CEE 150 Quarter Notes All

CEE 150CEE 150

Mechanics of MaterialsMechanics of Materials

Instructor:

Dr. Farzin Zareian 1 1

Dr. Farzin Zareian

Course outline

Course units: 4 units

Prerequisites: CEE 30 or ENGR 30, or MAE 30 (Statics)

Text Book: F.P. Beer, E.R. Johnston, J.T. DeWolf, Mechanics of Materials, Other Books:

Hibbeler, Mechanics of Materials, PEARSON (Prentice Hall)

J.M. Gere, S.P. Timoshenko, Mechanics of Materials

E P P T A B l E i i M h i f S lidE.P. Popov, T.A. Balan, Engineering Mechanics of Solids,

AND MANY OTHERS

Dr. Farzin Zareian

Farzin ZareianTypewritten TextMcGrawHill

Farzin ZareianTypewritten Text




Course outlineMeetings: Lecture: Tuesday and Thursdsday. 3:30 PM 4:50 PM.

DBH 1600

Office hours: Tuesday 2:00 PM 3:00 PM. E/4141 Engineering

Gateway

Work Credit (CEE 150): Home Work & Quizzes 20% (once a week)

Midterm exam 40% (Nov. 17th)Final exam 40% (Dec 8 4:00 - 6:00)

Total 100%

Dr. Farzin Zareian




Course outlineCourse Outcomes:

1- Apply knowledge of mathematics, science, and engineering dealing with mechanics of materials under axial loading, torsion, bending, and combined loading.

2- Draw axial force, torque, shear and moment diagrams of simple members subject to combined loading.

3- Compute stresses and strains in simple members subject to axial loading, torsion, bending, and combined loading.

4- Compute deflection of beams.

5- Compute buckling load of compressive members.

6- Design components to meet desired needs in terms of strength and deflection.

Dr. Farzin Zareian

Course outline

Topic ChaptersDateTopic ChaptersDate

Introduction & Concept of Stress and Strain

Review of Statistics, Stress, Analysis and Design of members, Different types of stresses, General

Chapter 1 (B&J)Chapter 1 -3 (Hibb.)

Week 1

Axial Loading

yp ,loading conditions and components of stress

Axial Loading

Normal Strain, Stress-Strain Diagram, Engineering vs. True Stress and Strain, Hooks Law, Elastic vs.

Week 1 Chapter 2 (B&J)Chapter 4 (Hibb.)and Strain, Hooks Law, Elastic vs.

Plastic Behavior, Axial Deformation, Poison Ratio, Shear Strain, Stress Concentration, Plastic Deformation

Chapter 4 (Hibb.)

Dr. Farzin Zareian

Course outline

Topic DateBeer & Johnson

Topic Date

Torsion

Section Numbers

Deformations in circular shaftsDeformations in circular shafts, Stress and strain in circular members, Polar moment of inertia, Stress concentration in circular

Week 2 Chapter 3 (B&J)Chapter 5 (Hibb.)

P B di

shafts, design of circular shafts for torsion

Pure BendingDeformation and stress/strain in symmetric members in pure bending (El ti d Pl ti b h i ) Stress

Week 3 Chapter 4 (B&J)Chapter 6 (Hibb )(Elastic and Plastic behavior), Stress

concentration, effect of eccentric loading on distribution of stress and strain design considerations

Chapter 6 (Hibb.)

Dr. Farzin Zareian

and strain, design considerations

Course outline


Topic Date

Analysis of Beams

Section Numbers

Sh d B di M tWeek 4 Chapter 4 (B&J)

Shear

Shear and Bending Moment Diagrams, Design for bending

Chapter 6 (Hibb.)

Shear

Distribution and determination of shear stress on the horizontal face of a beam element (rectangular

Week 5 Chapter 5 (B&J)Chapter 7 (Hibb )of a beam element (rectangular,

arbitrary shape, and thin walled members), design considerations.

Chapter 7 (Hibb.)

Dr. Farzin Zareian

Course outline


Topic Date

Transformation of Stress

Section Numbers

C bi d l di Pl tCombined loading, Plane stress, principal stresses and maximum shearing stress, Mohrs circle for plane stress, Design of beams for

Week 6Chapter 7-8 (B&J)Chapter 8-9 (Hibb.)

Deflection of Beams

plane stress, Design of beams for strength requirements

Deformation under transverse loading, curvature vs. rotation. Design of beams for deflection requirements Superposition

Week 7 Chapter 9 (B&J)Chapter 12 (Hibb.)requirements. Superposition Chapter 12 (Hibb.)

Dr. Farzin Zareian

Course outline

TopicBeer & Johnson

DateTopic

Columns

Section NumbersDate

Stability, Eulers formula for columns with different boundary conditions, design of columns s bjected to eccentric and

Week 9 Chapter 10 (B&J)Chpater 13 (Hibb.)

subjected to eccentric and concentric loads

Review Week 10Review Week 10

Dr. Farzin Zareian

Course outlineHomework Policy:

No Due date. However, there dates for quizzes (see each homework set)

One problem from two homework assignments (with changed numbers) will be test during lecture time as quiz.

You get only few minutes to do a quiz.

Your submitted work must be clean, show all work, and include units.

Submitted work that does not abide to these simple yet crucial rules will get almost no credit.

Dr. Farzin Zareian

HW 1 & 2on 10/13

HW 3 & 4on 10/27

HW 5 & 6on 11/10

HW 7 & 8on 11/24

HW 9 & 10on 12/3

Course outline

Exam Policy:

No make-up exam. Unless in extreme circumstances.

Please bring only a reliable calculator(s) an UCI Examination Blue Book(s) andPlease bring only a reliable calculator(s), an UCI Examination Blue Book(s), and use pencil.

The exams are closed book closed notes and closed discussionThe exams are closed-book, closed-notes, and closed-discussion

Class Attendance Policy:

I strongly recommend that you attend the class and join the discussions.

Quizzes are given during the class. MAY BE IN THE FIRST MINUTE

Dr. Farzin Zareian

Course outlinePoints for success:

Start early and keep up with the pace of the class Due to the heavy course loadStart early and keep up with the pace of the class. Due to the heavy course load we will be moving fast.

Invest time in doing your homeworks and do plenty of exerciseInvest time in doing your homeworks and do plenty of exercise.

Prepare before attending each class session.

Take full advantage of your discussion sessions and office hours We areTake full advantage of your discussion sessions and office hours. We are here to help, but we can not help you unless we know what is the

blproblem.

Dr. Farzin Zareian

Course outlineAcademic Dishonesty:Please review the UCIs academic dishonesty policy at: y p yhttp://www.senate.uci.edu/senateweb/default2.asp?active_page_id=754

You are required to read the complete description provided inYou are required to read the complete description provided in the academic senate website regarding academic dishonesty.y

A i id f d i di h i ENGRCEE 150 illAny incident of academic dishonesty in ENGRCEE 150 will be followed with disciplinary action to the fullest degree.

Dr. Farzin Zareian

Chapter 1

Introduction Concept of Stressp

Dr. Farzin Zareian 11

Introduction - Concept of Stress

Why do we need to know mechanics of materials?

Analysis of structuresFind stresses and deformations in structural members due to applied loadingapplied loading

Design of structuresFind dimensions and arrangement of structural members to withstand applied loads safely.

You are responsible for providing a reliable and safe path for loads applied to your structure fromsafe path for loads applied to your structure from their point of initiation up to the structures supports.

Dr. Farzin Zareian

supports.


Review of Statics

The structure is designed to support a 30 kN load.

The structure consists of a boom and rod joined by pins (zero

i ) hmoment connections) at the junctions and supports.

Perform a static analysis to determine the internal force in each structural member and theeach structural member and the reaction forces at the supports.

Di i b t t ti ll d t i t t t

Dr. Farzin Zareian

Discussion about statically determinate structure


Structure Free-Body DiagramSt t i d t h d f t d Structure is detached from supports and the loads and reaction forces are indicated.

Conditions for static equilibrium:

kN40

m8.0kN30m6.00 xCA

AM

kN400

kN40

xx

xxx

x

ACCAF

A

kN300kN300

yy

yyy

xx

CACAF

Ay and Cy can not be determined from these equations.

Dr. Farzin Zareian


Component Free-Body Diagram

I dditi t th l t t t h In addition to the complete structure, each component must satisfy the conditions for static equilibrium.

m8.00 yB AM Consider a free-body diagram for the boom:

0

y

yB

A

Substitute into the structure equilibrium

kN30yCequation

Results:

kN30kN40kN40 yx CCA

Reaction forces are directed along boom and rod

Dr. Farzin Zareian

g


Method of Joints

The boom and rod are 2-force members, i.e., the members are subjected to only two forces which are applied at member ends.

For equilibrium, the forces must be parallel to an axis between the force application points, equal in magnitude, and in opposite directionsin magnitude, and in opposite directions

Joints must satisfy the conditions for static equilibrium which may be expressed in the form

f f t i lof a force triangle:

kN300 BFF

F

kN50kN403kN30

54

BCABFF

FF

Dr. Farzin Zareian

kN50kN40 BCAB FF

Introduction - Concept of StressStress AnalysisCan the structure safely support the 30 kN load?Can the structure safely support the 30 kN load?

From a Statics analysis

FAB = 40 kN (compression) AB ( p )

FBC = 50 kN (tension)

At any section through member BC, the internal force is 50 kN with a force intensity or stress of

MPa159m10314N1050

26-

3

AP

BC m10314A

MPa165all From the material properties for steel, the

allowable stress is

Conclusion: the strength of member BC is adequate.

UNITS ??? (Lb N) ( 2 i 2) (P i)

Dr. Farzin Zareian

UNITS ??? (Lb vs. N), (m2 vs. in2), (Pa vs. psi)

Introduction - Concept of StressDesignDesign of new structures requires selection of appropriate materialsDesign 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 ( = 100 MPa) What is anto construct the rod from aluminum (all= 100 MPa). What is an appropriate choice for the rod diameter?

N1050 3PP m10500Pa10100N1050

2

266

3

dA

PAAP

allall

mm2.25m1052.2m10500444

226

Ad

A

An aluminum rod 26mm or more in diameter is adequate.

Dr. Farzin Zareian

Introduction - Concept of StressAxial Loading: Normal Stress The resultant of the internal forces for anThe resultant of the internal forces for an

axially loaded member is normal to a section cut perpendicular to the member axis.

AP

AF

ave lim

The force intensity on that section is defined as the normal stress.

AA aveA 0 The normal stress at a particular point may not be

equal to the average stress but the resultant of

ave dAdFAP

q gthe stress distribution must satisfy

A

The detailed distribution of stress is statically indeterminate, i.e., can not be found from statics

l

Dr. Farzin Zareian

alone.


Concentric & Eccentric Loading

A uniform distribution of stress in a section infers that the line of action for the resultant of th i t l f th h th t idthe internal forces passes through the centroid of the section.

A uniform distribution of stress is only possible y pif the concentrated loads on the end sections of two-force members are applied at the section centroids. This is referred to as concentric loading.

If a two-force member is eccentrically loaded, then the resultant of the stress distribution in athen the resultant of the stress distribution in a section must yield an axial force and a moment.

Dr. Farzin Zareian

Introduction - Concept of StressShearing Stress Forces P and P are applied transversely to theForces P and P are applied transversely to the

member AB. Corresponding internal forces act in the plane of section C and are called shearing forces.g

The resultant of the internal shear force distribution is defined as the shear of thedistribution is defined as the shear of the section and is equal to the load P. The corresponding average shear stress is,

APave

Shear stress distribution varies from zero at Shear stress distribution varies from zero at the member surfaces to maximum values that may be much larger than the average value.

Dr. Farzin Zareian


Shearing Stress Examples Double Shear

Single Shear

AF

AP ave FPave

Dr. Farzin Zareian

AAave AA 2ave


Bearing Stress in Connections

Bolts, rivets, and pins create stresses on the points of contact or bearing surfaces

f th b th tof the members they connect.

The resultant of the force distribution on the surface is equal and opposite to the force exerted on the pin.

PP

Corresponding average force intensity is called the bearing stress,

dtP

AP b

Dr. Farzin Zareian

Introduction - Concept of StressStress Analysis & Design Example 1

Determine the stresses in the members and connections of theconnections of the structure shown.

From a static analysis:

FAB = 40 kN (compression)

FBC = 50 kN (tension)

Must consider maximum normal stresses in AB and BC and thestresses in AB and BC, and the shearing stress and bearing stress at each pinned connection

Dr. Farzin Zareian

co ect o


Rod Stresses The rod is in tension with an axial The rod is in tension with an axial

force of 50 kN.

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

Dr. Farzin Zareian


Rod Stresses

At the flattened rod ends, the smallest cross-sectional area occurs at the pin centerline,

1050

m10300mm25mm40mm203

26

NPA

MPa167m10300

105026,

NAP

endBC

Dr. Farzin Zareian


Pin Shearing Stresses The cross-sectional area for pins at A B and C The cross-sectional area for pins at A, B, and C,

262

2 m104912mm25

rA

3

The force on the pin at C is equal to the force exerted by the rod BC,

MPa102m10491N1050

26

3

, A

PaveC

Dr. Farzin Zareian


Pin Shearing Stresses

Divide the pin at B into sections to determine the section with the largest shear force,

kN15EP(largest) kN25GP

Evaluate the corresponding average shearing stress,

MPa9.50m10491

kN2526, A

PGaveB

g ,

Dr. Farzin Zareian


Boom Normal Stresses

The boom is in compression with an axial force of 40 kN and average normal stress of

26 7 MP (A 30 X50 1 5 10 3 2)26.7 MPa.(A = 30mmX50mm=1.5x10-3m2)

Dr. Farzin Zareian


Pin Shearing Stresses

Th i t A i i d bl The pin at A is in double shear with a total force equal to the force exerted b the boom ABby the boom AB,

, 6 2

40 2 kN 40.7 MPa491 10 mA ave

PA

Dr. Farzin Zareian


Pin Bearing Stresses

To determine the bearing stress at A in the boom To determine the bearing stress at A in the boom AB, we have t = 30 mm and d = 25 mm,

MPa3.53kN40 Pb

To determine the bearing stress at A in the bracket, we

MPa3.53mm25mm30tdb

MPa0.32mm25mm50kN40

tdP

bhave t = 2(25 mm) = 50 mm and d = 25 mm,

Dr. Farzin Zareian

mm25mm50td

Introduction - Concept of StressExtra Stress Analysis & Design Examples Problems 1.15, 1.21, 1.25 from your book

Dr. Farzin Zareian


Dr. Farzin Zareian


Stress on an Oblique Plane Pass a section through the member forming Pass a section through the member forming

an angle with the normal plane. From equilibrium conditions, the q ,

distributed forces (stresses) on the plane must be equivalent to the force P.

sincos PVPF

Resolve P into components normal and tangential to the oblique section,

sincos PVPF

coscos 200 A

PAP

AF

The average normal and shear stresses on the oblique plane are

cossinsincos

00

00

AP

AP

AV

oblique plane are

Dr. Farzin Zareian

cos 0


Maximum Stresses

Normal and shearing stresses on an oblique plane

cossincos2 PP

The maximum normal stress occurs when h f l i di l h

00 AA

0m AP

the reference plane is perpendicular to the member axis,

0m A

The maximum shear stress occurs for a plane at + 45o with respect to the axis,

00 2

45cos45sinAP

AP

m

p p ,

Dr. Farzin Zareian


Stress Under General Loadings

A member subjected to a general combination of loads is cut into twosegments by a plane passing through Q.

The distribution of internal stress components may be defined as,

VVA

F

xx

x

Ax

lim0

AV

AV

z

Axz

y

Axy

limlim 00 For equilibrium, an equal and opposite

internal force and stress distribution must be exerted on the other segment of the member.

Dr. Farzin Zareian


State of Stress

The combination of forces generated by the stresses must satisfy the conditions for equilibrium:

00

zyx

zyx

MMMFFF

AAM 0 Consider the moments about the z axis:

yxxy

yxxyz aAaAM

andsimilarly,

0

zyyzzyyz andsimilarly,It follows that only 6 components of stress are

required to define the complete state of

Dr. Farzin Zareian

stress.


Factor of SafetyFactor of safety considerations:

Structural members or machines must be

Factor of safety considerations: uncertainty in material properties uncertainty of loadings

designed such that the working stresses are less than the ultimate

uncertainty of analyses number of loading cycles types of failure

strength of the material.

safetyofFactorFS

types of failure maintenance requirements and

deterioration effectsi t f b t i t it

stress allowablestress ultimate

safety ofFactor

all

u

FS

FS importance of member to integrity of whole structure

risk to life and property influence on machine function

Dr. Farzin Zareian

Stress and Strain Axial Loadingg


Stress & Strain Axial Loading

Stress & Strain: Axial Loading

Suitability of a structure or machine may depend on the d f i i h ll hdeformations in the structure as well as the stressesinduced under loading. Statics analyses alone are notsufficientsufficient.

This chapter is concerned with deformation of a pstructural member under axial loading. Later chapters will deal with torsional and pure bending loads.

Dr. Farzin Zareian


Normal StrainOBSERVATION

P PP2 P

strain normal

stress

L

AP

L

AP

AP

22

LL

AP

22

Dr. Farzin Zareian

L L LL2


Normal Strain

Uniform Cross section

L

General Cross section

li d 0

limx x dx

Strain is dimensionless 150 06 250 06 / 2500 600

E m E m m Dr. Farzin Zareian

0.600L m


Stress-Strain Diagram:D til M t i lDuctile Materials

Dr. Farzin Zareian

Brittle Materials


Stress-Strain Diagram:

Compression vs. Tension

Dr. Farzin Zareian


Stress-Strain Diagram:

Engineering Stress and Strain True Stress and Strain

0L LL L

L

tL dL LLn

L L L

0 0L LP

0 0t

LL L L P

Dr. Farzin Zareian

0A t A


Hookes Law: Modulus of Elasticity

Below the yield stress

E

ElasticityofModulus or Modulus Youngs

E

E

y

Strength is affected by alloying, heatStrength is affected by alloying, heat treating, and manufacturing process but stiffness (Modulus of Elasticity) is not.

Dr. Farzin Zareian


Elastic vs. Plastic Behavior

If the strain disappears when the stress is removed, thethe stress is removed, the material is said to behave elastically.

The largest stress for which this occurs is called the elastic limit.

When the strain does not return to zero after the stress is removed, thezero after the stress is removed, the material is said to behave plastically.

Dr. Farzin Zareian

Stress & Strain Axial LoadingFatigue

Fatigue properties are shown Fatigue properties are shown on -N diagrams.

A member may fail due to fatigueA member may fail due to fatigue at stress levels significantly below the ultimate strength if subjected to many loading cyclesto many loading cycles.

For ferrous material, when the t i d d b l thstress is reduced below the

endurance limit, fatigue failures do not occur for any number of cycles.

For nonferrous material, fatigue limit is the stress corresponding to failure at a specific number of

l

Dr. Farzin Zareian

cycles.


Deformations Under Axial Loading

AEP

EE

From Hookes Law:

From the definition of strain:

AEE

L

Equating and solving for the deformation,

AEPL

With variations in loading cross section or With variations in loading, cross-section or material properties,

( )( ) ( )

Li iPL P x dx

Dr. Farzin Zareian

0 ( ) ( )i i iA E A x E x


Sample ProblemDetermine the deformation of steel rod

shown below. Assume E = 29 E06 psi

Dr. Farzin Zareian


Sample Problem

The rigid bar BDE is supported by two links AB and CD.AB and CD.

Link AB is made of aluminum (E = 70 GPa) and has a cross-sectional area of 500 mm2and has a cross-sectional area of 500 mm .

Link CD is made of steel (E = 200 GPa) and has a cross-sectional area of 600 mm2.

For the 30-kN force shown, determine the deflection a) of B, b) of D, and c) of E.

Dr. Farzin Zareian


SolutionDisplacement of B:

APL

BFree body: Bar BDE Pa1070m10500 m3.0N1060 926-

3

AEB

m10514 6 mm514.0B

FM

CD

B

m2.0m6.0kN3000

B

PLDisplacement of D:

tensionFCD

CD

0M kN90

D Pa10200m10600 m4.0N1090 926-

3

AED

ncompressioF

F

AB

AB

kN60m2.0m4.0kN300

m10300Pa10200m10600

6

mm300.0D

Dr. Farzin Zareian

mm300.0D


Solution (Cont.)

BHBBDisplacement of E:

mm 200mm0 300mm 514.0

xx

HDDD

mm 7.73mm 0.300

xx

HEEE

mm773

mm7.73400mm3000

E

HDDD

mm 928.1mm 7.73mm 300.0

E

9281Dr. Farzin Zareian

mm928.1E


Static Indeterminacy

A ill b i ll i d i A structure will be statically indeterminatewhenever it is held by more supports than are required to maintain its equilibrium.

Redundant reactions are replaced with unknown loads which alongwith unknown loads which along with the other loads must produce compatible deformations.

Deformations due to actual loads and redundant reactions are determined separately and then added or superposed.

0 RL

Dr. Farzin Zareian


Sample Problem

What is the deformation of rod and tube when force P is exerted?tube when force P is exerted?

Dr. Farzin Zareian


Sample Problem

Determine the reactions at A and B for the steel bar and loading shown,the steel bar and loading shown, assuming a close fit at both supports before the loads are applied.

Dr. Farzin Zareian


Solution

S l f h di l B d h li dSolve for the displacement at B due to the applied loads with the redundant constraint released,

PPPP 343

321 N10900N106000

LLLLAAAAPPPP

4321

2643

2621

4321

m 150.0m10250m10400N10900N106000

EEALP

iii

ii9

L

10125.1 Solve for the displacement at B dueSolve for the displacement at B due

to the redundant constraint,

B

AARPP

262

261

21

m10250m10400

BiiR

RLP

LL3

21

21

1095.1m 300.0

Dr. Farzin Zareian

i

iiR EEA


Solution (Cont.)

R i h h di l d h l dRequire that the displacements due to the loads and due to the redundant reaction be compatible, 0 RL

kN577N10577

01095.110125.10

3

39

BRL

RE

RE

kN577N10577 3 BR

Find the reaction at A due to theFind the reaction at A due to the loads and the reaction at B

kN577kN600kN3000 RFkN323

kN577kN600kN3000

A

Ay

RRF

kN577kN323

B

A

RR

WHAT IF YOU HAD A GAP???

Dr. Farzin Zareian

BWHAT IF YOU HAD A GAP???


Thermal Stresses A temperature change results in a change in A temperature change results in a change in

length or thermal strain. There is no stress associated with the thermal strain unless the l i i i d b h

Treat the additional support as redundant and apply the principle of superposition

elongation is restrained by the supports.

AEPLLT PT

and apply the principle of superposition.

coef.expansion thermal The thermal deformation and the

deformation from the redundant support AEPLLT 0

0 PT deformation from the redundant support must be compatible.

TEPTAEP

AE

T T

Dr. Farzin Zareian

TEA

TPlease make sure that you do Example 2.06 of your book


Complementary Problems

Problem 2.48

200sE GPa 70sE GPa11.7 06 /s E C 23.6 06 /s E C

Unstressed at 20 degrees

What is the stress in Aluminum

shell at T = 180 degrees?

Dr. Farzin Zareian


Poissons Ratio

For a slender bar subjected to axial loading:0 zyxx E

The elongation in the x-direction is accompanied by a contraction in the other

0

directions. Assuming that the material is isotropic (no directional dependence),

0 zy

Poissons ratio is defined as

x

z

x

y

strain axialstrain lateral

Dr. Farzin Zareian


Generalized Hookes Law

For an element subjected to multi-axial loading, the normal strain components resulting from the stress components mayresulting from the stress components may be determined from the principle of superposition. This requires:

1) strain is linearly related to stress

2) deformations are small

EEEzyx

zyxx

With these restrictions:

EEE

EEEzyx

z

zyxy

Dr. Farzin Zareian

EEE



Problem 2.63 800Axial elongation 0.45"Diameter shrink 0.025"

P lb

? ? ?E G

Dr. Farzin Zareian


Dilatation: Bulk Modulus Relative to the unstressed state the dilatation Relative to the unstressed state, the dilatation

(change in volume per unit volume) is

zyxzyxe 111111 zyxzyx

zyxzyx

E 21

213 ppe For element subjected to uniform

hydrostatic pressure,

modulusbulk 213 Ek

kEpe

10 Subjected to uniform pressure, dilatation

must be negative, therefore

Dr. Farzin Zareian

210


Shearing Strain

A cubic element subjected to a shear stress will deform into a rhomboid. The corresponding shear strain is quantified in terms of the changeq gin angle between the sides, xyxy f

A plot of shear stress vs. shear strain is similar to the previous plots of normal stress vs. normal strain except that the

GGG

stress vs. normal strain except that the strength values are approximately half. For small strains,

zxzxyzyzxyxy GGG where G is the modulus of rigidity or shear modulus.

Dr. Farzin Zareian


Sample Problem

A rectangular block of material with modulus of rigidity G = 90 ksi is bonded to two rigid horizontal plates The lowerto two rigid horizontal plates. The lower plate is fixed, while the upper plate is subjected to a horizontal force P.

Knowing that the upper plate moves through 0.04 in. under the action of the force, determine

a) the average shearing strain in the material, and

b) the force P exerted on the plate.

Dr. Farzin Zareian


Solution

Determine the average angular deformation or shearing strain of the block.

i040 rad020.0in.2

in.04.0tan xyxyxy

Apply Hookes law for shearing stress and strainApply Hooke s law for shearing stress and strain to find the corresponding shearing stress.

psi1800rad020.0psi1090 3 xyxy G xyxyUse the definition of shearing stress to find the force P.

3 lb1036in.5.2in.8psi1800 3 AP xykips0.36P

Dr. Farzin Zareian


Relation Among E, , and G An axially loaded slender bar will An axially loaded slender bar will

elongate in the axial direction and contract in the transverse directions.

An initially cubic element oriented as in top figure will deform into a rectangular parallelepiped The axial load producesparallelepiped. The axial load produces a normal strain.

If the cubic element is oriented as in the bottom figure, it will deform into a rhombus. Axial load also results in a shear strain.

12GE Components of normal and shear strain are related,

strain.

Dr. Farzin Zareian

2G


Sample Problem

A circle of diameter d = 9 in. is scribed on an unstressed aluminum plate of thickness t = 3/4 in Forces acting in thethickness t = 3/4 in. Forces acting in the plane of the plate later cause normal stresses x = 12 ksi and z = 20 ksi.

For E = 10x106 psi and = 1/3, determine the change in:

the length of diameter AB, the length of diameter CD, the thickness of the plate, and pthe volume of the plate.

Dr. Farzin Zareian


Solution Evaluate the deformation components. 3Apply the generalized Hookes

Law to find the three components of normal strain.

in.9in./in.10533.0 3 dxAB in.108.4 3AB

11

EEEzyx

x

in.9in./in.10600.13 dzDC

in.104.14 3DC in./in.10533.0

ksi20310ksi12

psi10101

3

6

in.75.0in./in.10067.1 3 tyt in.10800.0 3t

in /in100671 3

EEEzyx

y

Find the change in volume333 /inin100671 e in./in.10067.1

3

EEEzyx

z

33 in75.0151510067.1/inin10067.1

eVV

e zyx

3in187.0V

Dr. Farzin Zareian

in./in.10600.1 3 in187.0V


Stress and Strain DistributionSaint-Venants Principlep

Except in the immediate vicinity of point loads, the stress distribution may be assumed independent of the actual mode of application of the loads

Dr. Farzin Zareian

Torsion


Torsion

What is Torsion?

This chapter aims at:Analyzing the stresses and strains in members of circular cross section subjected to twisting couples, or torques

Dr. Farzin Zareian

Torsion

Torsion; Engineering Applications

In Civil Engineering

In Mechanical Engineering

Dr. Farzin Zareian

Torsion

Stresses in a Shaft

Net of the internal shearing stresses is an internal torque, equal and opposite to the

dAdFT applied torque,

Unlike the normal stress due to axial loads, the distribution of shearing stresses

Free-body Diagramdue to torsional loads can not be assumed uniform.

Dr. Farzin Zareian

Torsion

Axial Shear Component

Very small element of shaft

Conditions of equilibrium require theConditions of equilibrium require the existence of equal stresses on the faces of the two planes containing the axis of the shaft.

Dr. Farzin Zareian

Torsion

Deformations in a Circular Shaft

From observation, the angle of twist of the shaft is proportional

h li d d hto the applied torque and to the shaft length.

T When subjected to torsion, every

cross-section of a circular shaft L remains plane and undistorted.

Cross-sections of noncircular(non axisymmetric) shafts are

Circular shaft attached to a fixed support

(non-axisymmetric) shafts are distorted when subjected to torsion.

Dr. Farzin Zareian

Torsion

Deformations in a Circular Shaft

No Equal End rigid plates to keep all

sections plane and

Dr. Farzin Zareian

Distortion Rotation undistorted

Torsion

Shearing Strain Consider an interior section of the

shaft. As a torsional load is applied, an element on the interior cylinderan element on the interior cylinder deforms into a rhombus.

Since the ends of the element remain planar, the shear strain is equal to angle of twist.

It follows that:L

L or

Shear strain is proportional toShear strain is proportional to twist and radius:

maxmax and cLc

Dr. Farzin Zareian

cL

Torsion

Stresses in the Elastic Range

Multiplying the previous equation by the shear modulus,

max GcG c From Hookes Law,

max cG The shearing stress varies linearly with the radialThe shearing stress varies linearly with the radial

position in the section. Sum of the moments from the internal stress distribution

is equal to the torque on the shaft at the section,J

cdA

cdAT max2max

The results are known as the elastic torsion formulas,

and TTc

Polar moment of inertia

Dr. Farzin Zareian

and max JJ

Torsion

Sample Problem

max

max min

120 MPa? ?T

Repeat and assume the shaft is not hollow.

max min

Dr. Farzin Zareian

Torsion

Normal Stresses Elements with faces parallel and

perpendicular to the shaft axis are subjected to shear stresses onlysubjected to shear stresses only. Normal stresses, shearing stresses or a combination of both may be found f th i t tifor other orientations.

Consider an element at 45o to the shaft axisshaft axis,

0max

0max0max

2

245cos2

AFAAF

max0

0max45 2

o AA

Dr. Farzin Zareian

Torsion

Normal Stresses (Cont.) Element a is in pure shear.

Element c is subjected to a tensile stress on two faces and compressive stress on the other two.

N t th t ll t f l t Note that all stresses for elements aand c have the same magnitude

Failure Modes

Ductile Fracture Brittle Fracture

Dr. Farzin Zareian

Torsion

Sample Problem

Shaft BC is hollow with inner and outer diameters of 90mm and 120mm,

i lrespectively.

Shafts AB and CD are solid of diameter d For the loading shown determined. For the loading shown, determine

(a) the minimum and maximum shearing stress in shaft BC, g ,

(b) the required diameter d of shafts ABand CD if the allowable shearing stress in these shafts is 65 MPa.

Dr. Farzin Zareian

Torsion

Solution

Cut sections through shafts AB and BCand perform static equilibrium

l i fi d l dianalysis to find torque loadings.

BCx TM mkN14mkN60 ABx TM mkN60CDAB TT mkN6 mkN20 BCT

Dr. Farzin Zareian

Torsion

Solution (Cont.)

Apply elastic torsion formulas to find minimum and maximum stress on h f BCshaft BC.

444142 045.0060.022 ccJ m060.0mkN202 cTBC

46 m1092.1322

MPa2.86m1092.13 46

22max

J

BC

mm45 c

MPa7.64mm60mm45

MPa2.86

min

min

2

1

max

min

cc

MPa7.64MPa2.86

min

max

Dr. Farzin Zareian

Torsion

Solution (Cont.)

Given allowable shearing stress and applied torque, invert the elastic torsion formula

fi d h i d dito find the required diameter.

mkN665 MPaTcTc

m109.3822

3

34max

cccJ

mm8.772 cd

Dr. Farzin Zareian

Torsion


Dr. Farzin Zareian

Torsion

Angle of Twist in the Elastic Range

Recall that the angle of twist and maximum shearing strain are related,

L

c max In the elastic range, the shearing strain g , g

and shear are related by Hookes Law,

JGTc

G maxmax JGG

Equating the expressions for shearing strain and solving for the angle of twist, JG

TL

Dr. Farzin Zareian

Torsion

Angle of Twist in the Elastic Range (Cont.)

If the torsional loading or shaft cross-section changes along the length, the angle of rotation is found as the sum of segment rotations

iiGJLT

iiiGJ

L

JGTdx

JGTdxd

0

Dr. Farzin Zareian

Torsion

Statically Indeterminate Shafts

Given the shaft dimensions and the applied torque, we would like to find the torque reactions at A and B.

F f b d l i f th h ftftlb90 BA TT

From a free-body analysis of the shaft,

which is not sufficient to find the end torqueswhich is not sufficient to find the end torques. The problem is statically indeterminate.

Divide the shaft into two components which must have compatible deformations,

ftlb90&012

21

12

21

2

2

1

121 AAABBA TJL

JLTTJLJLT

GJLT

GJLT

Dr. Farzin Zareian

Torsion

Sample Problem

Two solid steel shafts are connected by gears. Knowing that for each shaft G = 11.2 x 106Knowing that for each shaft G 11.2 x 10psi and that the allowable shearing stress is 8 ksi, determine

(a) the largest torque T0 that may be applied to the end of shaft AB,

(b) th di l th h hi h(b) the corresponding angle through which end A of shaft AB rotates.

Dr. Farzin Zareian

Torsion

SolutionApply a static equilibrium analysis onApply a static equilibrium analysis on

the two shafts to find a relationship between TCD and T0 .

CDC

B

TFMTFM

in.45.20in.875.00 0

08.2 TTCD

Apply a kinematic analysis to relate the angular rotations of the gears

C

CCBB

rrr

in.45.2

Apply a kinematic analysis to relate the angular rotations of the gears.

CCB

B r

in.875.0

CB 8.2

Dr. Farzin Zareian

Torsion

Solution (Cont.)

Find the T0 for the maximum allowable torque on each shaft choose the smallestsmallest.

in3750

in.375.080004

0max TpsiJ

cTAB

AB in.lb663

in.375.020 T

J AB

in.5.02

in.5.08.280004

0max TpsiJ

cTCD

CD

in.lb5610 T

Dr. Farzin Zareian

Torsion

Solution (Cont.)

Find the corresponding angle of twist for each shaft and the net angular rotation of end Arotation of end A.

642/ psi102.11in.375.0

.in24in.lb561

AB

ABBA GJ

LT

64/o

2

.in24in.lb5618.22.22rad387.0

ps0..375.0

CDDC

AB

LT

GJ

oo

o

642

/

268952828295.2rad514.0

psi102.11in.5.0

CDDC GJ

oo

/ 2.2226.826.895.28.28.2

BABACB

o48.10A

Dr. Farzin Zareian

Torsion


Dr. Farzin Zareian

Pure Bendingg


Pure Bending

What is Pure Bending?

This chapter aims at:Analyzing the stresses and strains in prismatic members subjected to equal and opposite couples acting in the same longitudinal plane

Dr. Farzin Zareian

Pure Bending

Other Loading Types Eccentric Loading: Axial loading which

does not pass through section centroid produces internal forces equivalent to an axial force and a couple

Transverse Loading: Concentrated or distributed transverse load producesdistributed transverse load produces internal forces equivalent to a shear force and a couple

Principle of Superposition: The normal stress due to pure bending may be

bi d i h h l dcombined with the normal stress due to axial loading and shear stress due to shear loading to find the complete state of stress.

Dr. Farzin Zareian

Pure Bending

Symmetric Member in Pure Bending

Internal forces in any cross section are equivalent to a couple. The moment of the couple is the section bending moment.

From Statics, a couple M consists of two l d it fequal and opposite forces.

The sum of the components of the forces in any direction is zero.any direction is zero.

The moment is the same about any axis perpendicular to the plane of the couple and zero about any axis contained in the plane.

Dr. Farzin Zareian

Pure Bending

Symmetric Member in Pure Bending

These requirements may be applied to the sums of the components and moments of the statically indeterminate elementary internal forces.

dAMdAF xx

00

MdAyMdAzM

x

xy

0

y xz

Dr. Farzin Zareian

Pure BendingDeformation in a Symmetric Member in Pure BendingBeam with a plane of symmetry in pure bending:Beam with a plane of symmetry in pure bending:

member remains symmetric, and bends uniformly to form a circular arc

cross-sectional plane passes through arc center and remains planar

length of top decreases and length of bottom increases

a neutral surface must exist that is parallel to the upper and lower surfaces and for which the length does not changeg g

stresses and strains are negative (compressive) above the neutral plane and positive (tension) b l i

Dr. Farzin Zareian

below it

Pure Bending

Strain due to Bending

Consider a beam segment of length L. After deformation, the l h f h l flength of the neutral surface remains L. At other sections,

yyLL

yL

x

cc

yyL

linearly) ries(strain va 1 = x my c

mx

mm

y

c

c

or Curvature

Dr. Farzin Zareian

mx c

Pure Bending

Stress due to BendingF li l l i i l For a linearly elastic material,

mxx EcyE

For static equilibrium,linearly) varies(stressmc

y

dAF 0

yM y dA y dA For static equilibrium,

MdAyMdAzM

dAF

xz

xy

xx

0

0

dAyc

dAcydAF

m

mxx

0

0

2

x m

m m

M y dA y dAc

IM y dAc cM M

cSubstituting m x m

x

Mc M yI S cMyI

First moment with respect to neutral plane is zero. Therefore, the neutral surface must pass

Dr. Farzin Zareian

Ipthrough the section centroid.

Pure Bending

Beam Section Properties

The maximum normal stress due to bending,

SM

IMc

m

modulussection

inertia ofmoment section

IS

ISI A beam section with a larger

section modulus will have a lower maximum stressmodulussection

cS

Consider a rectangular beam cross section,

Ahbhhbh

cIS 6

161

22

3121

Dr. Farzin Zareian

Pure Bending

First Moment of Inertia

xA

Q ydA Ay

A

ydAy

Ay

A

1

N

i iiA

ydA A y1

1

iAN

ii

yA A

Dr. Farzin Zareian

1iRequired reading: Appendix A

Pure Bending

Second Moment of Inertia

2x

A

I y dA 2x xI I Ad Parallel Axes TheoremParallel Axes Theorem

Dr. Farzin Zareian

Pure Bending

Beam Section Properties

Between two beams with the same cross sectional area, the beam with the greater depth will be more effective in resisting bendingwill be more effective in resisting bending.

Structural steel beams are designed to have a largeStructural steel beams are designed to have a large section modulus.

Dr. Farzin Zareian

Pure Bending

Properties of American Standard Shapes

Dr. Farzin Zareian

Pure Bending

Sample ProblemA cast-iron machine part is acted upon

by a 3 kN-m couple. Knowing E = 165 GPa and neglecting the effects of165 GPa and neglecting the effects of fillets, determine

(a) the maximum tensile and ( )compressive stresses,

(b) the radius of curvature.

(c) Curvature.

Dr. Farzin Zareian

Pure Bending

Solution

Based on the cross section geometry, calculate the location of the section

id d f i icentroid and moment of inertia.

32 mm ,mm ,mm Area, Ayy

3

3

3

101143000104220120030402109050180090201

AyA mm38101143

AyY 101143000 AyA mm383000

AY

231212 dAbhdAIIx

4-943 m10868mm10868 I 2312123121 18120040301218002090

Dr. Farzin Zareian

Pure Bending

Solution (Cont.)

Apply the elastic flexural formula to find the maximum tensile and

icompressive stresses.

I

Mcm

49 m10868m022.0mkN 3

IcM

IA

A MPa0.76A

49 m10868m038.0mkN 3

IcM B

B

Calculate the curvature

MPa3.131B

49- m10868GPa 165 mkN 31 EIMm747

m1095.201 1-3

Calculate the curvature

Dr. Farzin Zareian

m7.47

Pure Bending


M = 25 KN.m, Determine the stresses at points C, D d ED, and E

Dr. Farzin Zareian

Pure Bending


Dr. Farzin Zareian

Pure Bending

Bending of Members Made of Several Materials

Very significant in analyzing Reinforced Concrete Members

Dr. Farzin Zareian

Always start with strain distribution diagram. Stress distribution diagram will follow

Pure Bending

Bending of Members Made of Several Materials

Transformed Section Method

2EnE

Develop the Transformed S ti (TS)

Find the Neutral Axis

i TSx

MyI

1E Section (TS) in TS transformI

If all E2 If all E1

2 11 & x x 1 2 & x xn

Dr. Farzin Zareian

n

Pure Bending

Sample ProblemEwood = 12.5GPa & Esteel = 200GPa. If M = 50kN.m, then find maximum stress in wood and steel.

Dr. Farzin Zareian

Pure Bending


Ec = 3.75E06 psi & Es = 30.0E06 psi. If M = 150 kip.ft, then find maximum stress i l din steel and concrete.

Repeat, assuming that top thickness is 12

Dr. Farzin Zareian

Pure Bending


Ec = 3.75E06 psi & Es = 30.0E06 psi. If M = 150 kip.ft, then find maximum stress i l din steel and concrete.

Repeat, assuming that top thickness is 5

Dr. Farzin Zareian

Pure Bending

General Section, General relationship

0 x dA xM y dA

Dr. Farzin Zareian

Pure Bending

yc

x Y

Plastic Deformationx

cx Y

yc

x Y mMcI

I I

x

cx Y m xI IM

c y

yc

x Y 22I S bc x

cx Y 3

S bcc

223 m

M bc yc

x Y

3The same in negative

Dr. Farzin Zareian

x

cx Y

gaxes

Pure Bending

yc

x Y


cx Y

yc

x Y Y

YM c

I

x

cx Y xY

YycM

I I

yc

x Y 22I S b x

cx Y 3

S bcc

223Y Y

M bc yc

x Y

3Y YThe same in negative

Dr. Farzin Zareian

x

cx Y

gaxes

Pure Bending

yc

x Y


cx Y

M

yc

x Y m Y

McI

If I know y x

cx Y

2

x YY

c

yy

M b y dy

If I know yY

yc

x Y 0

0

2

2 2 Y

Y

Y

x

y c

x xy

y c

M b y dy

b y dy b y dy

y

x

cx Y

0

2

0

2 2

2 2

Y

Y

Y

Y YY y

y cY

YY y

yb y dy b y dyy

b y dy b y dyy

yc

x Y

3 2

0

22

2

2 23 2

113

Y

Y

y c

YY

Y y

YY

b y yby

yM bc

The same in negative

Dr. Farzin Zareian

x

cx Y

23Y c g

axes

Pure Bending

yc

x Y


cx Y

yc

x Y 2

22

113

YY

yM bcc

x

cx Y 2

2

3 112 3

Yy

yM Mc

yc

x Y 2 &

3 11

Y Y Y Yy c

M M

x

cx Y

212 3y YM M

yc

x Y The same in negative

Dr. Farzin Zareian

x

cx Y

gaxes

Pure Bending

yc

x Y


cx Y

0

yc

x Y 0

0

2

Yc

P x

y

M b y dy

x

cx Y

0

0

2 c

Yb y dy

yc

x Y 02

2

3

c

Yb y dy

b

x

cx Y

2 32P Y Y

M bc M

& P PM Mk ZM

yc

x Y

Y YM Shape Factor

Plastic Section M d l

Dr. Farzin Zareian

x

cx Y Modulus

Pure Bending

Sample Problem

M = 36.8 kN.m is applied. E = 200GPa, and y = 240MPa. Determine: Thicknessand y 240MPa. Determine: Thickness of elastic core (yY = ?), radius of curvature of neutral surface? Draw the Moment-Curvature DiagramMoment-Curvature Diagram.

Dr. Farzin Zareian

Pure Bending

Sample Problem

E = 29E06 psi, and y = 50ksi. D h M C DiDraw the Moment-Curvature Diagram with 3 points: a) first yield, b) plastic flanges, c) fully plastic

Dr. Farzin Zareian

Pure Bending

Eccentric Axial Loading in a Plane of Symmetry

Stress due to eccentric loading found by superposing the uniform stress due to a centric load and linear stress distribution due a pure bending moment

xxx bendingcentric I

MyAP

E t i l diEccentric loading

PdMPF

Dr. Farzin Zareian

Pure Bending

Sample Problem

The largest allowable stresses for the cast iron link are 30 MPa in tension andiron link are 30 MPa in tension and 120 MPa in compression. Determine the largest force P which can be applied to the linkapplied to the link.

From previous Sample Problem,p p ,23

m038.0m103

YA

49 m10868 I

Dr. Farzin Zareian

Pure Bending

SolutionDetermine equivalent centric and

bending loads.m028.0010.0038.0 d

moment bending 028.0load centric

PPdMP

Superpose stresses due to centric and bending loads

3 9

0.028 0.022377

3 10 868 10A

A

PMcP P PA I

Superpose stresses due to centric and bending loads

3 9

3 10 868 100.028 0.038

15593 10 868 10

BB

A IPMcP P P

A I

Dr. Farzin Zareian

Pure Bending

Solution (Cont.)

Evaluate critical loads for allowable stressesallowable stresses.

kN077MPa1201559kN6.79MPa30377

PPPPA

kN0.77MPa1201559 PPB

The largest allowable load kN0.77P

Dr. Farzin Zareian

Pure Bending

General Case of Eccentric Axial Loading

Consider a straight member subject to equal and opposite eccentric forces. The eccentric force is equivalent to the system of a centric force and two couples.

PbMPaMP

forcecentric

By the principle of superposition, the combined stress distribution is

PbMPaM zy

combined stress distribution is

y

y

z

zx I

zMI

yMAP

If the neutral axis lies on the section, it may be found from A

PzI

My

IM

y

y

z

z

Dr. Farzin Zareian

Pure Bending


Problems 4.109 and Problem 4.145

P = 50kN. Section W 150 X 24. Determine afor max < 90MPa

Determine largest P for this member. all = 10ksi

Dr. Farzin Zareian

for max 90MPa0 s

Analysis and Design of Beams for Bending


Analysis and Design of Beams


This chapter aims at analyzing and designing of beams.

Beams: structural members supporting loads at various points along the member

s c apte a s at a a y g a d des g g o bea s.

Dr. Farzin Zareian

points along the member


Introduction Transverse loadings of beams are classified as concentrated loads or distributed loads

Applied loads result in internal forces consisting of a shear force (from the shear stress distribution) and a bending couplestress distribution) and a bending couple(from the normal stress distribution)

Normal stress is often the critical Normal stress is often the critical design criteria

SM

IcM

IMy

mx Requires determination of the location and magnitude of largest bending moment

Dr. Farzin Zareian


Classification of Beam Supports

Statically Determinate Beamsy

Statically Indeterminate Beams

Dr. Farzin Zareian

Analysis and Design of BeamsShear and Bending Moment Diagrams

Determination of maximum normal andDetermination of maximum normal and shearing stresses requires identification of maximum internal shear force and bending couplebending couple.

Shear force and bending couple at a point are determined by passing a section y p gthrough the beam and applying an equilibrium analysis on the beam portions on either side of the sectionon either side of the section.

Sign conventions for shear forces V and Vand bending couples M and M

Dr. Farzin Zareian


Relations Among Load, Shear, and Bending Moment

Relationship between load and shear:

xwVVVF 0:0 xwV

xwVVVFy

0:0

Dx

CD dxwVVwdV

CxCD dxwVVwdx

Relationship between shear and bending moment:

2210

2:0

xwxVM

xxwxVMMMM C

DC

x

xCD dxVMMVdx

dM

Dr. Farzin Zareian




0 : 0F V V V w x P Mcon

0 : 0y concon

F V V V w x PV w x P

Dx

D CdV w V V P w dx Pcon

C

D C conx

w V V P w dxdx


2

0 : 02

1

C conxM M M M M V x w x

M M V x w x

Mcon

2con

M M V x w x Dx

D C condM V M M M V dxdx

PconDr. Farzin Zareian

Cxdx




Dx

Mcon

Dx

D C conx

V V P w dx Pcon

Cx


Dx

M M M V dx Mcon

C

D C conx

M M M V dx Pcon

Dr. Farzin Zareian



Concentrated LoadPPVC,MC

VD,MD

V

VD,MDDx

D C conx

V V P w dx Cx

Dx

M

D

C

D C conx

M M M V dx Dr. Farzin Zareian



Concentrated Moment

MVC,MC

VD,MD

V

VD,MDDx

D C conx

V V P w dx Cx

Dx

M

D

C

D C conx




Uniform Distributed Load

wVC,MC

VD,MD

V

VD,MDDx

D C conx

V V P w dx Cx

Dx

M

D

C

D C conx




Triangular Distributed Load

wVC,MC

VD,MD

V

VD,MDDx

D C conx

V V P w dx Cx

Dx

M

D

C

D C conx



Sample Problem

Draw the shear and bending moment diagrams for the beam and loading shown.

Dr. Farzin Zareian


SolutionTaking the entire beam as a free body,

determine the reactions at A and D. ft28kips12ft14kips12ft6kips20ft240 A DM

kips12kips26kips12kips200Fkips26

ft28kips12ft14kips12ft6kips20ft240

y

y

A

AD

DM

kips18yAApply the relationship between shear and load to

develop the shear diagram.

dV w dx - zero slope between concentrated loads

- linear variation over uniform load segment

Dr. Farzin Zareian

linear variation over uniform load segment


Solution (Cont.)Apply the relationship between bending momentApply the relationship between bending moment

and shear to develop the bending moment diagram. dM V dx

- bending moment at A and E is zero

b di t i ti b t A B C- bending moment variation between A, B, Cand D is linear

- bending moment variation between D and Ebending moment variation between D and Eis parabolic

- net change in bending moment is equal to areas under shear distribution segments

- total of all bending moment changes across th b h ld b

Dr. Farzin Zareian

the beam should be zero


Design of Prismatic Beams for Bending

The largest normal stress is found at the surface where the maximum bending moment occurs

SM

IcM

mmaxmax

moment occurs.

A safe design requires that the maximum normal stress be less than the allowable stress for the material used. This criteria leads to the determination of the minimum acceptable section modulus.

allm

MS

max

i

modulus.all

S min Among beam section choices which have an

acceptable section modulus, the one with the smallestacceptable section modulus, the one with the smallestweight per unit length or cross sectional area will be the least expensive and the best choice.

Dr. Farzin Zareian


Sample Problem

A simply supported steel beam is to carry the distributed and concentrated loads shown. Knowing that the allowable normal stress for the grade of steel to be used is 160 MPa, select the wide-flange shape that should be used.

Dr. Farzin Zareian


SolutionConsidering the entire beam as a free-body,

determine the reactions at A and D. m4kN50m5.1kN60m50 A DM

kN50kN60kN0.580kN0.58

m4kN50m5.1kN60m50

yy

A

AFD

DM

kN0.52yADevelop the shear diagram and determine the

maximum bending moment.

kN60kN0.52 yA

curveloadunderareaVVAV

kN8

kN60

B

AB

VcurveloadunderareaVV

Dr. Farzin Zareian


Solution (Cont.)Maximum bending moment occurs at

V = 0 or x = 2.6 m. kN6.67, EtoAcurveshearunderareaM kN6.67,

maxEtoAcurveshearunderareaM

Determine the minimum acceptable beam section modulus. M

3336

maxmin

mm105.422m105.422MPa160

mkN6.67

all

MS

mm10 33SShape

Choose the best standard section which meets this criteria. 5497.38W310

4749.32W36063738.8W410

ee s s c e .

9.32360W 4481.46W2005358.44W250

Dr. Farzin Zareian

Shearing Stresses in Beams and Thin-Walled

Members


Shearing Stresses

Shearing Stresses in Beams and Thin-Walled Members

Shearing Stresses are important, particularly in the design S ea g St esses a e po ta t, pa t cu a y t e des gof short, stubby beams. Their analysis will be the subject of this chapter.

Dr. Farzin Zareian

Shearing Stresses

Introduction Transverse loading applied to a beam results in normal and shearingstresses in transverse sectionsstresses in transverse sections.

Distribution of normal and dAMdAF 00shearing stresses satisfies MdAyMdAF

dAzMVdAFdAzyMdAF

xzxzz

xyxyy

xyxzxxx

00

00

When shearing stresses are exerted on the

y xzxzz

When shearing stresses are exerted on the vertical faces of an element, equal stresses must be exerted on the horizontal faces

Dr. Farzin Zareian

Shearing Stresses

Shear on the Horizontal Face of a Beam Element Consider prismatic beam; For equilibrium of beam element

dAHF 0

A

CD

ACDx

dAyI

MMH

dAHF 0

Note,dAyQ

A

AI

xVxdx

dMMM CD Substituting, VQSubstituting,

flowshearVQHq

xI

VQH

Dr. Farzin Zareian

fIx

q

Shearing Stresses

Shear on the Horizontal Face of a Beam Element Shear flow,

flowshearVQHq where

aboveareaofmomentfirst ydAyQ

fIx

q

section cross full ofmoment second

aboveareaofmoment first

'

2

1

AA

A

dAyI

ydAyQ

Same result found for lower areaQVH

HHQQ

qIQ

xq

axis neutral respect toh moment witfirst 0

Dr. Farzin Zareian

HH

Shearing Stresses

Sample Problem

A beam is made of three planks, nailed together Knowing that thenailed together. Knowing that the spacing between nails is 25 mm and that the vertical shear in the beam i V 500 N d t i th his V = 500 N, determine the shear force in each nail.

Dr. Farzin Zareian

Shearing Stresses

SolutionDetermine the horizontal force

per unit length or shear flow qon the lower surface of the upper plank.

N3704m1016.20

)m10120)(N500(46-

36

IVQq

m0600m1000m0200 yAQ Calculate the corresponding m

N3704

312136

m100.0m020.0

m10120m060.0m100.0m020.0

I

yAQ p gshear force in each nail for a nail spacing of 25 mm.

46

2

3121

]m060.0m100.0m020.0

m020.0m100.0[2

N6.92F)3704)(m025.0()m025.0( mNqF

Dr. Farzin Zareian

46 m1020.16

Shearing Stresses

Determination of the Shearing Stress in a Beam The average shearing stress on the horizontal face of the element is obtained by dividing the shearingobtained by dividing the shearing force on the element by the area of the face.

VQxtx

IVQ

Axq

AH

ave

ItVQ

On the upper and lower surfaces of the beam, yx= 0. It yfollows that xy= 0 on the upper and lower edges of the transverse sections.

Dr. Farzin Zareian

Shearing Stresses

Shearing Stresses xy in Common Types of Beams For a narrow rectangular beam,

yVVQ 132

AV

cy

AIbQ

xy

23

12

max

2

A2

For American Standard (S-beam) and ( )wide-flange (W-beam) beams

VQ

web

ave

AVIt

max

Dr. Farzin Zareian

Shearing Stresses

Sample Problem

A timber beam is to support the three concentrated loads shownthree concentrated loads shown. Knowing that for the grade of timber used,

psi120psi1800 allall determine the minimum required qdepth d of the beam.

Dr. Farzin Zareian

Shearing Stresses

Solution

Develop shear and bending moment diagrams Identify the maximumsdiagrams. Identify the maximums.

kips3max Vinkip90ftkip5.7

p

max

max

M

Dr. Farzin Zareian

Shearing Stresses

Solution (Cont.) Determine the beam depth based on allowable normal stress.

i269

in.5833.0in.lb1090psi 1800 2

3max

ddS

Mall

Determine the beam depth based ll bl h t

in.26.9d

3

121 dbI

on allowable shear stress.

in.3.5lb3000

23psi120

23 max

dAV

all

22

in53161

d

dbcIS

in.71.10d

Required beam depth is equal to 2in.5833.0

in.5.36d

d

in.71.10d

Required beam depth is equal to the larger of the two.

Dr. Farzin Zareian

in.71.10d

Shearing Stresses

Shearing Stresses in Thin-Walled Members

Consider a segment of a wide-flange beam subjected to the vertical shear V. The longitudinal shear force on the gelement is

xI

VQH The corresponding shear stress is

ItVQ

xtH

xzzx

Itxt Previously found a similar expression for the shearing stress in the web

VQIt

VQxy in the flanges 0xy

in the web 0Dr. Farzin Zareian

in the web 0xz

Shearing Stresses


The variation of shear flow across the section depends only on the variation of the first moment.

VQI

VQtq

F b b thl For a box beam, q grows smoothly from zero at A to a maximum at C and C and then decreases back to zero at E.

The sense of q in the horizontal i f h i b d d dportions of the section may be deduced

from the sense in the vertical portions or the sense of the shear V.

Dr. Farzin Zareian

Shearing Stresses


For a wide-flange beam, the shear flow increases symmetrically from zero at Aand A, reaches a maximum at C and then ,decreases to zero at E and E.

Th ti it f th i ti i d The continuity of the variation in q and the merging of q from section branches suggests an analogy to fluid flow.

The sense of q in the horizontal i f h i b d d dportions of the section may be deduced

from the sense in the vertical portions or the sense of the shear V.

Dr. Farzin Zareian

Shearing Stresses

Sample Problem

Knowing that the vertical shear is 50 kips in a W10x68 rolled-steel beam, determine the horizontal shearing stress in the top flange at the point a.

Dr. Farzin Zareian

Shearing Stresses

Solution

For the shaded area,

i8154i7700i314Q 3in98.15

in815.4in770.0in31.4Q

The shear stress at a,

in98.15kips50 3VQ in770.0in394 p 4 ItQksi63.2

Dr. Farzin Zareian

7UDQVIRUPDWLRQRI6WUHVV

Dr. Farzin Zareian

Stress and Strain Transformations


Thi h t d t i h th t f t dThis chapter determines how the components of stress and strain are transformed under a rotation of the coordinate axes.

Dr. Farzin Zareian


Introduction The most general state of stress at a point may be represented by 6 componentscomponents,

stressesshearingstresses normal,, zyx

),, :(Notestressesshearing,,

xzzxzyyzyxxy

zxyzxy

Same state of stress is represented by a different set of components if axes are rotated.axes are rotated.

Dr. Farzin Zareian


Introduction Plane Stress - state of stress in which two faces of the cubic element are free of stress. For the illustrated example, the state of stress is defined by

.0,, and xy zyzxzyx

State of plane stress occurs:

(a) in a thin plate subjected to forces(a) in a thin plate subjected to forces acting in the midplane of the plate.

(b) on the free surface of a structural ( )element or machine component, i.e., at any point of the surface not subjected to an external force

Dr. Farzin Zareian

subjected to an external force.


Transformation of Plane Stress Equilibrium conditions of a prismatic element Equilibrium conditions of a prismatic element with faces perpendicular to the x, y, and x, y axes.

sincoscoscos0 AAAF

coscossincos0cossinsinsin

sincoscoscos0

AAAFAA

AAAF

xyxyxy

xyy

xyxxx

The equations may be rewritten to yield

sinsincossin AA xyy The equations may be rewritten to yield

2sin2cos22 xy

yxyxx

2cos2sin

2sin2cos22

yx

xyyxyx

y

Dr. Farzin Zareian

2cos2sin2 xyyx


Principal Stresses The previous equations are combined to yield parametric equations for a circle,

222 R 2

2

22 xyyxyx

ave

yxavex

R

R

Principal stresses occur on the principal planes of stress with zero shearing stresses.

p g

2

2

minmax, 22 xyyxyx

o90byseparatedanglestwodefines:Note

22tan

yx

xyp

Dr. Farzin Zareian

90by separatedanglestwodefines :Note


Maximum Shearing Stress

Maximum shearing stress occurs for

22

2

max xyyxR

90by separated angles twodefines :Note

22tan

o

xy

yxs

45by fromoffset and o

yxave

p

2ave

Dr. Farzin Zareian


Mohrs Circle for Plane Stress

Mohrs circle is applied with simple geometric considerations. Critical valuesare estimated graphically or calculated.

For a known state of plane stress, plot the points X and Y and

are estimated graphically or calculated.

plot the points X and Y and construct the circle centered at C.

2

2

yxyx R 22 xyave

R

The principal stresses are obtained at A and B.

yx

xypave R

2

2tanminmax,

Dr. Farzin Zareian

yx


Mohrs Circle for Plane Stress For the state of stress at an angle with respect to

With Mohrs circle uniquely defined, the state of stress at other axes orientations may be depicted.

g pthe xy axes, construct a new diameter XY at an angle 2 with respect to XY.2 with respect to XY.

Normal and shear stresses are obtained from the coordinates XY.

Dr. Farzin Zareian


Mohrs Circle for Plane Stress

Mohrs circle for t i i l l dicentric axial loading:

Mohrs circle for torsional loading:

Dr. Farzin Zareian


Sample Problem

For the state of stress shown, determinedetermine

(a) the principal planes and the i i l tprincipal stresses,

(b) the stress components exerted on the element obtained by rotating the given element counterclockwise through 30 degrees.

Dr. Farzin Zareian


Solution

C M h i lConstruct Mohrs circle

MPa802

601002

yxave

P i i l l d t

MPa52482022

2222 FXCFR

Principal planes and stresses

4.6724.220482tan pp CF

XF clockwise7.33 p

5280 CAOCOA

MPa28min

5280max CAOCOA

MPa132max 5280min BCOCOA

Dr. Farzin Zareian


Solution (Cont.)Stress components after rotation by 30o

Points X and Y on Mohrs circle that correspond to stress components on thecorrespond to stress components on the rotated element are obtained by rotating XY counterclockwise through

652cos52806.52cos5280

6.524.6760180

CLOCOLKCOCOKx

MP448

6.52sin526.52cos5280

XKCLOCOL

yx

y

MPa3.41MPa6.111

MPa4.48

yx

y

x

Dr. Farzin Zareian

yx


Sample Problem

A single horizontal force P of 150 lb magnitude is applied to end D ofmagnitude is applied to end D of lever ABD. Determine

( ) th l d h i t(a) the normal and shearing stresses on an element at point H having sides parallel to the x and y axes,

(b) the principal planes and principal stresses at the point H.

Dr. Farzin Zareian


SolutionDetermine an equivalent force-coupleDetermine an equivalent force couple

system at the center of the transverse section passing through H.

lb150P inkip5.1in10lb150

inkip7.2in18lb150lb150

xMTP

Evaluate the normal and

px

Evaluate the normal and shearing stresses at H.

in6.0inkip5.1 Mc

ksi96.7ksi84.80 yyx

421

441

in6.0in6.0inkip7.2

in6.0

JTc

I

xy

y

Dr. Farzin Zareian

2 in6.0J


Solution (Cont.)Determine the principal planes and

calculate the principal stresses. 8196.7222tan xy

119,0.612

8.184.80

2tan

p

yx

yp

2

5.59,5.30p

2

2

2

minmax,

84808480

22

xyyxyx

ksi68.4ksi52.13 i

296.72

84.802

84.80

Dr. Farzin Zareian

ksi68.4ksi52.13 minmax

Principal Stresses Under a Given Loading

Dr. Farzin Zareian

Principal Stresses

Principle Stresses in a Beam Prismatic beam subjected to transverse loading McMy

Principal stresses determined from methodsIt

VQIt

VQII

mxy

mx

determined from methods of previous chapter.

Can the maximumCan the maximum normal stress within the cross-section be larger thanlarger than

IMc

m

Dr. Farzin Zareian

Principal Stresses

Principle Stresses in a Beam

Dr. Farzin Zareian

Principal Stresses

Principle Stresses in a Beam

C ti h lt i Cross-section shape results in large values of xy near the surface where x is also large.

max may be greater than mmax y g m

Dr. Farzin Zareian

Principal Stresses

Sample Problem

A 160-kN force is applied at the end of a W200x52 rolled-steel beam.

Neglecting the effects of fillets and of stress concentrations, determine ,whether the normal stresses satisfy a design specification that they be equal to or less than 150 MPa atequal to or less than 150 MPa at section A-A.

Dr. Farzin Zareian

Principal Stresses

SolutionDetermine shear and bending moment in

Section A-A

kN160

m-kN60m375.0kN160 AVM

kN160AV

Calculate the normal stress at top surface and at flange-web junctionsurface and at flange web junction.

MPa2.117m10512

mkN6036

SM A

a

MPa9.102mm103mm4.90MPa2.117

cy

bab

Dr. Farzin Zareian

Principal Stresses

Solution (Cont.)Evaluate shear stress at flange-web junction.

m106248

mm106.2487.966.1220436

33

Q

m0079.0m107.52 m106.248kN160m106.248

46

36

ItQVA

b

The principal stress at flange-web junction:MPa5.95

22 5.95

29.102

29.102 2

2

2221

21

max

bbb

MPa 150MPa9.15922

Design specification is not satisfied

Dr. Farzin Zareian

Design specification is not satisfied.

Principal Stresses

Sample Problem

The overhanging beam supports a uniformly distributed load and auniformly distributed load and a concentrated load. Knowing that for the grade of steel to used all = 24 ksi

d 14 5 k i l t th idand all = 14.5 ksi, select the wide-flange beam which should be used.

Dr. Farzin Zareian

Principal Stresses

SolutionDetermine reactions at A and DDetermine reactions at A and D.

kips410kips590

AD

DA

RMRM

Determine maximum shear and bending moment from shear and bending gmoment diagrams.

kips 2.12withinkip4.239max

VMkips43

maxV

Calculate required section modulus and qselect appropriate beam section.

3maxmin

239.4 kip ft 119.7 in24 k i

MS section!beam62Select W21

Dr. Farzin Zareian

min 24 ksiall

Principal Stresses

Solution (Cont.)Find maximum shearing stressFind maximum shearing stress.

Assuming uniform shearing stress in web,

ksi14.5ksi12.5i

kips 432

maxmax V

Find maximum normal stress.239 4 12 kip inM

in 8.40 2max webA

max

3

239.4 12 kip in 22.6 ksi127in

9.8822.6 ksi 21.3 ksi10 5

a

bb a

MS

y

c

b 2

10.512.2 kips 1.45 ksi8.40 inweb

cV

A

ksi45.12

ksi3.212

ksi3.21 22

max

Dr. Farzin Zareian

ksi24ksi4.21

Deflection of Beams

Dr. Farzin Zareian

Deflection of Beams

Deflection of Beams

The main interest of this chapter is the determination of the maximum deflection of a beam under a given loading.

Dr. Farzin Zareian

Deflection of Beams

Deformation of a Beam Under Transverse Loading Relationship between bending moment and curvature for pure bending remains valid for generalbending remains valid for general transverse loadings.

EIxM )(1

Cantilever beam subjected to concentrated load at the free end,

EIPx

1

Curvature varies linearly with x

At the free end A, A,01 AA

,0

At the free end B, PLEI

BB

,01

Dr. Farzin Zareian

PLB

Deflection of Beams

Deformation of a Beam Under Transverse Loading

O h i b Overhanging beam Reactions at A and C Bending moment diagram

Curvature is zero at points where the bending moment is zero, i.e., at

M )(1g , ,

each end and at E.EI

xM )(1 Beam is concave upwards where theBeam is concave upwards where the bending moment is positive and concave downwards where it is negative.

An equation for the beam shape or elastic curve is required to determine

i d fl ti d l

Dr. Farzin Zareian

maximum deflection and slope.

Deflection of Beams

Equation of the Elastic Curve From elementary calculus, simplified for beam parameters,

2 yd

2

2

232

2

1

1dx

yd

dydx

yd

Substituting and integrating,

1dx

2

21

CdMdyEIEI

xMdx

ydEIEI

x

21

00

10

CxCdxxMdxyEI

CdxxMdxdyEIEI

xx

Dr. Farzin Zareian

00

Deflection of Beams

Equation of the Elastic Curve Constants are determined from boundary conditions

2100

CxCdxxMdxyEIxx

Three cases for statically determinant beams,

- Simply supported beam 0,0 BA yy

- Overhanging beam 0,0 BA yyg g BA yy

- Cantilever beam 0,0 AAy

Dr. Farzin Zareian

Deflection of Beams

Direct Determination of the Elastic Curve From the Load DistributionFrom the Load Distribution

For a beam subjected to a distributed load,

E ti f b di l t b

xwdxdV

dxMdxV

dxdM 2

2

Equation for beam displacement becomes

xwdx

ydEIdx

Md 44

2

2

Integrating four times yields dxxwdxdxdxxyEI

432

2213

161 CxCxCxC

y

Constants are determined from boundary conditions

Dr. Farzin Zareian

Constants are determined from boundary conditions.

Deflection of Beams

Sample Problems

Problem 9.4

and

L/2

Dr. Farzin Zareian

Deflection of Beams

Method of Superposition

Principle of Superposition:

Deformations of beams subjected to combinations of loadings may be

Procedure is facilitated by tables of solutions for common

obtained as the linear combinationof the deformations from the individual loadings.

types of loadings and supports.

Dr. Farzin Zareian

g

Deflection of Beams

Sample Problem

For the beam and loading shown, determine the slope and deflectiondetermine the slope and deflection at point B.

Superpose the deformations due to Loading I and Loading II as shown.

Dr. Farzin Zareian

Deflection of Beams

SolutionL3 L4Loading I EI

wLIB 6

3

EI

wLy IB 8

4

Loading II EI

wLIIC 48

3

EI

wLy IIC 128

4

In beam segment CB, the bending moment is zero and the elastic

i t i ht licurve is a straight line.

EI

wLIICIIB 48

3

EI48

EI

wLLEI

wLEI

wLy IIB 3847

248128

434

Dr. Farzin Zareian

Deflection of Beams

Solution (Cont.)

Combine the two solutions,

EI

wLEI

wLIIBIBB 486

33

EI

wLB 48

7 3

EI

wLEI

wLyyy IIBIBB 3847

8

44

EI

wLyB 38441 4

Dr. Farzin Zareian

Deflection of Beams


Problem 9.7, 9.15

Dr. Farzin Zareian

Deflection of Beams

Statically Indeterminate Beams From free-body diagram, note that there are four unknown reaction components.

Conditions for static equilibrium yield

The beam is statically indeterminate.000 Ayx MFF

Conditions for static equilibrium yield

y

CCdMdEI xx Also have the beam deflection equation,

2100

CxCdxxMdxyEI which introduces two unknowns but provides three additional equations from

0,At 00,0At yLxyx

provides three additional equations from the boundary conditions:

Dr. Farzin Zareian

Deflection of Beams

Application of Superposition to Statically Indeterminate Beams

Method of superposition may be applied to determine the reactions

(2) Determine the beam deformation without the redundant support.pp

at the supports of statically indeterminate beams.

pp

(3) Treat the redundant reaction as an unknown load which, together

(1) Designate one of the reactions as redundant and eliminate or modify the support.

with the other loads, must produce deformations compatible with the original supports.

Dr. Farzin Zareian

Deflection of Beams

Sample Problem

For the uniform beam, determine the reaction at A derive thethe reaction at A, derive the equation for the elastic curve, and determine the slope at A.

(Note that the beam is statically indeterminate to the first degree)

Dr. Farzin Zareian

Deflection of Beams

Solution

Consider moment acting at section D,

M D 0 Mx

LxwxRA 032

1

3

20

The differential equation for

LxwxRM A 6

30

The differential equation for the elastic curve,

xwxRMydEI3

02

L

xRMdx

EI A 62

Dr. Farzin Zareian

Deflection of Beams

Solution (Cont.)

Integrate twice

1

402

2421 C

LxwxREI

dxdyEI A

21

503

12061

242

CxCL

xwxRyEI

Ldx

A LxwxRM

dxydEI A 6

30

2

2

Apply boundary conditions:

0:0,0at 2 Cyx

Ldx 6

Solve for reaction at A

1

0242

1:0,at

4

1

302 Lw

CLwLRLx A 0301

31 4

03 LwLRA

01206

1:0,at 2103 CLCLwLRyLx A

LwRA 0101

Dr. Farzin Zareian

10

Deflection of Beams

Solution (Cont.)

Substitute for C1, C2, and RA in th

CEE 150 Quarter Notes All

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