Theory of Elasticity

Theory of Elasticity

Ph.D. in Metallurgical and Materials Science Engineering Dr. Reham Reda Abbas

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Contents 1. Definition Course’s Science.

of

Area

the 6. Theory of Elasticity. 6.1 Description of a Stress of at a Point

2. Introduction.

3. Stress/Strain Relations 4. The Concept of Stress. 5. Aim of the Course. 2

Dr. Reham Reda Abbas

6.2 Stress Under General Loading

Conditions

(Stress Tensor).

6.3 Basic Assumptions of the Theory of Elasticity

6.4Uni-Axial Stress State.

6.5 Bi-Axial Stress State (2D). I. 1. Stresses Elements on Oblique Plane (Stress Transformation). 2. Principal Stresses in 2D & Max. Shear Stress. Mohr’s Circle in Plane Stress 1. Strain on Oblique Plane (Strain Transformation). 2. Principal Strain in 2D & Max. Shear Strain. Mohr’s Circle in Plane Strain) 3


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1. Definition of the Course’s Area of Science  Mechanics of materials is an area of science concerned with the behavior of bodies when subjected to forces or displacements.  Solid mechanics is the branch of mechanics of materials that studies the behavior of solid materials, especially their motion and deformation under the action of forces, temperature changes, phase changes, and other external or internal agents.  Solid mechanics is fundamental for civil, aerospace, mechanical engineering, ..etc. Dr. Reham Reda Abbas

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2. Introduction  Elasticity It is defined as the ability of a material to come back its original shape after deformation when external forces are removed, (σapplied < σy of the material).  In elastic deformation, the shape change is a result of stretching of inter-atomic bonds.

 Plasticity It is defined as the ability of the material to change its shape and dimensions under load and to keep the new shape and dimensions after the load is releases, (σapplied > σy of the material).  Plastic deformation process must involve rupturing of the inter-atomic bonds and then reforming.  Plastic deformation is permanent. Strength and hardness are measures of a material’s resistance to this deformation.

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3. Stress/Strain Relations  Stress (σ) • Consider a free-body for a cylindrical bar, as shown. The external load (F) is balanced by the internal resisting force (∫σ.dA), where (σ) is the stress normal to the cutting plane and (A) is the cross-sectional area of the bar. The equilibrium equation is: F= ∫σ.dA • If the stress is distributed uniformly over the area (A), that is, if (σ) is constant, the equation becomes: F= ∫σ.dA = σ. A • Average Stress is the ratio of the force to the cross sectional area. F

σ

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A

 Unit Pa = N/m2 = kg/m s2 Dr. Reham Reda Abbas

 Strain (ε)

• For the stress to be absolutely uniform, every longitudinal element in the bar would have to experience exactly the same strain, and the proportionality between stress and strain would have to be identical for each element.

• Strain is the displacement compared to the length.  Measures a fractional change. L   Unitless quantity.

L

• Average strain can be obtained from Hook’s law: E= σ ̸ ε 8 Dr. Reham Reda Abbas

 Shear Stress

 On a microscopic scale, plastic deformation corresponds to the net movement of large numbers of atoms in response to an applied shear stress.  Whatever the type of the applied stress on the metal (tensile or compression,..etc.), shear components exist at all.

F  A

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 Shear Strain

  tan  

x L

 Note: Materials also have a modulus from shear forces. Shear modulus (G) also relates to Hooke’s law: τ=Gγ G = E/ 2(1+ν) 10


 Poisson Ratio (ν)

 In all engineering materials, the elongation produced by an axial tensile force (P) in the direction of the force is accompanied by a contraction in any transverse direction, as shown. The ratio of the lateral strain over the axial strain is called Poisson’s ratio:

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4. Concept of Stress  The value of stress for a particular material (used in a particular way) which is considered to be a safe stress is usually called the allowable (working) stress (σw).  For static application, the allowable stress of ductile metals is usually based on the yield strength (σ0) and for brittle metals on the ultimate tensile strength (σu).  The values of allowable stress are established by technical organization such as the American Society of Mechanical Engineers (ASME).  The allowable stress (σw) may be considered as either the yield strength or the tensile strength divided by a number called “the safety factor”.  The value of the safety factor depends on the type of applied load (static or dynamic), type of material, ….etc.

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 In general, the stress will not be uniform over the area. This may be due to:  The anisotropy between grains in a polycrystalline metal.  The presence of more than one phase also give rise to non-uniformity of stress on a microscopic scale.  If the bar is not straight or not centrally loaded, the strains will be different for certain longitudinal elements and the stress will not be uniform  When there is an abrupt change in cross section. This results in a stress raiser of stress concentration. Up to now, we have said little about how these normal and shear stresses might vary with position throughout a solid. We will address a question: o How might stresses vary from one point to another throughout a body? Dr. Reham Reda Abbas

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Example

 A load applied to any mechanical member will induce internal stresses within the member.  The stresses acting on the material cause deformation of the material in various manner (elastic or plastic).  Deformation of the material is measured by strain value.  Internal force lines are denser near the hole, a common stress concentration. 14

How to measure the stress around the hole? Dr. Reham Reda Abbas

5. Aim of This Course

 The main aim of this course is to:  Calculate the acting values of the load at any point of a material (when used as a member in a structure or machine) and decide if the member can withstand without failure or not.  Predict the limiting values of the load at any point of a body.  Predict the load required at any point in a body to cause plastic flow to a chosen amount of deformation.  Both plasticity and elasticity play major roles in fields such as analyzing metal forming operations, mathematical modeling and any other field involving designing and developing mechanical objects.

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6. Theory of Elasticity

6.1 Description of a Stress at a Point  Two basic types of external forces which may act on a body: forces are Body forces are forces  Surface distributed throughout the forces distributed over the volume of the body, such surface of the body, such as centrifugal, gravitational as hydrostatic pressure or the force exerted by one forces, magnetic forces. body on another.

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Note: Theory of elasticity deals only with surface forces. Dr. Reham Reda Abbas

 To define the stress at a given point Q of the cross section, we should consider a small area ΔA (Fig). Dividing the magnitude of ΔF by ΔA, we obtain the average value of the stress over ΔA. Letting ΔA approach zero, we obtain the stress at point Q:

 In general, the value obtained for the stress σ at a given point Q of the section is different from the value of the average stress and σ is found to vary across the section.

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6.2 Stress Under General Loading Conditions (Stress Tensor)

Stress tensor is used to determine the equivalent stress at any point of the body. There are nine stress components are sufficient to determine the stress condition at a point of a body. Their totally being termed the stress tensor.

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 T   

         

XX

YX

ZX

  

XY

YY

ZY


τ

  

XZ

YZ

ZZ

   

τ

τ

τ

τ

τ

X- plane Y- plane Z- plane

 Stress components are defined with two subscripts. The first denotes the normal to the plane on which the force acts and the second is the direction of the force. For example, σxx is a stress acting on the x-plane in the x-direction.  Three of them are normal stress (σxx, σyy, σzz). They may be tensile or compressive.  Six of them are tangential stresses (τxy, τxz, τyx, τyz, τzx, τzy). A shear stress component is one in which the force acts parallel to the plane. A long the main diagonal of the tensor, the normal stresses are located, and there is symmetrically with respect to the diagonal. Note: The use of the opposite convention should cause no problem because τxy = τyx .

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 If the stress tensor is known at a point of a body, the stresses state of the whole body can be determined. 6.3 Basic Assumptions of the Theory of Elasticity  Four assumptions are considered for a body that is being analyzed by elasticity theory. The member must be: 1. In equilibrium. 2. A continuous body. 3. A homogeneous body. 4. An Isotrobic. 20


6.4 Uni-Axial Stress State I. Stresses Elements on Oblique Plane (Stress Transformation)

 Stress elements are a useful way to represent stresses acting at some point on a body. Isolate a small element and show stresses acting on all faces. Dimensions are “infinitesimal”, but are drawn to a large scale.  To obtain a complete picture of the stresses in a bar, we must consider the stresses acting on an “inclined or oblique” (as opposed to a “normal”) section through the bar. 21


 For oblique plane, we are really just rotating axes to represent stresses in a new coordinate system (x1y1).  The force P can be transformed/resolved into two components:  Normal force: perpendicular to the inclined plane, N = P cos θ (using equilibrium condition ∑Fx1=0)  Shear force: tangential to the inclined plane, V = - P sin θ (using equilibrium condition ∑Fy1=0)  If we know the areas on which the forces act, we can calculate τ the associated stresses.

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Note: 23


6.5 Bi-Axial Stress State (2D).

 In a number of engineering applications, the geometry of the body and loading allow us to model the problem using 2-D approximation. Such a study is called ''Plane elasticity''. There are two categories of plane elasticity: plane stress and plane strain. I. Plane Stress.  When an element is in plane stress in the xy plane, only the x and y faces are subjected to stresses (σz = 0 and τzx = τxz = τzy = τyz = 0). 24


Such situation occurs in a thin plate subjected to forces acting in the mid-plane of the plate (Fig.).  Thin-walled pressure vessels provide an important application of the analysis of plane stress.  It also occurs on the free surface of a structural element or machine component, i.e., at any point of the surface of that element or component that is not subjected to an external force (Fig.).

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Examples of plane stress

1. Stresses Elements on Oblique Plane (Stress Transformation)  Plane Stress is the stress system that is known in terms of coordinate system xy. We want to find the stresses in terms of the rotated coordinate system x1y1.  We are really just rotating axes by angle (θ) to represent stresses in a new coordinate system.  Why? A material may yield or fail at the maximum value of σ or τ. This value may occur at some angle other than θ = 0. (Remember that for uniaxial tension the maximum shear stress occurred when θ = 45 degrees while the maximum normal stress occurred when θ = o)

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 Let us assume that a state of plane stress exists at point Q (with σz = τzx = τzy = 0), and that it is defined by the stress components σx, σy, and τxy Q associated with the element shown in Fig.  We propose to determine the stress components σx1, σy1, and τx1y1 associated with the element after it has been rotated through an angle θ.  In order to determine the normal stress σx1 and the shearing stress τx1y1 exerted on the face perpendicular to the x1 axis, we consider an element with faces respectively perpendicular to the x, y, and x1 axes (Fig.). 27


Forces can be found from stresses if the area on which the stresses act is known. Force components can then be summed. 28


Left face has area A. Bottom face has area A tan θ. Inclined face has area A sec θ. Note: sec θ =1/cos θ

 We observe that, if the area of the oblique face is denoted by A, the areas of the vertical and horizontal faces are respectively equal to A cos θ and A sin θ.  Using components along the x1 and y2 axes, we write the following equilibrium equations: Sum forces in the x1 direction, ∑Fx1=0

Sum forces in the y1 direction, ∑Fy1=0

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Using τxy=τyx and simplifying gives:

 Using the following trigonometric identities

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Hence; transformation equations for plane stress:

Eq. (1)

Eq. (2)

For stresses on the y1 face, substitute θ+90 for θ:

Cos(180+ 2θ)=- Cos θ Sin(180+2θ)=- Sin θ

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Summing the equations of σx1 and σy1 gives : Note: Equations (1) & (2) describe the normal stress and shear stress on any plane through a point in a body subjected to a plane-stress situation.

Note:  The sum of the normal stresses on two perpendicular planes is an invariant quantity that is independent of orientation or angle θ.  The variation of normal stress and shear stress occur in the form of a sine wave, with a period of θ=180°. These relationships are valid for any state of stress. 32


2. Principal Stresses in 2D & Max. Shear Stress

Principal Stresses in 2D  The maximum and minimum normal stresses (σ1 and σ2) are known as the principal stresses. To find the principal stresses, we must differentiate the transformation equations.

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dCos(2θ)/dθ=- 2Sin2θ dSin (2θ)/dθ =2Cos2θ

 θp are principal angles associated with the principal stresses Note:  There are two values of θp in the range 0-180°, with values differing by 90°.  So, the planes on which the principal stresses act are mutually perpendicular.  To find out which principal stress goes with which principal angle, we could use the equations for sin θp and cos θp or for σx1. 34


 We can now solve for the principal stresses by substituting for θp in the stress transformation equation for σx1. This tells us which principal stress is associated with which principal angle.

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 Substituting for R and re-arranging gives the larger of the two principal stresses:

 To find the smaller principal stress, use: σ1 + σ2 = σx + σy

These equations can be combined to give:

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Eq. (3) Dr. Reham Reda Abbas

 Note: The planes on which the principal stresses act are called the principal planes. What shear stresses act on the principal planes?  Compare the equations for τx1y1 (=τθ) and dσx1/dθ:

 Solving either equation gives the same expression for tan 2θp.  Hence, the shear stresses are zero on the principal planes. 37


Maximum Shear Stress To find the maximum shear stress, we must differentiate the transformation equation for shear.

Eq. (4)

Note: There are two values of θs in the range 0-180°, with values differing by 90°.  So, the planes on which the maximum shear stresses act are mutually perpendicular

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Because shear stresses on perpendicular planes have equal magnitudes, the maximum positive and negative shear stresses differ only in sign. We can now solve for the maximum shear stress by substituting for θs in the stress transformation equation for τx1y1.

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 Use equations for sin θs and cos θs or τx1y1 to find out which face has the positive shear stress and which the negative.  What normal stresses act on the planes with maximum shear stress? Substitute for θs in the equations for σx1 and σy1 to get:

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 Finally, we can ask how the principal stresses and maximum shear stress are related and how the principal angles and maximum shear angles are related.

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 So, the planes of maximum shear stress (θs) occur at 45° to the principal planes (θp). 42


Mohr’s Circle in Plane Stress

 A very useful graphical method for representing the state of stress at a point on an oblique plane through a point (element) was suggested by Mohr.  An alternative method for the solution of problems involving the transformation of plane stress, based on the use of Mohr’s circle, will be presented.  Transformation equations (Algebraic Solution) for plane stress, equations (1) & (2), can be rearranged to be as following: 43


 By squaring each of these above equations and adding: Eq. (5)

 Note:

Cos2(x) + Sin2(x)= 1

 The above equation is the equation of a circle of the (x-h)2 + y2 = R2 form:

 Thus, Mohr’s circle is a circle in σx1 and τx1y1 coordinates with a radius (R) equal to: 44


 (h) is the center (C) displacement from the origin (0) to the right, from equation (5), h = σave:

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Fig. Mohr’s circle. Dr. Reham Reda Abbas

The two points A and B where the circle of Fig. intersects the horizontal axis are of special interest: Point A corresponds to the maximum value of the normal stress σx1, while point B corresponds to its minimum value (principle stresses). Substituting in the above equation for σave and R, will obtain equation (3) for the principle stresses. Besides, both points correspond to a zero value of the shearing stress τx1y1. Thus, the values θp of the parameter θ which correspond to points A and B can be obtained by setting τx1y1 = 0.

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The points D and E located on the vertical diameter of the circle correspond to the max. and min. numerical value of the shearing stress τx1y1. Since the abscissa of points D and E is σave = (σx+σy)/2, the values θs of the parameter θ corresponding to these points are obtained by setting σx1= (σx+σy)/2 in Eq. (1). It follows that the sum of the last two terms in that equation must be zero. Thus, for θ=θs, we write: Rewrite the above equation we obtain the eq. (4). 47


 Procedure for Constructing Mohr’s Circle. 1. Draw a set of coordinate axes with σθ (or σx1) as abscissa (positive to the right) and τθ (or τx1y1) as Yordinate (positive upward). 2. Calculate h value and locate the centre of the circle (c) at the point having coordinates σx1 = σavg and τx1y1 = 0. 3. Calculate the radius (R) and draw the circle from the center (c).  To specify a given plane: 1. Locate point X, representing the stress conditions on the x face of the element by plotting its coordinates σx1 = σx and τx1y1 = - τxy. Note that point A on the circle corresponds to θ = 0°. 48 Dr. Reham Reda Abbas

2. Locate point Y, representing the stress conditions on the y face of the element by plotting its coordinates σx1 = σy and τx1y1 = τxy. Note that point B on the circle corresponds to θ = 90°. 3. Draw a line from point X to point Y, a diameter of the circle passing through point C. Points X and Y (representing stresses on planes at 90° to each other) are at opposite ends of the diameter (and therefore 180° apart on the circle).

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Fig. Circular relationship of transformed stresses. Dr. Reham Reda Abbas

 Note:  A point on Mohr’s circle gives the magnitude and direction of the normal and shear stresses on any plane in the element, such as at point (M). Mohr’s Circle on an Oblique Plane

c

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Note that:  Notice that although 2θ appears in Mohr’s circle, θ appears on the stress element.  2θ is then positive counterclockwise, which agrees with the direction of 2θ used in the derivation of the transformation equations and the direction of θ on the stress element.  Procedure To Find Stresses on an Oblique Plane Using Mohr’s Circle. 1. On Mohr’s circle, measure an angle (2θ) counterclockwise from radius CX, because point X corresponds to θ = 0 and hence is the reference point from which angles are measured. 52


2. The angle (2θ) locates the point X' on the circle, which has coordinates (σx1,τx1y1). Point X' represents the stresses on the x1 (or x') face of the inclined element. 3. Point Y', which is diametrically opposite point X' on the circle, is located at an angle 2θ + 180° from CX (and 180° from CX'). Thus point Y' gives the stress on the y1 face of the inclined element. τ

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Note: Plane Elasticity

 Many problems in elasticity may be treated by a two dimensions or plane theory of elasticity.  There are two general types of problems involved in this plane analysis: (1) plane stress and (2) plane strain.  Each type can be defined by certain restrictions and assumption on the stress and displacement fields.  Plane stress and plane strain do not ordinarily occur simultaneously.

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II. Plane Stress Versus Plane Strain 1) Plane stress: 1) Plane strain:

Stresses σz=0, τxz=0, τyz=0 σx, σy, τxy may be non-zero. Strains γxz=0, γyz=0 εx, εy, εz, γxy may be nonzero.

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Stresses τxz=0, τyz=0 σx, σy, σz, τxy may be nonzero. Strains εz=0, γxz=0, γyz=0 εx, εy, γxy may be non-zero.

2) Plane stress is defined to be a state of stress in which the normal stress (σz) and shear stresses (τxz) and (τyz), directed perpendicular to xy plane are assumed to be zero. 3) Plane stress deals with a geometry of the body with one dimension (z coordinate) much small in comparison with the dimensions of the body in the other two dimensions (x and y coordinates).

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2) Plane strain is defined to be a state of strain in which the normal strain (εz) and shear strains (γxz) and (γyz), directed perpendicular to xy plane are assumed to be zero. 3) Plane strain deals with a geometry of the body that has a dimension in one (z coordinate) is very large in comparison with the dimensions of the body in the other two dimensions (x and y coordinates).

Plane Stress Versus Plane Strain  Members, e.g. plate, that  The assumptions of plane are thin (those with a small strain are realistic for long z dimension compared to or very thick bodies, (in z the in-plane x and y direction), e.g. prismatic dimensions) so the stresses cylinder, held between of z direction can be fixed, smooth and rigid neglected; hence loads act planes, subjected to loads only in the x-y plane can that act in x and/or y be considered to be under and/or z directions. The value of the strains of z plane stress condition. direction is very small compared to the original length and can be Dr. Reham Reda Abbas 57 neglected.

Plane Stress Versus Plane Strain 4) The loads are applied 4) The applies forces acts uniformly over the in x-y plane and don’t vary thickness of the plate and with z direction, i.e. the act in the plane of the plate loads are uniformly as shown. distributed with respect to The loading and boundary the large dimension and condition in plane stress act perpendicular to it analysis: (i) the loads may be point forces or distributed forces applied over the thickness of the plate, (ii) supports may be fixed points or fixed edges Dr. Reham Reda Abbas 58 or roller support.

Examples of Plane Elasticity Problems

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Plane Stress Versus Plane Strain

5) Plane stress analysis includes problems such as: (a) plates with holes, fillets, or other changes in geometry that are loaded in their plane resulting in local stress concentrations. (b) Thin plate, such as pressure vessels and fuselage. 60


5) Plane strain analysis includes problems such as: (a) long underground box culvert subjected to a uniform load acting constantly over its length. (b) long cylindrical control rod subjected to a load that remains constant over the rod length (or depth). (c) dams, tunnels and other geotechnical works.

Examples of Plane Elasticity Problems

Plane Stress Problems

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Plane Strain Problems

Long Cylinders Under Uniform Loading Dr. Reham Reda Abbas

Plane Stress Problems

Thin Plate With Central Hole

Plane Strain Problems

Dams

Circular Plate Under Edge Loadings 62


Solid Propellant Rockets

III. Plane Strain 1. Strain on Oblique Plane (Strain Transformation)  We want to derive equations for the normal strains εx1 and εy1 and the shear strain γx1y1 associated with the x1y1 axes, which are rotated counterclockwise through an angle θ from the xy axes.  Consider a rectangular element in the xy plane after strains εx, εy, and γxy are applied. 63


 The total train in x1 direction is the sum of the change of diagonal length (as a result of εx, εy, and γxy, separately) in x1 direction. Diagonal increases in length in the x1 direction by εxdx cosθ, as a result of εx dx in the x direction . Diagonal rotates clockwise by α1. Let: tanα1= α1 Diagonal increases in length in the x1 direction by εydy sinθ. Diagonal rotates counterclockwise by α2. 64


Let: tanα2= α2 Diagonal increases in length in the x1 direction by γxydx cosθ. Diagonal rotates clockwise by α3. Let: tanα3= α3

 The total increase in the length of the diagonal is:  The normal strain εx1 is the change in length over the original length. 65


 So, the normal strain εx1 is:

 The normal strain εy1 can be found by substituting θ+90° into the equation for εx1.  To find the shear strain γx1y1, we must find the decrease in angle of lines in the material that were initially along the x1y1 axes. γx1y1= tan α+ tanβ ~ α+ β 66 Dr. Reham Reda Abbas

 To find α, we just sum α1, α2, and α3, taking the direction of the rotation into account.

 To find the angle β, we can substitute θ+90 into the equation for α, but we must insert a negative sign, since α is counterclockwise and β is clockwise. 67


 So, the shear strain γx1y1 is:

 Using trigonometric identities for sinθ, cosθ, sin2θ, and cos2θ gives the strain transformation equations

 The transformation of shear strain with respect to the {x,y} coordinates to the strains with respect to {x',y'} is performed via the equations. 68


 Note: Now, if compare the strain transformation equations to the stress transformation equations, we will find that the equations have the same form, but with different variables:

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2. Principal Stresses in 2D & Max. Shear Stress Principal Strain in 2D

Maximum Shear Strain

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 Note:  Shear strains are zero on the principal planes.  Principal stresses and principal strains occur in the same directions.  The maximum shear strains are associated with axes at 45° to the directions of the principal strains.

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Mohr’s Circle in Plane Strain

 Plot εx1 instead of σx1.  Plot (γx1y1/2) instead of τx1y1 Principal strains ε1, ε2.

Maximum shear strain γmax with associated normal strain εs. 72


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6.6 General Three-Dimensional State  3D- Stress State.  In this case, we will consider the three faces of the elements. Since there are three faces, there are three principal directions and three principal stresses σ1, σ2, and σ3. Note that for plane stress, the stress-free surface contains the third principal stress which is zero and we were able to specify any stress state σx , σy , and τxy and find the principal stresses and directions.  But six components of stress (σx, σy, σz, τxy, τyz, and τzx) are required to specify a general state of stress in three dimensions, and the problem of determining the principal stresses and directions is more difficult.

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 In design, three-dimensional transformations are rarely performed since most maximum stress states occur under plane stress conditions. One notable exception is contact stress, which is not a case of plane stress, where the three principal stresses are given.  In fact, all states of stress are truly three-dimensional, where they might be described one- or twodimensionally with respect to specific coordinate axes.  Here it is most important to understand the relationship among the three principal stresses.  The process in finding the three principal stresses from the six stress components are mathematically very complex. 75


 In plotting Mohr’s circles for three-dimensional stress, the principal normal stresses are ordered so that σ1 ≥ σ2 ≥ σ3 (Fig). The stress coordinates σ , τ for any arbitrarily located plane will always lie on the boundaries or within the shaded area.  In this case, there are three max. shear stresses τ1/2, τ2/3, and τ1/3. Each of these occurs on the two planes, one of which is shown in Fig.

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Of course, τmax = τ1/3 when the normal principal stresses are ordered (σ1 > σ2 > σ3). 77


 3D- Strain State.  In order to determine the total strain produced along a particular direction, we can apply the principle of superposition.

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 For Example, the resultant strain along the x-axis, comes from the strain contribution due to the application of x, y and z. The situation can be summarized by the as following. ______________________ _______________________________ Stress Strain in the Strain in the Strain in the x direction y direction z direction

x y z

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x  x x



x

E v y   E v z   E


v x E

z  

E v z y   E



y   y 

y

v x E v y z   E

z 

z

E

 By superposition of the components of strain in the x, y, and z directions, the strain along each axis can be written as: 1  x   x   ( y   z ) E

1  y   y  v z   x  E

1  z   z  v x   y  E

 The shearing stresses acting on the unit cube produce shearing strains.

xy  G xy

 yz  G yz

xz  G xz

 The proportionality constant G is the modulus of elasticity in shear, or the modulus of rigidity. Values of G are usually determined from a torsion test. 80


 Mohr’s circles for three-dimensional strain.

81


6.7 Stress Concentration.  Machine members often have regions at which the state of stress is concentrated, i.e. significantly greater than theoretical predictions or the average value of the stress in the member. These regions such as: (1) The presence of discontinuities or a sudden change in cross section alters the stress distribution causing higher and localized stresses near these areas.  Any type of discontinuity (hole, fillet, shoulder, notch) serve as a stress raiser where it increases the stress in the vicinity of the discontinuity. (2) Internal microscopic irregularities (nonhomogeneities like inclusion) of the material created by such manufacturing processes as casting and molding.

82


Stress distribution near circular hole in flat bar under axial loading.

83


Stress distribution near fillets in flat bar under axial loading.

 The designer is more interested in the maximum value of the stress in a given section, than in the actual distribution of stresses in that section (which contains stress raisers), since the main concern is to determine whether the allowable stress will be exceeded under a given loading, and not where this value will be exceeded. For this reason, one defines the ratio:

 Stress concentration factor (kt) is used to relate the actual maximum stress at the discontinuity to the nominal stress without the discontinuity. 84


 Similarly, we can also estimate the highly localized amplification of shear stress in the vicinity of a geometric stress concentration: τ max = Ktsτ0 where, τ0 = Nominal shear stress. Kts = Theoretical stress-concentration factor for shear. Characteristics of Stress Concentration Factors (kt). 1. Function of the geometry or shape of the part, but not its size or material properties; 2. Function of the type of loading applied to the part (axial, bending or torsional in elastic range); 85


3. Function of the specific geometric stress raiser in the part, i.e. They depend only on the type of discontinuity and the geometry (e.g. fillet radius, notch, or hole); 4. Always defined with respect to a particular nominal stress; 5. Typically assumes a linear elastic, homogeneous, isotropic material.  Stress concentration factors can be computed once and for all in terms of the ratios of the geometric parameters involved, and the results obtained can be expressed in the form of tables or of graphs, as shown in Fig. 86


 Thin plate in tension or simple compression with a transverse hole.

87


To determine the maximum stress occurring near a discontinuity in a given member subjected to a given axial load P, the designer needs only to compute the average stress or the nominal stress (σ0=P/A), in the critical section, and multiply the result obtained by the appropriate value of the stress-concentration factor (K).  You should note, however, that this procedure is valid only as long as σmax does not exceed the proportional limit of the material, since the values of K plotted in Fig. were obtained by assuming a linear relation between stress and strain. 88


 Determination of Kt Value The stress-concentration factor, associated with a specific geometry and loading condition of a part, can be derived through experimental, analysis or computational methods. 1. Experimental Methods. Optical methods, such as photoelasticity, are very dependable and widely used for experimentally determining the stress concentration at a point on a part.

89


2. Analytical Methods. The theory of elasticity can be used to analyze certain geometrical shapes to calculate stress-concentration factors. 3. Computational Methods. Finite-element techniques provide a powerful and inexpensive computational method of assessing stressconcentration factors. Following are comparisons of stress-concentration factors derived using experimental, analytical and computational methods for a rectangular filleted bar in tension. 90


 Flat bars with fillets. Rectangular filleted bar in tension or simple compression. σ0=F/A, where A = d.t, t: thickness

91


Stress distribution in a rectangular filleted bar in simple tension obtained through photoelastic procedures.

Stress contours of stress generated by a finite element model of one half of a rectangular filleted bar in tension. 92


 Application to Ductile and Brittle Materials for Static Loading. When dealing with brittle materials it is very important to consider the stress concentration areas because rupture will initiate there and the entire part will fail, while for ductile materials stress concentrations are usually not considered because the material will yield at the high stress location and this relieves the stress concentration. I. Ductile Materials. • While stress concentration must be considered for fatigue and impact loading of most materials, stressconcentration factors are seldom applied to ductile materials under static loading. 93


•This design practice is justified by four points: If the magnitude of the loading is large enough to cause yielding due to the stress concentration, the localized area will plastically deform immediately upon loading; Ductile materials typically work-harden (strainstrengthen) on yielding, resulting in a localized increase in material strength.  It is important to note, that even though the stressconcentration factor is not usually applied to estimate the stresses at a stress raiser in a ductile material, the higher state of stress does in fact exist. Ductile Material Practice: σ max =σ0 94


II. Brittle Materials. • Stress-concentration factors are always required for brittle materials, regardless of the loading conditions, since brittle behavior results in failure/fracture. This type of failure is characteristic of brittle materials which do not exhibit a yielding or plastic range. • To avoid such catastrophic failure, the design practice is to always use a stress-concentration factor for brittle materials to ensure that the state of stress is accurately represented. Brittle Material Practice: σmax = kt σ0 95


 Problem Statement. A bar machined from cast iron (a brittle material) is subjected to a static axial load, as shown. Find: The critical section of the bar. Given: σUTS = 20 kpsi.

96


6.8 Strain Measurement Methods. • In designing any application, for example if new model of an airplane, automobile or railroad vehicle will be introduced, it is significant to know the stress borne by each material part. However, at the present scientific level, there is no technology which enables direct measurement and judgment of stress. It is very difficult to measure normal and shear stresses in a body, particularly stresses at a point. It is relatively easy to measure the strains on the surface of a body (normal strains, that is, not shear strains). • So, the strain on the surface, of a structural element or machine component, is measured in order to know the internal stress. 97


 Note:  Classification of strain measuring methods:  Based on mounting: 1. Un-bonded strain gauge. 2. Bonded strain gauge.  Based on principle of working : 1. Mechanical methods: such as extensometers and other mechanical devices. 2. Optical methods: such as photoelasticity, or other optical techniques. 3. Electrical methods: such as resistive, piezoelectric, piezoresistive stain gauges and similar electronic techniques. 98


 Resistance Strain Gauges.  Note: Measurement of fluctuating stress on parts of running vehicles or flying aircraft was made possible using a load cell. A load cell is generally comprised of three parts: (1) a mechanical system, (2) a strain gauge, and (3) an electronic amplification device.  Strain gage is the most common sensing element to measure surface strain. It is a type of electric transducer that converts a mechanical displacement (due to the applied load) into a change of resistance.  This change in resistance of the conductor can be measured easily against the applied force using strain gauge. 99


 A fundamental parameter of the strain gauge is its sensitivity to strain, expressed quantitatively as the gauge factor (GF).  Gauge factor is defined as the ratio of fractional change in electrical resistance to the fractional change in length (strain).  As the length increases, diameter decreases; hence the resistance increase. GF = gauge factor (related to Poisson’s ratio and resistivity) R = Original resistance of strain gage, (ohm) ΔR: resistance change due to the strain ε, Ω

o The gage factor, differs depending on the materials 0f the gauge. o The Gauge Factor for metallic strain gauges is typically around 2.

100


 Note: The metal of which the filament element is made is the principle factor determining the magnitude of this gauge factor, for example: an alloy of 60%Cu40%Ni produces wire or foil gages factor of approximately 2. Thus, a strain gage using this alloy for the sensing element enables conversion of mechanical strain to a corresponding electrical resistance change.  Note: Performance of bonded metallic strain gauges (the detected resistance) depends on: *grid material and configuration, *backing material, *bonding material and method, *gauge protection, * temperature, and *associated electrical circuitry. 101


 Resistance Strain Gauge Types 1- Un-bonded Strain Gauge  The earliest strain gauges were of the “un-bonded” type and used pillars, separated by the gauge length, with wires stretched between them.

2- Bonded Strain Gauge  Later gauges were “bonded”, with the resistance element applied directly to the surface of the strained member. 102


 The methods in widest use employs the bonded electrical wire or foil resistance strain gages. (1) The bonded wire gage consists of a grid fine filament cemented between two sheets of treated paper or plastic backing. The backing serves to insulate the grid from the metal surface on which it is to be bonded, another function serves as a carrier so that the filament may be conveniently handle. Generally, 0.025mm diameter wire is used.  The gauge is bonded to the strained surface of measured material by a thin layer of a special glue. 103


Electrical Resistance Gauge 104


(2) The grid in the case of bonded foil gage is constructed of very thin metal foil (approximately 0.0025mm) rather than the wire.  The foil-type gauges typically consist of a metal film element on a thin epoxy support and are made using printed-circuit techniques.  Because of filament cross section of a foil gage is rectangular, the ratio of surface area to cross-sectional area is higher that of a round wire. This results in increasing heat dissipation and improving adhesion between the grid and the backing material.  Foil-type gauges can be made in a number of configurations. The selection of a particular bonded gage will depend upon the specific service application.

105


How Do Resistance Strain Gauges Work?  In order to measure strain with a bonded resistance strain gauge, it must be connected to an electric circuit that is capable of measuring the small changes in resistance corresponding to strain.  An electrical bridge circuit attached the gage by means of lead wires, is then used to translate electrical changes into strains.  Minute resistance changes are measured with a dedicated strain amplifier using an electric circuit called a Wheatstone bridge.  The Wheatstone bridge, one of the most accurate and convenient systems used, is capable of measuring the resistance resulted from strain as small as 1µm. This circuit is also well suited for temperature compensation.

106


107


Rosette Gages and Strain state

 Special combination gages are available for measurement of the state of strain at a point on a surface simultaneously in three or more directions.  Generally, these consist of three gages whose axes are either 45° or 60° apart forming a strain rosette.

108


 Using the reading of the three independent gages (a strain rosette) together with Mohr’s circle for strain, the principle strains and their orientations may readily be obtained.  From these three independent measurements of normal strain at a point, three equations of strains at different angles develop, by solving these equations, εx- εy- γxy at any direction can be obtained.

End of Elasticity Part 109


Theory of Elasticity

Theory of Elasticity

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