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Transcript of WK1 Introduction(010111)
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8/21/2019 WK1 Introduction(010111)
1/21
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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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- 2
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 -
8/21/2019 WK1 Introduction(010111)
3/21
2009 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALSFifth
Edition
Beer Johnston DeWolf Mazurek
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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8/21/2019 WK1 Introduction(010111)
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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
2009 The McGraw-Hill Companies, Inc. All rights reserved.
1IntroductionConcept of Stress
MECHANICS OF MATERIALSFE
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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- 5
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.
MECHANICS OF MATERIALSFE
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8/21/2019 WK1 Introduction(010111)
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2009 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALSFifth
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1- 9
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
MECHANICS OF MATERIALSFE
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8/21/2019 WK1 Introduction(010111)
7/21 2009 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALSFifth
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Beer Johnston DeWolf Mazurek
1- 10
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:
MECHANICS OF MATERIALSFE
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8/21/2019 WK1 Introduction(010111)
8/21 2009 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALSFifth
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1- 11
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
MECHANICS OF MATERIALSFE
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8/21/2019 WK1 Introduction(010111)
9/21 2009 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALSFifth
Edition
Beer Johnston DeWolf Mazurek
1- 12
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:
MECHANICS OF MATERIALSFE
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8/21/2019 WK1 Introduction(010111)
10/21 2009 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALSFifth
Edition
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1- 13
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
MECHANICS OF MATERIALSFE
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8/21/2019 WK1 Introduction(010111)
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MECHANICS OF MATERIALSFifth
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1- 14
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
MECHANICS OF MATERIALSFE
B J h t D W lf M k
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8/21/2019 WK1 Introduction(010111)
12/21 2009 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALSFifth
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1- 15
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
MECHANICS OF MATERIALSFE
B J h t D W lf M k
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8/21/2019 WK1 Introduction(010111)
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MECHANICS OF MATERIALSFifth
Edition
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1- 16
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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MECHANICS OF MATERIALSFifth
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1- 17
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
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MECHANICS OF MATERIALSFifth
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Shearing Stress Examples
Single Shear
A
F
A
Pave
Double Shear
A
F
A
P
2ave
MECHANICS OF MATERIALSFi
Ed
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1- 19
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 MATERIALSFi
Ed
Beer Johnston DeWolf Mazurek
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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)
MECHANICS OF MATERIALSFi
Ed
Beer Johnston DeWolf Mazurek
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8/21/2019 WK1 Introduction(010111)
18/21 2009 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALSfth
dition
Beer Johnston DeWolf Mazurek
1- 21
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
Beer Johnston DeWolf Mazurek
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8/21/2019 WK1 Introduction(010111)
19/21 2009 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALSfth
dition
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1- 23
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
MECHANICS OF MATERIALSFif
Ed
Beer Johnston DeWolf Mazurek
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8/21/2019 WK1 Introduction(010111)
20/21 2009 The McGraw-Hill Companies, Inc. All rights reserved.
MECHANICS OF MATERIALSfth
dition
Beer Johnston DeWolf Mazurek
1- 24
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
MECHANICS OF MATERIALSFif
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MECHANICS OF MATERIALSfth
dition
Beer Johnston DeWolf Mazurek
1 25
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