*Todays Objectives:Students will be able to:explain the concepts of normal, shear, bearing and thermal stress.Topics:

Stress Normal Stress Shear StressStrainNormal strain Failure theories

CHAPTER 1 : WORKING STRESSES (PART I)

ApplicationThis steel tie rod is used to suspend a portion of a staircase,and as a result, it is subjected totensile stress.The pin on this tractor is subjected to double shear.

Concepts of Stress DEFINITION : Stress is the internal resistance offered by a unit area of the material from which a member is made to an externally applied load. FORMULA :

UNIT (SI) : N/mm2 or MPa N/m2 or Pa

Type of Stress Normal Stress : stress which acts perpendicular, or normal to, the () cross section of the load-carrying member. : can be either compressive or tensile. Shear Stress : stress which acts tangent to the cross section of () the load-carrying member. : refers to a cutting-like action.

General State of Stress Figure shows the state of stress acting around a chosen point in a bodyStress components are defined for the planes cut parallel to the x, y and z axes. For equilibrium, equal and opposite stresses are exerted on the hidden planes.

Examples of Axially Loaded BarUsually long and slender structural membersTypical examples : truss members, hangers, bolts

Assumptions :Uniform deformation: Bar remains straight before and after load is applied, and cross section remains flat or plane during deformationIn order for uniform deformation, force P be applied along centroidal axis of cross section C

Average Normal Stress Distribution = average normal stress at any point on cross sectional areaP = internal resultant normal forceA = x-sectional area of the bar

Procedure of Analysis Use equation of = P/A for x-sectional area of a member when section subjected to internal resultant force P

Internal Loading Section member perpendicular to its longitudinal axis at pt where normal stress is to be determined Draw free-body diagram Use equation of force equilibrium to obtain internal axial force P at the section Determine members x-sectional area at the section Compute average normal stress = P/AAverage Normal Stress

ExampleGiven:Bar ABCD with a constant width of 35 mm and a thickness of 10 mm.Find:The maximum average normal stress in the bar when it is subjected to the loading as shown.

SolutionInternal loadingNormal force diagramBy inspection, largest loading area is BC, where PBC = 30 kN

Maximum average normal stress

Average Shear StressForces P and P are applied transversely to the member AB.The corresponding average shear stress is,The resultant of the internal shear force distribution is defined as the shear of the section and is equal to the load P.Corresponding internal forces act in the plane of section C and are called shearing forces.

EquilibriumFor the equilibrium condition, force and moment equilibrium requires the shear stress acting on the top face of the element, to be accompanied by shear stress action on the three other faces.

Example of Shear StressSingle ShearDouble Shear

Procedure of AnalysisThe equation avg=V/A is used to compute only the average shearstress in the material. Internal Shear Section member at the point where the avg is to be determined Draw free-body diagram Calculate the internal shear force VAverage Shear Stress Determine sectioned area A Compute average shear stress avg = V/A

ExampleGiven:The bar with a square cross section for which the depth and thickness are 40 mm. An axial force of 800 N is applied along the centroidal axis of the bars cross-sectional area.Find:The average normal stress and average shear stress acting on the material along (a) section plane a-a and (b) section plane b-b

SolutionPart (a)Internal loadingBased on free-body diagram, resultant loading of axial force, P = 800 NAverage stressAverage normal stress, No shear stress on section, since shear force at section is zero.

Part (b)Internal loading

Or directly using x, y axes,Average normal stressAverage shear stress

Bearing Stress on FloorA floor on which a leg of a heavymachine is set, tending to causean indentation in the floor (calleda bearing load)DEFINITION : Bearing stress is normal stress produced by the compression of one surface against another.

*Bearing Stress in ConnectionsCorresponding average force intensity is called the bearing stress,Bolts, rivets, and pins create stresses on the points of contact or bearing surfaces of the members they connect.The resultant of the force distribution on the surface is equal and opposite to the force exerted on the pin.

*Normal Strain DEFINITION : the elongation or contraction of a line segment per unit of length Consider line AB in figure below After deformation, s changes to s FORMULA (average normal strain) : If normal strain is known, final length of a short line segment in direction of n after deformation.

Hence, when is positive, initial line will elongate, if is negative, the line contractsunit : dimensionless=total deformation

original length

Example of Normal Strain

Shear Strain DEFINITION : the change in angle that occurs between two line segments that were originally perpendicular to one another This angle is denoted by (gamma) and measured in radians (rad). Consider line segments AB and AC originating from same point A in a body, and directed along the perpendicular n and t axes After deformation, lines become curves, such that angle between them at A is FORMULA : If is smaller than /2, shear strain is positive, otherwise, shear strain is negative2.2 Strain

Factor of SafetyStructural members or machines must be designed such that the working stresses are less than the ultimate strength of the material.Factor of safety considerations:uncertainty in material properties uncertainty of loadingsuncertainty of analysesnumber of loading cyclestypes of failuremaintenance requirements and deterioration effectsimportance of member to integrity of whole structurerisk to life and propertyinfluence on machine functionIn all the equations, F.S. is chosen to be greater than 1, to avoid potential for failure.Specific values will depend on types of material used and its intended purpose

Offset Method to Determine Yield StrengthNormally, a 0.2 % strain is chosen.2. From this point on the axis, a line parallel to initial straight-line portion of stress-strain diagram is drawn.3. The point where this line intersects the curve defines the yield strength.

Theory Most engineering materials exhibit a linear relationship between stress and strain within the elastic region Discovered by Robert Hooke in 1676 using springs, known as Hookes lawE : modulus of elasticity or Youngs modulus [Unit : Pa] Modulus of elasticity is a mechanical property that indicates the stiffness of a material Materials that are very stiff, such as steel [Est = 200 GPa] have large E values, while spongy materials such as vulcanized rubber [Er = 0.70 MPa] have low valuesIMPORTANTModulus of elasticity E, can be used only if a material has linear-elastic behavior.2) Also, if stress in material is greater than the proportional limit, the stress-strain diagram ceases to be a straight line and the equation is not valid

Stress-Analysis is performed on a component to determineThe required size or geometry (design) an allowable load (service) cause of failure (forensic)For all of these, a limit stress or allowable stress value for the component material is required.Hence, a Failure-Theory is needed to define the onset or criterion of failure

Occurs if a component can no longer function as intended.

Failure Modes:

yielding: a process of global permanent plastic deformation. Change in the geometry of the object.

low stiffness: excessive elastic deflection.

fracture: a process in which cracks grow to the extent that the component breaks apart.

buckling: the loss of stable equilibrium. Compressive loading can lead to bucking in columns.

The failure of a statically loaded member in uni-axial tension or compression is relatively easy to predict. One can simply compare the stress incurred with the strength of the material. When the loading conditions are Complex (i.e. biaxial loading, sheer stresses) then we must use some method to compare multiple stresses to a single strength value. These methods are known failure theories

NEED FOR FAILURE THEORIES

To design structural components and calculate margin of safety.To guide in materials development.To determine weak and strong directions.

MAXIMUM NORMAL STRESS THEORYthis theory postulates, that failure will occur in the structural component if the maximum normal stress in that component reaches the ultimate strength, u obtained from the tensile test of a specimen of the same material. Thus, the structural component will be safe as long as the absolute values of the principle stresses 1 and 2 are both less than u:1 = U and 2 = U

This theory deals with brittle materials only. The maximum normal stress theory can be expresses graphically as shown in the figure. If the point obtained by plotting the values 1 and 2 of the principle stress fall within the square area shown in the figure, the structural component is safe.If it falls outside that area, the component will fail.

1 2 uu-u-u

MAXIMUM SHEARING STRESS THEORYThis theory is based on the observation that yield in ductile materials is caused by slippage of the material along oblique surfaces and is due primarily to shearing stress.A given structural component is safe as long as the maximum value max of the shearing stress in that component remains smaller than the corresponding value of the shearing stress in a tensile test specimen of the same material as the specimen starts to yield.For a 3D complex stress system, the max shear stress is given by: max = (1-2)On the other hand, in the 1D stress system as obtained in the tensile test, at the yield limit, 1= Y and 2=0, therefore:-max= Y

MAXIMUM SHEARING STRESS THEORY (CONT.) Thus,

Graphically, the maximum shear stress criterion requires that the two principal stresses be within the green zone as shown in the figure.

MAXIMUM DISTORTION ENERGY THEORYThis theory is based on the determination of the distortion energy in a given material, i.e. of the energy associated with changes in shape in that material (as opposed to the energy associated with changes in volume in the same material).A given structural component is safe as long as the maximum value of the distortion energy per unit volume in that material remains smaller than the distortion energy per unit volume required to cause yield in a tensile test specimen of the same material. The distortion energy per unit volume in an isotropic material under plane stress is:

MAXIMUM DISTORTION ENERGY THEORY (CONT)In the particular case of a tensile test specimen that is starting to yield, we have:-

This equation represents a principal stress ellipse as illustrated in the figureVon Mises criterion also gives a reasonable estimation of fatigue failure, especially in cases of repeated tensile and tensile-shear loading

PROBLEM 1The solid shaft shown in Figure has a radius of 0.5 cm and is made of steel having a yield stress of 360 MPa. Determine if the loadings cause the shaft to fail according to Tresca and von mises theories.

SOLUTIONCalculating the stresses caused by axial force and torque

The Principal stresses

SOLUTION ( CONTD..)Applying Maximum Shear stress theory

So shear failure occursApplying Maximum distortion theory

No Failure

*