157635065-strength-of-materials-notes.pdf

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STRENGTH OF MATERIALS SUBJECT CODE: CE 43

EINSTEIN COLLEGE OF ENGG DEPT OF CIVIL ENGG

EINSTEIN

COLLEGE OF ENGINEERINGSir.C.V.Raman Nagar, Tirunelveli-12

Department of Civil Engineering

CE 43- STRENGTH OF MATERIALS

Lecture notes

Prepared by

V.TAMILARASI


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UNIT - I

Stress Terms

Stress is defined as force per unit area. It has the same units as pressure, and in fact

pressure is one special variety of stress. However, stress is a much more complexquantity than pressure because it varies both with direction and with the surface it acts on.

Compression Stress that acts to shorten an object.

Tension Stress that acts to lengthen an object.

Normal Stress Stress that acts perpendicular to a surface. Can be either compressional ortensional.

Shear

Stress that acts parallel to a surface. It can cause one object to slide over another.It also tends to deform originally rectangular objects into parallelograms. Themost general definition is that shear acts to change the angles in an object.

Hydrostatic Stress (usually compressional) that is uniform in all directions. A scuba diver

experiences hydrostatic stress. Stress in the earth is nearly hydrostatic. The termfor uniform stress in the earth is lithostatic.

Directed Stress Stress that varies with direction. Stress under a stone slab is directed; there is a

force in one direction but no counteracting forces perpendicular to it. This is whya person under a thick slab gets squashed but a scuba diver under the same

pressure doesn't. The scuba diver feels the same force in all directions.

In geology we never see stress. We only see the results of stress as it deforms materials.

Even if we were to use a strain gauge to measure in-situ stress in the rocks, we would not

measure the stress itself. We would measure the deformation of the strain gauge (that'swhy it's called a " strain gauge") and use that to infer the stress.

Strain Terms

Strain is defined as the amount of deformation an object experiences compared to itsoriginal size and shape. For example, if a block 10 cm on a side is deformed so that it

becomes 9 cm long, the strain is (10-9)/10 or 0.1 (sometimes expressed in percent, in thiscase 10 percent.) Note that strain is dimensionless.

Longitudinal or Linear Strain Strain that changes the length of a line without changing its direction. Can be

either compressional or tensional.

Compression


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Longitudinal strain that shortens an object.

Tension Longitudinal strain that lengthens an object.

Shear Strain that changes the angles of an object. Shear causes lines to rotate.

Infinitesimal Strain Strain that is tiny, a few percent or less. Allows a number of useful mathematicalsimplifications and approximations.

Finite Strain Strain larger than a few percent. Requires a more complicated mathematical

treatment than infinitesimal strain.

Homogeneous Strain Uniform strain. Straight lines in the original object remain straight. Parallel linesremain parallel. Circles deform to ellipses. Note that this definition rules out

folding, since an originally straight layer has to remain straight.

Inhomogeneous Strain

How real geology behaves. Deformation varies from place to place. Lines may bend and do not necessarily remain parallel.

Terms for Behavior of Materials

Elastic Material deforms under stress but returns to its original size and shape when thestress is released. There is no permanent deformation. Some elastic strain, like in

a rubber band, can be large, but in rocks it is usually small enough to beconsidered infinitesimal.

Brittle

Material deforms by fracturing. Glass is brittle. Rocks are typically brittle at lowtemperatures and pressures.

Ductile Material deforms without breaking. Metals are ductile. Many materials show bothtypes of behavior. They may deform in a ductile manner if deformed slowly, but

fracture if deformed too quickly or too much. Rocks are typically ductile at hightemperatures or pressures.

Viscous Materials that deform steadily under stress. Purely viscous materials like liquids

deform under even the smallest stress. Rocks may behave like viscous materialsunder high temperature and pressure.

Plastic Material does not flow until a threshhold stress has been exceeded.

Viscoelastic


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Combines elastic and viscous behavior. Models of glacio-isostasy frequentlyassume a viscoelastic earth: the crust flexes elastically and the underlying mantle

flows viscously.

Beams

A beam is a structural member which carries loads. These loads are most often perpendicular to its longitudinal axis, but they can be of any geometry. A beam

supporting any load develops internal stresses to resist applied loads. These internalstresses are bending stresses, shearing stresses, and normal stresses.

Beam types are determined by method of support, not by method of loading. Below are

three types of beams that will be investigated in this course:

The first two types are statically determinate, meaning that the reactions, shears and

moments can be found by the laws of statics alone. Continuous beams are staticallyindeterminate. The internal forces of these beams cannot be found using the laws of

statics alone. Early structures were designed to be statically determinate because simpleanalytical methods for the accurate structural analysis of indeterminate structures were

not developed until the first part of this century. A number of formulas have been derivedto simplify analysis of indeterminate beams.

The three basic beam types can be combined to create larger beam systems. These

complex systems can inevitably be distilled to the simple beam types for analysis. The beams shown immediately below are combinations of the first two beam types; these

systems are all statically determinate.


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The two beam loading conditions that either occur separately, or in some combination,

are:

CONCENTRATED Either a force or a moment can be applied as a concentrated load. Both are applied at asingle point along the axis of a beam. These loads are shown as a "jump" in the shear or

moment diagrams. The point of application for such a load is indicated in the diagramabove. Note that this is NOT a hinge! It is a point of application. This could be point at

which a railing is attached to a bridge, or a lampost on the same.

DISTRIBUTED Distributed loads can be uniformly or non-uniformly distributed. Both types are

commonly found on all kinds of structures. Distributed loads are shown as an angle orcurve in the shear or moment diagram. A uniformly distributed load can evolve into a n

on-uniformly distributed load (snow melting to ice at the edge of a roof), but are normallyassumed to act as given. These loads are often replaced by a singular resultant force in

order to simplify the structural analysis.


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Introduction

Normally a beam is analysed to obtain the maximum stress and this is

compared to the material strength to determine the design safety margin. Itis also normally required to calculate the deflection on the beam under themaximum expected load. The determination of the maximum stress results

from producing the shear and bending moment diagrams. To facilitate thiswork the first stage is normally to determine all of the external loads.

Nomenclature

e = strain

σ = stress (N/m2)

E = Young's Modulus = σ /e (N/m2)y = distance of surface from neutral surface (m).

R = Radius of neutral axis (m).I = Moment of Inertia (m

4 - more normally cm

4)

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

M = Moment (Nm)

w = Distrubuted 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)S= Shear Force on Section (N)

L = length of beam (m)x = distance along beam (m)

Calculation of external forces

To allow determination of all of the external loads a free-body diagram is

construction with all of the loads and supports replaced by their equivalent

forces. A typical free-body diagram is shown below.

The unknown forces (generally the support reactions) are then determined

using the equations for plane static equilibrium.


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For example considering the simple beam above the reaction R 2 isdetermined by Summing the moments about R 1 to zero

R 2. L - W.a = 0 Therefore R 2 = W.a / L

R 1 is determined by summing the vertical forces to 0

W - R 1 - R 2 = 0 Therefore R 1 = W - R 2

Shear and Bending Moment Diagram

The shear force diagram indicates the shear force withstood by the beam

section along the length of the beam.The bending moment diagram indicates the bending moment withstood by

the beam section along the length of the beam.It is normal practice to produce a free body diagram with the shear diagram

and the bending moment diagram position below

For simply supported beams the reactions are generally simple forces. Whenthe beam is built-in the free body diagram will show the relevant support

point as a reaction force and a reaction moment....

Sign Convention

The sign convention used for shear force diagrams and bending moments is

only important in that it should be used consistently throughout a project. The sign convention used on this page is as below.

Typical Diagrams


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A shear force diagram is simply constructed by moving a section along the beam from (say)the left origin and summing the forces to the left of the

section. The equilibrium condition states that the forces on either side of asection balance and therefore the resisting shear force of the section is

obtained by this simple operation

The bending moment diagram is obtained in the same way except that themoment is the sum of the product of each force and its distance(x) from the

section. Distributed loads are calculated buy summing the product of thetotal force (to the left of the section) and the distance(x) of the centroid of

the distributed load.

The sketches below show simply supported beams with on concentratedforce.

The sketches below show Cantilever beams with three different load

combinations.

Note: The force shown if based on loads (weights) would need to beconverted to force units i.e. 50kg = 50x9,81(g) = 490 N.

Shear Force Moment Relationship


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Consider a short length of a beam under a distributed load separated by adistance δx.

The bending moment at section AD is M and the shear force is S. The bending moment at BC = M + δM and the shear force is S + δS.

The equations for equilibrium in 2 dimensions results in the equations..

Forces.

S - w.δx = S + δS

Therefore making δx infinitely small then.. dS /dx = - w

Moments.. Taking moments about C

M + Sδx - M - δM - w(δx)2 /2 = 0

Therefore making δx infinitely small then.. dM /dx = S

Therefore putting the relationships into integral form.

The integral (Area) of the shear diagram between any limits results in thechange of the shearing force between these limits and the integral of the

Shear Force diagram between limits results in the change in bendingmoment...

Torsion (mechanics)

In solid mechanics, torsion is the twisting of an object due to an applied torque. In

circular sections, the resultant shearing stress is perpendicular to the radius.

For solid or hollow shafts of uniform circular cross-section and constant wall thickness,the torsion relations are:

http://en.wikipedia.org/wiki/Solid_mechanicshttp://en.wikipedia.org/wiki/Solid_mechanicshttp://en.wikipedia.org/wiki/Solid_mechanicshttp://en.wikipedia.org/wiki/Torquehttp://en.wikipedia.org/wiki/Torquehttp://en.wikipedia.org/wiki/Torquehttp://en.wikipedia.org/wiki/Shear_stresshttp://en.wikipedia.org/wiki/Shear_stresshttp://en.wikipedia.org/wiki/Shear_stresshttp://en.wikipedia.org/wiki/Shear_stresshttp://en.wikipedia.org/wiki/Torquehttp://en.wikipedia.org/wiki/Solid_mechanics


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where:

R is the outer radius of the shaft. τ is the maximum shear stress at the outer surface. φ is the angle of twist in radians.

T is the torque ( N·m or ft·lbf ). ℓ is the length of the object the torque is being applied to or over.

G is the shear modulus or more commonly the modulus of rigidity and is usuallygiven in gigapascals (GPa), lbf/in2 (psi), or lbf/ft2.

J is the torsion constant for the section . It is identical to the polar moment ofinertia for a round shaft or concentric tube only. For other shapes J must be

determined by other means. For solid shafts the membrane analogy is useful, andfor thin walled tubes of arbitrary shape the shear flow approximation is fairly

good, if the section is not re-entrant. For thick walled tubes of arbitrary shapethere is no simple solution, and FEA may be the best method.

the product GJ is called the torsional rigidity.

The shear stress at a point within a shaft is:

where:

r is the distance from the center of rotation

Note that the highest shear stress is at the point where the radius is maximum, the surfaceof the shaft. High stresses at the surface may be compounded by stress concentrations

such as rough spots. Thus, shafts for use in high torsion are polished to a fine surfacefinish to reduce the maximum stress in the shaft and increase its service life.

The angle of twist can be found by using:

Polar moment of inertia

Main article: Polar moment of inertia

The polar moment of inertia for a solid shaft is:

http://en.wikipedia.org/wiki/Shear_stresshttp://en.wikipedia.org/wiki/Shear_stresshttp://en.wikipedia.org/wiki/Shear_stresshttp://en.wikipedia.org/wiki/Radianhttp://en.wikipedia.org/wiki/Radianhttp://en.wikipedia.org/wiki/Radianhttp://en.wikipedia.org/wiki/Newton_metrehttp://en.wikipedia.org/wiki/Newton_metrehttp://en.wikipedia.org/wiki/Newton_metrehttp://en.wikipedia.org/wiki/Foot-pound_forcehttp://en.wikipedia.org/wiki/Foot-pound_forcehttp://en.wikipedia.org/wiki/Foot-pound_forcehttp://en.wikipedia.org/wiki/Modulus_of_rigidityhttp://en.wikipedia.org/wiki/Modulus_of_rigidityhttp://en.wikipedia.org/wiki/Modulus_of_rigidityhttp://en.wikipedia.org/wiki/Gigapascalhttp://en.wikipedia.org/wiki/Gigapascalhttp://en.wikipedia.org/wiki/Gigapascalhttp://en.wikipedia.org/wiki/Pounds_per_square_inchhttp://en.wikipedia.org/wiki/Pounds_per_square_inchhttp://en.wikipedia.org/wiki/Pounds_per_square_inchhttp://en.wikipedia.org/wiki/Pounds_per_square_inchhttp://en.wikipedia.org/wiki/Torsion_constanthttp://en.wikipedia.org/wiki/Torsion_constanthttp://en.wikipedia.org/wiki/Torsion_constanthttp://en.wikipedia.org/wiki/Polar_moment_of_inertiahttp://en.wikipedia.org/wiki/Polar_moment_of_inertiahttp://en.wikipedia.org/wiki/Polar_moment_of_inertiahttp://en.wikipedia.org/wiki/Polar_moment_of_inertiahttp://en.wikipedia.org/wiki/Finite_element_methodhttp://en.wikipedia.org/wiki/Finite_element_methodhttp://en.wikipedia.org/wiki/Finite_element_methodhttp://en.wikipedia.org/wiki/Torsional_rigidityhttp://en.wikipedia.org/wiki/Torsional_rigidityhttp://en.wikipedia.org/wiki/Torsional_rigidityhttp://en.wikipedia.org/wiki/Stress_concentrationshttp://en.wikipedia.org/wiki/Stress_concentrationshttp://en.wikipedia.org/wiki/Stress_concentrationshttp://en.wikipedia.org/wiki/Polar_moment_of_inertiahttp://en.wikipedia.org/wiki/Polar_moment_of_inertiahttp://en.wikipedia.org/wiki/Polar_moment_of_inertiahttp://en.wikipedia.org/wiki/Polar_moment_of_inertiahttp://en.wikipedia.org/wiki/Stress_concentrationshttp://en.wikipedia.org/wiki/Torsional_rigidityhttp://en.wikipedia.org/wiki/Finite_element_methodhttp://en.wikipedia.org/wiki/Polar_moment_of_inertiahttp://en.wikipedia.org/wiki/Polar_moment_of_inertiahttp://en.wikipedia.org/wiki/Polar_moment_of_inertiahttp://en.wikipedia.org/wiki/Torsion_constanthttp://en.wikipedia.org/wiki/Pounds_per_square_inchhttp://en.wikipedia.org/wiki/Gigapascalhttp://en.wikipedia.org/wiki/Modulus_of_rigidityhttp://en.wikipedia.org/wiki/Foot-pound_forcehttp://en.wikipedia.org/wiki/Newton_metrehttp://en.wikipedia.org/wiki/Radianhttp://en.wikipedia.org/wiki/Shear_stress


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where r is the radius of the object.

The polar moment of inertia for a pipe is:

where the o and i subscripts stand for the outer and inner radius of the pipe.

For a thin cylinder

J = 2π R3 t

where R is the average of the outer and inner radius and t is the wall thickness.

Failure mode

The shear stress in the shaft may be resolved into principal stresses via Mohr's circle. If

the shaft is loaded only in torsion then one of the principal stresses will be in tension andthe other in compression. These stresses are oriented at a 45 degree helical angle around

the shaft. If the shaft is made of brittle material then the shaft will fail by a crackinitiating at the surface and propagating through to the core of the shaft fracturing in a 45

degree angle helical shape. This is often demonstrated by twisting a piece of blackboardchalk between one's fingers.

Deflection of Beams

The deformation of a beam is usually expressed in terms of its deflection from its original

unloaded position. The deflection is measured from the original neutral surface of the

beam to the neutral surface of the deformed beam. The configuration assumed by the

deformed neutral surface is known as the elastic curve of the beam.

http://en.wikipedia.org/wiki/Radiushttp://en.wikipedia.org/wiki/Radiushttp://en.wikipedia.org/wiki/Radiushttp://en.wikipedia.org/wiki/Principal_stresshttp://en.wikipedia.org/wiki/Principal_stresshttp://en.wikipedia.org/wiki/Principal_stresshttp://en.wikipedia.org/wiki/Mohr%27s_circlehttp://en.wikipedia.org/wiki/Mohr%27s_circlehttp://en.wikipedia.org/wiki/Mohr%27s_circlehttp://en.wikipedia.org/wiki/Brittlehttp://en.wikipedia.org/wiki/Brittlehttp://en.wikipedia.org/wiki/Brittlehttp://en.wikipedia.org/wiki/Brittlehttp://en.wikipedia.org/wiki/Mohr%27s_circlehttp://en.wikipedia.org/wiki/Principal_stresshttp://en.wikipedia.org/wiki/Radius


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Methods of Determining Beam Deflections

Numerous methods are available for the determination of beam deflections. These

methods include:

1. Double-integration method

2. Area-moment method

3. Strain-energy method (Castigliano‟s Theorem)

4. Three-moment equation

5. Conjugate-beam method

6. Method of superposition

7. Virtual work method

Of these methods, the first two are the ones that are commonly used.

Introduction

The stress, strain, dimension, curvature, elasticity, are all related, under

certain assumption, by the theory of simple bending. This theory relates to

beam flexure resulting from couples applied to the beam without

consideration of the shearing forces.

Superposition Principle

The superposition principle is one of the most important tools for solving beam loading problems allowing simplification of very complicated design

problems..

http://www.mathalino.com/reviewer/mechanics-and-strength-of-materials/double-integration-method-beam-deflectionshttp://www.mathalino.com/reviewer/mechanics-and-strength-of-materials/double-integration-method-beam-deflectionshttp://www.mathalino.com/reviewer/mechanics-and-strength-of-materials/area-moment-method-beam-deflectionshttp://www.mathalino.com/reviewer/mechanics-and-strength-of-materials/area-moment-method-beam-deflectionshttp://www.mathalino.com/reviewer/mechanics-and-strength-of-materials/area-moment-method-beam-deflectionshttp://www.mathalino.com/reviewer/mechanics-and-strength-of-materials/double-integration-method-beam-deflections


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For beams subjected to several loads of different types the resulting shear

force, bending moment, slope and deflection can be found at any location bysumming the effects due to each load acting separately to the other loads.

Nomenclature

e = strain

E = Young's Modulus = σ /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/ymax(m3 - more normally cm

3)

F = Force (N)

x = Distance along beamδ = deflection (m)

θ = Slope (radians) σ = stress (N/m

2)

Simple Bending

A straight bar of homogeneous material is subject to only a moment at one

end and an equal and opposite moment at the other end...

Assumptions

The beam is symmetrical about Y-YThe traverse plane sections remain plane and normal to the longitudinal

fibres after bending (Beroulli's assumption)The fixed relationship between stress and strain (Young's Modulus)for the

beam material is the same for tension and compression ( σ= E.e )


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Consider two section very close together (AB and CD).After bending the sections will be at A'B' and C'D' and are no longer

parallel. AC will have extended to A'C' and BD will have compressed toB'D'

The line EF will be located such that it will not change in length. This

surface is called neutral surface and its intersection with Z_Z is called theneutral axisThe development lines of A'B' and C'D' intersect at a point 0 at an angle of θ

radians and the radius of E'F' = RLet y be the distance(E'G') of any layer H'G' originally parallel to EF..Then

H'G'/E'F' =(R+y)θ /R θ = (R+y)/R

And the strain e at layer H'G' =

e = (H'G'- HG) / HG = (H'G'- HG) / EF = [(R+y)θ - R θ] /R θ = y /R

The accepted relationship between stress and strain is σ= E.e Therefore

σ = E.e = E. y /Rσ / E = y / R

Therefore, for the illustrated example, the tensile stress is directly related tothe distance above the neutral axis. The compressive stress is also directly

related to the distance below the neutral axis. Assuming E is the same forcompression and tension the relationship is the same.

As the beam is in static equilibrium and is only subject to moments (novertical shear forces) the forces across the section (AB) are entirelylongitudinal and the total compressive forces must balance the total tensile

forces. The internal couple resulting from the sum of ( σ.dA .y) over thewhole section must equal the externally applied moment.


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This can only be correct if Σ(yδa) or Σ(y.z.δy) is the moment of area of thesection about the neutral axis. This can only be zero if the axis passes

through the centre of gravity (centroid) of the section.

The internal couple resulting from the sum of ( σ.dA .y) over the whole

section must equal the externally applied moment. Therefore the couple ofthe force resulting from the stress on each area when totalled over the wholearea will equal the applied moment

From the above the following important simple beam bending relationship

results

It is clear from above that a simple beam subject to bending generates a

maximum stress at the surface furthest away from the neutral axis. Forsections symmetrical about Z-Z the maximum compressive and tensile stressis equal.

σmax = ymax. M / I

The factor I /ymax is given the name section Modulus (Z) and therefore

σmax = M / Z

Values of Z are provided in the tables showing the properties of standard

steel sections

Deflection of Beams

Below is shown the arc of the neutral axis of a beam subject to bending.


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For small angle dy/dx = tan θ = θ The curvature of a beam is identified as dθ /ds = 1/R

In the figure δθ is small and δx; is practically = δs; i.e ds /dx =1

From this simple approximation the following relationships are derived.

Integrating between selected limits.

The deflection between limits is obtained by further integration.

It has been proved ref Shear - Bending that dM/dx = S and dS/dx = -w =d2M /dx

Where S = the shear force M is the moment and w is the distributed load/unit length of beam. therefore

If w is constant or a integratatable function of x then this relationship can be

used to arrive at general expressions for S, M, dy/dx, or y by progressiveintegrations with a constant of integration being added at each stage. The

properties of the supports or fixings may be used to determine the constants.(x= 0 - simply supported, dx/dy = 0 fixed end etc )

http://www.roymech.co.uk/Useful_Tables/Beams/Shear_Bending.html#shearbendhttp://www.roymech.co.uk/Useful_Tables/Beams/Shear_Bending.html#shearbendhttp://www.roymech.co.uk/Useful_Tables/Beams/Shear_Bending.html#shearbendhttp://www.roymech.co.uk/Useful_Tables/Beams/Shear_Bending.html#shearbend


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In a similar manner if an expression for the bending moment is known thenthe slope and deflection can be obtained at any point x by single and double

integration of the relationship and applying suitable constants of integration.

Singularity functions can be used for determining the values when theloading a not simple ref Singularity Functions

Example - Cantilever beam

Consider a cantilever beam (uniform section) with a single concentrated load

at the end. At the fixed end x = 0, dy = 0 , dy/dx = 0

From the equilibrium balance ..At the support there is a resisting moment -FL and a vertical upward force F.

At any point x along the beam there is a moment F(x - L) = Mx = EI d2y /dx

2

Example - Simply supported beam

Consider a simply supported uniform section beam with a single load F at

the centre. The beam will be deflect symmetrically about the centre linewith 0 slope (dy/dx) at the centre line. It is convenient to select the origin at

the centre line.

http://www.roymech.co.uk/Useful_Tables/Beams/Singularity.htmlhttp://www.roymech.co.uk/Useful_Tables/Beams/Singularity.htmlhttp://www.roymech.co.uk/Useful_Tables/Beams/Singularity.htmlhttp://www.roymech.co.uk/Useful_Tables/Beams/Singularity.html


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Moment Area Method

This is a method of determining the change in slope or the deflection

between two points on a beam. It is expressed as two theorems...

Theorem 1If A and B are two points on a beam the change in angle (radians) between

the tangent at A and the tangent at B is equal to the area of the bendingmoment diagram between the points divided by the relevant value of EI (the

flexural rigidity constant).

Theorem 2If A and B are two points on a beam the displacement of B relative to the

tangent of the beam at A is equal to the moment of the area of the bendingmoment diagram between A and B about the ordinate through B divided by

the relevant value of EI (the flexural rigidity constant).

Examples ..Two simple examples are provide below to illustrate these theorems

Example 1) Determine the deflection and slope of a cantilever as shown..


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The bending moment at A = MA = FLThe area of the bending moment diagram AM = F.L

2 /2

The distance to the centroid of the BM diagram from B= xc = 2L/3The deflection of B = y b = A M.x c /EI = F.L

3 /3EI

The slope at B relative to the tan at A = θ b =AM /EI = FL2 /2EI

Example 2) Determine the central deflection and end slopes of the simply

supported beam as shown..

E = 210 GPa ......I = 834 cm4...... EI = 1,7514. 10

6 Nm

2

A1 = 10.1,8.1,8/2 = 16,2kNm

A2 = 10.1,8.2 = 36kNmA2 = 10.1,8.2 = 36kNm

A1 = 10.1,8.1,8/2 = 16,2kNmx1 = Centroid of A1 = (2/3).1,8 = 1,2

x2 = Centroid of A2 = 1,8 + 1 = 2,8x3 = Centroid of A3 = 1,8 + 1 = 2,8

x4 = Centroid of A4 = (2/3).1,8 = 1,2


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The slope at A is given by the area of the moment diagram between A and Cdivided by EI.

θA = (A1 + A2) /EI = (16,2+36).103 / (1,7514. 10 6)

= 0,029rads = 1,7 degrees

The deflection at the centre (C) is equal to the deviation of the point A abovea line that is tangent to C.

Moments must therefore be taken about the deviation line at A.

δC = (AM.xM) /EI = (A1 x1 +A2 x2) / EI = 120,24.103/ (1,7514. 10

6)

= 0,0686m = 68,6mm

: Force Method - Introduction and applications

: Three Moment Equation

5.5 Three Moment Equation

The continuous beams are very common in the structural design and it is necessary to developsimplified force method known as three moment equation for their analysis. This equation is a

relationship that exists between the moments at three points in continuous beam. The points areconsidered as three supports of the indeterminate beams. Consider three points on the beam marked as

1, 2 and 3 as shown in Figure 5.25(a). Let the bending moment at these points is , and and

the corresponding vertical displacement of these points are , and , respectively. Let and

be the distance between points 1 – 2 and 2 – 3, respectively.


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The continuity of deflected shape of the beam at point 2 gives

(5.4)From the Figure 5.25(d)

and

(5.5)Where

and(5.6)


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Energy Method

First law of thermodynamics

Adiabatic ProcessWork of Gravity


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Strain energy due to Normal stress

Strain energy due to shear stress


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Similarly by other components of stresses

Super position


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Ex.1 Bar in tension

Ex.2 Torsion of circular shaft (r , L)


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Ex.3 Bending strain Energy – Normal stress


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Strain energy due to shear in Beam

Castigliano Theorem


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Castig liano Theorem

Are corresponding deflection,twists

and rotation due to

Fictitions loads are applied at the point wherethere is no load and deflection is songat

at that point where there is no load

Ex.1


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Ex.2


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Ex.3


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Ex.4


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Ex.5


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SHORT COLUMNSINTRODUCTION: AXIAL COMPRESSION

In this chapter the term column will be used interchangeably with the term

comprerssion member, for brevity and in conformity with general usage.

Three types of reinforced concrete compression members are in use:

Members reinforced with longitudinal bars and lateral ties.

Members reinforced with longitudinal bars and continuous spirals.

Composite compression members reinforced longitudinally with structural

steel shapes, pipe, or tubing, with or without additional longitudinal bars,

and various types of lateral reinforcement.

The main reinforcement in columns is longitudinal, parallel to the direction

of the load, and consist of bars arranged in a square, rectangular, or circular

column.

Columns may be divided into two broad categories: short columns, for

which the strength is governed by the strength of the materials and the geometry

of the cross section, and slender columns, for which the strength may be

significantly reduced by lateral deflections. A number of years ago, an ACI -

ASCE survey indicated that 90 percent of columns braced against sidesway and

40 percent of unbraced columns could be designed as short columns. Only short

columns will be discussed in this chapter.

The behavior of short, axially loaded compression members, for lower

loads for which both materials remain in their elastic range of response, the steel

carries a relatively small portion of the total load. The steel stress fs is equal to

n times the concrete stress:


36/88




37/88



About c :-

About x :-

We know :-

Pacting parallel to (underformed axis to positive column)


38/88



Case1: CantileverColumn

Let EI is conott and q =0.

order DE can besolved

solution

are evaluated from the B.C

B.C

x=0, deflection = 0slope = 0

x=L, moment = 0shear force=0

Mechanism of membrane energy transfer to bending strain energy gives rise to Phenomenon of instability(Buckling).

Buckling Of Long slender column's


39/88



Then Trival Solution

For Non -Trival Solution

then

eflection Curve


40/88



case II:

Substituting in last two equation we get

Examine the ways ofsatisfying these equations

Objective =

Lowest critical load

Let c1 = 0 for non trivial solution we get


41/88



Both the conditions are simultaneously satisfying

when

then lowest Critical load

Other possibilitieseliminating c2 from equation (A) & (B)

Multiplying by and expanding in the expression and using the

trigonometric identity.

We get

Using trigonometric identity i.e.

Again trigonometric

identity

identical to the privious one


42/88



and

Which is larger than therefore

Case III:

x = 0 y = 0 => c2 + c4 = 0 => c2 = 0 => c4 = 0

x = L

Lowest

n =

1Deflection Shape

cn - any arbitrary constant which of course is very small does not violate the condition of

linearly i.e. small deflection.

Principle of SuperpositionThe principle of superposition is a central concept in the analysis of structures.

This is applicable when there exists a linear relationship between external forcesand corresponding structural displacements. The principle of superposition may

be stated as the deflection at a given point in a structure produced by several


43/88



loads acting simultaneously on the structure can be found by superposingdeflections at the same point produced by loads acting individually. This is illustrated

with the help of a simple beam problem. Now consider a cantilever beam of length L and having constant flexural rigidity EI subjected to two

externally applied forces and as shown in Fig. 2.1. From moment-area

theorem we can evaluate deflection below , which states that the tangentialdeviation of point from the tangent at point1P 2PCc A is equal to the first moment of the area of the EI

Mdiagram between A and C about . Hence, the deflection below

due to loads and acting simultaneously is (by moment-area theorem),

DEFLECTION OF BEAMS

Structures undergo deformation when subjected to loads. As a result of thisdeformation, deflection and rotation occur in structures. This deformation will disappear

when the loads are removed provided the elastic limit of the material is not exceeded.Deformation in a structure can also occur due to change in temperature & settlement of

supports.

Deflection in any structure should be less than specified limits for satisfactory performance. Hence computing deflections is an important aspect of analysis of

structures.There are various methods of computing deflections. Two popular methods are

i) Moment area Method, andii) Conjugate beam method

In both of these methods, the geometrical concept is used. These methods are

ideal for statically determinate beams. The methods give a very quick solution when the beam is symmetrical.

Moment Area Method

This method is based on two theorems which are stated through an example.Consider a beam AB subjected to some arbitrary load as shown in Figure 1.

Let the flexural rigidity of the beam be EI. Due to the load, there would be

bending moment and BMD would be as shown in Figure 2. The deflected shape of the beam which is the elastic curve is shown in Figure 3. Let C and D be two points

arbitrarily chosen on the beam. On the elastic curve, tangents are drawn at deflected positions of C and D. The angles made by these tangents with respect to the horizontal

are marked as Cθ and Dθ . These angles are nothing but slopes. The change is the angle

between these two tangents is demoted as CDθ . This change in the angel is equal to the

area of theEI

M diagram between the two points C and D. This is the area of the shaded

portion in figure 2.


44/88



Hence CDθ = Cθ Dθ = Area ofEI

M diagram between C and D

CDθ = Area BM 1 (a)

EI

It is also expressed in the integration mode as

CDθ = dxCD

EIM

1 (b)

Equation 1 is the first moment area theorem which is stated as follows:

Statement of theorem I:

The change in slope between any two points on the elastic cur ve for a member

subjected to bending is equal to the area ofEI

M diagram between those two points.


45/88



In figure 4, for the elastic curve a tangent is drawn at point C from which the

vertical intercept to elastic curve at D is measured. This is demoted as K CD. This verticalintercept is given by

K CD = (Area BM X)CD 2 (a)EI

Fig. 1

Fig. 2

Fig. 3

Fig. 4


46/88



Where X is the distance to the centroid of the shaded portion ofEI

M diagram

measured from D. The above equation can be expressed in integration mode as

K CD =

EIMxdx

CD

2 (b)

Equation (2) is the second moment area theorem which is stated as follows.

Statement of theorem II :

The vertical intercept to the elastic cur ve measured from the tangent drawn to

the elastic cur ve at some other point is equal to the moment ofEI

M diagram, moment

being taken about that point where verti cal intercept is drawn.

Sign Convention:

While computing Bending moment at a section, if free body diagram of Left HandPortion (LHP) is considered, clockwise moment is taken as positive. If free bodydiagram of Right Hand Portion (RHP) is considered, anticlockwise moment is taken as

positive. While sketching the Bending Moment Diagram (BMD), Sagging moment istaken as positive and Hogging moment is taken as negative.

Proof of Moment Area Theorems:

Figure 5 shows the elastic curve for the elemental length dx of figure 2 to an

enlarged scale. In this figure, R represents the radius of curvature. Then from equationof bending, with usual notations,

I

M =

R

E (3)

From figure 5,

Rdθ = dx

Hence R =dθ

dx

Substituting this value of R in equation (3),

I

M =

dθ

dx

E

I

M = E

dx

dθ


47/88



dθ =EI

M dx

dθ is nothing but change in angle over the elemental length dx. Hence to compute

change in angle from C to D,

θCD =

CD

dθ =

CDEIM

dx

Hence the proof.

C

1 2D

Figure 6 shows the elastic curve from C to D. Change in slope from 1 to 2 is dθ.Distance of elemental length from D is x.

dΔ = xdθ = xEI

M dx

Therefore, Δ from C to D = CDEI

M xdx

R

Fig. 5Fig. 6

K

dθ

dθ


48/88



Problem 1 : Compute def lections and slopes at C,D and E. Also compute slopes at A

and B.

To Compute Reactions:

00fxA

0WWVV0fy

BA


49/88



2WVV BA

+

03

2LW

3

WLLV0M

AB

WL3

2WL

3

WLLV

A

VA = W ; VB =W

Bending Moment Calculations:

Section (1) – (1) (LHP, 0 to L/3)+ Mx-x = Wx

At x = 0; BM at A = 0

x = 3L ; BM @ C = 3WL

Section (2) – (2) (LHP, 3L

to 32L

)+ Mx-x = Wx – W(x - 3L )

At x = 3L , BM @ C = 3L3L3L WWW = 3LW

At x =3

2L, BM @ D =

3

L

3

2LW

3

2LW

=

3

WL

3

2WL

3

2WL

= 3

WL

Section (3) – (3) RHP (0 to 3L )

+ Mx-x = Wx

At x = 0; BM @ B = 0

At x = 3L , BM @ D =3

WL

This beam is symmetrical. Hence the BMD & elastic curve is also symmetrical.

In such a case, maximum deflection occurs at mid span, marked as δ E. Thus, the tangentdrawn at E will be parallel to the beam line and θE is zero.

Also, δc = δD, θA = θB and θC = θD

To compute θC

From first theorem,

θCE = Area of BMD between E&CEI


50/88



θC~ θE =

EI

W 6L3L

=

18EI

WL2

θE being zero, θC = WL2

( )

18EI

To compute θΔ

From First theorem,

θΔE = Area of BMD between A&EEI

θA~ θE =EI

6

L

3

WL

3

WL

3

L2

1

=EI

18

WL

18

WL 22

θE being zero, θA =9EI

WL2

( )

θB =9EI

WL2

( )

To compute δE

From 2nd

theorem

K EA =

EI

XBMof AreaEA

K EA = EI

12

L

3

L

6

L

3

WL

3

L

3

2

3

WL

3

L

2

1


51/88



= EI

216

5WL

81

WL 33

=

64815WL8WL

EI1

33

=648EI

23WL3

From figure, K EA is equal to δE.

Therefore δE =648EI

23WL3

To compute θC

From 2nd

theorem

K EC =

EI

XBMDof AreaCE

=

EI

W 12L6L3L

=

216

1

EI

WL3

=216EI

WL3

δc = δE - K EC

216EI

WL

648EI

23WLδ

33

C

=648EI

3WL23WL 33


52/88



=648EI

20WL3

=162EI5WL

3

=162EI

5WLδδ

3

CD


53/88



Problem 2. For the canti lever beam shows in f igure, compute def lection and slope at

the free end.

Consider a section x-x at a distance x from the free end. The FBD of RHP is taken intoaccount.

(RHP +) BM @ X-X = MX-X = -10 (x) (x/2) = -5x2

At x = 0; BM @ B = 0


54/88



At x = 4m; BM @ A = -5(16) = -80 kNm

The BMD is sketched as shown in figure. Note that it is Hogging BendingMoment. The elastic curve is sketched as shown in figure.

To compute θB

For the cantilever beam, at the fixed support, there will be no rotation and hencein this case θA = 0. This implies that the tangent drawn to the elastic curve at A will be

the same as the beam line.From I theorem,

θAB = θA ~ θB = 4

0EI

Mdx

= dx5XEI

14

0

2

= 403x3

EI

5

= 3EI

32064

3EI

5

θA being zero,

θB =3EI

320 ( )

To compute δB

From II theorem

K AB = 4

0EI

Mxdx

= xdx5XEI

14

0

2

= 2564EI

5

EI

5 404

x4

=

EI

320

From the elastic curve,

K AB = δB = EI

320


55/88



Problem 3: Find def lection and slope at the fr ee end for the beam shown in f igure by

using moment area theorems. Take EI = 40000 KNm -2

Calculations of Bending Moment:

Region AC: Taking RHP +Moment at section = -6x

2/2

= -3x2


56/88



At x = 0, BM @ A = 0

x = 4m; BM @ C = -3(16) = - 48kNm

Region CB: (x = 4 to x = 8)

Taking RHP +, moment @ section = -24 (x-2)= -24x+48;

At x = 4m; BM @ C = -24(4) + 48 = -48kNm;x = 8 m BM @ B = -144 kNm;

To compute θB:

First moment area theorem is used. For the elastic curve shown in figure. Weknow that θA = 0.

θAB = θA ~ θB = EIMdx

= 8

4

4

0

2 dx4824xEI

1dx3x

EI

1

842x4

03x

A48x24

EI

1

EI

3θ

23

= 4848166412EI

1

EI

64

= -0.0112 Radians

= 0.0112 Radians ( )

To compute δB

EIMxdx

K AB

= 8

4

4

0

2 xdx4824x

EI

1xdx3x

EI

1

= 842x843x404x 234 4824EI

1

EI

3

=

16642464512

3

24

EI

1256

4EI

3


57/88



= 11523584EI

1

EI

192

=

0.0656m0.0656mEI

2624

Problem 4: For the cantil ever shown in f igure, compute defl ection and at the points

where they are loaded.


58/88



To compute θB :

θBA = θB ~ θA = 151.537.52.5EI

121

21

θB = EI

58.125 ( )

θC = 151.51537.51.5EI

121

21

=EI

50.625 ( )

δB =

1451.5EI

1

EI

2.537.52.521

32

21

=EI

100.625

= EI

100.625

δC = 1451.50.8571537.51.5EI

121

21

δC = EI

44.99

STRAIN ENERGY

Introduction

Under action of gradually increasing external loads, the joints of a structuredeflect and the member deform. The applied load produce work at the joints to which

they are applied and this work is stored in the structure in the form of energy known asStrain Energy. If the material of structure is elastic, then gradual unloading of the

structure relieves all the stresses and strain energy is recovered.The slopes and deflections produced in a structure depend upon the strains

developed as a result of external actions. Strains may be axial, shear, flexural or torsion.Therefore, ther is a relationship can be used to determine the slopes and deflections in a

structure.

4.2 Strain energy and complementary strain energy


59/88



When external loads are applied to a skeletal structure, the members developinternal force „F‟ in the form of axial forces („P‟), shear force („V‟) , bending moment

(M) and twisting moment (T). The internal for „F‟ produce displacements „e‟. Whileunder goint these displacements, the internal force do internal work called as Strain

Energy

Figure 1 shows the force displacement relationship in which F j is the internalforce and e j is the corresponding displacement for the jth element or member of thestructure.

The element of internal work or strain energy represented by the area the stripwith horizontal shading is expressed as:

Strain energy stored in the jth

element represted by the are under force-

displacement curve computed as :

For m members in a structure, the total strain energy is

The area above the force-displacement curve is called Complementary Energy.

For jth element, the complementary strain energy is represented by the area of the strip

with vertical shading in Fig.1 and expressed as

F j

Strain Energy(Ui) j

e j e j e j+e j

F j

F j+F j

Complementry SE(Ui) j

Fig.1 FORCE-DISPLACEMENT RELATIONSHIP

.....(1) eFU j ji

.....(2) deF)U( j j ji

.....(3) deF)(UUm

1 j

m

1 j

j j jii


60/88



Complementary strain energy of the entire structure is

Complementary strain energy of the entire structure is

When the force-displacement relationship is linear, then strain energy and

complimentary energies are equal

4.3 Strain energy expressions

Expression for strain energy due to axial force, shear force and bending moment

is provided in this section

4.3.1 Strain energy due to Axi al force

A straight bar of length „L‟ , having uniform cross sectional area A and E is the Young‟s

modulus of elasticity is subjected to gradually applied load P as shown in Fig. 2. The bar

LA,E

dL

.....(4) FeU j ji

.....(5) dFe)U( j j j

i

.....(6) dFe Um

1 j

j ji

.....(7) UU ii

Fig.2


61/88



deforms by dL due to average force 0+(P/2) = P/2. Substituting F j = P/2 and de j = dl in

equation 2, the strain energy in a member due to axial force is expressed as

From Hooke‟s Law, strain is expressed as

Hence

Substituting equation 9 in 8, strain energy can be expressed as

For uniform cross section strain energy expression in equation 10 can be modified as

If P, A or E are not constant along the length of the bar, then equation 10 is used instead

of 10a.

4.3.1 Strain energy due to Shear force

dydxdy

dx

(8).......dL2

P )(U Pi

A

P where,

E

dx

dL

...(9).......... AE

PLx dL

(10)....... 2AE

dxP )(UL

0

2

Pi

a)(10.......2AE

LP )(U

2

Pi

Fig.3Fig.4


62/88



A small element shown in Fig.3 of dimension dx and dy is subjected to shear force V x .

Shear stress condition is shown in Fig. 4. Shear strain in the element is expressed as

Where, Ar = Reduced cross sectional area and G= shear modulus

Shear deformation of element is expressed as

Substituting F j = Vx/2, de j = dev in equation (2) strain energy is expressed as

4.3.2 Strain energy due to Bending Moment

An element of length dx of a beam is subjected to uniform bending moment „M‟.

Application of this moment causes a change in slope d is expressed as

Where ,EI

M

R

1 x , Substituting F j = Mx/2, de j= deM in equation (2), Strain energy due to

bending moment is expressed as

Theorem of minimum Potential Energy

......(11)

GA

V

r

x

...(12).......... GA

dxV de

r

xv

(13)....... G2A

dxV )(U

L

0 r

2

xVi

......(14) EIdxM

R dx dde xM θ

(15)....... 2EIdxM )(U

L

0

2

xMi


63/88



Potential energy is the capacity to do work due to the position of body. A body of weight

„W‟ held at a height „h‟ possess an energy „Wh‟. Theorem of minimum potential energy

states that “ Of all the displacements which satisfy the boundary conditions of a

structural system, those corresponding to stable equilibrium configuration make the

total potential energy a relative minimum”. This theorem can be used to determine the

critical forces causing instability of the structure.

Law of Conservation of Energy

From physics this law is stated as “Energy is neither created nor destroyed”. For the

purpose of structural analysis, the law can be stated as “ If a structure and external

loads acting on it are isolated, such that it neither receive nor give out energy, then

the total energy of the system remain constant”. With reference to figure 2, internal

energy is expressed as in equation (9). External work done We = -0.5 P dL. From law of

conservation of energy Ui+We =0. From this it is clear that internal energy is equal to

external work done.

Principle of Virtual Work:

Virtual work is the imaginary work done by the true forces moving through imaginary

displacements or vice versa. Real work is due to true forces moving through true

displacements. According to principle of virtual work “ The total virtual work done by

a system of forces during a virtual displacement is zero”.

Theorem of principle of virtual work can be stated as “If a body is in equilibrium under

a Virtual force system and remains in equilibrium while it is subjected to a small

deformation, the virtual work done by the external forces is equal to the virtual

work done by the internal stresses due to these forces”. Use of this theorem for

computation of displacement is explained by considering a simply supported bea AB, of

span L, subjected to concentrated load P at C, as shown in Fig.6a. To compute deflection

at D, a virtual load P‟ is applied at D after removing P at C. Work done is zero a s the

load is virtual. The load P is then applied at C, causing deflection C at C and D at D, as

shown in Fig. 6b. External work done We by virtual load P‟ is . If the virtual

load P‟ produces bending moment M‟, then the internal strain energy stored by M‟ acting

on the real deformation d in element dx over the beam equation (14)

2

δP' W De

L

0i

U

0

L

0i

EI2

dxMM' U;

2

dθM' dU


64/88



Where, M= bending moment due to real load P. From principle of conservation of energy

We=Wi

If P‟=1 then

Similarly for deflection in axial loaded trusses it can be shown that

Where,

= Deflection in the direction of unit load

P‟ = Force in the ith

member of truss due to unit load

P = Force in the ith

member of truss due to real external load

A BC D

a

xL

P

A BC D

a

xL

P P’

C D

Fig.6a

Fig.6b

L

0

D

EI2

dxMM'

2

δP'

(16) EI

dxMM' δL

0D

(17) AEdxPP' δ

n

0


65/88



n = Number of truss members

L = length of ith truss members.

Use of virtual load P‟ = 1 in virtual work theorem for computing displacement is called

Unit Load Method

Castigliano‟s Theorems:

Castigliano published two theorems in 1879 to determine deflections in structures and

redundant in statically indeterminate structures. These theorems are stated as:

1st Theorem : “If a linearly elastic structure is subjected to a set of loads, the partial

derivatives of total strain energy with respect to the deflection at any point is equal

to the load applied at that point”

2nd Theorem : “If a linearly elastic structure is subjected to a set of loads, the partial

derivatives of total strain energy with respect to a load applied at any point is equal

to the deflection at that point”

The first theorem is useful in determining the forces at certain chosen coordinates. The

conditions of equilibrium of these chosen forces may then be used for the analysis of

statically determinate or indeterminate structures. Second theorem is useful in computing

the displacements in statically determinate or indeterminate structures.

Betti‟s Law:

It states that If a structure is acted upon by two force systems I and II, in equilibrium

separately, the external virtual work done by a system of forces II during the

deformations caused by another system of forces I is equal to external work done by

I system during the deformations caused by the II system

(18) N.....1,2, j Pδ

U j

j

(19) N.1,2,...... j δP

U j

j

I II


66/88



A body subjected to two system of forces is shown in Fig 7. W ij represents work done by

ith system of force on displacements caused by jth

system at the same point. Betti‟s law

can be expressed as Wij = W ji, where W ji represents the work done by jth

system on

displacement caused by ith

system at the same point.

Numerical Examples

1. Derive an expression for strain energy due to bending of a cantilever beam of

length L, carrying uniformly distributed load „w‟ and EI is constant

Solution:

Bending moment at section 1-1 is

Strain energy due to bending is

w

x

1

1

Fig. 7

2

wx-M

2

x

2EI

dxM )(UL

0

2

xMi

L

0

L

0

5242L

0

22

i 40EI

xwdx

8EI

xw

2EI

dx2

wx-

U

Answer 40EI

Lw U

52

i


67/88



2. Compare the strain energies due to three types of internal forces in the

rectangular bent shown in Fig. having uniform cross section shown in the same Fig.

Take E=2 x 105 MPa, G= 0.8 x 105 MPa, Ar= 2736 mm2

Solution:

Step 1: Properties

A=120 * 240 – 108 * 216 = 5472 mm2,

E= 2 * 105 MPa ; G= 0.8 * 105 MPa ; Ar = 2736 mm2

Step 2: Strain Energy due to Axial Forces

Member AB is subjected to an axial comprn.=-12 kN

Strain Energy due to axial load for the whole str. is

Step 3: Strain Energy due to Shear Forces

Shear force in AB = 0; Shear force in BC = 12 kN

Strain Energy due to Shear for the whole str. Is

Step 4: Strain Energy due to Bending Moment

Bending Moment in AB = -12 * 4 = -48 kN-m

120 mm

240 mm12 mm5

m

4m

12kN

A

B C

4633

mm10x47.5412

216*108 -

12

240*120 I

mm- N328.9410*2*5472*2

5000*)10*(-12

2AE

LP )(U

5

232n

1i

2

Pi

mm- N78.1315 x100.8*2736*2

4000*)10*(12

G2A

LV )(U

5

232n

1ir

2

xVi


68/88



Bending Moment in BC = -12 x

Strain Energy due to BM for the whole structure is

Step 5: Comparison

Total Strain Energy = (Ui)p + (Ui)V+ (Ui)M

Total Strain Energy =328.94 +1315.78 +767.34 x 103

= 768.98 x 103 N-mm

Strain Energy due to axial force, shear force and bending moment are 0.043%, 0.17% &

99.78 % of the total strain energy.

3. Show that the flexural strain energy of a prismatic bar of length L bent into a complete

circle by means of end couples is

Solution:

Circumference = 2 R =L or

From bending theory

M M

L

R

10*47.54*2x10*2

5000*)10*(-48

2EI

dxM )(U

65

262n

1i

2

xMi

mm- N10*767.34

10*47.54*10*2*2

dxx)*10*(-12 34000

065

23

2L

πEI2

appliedcoupleMwhereL

EI2

2πL

EIM

Answer

L

EI2π

2EI

LL

πEI2

2EI

LM )(U

22

2

Mi


69/88



4. Calculate the strain energy in a truss shown in Fig. if all members are of same cross-

sectional area equal to 0.01m2 and E=200GPa

Solution: To calculate strain energy of the truss, first the member forces due to external

force is required to be computed. Method of joint has been used here to compute member

forces. Member forces in the members AB, BC, BD, BE, CE and DE are only computed

as the truss is symmetrical about centre vertical axis.

Step1: Member Forces:

i) Joint A: From triangle ACB, the angle = tan-1(3/4)=36052‟

The forces acting at the joint is shown in Fig. and the forces in members are computed

considering equilibrium condition at joint A

4m

AGC E 4m4m4m

30 kN

B D

30 kN

3

m

F

H


70/88



Fy=0; FABsin+30=0; FAB=-50kN (Compression)

Fx=0; FABcos + FAC=0; FAC=40kN (Tension)

ii) Joint C: The forces acting at the joint is shown in Fig. and the forces in

members are computed considering equilibrium condition at joint C

Fy=0; FCB=0;

Fx=0; FCE - 40=0; FCE=40kN(Tension)

iii) Joint B: The forces acting at the joint is shown in Fig. and the forces in

members are computed considering equilibrium condition at joint B

FAC

FAB

RA= 30 kN

FCE FAC=40kN

FCB


71/88



Fy=0; -30+50 sin-FBEsin=0; FBE=0

Fx=0; 50 cos - FBD=0; FBD=-40kN (Compression)

iv) Joint D: The forces acting at the joint is shown in Fig. and the forces in

members are computed considering equilibrium condition at joint D

Fy=0; FDE=0; Fx=0; FDF + 40=0; FDF=-40kN (Comprn.)

Forces in all the members are shown in Fig.

00000

-50

40 40 40 40

-50

-40-40

H

FDB

EC G

A

Step 2: Strain Energy

A= 0.01m2; E=2*105 N/mm2 = 2*108 kN/m2, AE = 2*106 kN

(Ui)p=15.83*10-3

kN-m

FBD

30kN

FAB= 50kN FCB=0FBE

FDF FBD=40 FDE

4*(-40)*25)50(24*40*410*2*2

1

2AE

LP )(U 222

6

13n

1i

2

Pi


72/88



5. Determine the maximum slope and maximum deflection in a cantilever beam of span

L subjected to point load W at its free end by using strain energy method. EI is constant

Solution:

i) Maximum Deflection

BM at 1-1 Mx= -Wx

From 2nd theorem of Castigliaino

ii) Maximum Slope

Maximum slope occurs at B, Virtual moment M‟ is applied at B

Bending moment at 1-1 is Mx = -Wx – M‟

L

x

1

1 W

A M’

B

L

x

1

1 W

A B

B

L

0

xx

δ EI

dxMW

M

M

U

L

0B

EI

dx(-Wx)(-x)δ x,-

W

M

Answer

3EI

WL

3

x

EI

W dxx

EI

Wδ

3L

0

L

0

32

B

EI

dxMM'

M

M'

Uθ

L

0

xx

B


73/88



From 2nd theorem of Castiglano

Substituting M‟=0

6. Calculate max slope and max deflection of a simply supported beam carrying udl of

intensity w per unit length throughout its length by using Castigliano‟s Theorem

L

w

L

0B

x

EI

dx)M'-(-Wx(-1)θ 1,-

M'

M

Answer 2EI

WL

2

x

EI

W dxx

EI

Wθ

2L

0

L

0

2

B


74/88



i) Maximum Slope:

Maximum slope occurs at support. A virtual moment M‟ is applied at A.

Reactions:

BM at 1-1

Put M‟=0

ii) Maximum Deflection:

Maximum Deflection occurs at mid span. A virtual downward load W‟ is applied at

mid-span.

L

w

1

1

x

M’

RA RB


75/88



Reactions:

BM at 1-1

Put W‟=0

L

w

1

1

x

M’

RA RB

L/2

2

W'

2

wLR ;

2

W'

2

wLR BA

ACRegionfor2

x

W'

M and

2

wx -x

2

W'

2

wLM x

2

x

L/2

0

2

C dx2x

2wx -x)

LW'

2wL(

EI2

W'U

L/2

0

43L/2

0

32

C4

x -

3

Lx

2EI

w dxx-(Lx

4EI

w2δ

64

L -

24

L

2EI

w δ

L/2

0

44

C

Answer 384EI

5wL δ

4

Cmax


76/88



CONJUGATE BEAM METHOD

This is another elegant method for computing deflections and slopes in beams.The principle of the method lies in calculating BM and SF in an imaginary beam called as

Conjugate Beam which is loaded with M/EI diagram obtained for real beam. ConjugateBeam is nothing but an imaginary beam which is of the same span as the real beam

carrying M/EI diagram of real beam as the load. The SF and BM at any section in theconjugate beam will represent the rotation and deflection at that section in the real beam.

Following are the concepts to be used while preparing the Conjugate beam.

It is of the same span as the real beam.

The support conditions of Conjugate beam are decided as follows:

Some examples of real and conjugate equivalents are shown.


77/88



Problem 1 : For the Cantil ever beam shown in f igure, compute def lection and rotation

at (i) the free end (i i) under the


78/88



load

Conjugate Beam:

By taking a section @ C´ and considering FBD of LHP,

EI

2253

EI

150-f SF 21x


79/88



BM @ C´= ;EI

45023

EI

150-21

Similarly by taking a section at A‟ and considering FBD of LHP;

SF @ A‟ =

EI

225

BM @ A‟ = EI

90022

EI

225

SF @ a section in Conjugate Beam gives rotation at the same section in Real Beam

BM @ a section in Conjugate Beam gives deflection at the same section in Real Beam

Therefore, Rotation @ C =

EI

225 ( )

Deflection @ C= EI

450

Rotation @ A =EI

225 ( )

Deflection @ A = EI

900


80/88



Problem 2: For the beam shown in f igure, compute def lections under the loaded

poin ts. Also compute the maximum def lection. Compute, also the slopes at suppor ts.

Note that the given beam is symmetrical. Hence, all the diagrams for this beam should besymmetrical. Thus the reactions are equal & maximum deflection occurs at the mid span.

The bending moment for the beam is as shown above. The conjugate beam is formed and

it is shown above.For the conjugate beam:

]BeamConjugateonload[TotalVV 21'

B

'

A

= EI30EI602121 432


81/88



=EI

150

EI

120

EI

18021

To compute δC :A section at C‟ is placed on conjugate beam. Then considering FBD of LHP;

+ BM @ C‟= 1EI

6033

EI

15021

=EI

360

EI

90

EI

450

;EI

360δC

δD = δC (Symmetry)

To compute δE:

A section @ E‟ is placed on conjugate beam. Then considering FBD of

LHP;

+ BM @ E‟= 12EI

303

EI

6035

EI

15021

i.e δE = EI

420

EI

60

EI

270

EI

750

θA =EI

150 ( ) θB =

EI

150 ( )


82/88



Problem 3: Compute def lection and slope at the loaded point for the beam shown in

f igure. Given E = 210 Gpa and I = 120 x 10 6 mm

4 . Al so calcul ate slopes at A and B.

Note that the reactions are equal. The BMD is as shown above.


83/88



To Compute reactions in Conjugate Beam:

EI

720

EI

360

EI

3606V '

A

;

SF and BM at C‟ is obtained by placing a section at C‟ in the conjugate beam.

SF @ C‟ =

+ BM @ C‟ =

Given E = 210 x 109 N/m2 = 210 x 10

6 kN/m

2

I = 120 x 106

mm4

= 120 x 10

6 (10

-3 m)

4

= 120 x 106 (10-12)= 120 x 10

-6 m

4;

EI = 210 x 106 (120 x 10

-6) = 25200 kNm

-2

Rotation @ C =25200

30 = 1.19 x 10

-3 Radians ( )

03

EI

120

2

13

EI

60

2

1VV0fy 'B

'

A

0;EI

180

EI

90

VV

'

B

'

A

EI

270VV 'B

'

A

023EI

120

2

143

EI

60

2

16V 0m '

AB'

EI

120V 'A

EI

150V 'B

3EI

60

2

1

EI

120

EI

30

13EI

60

2

13

EI

120

EI

270

EI

90

EI

360


84/88



Deflection @ C =25200

270 = 0.0107 m

= 10.71 mm ( )

θA = 4.76 X 10-3 Radians

θB = 5.95 X 10-3

Radians:

Problem 4: Compute slopes at suppor ts and deflections under loaded points for the

beam shown in f igure.


85/88



To compute reactions and BM in real beam:

+ 0M

B 031006509VA

66.67kN9

600VA 83.33kNVB

BM at (1) – (1) = 66.67 xAt x = 0; BM at A = 0, At x = 3m, BM at C = 200 kNm

BM at (2) – (2) = 66.67 x – 50 (x-3) = 16.67 x + 150

At x = 3m; BM at C = 200 kNm, At x = 6m, BM at D = 250 kNm

BM at (3) – (3) is computed by taking FBD of RHP. ThenBM at (3)-(3) = 83.33 x (x is measured from B)

At x = 0, BM at B = 0, At x = 3m, BM at D = 250 kNm

To compute reactions in conjugate beam:

+ 0M 'B

i.e 02EI

83.333

2

14

EI

253

2

14.5

EI

10037

EI

2003

2

19V 'A

EI

38509V '

A

EI

1003

EI

2003

2

1VV0fy 'B

'

A

EI83.333

21

EI253

21

EI

762.5

EI

427.77V 'A

EI

334.73V 'B

150VV0fy BA


86/88



( ) ( )

To Compute δC :

A Section at C‟is chosen in the conjugate beam:

+ BM at C‟ = 1EI

2003

2

13

EI

427.77

=EI

983.31

δC = EI

983.31

To compute δD:

Section at D‟ is chosen and FBD of RHP is considered.

+ BM at D‟ = 1EI

83.333

2

13

EI

334.73

=EI

879.19

EI

879.19δD

EI

427.77θA

EI

334.73θB


87/88



Problem 5: Compute to the slope and def lection at the fr ee end for the beam shown in

figure.


88/88


The Bending moment for the real beam is as shown in the figure. The conjugate beam

also is as shown.

Section at A‟ in the conjugate beam gives

SF @ A‟ = 4

0

2

dxEI

5x

= 643EI

5

EI

5 403

x3

= 3EI

320

θA =3EI

320 ( )

BM @ A‟ = dxx5xEI

1 4

0

2

= 2564EI5

4

x

EI

5-4

0

4

δA = EI

320

157635065-strength-of-materials-notes.pdf

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