WK1 Introduction(010111)

8/21/2019 WK1 Introduction(010111)

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2009 The McGraw-Hill Companies, Inc. All rights reserved.

MECHANICS OF MATERIALSFifth

Edition

Beer Johnston DeWolf Mazurek

1- 1

BMM 1533 Strength Of Materials 1


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Edition


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Test 1 ( 7th week 2 hrs) 15 %

Test 2 (12th week. 2 hrs) 15 %

Assignments to be submitted within a week after each chapterQuestions

from lecturer (2 week once or 1 chapter per assignment.)15 %

Quizzes (Once every two weeks/every chapter) 15 %

Final Examination (Answer All four questions)Sample exam question

Fundamental Equations of Mechanic Of Materials and Geometric properties of Area elements will begiven)

40 %

Total 100%
http://localhost/var/www/apps/conversion/tmp/scratch_4/BMM1543_EXAM_FINAL_20607_SET_1.dochttp://localhost/var/www/apps/conversion/tmp/scratch_4/Exam%20Equations.ppthttp://localhost/var/www/apps/conversion/tmp/scratch_4/Exam%20Equations.ppthttp://localhost/var/www/apps/conversion/tmp/scratch_4/Exam%20Equations.ppthttp://localhost/var/www/apps/conversion/tmp/scratch_4/Exam%20Equations.ppthttp://localhost/var/www/apps/conversion/tmp/scratch_4/BMM1543_EXAM_FINAL_20607_SET_1.doc


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COs Chapters Assessments Lecturer

CO1

& 2

1 & 2 Test 1 & Exam En. Nasrul Test Week 7th

Answer all two (2)

questions.CO3 3 Test 1 & Exam Mr. Lee

CO4 4 Test2 Mr. Lee Test Week 12thAnswer all two (2)

questions.C05 5 & 6 Test2 & Exam Mr. Lee

CO6 7 Exam Dr. Shar

1- 3

Teaching is on the parallel basis. Lecturer must key in marks for his/her class.

Each chapter must be assess with one quiz and an assignment

COs Chapters Assessments Lecturer

CO1

& 2

1 & 2 Exam .. One Question En. Nasrul (Q1)

Answer all four

questions.

Need two sets of

questions.

CO3

& 4

3 & 4

(composite

beam only)

Exam One Question Mr. Lee (Q2)

C05 5 & 6 Exam . One Question En. Lee (Q3)

CO6 7 Exam One Question Dr. Shar (Q4)


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MECHANICS OF

MATERIALS

Fifth SI Edition

Ferdinand P. Beer

E. Russell Johnston, Jr.

John T. DeWolf

David F. Mazurek

Lecture Notes:

J. Walt Oler

Texas Tech University

CHAPTER


1IntroductionConcept of Stress

MECHANICS OF MATERIALSFE


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Concept of Stress

The main objective of the study of the mechanicsof 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 structureinvolve the determination ofstressesand

deformations. This chapter is devoted to the

concept of stress.



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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



7/21 2009 The McGraw-Hill Companies, Inc. All rights reserved.


Edition


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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:





Edition


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Component Free-Body Diagram

In addition to the complete structure, each

component must satisfy the conditions forstatic equilibrium

0

m8.00

y

yB

A

AM

Consider a free-body diagram for the boom:

kN30yC

substitute into the structure equilibrium

equation

Results:

kN30kN40kN40 yx CCA

Reaction forces are directed along boom

and rod





Edition


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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

For equilibrium, the forces must be parallel to

to an axis between the force application points,

equal in magnitude, and in opposite directions

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:





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Stress Analysis

Conclusion: the strength of memberBCisadequate

MPa165all

From the material properties for steel, theallowable stress is

From a statics analysis

FAB= 40 kN (compression)

FBC= 50 kN (tension)

Can the structure safely support the 30 kN

load?

dBC= 20 mm MPa159m10314

N105026-

3

A

PBC

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

or stress of





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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?

mm2.25m1052.2

m1050044

4

m10500Pa10100

N1050

226

2

26

6

3

Ad

dA

PA

A

P

allall

An aluminum rod 26 mm or more in diameter is

adequate


B J h t D W lf M k




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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

Fave

A

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.

The normal stress at a particular point may not be

equal to the average stress but the resultant of the

stress distribution must satisfy

A

ave dAdFAP


B J h t D W lf M k




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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 centroid

of 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.

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.

MECHANICS OF MATERIALSF

E

B J h t D W lf M k




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Shearing Stress

ForcesPand Pare applied transversely to the

memberAB.

A

Pave

The corresponding average shear stress is,

The resultant of the internal shear force

distribution is defined as theshearof the section

and is equal to the loadP.

Corresponding internal forces act in the plane

of section Cand are calledshearingforces.

Shear stress distribution varies from zero at themember surfaces to maximum values that may be

much larger than the average value.

The shear stress distribution cannot be assumed to

be uniform.

MECHANICS OF MATERIALSFi

E

Beer Johnston DeWolf Ma rek




Edition


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Shearing Stress Examples

Single Shear

A

F

A

Pave

Double Shear

A

F

A

P

2ave


Ed





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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.


Ed




MECHANICS OF MATERIALSifth

Edition


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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)


Ed




MECHANICS OF MATERIALSfth

dition


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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 40kN and average normal stress of26.7 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.

MECHANICS OF MATERIALSFif

Ed





dition


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Pin Shearing Stresses

The cross-sectional area for pins atA,B,

and C,26

22 m10491

2

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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dition


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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,

Pin Shearing Stresses

22 2015


Ed



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dition


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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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