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    Dr. R. Chandra SekaranPh.D(C.Engg)

    Professor, Harbour Engineering

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    S.No. Determinate Structures Indeterminate Structures

    1

    Equilibrium conditions are fully adequate

    to analyse the structure.

    Conditions of equilibrium are not

    adequate to fully analyse the structure.

    2

    Bending moment or shear force at anysection is independent of the materialproperty of the structure.

    Bending moment or shear force at anysection depends upon the materialproperty.

    3

    The bending moment or shear force at anysection is independent of the cross-section or moment of inertia.

    The bending moment or shear force atany section depends upon thecross-section or moment of inertia.

    4Temperature variations do not causestresses.

    Temperature variations causestresses.

    5 No stresses are caused due to lack of fit. Stresses are caused due to lack of fit.

    6

    Extra conditions like compatibility of

    displacements are not required to analysethe structure.

    Extra conditions like compatibility ofdisplacements are required to analyse

    the structure along with the equilibriumequations.

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    Analysis of Statically Determinate Structures

    Idealised Structure

    Principle of Superposition

    Equations of Equilibrium

    Determinacy and Stability

    Beams

    Frames

    Gable Frames

    Application of the Equations of Equilibrium

    Analysis of Simple Diaphragm and Shear

    Wall Systems Problems

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    Statically Determinate Structures:

    Figure shows a rigid beam BD supported by two vertical wires BF andDG; the beam carries a force of 4W at C. We suppose the wires extend bynegligibly small amounts, so that the geometrical configuration of thestructure is practically unaffected; then for equilibrium the forces in thewires must be 3 W in BF and W in DG. As the forces in the wires areknown, it is a simple matter to calculate their extensions and hence todetermine the displacement of any point of the beam.

    The calculation of the forces in the wires and structure of Figure is said tobe statically determinate.

    If, however, the rigid beam be supported by three wires,with an additional wire, say, between H and J, thenthe forces in the three wires cannot be solved byconsidering statical equilibrium alone;such a structure is statically indeterminate.

    If the frame has just sufficient bars or rods to prevent collapse without the application of externalforces, it is said to be simply-stiff, when there are more bars or rods than this, the frame is saidto be redundant.

    Ifm be the total number of members and j is the total number ofjoints, we must have m = 2j 3, if the frame is to be simply-stiffor statically determinate.

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    A type of stress analysis problem in which internal stresses are not calculable on consideringstatical equilibrium alone; such problems are statically indeterminate. Consider the rigid beam BDof Figure which is supported on three wires; suppose the tensions in the wires are T1, T2 and T3.Then by resolving forces vertically, we have T 1 + T 2 + T 3 = 4W .. (1)and by taking moments about the point C, T 1 T 2 3 T 3 = 0 .. (2)From these equilibrium equations alone we cannot derive the values of the three tensile forces;a third equation is found by discussing the extensions of the wires or considering compatibility.If the wires extend by amounts e 1, e 2 , e 3 we must have from Figure (ii) thate 1 + e 3 = 2e 2 . (3) because the beam BD is rigid.

    Suppose the wires are all of the same material andcross-sectional area, and that they remain elastic.Then we may write , , .Where is a constant common to the three wires.

    Then equation (3) becomes T 1 + T 3 = 2 T 2 .. (4)The three equations (1), (2) and (4) then give

    Equation (3) is a condition which the extensions of the wires must satisfy.It is called a strain compatibility condition.

    Statically indeterminate problems are soluble ifstrain compatibilities are considered as well as staticalequilibrium.

    Statically Indeterminate Structures:

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    Static Indeterminacy

    The aim of structural analysis is to evaluate the external reactions, the deformed shape and

    internal stresses in the structure.

    If this can be accomplished by equations of equilibrium, then such structures are

    known as determinate structures.

    However, in many structures it is not possible to determine either reactions or internal

    stresses or both using equilibrium equations alone. Such structures are known as the statically

    indeterminate structures. The indeterminacy in a structure may be external, internal or both. A

    structure is said to be externally indeterminate if the number of reactions exceeds the number

    of equilibrium equations.

    Beams shown in Fig.(a) and (b) have four reaction components, whereas we have only

    3 equations of equilibrium. Hence the beams in Figs. (a) and (b) are externally indeterminate tothe first degree.

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    So far, we have determined the degree of indeterminacy by inspection. Such an

    approach runs into difficulty when the number of members in a structure increases.

    Hence, let us derive an algebraic expression for calculating degree of static

    indeterminacy. Consider a planar stable truss structure having m members and n joints. Let

    the number of unknown reaction components in the structure be r.

    Now, the total number of unknowns in the structure is (m + r).

    At each joint we could write two equilibrium equations for planar truss structure,

    Hence total number of equations that could be written is 2j.

    If then the structure is statically determinate as the number of

    unknowns are equal to the number of equations available to calculate them.

    The degree of indeterminacy may be calculated as

    Plane Frame

    Frame has 15 members, 12 joints and 9 reaction

    components. Hence, the degree of indeterminacy of

    the structure is

    Whereas for Plane Frameateach jointwe

    could write 3 equations ofequilibrium.

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    Kinematic IndeterminacyWhen the structure is loaded, the joints undergo displacements in the form of

    translations and rotations.

    In the displacement based analysis, these joint displacements are treated as unknown

    quantities. Consider a propped cantilever beam shown in Fig. (a).

    Usually, the axial rigidity of the beam is so high that the change in its length along axial

    direction may be neglected. The displacements at a fixed support are zero. Hence, for a propped

    cantilever beam we have to evaluate only rotation at B and this is known as the kinematic

    indeterminacy of the structure.

    A Fixed - Fixed beam is kinematically determinate but statically indeterminate to 3rd degree.A simply supported beam and a cantilever beam are kinematically indeterminate to 2nd degree.

    Pr til r

    til r

    Si ly Su rt

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    STIFFNESS

    Force-Displacement Relationship

    Consider linear elastic spring as shown in Fig. Let us do a simple experiment. Apply a force P at

    the end of spring and measure the deformation . Now increase the load to and measure the

    deformation u. Likewise repeat the experiment for different values of load P . Result may be

    represented in the form of a graph as shown in the above figure where load is shown on Y-axis

    and deformation on abscissa. The slope of this graph is known as the stiffness of the spring and

    is represented by k

    and is given by

    The inverse of the stiffness is known asflexibility. It is usually denoted by a andit has a unit of displacement per unit force.

    tiffness has a unit of force

    per unit elon ation.

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    Built -in (Encastre) beams are fixed at both ends. Continuous beams, which

    are beams with more than two supports and covering more than one span, are not

    statically determinate using the static equilibrium laws.

    Nomenclature

    = strain, = stress (N/m2

    ), E = Young's Modulus = /e (N/m2

    ),y = distance of surface from neutral surface (m), R = Radius of neutral axis (m).

    I = Moment of Inertia (m4 - more normally cm4)

    Z = section modulus = I/y max(m3 - more normally cm3)

    M = Moment (Nm), w = Distributed load on beam (kg/m) or (N/m as force units)

    W = total load on beam (kg ) or (N as force units),

    F= Concentrated force on beam (N)L = length of beam (m), x = distance along beam (m)

    Built in beams

    A built in beam is normally considered to be horizontal with both ends built-in at the same level

    and with zero slope at both ends. A loaded built in beam has a moment at both ends and

    normally the maximum moments at one or both of the two end joints.

    A built in beam is generally much stronger than a simply supported beam of the same

    geometry. The bending moment reduces along the beam and changes sign at points of contra

    flexure between the supports and the load.

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    As an example consider a simply supported beam subjected to a unit concentrated load at the

    centre. Now the deflection at the centre is given by

    The stiffness of a structure is defined as the force required for the unit deformation of the

    structure. Hence, the value of stiffness for the beam is equal to

    Second example, consider a cantilever beam subjected to a concentrated load (P) at its tip.

    Under the action of load, the beam deflects and from first principles the deflection below the

    load (u) may be calculated as,

    For a given beam of constant cross section, length L, Youngs modulus E, and moment of inertia

    Izz the deflection is directly proportional to the applied load. Eqn. may be written as

    Hence, the stiffness of the beam is

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    Beam Deflecti er iti :

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    Definition of Stress

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    Definition of StressThe concept ofstress originated from the study of strength and failure of solids. Thestress field is the distribution of internal "tractions" that balance a given set of externaltractions and body forces. First, we look at the external traction T that represents theforce per unit area acting at a given location on the body's surface. Traction T is a

    bound vector, which means T cannot slide along its line of action or translate to anotherlocation and keep the same meaning.

    In other words, a traction vector cannot be fully described unless both the force and thesurface where the force acts on has been specified. Given both Fand s, the tractionT can be defined as

    The internal traction within a solid or stress can be defined in a similar manner

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    The internal traction within a solid, or stress, can be defined in a similar manner.Suppose an arbitrary slice is made across the solid shown in the above figure, leading tothe free body diagram shown at right. Surface tractions would appear on the exposedsurface, similar in form to the external tractions applied to the body's exterior surface.The stress at point P can be defined using the same equation as was used for T.

    Stress therefore can be interpreted as internal tractions that act on a defined internaldatum plane. One cannot measure the stress without first specifying the datum plane.

    Surface tractions, or stresses acting on aninternal datum plane, are typicallydecomposed into three mutuallyorthogonal components.

    One component is normal to the surfaceand represents direct stress. The othertwo components are tangential to thesurface and representshear stresses.

    What is the distinction between normaland tangential tractions, or equivalently,direct and shear stresses?

    The Stress Tensor (or Stress Matrix)

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    Direct stresses tend to change the volume of the material (e g hydrostatic pressure)

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    Direct stresses tend to change the volume of the material (e.g. hydrostatic pressure)and are resisted by the body's bulk modulus (which depends on the Young's modulusand Poisson ratio). Shear stresses tend to deform the material without changing itsvolume, and are resisted by the body's shear modulus.

    For example, the stress state at point P can berepresented by an infinitesimal cube with threestress components on each of its six sides(one direct and two shear components).Since each point in the body is under staticequilibrium (no net force in the absence of any

    body forces), only nine stress components fromthree planes are needed to describe thestress state at a point P.These nine components can beOrganised into the matrix:

    where shear stresses across the diagonal are identical (i.e. xy = yx, yz= zy, andzx = xz) as a result of static equilibrium (no net moment). This grouping of the ninestress components is known as the stress tensor (or stress matrix).

    The subscript notation used for thenine stress components have the

    following meaning:

    Note: The stress state is a second order tensor since it is a quantity associated with

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    Note: The stress state is a second order tensor since it is a quantity associated withtwo directions. As a result, stress components have 2 subscripts.

    A surface traction is a first order tensor (i.e. vector) since it a quantity associated withonly one direction. Vector components therefore require only 1 subscript.

    Mass would be an example of a zero-order tensor (i.e. scalars), which have norelationships with directions (and no subscripts).

    Consider the static equilibrium of a solid subjected to the body force vector field b.Applying Newton's first law of motion results in the following set of differential equations

    which govern the stress distribution within the solid,

    In the case of two dimensional stress, the above equations reduce to,

    id d i h i i i l l h hi h i h dGl b l S i

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    Consider a rod with initial length L which is stretched to alength L'. The strain measure , a dimensionless ratio, isdefined as the ratio of elongation with respect to theoriginal length,

    Global 1D Strain

    Consider an arbitrary point in the bar P, which has aposition vector x, and its infinitesimal neighbour dx.Point P shifts to P', which has a position vector x',after the stretch. In the meantime, the small "step" dxis stretched to dx'.

    The strain at point p can be defined

    Since the displacement ,

    the strain can hence be rewritten as,

    Infinitesimal 1D Strain

    The above strain measure is defined in a global sense. Thestrain at each point may vary dramatically if the bar'selastic modulus or cross-sectional area changes.

    Tensile and compressive strains

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    Tensile and compressive strains

    In the steel wire experiment we discussed the extension of the whole wire.

    If we measure the extension of, say, the lowest quarter-length of the wire we find that

    for a given load it is equal to a quarter of the extension of the whole wire.

    In general we find that, at a given load, the ratio of the extension of any length to thatlength is constant for all parts of the wire; this ratio is known as the tensile strain.

    Suppose the initial unstrained length of the wire is Lo, and e is the extension due to

    straining; the tensile strain is defined as

    This definition of strain is useful only for small distortions, in which the extension e issmall compared with the original length Lo;this definition is adequate for the study of most engineering problems, where we are

    concerned with values of of the order 0.001, or so.

    If a material is compressed the resulting strain is defined in a similar way, except thate is the contraction of a length.

    We note thatstrain is a non-dimensional quantity, being the ratio ofthe extension, or contraction, of a bar to its original length.

    3D Strain Matrix

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    There are a total of 6strain measures whichcan be organised intoa matrix.

    3D Strain Matrix

    Engineering Shear Strain

    Focus on the strain xy for a moment. The expression inside the parentheses can berewritten as, where called the

    engineering shear strain, xy is a total measure of shear strain in the x-y plane.

    In contrast, the shear strain xy is the average of the shear strain on the x face alongthe y direction, and on the y face along the x direction.Engineering shear strain is commonly used in engineering reference books. However, pleasebeware of the difference between shear strain and engineering shear strain, so as to avoid errors inmathematical manipulations.

    Compatibility Conditions

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    In the strain-displacement relationships, there are six strain measures but only threeindependent displacements. That is, there are 6 unknowns for only 3 independentvariables. As a result there exist 3 constraint, or compatibility, equations. These

    compatibility conditions for infinitesimal strain referred to rectangular Cartesiancoordinates are,

    In two dimensional problems (e.g. plane strain), all z terms are set to zero. Thecompatibility equations reduce to,

    Note that some references use engineering shear strain ( ) whenreferencing compatibility equations.

    p y

    General Definition of 3D Strain

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    As in the one dimensional strainderivation, suppose that pointP in abody shifts to pointPafterdeformation. The infinitesimal strain-displacement relationships can besummarised as,

    where u is the displacement vector,x is coordinate, and the two indices iandj can range over the threecoordinates {1, 2, 3} in threedimensional space.Expanding the above equation for eachcoordinate direction gives,

    where u, v, and w are the displacements inthe x, y, and z directions respectively(i.e. they are the components ofu).

    Bulk Elastic Properties

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    pThe bulk elastic properties of a material determine how much it will compress under agiven amount of external pressure. The ratio of the change in pressure to the fractionalvolume compression is called the bulk modulus of the material.

    A representative value for the bulk modulus for steel is and that for water is

    The reciprocal of the bulk modulus is called the compressibility of the substance. Theamount of compression of solids and liquids is seen to be very small.

    The bulk modulus of a solid influences the speedof sound and other mechanical waves in thematerial. It also is a factor in the amount ofenergy stored in solid material.

    A common statement is that water is an incompressible fluid.

    This is not strictly true, as indicated by its finite bulk modulus,but the amount of compression is very small.

    At the bottom of the Pacific Ocean at a depth of about 4000meters, the pressure is about 4 x 107 N/m2. Even under thisenormous pressure, the fractional volume compression is onlyabout 1.8% and that for steel would be only about 0.025%.

    So it is fair to say that water is nearly incompressible.

    Young's Modulus

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    ou g s o u usIn solid mechanics, Young's modulus (E) is a measure of the stiffness of an isotropicelastic material. It is also known as the Young modulus, modulus of elasticity,elastic modulus (though Young's modulus is actually one of several elastic moduli such as the bulkmodulus and the shear modulus) or tensile modulus.

    It is defined as the ratio of the uni-axial stress over the uni-axial strain in the range ofstress in which Hooke's Law holds.

    Young's modulus is named after Thomas Young, the 18th century British scientist. However, the concept wasdeveloped in 1727 by Leonhard Euler.

    For the description of the elastic properties of linear objects like wires, rods, columnswhich are either stretched or compressed, a convenient parameter is the ratio of thestress to the strain, or Young's modulus of the material. Young's modulus can beused to predict the elongation or compression of an object as long as the stress is lessthan the yield strength of the material.This can be experimentally determined from the slope of a stress-strain curve created during tensile tests

    conducted on a sample of the material.

    Material

    e it

    (kg/ 3)

    Y g'

    Modulus109 N/

    Ulti ateStre gth

    Su106 N/

    YieldStre gth

    S106 N/

    Steela 7860 200 400 250

    Aluminum 2710 70 110 95

    a Structural steel

    shear modulus or modulus of rigidity, denoted by G, or sometimes S or , is defined as the

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    ratio of shear stress to the shear strain:

    Where = shear stress;F is the force which acts andA is the area on which the force acts

    = shear strain;x is the transverse displacement ,

    I is the initial lengthShear modulus is usually measured in GPa (giga pascal)

    The shear modulus is concerned with the deformation of a solid when it experiences a force parallel

    to one of its surfaces while its opposite face experiences an opposing force (such as friction). In thecase of an object that's shaped like a rectangular prism, it will deform into a parallelepiped.

    Anisotropic materials such as wood and paper exhibit differing material response to stress or strainwhen tested in different directions. In this case, when the deformation is small enough so that thedeformation is linear, the elastic moduli, including the shear modulus, will then be a tensor, ratherthan a single scalar value.

    In homogeneous and isotropic solids, there are two kinds of waves, pressure waves and shearwaves. The velocity of a shear wave, (vs) is controlled by the shear modulus,

    whereG is the shear modulus

    is the solid's density.

    shear modulus describes the material's response to shearing strains.

    Hooke's law

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    Hooke's law is named after the 17th century British physicist Robert Hooke.He first stated this law in 1676 as a Latin anagram, whose solution he published in 1678 as Ut tensio, sic vis,meaning: As the extension, so the force.

    Objects that quickly regain their original shape after being deformed by a force, with themolecules or atoms of their material returning to the initial state of stable equilibrium,often obey Hooke's law. We may view a rod of any elastic material as a linear spring.The rod has length L and cross-sectional area A. Its extension (strain) is linearlyproportional to its tensile stress by a constant factor, the inverse of its modulus ofelasticity E, hence, or

    Hooke's law only holds for some materials under certain loading conditions. Steel exhibits linear-elastic behaviorin most engineering applications; Hooke's law is valid for it throughout its elastic range (i.e., for stresses belowthe yield strength). For some other materials, such as aluminium, Hooke's law is only valid for a portion of theelastic range. For these materials a proportional limit stress is defined, below which the errors associated with the

    linear approximation are negligible.

    For systems that obey Hooke's law, the extension produced is directly proportional to

    the load: where:is the distance that the spring has been stretched or compressed away from the equilibriumposition, (meters),is the restoring force exerted by the material (Newtons), and

    is the force constant (or spring constant). The constant has units of force per unit length

    (newtons per meter). When this holds, we say that the behavior is linear.negative sign on the right hand side of the equation because the restoring force always acts in the opposite

    direction of the x displacement.

    Poisson's ratio ( ), named after Simeon Poisson, is the ratio of the relative contraction

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    strain, or transverse strain (normal to the applied load), to the relative extensionstrain, or axial strain (in the direction of the applied load).When a sample of material is stretched in one direction,it tends to contract (or rarely, expand) in the other two directions.Conversely, when a sample of material is compressed in one direction,it tends to expand (or rarely, contract) in the other two directions.Poisson's ratio ( ) is a measure of this tendency.The Poisson's ratio of a stable material cannot beless than 1.0 nor greater than 0.5 due to therequirement that the elastic modulus,the shear modulus and bulk modulus

    have positive values.Assuming that the material is compressed along the axial direction:

    where, is the resulting Poisson's ratio,is transverse strain (negative for axial tension, positive for axial compression)is axial strain (positive for axial tension, negative for axial compression).

    On the molecular level, Poissons effect is caused by slight movements between molecules and the stretching of

    molecular bonds within the material lattice to accommodate the stress. When the bonds elongate in the stressdirection, they shorten in the other directions. This behavior multiplied millions of times throughout the materiallattice is what drives the phenomenon.

    Most materials have between 0.0 and 0.5. Cork is close to 0.0, showing almost no Poisson contraction, moststeels are around 0.3, and rubber is nearly incompressible and so has a Poisson ratio of nearly 0.5. A perfectlyincompressible material deformed elastically at small strains would have a Poisson's ratio of exactly 0.5. Somematerials, mostly polymer foams, have a negative Poisson's ratio; if these auxetic materials are stretched in one

    direction, they become thicker in perpendicular directions.

    Generalised Hoo e's law : For an isotropic material, the deformation of a material in the

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    direction of one axis will produce a deformation of the material along the other axes in threedimensions. Thus it is possible to generalise Hooke's Law into three dimensions:

    Where , and are strain in the direction of x, y and z axisx , y and zare stress in the direction of x, y and z axisE is Young's modulus (the same in all directions: x, y and z for isotropic materials)

    is Poisson's ratio (the same in all directions: x, y and z for isotropic materials)

    Volumetric change: The relative change of volume V/Vdue to the stretch of the material can becalculated using a simplified formula (only for small deformations):

    Where V is material volumeV is material volume change

    L is original length, before stretch

    L is the change of length:L =Lnew Lold

    If a rod with diameter (or width, or thickness) d and length L is subject to tension so that its length will changeby L then its diameter d will change by (the value is negative, because the diameter will decrease with

    increasing length):

    The above formula is true only in the case of small deformations; if deformations are large then the following(more precise) formula can be used

    For Orthotropic material, such as wood in which Poisson's ratio is different in each direction (x, y and z axis) therelation between Young's modulus and Poisson's ratio is described as

    Ei is a Young's modulus along axis i and jk is a Poisson's ratio in plane jk

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    PLANE STATE of STRESS

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    Common engineering problems involving stresses in a thin plate or on the free surfaceof a structural element, such as the surfaces of thin-walled pressure vessels underexternal or internal pressure, the free surfaces of shafts in torsion and beams under

    transverse load, have one principal stress that is much smaller than the other two.

    By assuming that this small principal stress is zero, the three-dimensional stress statecan be reduced to two dimensions. Since the remaining two principal stresses lie in aplane, these simplified 2D problems are called plane stress problems.

    Assume that the negligible principal stressis oriented in the z-direction. To reduce the3D stress matrix to the 2D plane stress matrix,remove all components with z subscripts to get,

    Where xy = yx for static equilibrium.The sign convention for positive stress components

    in plane stress is illustrated in the above figure on the 2D element.

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    Principal Directions, Principal Stress

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    The normal stresses (x' and y') and the shear stress (x'y') vary smoothly with respectto the rotation angle , in accordance with the coordinate transformation equations.There exist a couple of particular angles where the stresses take on special values. First,

    there exists an angle p where the shear stress x'y'becomes zero. That angle is foundby setting x'y'to zero in the above shear transformation equation and solving for (setequal to p). The result is,

    The angle p defines the principal directions where the only stresses are normalstresses. These stresses are called principal stresses and are found from the originalstresses (expressed in the x,y,z directions) via,

    Coordinate Transformations

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    The coordinate directions chosen to analyse a structure are usually based on the shapeof the structure. As a result, the direct and shear stress components are associated withthese directions. For example, to analyse a bar one almost always directs one of the coordinate directions

    along the bar's axis.Stresses in directions that do not line up with the original coordinate set are alsoimportant. For example, the failure plane of a brittle shaft under torsion is often at a 45 angle with respect tothe shaft's axis. Stress transformation formulas are required to analyse these stresses.

    The transformation of stresses with respect to the {x,y,z} coordinates to the stresses

    with respect to {x',y',z'} is performed via the equations, where U is the rotation anglebetween the two coordinate sets (positive in the counterclockwise direction).

    The transformation to the principal directions can be illustrated as:

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    Maximum Shear Stress DirectionAnother important angle is where the maximum shear stress occurs This is found by finding the

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    Another important angle s is where the maximum shear stress occurs. This is found by finding themaximum of the shear stress transformation equation, and solving for . The result is,

    The maximum shear stress is equal to one-half the difference between the two principal stresses,

    The transformation to the maximum shear stress direction can be illustrated as:

    Mohr's Circle

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    Introduced by Otto Mohr in 1882, Mohr's Circle illustrates principal stresses and stresstransformations via a graphical format,

    The two principal stresses are shown in red, and the maximum shear stress is shown in orange.Recall that the normal stresses equal the principal stresses when the stress element is aligned with

    the principal directions, and the shear stress equals the maximum shear stress when the stresselement is rotated 45 away from the principal directions.

    As the stress element is rotated away from the principal (or maximum shear) directions, the normaland shear stress components will always lie on Mohr's Circle.

    Derivation of Mohr's Circle

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    To establish Mohr's Circle, we first recall the stress transformation formulas for planestress at a given location,

    Using a basic trigonometric relation (cos2 2 + sin2 2 = 1) to combine the two aboveequations we have,

    This is the equation of a circle, plotted on a graph where the abscissa is the normalstress and the ordinate is the shear stress. This is easier to see if we interpretx and yas being the two principal stresses, and xy as being the maximum shear stress. Thenwe can define the average stress, avg and a "radius" R (which is just equal to themaximum shear stress),

    The circle equation above now takes on a more familiar form,

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    The circle is centered at the average stress value, and has a radius R equal tothe maximum shear stress, as shown in the figure below,

    Principal Stresses from Mohr's Circle

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    A chief benefit of Mohr's circle is that the principal stresses 1 and 2 and themaximum shear stress max are obtained immediately after drawing the circle,

    where,

    Principal Directions from Mohr's Circle

    Mohr's Circle can be used to find the directions of the principal axes To show this first

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    Mohr s Circle can be used to find the directions of the principal axes. To show this, firstsuppose that the normal and shear stresses, x , y and xy are obtained at a given pointO in the body. They are expressed relative to the coordinates XY, as shown in the stresselement at right below.

    The Mohr's Circle for this general stress state is shown at left above. Note that it'scentered atavgand has a radius R, and that the two points {x , xy} and {y , -Txy} lieon opposites sides of the circle. The line connectingx and y will be defined as Lxy.

    The angle between the current axes (X and Y) and the principal axes is defined as

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    g ( ) p pp and is equal to one half the angle between the line Lxy and the -axis as shown inthe schematic below,

    Note that the coordinate rotation angle p is defined positive when starting at the XY coordinates

    and proceeding to the XpYp coordinates. In contrast, on the Mohr's Circle p is defined positivestarting on the principal stress line (i.e. the -axis) and proceeding to the XY stress line (i.e. lineLxy). The angle p has the opposite sense between the two figures, because on one it starts on the

    XYcoordinates, and on the other it starts on the principal coordinates.

    Rotation Angle on Mohr's Circle

    To do this we proceed as follows:

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    Step 1: Draw Mohr's circle for the given stress state ( x , y and xy).

    Step 2: Draw the line Lxy across the circle from (x , xy) to (y , - xy).

    Step 3: Rotate the line Lxy by 2* (twice as much as the angle between XYand X'Y')and in the opposite direction of.

    Step 4: The stresses in the new coordinates (x' , y' and x'y) are then read offthe circle.

    Mohr's Circle For PLANE STRAIN

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    Strains at a point in the body can be illustrated by Mohr's Circle. The idea andprocedures are exactly the same as for Mohr's Circle for plane stress.

    The two principal strains are shown in red, andthe maximum shear strain is shown in orange.Recall that the normal strains are equal to theprincipal strains when the element is alignedwith the principal directions, and the shear

    strain is equal to the maximum shear strainwhen the element is rotated 45 away from theprincipal directions.

    As the element is rotated away from theprincipal (or maximum strain) directions, the

    normal and shear strain components willalways lie on Mohr's Circle.

    Derivation of Mohr's Circle

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    To establish the Mohr's circle, we first recall the strain transformation formulas for planestrain,

    Using a basic trigonometric relation (cos22+ sin22 = 1) to combine the above twoformulas we have,

    This equation is an equation for a circle. To make this more apparent, we canrewrite it as,

    where,

    The circle is centered at the average strain valueAvg and has a radius R equal to the maximumshear strain, as shown in the figure.

    Principal Strains from Mohr's Circle

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    A chief benefit of Mohr's circle is that the principal strains 1 and 2 and the maximumshear strain xyMax are obtained immediatelyafter drawing the circle,

    where,

    Principal Directions from Mohr's Circle

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    Mohr's Circle can be used to find the directions of the principal axes. To show this, firstsuppose that the normal and shear strains, x , y and xy are obtained at a given pointO in the body. They are expressed relative to the coordinates XY, as shown in the strain

    element at right below.

    The Mohr's Circle for this general strain state is shown atleft above. Note that it's centered atAvgand has a radiusR, and that the two points (x , xy) and (y , - xy) lie onopposites sides of the circle. The line connecting x and ywill be defined as Lxy.

    The angle between the current axes (X and Y) and the principal axes is defined as pd i l h lf h l b h li d h i h

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    p

    and is equal to one half the angle between the line Lxy and the -axis as shown

    Note that the coordinate rotation anglep is defined positive when starting atthe X Y coordinates and proceeding tothe XpYp coordinates. In contrast, on theMohr's Circle

    pis defined positive

    starting on the principal strain line (i.e.the -axis) and proceeding to the X Ystrain line (i.e. line Lxy). The angle phas the opposite sense between the twofigures, because on one it starts on theXY coordinates, and on the other it starts on the principal coordinates.

    Rotation Angle on Mohr's Circle

    Strain Transform by Mohr's Circle

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    Mohr's Circle can be used to transform strains from one coordinate set to another,similar that that described on the plane strain page. Suppose that the normal and shearstrains, x , y and xy are obtained at a pointO in the body, expressed with respect to

    the coordinates XY. We wish to find the strains expressed in the new coordinate set X'Y',rotated an angle from XY, as shown below:

    To do this we proceed as follows:

    St 1 D M h ' i l f th i t i t t ( d h b l )

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    Step 1: Draw Mohr's circle for the given strain state (x , y and xy shown below).

    Step 2: Draw the line Lxy across the circle from (x, xy) to (y, - xy).

    Step 3: Rotate the line Lxy by2* (twice as much as the angle between XYand X'Y')and in the opposite direction of.

    Step 4: The strains in the new coordinates (x' , y'and x'y) are then read.

    Beams:

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    Consider first the simple case of a beam which is fixed rigidly at one end B and is quite free atits remote end D, Figure 1 ; such a beam is called a cantilever. Imagine that the cantilever ishorizontal, with one end B embedded in a wall, and that a lateral force W is applied at the free endD. Suppose the cantilever is divided into two lengths by an imaginary section C; the lengths BCand CD must individually be in a state ofstatical equilibrium. If we neglect the mass of the

    cantilever itself, the loading actions over the section C ofCD balance the actions of the force W atC. The length CD of the cantilever is in equilibrium if we apply an upwards vertical force F and ananti-clockwise couple M atC; F is equal in magnitude to W, and M is equal to W(L - z),where z ismeasured from B. The force F atC is called a shearing force, and the couple M is a bendingmoment. But at the imaginary section Cof the cantilever, the actions F and M on CD are provided

    by the length BC of the cantilever. In fact, equal and opposite actions F and M are applied by CDto BC. For the length BC, the actions atCare a downwards shearing force F, and a clockwise

    couple M.

    When the cantilever carries external loads which are not appliednormally to the axis of the beam, Figure 2, axial forces are set up inth b If W i i li d t l t th i f th b th

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    the beam. IfW is inclined at an angle to the axis of the beam theaxial thrust in the beam at any section is

    The bending moment and shearing force at a section a distance zfrom the built-in end are

    Relation between the intensity of loading, SF and BM:

    Consider a straight beam under any system of lateral loads andexternal couples, Figure 3; an element length z of the beam at a

    distance z from one end is acted upon by an external lateral load,and internal bending moments and shearing forces.

    Suppose external lateral loads are distributed so that the intensity ofloading on the elemental length z is w.

    Then the external vertical force on the element is w z ; this is reacted by an internal bendingmomentM and shearing force F on one face of the element, and M +M and F +F on

    th th f f th l t

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    the other face of the element.For vertical equilibrium of the element we have,

    If z is infinitesimally small,

    Suppose this relation is integrated between the limits z1 and z2 then

    IfF1 and F2 are the shearing forces atz = z1 and z = z2 respectively, then

    Furthermore, for rotational equilibrium ofthe elemental length z,

    neglecting higher orders Then, in the limit as zapproaches zero,

    On integrating between the limits z = z1 and z2 we have

    Thus

    The shearing force F at a section distance z from one end of the beam is

    On substituting this value of F into equation,

    From equations & we have that the bending moment M has a stationaryvalue when the shearing force F is zero. Thus, we get

    Relations developed are merely statements of statical equilibrium, are

    therefore true independently of the state of the material of the beam.

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    A cantilever 5 m long carries a uniformly distributed vertical load 480 N per metre from C from H,and a concentrated vertical load of 1000 N at its mid-length,D. Construct the SFD and BMD.

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    The SF due to the distributed load increases uniformly fromzero at H to + 1920 N at C, and remains constant at+1920N from C to B; this is shown by the lines (i). Due to theconcentrated load at D, the shearing force is zero from H to

    D, and equal to +1000 N from D to B, as shown by lines (ii).Adding the two together we get the total shearing forceshown by lines (iii).

    The bending moment due to the distributed load increasesparabolically from zero at H to

    at C. The total load on CH is 1920 N with its centre ofgravity 3 m from B; thus the bending moment at B due tothis load is

    From C to B the bending moment increases uniformly,giving lines (i). The bending moment due to theconcentrated load increases uniformly from zero at D to

    at B, as shown by lines (ii). Combining (i) and (ii), the totalbending moment is given by (iii).

    The method used here for determining shearing-force and bending-moment diagrams is known as the principle of superposition.

    m

    1. Cantilever beam with end moment, m

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    m

    L

    x

    m

    x

    ( )zM x

    x=L-x

    ( )yV xcu beam and findin e na forcesandmomen

    ( )P xm

    x

    ( ) 0yV x !

    ( )z M x m!

    x

    L

    zM

    Sum forces in x and y direction

    to obtain P(x)=0 and

    Sum moments at any point x

    to obtain( )z M x m!

    2. Cantilever beam with end load, F. Sum forces in y direction to find

    ( )V F

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    L

    x

    x

    ( )zM x

    x=L x( )yV xcu b m nd nd

    n rn orces andmomen

    ( )P x

    x

    F

    F

    F

    0 ( ) ( )zM M x F L x! !

    ( ) ( )z M x F L x!

    or

    ( ) ( )z M x F L x!

    L

    FL( )yV x F!

    zM

    yV

    F

    x

    L

    x

    The shear and moment diagrams:

    ( )yV x F!

    0p

    3. Cantilever beam with uniform distributed load, po

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    x

    ( )zM

    x=L-x

    ( )yV xcut m ii ternal rcesandm ment

    ( )P x

    x

    L

    x

    0p

    0p 0p

    Sum forces in y direction to obtain

    0( ) ( )yV x p L x!

    L

    zMyV

    x

    L

    x

    0( ) ( )y x p x L! 0p L 20( ) ( ) /2zM x p L x!

    2/ 2p L

    Sum moments at any point x to obtain

    00 ( ) [ ( )]( ) / 2zM M x L x L x! !

    2

    0( ) ( ) /2z M x p L x!

    or

    Shear and bending moment diagrams :

    3a. Cantilever beam with uniform distributed load, po. Use the integration

    method to obtain V&M diagrams.

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    L

    x

    0p0

    yy

    Vp p

    x

    x

    x! !

    We t e earat =L i , let i tegratefr m =L t any int :

    0

    x x

    yL LdV dx!

    ( ) ( )y yV x V L0

    0 0 ( )x

    L p p x L! !

    0( ) ( )yV x p L x!

    Next, integrate the shear equation to obtain the moment equation.

    0 ( )

    z

    y

    M

    V pL

    xx

    x

    x

    We now t emoment i zeroat t e freeend,

    =L, o integrate from =L toany oint :

    0( )x x

    zL LdM p L x dx! ( ) ( )z zx L

    0 2

    0

    2 2 2

    0 0

    ( / 2)

    ( / 2 / 2) ( ) / 2

    x

    L p Lx x

    p Lx x L p x L

    !

    ! !

    4. Simply Supported Beam with Point oad

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    From equilibrium: R1 + R2 = P

    5. Simply Supported Beam with istributed Normal oad

    po

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    )

    )

    '= -

    R RR

    NO MOMENTS AT ENDS

    WITHSIMPLE SUPPORTS

    o

    po

    po

    From equilibrium: R = R = po /2.

    M(x)

    xL

    V(x)

    xL

    L/ p

    o

    M(x)=(po/2) x (x-L)

    ds

    d1 U

    VO !!

    Small deflection and

    slope dx

    d

    ds

    d1 UU

    VO }!!

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    dsV slope dxdsV

    V

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    We assume that the predominate deflection is normal to the x axis. Predominate strainis in the axial (x) direction. Assume small strain and rotations of beam so that axialt i li l ti l ti i l

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    strains vary linearly over cross-section or plane sections remain plane.In terms of the general elasticity problem, the above can be stated as1. If the cross-sectional dimensions are small compared to the beam length, then

    applied transverse tractions (in y and z directions) will be small compared to theresultant internal stress in the x direction. --> small transverse loads producelarge axial stresses. Thus we assume that the only major stress is Txx (all otherstresses are zero or negligible).

    0 0

    [ ] 0 0 0

    0 0 0

    Txx

    T !

    The stress tensor reduces to

    .

    Equilibrium Conservation of Linear Momentum reduces to

    This implies that Txx=Txx(y, ) for any point x.

    0

    Txx

    x

    x

    x

    !

    2. Stress-Strain. We assume a linear isotropic material so that stress and strain arelinearly related to each other: Txx = E Ixx and Iyy = Izz= - R Ixx = - (R/E) Txx.If Txx=Txx(y,z), then Ixx=Ixx(y,z) also.

    3. Strain-Displacement uxxxx

    x

    Ix

    !

    4. Kinematic assumptions: All deformation is described by the displacement,( )u xo

    and rotation of the centroidal axis.Vertical displacement is function of x only:

    ( , ) ( )

    ( , )

    u x y u x y oy

    duoy

    u x y yx dx

    !

    !

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    TYPES of BENDING:Pure or circular bending: This type of bending occurs when the only internal force in the bar is aconstant bending moment, i.e., the axial (N) and shear (V ) forces and the torsional moment (T)

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    constant bending moment, i.e., the axial (N) and shear (V ) forces and the torsional moment (T)are zero. The designation of circular bending is suggested by the fact that the deformed axis of theinitially prismatic bar is an arc of circumference, when the bending moment is constant (a constantmoment implies a constant curvature).

    Non-uniform bending: This designation is normally used for a loading causing bending momentand shear force, that is, for a non-constant bending moment. The axial force and the torsionalmoment are zero.Composed bending: This designation is used for a loading causing bending moment and axialforce. Constant(circular composed bending):

    Variable (non uniform composed bending):

    Each of these three types of bending may besub-divided into plane and inclined bending.

    An elementary bending problem is that of a rectangular beam under end couples.

    Consider a straight uniform beam having a rectangular cross-section of breadth b and depth h,

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    Co s de a s a g u o bea a g a ec a gu a c oss sec o o b ead b a d dep ,figure a, the axes of symmetry of the cross-section are Cx ,Cy. A long length of the beam is bentin the yz-plane, Figure b, in such a way that the longitudinal centroidal axis, Cz, remainsunstretched and takes up a curve of uniform radius of curvature, R.

    We consider an elemental length z of the beam, remote from the ends; in the unloadedcondition,AB and FD are transverse sections at the ends of the elemental length, and thesesections are initially parallel. In the bent form we assume that planes such as AB and FD remain flatplanes;ABand F Din Figure b are therefore cross-sections of the bent beam, but are nolonger parallel to each other.

    In the bent form, some of the longitudinal

    fibres, such as A F ; are stretched, whereasothers, such as B D are compressed.The unstrained middle surface of the beam isknown as the neutral axis.

    Now consider an elemental fibre HJ of the beam, parallel to the longitudinal axis Cz, Figure c;this fibre is at a distance y from the neutral surface and on the tension side of the beam.The original length of the fibre HJ in the unstrained beam is z; the strained length is

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    g g J ; g

    because the angle betweenAB and F Din Figure b and c is

    (zR). Then during bending HJ stretches an amount

    The longitudinal strain of the fibre HJ is therefore

    Then the longitudinal strain at any fibre is proportional to thedistance of that fibre from the neutral surface; over thecompressed fibres, on the lower side of the beam, the strains areof course negative.If the material of the beam remains elastic during bending then

    the longitudinal stress on the fibre HJ is.. 1

    The distribution of longitudinal stresses over the cross-section takes the form shown in Figure d;because of the symmetrical distribution of these stresses aboutCx, there is no resultantlongitudinal thrust on the cross-section of the beam. The resultant hogging moment is

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    g gg g

    .. 2

    On substituting for from equation (1), we have

    .. 3

    where Ix is the second moment of area of the

    cross-section aboutCx. From equations (1) and(3), we have.. 4

    Equation (3) implies a linear relationship between M, the applied moment, and (l/R), the curvature of the beam.

    The constant EI, in this linear relationship is called the bending stiffness or flexural stiffness of the beam;E I is also known as Flexural Rigidity.

    Bending of a beam about a principal axisWe considered the bending of a straight beam of rectangular cross-section; this formof cross-section has two axes of symmetry. More generally we are concerned with sections having

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    y y g y gonly one, or no, axis of symmetry.Consider a long straight uniform beam having any cross-sectional form; the axes Cx and Cy areprincipal axes of the cross-section. The principal axes of a cross-section are those

    centroid axes for which the product second moments of area are zero. In Figure a, C is thecentroidal of the cross-section; Cz is the longitudinal centroidal axis.

    When end couples M are applied to thebeam, we assume as before thattransverse sections of the beam remainplane during bending. Suppose further

    that, if the beam is bent in the yz-planeonly, there is a neutral axis C ' x ;Figure a, which is parallel to Cx and isunstrained; radius of curvature of thisneutral surface is R, Figure b.

    As before, the strain in a longitudinal fibre at a distance y from Cx is

    If the material of the beam remains elastic during bendingthe longitudinal stress on this fibre is

    If there is to be no resultant longitudinal thrust on the beam at anytransverse section we must have

    Where b is the breadth of an elemental strip of the cross-section parallel to Cx, and the integrationis performed over the whole cross-sectional area, A. But

    This can be zero only ifC ' xis a centroidal axis; now, Cx is a principal axis, and is therefore acentroidal axis, so thatCx and Cx are coincident, and the neutral axis is Cx in any cross-sectionof the beam. The total moment aboutCx of the internal stresses is

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    .. 5

    The stress in any fibre a distance y from Cx is.. 6

    No moment about Cy is implied by this stress system, for

    because Cx and Cy are principal axes for which , or the product second moment of area,is zero; A is an element of area of the cross-section.

    Beams having two axes of symmetry in the cross-section:

    I-section

    beam

    Solid circularcross-section

    Hollow circularcross-section.

    For bending about the axis Cx

    Similarly for bending by a couple My aboutCy,

    From above equations (7) and (8) we see that thegreatest bending stresses occur in the extreme

    longitudinal fibres of the beams.

    .. 7

    .. 8

    A steel bar of rectangular cross-section, 10 cm deep and 5 cm wide, is bent in the planes of thelonger sides. Estimate the greatest allowable bending moment if the bending stresses are not toexceed 150 MN/m2 in tension and compression.

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    The bending moment is applied about Cx. The second moment of areaabout this axis is

    The bending stress at a fibre a distance y from Cx is,

    where M is the applied moment. If the greatest stresses are notto exceed 150 MN/m2, we must have

    The greatest bending stresses occur in the extreme fibres where y = 5 cm.Then

    The greatest allowable bending moment is therefore 12 500 Nm.

    The second moment of area aboutCy is

    The greatest allowable bending moment aboutCy is

    which is only half that aboutCx.

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    Normal ( x-direction) Sresses in Beams of linearly elastic material(a) Normal stress distribution

    (b) z-axis: neutral axis

    Max. normal stress at z= c1 & c2

    zzx

    I

    yM!W

    yM 11S

    M

    I

    cM!!W

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    zzx

    I

    y!W

    11

    SIzz

    2

    22

    S

    M

    I

    cM

    zz

    !!

    1

    11

    S

    M

    I

    cM

    zz

    !!W

    2

    22

    S

    M

    I

    cM

    zz

    W

    S1 and S2 are section

    moduli

    ii

    c

    I!

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    Relative sliding oftwo separate beams

    due to shear

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    2/h 2/hy

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    !!

    !

    2/hy

    yy

    11

    1

    dAF W !!

    !

    2/hy

    yy

    22

    1

    dAF W

    zz1

    I

    yM!W

    zz2

    I

    ydMM

    xx!W

    0!X

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    1F 2F

    !!

    !

    2/hy

    yy 1

    ydAQ

    dAFFdxb 2/hyyy

    1212

    1

    !! !!

    WWX

    zzbIQV!X

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    Distribution of shear stresses in a

    beam of rectangular cross section:

    (a) cross section of beam, and (b)

    diagram showing the parabolicdistribution of shear stresses over

    the height of the beam.

    The first step in determining the shear stress at any location is to look at a section in a small slicefrom the beam. Summing the forces due to the normal bending stresses in the horizontal direction

    Horizontal Shear Stress:

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    givesFx = 0 - P+ (P+ dP) + b dx = 0 dP/dx = b (1)

    where b is the beam depth at the location of the shear stress being calculated.

    P can be found by integrating the normal stress over a sectionA(shown in diagram), giving

    But the bending stress is b = My/I. Substituting and simplifying gives

    Since M and I do not change, they can be moved outside the integral.

    The integral now is just the first moment of the area that is commonly used to find the centroid ofan area, and is called "Q". Substituting P into the equation (1) and using Q, gives

    Recall, the derivative, dM/dx is equal to the vertical shear load V.This gives,

    The final horizontal shear stress equation is

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    Shear stress acts on two different parallel surfaces of any element as shown in the diagram. Oneside cannot be under a different shear stress magnitude than the other. If a small element is takenf d h ll l id ill h h l di i h i di i

    Vertical Shear Stress:

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    from a structure under shear, parallel sides will have shear stress loading in the opposite direction,causing it to shear as shown in the diagram. Notice, the other two sides try to resist the slidingmotion, and the stress element stays in equilibrium.

    Shear Stress on Element

    Horizontal and Vertical Shear Stressat the Same Location in a Beam

    Similarly, a small element taken from a beam under a shear loading willhave equal shear stresses in the vertical and horizontal directions as shownin the diagram at the left. The magnitude of the shear stress will depend onthe location of the stress element.

    There are three possible shear stresses on a three dimensional cube. Thissection has only examined one dimension since shear loading in beams isgenerally only in one direction. But just like uni-axial loading, shear loadingcan be in three directions.

    Rectangular beams are so common, it is helpful to plot the shear stressfrom top to bottom. The resulting equation is

    Rectangular Beams

    Shear Stress Distribution in Rectangular Beam

    It is a parabolic shape with the maximumat the center. The center shear stress is

    max =1.5 V/A

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    Axial load associated with

    Inclined end load

    Stress resultants N, V, ;and M

    Acting on a cross section at

    distance x

    Tensile stress due to axial

    force NTensile and compressive stress due to

    bending moment M

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    A

    N

    !W I

    My!W

    Possible stress distributions due to combined

    N and M

    I

    My

    A

    N!W

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    Eccentric axial end load

    Axial load and associated

    bending

    Shift of the neutral axis

    I

    Pey

    A

    P!W

    Ae

    Iyo !

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    ` For beams with axial force, bending iscalculated without consideration of the

    deflections Original configuration is used

    ` Valid for small deflections Stocky beams

    x Length to height ratio < 10

    ` Bending moment due axial force Buckling of columns

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    DEFLECTION of BEAMS:The loading actions at any section of a simply-supported beam or cantilever can be resolved into abending moment and a shearing force. There are ways of estimating the stresses due to thesebending moments and shearing forces There is however another aspect of the problem of

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    bending moments and shearing forces. There is, however, another aspect of the problem ofbending which remains to be treated, namely, the calculation of the stiffnessof a beam. In mostpractical cases, it is necessary that a beam should be not only strong enough for its purpose, but

    also that it should have the requisite stiffness, that is, it should not deflect from its original positionby more than a certain amount.

    Again, there are certain types of beams, such as those carried by more than two supports andbeams with their ends held in such a way that they must keep their original directions, for whichwe cannot calculate bending moments and shearing forces without studying the deformations ofthe axis of the beam; these problems are in factstatically indeterminate. In this module we

    consider methods of finding the deflected form of a beam under a givensystem of external loads and having known conditions of support.

    Elastic bending of straightbeams:From Beam Equation a straight beam of uniform cross-section, when subjected to end couples Mapplied about a principal axis, bends into a circular arc of radius R, given bywhere EI, the product of Young's modulus E and the second moment (1)

    of area I about the relevant principal axis, is the flexural stiffnessofthe beam; equation (1) holds only for elastic bending.

    Where a beam is subjected to shearing forces, as well as bending moments, the axis of the beamis no longer bent to a circular arc. To deal with this type of problem, we assume that equation stilldefines the radius of curvature at any point of the beam where the bending moment isM. This implies that where the bendingmomentvariesfrom one sectionof the beam to

    another, the radiusof curvature also varies from section to section, in accordance with eqn.(1)

    Differential Equation of elastic curve:In the unstrained condition of the beam, Cz is the longitudinal centroidal axis, Figure.1, and Cx,Cyare the principal axes in the cross-section. The co-ordinate axes Cx,Cy are so arranged that the y-axis is vertically downwards. (This is convenient as most practical loading conditions give rise to vertically

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    y ( p g g ydownwards deflections.)Suppose bending moments are applied about axes parallel to Cx, so that bending is restricted to

    the yz-plane, because Cx and Cy are principal axes.

    Fig.2 Displacements of the longitudinal

    axis of the beam.

    Fig.1 Longitudinal and principal centroidalaxes for a straight beam.

    Consider a short length of the unstrained beam,corresponding with DF on the axis Cz, Figure 2.In the strained condition D and F are displaced to D'and F', respectively, which lies in the yz plane.Any point such as D on the axis Cz is displaced by an

    amountv parallel to Cy; it is also displaced a small,but negligible, amount parallel to Cz.The radius of curvature R at any section of the beamis then given by

    We are concerned generally with only smalldeflections, in which v is small; this implies that(dv/dz) is small, and that (dv/dz)2 is negligiblecompared with unity.

    (2)

    (3)

    Sign Convention: we have already adopted the convention that sagging bending moments are

    Hence from 1 and 3(4)

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    positive. When a length of the beam is subjected to sagging bending moments, as in Figure 3, thevalue of(dv/dz) along the length diminishes as z increases; hence a sagging moment implies that

    the curvature is negative. Then (5) where M is the sagging bending moment.

    Where the beam is loaded on its axis of shear centres, so that no twisting occurs, M may be writtenin terms of shearing force F and intensity w of vertical loading at any section.

    On substituting for M (6)

    This relation is true ifEI varies from one section of a beamto another. Where El is constant along the length of a beam,

    (7)

    Fig 3.Deflected form of

    a beam in pure bending.

    As an example of the use of equation (4), consider the caseof a uniform beam carrying couples M at its ends, Fig.3. Thebending moment at any section is M, so the beam is under aconstant bending moment. Equation (5) gives

    On integrating once, we have

    (8) where A is a constant.

    On integrating once more (9)

    where B is another constant. If we measure v relative to a line CD joining the ends of the beam,v is zero at each end. Then v = 0, for z = 0 and z = L.

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    v is zero at each end. Then v 0, for z 0 and z L.On substituting these two conditions into equation (9), we have and

    Therefore

    At the mid-length, z = , and

    (1

    0)

    (11) which is the greatest deflection.

    (12)

    It is important to appreciate that equation (3), expressing the radius of curvature R in terms ofv,is only true if the displacementv is small.

    Fig.4 Distortion of a beam in pure bending.

    We can study more accurately the pure bending of a beam byconsidering it to be deformed into the arc of a circle, Figure 4; asthe bending momentM is constant at all sections of the beam,the radius of curvature R is the same for all sections.

    IfL is the length between the ends, and D is the mid-point,

    Thus the central deflection v is

    Suppose L/R is considerably less than unity; then

    which can be written

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    But

    Clearly, if is negligible compared with unity we have, approximately,

    which agrees with equation (11). The more accurate equation (13) shows that, when is notnegligible, the relationship between v and M is non-linear; for all practical purposes thisrefinement is unimportant, and we find simple linear relationships of the type of equation (11)are sufficiently accurate for engineering purposes.

    (13)

    Simply-supported beam carrying a uniformly distributed load

    A beam of uniform flexural stiffness EI and span L is simply-supported at its ends, Figure 5;it carries a uniformly distributed lateral load of w per unit length, which induces bending in the yz

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    it carries a uniformly distributed lateral load ofw per unit length, which induces bending in the yzplane only. Then the reactions at the ends are each equal to wL; ifz is measured from the endC, the bending moment at a distance z from C is

    Fig. 5 Simply-supported beamcarrying a uniformly supported load.

    On integrating twice,

    From equation (5)

    Then equation (14) becomes

    The deflection at the mid-length, z = L , is

    (14)

    (15)

    (16)

    Cantilever with a concentrated load

    A uniform cantilever of flexural stiffness El and length L carries a vertical concentrated load W atthe free end, Figure 6. The bending moment a distance z from the built-in end is

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    Fig. 6 Cantilever carrying a verticalload at the remote end.

    At the endz

    = 0, there iszero slope in the deflectedform, so that dv/dz = 0; then equation (17) gives

    A = 0. Furthermore, atz = 0 there is alsono deflection, so that B = 0. Then

    (17)

    From equation (5)

    On integrating twice,

    At the free end, z = L (18)

    The slope of the beam at the free end is(19)

    When the cantilever is loaded at some point between the ends, at a distance a , say, from thebuilt-in support, Figure 7, the beam between G and D carries no bending moments and thereforeremains straight. The deflection atG can be deduced from equation (18); for z = a,

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    (21)

    (20)

    and the slope at z = a is

    Then the deflection at the free end D of the cantilever is

    Fig. 7 Cantilever with a loadapplied between the ends.

    (22)

    Cantilever with a uniformly distributed load

    A uniform cantilever, Figure 8, carries a uniformly distributed load ofw per unit length over thewhole of its length. The bending moment at a distance z from C is

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    Fig. 8 Cantilever carryinga uniformly distributed load.

    At the built end, z = 0, and we have

    From equation (5)

    On integrating twice,

    Thus A = B = 0. Then

    At the free end, D, the vertical deflection is(23)

    Propped cantilever with distributed load

    The uniform cantilever of Figure 9 (i) carries a uniformly distributed load w and is supportedon a rigid knife edge at the end D. Suppose P is the force on the support atD. Then we regard

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    Figure (i) as the superposition of the effects ofP and w acting separately.

    Fi 9 (i) Uniformly loaded cantileverpropped at one end.(ii) Deflections due to w alone.(iii) Deflections due to alone.

    Ifw acts alone, the deflection atD is given byequation (23), and has the value

    If the reaction P acted alone, there would be anupward deflection

    atD. If the support maintains zero deflection atD,

    This gives

    or (24)

    Simply-supported beam carryin a concentrated lateral loadConsider a beam of uniform flexural stiffness EI and length L, which is simply-supported at itsends C and G, Figure 13.1 1. The beam carries a concentrated lateral load W at a distance a fromC. Then the reactions at C and G are

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    Fig 10. Deflections of a simply-supportedBeam carrying a concentrated lateral load.

    Now consider a section of the beam a distance z from C;

    ifz < a, the bending moment at the section is M = Vczand if z > a, M = Vcz W (z -a)

    In these equationsA,B, A'and B' are arbitrary constants. Now for z = a the values ofv given byequations (27) and(28) are equal, and the slopes given by equations (25) and (26) are also equal,as there is continuity of the deflected form of the beam through the pointD. Then

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    At the extreme ends of the beam v = 0, so that when z = 0equation (27) gives B = 0, and when z =L, equation (28)gives

    Then equations (27) and (28) may be written

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    The second relation, for z > a, may be written

    Then equations (32) and (33) differ only by the last term of equation (34); if the last term ofequation (34) is discarded when z < a, then equation (34) may be used to define the deflected

    form in all parts of the beam. On putting z = a, the deflection at the loaded pointD is

    When W is at the centre of the beam, a = L/2 , and

    This is the maximum deflection of the beam only when a = L.

    A steel rod 5 cm diameter protrudes 2 m horizontally from a wall. (i) Calculate the deflection due toa load of 1 kN hung on the end of the rod. The weight of the rod may be neglected. (ii) If a verticalsteel wire 3 m long, 0.25 cm diameter, supports the end of the cantilever, being taut butunstressedb f th l d i li d l l t th d d fl ti li ti f th l d

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    before the load is applied, calculate the end deflection on application of the load.Take E = 200GN/m2.

    The deflection at the end is then

    If this equals the stretching of the wire, then

    This gives T = 934 N, and the deflection of the cantilever becomes

    The shear load and bending moment diagrams

    are constructed by integrating the distributed

    load to get the shear diagram

    Drawing shear force and bending moment diagrams

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    load to get the shear diagram

    (adding jumps at all point loads), and

    integrating the shear diagram to get the bending

    moment (adding jumps at all point couples).

    Example of one shear load and bending moment diagram.

    1. First draw the free-body-diagram of the beam

    with sufficient room under it for the shear and

    moment diagrams (if needed, solve for support

    reactions first).

    2. Draw the shear diagram under the free-body-

    diagram. The distributed load is the slope of the shear

    diagram and each point load represents a jump in the

    shear diagram. Label all the loads on the shear

    diagram

    3. Draw the moment diagram below the shear

    diagram. The shear load is the slope of the moment

    and point moments result in jumps in the moment

    diagram. The area under the shear diagram equals the

    change in moment over the segment considered (up

    to any jumps due to point moments). Label the value

    of the moment at all important points on the moment

    diagram.

    For the beam and loading shown, (a) draw

    the shear and bending moment diagrams,

    (b) determine the maximum absolute

    values of the shear and bending moment.

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    g

    (a) FBD Beam:

    Along AB:

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    (b) From diagrams:

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    Slope and Displacement by the Moment-Area Method:The moment-area method provides a semi graphical technique for finding the slope anddisplacement at specific points on the elastic curve of a beam or shaft. Application of

    the method requires computing areas associated with the beams moment diagram; so if

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    heorem I: The angle between the tangents at any two points on the elastic curveequals the area under the M/EI diagram between these points.

    A/B = (M/EI) dxIfAand B are two points on a beam the change in angle (radians) between the tangent at A and

    the tangent at B is equalto thearea of the bendingmoment diagram between the pointsdivided by therelevant value of EI(the flexural rigidity constant).

    heorem II: The vertical deviation of the tangent at a point (A) on the elastic curve withrespect to the tangent extended from another point (B) equals the moment of the area

    under the M/EIdiagram between these two points (A and B). This moment is computed

    about point (A) where the vertical deviation (tA/B) is to be determined.IfA and B are two points on a beam the displacement ofB relative to the tangent of the beam at Ais equal to the moment of thearea of the bendingmoment diagram betweenAandBaboutthe ordinate through B divided by therelevant value of EI(the flexural rigidity constant).

    tA/B = x(M/EI)dx or tA/B =x (M/EI) dx

    this diagram consists of simple shapes, the method is very convenient to use. Normally

    this is the case when the beam is loaded with concentrated forces and couple moments.

    Determine the deflection and slope of a cantilever as shown.

    The bending moment at A = MA = WL

    The area of the bending moment diagram

    AM = W.L2 /2

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    AM W.L /2

    The distance to the centroid of the BM

    diagram from B = xc = (2 L/3)

    The deflection of B = y b= A M. x c /EI = W.L

    3 /3EI

    The slope at B relative to the tan at A =b

    =AM/EI = W

    L2 /2EI