

21 F

which acts from C m2631

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Chapter 1: Stress


Solution:Equations of Equilibrium

(Ans)mN0108

02540;0

(Ans)540

0540;0

(Ans)0

0;0

C

CC

C

Cy

C

Cx

M

MM

V

VF

N

NF

Applying the equations of equilibrium we have

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Chapter 1: Stress


Example 1.5Determine the resultant internal loadings acting on the cross section at B of the

pipe. The pipe has a mass of 2 kg/m and is subjected to both a vertical force of50 N and a couple moment of 70 Nm at its endA. It is fixed to the wall at C.

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Chapter 1: Stress


SolutionFree-Body Diagram

N525.2481.925.12

N81.981.95.02

AD

BD

WW

Calculating the weight of each segment of pipe,

Applying the six scalar equations of equilibrium,

(Ans)N3.84

050525.2481.9;0

(Ans)0;0

(Ans)0;0

xB

zBz

yBy

xBx

F

FF

FF

FF

(Ans)0;0

(Ans)mN8.77

025.150625.0525.24;0(Ans)mN3.30

025.081.95.0525.245.05070;0

zBzB

yB

yByB

xB

xBxB

MM

M

MMM

MM

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Chapter 1: Stress


Stress

Distr ibut ionof internal loading is important inmechanics of materials.

We will consider the material to be con t inuous.

Thisin tensi tyof internal force at a point is called

stress.

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Chapter 1: Stress


Stress

Normal Stress Force per unit area acting normal toA

Shear Stress

Force per unit area acting tangent toA

A

FzA

z

0lim

A

F

A

F

y

Azy

x

Azx

0

0

lim

lim

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Chapter 1: Stress


Average Normal Stress in an Axially Loaded Bar

When a cro ss -sect io nal area bar is subjected toaxial force through the centroid, it is only subjected

to normal stress.

Stress is assumed to be averaged over the area.

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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 componentsthat are equal in magnitude

but opposite in direction.

A

P

AP

dAdFA

= average normal stress

P = resultant normal force

A = cross sectional area of bar

MECHANICS OF MATERIALSTh

Ed

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MECHANICS OF MATERIALShird

dition


1 - 23

The normal stress at a particular point may not be

equal to the average stress but the resultant of the

stress distribution must satisfy

Aave dAdFAP

Axial Loading: Normal Stress

The resultant of the internal forces for an axially

loaded member is normalto a section cutperpendicular to the member axis.

A

P

A

FaveA

0lim

The force intensity on that section is defined as

the normal stress.

The detailed distribution of stress is statically

indeterminate, i.e., can not be found from statics

alone.

MECHANICS OF MATERIALSTh

Ed

B J h t D W lf

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MECHANICS OF MATERIALSirddition


1 - 24

If a two-force member is eccentrically loaded,

then the resultant of the stress distribution in asection must yield an axial force and a

moment.

Centric & Eccentric Loading

The stress distributions in eccentrically loaded

members cannot be uniform or symmetric.

A uniform distribution of stress in a section

infers that the line of action for the resultant of

the internal forces passes through the centroidof the section.

A uniform distribution of stress is only

possible if the concentrated loads on the end

sections of two-force members are applied atthe section centroids. This is referred to as

centric loading.

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Chapter 1: Stress


Example 1.6The 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 loadingshown.

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

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Chapter 1: Stress


Graphically, the normal force diagram is as shown.

Solution:

By inspection, the largest loading is in region BC,

kN30BCP

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

the largest average normal stress is

(Ans)MPa7.8501.0035.0

10303

A

PBCBC

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Chapter 1: Stress


3kN/m80st

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

. Determine the average compressive stressacting at pointsA and B.

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

the internal axial force P at the section is

kN042.8

02.08.080

0;0

2

P

P

WPF stz

The average compressive stress becomes

(Ans)kN/m0.64

2.0

042.8 22

A

P

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Chapter 1: Stress


Average Shear Stress

The average shear stress distributed over eachsectioned area that develops a shear force.

2 different types of shear:

A

Vavg

= average shear stressP = internal resultant shear force

A = area at that section

a) Single Shear b) Double Shear

MECHANICS OF MATERIALSThi

Edi


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MECHANICS OF MATERIALSrdtion


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Shearing Stress Examples

A

F

A

Pave

Single Shear

A

F

A

P

2ave

Double Shear

MECHANICS OF MATERIALSThi

Edi


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Bearing Stress in Connections

Bolts, rivets, and pins create

stresses on the points of contactor bearing surfacesof the

members they connect.

dt

P

A

Pb

Corresponding average force

intensity is called the bearing

stress,

The resultant of the force

distribution on the surface isequal and opposite to the force

exerted on the pin.

MECHANICS OF MATERIALSThir

Edit Beer Johnston DeWolf

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Would like to determine the

stresses in the members and

connections of the structure

shown.

Stress Analysis & Design Example

Must consider maximum

normal stresses inABand

BC, and the shearing stress

and bearing stress at each

pinned connection

From a statics analysis:

FAB= 40 kN (compression)

FBC= 50 kN (tension)



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Rod & Boom Normal Stresses

The rod is in tension with an axial force of 50 kN.

The boom is in compression with an axial force of 40

kN and average normal stress of 26.7 MPa.

The minimum area sections at the boom ends are

unstressed since the boom is in compression.

MPa167m10300

1050

m10300mm25mm40mm20

26

3

,

26

N

A

P

A

endBC

At the flattened rod ends, the smallest cross-sectional

area occurs at the pin centerline,

At the rod center, the average normal stress in thecircular cross-section (A= 314x10-6m2) is BC= +159

MPa.



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1 - 33


The cross-sectional area for pins atA,B,

and C,26

22

m104912

mm25

rA

MPa102m10491

N105026

3

,

A

PaveC

The force on the pin at Cis equal to the

force exerted by the rodBC,

The pin atAis in double shear with atotal force equal to the force exerted by

the boomAB,

MPa7.40m10491

kN2026,

A

PaveA



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1 - 34

Divide the pin atBinto sections to determine

the section with the largest shear force,

(largest)kN25

kN15

G

E

P

P

MPa9.50m10491

kN2526,

A

PGaveB

Evaluate the corresponding averageshearing stress,




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Pin Bearing Stresses

To determine the bearing stress atAin the boomAB,

we have t= 30 mm and d= 25 mm,

MPa3.53

mm25mm30

kN40

td

Pb

To determine the bearing stress atAin the bracket,

we have t= 2(25 mm) = 50 mm and d= 25 mm,

MPa0.32

mm25mm50

kN40

td

Pb



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MECHANICS OF MATERIALSrdion


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Stress in Two Force Members

Will show that either axial or

transverse forces may produce bothnormal and shear stresses with respect

to a plane other than one cut

perpendicular to the member axis.

Axial forces on a two force

member result in only normal

stresses on a plane cut

perpendicular to the member axis.

Transverse forces on bolts and

pins result in only shear stresseson the plane perpendicular to bolt

or pin axis.



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MECHANICS OF MATERIALSdion

1 - 37

Pass a section through the member forming

an angle q with the normal plane.

qq

q

q

q

q

q

q

q

cossin

cos

sin

cos

cos

cos

00

2

00

A

P

A

P

A

V

A

P

A

P

A

F

The average normal and shear stresses on

the oblique plane are

Stress on an Oblique Plane

qq sincos PVPF

ResolvePinto components normal andtangential to the oblique section,

From equilibrium conditions, the

distributed forces (stresses) on the plane

must be equivalent to the forceP.


Editi Beer Johnston DeWolf

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1 - 38

The maximum normal stress occurs when the

reference plane is perpendicular to the member

axis,

00

m A

P

The maximum shear stress occurs for a plane at

+ 45owith respect to the axis,

00 2

45cos45sinA

P

A

Pm

Maximum Stresses

qqq cossincos0

2

0 A

P

A

P

Normal and shearing stresses on an oblique

plane



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1 - 39

Stress Under General Loadings

A member subjected to a general

combination of loads is cut into

two segments by a plane passing

through Q

For equilibrium, an equal and

opposite internal force and stress

distribution must be exerted on

the other segment of the member.

A

V

A

V

A

F

xz

Axz

xy

Axy

x

Ax

limlim

lim

00

0

The distribution of internal stress

components may be defined as,



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MECHANICS OF MATERIALSdon

1 - 40

Stress components are defined for the planes

cut parallel to thex,yandzaxes. For

equilibrium, equal and opposite stresses areexerted on the hidden planes.

It follows that only 6 components of stress are

required to define the complete state of stress

The combination of forces generated by the

stresses must satisfy the conditions for

equilibrium:

0

0

zyx

zyx

MMM

FFF

yxxy

yxxyz aAaAM

0

zyyzzyyz andsimilarly,

Consider the moments about thezaxis:

State of Stress



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MECHANICS OF MATERIALSdon

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Factor of Safety

stressallowablestressultimate

safetyofFactor

all

u

FS

FS

Structural members or machines

must be designed such that theworking stresses are less than the

ultimate strength of the material.

Factor of safety considerations:

uncertainty in material properties uncertainty of loadings

uncertainty of analyses

number of loading cycles

types of failure

maintenance requirements and

deterioration effects

importance of member to structures

integrity

risk to life and property

influence on machine function

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Chapter 1: Stress


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

the average compressive stress along the smooth areas of contact defined byABand BC, and the average shear stress along the horizontal plane defined by

EDB.

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

N240003000;0

N180003000;0

54

53

BCBCy

ABABx

FFF

FFF

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Chapter 1: Stress


The shear force acting on the sectioned horizontal plane EDB is

Solution:

N1800;0 VFx

Average compressive stresses along the AB and BC planes are

(Ans)N/mm20.1

4050

2400

(Ans)N/mm80.14025

1800

2

2

BC

AB

(Ans)N/mm60.0

4075

1800 2avg

Average shear stress acting on the BD plane is

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Chapter 1: Stress


Allowable Stress

Many unknown factors that influence the actualstress in a member.

A factor of safety is needed to obtained allowable

load.

The facto r o f safety (F.S.) is a ratio of the failureload divided by the allowable load

allow

fail

allow

fail

allow

fail

SF

SF

F

FSF

.

.

.

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Chapter 1: Stress


Example 1.14The control arm is subjected to the loading. Determine to the nearest 5 mm the

required diameter of the steel pin at C if the allowable shear stress for the steel is. Note in the figure that the pin is subjected to double shear.

Solution:For equilibrium we have

MPa55allowable

kN3002515;0

kN502515;0kN150125.025075.0152.0;0

53

54

5

3

yyy

xxx

ABABC

CCF

CCFFFM

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Chapter 1: Stress


Solution:

The pin at C resists the resultant force at C. Therefore,

kN41.30305 22 CF

mm8.18

mm45.2462

m1045.2761055

205.15

2

26

3

2

d

d

VA

allowable

The pin is subjected to double shear, a shear force of 15.205 kN acts over its cross-

sectional area between the arm and each supporting leaf for the pin.

The required area is

Use a pin with a diameter of d = 20 mm. (Ans)

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Chapter 1: Stress


Example 1.17The rigid barAB supported by a steel rodAC having a diameter of 20 mm and an

aluminum block having a cross sectional area of 1800 mm2. The 18-mm-diameterpins atA and C are subjected to single shear. If the failure stress for the steel and

aluminum is and respectively, and the failure

shear stress for each pin is , determine the largest load P that can be

applied to the bar. Apply a factor of safety of F.S. = 2.

Solution:The allowable stresses are

MPa680failst

MPa4502

900

..

MPa352

70

..

MPa3402

680

..

SF

SF

SF

fail

allow

failal

allowal

failst

allowst

MPa70failal

MPa900fail

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Chapter 1: Stress


There are three unknowns and we apply the equations of equilibrium,

Solution:

(2)075.02;0

(1)0225.1;0

PFM

FPM

BA

ACB

We will now determine each value of P that creates the allowable stress in the rod,

block, and pins, respectively.

For rod AC, kN8.10601.010340 26 ACallowstAC AF

Using Eq. 1,

kN17125.1

28.106P

For block B, kN0.63101800103566

BallowalB AF

Using Eq. 2,

kN16875.0

20.63P

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Chapter 1: Stress

Solution:

For pin A or C,

kN5.114009.010450 26 AFV

allowAC

Using Eq. 1,

kN18325.1

25.114P

When P reaches its smallest value (168 kN), it develops the allowable normal

stress in the aluminium block. Hence,

(Ans)kN168P

L1- June 14 MOM

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