EG-120

COLLEGE OF ENGINEERING

STRENGTH OF MATERIALS

MODULE EG-120

Prof. J. Bonet

d:\documents\teaching\strength\cover.doc, 19/01/12

College College of Engineeringof Engineering

STRENGTH OF MATERIALSSTRENGTH OF MATERIALS

Prof J BonetProf J BonetProf. J. BonetProf. J. Bonet

MODULE EGMODULE EG--120120

Introduction 11/19/2012 Introduction 11/19/2012

STRENGTH OF MATERIALSSTRENGTH OF MATERIALSSTRENGTH OF MATERIALSSTRENGTH OF MATERIALS

The study of how materials can sustain external College College of of

EngineeringEngineering

Prof. J. Prof. J. BonetBonet

yactions without failure by using simplified mathematical models.

A tiEGEG--120120

Strength of Strength of MaterialsMaterials

Actions: Forces

Temperature Changesp g

Settlements

Failure:

Introduction Rupture

Excessive deformation

St th f M t i l i lifi d d l f1/19/2012

Strength of Materials uses very simplified models of complex structures to obtain useful engineering results for important classes of problems such as

2 Beam Theory, Simple Torsion,..

SYLLABUSSYLLABUSSYLLABUSSYLLABUS

Basic Concepts: Governing principles, Stress; College College of of



g p pStrain; Elasticity; Strain Energy; Material Failure; Time Effects.

B i B th I t d ti A i l FEGEG--120120


Basic Beam theory: Introduction; Axial Forces, Bending Moments and Shear Force Diagrams; Longitudinal Stresses; Deflection of Beams

Analysis of Stress and Strain: Shear Stress and Direct Stress; Mohr’s Circle; Principal Stresses; Strain; Stress Strain Relationships; Pressurised

IntroductionStrain; Stress-Strain Relationships; Pressurised vessels; 3-D stress and strain.

Advanced Beam Theory: Direct stresses in General

1/19/2012

Advanced Beam Theory: Direct stresses in General Beam Sections; Shear Stresses in a Beam Section; Simple Torsion; Stability and Buckling of Beams.

3

ASSESSMENT & TEXTSASSESSMENT & TEXTSASSESSMENT & TEXTSASSESSMENT & TEXTS

Written examination:Written examination: 80%80%College College of of



Blackboard TestsBlackboard Tests 20%20%

EGEG--120120


TEXTBOOKSTEXTBOOKS

D Gross W Hauger J Shroder W Wall J Bonet D. Gross, W. Hauger, J. Shroder, W. Wall, J. Bonet, Engineering Mechanics 2: Mechanics of Materials, Springer.

Introduction Hibbeler, RC, Mechanics of Materials, Prentice Hall, SI Second Edition.

C Chil & R “St th f M t i l &1/19/2012

Case, Chilver & Ross, “Strength of Materials & Structures”, 4th Edition.

Class notes4

Class notes

BASIC CONCEPTSBASIC CONCEPTS

College of

BASIC CONCEPTSBASIC CONCEPTS

PRINCIPLES: Governing principles

Engineering


EGEG--120120

Governing principles St. Venant's principle

INTERNAL FORCES AND STRESSES: Internal forcesEGEG 120120


Internal forces Direct stress Shear stress

DISPLACEMENTS AND STRAINS: Displacements Linear strain Sh t i

Basic Concepts

Shear strain

MATERIAL BEHAVIOUR: Stress-Strain relationships

1/19/2012

p Superposition principle Material properties Strain energy

4 Material failure Time effects

GOVERNING PRINCIPLESGOVERNING PRINCIPLES

College of

GOVERNING PRINCIPLESGOVERNING PRINCIPLES

The way in which materials transmit loads is Engineering


EGEG--120120

ygoverned by two basic principles:

Equilibrium: the sum of forces and moments on a b d t f th b d t b l t

EGEG 120120


body or any part of the body must be equal to zero.

F F Mx y 0 0 0 ; ;

Certain problems can be solved using only equilibrium considerations These are known as

x y

Basic Concepts

equilibrium considerations. These are known as statically determinate.

Compatibility: the movements resulting from the

1/19/2012

Compatibility: the movements resulting from the external loads must be internally compatible (i.e. the material must not break) and compatible with the external support conditions

5external support conditions.

ST. VENANT'S PRINCIPLEST. VENANT'S PRINCIPLE

College of

ST. VENANT S PRINCIPLEST. VENANT S PRINCIPLE

A useful further principle is St. Venant’s Principle: no Engineering


EGEG--120120

p p pmatter how complex the distribution of external forces at a small region on the surface of a body is, the resulting effect at a small distance away will onlyEGEG 120120


the resulting effect at a small distance away will only depend on the statically equivalent force.

Basic Concepts

1/19/2012

6

INTERNAL FORCESINTERNAL FORCES

College of

INTERNAL FORCESINTERNAL FORCES

Consider a bar with an external load and its Free Engineering


EGEG--120120

Body Diagram:F F

EGEG 120120


Taking a cut through a section of a bar, equilibrium

F

and Newton’s third law of action and reaction imply the existence of an equal internal force acting on each section of the bar:

Basic Concepts

F F F F

1/19/2012 On a given slice we have:

7 F F

DIRECT STRESSDIRECT STRESS

College of

DIRECT STRESSDIRECT STRESS

Stress is the amount of internal force per unit area:Engineering


EGEG--120120

p

F FA

FAEGEG 120120


Units: Newtons/metre2 or N/m2 or Pascal Typically

A

Units: Newtons/metre or N/m or Pascal. Typically engineers use MN/m2, i.e. 106 N/m2 or N/mm2.

Stress can be tensile (+) or compressive (-):

Basic Concepts

( ) p ( )

1/19/2012

Tension (+) Compression (-)8

Tension (+) Compression ( )

SHEAR STRESSSHEAR STRESS

College of

SHEAR STRESSSHEAR STRESS

The force acting on an area may be normal or Engineering


EGEG--120120

g ytangential to the area. The direct stress is then the normal force per unit area and the shear stress is the tangential force per unit area:EGEG 120120


the tangential force per unit area:

F Fn t

FFt

FA

FA

n t and Fn

A

Basic Concepts Signs:

F

1/19/2012

x

y

+

9x

DISPLACEMENTSDISPLACEMENTS

College of

DISPLACEMENTSDISPLACEMENTS

As a result of the external actions materials will Engineering


EGEG--120120

deform. This deformation manifests itself in small movements or displacements of material points. It has units of length (m or mm):EGEG 120120


has units of length (m or mm):

u F

In the case of a beam where the displacement isBasic

Concepts

In the case of a beam where the displacement is perpendicular to the structure, it is known as deflection:

1/19/2012 d

10 F

LINEAR STRAINLINEAR STRAIN

College of

LINEAR STRAINLINEAR STRAIN

All materials deform when subject to external actions Engineering


EGEG--120120

jsuch as loads or temperature changes. The deformation, i.e. change in shape is measured by the strain at a point:EGEG 120120


strain at a point:

Linear Strain is defined as the change in length over the initial length:

Basic Concepts

1/19/2012

Strain is dimensionless It is often given as a %

11 Strain is dimensionless. It is often given as a %.

SHEAR STRAINSHEAR STRAIN

College of

SHEAR STRAINSHEAR STRAIN

Deformation can also imply distortion which is d b h h i h h iEngineering


EGEG--120120

measured by the shear strain as the change in angle:

EGEG 120120


Basic Concepts

The shear strain is dimensionless and often given as a percentage %

1/19/2012

a percentage %.

12

STRESSSTRESS -- STRAIN RELATIONSHIPSSTRAIN RELATIONSHIPS

College of

STRESS STRESS STRAIN RELATIONSHIPSSTRAIN RELATIONSHIPS

Derived from tensile tests:Engineering


EGEG--120120

Strain Gauge

F FEGEG 120120


Strain is related to stress via the stress-strain curve :

F F

Strain is related to stress via the stress-strain curve :

Breaking Point

Basic Concepts

P ti lit

Point

1/19/2012

ProportionalityLimit

Linear Elastic

13 Range

SUPERPOSITION PRINCIPLESUPERPOSITION PRINCIPLE

College of

SUPERPOSITION PRINCIPLESUPERPOSITION PRINCIPLE

In the linear elastic range the effect of more than one Engineering


EGEG--120120

gload can be obtained by adding the effect of each individual load acting alone:

EGEG 120120


F1

F2

=Basic

Concepts F1

1/19/2012F2+

14

MATERIAL PROPERTIESMATERIAL PROPERTIES

College of

MATERIAL PROPERTIESMATERIAL PROPERTIES

In the elastic range, direct stress is proportional to Engineering


EGEG--120120

g p plinear strain. The proportionality coefficient is Young’s Modulus E of the material:

EEGEG 120120

Strength of Strength of MaterialsMaterials Shear stress is proportional to shear strain. The

proportionality coefficient is the Shear Modulus G:

E

proportionality coefficient is the Shear Modulus G:

G

Basic Concepts Thermal effects. Changes in temperature lead to a

linear strain which is proportional to the temperature

1/19/2012

linear strain which is proportional to the temperature change. The proportionality coefficient is the Coefficient of Thermal Expansion :

15 T T

POISSON'S RATIOPOISSON'S RATIO

College of

POISSON S RATIOPOISSON S RATIO

The result of a direct stress in one direction is a Engineering


EGEG--120120

direct strain in the same direction plus a lateral strain:

dEGEG 120120


dd

d

Basic Concepts

The ratio between direct and lateral strain is given by P i ’ ffi i ( i ll 0 3)

1 2

and dd

1/19/2012

Poisson’s coefficient (typically 0.3):

2

16

1

MATERIAL PARAMETERS: Typical ValuesMATERIAL PARAMETERS: Typical Values

College of

MATERIAL PARAMETERS: Typical ValuesMATERIAL PARAMETERS: Typical Values

Material E(GN/m2) (oC-1) u(MN/m2) l(MN/m2)Engineering


EGEG--120120

Mild Steel 200 1.2 10-5 370 280

High Steel 200 1 3 10-5 1550 770EGEG 120120


High Steel 200 1.3 10 1550 770

Concrete T 14 1.2 10-5 3 -

Concrete C 14 1.2 10-5 30 -

Carbon Fibre 170 - 1400 -Basic

Concepts Glass Fibre 60 - 1600 -

-5

1/19/2012

Aluminium 70 2.3 10-5 430 280

Titanium 120 0.9 10-5 690 385

17Magnesium 45 2.7 10-5 280 155

STRAIN ENERGYSTRAIN ENERGY

College of

STRAIN ENERGYSTRAIN ENERGY

FFEngineering


EGEG--120120

When a material is deformed, the work done by the external forces is accumulated as elastic strain

i th t i l

FF

EGEG 120120


energy in the material.

W F A V z zzd d d

The strain energy per unit volume w is the area under the stress-strain relationship:

Basic Concepts

under the stress-strain relationship:

w z ( ) d

1/19/2012 For linear elastic materials w is:

2 218 w E E 1

212

2 12

2

MATERIAL FAILUREMATERIAL FAILURE

College of

MATERIAL FAILUREMATERIAL FAILURE

All materials fail at different values of stress. Engineering


EGEG--120120

Depending on the amount of strain (or strain energy) before failure, the material is said to be brittle or ductile:EGEG 120120


ductile:

DUCTILE MATERIAL BRITTLE MATERIAL

Basic Concepts

Breaking Point Breaking Point

1/19/2012

19

TIME EFFECTSTIME EFFECTS

College of

TIME EFFECTSTIME EFFECTS

Creep: the deformation of materials under load increases with time:Engineering


EGEG--120120

increases with time:

tertiary creepEGEG 120120


secondary creep

primary creep

Fatigue: materials subject to cyclic loads eventually fail at a lower than the short term failure stress:

primary creep t

Basic Concepts

a at a o e t a t e s o t te a u e st ess

Mild SteelEndurance limit

u

1/19/2012Aluminium

20 No. Cycles104 105 107106

BASIC BEAM THEORYBASIC BEAM THEORYBASIC BEAM THEORYBASIC BEAM THEORY

INTRODUCTION:College of

Engineering


Definition Support Conditions Types of Beams


INTERNAL FORCES: Axial force, shear force and bending moment Sign convention Equilibrium of a section

DIRECT STRESSES: Assumptions

Basic Beam Theory

Assumptions Stress and strain due to axial forces Stresses and strains due to bending Second moment of area

1/19/2012

Second moment of area

BEAM DEFLECTION: Introduction Cantilever with a point load

21 Cantilever with a point load Statically indeterminate beams

BEAMSBEAMS -- DEFINITIONDEFINITIONBEAMS BEAMS DEFINITIONDEFINITION

A Beam is a member with one dimension, its length, College of

Engineering


gmuch larger than the other two (width and depth):

P1 PStrength of Strength of MaterialsMaterials

P1 P2

P

wM

P3

Basic Beam Theory

A beam can be loaded transversely with point loads

1/19/2012

A beam can be loaded transversely with point loads, moments or distributed loads w (N/m). These are often constant or uniformly distributed loads (UDL).

22

SUPPORT CONDITIONSSUPPORT CONDITIONSSUPPORT CONDITIONSSUPPORT CONDITIONS

Beam supports can be:College of

Engineering


pp

Built-in or “encastrated” or “clamped”:


VMH

Simple supports:Basic Beam

Theory

Simple supports:

V

1/19/2012 H

V23 V

TYPES OF BEAMSTYPES OF BEAMSTYPES OF BEAMSTYPES OF BEAMS

Depending on the supports beams are classified as:College of

Engineering


p g pp

Cantilever:


Simply supported:

Propped cantilever:Basic Beam

Theory

Propped cantilever:

1/19/2012 Continuous Beam:

24

INTERNAL BEAM FORCESINTERNAL BEAM FORCESINTERNAL BEAM FORCESINTERNAL BEAM FORCES

Consider a beam under general load conditions and College of

Engineering


gtake a cut through a general section. Equilibrium of each resulting component implies the existence of an internal axial force N, shear force S and bending


internal axial force N, shear force S and bending moment M acting on each section along the beam:

P

x

Basic Beam Theory

Py

S

1/19/2012 N

S

N

25 MSM

A.F., S.F. AND B.M. DIAGRAMSA.F., S.F. AND B.M. DIAGRAMSA.F., S.F. AND B.M. DIAGRAMSA.F., S.F. AND B.M. DIAGRAMS

In general, the internal axial & shear forces and the College of

Engineering


gbending moment are not constant but change from section to section along the beam, i.e. they are functions of x. The corresponding graphs are the


functions of x. The corresponding graphs are the axial force, shear force and bending moment diagrams. For the case above:

M P x ( ) cos

xBasic Beam

Theory

x

xx

M

1/19/2012

xx

26 SN sinN P q= cosS P q=

SIGN CONVENTIONSSIGN CONVENTIONSSIGN CONVENTIONSSIGN CONVENTIONS

Due to Newton’s third law of action and reaction the College of

Engineering


internal forces on left facing and right facing sections are in opposite directions. Hence to establish a sign convention the pair acting on opposing faces of a


convention the pair acting on opposing faces of a slice is considered:

Basic Beam Theory

Positive BM SF and AF are:

1/19/2012

Positive BM, SF and AF are:

M M S S N N27

S

EQUILIBRIUM OF A BEAM SECTIONEQUILIBRIUM OF A BEAM SECTIONEQUILIBRIUM OF A BEAM SECTIONEQUILIBRIUM OF A BEAM SECTION

Consider a beam section with a vertical distributed College of

Engineering


load w and a horizontal distributed load wx:

w M dMStrength of Strength of MaterialsMaterials

M MSN N

N dN

S dx

w

S dS

Basic Beam Theory Equilibrium gives:

wx

1/19/2012dNdx

wdSdx

wdMdx

Sd Mdx

wx 2

2

28

DIRECT BEAM STRESSESDIRECT BEAM STRESSES -- ASSUMPTIONSASSUMPTIONSDIRECT BEAM STRESSES DIRECT BEAM STRESSES ASSUMPTIONSASSUMPTIONS

In order to determine the stresses and strains inside College of

Engineering


beams the following basic assumptions are made:

Beam sections remain undeformed:Strength of Strength of MaterialsMaterials

The material behaviour in tension and compression is

Basic Beam Theory

identical and remains inside the linear elastic range.

Additionally the superposition principle will also be used The resulting theory was first proposed by Euler

1/19/2012

used. The resulting theory was first proposed by Euler and is known as Euler Beam Theory.

29

Stress And Strain Due To Axial ForceStress And Strain Due To Axial ForceStress And Strain Due To Axial ForceStress And Strain Due To Axial Force

Consider a beam with a simple (i.e. doubly symmetric) College of

Engineering


p ( y y )section with area A under an axial force N.

Beam cross-sectionu

Strength of Strength of MaterialsMaterials N N x x

The resulting stress and strain in a section will be t t d l t

u dudx y

Basic Beam Theory

constant and equal to:

NA

ENEA

1/19/2012 The strain is related to the horizontal displacement as:

A

du du N30 du

dxdudx

NEA

STRESSES AND STRAINS IN BENDINGSTRESSES AND STRAINS IN BENDINGSTRESSES AND STRAINS IN BENDING STRESSES AND STRAINS IN BENDING

Consider a beam with a rectangular cross-section College of

Engineering


gunder pure bending:


M M x za b

After deformation: Neutral fibre

dxy

Basic Beam Theory

R

Neutral fibre

yR

ER

y

1/19/2012a’ b’

E

31 y R

STRESSES DUE TO A BENDING MOMENTSTRESSES DUE TO A BENDING MOMENTSTRESSES DUE TO A BENDING MOMENTSTRESSES DUE TO A BENDING MOMENT

The stresses due to pure bending vary linearly College of

Engineering


p g y yacross the section:

M


zy

dA

Noting that the resultant moment of these stresses must be M gives:

Basic Beam Theory

y

MI

ER

MI

yR

MEI

or = and 1

1/19/2012where the second moment of area I of the section about the z axis is:

I dA2z32 I y dAA

= 2z

SECOND MOMENTS OF AREASECOND MOMENTS OF AREASECOND MOMENTS OF AREASECOND MOMENTS OF AREA

bt

College of Engineering


2r 2r


Ibh

3

12I

r 4

4I t r 3

12

P ll l A i ThP ll l A i ThBasic Beam

Theory

Parallel Axis Theorem:Parallel Axis Theorem:0

1/19/2012I I AdC0

2 cc

33

BEAM DEFLECTIONBEAM DEFLECTIONBEAM DEFLECTIONBEAM DEFLECTION

The curvature radius R of a beam is related to the College of

Engineering


bending moment M by:1R

MEI

dx


R EIdxd

x

yR

Basic Beam Theory

For small deflections y(x) the slope = dy/dx and

y

d

1/19/2012

For small deflections y(x) the slope dy/dx and therefore:

1 2 2d d y d y M

hence34 2 2R dx dx dx E

hence I

CANTILEVER WITH POINT LOAD: DeflectionCANTILEVER WITH POINT LOAD: DeflectionCANTILEVER WITH POINT LOAD: DeflectionCANTILEVER WITH POINT LOAD: Deflection

Consider a cantilever beam with an point load:College of

Engineering


p

M x P x( ) ( ) P


x

y

The deflection is:

d P d2

y

Basic Beam Theory

d y

dx

PEI

x ydydx

P Py

2

20

2 3

0 0 0

F I

zz( ) ( ) and and

1/19/2012

y xPEI

x dx dxPEI

x x2 3

63 F

HIK zz( ) ( ) ( )

P3

35 and PE

3 I

STATICALLY INDETERMINATE BEAMSSTATICALLY INDETERMINATE BEAMSSTATICALLY INDETERMINATE BEAMSSTATICALLY INDETERMINATE BEAMS

Knowing the expressions for the deflection and slope College of

Engineering


g p pof a beam it is possible to solve for redundant reactions using the superposition principle:


w

V (?)w

1Basic Beam

Theory V

w

1

V

1/19/2012

w

wV V 1 Hence 36

1

ANALYSIS OF STRESS AND STRAINANALYSIS OF STRESS AND STRAIN

2-DIMENSIONAL STATES OF STRESS: Shear stress in a tensile test Mohr’s circle for the tensile test

College of College of EngineeringEngineering


Mohr s circle for the tensile test General 2-D stress system Principal stresses Mohr’s circle

EGEG--120120


2-DIMENSIONAL STRAIN: Direct strain Shear strain G l 2 D t i General 2-D strain

ELASTIC STRESS-STRAIN RELATIONSHIPS: Strain due to uni-axial stress Direct stress strain relationships

Stress and Strain

Direct stress-strain relationships Shear stress-strain relationship

PRESSURISED VESSELS: Longitudinal and Hoop Stresses

1/19/2012

Longitudinal and Hoop Stresses Cylindrical shells Spherical shells

3-D STRESS AND STRAIN:37

3 D STRESS AND STRAIN: 3-D stresses & Strains General elastic stress-strain equations

SHEAR STRESSES IN A TENSILE TESTSHEAR STRESSES IN A TENSILE TESTSHEAR STRESSES IN A TENSILE TESTSHEAR STRESSES IN A TENSILE TEST

Consider a simple tensile test situation:College of College of



p

x xx

y

EGEG--120120


and obtain the direct and shear stress acting on the

x

and obtain the direct and shear stress acting on the

cut shown using equilibrium:

Stress and

Strain

x

1/19/2012 1

212 2x x cos

38 12 2xt sin

MOHR'S CIRCLE FOR TENSILE TESTMOHR'S CIRCLE FOR TENSILE TESTMOHR S CIRCLE FOR TENSILE TESTMOHR S CIRCLE FOR TENSILE TEST

The direct and shear stress in a tensile test given by:College of College of



g y

12

12

12

2

2x x

xt

cos

sinEGEG--120120


can be graphically interpreted as points in a circle known as Mohr’s Circle:

2 x

= 45

Stress and Strain

½x x2

1/19/2012 = 90 = 0

39

GENERAL 2GENERAL 2--D STRESS SYSTEMD STRESS SYSTEMGENERAL 2GENERAL 2 D STRESS SYSTEMD STRESS SYSTEM

In general materials are subject to direct and shear stress:College of College of



stress:

yx xy

y

EGEG--120120


xy

xy

x x

Due to rotational equilibrium the shear stresses are equal:

yx y

Stress and Strain

equal:

Th b i i i

yx xy

1/19/2012

The stresses can be written as a symmetric matrix:

LM OP x xy

40 NM QP yx y

STRESSES ON AN INCLINED SECTIONSTRESSES ON AN INCLINED SECTIONSTRESSES ON AN INCLINED SECTIONSTRESSES ON AN INCLINED SECTION

Consider a general stress case, the stresses on an inclined plane are:College of College of



inclined plane are:

x

xy

EGEG--120120


E ilib i i

x

xy

Equilibrium gives: y

12

12 2 2( ) ( ) cos sinx y x y xy

Stress and Strain

There are two angles for which = 0, these define

12 2 2( ) sin cosx y xy

1/19/2012

the principal directions of stress:

1 1 1 12

902

tan tanxy xyand41

1 2 2 290

tan tan

x y x y and

PRINCIPAL STRESSESPRINCIPAL STRESSESPRINCIPAL STRESSESPRINCIPAL STRESSES

The direct stresses on the principal directions are the College of College of



p pprincipal stresses: y

yxEGEG--120120


2 1

1

x

xy

x

y xy

2

y

yx

Stress and Strain The principal stress values are:

y

2 2

1/19/2012

112

12

2 2

212

12

2 2

4

4

( ) ( )

( ) ( )

x y x y xy

x y x y xy

42

MOHR'S CIRCLEMOHR'S CIRCLEMOHR S CIRCLEMOHR S CIRCLE

The direct and shear stress at an angle from the College of College of



gprincipal directions define a circle:

12 1 2

12 1 2 2( ) ( ) cos 1

EGEG--120120


12 1 2 2( ) sin

2

max

xyMaximum shear:Maximum shear:

Stress and Strain

2 1

y max 12 1 2( )2

1/19/2012

xy

12 1 2( )

x

43y

DIRECT STRAINDIRECT STRAINDIRECT STRAINDIRECT STRAIN

Direct strain along the x axis:College of College of



gy

EGEG--120120


xx x

xx

x

Direct strain along the y axis:

yStress and

Strain

y

y yy

1/19/2012y

y

44 x

SHEAR STRAINSHEAR STRAINSHEAR STRAINSHEAR STRAIN

The shear strain measures the distortion as the College of College of



change in angle. This can be measured with reference to the x or y axis:

EGEG--120120


y y

xy

yx

xxStress and

Strain

A simple rotation shows that:

1/19/2012 xy yx

45

GENERAL 2GENERAL 2--D STRAIND STRAINGENERAL 2GENERAL 2 D STRAIND STRAIN

In general shear and direct strains will take place College of College of



g psimultaneously:

y

EGEG--120120


xyy

y y

x

Stress and Strain

The strains can be written as a symmetric matrix:

x x x

1/19/2012

y

LNM

OQP

x xy

46 NM QP yx y

STRAINS DUE TO UNISTRAINS DUE TO UNI--AXIAL STRESSAXIAL STRESSSTRAINS DUE TO UNISTRAINS DUE TO UNI AXIAL STRESSAXIAL STRESS

Consider a 1x1 block of material under uniaxial stress College of College of



x :

1y

y

EGEG--120120

Strength of Strength of MaterialsMaterials 1 x

x x

x

If E is the Young’s modulus and the Poisson’s ratiothe strains in the x and y directions resulting from xare:

Stress and Strain

are:

Si il l th t i lti f t i th

x x y x xE E

1 and

1/19/2012

Similarly the strains resulting from a stress in the y direction y are:

1 d47 y y x y yE E

1 and

DIRECT STRESSDIRECT STRESS--STRAIN RELATIONSHIPSSTRAIN RELATIONSHIPSDIRECT STRESSDIRECT STRESS STRAIN RELATIONSHIPSSTRAIN RELATIONSHIPS

Consider a unit block subject to direct stresses in both College of College of



jaxes:

y

yy

EGEG--120120

Strength of Strength of MaterialsMaterials 1

y

x x

1 x

x

Stress and Strain Using the superposition principle gives:

y

1 E ( )1/19/2012

x x y

y y x

E

1

1

( )

( )

x x y

y y x

E

E

1 2

2

( )

( )or

48 y y xE

( )

y y x

1 2 ( )

SHEAR STRESSSHEAR STRESS--STRAIN RELATIONSHIPSTRAIN RELATIONSHIPSHEAR STRESSSHEAR STRESS STRAIN RELATIONSHIPSTRAIN RELATIONSHIP

Consider a unit block subject to shear stress xy:College of College of



j xy

yx

y xy

EGEG--120120

Strength of Strength of MaterialsMaterials x

xy xy

Noting that the principal axes are at 45, that:

x xy

Stress and Strain

i l i hi b d i f h

1 2 2 2 xy and 1

1/19/2012

gives a relationship between and in terms of the shear modulus G of the material as:

G GEh49 xy xyG G

E

where

2 1( )

PRESSURISED VESSELSPRESSURISED VESSELSPRESSURISED VESSELS PRESSURISED VESSELS

Many engineering applications involve thinthinli d i l h i l l dCollege of College of



cylindrical or spherical vessels under pressure:

p2REGEG--120120


p

pp

Stress and Strain

We can identify Longitudinal Longitudinal stresses and HoopHoopor circumferential stresses in a cylinder:

lshs

s1/19/2012

hs

lsls

50

hs

STRESSES IN THIN CYLINDRICAL SHELLSSTRESSES IN THIN CYLINDRICAL SHELLSSTRESSES IN THIN CYLINDRICAL SHELLSSTRESSES IN THIN CYLINDRICAL SHELLS

Equilibrium shows that the hoop stresses are:College of College of


Prof. J. Prof. J. BonetBonet R

t

EGEG--120120


hshs p

For a closed cylinder, the longitudinal stress due to internal pressure is:

Stress and Strain ls

1/19/2012

p

ls51

STRESSES IN SPHERICAL SHELLSSTRESSES IN SPHERICAL SHELLSSTRESSES IN SPHERICAL SHELLSSTRESSES IN SPHERICAL SHELLS

In a thin sphere under pressure, there are equal College of College of



p p qspherical stresses in all directions:ss

EGEG--120120


ss ss

Equilibrium across a maximum circle gives

ssss

Stress and Strain

q g

R

t

1/19/2012ssss p

R

52

33--D STRESS SYSTEMD STRESS SYSTEM33 D STRESS SYSTEMD STRESS SYSTEM

In general: z



g zzy

yzzxEGEG--120120


xz yx

yy

xyx

Stress and Strain

The stresses can be written as:x

1/19/2012

LMMOPP

x xy xz

yx x yz

xy yx

xz zx

and

53 NMM Q

PP y y

zx zy z zy yz

There are 6 independent strains in 3-D, 3 direct College of College of



pstrains:

yzzz

EGEG--120120


yy

y

x

x

z

and 3 shear strains:

x xx

Stress and Strain zzz

zx

yx

1/19/2012y y yzy

zx

54 x x x

33--D STRESSD STRESS--STRAIN RELATIONSHIPSSTRAIN RELATIONSHIPS33 D STRESSD STRESS STRAIN RELATIONSHIPSSTRAIN RELATIONSHIPS

The general stress-strain relationships for an elastic College of College of



g pmaterial are:

E ( ) G xy xy EGEG--120120


E

E

E

x x y z

y y x z

( )

( )

( )

G

G

G

xy xy

xz xz

yz yz

The direct stress-strain equations can be inverted

E z z x y ( ) yz yz

Stress and Strain

qand expressed in terms of the Lamè coefficients:

E 2 ( )1/19/2012

GE

E2 1( )

x x x y z

y y x y z

2

2

2

( )

( )

( )55

1 1 2( )( ) z z x y z 2 ( )

ADVANCED BEAM THEORY (I)ADVANCED BEAM THEORY (I)ADVANCED BEAM THEORY (I)ADVANCED BEAM THEORY (I)

COMBINED STRESSES IN BEAMS:COMBINED STRESSES IN BEAMS:College of College of



Stress due to bending moment and axial force

Eccentric thrust, section core

Centroid of a symmetric sectionEGEG--120120


Centroid of a symmetric section

Centroid of a general section

Stress due to bending in two planes

Bending of beams with general sections

Double bending of beams

Principal axesBeam Theory

Principal axes

SHEAR STRESSES IN BEAM SECTIONS:SHEAR STRESSES IN BEAM SECTIONS: Warping of sections due to shear

1/19/2012 Shear Stresses in a beam section

Shear stresses in a rectangular section

Shear stress flow in thin sections53

Shear stress flow in thin sections

ADVANCED BEAM THEORY (II)ADVANCED BEAM THEORY (II)ADVANCED BEAM THEORY (II)ADVANCED BEAM THEORY (II)

TORSION OF BEAMS:TORSION OF BEAMS:College of College of



Torsion of beams

Torsion of circular beamsEGEG--120120


Shear stresses due to torsion

Torsion of thin walled tubes

INSTABILITY OF BEAMS AND COLUMNS:INSTABILITY OF BEAMS AND COLUMNS:Beam Theory

Buckling of beams and columns

Buckling modes

1/19/2012 Buckling of beams with initial deflections

54

STRESSES DUE TO COMBINED M & NSTRESSES DUE TO COMBINED M & NSTRESSES DUE TO COMBINED M & NSTRESSES DUE TO COMBINED M & N

Consider a section under a combined axial force N College of College of



and bending moment M. Using the superposition principle: M

EGEG--120120


z

y

N

y

N MBeam Theory

N M

NNA

MMI

y

1/19/2012

N A I

N M

y55

A I

y

ECCENTRIC THRUSTECCENTRIC THRUST -- SECTION CORESECTION COREECCENTRIC THRUST ECCENTRIC THRUST SECTION CORESECTION CORE

Consider an axial force N acting at a distance e from College of College of



gthe section centre:

N

M Ne

EGEG--120120


e

y

NN

F INey1

The region where a compressive force N can be

FHIKN

AyI

Beam Theory

g papplied without resulting in tensile stresses is known as the section corethe section core. For a circular section:

1/19/2012R 4

R

56

CENTROID OF A SYMMETRIC SECTIONCENTROID OF A SYMMETRIC SECTIONCENTROID OF A SYMMETRIC SECTION CENTROID OF A SYMMETRIC SECTION

Consider a section only symmetric about the vertical College of College of



y yaxis: z

ycMN

EGEG--120120

Strength of Strength of MaterialsMaterials y

M

N A M y y Ic( )

MNA

y y dA NMI

y y dAcA

cA

z z( ) ( )0 0 or

Beam Theory

Both equations lead to the definition of the centroidcentroid:

A A

1/19/2012y

AydAc z1

57A A

CENTROID OF A GENERAL SECTIONCENTROID OF A GENERAL SECTIONCENTROID OF A GENERAL SECTIONCENTROID OF A GENERAL SECTION

Consider a general section with uniform stress:College of College of



g

zyc

EGEG--120120


yc

z

Enforcing that: yzc

z zBeam Theory

Gi

( ) ( )y y dA z z dAcA

cA

z z0 0 and

1/19/2012

Gives:

yA

ydA zA

z dAc cAA

zz1 1 and

58A A AA

STRESSES DUE TO BENDING IN 2 PLANESSTRESSES DUE TO BENDING IN 2 PLANES

Consider the case of a beam bending in 2 directions:

STRESSES DUE TO BENDING IN 2 PLANESSTRESSES DUE TO BENDING IN 2 PLANES



g

MMEGEG--120120


z z z

y

M y

y

M z

x

Using the superposition principle:y yy

E

yE

zBeam Theory and for doubly symmetric sections:

R

yR

zy z

1 1M M M M

1/19/2012

h zz z2 2

1 1R

MEI R

M

EIMI

yM

Iz

y

z

z

y

y

z

z

y

y

z and hence

59 where: I y dA I z dA I yz dAz yAA

yzA

zz z2 2 0 ; and

Bending Of Beams With General SectionsBending Of Beams With General SectionsBending Of Beams With General SectionsBending Of Beams With General Sections

Consider a beam with a general cross section:College of College of



g

zx

w w

EGEG--120120


Assuming gives: E

y M z dAE

I z 0 ?

yy

z

Assuming gives:

To support the vertical loads the beam has to bend in the vertical and horizontal planes:

R

yy

M z dAR

Iyy

yz z 0 ?

Beam Theory

p

z

ER

yER

zy z

1/19/2012

1 1 I M Iy

Izyz z y zy

FG

IJand

where the conditions My=0 and give:M ydAz z60 2R R I I I I

yI

zz y y y z yz y

HG KJ and

Double Bending Of General SectionsDouble Bending Of General SectionsDouble Bending Of General SectionsDouble Bending Of General Sections

Consider a general section under bending moments College of College of



g gMy and Mz:

centroid

x MEGEG--120120


zx M z

M y

Combining:y

Beam Theory

Gives:

z zER

yER

z M ydA M z dAy z

zA

yA

with and

1/19/2012

Gives:

M

M

I I

I IE R

E Ry z

I I

I I

M

Mz z yz y z yz zLNMOQP LNM

OQPLNMOQP

LNM

OQPLNMOQP

//

and 1

61M I I E R

yI I My yz y z yz y yNM QP NM QPNM QP NM QP NM QP/

PRINCIPAL AXES OF A SECTIONPRINCIPAL AXES OF A SECTIONPRINCIPAL AXES OF A SECTIONPRINCIPAL AXES OF A SECTION

Consider a general section:College of College of


Prof. J. Prof. J. BonetBonetz

y y zcos sini EGEG--120120

Strength of Strength of MaterialsMaterials New axes can be defined so that Iy’z’ = 0. These are

zyy

z y zsin cos

y z

known as Principal AxesPrincipal Axes. Noting that:

I y z dA y z y z dAy z z z ( cos sin )( sin cos ) Beam Theory

G

y y y

I I I

y zA A

y z zy

z z ( )( )

sin cos sin cos (cos sin ) 2 2 0

1/19/2012

Gives:

tan22

I My

Mzzy z ythen

62tan2

I I Iy

Iz

z y z y then

Warping Of Beam Sections Due To ShearWarping Of Beam Sections Due To ShearWarping Of Beam Sections Due To Shear Warping Of Beam Sections Due To Shear

Consider a beam section with a shear force S:College of College of


Prof. J. Prof. J. BonetBonetx z

S SEGEG--120120


Note that the assumption of un-deformable sectionsy

S S

dx Note that the assumption of un-deformable sections

leads to constant shear strain and stress:NOT POSSIBLE

Beam Theory S

A

1/19/2012

Hence the section has to warp:

63 Hence the section has to warp:

SHEAR STRESSES IN A BEAM SECTIONSHEAR STRESSES IN A BEAM SECTIONSHEAR STRESSES IN A BEAM SECTIONSHEAR STRESSES IN A BEAM SECTION

Consider a portion of a beam section with a shear College of College of



pforce S and bending moment M:

xMM dM

SEGEG--120120


zx

y

M S

A d

Using equilibrium and:

ydxby

Ay

Beam Theory

g q

gives:

MyI

dMdX

S and

1/19/2012

gives:

FHGIKJ zSQ

Ib Q y dA

Q A

IbSA

yy

y where or ; 64 HG KJzIb

yIb Ay

yA y

y

Shear Stress In A Rectangular SectionShear Stress In A Rectangular SectionShear Stress In A Rectangular SectionShear Stress In A Rectangular Section

Consider the particular case of a rectangular section:College of College of



p g

I bh 112

3zh

EGEG--120120


b

A bh

Q y bdybh by

y

h

z2 2 2

8 2

/

h y

The shear stress distribution is therefore:

b Q y yyyz 8 2y

Beam Theory

The shear stress distribution is therefore:

F IL O32

S y1/19/2012

FHGIKJ

LNMM

OQPP

32

12

SA

yh

6515. ( )S A

SHEAR STRESS FLOW IN THIN SECTIONSSHEAR STRESS FLOW IN THIN SECTIONSSHEAR STRESS FLOW IN THIN SECTIONS SHEAR STRESS FLOW IN THIN SECTIONS

Consider a thin-walled I beam section with a shear College of College of



force S: Along the flange:

Q hts b ty y 12 1 and t

EGEG--120120


Along the web: y y2 1

Q t bh s h sy 12 2 2[ ( )]

h

S1S2

b ty

b

1S2

Beam Theory Along the flange:

ShI

s2 1

1/19/2012Along the web:

s1

s2

I2

S

bh s h s2 2( )66 2 2 2I

( )

TORSION OF BEAMSTORSION OF BEAMSTORSION OF BEAMSTORSION OF BEAMS

Beams can be subject to twisting moments:College of College of



j g

EGEG--120120

Strength of Strength of MaterialsMaterials z zx x

The simplest case is the torsion of a circular shaft:

xy y

Beam Theory

1/19/2012

67

TORSION OF A CIRCULAR SHAFTTORSION OF A CIRCULAR SHAFTTORSION OF A CIRCULAR SHAFTTORSION OF A CIRCULAR SHAFT

Consider the torsion of a circular shaft:College of College of


Prof. J. Prof. J. BonetBonetNoting that:

dEGEG--120120


d and

rddx

G ;

d

rdx T r dA

A

z Beam Theory

gives:R

TI

rddx

TGI0 0

;

1/19/2012 where the Polar second moment of area isPolar second moment of area is::

I dx GI0 0

I r dAz 268

I r dAA

0 z

SHEAR STRESSES DUE TO TORSIONSHEAR STRESSES DUE TO TORSIONSHEAR STRESSES DUE TO TORSIONSHEAR STRESSES DUE TO TORSION

Torsion results in a shear stress distribution as:College of College of


Prof. J. Prof. J. BonetBonet Tr

EGEG--120120


I

r0

T

R

For a circular section the polar moment of area is:

Beam Theory I r dA r r d dr R

R

A0 zzz 2 2

0

24

0 2

1/19/2012so the maximum shear stress is:

A 00

max 2

3T

R69

R

TORSION OF THIN WALLED TUBESTORSION OF THIN WALLED TUBESTORSION OF THIN WALLED TUBESTORSION OF THIN WALLED TUBES

Consider a thin walled tubular section:College of College of



T tr ds z t

AT

EGEG--120120


T tr ds

t dA

zz

2dA r

AT

Equilibrium implies that t must be constant and

ds

Beam Theory

Equilibrium implies that t must be constant and hence:

T

1/19/2012

Where A is the area (in blue) contained inside the

tA2

70Where A is the area (in blue) contained inside the hollow section.

BUCKLING OF BEAMS AND COLUMNSBUCKLING OF BEAMS AND COLUMNSBUCKLING OF BEAMS AND COLUMNSBUCKLING OF BEAMS AND COLUMNS

Consider a simply supported column under College of College of



p y ppcompression: x

N M Ny

d y M

2

EGEG--120120


d y

dx

ME

y y

2

2

0 0I

( ) ( )

For certain high values of N, the beam becomes

yy y( ) ( )

Beam Theory

g ,unstable and bends significantly. This is known as bucklingbuckling. Assuming a sinusoidal deflection after buckling gives:y x( ) sin

1/19/2012

buckling gives:y x( ) sin

k NEI

Nk EI

hence 2 2

2

71Each value of k represents a buckling mode.buckling mode.

EI

BUCKLING MODESBUCKLING MODESBUCKLING MODESBUCKLING MODES

The first 3 buckling modes of a simply supported College of College of



beam are:N 1

N 2 N 3

EGEG--120120


/ 2 / 3

The corresponding buckling loads are:

Mode 1 Mode 2 Mode 3

Beam Theory

The first value is the Euler buckling load:Euler buckling load:

NE

kkk

k 2

2I

where (length between inflections)

1/19/2012

The first value is the Euler buckling load:Euler buckling load:

NE

E 2

2I

72 For values below this, the beam is stable.

Buckling Of Beam With Initial DeflectionBuckling Of Beam With Initial DeflectionBuckling Of Beam With Initial DeflectionBuckling Of Beam With Initial Deflection

Consider a beam with a n initial imperfection in the College of College of



pform of a deflection y x y xi ( ) sin 0 a f

M N y yi ( )Nx

EGEG--120120


y y

d y

dx

ME

i

( )2

2 Iy0 yc

y y ( ) ( )0 0

y

Beam Theory Assuming a final buckled deflection

gives:y x y xc( ) sin a f

1/19/2012 y yN

N NcE

0

NE

73E

y yc / 0

Prof. J. Bonet, 11/01/11 1c:\users\cgbonet\documents\teaching\strength\example_sheets.docx

STRENGTH OF MATERIALS EG-120 Example Sheet No. 1

Basic Concepts (For steel assume E = 200 GN/m2 and ν = 0.3)

1- The steel bolt shown carries a tensile load of 45 kN. The smooth section of the shaft has a diameter of 26 mm and the threads are 2 mm deep. (a) Determine the tensile stress in the smooth part of the shaft and in the threaded section. (Ans.: 84.75 MN/m2 & 118.4 MN/m2) (b) Calculate the increase in length of an original 30 mm length of the smooth shaft and the resulting reduction in section. (Ans.: .0127 mm & -.135 mm2)

2- The piston rod of a double acting cylinder is 3 m long and has a diameter of 150 mm. The piston has a diameter of 420 mm and is subject to an oil pressure of 12 MN/m2 on one side and 4 MN/m2 on the other. On the return stroke these pressures are reversed. (a) Obtain the stress in the piston rod. (Ans.: -66.7 MN/m2) (b) Determine the change in length of the rod between strokes. (Ans.: 1.76 mm)

3- A plank of wood is attached to a wall by means of two nails of diameter 3 mm placed as shown in the figure. Determine the shear stress in the nails if a vertical force of 1 kN is applied at the end of the plank. (Ans.: 212 MN/m2 & 354 MN/m2)

4- A manual signal is operated remotely by means of a 750 m long steel cable with a diameter of 0.5 cm. It is know that in order to move the signal end of the cable by 17.5 cm, a movement of 46.2 cm is required at the remote end. (a) Determine the strain in the cable. (Ans.: 3.83 10-4) (b) Determine the force required to operate the signal. (Ans.: 1.5 kN)

45 kN

60 cm 40 cm

46.2 cm

17.5 cm


5- A rigid rod 2 m in length is suspended by two steel wires as shown in the figure. Both wires are 1.5 m long but the left wire has a diameter of 1.5 mm whereas the right wire is 2.5 mm in diameter. (a) If a 200 N weight is suspended from the middle of the rod, determine the stresses in the wires and the resulting slope of the rod. (Ans.: 56.6 MN/m2, 20.4 MN/m2 & 0.008°) (b) Determine the point from which weight has to be suspended in order to keep the rod horizontal. (Ans.: 1.47 m from left)

6- A 1 m long aluminium rod with a Young’s modulus of 70 GN/m2 and a diameter of 2.5 cm is placed inside an equally long hollow steel cylinder with an internal diameter of 4 cm and 0.2 cm in thickness. Rod and cylinder are then welded together by means of rigid plates at either end (see figure). Determine the stress and strain in both the rod and cylinder when a 10 kN tensile force is applied at the end plates. (Ans.: ε =1.18 10-4, σstee l= 23.6 MN/m2 & σaluminium = 8.26 MN/m2)

7- A 25 mm long steel rod and a 50 mm long copper rod are joined together as shown. Both rods have a diameter of 10 mm. The combined rod is placed between rigid surfaces and its temperature is raised by 50 °C. The Young’s modulus of copper is 110 GN/m2 and the coefficients of thermal expansion are 1.2x10-5 per °C for steel and 1.9x10-5 per °C for copper. (a) Determine the stresses in each rod. (Ans.: 107.8 MN/m2) (b) Find the displacement of the copper-steel interface. (Ans.: 1.53 10-3 mm to right)

8- A weight of 200 N is suspended 2 m below a horizontal beam by means of two steel wires 2 mm in diameter at angles of 30° and 60° with the vertical (see figure). (a) Find the forces in the wires. (Ans.: 100 N and 173 N) (b) Obtain the vertical and horizontal displacements of the weight. (Ans.: 0.87 mm downwards and .23 mm leftwards)

30°

2 m

1.5 m

2 m

1.5 m

200 N

Steel Copper

25 mm 50 mm

200 N

60°


STRENGTH OF MATERIALS Example Sheet No. 2

Beam Theory For steel take E=200 GN/m2, ν=0.3 and ρ=7840 kg/m3. Take gravity g=10 m/s2.

9- Consider a 250 m long uniform steel rod with a section of 100 mm2 hanging under the action of gravity. (a) If the rod hangs from a single point at the top as shown in figure (a): sketch the axial force diagram; determine the axial strain at any point of the rod and the vertical displacement at the lowest and middle points. (Ans.: umid= 9.2 mm, ulow=12.25 mm.) (b) If the rod hangs from two points as shown in figure (b): determine the reaction at bottom of the rod; sketch the axial force diagram; and obtain the vertical displacement at the middle point. (Ans.: umid=3 mm.)

10- An infinitely long steel wire rests on a surface with a uniform friction coefficient µ=0.4. The wire has a section of 300 mm2 and is pulled on one end by a force of 1000 N. Determine: (a) the axial force along the wire; (b) the stress and strain along the wire; and (c) the horizontal displacement at the end point where the load is applied. (Ans.: u=0.89 mm)

11- Draw the bending moment and shear force diagrams for the simply supported beams of length shown (=12 m, P=6kN, w=2 kN/m and M=1 kNm)

12- Draw the bending moment and shear force diagrams for the cantilevered beams of length shown (use ,P,M & w as above):

/2

/2

P P P

(a) (b)

1000 N

/2 /2

M

/3 /3

P /3

/3 P /3 /3

/3 /3 w

P /3

/3 w

/3 /3 w

M

w

M M w

P

/2 P

w

w

/2 P

/2


13- A steel beam is to be constructed by welding together three equal plates with rectangular cross sections of dimensions 10 x 100 mm2. (a) Determine the second moment of area of each of the possible welded configurations shown in the figure with respect to their middle horizontal axes. (Ans. in mm4: I(a)=22.5 106, I(b)=25 103, I(c)=2.5 106, I(d)=22.5 104, I(e)=1.68 106, I(f)=8.5 105, I(g)=6.9 106, I(h)=7.73 106) (b) Determine the maximum stress that will result from the application of a bending moment of 1 KNm on each of the sections shown. (Ans. (a)-(h) in N/mm2: 6.7, 200, 20, 67, 30, 59, 8.7, 14)

14- For the simply supported beams of length , Young modulus E and second moment of area I shown, draw the shear force and bending moment diagrams and determine the deflection at the centre of the beam and the slope at the supports. (Ans. clockwise: ymid=M2/16EI, P3/48EI, 23P3/648EI, 5w3/384EI)

15- For the cantilevered beams of length , Young modulus E and second moment of area I shown, draw the SF & BM diagrams and determine the deflection and slope at the free end. (Ans.: y=5P3/48EI, w4/8EI ; θ=P2/8EI,

w3/6EI)

16- Consider the propped cantilevered beams of length , Young modulus E and second moment of area I shown. (a) Using the superposition principle, determine the redundant vertical reaction. (Ans.: V=5P/16, 3w/8)

(b) Obtain expressions for both the position and magnitude of the maximum deflection. (Ans.: y=P3/48 3EI, w2/185EI)

(c) Sketch the deflected shape of the beam.

M /2

/2 P

/3 /3 w

P w

/2

w

P /2

(a)

(b)

(c) (d)

(e)

(f)

(g)

(h)

P P



Stress and Strain

17- Consider the element of material shown in the figure subject to biaxial direct and shear stresses. For each of the cases given below, draw the Mohr circle; determine the principal stresses and their direction, as well as the maximum shear stress and its direction.

(a) σx=6 N/mm2, σy=8 N/mm2, τxy=3 N/mm2 (Ans.: σ1=10.2, σ2=3.8 at -36°, τmax=3.2 )

(b) σx= -5 N/mm2, σy=7 N/mm2, τxy=4 N/mm2 (Ans.: σ1=8.2, σ2=-6.2 at -16.8°, τmax=7.2 )

(c) σx=10 N/mm2, σy= -6 N/mm2, τxy=4 N/mm2 (Ans.: σ1=-11, σ2=-7 at 13.3°, τmax=9 )

18- For the rectangular stressed piece of material

shown, draw the Mohr circle and determine the direct and shear stress on the plane shown for:

(a) α = 30o , (b) α = 45o , (c) α = 60° (Ans.: (a) σ=116 τ=-37 (b) σ=90 τ=-60 (c) σ=56 τ=-67)

19- A crack has been detected at 75° in a wall subject to a vertical stress of 60 MN/m2. If the maximum friction coefficient on the crack’s surface is 0.5, determine the minimum horizontal compressive stress required to ensure the stability of the wall.

(Ans.: σ = -18.13 N/mm2)

20- A steel plate is subject to two mutually

perpendicular stresses, one compressive of 45 MN/m2 , the other tensile of 75 MN/m2, and a shearing stress, parallel to these directions, of 45 MN/m2. (a) Find the principal stresses. (b) Find the strains in the horizontal and vertical axes taking E = 200 GN/m2 and ν = 0.3. (c) Find the principal strains. (Ans.: (a) 90 MN/m2 and – 60 MN/m2 (c) 5.4 10-4 and –4.35 10-4)

45

45

45

60 N/mm2

120 N/mm2 30 N/mm2

30 N/mm2

30 N/mm2

σx

σy

τxy

τxy

τxy

τxy

σx

σy

α

120 N/mm2

30 N/mm2

45

45

75

45

75


21- Consider a long closed cylindrical shell of length , radius r and small thickness t under internal pressure p as shown in the figure. (a) Determine, at a point away from the edges, the longitudinal and circumferential stresses and the maximum shear stress. (Ans.: σ

=pr/2t,σc=pr/t, τmax=pr/4t)

(b) Assuming that the material has a Young’s modulus E and Poisson’s ratio ν, determine the change in length and diameter of the tube. (Ans.: ∆ = pr(1-2ν)/2Et, ∆d = pr2(2-ν)/Et)

22- An open thin steel pipe (E=200 GN/m2, ν=0.3) with an internal diameter of 10 cm and thickness of 1 mm is subject to an internal fluid pressure of 3 N/mm2 and a total compressive load of 15.7 kN. Determine the principal stresses on the tube, the maximum shear stress and the change in diameter. (Ans.:σ1 = 150 MN/m2 , σ2 = -50 MN/m2, τmax = 100 MN/m2)

23- An open thin tube of thickness t and diameter d of a given material (E1, ν, α1) is lined with a second tube of identical thickness but made of a different material (E2, ν, α2). (Assume that the diameters of the tubes are identical.) Determine the principal stresses in each tube when: (a) An internal pressure p is applied

(b) An increment of temperature ∆T is applied. (Assume that α1>α2.)

24- A block of elastic material is placed between

two horizontal rigid surfaces and stressed along two orthogonal horizontal directions as shown. (a) Determine the stress in the vertical direction σz in terms of σx, σy and the elastic constants of the material. (Ans.:σz = ν(σx + σy )) MN/m2) (b) Determine the stress-strain relationships between σx, σy, τxy and εx, εy, γxy. (Ans.: τxy= Gγxy,

σν ε νε

ν νxx yE=

− +

+ −

( )( )( )11 1 2

, σν ε νε

ν νyy xE=

− +

+ −

( )( )( )11 1 2

)

p

σy

σx

σx

σy

τxy

σz

σz

2r

p

d



Advanced Beam Theory

25- For the homogenous sections shown, determine: (a) the positions of the centroids; (b) the second moment of area with respect to horizontal and vertical axes passing through the centroids;

26- A tensile load of 10 kN and a vertical load of 2 kN are applied on the free end of a 10 m long cantilevered beam as shown in the figure. The beam is made of wood with a Young’s modulus of 10 GN/m2 and has a rectangular cross section of 0.6 x 0.3 m2. (a) Determine the direct and shear stresses at any point on the section at the built-in end of the beam. (b) Determine the vertical and longitudinal displacements at the corner point where the tensile load is applied.

27- A hollow circular beam with external and internal radii of 0.2 m and 0.1 m respectively has three point loads of 1 kN each acting on its free end as shown in the figure. (a) Determine the direct and shear stresses on any section of the beam. (b) Draw the Mohr’s circle of stress at point A and find the maximum shear stress at this point.

50 5

40 40

All dimensions in mm

100 60 250

100

3

2 kN

10 kN

1 kN

1 kN

1 kN

A


28- An open steel pipe has a length of 1 m, radius of 100 mm and thickness of 1 mm. The pipe is under the action of a torsional moment of 2 kNm and a tensile load of 100 kN acting at the open ends. Additionally, a uniform compressive pressure of 1N/mm2 is applied on the outer surface of the pipe. (a) Determine the direct and shear stresses along the axial and circumferential directions of the pipe. (b) Draw the resulting Mohr circle of stress. (c) Determine the principal stresses and the maximum shear stress. (d) Taking a Young’s modulus of 200 GN/m2 and a Poisson’s ratio of 0.3, find the changes in length and diameter of the pipe.

100 kN 100 kN

2 kNm 2 kNm

1 m

1 N/mm2

Documents

EG-120