Shearing Strain

© 2006 The McGraw-Hill Companies, Inc. All rights reserved.

MECHANICS OF MATERIALS

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Beer • Johnston • DeWolf

2 - 1

Shearing Strain

• A cubic element subjected to a shear stress will deform into a rhomboid. The corresponding shear strain is quantified in terms of the change in angle between the sides,

xyxy f



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Given:A metallic rectangular plate of length L and width L/2 is shown in figure. When the plate is subjected to stresses acting along the edge faces, it distorts into a parallelogram.

Find:

Initial Plate Deformed Plate



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Let the origin of the coordinates be located at the corner A. The shearing strain equals the change in angle between the positive x and y direction.

In as much as deformations are small we can write,

Hence,

The plus sign indicates that the angle BAD has decreased.



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A thin triangular plate ABC is uniformly deformed into a shape of ABC’, as shown by dashed lines in fig.

The edge AB is rigidly attached to the frame .The deformed edges AC’=BC’ are straight lines.



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=

Referring to the figure

The change in length OC is

The length of deformed edges are-



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The angle ACB becomes-

The change in right angle is then 90-89.943=0.0570



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A 0.4m by 0.4 m square ABCD is drawn on a tin panel or flat plate of an aircraft prior to loading. Subsequent to loading ,the square has the dimensions shown by dashed lines.

The average values of two dimensional strain components at corner A

`



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For the square plate under consideration we have -



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Small-Strain Approximation



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• A plot of shear stress vs. shear strain is similar to the previous plots of normal stress vs. normal strain except that the strength values are approximately half. For small strains,

zxzxyzyzxyxy GGG

where G is the modulus of rigidity or shear modulus.

Shearing Strain



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2 - 17

Poisson’s Ratio

• For a slender bar subjected to axial loading:

0 zyx

x E

• The elongation in the x-direction is accompanied by a contraction in the other directions. Assuming that the material is isotropic (no directional dependence),

0 zy

• Poisson’s ratio is defined as

x

z

x

y

strain axial

strain lateral



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2 - 18

Generalized Hooke’s Law

• 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 stress2) deformations are small

EEE

EEE

EEE

zyxz

zyxy

zyxx

• With these restrictions:



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Dilatation: Bulk Modulus

• Relative to the unstressed state, the change in volume is

e)unit volumper in volume (change dilatation

21

111111

zyx

zyx

zyxzyx

E

e

• For element subjected to uniform hydrostatic pressure,

modulusbulk 213

213

Ek

k

p

Epe

• Subjected to uniform pressure, dilatation must be negative, therefore

210



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2 - 20

Shearing Strain

• A cubic element subjected to a shear stress will deform into a rhomboid. The corresponding shear strain is quantified in terms of the change in angle between the sides,

xyxy f

• A plot of shear stress vs. shear strain is similar to the previous plots of normal stress vs. normal strain except that the strength values are approximately half. For small strains,

zxzxyzyzxyxy GGG

where G is the modulus of rigidity or shear modulus.



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2 - 21

Example 2.10

A rectangular block of material with modulus of rigidity G = 620 MPa is bonded to two rigid horizontal plates. The lower plate is fixed, while the upper plate is subjected to a horizontal force P. Knowing that the upper plate moves through 1.00 mmunder the action of the force, determine a) the average shearing strain in the material, and b) the force P exerted on the plate.

SOLUTION:

• Determine the average angular deformation or shearing strain of the block.

• Use the definition of shearing stress to find the force P.

• Apply Hooke’s law for shearing stress and strain to find the corresponding shearing stress.



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2 - 22

• Determine the average angular deformation or shearing strain of the block.

rad020.0.50

mm.0.1tan xyxyxy mm

• Apply Hooke’s law for shearing stress and strain to find the corresponding shearing stress.

MPa4.12rad020.0MPa620 xyxy G

• Use the definition of shearing stress to find the force P.

kN 76.153mm.62.200MPa4.12 mmAP xy

kN 76.153P



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The above figure shows a square element ABCD, sides of unstrained length 2units under the action of equal normal stresses σ,tension and compression. Then it has been shown that the element EFGH is in pure shear of equal magnitude σ.



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Linear strain in direction EG=σ/E + vσ/E

From the above figure,

say, ε = (1+v)σ/E -------------------------→ (1)

Linear strain in direction HF = -σ/E - vσ/E = - ε

Hence the strained lengths of EO and HO are 1+ε and 1-ε respectively.



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The shear strain is ø = σ/G --------------------------------------→ (2)

on the element EFGH and the angle EHG will increase to ∏/2 +ø.

Angle EHO is half this value, i.e. ∏/4 +ø/2.

Considering the triangle EOH, tan EHO = EO/HO i.e. tan ( ∏/4+ø/2) = (1+ε)/(1-ε)



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Expanding, (1+ε)/(1-ε) = (tan ∏/4 +tan ø/2) / (1- tan ∏/4 . tan ø/2)

since tan ∏/4 = 1 and tan ø/2 = ø/2 for small angles Clearly ε = ø/2

= (1+ ø/2) / (1- ø/2) (approx.)

and by substitution for ε and ø from (1) and (2),we get

(1+v)σ/E = σ/2G

Rearranging , E = 2G(I+v)



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2 - 28

Relation Among E, and G

• An axially loaded slender bar will elongate in the axial direction and contract in the transverse directions.

12G

E

• Components of normal and shear strain are related,

• If the cubic element is oriented as in the bottom figure, it will deform into a rhombus. Axial load also results in a shear strain.

• An initially cubic element oriented as in top figure will deform into a rectangular parallelepiped. The axial load produces a normal strain.



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2 - 29

Sample Problem 2.5

A circle of diameter d = 225 mm is scribed on an unstressed aluminum plate of thickness t = 20 mm Forces acting in the plane of the plate later cause normal stresses x = 82.7 MPa and z = 138 MPa.

For E = 69 GPa and = 1/3, determine the change in:

a) the length of diameter AB,

b) the length of diameter CD,

c) the thickness of the plate, and

d) the volume of the plate.



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2 - 30

SOLUTION:

• Apply the generalized Hooke’s Law to find the three components of normal strain.

m/m10600.1

m/m.10066.1

m./m.10532.0

MPa1383

10MPa7.82

GPa 69

1

3

3

3

EEE

EEE

EEE

zyxz

zyxy

zyxx

• Evaluate the deformation components.

m10225m/m10533.0 33 dxAB

m10225m/m10600.1 33 dzDC

m1020m/m10066.1 33 tyt

mm 12.0AB

mm36.0DC

mm 02.0t

• Find the change in volume

33

3

mmmm20mm375mm37510066.1

10066.1

eVV

e zyx

3mm2998V



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2 - 31

Composite Materials

• Fiber-reinforced composite materials are formed from lamina of fibers of graphite, glass, or polymers embedded in a resin matrix.

z

zz

y

yy

x

xx EEE

• Normal stresses and strains are related by Hooke’s Law but with directionally dependent moduli of elasticity,

x

zxz

x

yxy

• Transverse contractions are related by directionally dependent values of Poisson’s ratio, e.g.,

• Materials with directionally dependent mechanical properties are anisotropic.



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2 - 32

Saint-Venant’s Principle

• Loads transmitted through rigid plates result in uniform distribution of stress and strain.

• Saint-Venant’s Principle: Stress distribution may be assumed independent of the mode of load application except in the immediate vicinity of load application points.

• Stress and strain distributions become uniform at a relatively short distance from the load application points.

• Concentrated loads result in large stresses in the vicinity of the load application point.



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Stress Concentration

• Before ripping a piece of cloth a tailor puts a small cut with scissors.

• Its easier to tear a piece of paper when it has a small cut in it.

• Why are the door and window openings in aero-planes and ships rounded ?



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2 - 34

Stress Concentration: Hole

Discontinuities of cross section may result in high localized or concentrated stresses. ave

max

K



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2 - 35

Stress Concentration: Fillet



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2 - 36

Example 2.12

Determine the largest axial load P that can be safely supported by a flat steel bar consisting of two portions, both 10 mm thick, and respectively 40 and 60 mm wide, connected by fillets of radius r = 8 mm. Assume an allowable normal stress of 165 MPa.

SOLUTION:

• Determine the geometric ratios and find the stress concentration factor from Fig. 2.64b.

• Apply the definition of normal stress to find the allowable load.

• Find the allowable average normal stress using the material allowable normal stress and the stress concentration factor.



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2 - 37

• Determine the geometric ratios and find the stress concentration factor from Fig. 2.64b.

82.1

20.0mm40

mm850.1

mm40

mm60

K

d

r

d

D

• Find the allowable average normal stress using the material allowable normal stress and the stress concentration factor.

MPa7.9082.1

MPa165maxave

K

• Apply the definition of normal stress to find the allowable load.

N103.36

MPa7.90mm10mm40

3

aveAP

kN3.36P



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• What are composites?

• Why fibers are strong ?



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2 - 39

Elastoplastic Materials

• Previous analyses based on assumption of linear stress-strain relationship, i.e., stresses below the yield stress

• Assumption is good for brittle material which rupture without yielding

• If the yield stress of ductile materials is exceeded, then plastic deformations occur

• Analysis of plastic deformations is simplified by assuming an idealized elastoplastic material

• Deformations of an elastoplastic material are divided into elastic and plastic ranges

• Permanent deformations result from loading beyond the yield stress



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2 - 40

Plastic Deformations

• Elastic deformation while maximum stress is less than yield stressK

AAP ave

max

• Maximum stress is equal to the yield stress at the maximum elastic loading

K

AP YY

• At loadings above the maximum elastic load, a region of plastic deformations develop near the hole

• As the loading increases, the plastic region expands until the section is at a uniform stress equal to the yield stress

Y

YU

PK

AP



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2 - 41

Residual Stresses

• When a single structural element is loaded uniformly beyond its yield stress and then unloaded, it is permanently deformed but all stresses disappear. This is not the general result.

• Residual stresses also result from the uneven heating or cooling of structures or structural elements

• Residual stresses will remain in a structure after loading and unloading if

- only part of the structure undergoes plastic deformation

- different parts of the structure undergo different plastic deformations



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2 - 42

Example 2.14, 2.15, 2.16

A cylindrical rod is placed inside a tube of the same length. The ends of the rod and tube are attached to a rigid support on one side and a rigid plate on the other. The load on the rod-tube assembly is increased from zero to 25 kN and decreased back to zero.

a) draw a load-deflection diagram for the rod-tube assembly

b) determine the maximum elongation

c) determine the permanent set

d) calculate the residual stresses in the rod and tube.

MPa248

GPa 208

mm48 2

Yr

r

r

σ

E

A

MPa310

GPa104

mm65 2

Yt

t

t

σ

E

A



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a) Draw a load-deflection diagram for the rod-tube assembly

mm9.0

m1075008MPa2

48MPa2

kN12mm4848MPa2

3

2

LE

Lδ

AP

r

YrYrYr

rYrYr

mm235.2

m1075004MPa1

10MPa3

kN20m1065MPa310

3

26

LE

Lδ

AP

t

YtYtYt

tYtYt

tr

tr PPP

Example 2.14, 2.15, 2.16



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2 - 44

b,c) determine the maximum elongation and permanent set

• At a load of P = 25 kN, the rod has reached the plastic range while the tube is still in the elastic range

m)10750(GPa104

00MPa2

MPa200mm56

kN13

kN13kN1225

kN12

3t

2t

LE

L

A

P

PPP

PP

t

tt

t

t

rt

Yrr

mm44.1max t

• The rod-tube assembly unloads along a line parallel to 0Yr

mm126.1440.1

mm126.1mmkN2.22

kN25

slopemmkN2.22.9mm.0

kN20

maxp

max

m

P

m

mm314.0p

Example 2.14, 2.15, 2.16



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2 - 45

• Calculate the residual stresses in the rod and tube.

Calculate the reverse stresses in the rod and tube caused by unloading and add them to the maximum stresses.

MPa52MPa156208

MPa64MPa312248

ksi15604GPa11050.1

MPa31208GPa21050.1

mmmm1050.1750

mm126.1

,

,

3

3

3

tttresidual

rrrresidual

tt

rr

E

E

L

Example 2.14, 2.15, 2.16



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Managing thermal stress

• Most structures, small or large, are made of two or more materials that are clamped, welded or otherwise bonded together. This causes problems when temperatures change.

• Railway track will bend and buckle in exceptionally hot weather (steel, high α, clamped to mother earth with a much lower α).

• Bearings seize, doors jam.

• Thermal distortion is a particular problem in equipment designed for precise measurement or registration like that used to make masks for high-performance computer chips, causing loss of accuracy when temperatures change.



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• All of these derive from differential thermal expansion, which, if constrained (clamped in a way that stops it happening) generates thermal stress.

• As an example, Many technologies involve coating materials with a thin

surface layer of a different material to impart wear resistance, or resistance to corrosion or oxidation. The deposition process often operates at a high temperature. On cooling, the substrate and the surface layer contract by different amounts because their expansion coefficients differ and this puts the layer under stress.



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• This residual stress calculated as follows.

• Think of a thin film bonded onto a component that is much thicker than the

film, as in Figure.

• First imagine that the layer is detached, as in Figure. A temperature drop of ∆T causes the layer to change in length by

1 1 0L L T Fig: Thermal stresses in thin films arise on cooling or heating when their expansion coefficients differ. Here that of the film is α1 and that of the substrate, a massive body, is α2.



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• Meanwhile, the substrate to which it was previously bonded contracts by . If the surface layer shrinks more than the substrate.

• If we want to stick the film back on the much-more-massive substrate, covering the same surface as before, we must stretch it by the strain

• This requires a stress in the film of

2 2 0L L T 1 2

1 21 2

0

( )L L

TL

1 1 1 2( )E T



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• The stress can be large enough to crack the surface film. The pattern of cracks seen on glazed tiles arises in this way.

• So if you join dissimilar materials you must expect thermal stress when they are heated or cooled. The way to avoid it is to avoid material combinations with very different expansion coefficients.

• Avoiding materials with α mismatch is not always possible. As example, joining glass to metal—Pyrex to stainless steel, say—a common combination in high vacuum equipment. Their expansion coefficients for Pyrex (borosilicate glass), α =4 x 10-6/K; for stainless steel, α =20x 10-6/K: a big mismatch.

• Vacuum equipment has to be ‘baked out’ to desorb gases and moisture, requiring that it be heated to about 150°C, enough for the mismatch to crack the Pyrex.



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• The answer is to grade the joint with one or more materials with expansion that lies between the two: first join the Pyrex to a glass of slightly higher expansion and the stainless to a metal of lower expansion—the property chart suggests soda glass for the first and a nickel alloy for the second—and then diffusion-bond these to each other, as in Figure.

• The graded joint spreads the mismatch, lowering the stress and the risk of damage.

• There is an alternative, although it is not always practical. It is to put a compliant layer—rubber, for instance—in the joint; it is the way windows are mounted in cars. The difference in expansion is taken up by distortion in the rubber, transmitting very little of the mismatch to the glass or surrounding steel.



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Material for force loop

Fig. Precision measuring device.

• Force loop will support heat sources like fingers of the operators , electronic components which generate heat.

• Material is selected by considering one dimensional heat flow through rod insulated except at its end , one of which is at ambient temperature and other connected to heat source.

• Fourier’s law:

Where q is heat input per unit area ,is thermal conductivity, thermal gradient

dTq

dx

• Thermal strain: Where is thermal expansion coefficient , ambient temperature

0( )T T



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• The distortion is proportional to gradient of strain.

d dTq

dx dx

• The material is selected having large value of index :



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The theory of pressure vessels, applies, with minor modifications, to things other than closed containers: that is, to ‘open’ membrane and fabrics which have to sustain pressure from the free movement of wind or water.

tents and kitesawnings and fabric covered aircraft and parachutes sails of ship and windmills eardrums fishes’ fins and the wings of bats and pterodactyls



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A stiff open framework of rods or spars or bones with some kind of flexible fabric or skin or membrane. As soon as any lateral force comes upon the membrane by reason of pressure of wind or water, it must deflect or bow into a curved shape which, to a first approximation, may be treated as a part or segment of a sphere or cylinder.

The force or tension in the membrane, per unit width, is pr, the product of the wind pressure (p) and the radius of curvature (r) of the membrane.



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The sail bulge between its supports so that, as the wind pressure increases, the radius of curvature diminishes, and so the tension force in the canvas remains roughly constant however hard the wind may blow.



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In bats the wings are constructed by stretching a membrane of very flexible skin over a framework of long, thin bones which are, in essence, the fingers of the hands.

Fruit-bats, for instance, are quite large animals with a wing-span of over a meter. In their native India, where they are a pest, they think nothing of flying thirty or forty miles in a night in order to rob an orchard. Since they can do this without becoming unduly exhausted they must therefore be efficient flying machines.



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When a fruit-bat is photographed in flight it can be seen that, on the down stroke of the wing, the skin membrane bulges upward into a form which is roughly semi-circular, thereby minimizing the mechanical load upon its bones. It is clear that there can be in practice be little or no aerodynamic loss as a consequence of this change in shape.



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Given the job of designing a flying animal, a modern engineer would perhaps produce something like a bat, or possibly some sort of flying insect. I do not think that it would occur to him to invent feathers. Yet presumably there are very good reasons for their existence. One imagines that both bats and pterodactyls tend to lose a good deal of energy in the form of heat from the skin of their wings; but then reasonable heat insulation could be provided by fur.



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feather, like horns and claws, developed from hair. However, hair is presumably better when it is soft, and so the keratin from which hair is made has quite a low Young’s modulus. In feather the keratin molecules has been made stiffer by cross linking the molecular chains with sulphur atoms (which accounts for the smell of the burnt feather).



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There are, no doubt, aerodynamic advantage in using feathers, since their employment extends the choice of outside shapes which an animal can make use of. For one thing, ‘thick’ wing-sections have often better aerodynamic efficiencies than the thin ones which result from membranes. It is easy to get an efficient ‘thick’ section by padding out the wing profile with feather at the cost of very little weight increase. Furthermore, feathers are better adapted than skin and bone for providing anti-stalling devices such as ‘slots’ and ‘flaps’.



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Beer • Johnston • DeWolfHowever, I am inclined to think that the main advantages of feathers to an animal may be structural. Anybody who has flown model aeroplanes knows, to their cost, how vulnerable any small flying machine must be to accidental damage from things like trees and bushes, or even from careless handling. Many birds fly constantly in and out of the trees and hedges and other obstacles. Indeed they use such cover as a refuge from their enemies. For most birds loss of reasonable number of feather is not a very serious matter. Besides, it is better to leave the cat with a mouthful of feather than to be eaten.



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When a hawk kills a bird in the air it does not usually do so by wounding it with its beak or talons - which would probably not penetrate the feather. It kills by striking the bird in the back with its outstretched feet so as to impart a violent acceleration to the bird as a whole which has the effect of breaking its neck – very much as happens in judicial hanging.

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Shearing Strain