Introduction to Elasticity Part 9 pptx

In general, a suitable stress function can be obtained from any two analytic functions Stresses around an elliptical hole In a development very important to the theory of fracture, Ingli

i.e any function of the form xψ, where ψ is harmonic, satisfies Eqn 12, and many thus be used

as a stress function Similarly, it can be shown that yψ and (x2+y2)ψ = r2ψ are also suitable, as

is ψ itself In general, a suitable stress function can be obtained from any two analytic functions

Stresses around an elliptical hole

In a development very important to the theory of fracture, Inglis5 used complex potential tions to extend Kirsch’s work to treat the stress field around a plate containing an ellipticalrather than circular hole This permits crack-like geometries to be treated by making the minoraxis of the ellipse small It is convenient to work in elliptical α, β coordinates, as shown in Fig 4,defined as

func-x = c cosh α cos β, y = c sinh α sin β (27)

5 C.E Inglis, “Stresses in a Plate Due to the Presence of Cracks and Sharp Corners,” Transactions of the Institution of Naval Architects, Vol 55, London, 1913, pp 219–230.

Figure 4: Elliptical coordinates.

where c is a constant If β is eliminated this is seen in turn to be equivalent to

x2cosh2α +

y2sinh2α = c

4 Re ψ0(z) = σ2[zψ00(z) + χ00(z)] = σ

)

These boundary conditions can be satisfied by potential functions in the forms

4ψ(z) = Ac cosh ζ + Bc sinh ζ4χ(z) = Cc2ζ + Dc2 cosh 2ζ + Ec2 sinh 2ζwhere A, B, C, D, E are constants to be determined from the boundary conditions When this

is done the complex potentials are given as

4ψ(z) = σc[(1 + e2α0) sinh ζ − e2α0cosh ζ]

The stresses σx, σy, and τxy can be obtained by using these in Eqns 26 However, the amount

of labor in carrying out these substitutions isn’t to be sneezed at, and before computers weregenerally available the Inglis solution was of somewhat limited use in probing the nature of thestress field near crack tips

Figure 5: Stress field in the vicinity of an elliptical hole, with uniaxial stress applied in direction (a) Contours of σy, (b) Contours of σx

y-Figure 5 shows stress contours computed by Cook and Gordon6 from the Inglis equations.

A strong stress concentration of the stress σy is noted at the periphery of the hole, as would

be expected The horizontal stress σx goes to zero at this same position, as it must to isfy the boundary conditions there Note however that σx exhibits a mild stress concentration(one fifth of that for σy, it turns out) a little distance away from the hole If the material hasplanes of weakness along the y direction, for instance as between the fibrils in wood or manyother biological structures, the stress σx could cause a split to open up in the y direction justahead of the main crack This would act to blunt and arrest the crack, and thus impart a mea-sure of toughness to the material This effect is sometimes called the Cook-Gordon tougheningmechanism

sat-The mathematics of the Inglis solution are simpler at the surface of the elliptical hole, sincehere the normal component σα must vanish The tangential stress component can then becomputed directly:

(σβ)α=α 0 = σe2α0

"

sinh 2α0(1 + e−2α0)cosh 2α0− cos 2β − 1



(33)This can also be written in terms of the radius of curvature ρ at the tip of the major axis as



(34)

6 J.E Gordon, The Science of Structures and Materials, Scientific American Library, New York, 1988.

This result is immediately useful: it is clear that large cracks are worse than small ones (thelocal stress increases with crack size a), and it is also obvious that sharp voids (decreasing ρ)are worse than rounded ones Note also that the stress σy increases without limit as the crackbecomes sharper (ρ → 0), so the concept of a stress concentration factor becomes difficult touse for very sharp cracks When the major and minor axes of the ellipse are the same (b = a),the result becomes identical to that of the circular hole outlined earlier.

Stresses near a sharp crack

Figure 6: Sharp crack in an infinite sheet

The Inglis solution is difficult to apply, especially as the crack becomes sharp A moretractable and now more widely used approach was developed by Westergaard7, which treats asharp crack of length 2a in a thin but infinitely wide sheet (see Fig 6) The stresses that actperpendicularly to the crack free surfaces (the crack “flanks”) must be zero, while at distancesfar from the crack they must approach the far-field imposed stresses Consider a harmonicfunction φ(z), with first and second derivatives φ0(z) and φ00(z), and first and second integralsφ(z) and φ(z) Westergaard constructed a stress function as

It can be shown directly that the stresses derived from this function satisfy the equilibrium,compatibility, and constitutive relations The function φ(z) needed here is a harmonic functionsuch that the stresses approach the far-field value of σ at infinity, but are zero at the crack flanksexcept at the crack tip where the stress becomes unbounded:

This gives the needed singularity for z = ±a, and the other boundary conditions can be verifieddirectly as well The stresses are now found by suitable differentiations of the stress function;for instance

where K = σ√

πa is the stress intensity factor, with units of Nm−3/2 or psi√

in (The factor πseems redundant here since it appears to the same power in both the numerator and denominator,but it is usually included as written here for agreement with the older literature.) We will see

in the Module on Fracture that the stress intensity factor is a commonly used measure of thedriving force for crack propagation, and thus underlies much of modern fracture mechanics Thedependency of the stress on distance from the crack is singular, with a 1/√

2 Consider a thick-walled pressure vessel of inner radius ri and outer radius ro, subjected

to an internal pressure pi and an external pressure po Assume a trial solution for theradial displacement of the form u(r) = Ar + B/r; this relation can be shown to satisfythe governing equations for equilibrium, strain-displacement, and stress-strain governingequations

(a) Evaluate the constants A and B using the boundary conditions

σr =−pi @ r = ri, σr=−po @ r = ro(b) Then show that

Prob 2

5 Show that stress functions in the form of quadratic or cubic polynomials (φ = a2x2+

b2xy + c2y2 and φ = a3x3 + b3x2y + c3xy2 + d3y3) automatically satisfy the governingrelation ∇4φ = 0

6 Write the stresses σx, σy, τxy corresponding to the quadratic and cubic stress functions ofthe previous problem

7 Choose the constants in the quadratic stress function of the previous two problems so as

to represent (a) simple tension, (b) biaxial tension, and (c) pure shear of a rectangularplate

The horizontal edges are not loaded, so we also have that τxy = 0 at y = ±h/2.(a) Show that these conditions are satisfied by a stress function of the form

φ = b2xy + d4xy3(b) Evaluate the constants to show that the stresses can be written

σx= P xy

I , σy = 0, τxy =

P2I

"

h2

Experimental Strain Analysis

David RoylanceDepartment of Materials Science and EngineeringMassachusetts Institute of Technology

to them One approach to this impasse is the experimental one, in which we seek to construct

a physical laboratory model that somehow reveals the stresses in a measurable way

It is the nature of forces and stresses that they cannot be measured directly It is the effect of

a force that is measurable: when we weigh an object on a spring scale, we are actually measuringthe stretching of the spring, and then calculating the force from Hooke’s law Experimental stressanalysis, then, is actually experimental strain analysis The difficulty is that strains in the linearelastic regime are almost always small, on the order of 1% or less, and the art in this field is that

of detecting and interpreting small displacements We look for phenomena that exhibit largeand measurable changes due to small and difficult-to-measure displacements There a number

of such techniques, and three of these will be outlined briefly in the sections to follow A gooddeal of methodology has been developed around these and other experimental methods, andboth further reading1 and laboratory practice would be required to put become competent inthis area

1856 by Lord Kelvin:

R = ρLA

1 Manual on Experimental Stress Analysis, Third Edition, Society of Experimental Stress Analysis (now Society

of Experimental Mechanics), 1978.

Figure 1: Wire resistance strain gage.

where here ρ is the material’s resistivity To express the effect of a strain  = dL/L in thewire’s long direction on the electrical resistance, assume a circular wire with A = πr2 and takelogarithms:

ln R = ln ρ + ln L − (ln π + 2 ln r)The total differential of this expression gives

r = dr

r =−νdL

Lthen

dρ

ρ = αR

dVVwhere αR is the constant of proportionality between resistance change and volume change.Writing the volume change in terms of changes in length and area, this becomes



= αR(1− 2ν)dL

LHence

dR/R

This quantity is called the gage factor, GF Constantan, a 45/55 nickel/copper alloy, has αR=1.13 and ν = 0.3, giving GF≈ 2.0 This material also has a low temperature coefficient ofresistivity, which reduces the temperature sensitivity of the strain gage

Figure 2: Wheatstone bridge circuit for strain gages.

A change in resistance of only 2%, which would be generated by a gage with GF = 2 at 1%strain, would not be noticeable on a simple ohmmeter For this reason strain gages are almostalways connected to a Wheatstone-bridge circuit as seen in Fig 2 The circuit can be adjusted

by means of the variable resistance R2 to produce a zero output voltage Vout before strain isapplied to the gage Typically the gage resistance is approximately 350Ω and the excitationvoltage is near 10V When the gage resistance is changed by strain, the bridge is unbalancedand a voltage appears on the output according to the relation

Vout

Vin =

∆R2R0where R0 is the nominal resistance of the four bridge elements The output voltage is easilymeasured because it is a deviation from zero rather than being a relatively small change su-perimposed on a much larger quantity; it can thus be amplified to suit the needs of the dataacquisition system

Temperature compensation can be achieved by making a bridge element on the opposite side

of the bridge from the active gage, say R3, an inactive gage that is placed near the active gagebut not bonded to the specimen Resistance changes in the active gage due to temperature willthen be offset be an equal resistance change in the other arm of the bridge

Figure 3: Cancellation of bending effects

It is often difficult to mount a tensile specimen in the testing machine without inadvertentlyapplying bending in addition to tensile loads If a single gage were applied to the convex-outward side of the specimen, its reading would be erroneously high Similarly, a gage placed onthe concave-inward or compressive-tending side would read low These bending errors can beeliminated by using an active gage on each side of the specimen as shown in Fig 3 and wiringthem on the same side of the Wheatstone bridge, e.g R1 and R4 The tensile component ofbending on one side of the specimen is accompanied by an equal but compressive component onthe other side, and these will generate equal but opposite resistance changes in R1 and R4 Theeffect of bending will therefore cancel, and the gage combination will measure only the tensilestrain (with doubled sensitivity, since both R1 and R4 are active).

Figure 4: Strain rosette

The strain in the gage direction can be found directly from the gage factor (Eqn 1) When thedirection of principal stress is unknown, strain gage rosettes are useful; these employ multiplegages on the same film backing, oriented in different directions The rectangular three-gagerosette shown in Fig 4 uses two gages oriented perpendicularly, and a third gage oriented at

45◦ to the first two.

Example 1

A three-gage rosette gives readings  0 = 150µ,  45 = 200 µ, and  90 = −100 µ (here the µ symbol indicates micrometers per meter) If we align the x and y axis along the 0◦and 90◦gage directions, then

 x and  y are measured directly, since these are  0 and  90 respectively To determine the shear strain

γ xy , we use the rule for strain transformation to write the normal strain at 45◦:

 45 = 200 µ =  x cos245 +  y sin245 + γ xy sin 45 cos 45Substituting the known values for  x and  y, and solving,

γ xy = 350 µ The principal strains can now be found as

 1,2=  x +  y

s

 x − y2

2+

γxy 2

2

= 240 µ, − 190 µ The angle from the x-axis to the principal plane is

tan 2θ p = γ xy /2

( x −  y )/2 → θ p = 27.2◦The stresses can be found from the strains from the material constitutive relations; for instance for steel with E = 205 GPa and ν = 3 the principal stress is

Wire-resistance strain gages are probably the principal device used in experimental stress ysis today, but they have the disadvantage of monitoring strain only at a single location Pho-toelasticity and moire methods, to be outlined in the following sections, are more complicated

anal-in concept and application but have the ability to provide full-field displays of the straanal-in bution The intuitive insight from these displays can be so valuable that it may be unnecessary

distri-to convert them distri-to numerical values, although the conversion can be done if desired

Figure 5: Light propagation

Photoelasticity employs a property of many transparent polymers and inorganic glasses calledbirefringence To explain this phenomenon, recall the definition of refractive index, n, which isthe ratio of the speed of light v in the medium to that in vacuum c:

n = v

2 M Hetenyi, ed., Handbook of Experimental Stress Analysis, Wiley, New York, 1950.

As the light beam travels in space (see Fig 5), its electric field vector E oscillates up and down

at an angular frequency ω in a fixed plane, termed the plane of polarization of the beam (Thewavelength of the light is λ = 2πc/ω.) A birefringent material is one in which the refractiveindex depends on the orientation of plane of polarization, and magnitude of the birefringence isthe difference in indices:

∆n = n⊥− nkwhere n⊥ and nk are the refractive indices on the two planes Those two planes that producethe maximum ∆n are the principal optical planes As shown in Fig 6, a birefringent materialcan be viewed simplistically as a Venetian blind that resolves an arbitrarily oriented electricfield vector into two components, one on each of the two principal optical planes, after whicheach component will transit the material at a different speed according to Eqn 4 The twocomponents will eventually exit the material, again traveling at the same speed but having beenshifted in phase from one another by an amount related to the difference in transit times

Figure 6: Venetian-blind model of birefringence

A photoelastic material is one in which the birefringence depends on the applied stress, andmany such materials can be described to a good approximation by the stress-optical law

where C is the stress-optical coefficient, and the quantity in the second parentheses is the ence between the two principal stresses in the plane normal to the light propagation direction;this is just twice the maximum shear stress in that plane The stress-optical coefficient is gen-erally a function of the wavelength λ

differ-The stress distribution in an irregularly shaped body can be viewed by replicating the actualstructure (probably scaled up or down in size for convenience) in a birefringent material such asepoxy If the structure is statically determinate, the stresses in the model will be the same as that

in the actual structure, in spite of the differences in modulus To make the birefringence effectvisible, the model is placed between crossed polarizers in an apparatus known as a polariscope.(Polarizers such as Polaroid, a polymer sheet containing oriented iodide crystals, are essentiallyjust birefringent materials that pass only light polarized in the polarizer’s principal opticalplane.)

The radiation source can produce either conventional white (polychromatic) or filtered(monochromatic) light The electric field vector of light striking the first polarizer with anarbitrary orientation can be resolved into two components as shown in Fig 7, one in the po-larization direction and the other perpendicular to it The polarizer will block the transversecomponent, allowing the parallel component to pass through to the specimen This polarizedcomponent can be written