13296872 chapter 2design against static load

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CHAPTER 2CHAPTER 2DESIGN AGAINST DESIGN AGAINST

STATIC LOADSTATIC LOAD

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INTRODUCTIONINTRODUCTION What is strength?What is strength? Strength is a property or characteristic of Strength is a property or characteristic of

mechanical element.mechanical element. What is Static Load?What is Static Load? Stationary force applied to a member. To be Stationary force applied to a member. To be

stationary the force should be unchanging in stationary the force should be unchanging in magnitude and directions.magnitude and directions.

A static load can produce axial tension or A static load can produce axial tension or compression, a shear load, a bending load, a compression, a shear load, a bending load, a torsional load or any combination of these.torsional load or any combination of these.

Purpose: relationship between strength and Purpose: relationship between strength and static loading in order to make decisions static loading in order to make decisions concerning material and its treatment, concerning material and its treatment, fabrication, geometry, safety, reliability, fabrication, geometry, safety, reliability, usability, manufacturability, etc.usability, manufacturability, etc.

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CONTENTCONTENT Modes of FailureModes of Failure Factor of SafetyFactor of Safety Representing Stress on a Stress ElementRepresenting Stress on a Stress Element Direct Stresses: Tension and CompressionDirect Stresses: Tension and Compression Deformation Under Direct Axial LoadingDeformation Under Direct Axial Loading Direct Shear StressDirect Shear Stress Torsional Shear StressTorsional Shear Stress Torsional DeformationTorsional Deformation Torsion in Members Having Noncircular Cross Torsion in Members Having Noncircular Cross

SectionsSections

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Mode of FailuresMode of Failures Ductile material is one which has Ductile material is one which has

relatively large tensile strain before relatively large tensile strain before fracture takes place. For example, fracture takes place. For example, steel and aluminum.steel and aluminum.

Brittle material has a relatively small Brittle material has a relatively small tensile strain before fracture. For tensile strain before fracture. For example, cast iron.example, cast iron.

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Factor of SafetyFactor of Safety Factor of safety (FS) is defined by Factor of safety (FS) is defined by

either of the equation;either of the equation;

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REPRESENTING STRESS REPRESENTING STRESS ON A STRESS ELEMENTON A STRESS ELEMENT

PositivePositive shear stresses tend to rotate the shear stresses tend to rotate the element in a clockwise directionelement in a clockwise direction

Negative Negative shear stresses tend to rotate shear stresses tend to rotate the element in a counterclockwise the element in a counterclockwise directiondirection

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DIRECT STRESSES: DIRECT STRESSES: TENSION AND TENSION AND

COMPRESSION COMPRESSION Stress: Internal resistance offered by a Stress: Internal resistance offered by a

unit area of material to an external loadunit area of material to an external load Perpendicular to elementPerpendicular to element Compressive stresses: Crushing action. Compressive stresses: Crushing action.

Negative by convention.Negative by convention. Tensile: Pulling action. Positive by Tensile: Pulling action. Positive by

convention.convention.

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DIRECT STRESSES: DIRECT STRESSES: TENSION AND TENSION AND

COMPRESSION COMPRESSION

Conditions:Conditions:• load-carry member must be straightload-carry member must be straight• line of action of the load must pass line of action of the load must pass

through the centroid of cross section of through the centroid of cross section of the memberthe member

• member must be of uniform cross sectionmember must be of uniform cross section• member must be short in the case of member must be short in the case of

compression memberscompression members

2, mNAF

areaforcestressnormal See exa

mple 3-1

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DEFORMATION UNDER DEFORMATION UNDER DIRECT AXIAL LOADING DIRECT AXIAL LOADING

Where:Where: = = total deformation of the member total deformation of the member carrying the axialcarrying the axialFF = direct axial load= direct axial loadLL = original load length of the member= original load length of the memberEE = modulus of elasticity of the material= modulus of elasticity of the materialAA= cross-sectional area of the member= cross-sectional area of the member = = direct/normal stressdirect/normal stress

mEL

EAFL See

example 3-2

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DIRECT SHEAR STRESS DIRECT SHEAR STRESS Occurs when the applied force tends to cut Occurs when the applied force tends to cut

through the member as scissors. Ex: tendency through the member as scissors. Ex: tendency for a key to be sheared off at the section for a key to be sheared off at the section between the shaft and the hub of a machine between the shaft and the hub of a machine element when transmitting torque (see next element when transmitting torque (see next slide).slide).

Apply force is assumed to be Apply force is assumed to be uniformlyuniformly distributed across the cross section.distributed across the cross section.

2mNAF

shearinareaforceshearing

s

See example 3-3

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DIRECT SHEAR STRESS DIRECT SHEAR STRESS

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TORSIONAL SHEAR TORSIONAL SHEAR STRESS STRESS

2max mNZT

JTc

p

See example 3-6

A torque will twist a member, causing a A torque will twist a member, causing a shear stress in the membershear stress in the member

A general shear stress formula:A general shear stress formula:

JTr

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TORSIONAL SHEAR TORSIONAL SHEAR STRESS STRESS

Where:Where:TT = torque= torquecc = = radius of shaft to its outside surfaceradius of shaft to its outside surfaceJJ = polar moment of inertia (= polar moment of inertia (Appendix 1Appendix 1))rr = radial distance from the center of the = radial distance from the center of the shaft to the point of interestshaft to the point of interestZZpp = polar section modulus (= polar section modulus (Appendix 1Appendix 1))

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TORSIONAL SHEAR TORSIONAL SHEAR STRESS STRESS

The distribution of stress is not uniform The distribution of stress is not uniform across the cross sectionacross the cross section

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TORSIONAL TORSIONAL DEFORMATION DEFORMATION

degGJTL

Where:Where:TT = torque= torqueLL = = length of the shaft over which the angle length of the shaft over which the angle of twist is being computedof twist is being computedGG= modulus of elasticity of the shaft = modulus of elasticity of the shaft material in shearmaterial in shearJJ = polar moment of inertia (= polar moment of inertia (Appendix 1Appendix 1))

See example 3-7

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TORSION IN MEMBERS TORSION IN MEMBERS HAVING NONCIRCULAR HAVING NONCIRCULAR

CROSS-SECTIONS CROSS-SECTIONS

radGKTL

mNQT

2max

See example 3-8

GJTL

ZT

JTc

p

max

CIRCULAR

SECTION

NONCIRCULAR SECTION

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TORSION IN TORSION IN MEMBERS MEMBERS

HAVING HAVING NONCIRCULANONCIRCULA

R CROSS-R CROSS-SECTIONSSECTIONS

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VERTICAL SHEAR STRESS VERTICAL SHEAR STRESS Beam carrying transverse loads experience Beam carrying transverse loads experience

shearing forces (shearing forces (VV) which cause shearing stress:) which cause shearing stress:

32 ; myAQmNItVQ

p See example 3-10

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VERTICAL SHEAR STRESS VERTICAL SHEAR STRESS

Where:Where:VV= shearing force= shearing forceQQ= = first momentfirst momentII = moment of inertia= moment of inertiatt = thickness of the section= thickness of the sectionAApp = area of the section above the place = area of the section above the place where the shearing force is to be computedwhere the shearing force is to be computedyy = distance from the neutral axis of the = distance from the neutral axis of the section to the centroid of the area section to the centroid of the area AApp

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VERTICAL SHEAR STRESS VERTICAL SHEAR STRESS In the analysis of beams, it is usual to In the analysis of beams, it is usual to

compute the variation in shearing force compute the variation in shearing force across the entire length of the beam and across the entire length of the beam and to draw the shearing force diagram.to draw the shearing force diagram.

Vertical shear stress = Horizontal shear Vertical shear stress = Horizontal shear stress, because any element of material stress, because any element of material subjected to a shear stress on one face subjected to a shear stress on one face must have a shear stress of the same must have a shear stress of the same magnitude on the adjacent face for the magnitude on the adjacent face for the element to be in equilibrium.element to be in equilibrium.

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STRESS DUE TO BENDING STRESS DUE TO BENDING A beam is a member that carries load A beam is a member that carries load

transverse to its axis. Such loads produce transverse to its axis. Such loads produce bending moments in the beam, which result in bending moments in the beam, which result in the development of bending stress.the development of bending stress.

Bending stress are normal stresses, that is, Bending stress are normal stresses, that is, either tensile or compressive.either tensile or compressive.

The maximum bending stress in a beam cross The maximum bending stress in a beam cross section will occur in the part farthest from the section will occur in the part farthest from the neutral axis of the section. At that point, the neutral axis of the section. At that point, the flexure formula gives the stress:flexure formula gives the stress: 2; mN

IMcformulaflexure

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STRESS DUE TO BENDING STRESS DUE TO BENDING

Where:Where:MM = magnitude of bending moment at = magnitude of bending moment at the sectionthe sectioncc = = distance from the neutral axis to the distance from the neutral axis to the outermost fiber of the beam cross sectionoutermost fiber of the beam cross sectionII = moment of inertia= moment of inertia

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STRESS DUE TO BENDING STRESS DUE TO BENDING The flexure formula was developed subject to the following conditions:The flexure formula was developed subject to the following conditions:

- Beam must pure bending. No shearing stress and axial loads.- Beam must pure bending. No shearing stress and axial loads.- Beam must not twist or be subjected to torsional load.- Beam must not twist or be subjected to torsional load.- Material of beam must obey Hooke’s law- Material of beam must obey Hooke’s law- Modulus of elasticity of the material must be the same in - Modulus of elasticity of the material must be the same in both tension and compression.both tension and compression.- Beam is initially straight and has constant cross section.- Beam is initially straight and has constant cross section.- No part of the beam shape fails because of buckling or - No part of the beam shape fails because of buckling or wrinkling.wrinkling.

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STRESS DUE TO BENDING STRESS DUE TO BENDING

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STRESS DUE TO BENDING STRESS DUE TO BENDING

][ 3mcIS

][ 2mNSM

For design, it is convenient to define the For design, it is convenient to define the term section modulus, term section modulus, SS::

The The flexure formulaflexure formula then becomes: then becomes:

Then, in design, it is usual to define a design Then, in design, it is usual to define a design stress, stress, d d ,and with the bending moment ,and with the bending moment known, then:known, then: ][ 3mMS d See

example 3-12

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FLEXURAL CENTER FOR FLEXURAL CENTER FOR BEAMSBEAMS

To ensure symmetrical bending i.e. no To ensure symmetrical bending i.e. no tendency to twist under loading, action of tendency to twist under loading, action of load pass through the line of symmetry:load pass through the line of symmetry:

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FLEXURAL CENTER FOR FLEXURAL CENTER FOR BEAMSBEAMS

If there is no vertical axis symmetry:If there is no vertical axis symmetry:

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BEAMS WITH BEAMS WITH CONCENTRATED BENDING CONCENTRATED BENDING

MOMENTSMOMENTS Beams with concentrated forces or Beams with concentrated forces or

distributed load, moment diagrams are distributed load, moment diagrams are continuouscontinuous

Machine elements that carry loads whose line Machine elements that carry loads whose line of action is offset from centroidal axis of of action is offset from centroidal axis of beams, a concentrated moment is exertedbeams, a concentrated moment is exerted

Examples: cranks, levers, helical gears, etc.Examples: cranks, levers, helical gears, etc.

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BEAMS WITH BEAMS WITH CONCENTRATED BENDING CONCENTRATED BENDING

MOMENTSMOMENTS

Bending on a Bending on a shaft carrying shaft carrying a cranka crank

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COMBINED NORMAL COMBINED NORMAL STRESSES: STRESSES:

SUPERPOSITION SUPERPOSITION PRINCIPLEPRINCIPLE

When the same cross section of a load-When the same cross section of a load-carrying member is subjected to both a carrying member is subjected to both a direct tensile and compressive stress and a direct tensile and compressive stress and a stress due to bending, the resulting normal stress due to bending, the resulting normal stress can be computed by the method of stress can be computed by the method of superposition:superposition: 2mN

AF

IMc

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STRESS STRESS CONCENTRATIONS CONCENTRATIONS

FACTORSFACTORS From figure below, the highest stress From figure below, the highest stress

occurs in the filletoccurs in the fillet

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CONCLUSION CONCLUSION

DIRECT STRESS: TENSION &

COMPRESSION SHEAR STRESSSHEAR STRESS

DIRECT SHEAR STRESS

TORSIONAL SHEAR STRESS

VERTICAL SHEARING STRESS

STRESS DUE TO BENDING

SUPERPOSITION PRINCIPLE

NORMAL STRESSNORMAL STRESS

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THANK YOUTHANK YOU