Contents ME 416

ME-416

STRESS ANALYSIS

Instructor: Dr. Khalid Rahman

Room No. G11 (FME Faculty Lobby)

Tel: 2351

Email: [email protected]

Teaching Assistant: Engr. Malik Abdul Wahab

◦ Engr. Umer

Text book:

•“Advanced Mechanics of Materials” 6th Edition by Arthur P. Boresi and Richard J. Schmidt

Reference book:

“Advanced Strength and Applied Stress Analysis” 2nd Edition by Richard G. Budynas

Course Material

COURSE LEARNING OUTCOME (CLOs)

CLO NO

CLO STATEMENT Upon Completion of the course student should be able to:

PLO Bloom Taxonomy

CLO1 Demonstrate the understanding of stress and strain from

the simple problems of state of stress at a point PLO2 C2

CLO2 Demonstrate the ability to apply the criterion for failure

in the design of structural members. PLO3 C3

CLO3 Analyze stresses on unsymmetrical beams, non-circular

shaft, thick wall cylinders and rotating disk. PLO2 C4

CLO4

Demonstrate the understating of advanced topics in

Stress Analysis (Stress Concentrations, Fracture

Mechanics, Fatigue).

PLO2 C3

Grading Policy Assignment 10%

Quizzes (announced) 10%

Project Assignment 5%

Midterm Exam 30%

Final 45%

No make-up quiz, assignment or exam

Attendance requirement as per institute policy

Course Contents

• Theories of Stress Strain

• Elastic and Inelastic Material Behavior

• Torsion

• Bending

• Thick walled Cylinders and Rotating Discs

• Failure criteria

• Advanced Topics in Stress Analysis

– Stress Concentration

– Fracture Mechanics

– Fatigue

Stress Analysis (ME-416) 6

Course Break down Lecture no Topics

1 Introduction

2 Introduction

3 Stress at a point 2D & 3D stress Tensors

4 Symmetry of stress components Stress acting on arbitrary plane Normal and shear stresses on oblique plane

5 Transformation of stresses Principal Stresses Principal values and directions

6 Octahedral stresses Mean and Deviator stress Plane stress

7 Mohr’s Circle (2D) Mohr’s Circle (3D)

8 Differential equation of motion of deformable body Other Coordinates

Lecture no Topics

9 Deformation of deformable body Strain theory (line element)

10 Relation between two lines (shear strain) Principal strain Plane strain

11 Small displacement theory

12 Linear stress strain temperature relations



15 Inelastic material behavior Uniaxial Stress strain Non linear material response

16 Yield Criteria Maximum Principal stress criterion Strain energy density criterion

17 Yield of ductile Metals Maximum shear stress (Tresca) criterion Distortional Energy density (Von Mises) criterion Comparison of failure criteria

Lecture no Topics

18 Torsion – Circular shaft Siant-Venant’s Semi inverse method

19 Linear Elastic solution

20 Soap film Analogy

21 Narrow Rectangular Cross section

22 Hollow thin walled torsion – Multiple sections

23 Hollow thin walled torsion members & multiply connected cross sections

24 Bending of straight bars Symmetric bending

25 Non Symmetric Bending

26 Bending stress in beam subjected to non symmetric bending

27 Shear center Shear flow in thin walled beam cross section

Lecture no Topics

28 Shear center for channel section

29 Curved beams

30 Curved Beams

31 Thick walled cylinders



34 Rotating disk

35 Rotating disk

36 Stress concentration

37 Stress Concentration

38 Stress Concentration

39 Fracture Mechanics


Lecture no Topics


42 Fatigue

43 Fatigue

44 Fatigue

45 Review

Office Visit Hours Monday Tuesday Wednesday Thursday

9:30AM ~ 11:00AM 10:00AM~11:30AM 10:00AM~11:30AM 9:00AM~11:30AM

Assignment Session, Project and Quiz Schedule Assignment Session and Quiz at FME QUIZ HALL (Time: 10:00 AM)

Assignment Sessions

1 15th Sept 2015

2 13th Oct 2015

3 3rd Nov 2015

4 17th Nov 2015

5 1st Dec 2015

6 15th Dec 2015

Quiz Schedule

1 8th Sept 2015

2 20th Sept 2015

3 10th Nov 2015

4 8th Dec 2015

Project Assignment due Date 11th Dec 2015

Chapter Problems

2 1,2,4,6,12,17,20,21,33,39,41,42,43,25,59,63,64

3 1,2,11,13,16

4 6,8,9,12,36,37,39,41,43,44

6 4,5,10,16,17,20,23,25,27,32,41,38,42,43,45,46,47,52,56,57

7 2,4,11,12,18,20,28,29,33,34,35,38

8 7,8,12,16,18,19,25

9 1,2,3,5,7,9,10,13

11 3,4,6,9,10,11,12,18,19,20,33,34,36,38

14 1,2,5,7,13,15,18,20

15 1,4,6,10,13,14,15,17,23

16 1,3,5,8,9,12,14,17,19,21,22

Assignments

Concept • The main objective of the study of Stress Analysis is to

provide the future engineer with the means of analyzing

and designing various machines and load bearing

structures.

• Both the analysis and design of a given structure involve the

determination of

•Stresses

•deformations

Introduction Any material or structure may

fail when it is loaded


1 Strength – The structure must be strong enough to carry the applied loads. 2 Stiffness – The structure must be stiff enough such that only allowable deformation occurs. 3 Stability – The structure must not collapse through buckling subjected to the applied compressive loads.

Stress analysis provides analytical, numerical and or experimental methods for determining the strength, stiffness and stability of load-carrying structural members

Examples


Stent expansion process

Artery after stent inflation

Examples


Boston Molasses Tank Failure

Boston Molasses Tank Failure • 15th August 1919 tank fail without warning

• Tank structure – Mild Steel

• 3 year old structure

• 15mm thick plate

• 15.25meter height and 27.5meter diameter

• During disaster 2.3 million gallon (14000 tons) stored molasses

• Molasses wave 10meter with speed at 40~55km/hr

• 30 years of effect as well as smell

• 21 died and 150 injured



• Leakage in the tank (manhole) was ignored and painted brown

• Failure due to sudden temperature change from -17⁰C (previous day) to 4.5⁰C on day of failure

• Investigation – Design was inadequate to withstand pressure and factor of safety was low

Liberty Ship Failure


• Initially ship building – mostly riveted joint

• WW-II mass production

• These ships first to have all welded joint

• 2700 Liberty ships were mass produced (some in 5 days)

• 1000 suffered significant failure

• Some broke suddenly into two because of low temperature


De Havilland Comet Failure


• World’s first commercial jetliner into service in 1952

• Previously propeller planes – low altitude

• During first year of service – 28000 passengers and 104 million miles

• US Civil Aeronautics Administration has refused to grant Comet airworthiness certificate. Reason Stress concentration

• Large size crew escape window




• After 9000 hours of equivalent flying failure in fuselage was found

• Small crack growth in the corner of escape latch window – cause of failure

• Fatigue Failure

• Boeing took lesson from Comet failure

• Major loss for De Havilland


Introduction • In stress analysis, a force can be categorized as

either external or internal – External Force (applied surface loads, force of gravity and support

reactions)

– Internal Force (resisting forces generated within loaded structural elements)

• The moment of a force is a measure of its tendency to cause a body to rotate about a specific point or axis


Point Load Distributed Load Moment about the beam–column connection

Introduction


Types of Forces 1. normal force, F, which is perpendicular to the cross-section; 2. shear force, V, which is parallel to the cross-section; 3. bending moment, M, which bends the material; and 4. twisting moment (torque), T, which twists the material about its

central axis.

Introduction

Equilibrium system 1. the resultant of all applied forces, including support

reactions, must be zero;

2. the resultant of all applied moments, including bending and twisting moments, must be zero.

The two equilibrium conditions are commonly used to determine support reactions and internal forces on cross-sections of structural members.


Stresses in the Members of a Structure

• Can the structure safely support the 30 kN load?

• The force per unit area, or intensity of the internal forces distributed over a given section, is called the stress on that section


Centric & Eccentric Loading • A uniform distribution of stress in a section

infers that the line of action for the resultant of the internal forces passes through the centroid of the section.

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

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

• The stress distributions in eccentrically loaded members cannot be uniform or symmetric


Shearing Stress • Forces 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. • The resultant of the internal shear force distribution is

defined as the shear of the section and is equal to the load P.

• The corresponding average shear stress is,

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

• The shear stress distribution cannot be assumed to be uniform.


Shearing Stress Examples


Stress in Two Force Members • Axial forces on a two force member

result in only normal stresses on a plane cut perpendicular to the member axis.

• Transverse forces on bolts and pins result in only shear stresses on the plane perpendicular to bolt or pin axis.

• Either axial or transverse forces may produce both normal and shear stresses with respect to a plane other than one cut perpendicular to the member axis.


Stress on an Oblique Plane • Normal and shearing stresses on an oblique

plane

• The maximum normal stress occurs when the reference plane is perpendicular to the member axis,

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


State of Stress • It follows that only 6 components of

stress are required to define the complete state of stress

• at a given point, shear cannot take place in one plane only; an equal shearing stress must be exerted on another plane perpendicular to the first one.


Generalized Hooke’s Law

L3-4 - 41

• For an element subjected to multi-axial

loading, the normal strain components

resulting 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

EEE

EEE

EEE

zyxz

zyxy

zyxx

• With these restrictions:

SHEARING STRAIN


• Plot of Shear stress vs. shear strain is

similar to normal stress vs. normal strain

except that the strength values are

approximately half.

zxzxyzyzxyxy GGG

Deformations of Members Under Axial Loading

• From Hooke’s Law:

• From the definition of strain:

• Equating and solving for the deformation,

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


6 - 44

Torsional Loads on Circular Shafts

• Interested in stresses and strains of

circular shafts subjected to twisting

couples or torques

• Generator creates an equal and

opposite torque T’

• Shaft transmits the torque to the

generator

• Turbine exerts torque T on the shaft

6 - 45

Stresses in Elastic Range

Jc

dAc

dAT max2max

• Recall that the sum of the moments from

the internal stress distribution is equal to

the torque on the shaft at the section,

421 cJ

41

422

1 ccJ

and maxJ

T

J

Tc

• The results are known as the elastic torsion

formulas,

• Multiplying the previous equation by the

shear modulus,

max

Gc

G

max

c

From Hooke’s Law, G , so

The shearing stress varies linearly with the radial position in the section.

6 - 46

Torsional Failure Modes

• Ductile materials generally fail in shear.

Brittle materials are weaker in tension

than shear.

• When subjected to torsion, a ductile

specimen breaks along a plane of

maximum shear, i.e., a plane

perpendicular to the shaft axis.

• When subjected to torsion, a brittle

specimen breaks along planes

perpendicular to the direction in which

tension is a maximum, i.e., along

surfaces at 45o to the shaft axis.

7 - 47

Pure Bending

Pure Bending: Prismatic members

subjected to equal and opposite couples

acting in the same longitudinal plane

7 - 48

Other Loading Types

• Principle of Superposition: The normal

stress due to pure bending may be

combined with the normal stress due to

axial loading and shear stress due to shear

loading to find the complete state of stress.

• 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 produces

internal forces equivalent to a shear force

and a couple

Elastic and Inelastic Response of a Solid

Modes of Failure

1. Failure by excessive deflection a) Elastic deflection b) Deflection caused by creep

2. Failure by general yielding 3. Failure by fracture

a) Sudden fracture of brittle materials b) Fracture of cracked or flawed members c) Progressive fracture (fatigue)

4. Failure by instability

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Contents ME 416