Strength of Materials

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0 INTRODUCTION 6

Introduction On the basis of time of action of load On the basis of direction of load On the basis of area of acting the load 9 1 Couple L0ADS Pure Bending torsion Free Body Diagram

Introduction Strength Classification of stresses Normal stresses 22 Shear stress

2 Stress tensor STRESSES Effects of various loads acting on the body

Introduction Classification of strain Normal strain Longitudinal and lateral strain 44 Volumetric strain

3 Shear strains STRAINS Sign conventions for shear strains

Properties of materials Young‟s modulus of elasticity Modulus of rigidity Bulk modulus 55

4 Poisson‟s ratio ELASTIC State of simple shear CONSTANTS Relationships between various constants

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Stress and strain diagram Limit of proportionality Elasticity Plasticity 73

5 Ductility, Brittleness, Malleability MECHANICAL Yield strength, ultimate strength, rapture PROPERTIES strength OF MATERIALS

Work done by load Strain energy due to torsion Strain energy due to bending Resilience 81

6 Toughness STRAIN Effect of carbon percentage on properties ENERGY, RESIELENCE AND TOUGHNESS

Principle of superposition Elongation of bar due to axial load Bar of varying cross-section 7 Uniformly tapering circular bar 98 NORMAL Uniformly tapering rectangular bar STRESSES Elongation of bar of uniform cross-section AND STRAIN due to self- weight Compound bars or parallel bars Statically indeterminate problems

Thermal effects Free expansion of bar Temperature stresses in bar fixed at the both 8 ends 108 THERMAL Temperature stresses in composite bars STRESSES

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Pure bending Theory of simple bending Moment of resistance 9 Bending equation 129 PURE Assumptions BENDING AND Design criteria BENDING Analysis of bending equation STRESSES

Distribution of shear stresses Assumptions Shear stress distribution –rectangular section 10 Circular section 144 SHEAR I-section STRESSES IN BEAMS

Pure torsion Moment of resistance Torsion equation 11 Assumptions 151 TORSION Shear stress distribution in shafts Analysis of torque equation Compound shafts

Introduction Stresses on inclined section pq State of stress at a point due to biaxial stress 12 State of stress due to simple shear 167 PRINCIPAL State of stress due to normal and shear stress STRESSES Normal and shear stress on a plane AND STRAINS perpendicular to oblique plane Mohr Circle

Type of support Types of beams Sign conventions 13 SFD and BMD 192 S.F.D Relationship b/w load, force and B.M. AND Cantilever beam B.M.D Simply supported beams

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Thin cylindrical shell subjected to internal pressure 14 Maximum shear stress in cylindrical shell 244 Volumetric strain of thin cylindrical shell THIN Design of thin cylinder CYLINDER Thin spherical shells subjected to internal pressure Volumetric strain in spherical shell Cylindrical shell with hemispherical ends

Lame‟s theory When only external pressure is zero 15 When only internal pressure is zero 261 When internal pressure is pr and external THICK pressure pR CYLINDER For solid circular shaft, subjected only to external pressure pr Graphical representation of lame‟s theory Compound cylinders Shrinking another cylinder over the cylinder Shrink fit allowance Thick spherical shell

Differential equation of the deflection curve of beam 16 Sign conventions Double integration method 268 DEFLECTION Steps for solving the problems OF BEAMS Macaulay‟s method Moment area method Conjugate beam method Strain energy method

Introduction Close coiled helical spring: axial pull Closed – coiled helical springs: axial couple or torque 307 Open – coiled helical spring: axial force 17 Open coiled helical spring: axial torque SPRINGS Series and parallel arrangement of springs Leaf or carriage springs Flat spiral springs

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Introduction Equilibrium of elastic body Buckling stress 18 Slenderess ratio 314 COLUMNS Euler‟s theory End conditions Rankine theory

Introduction Maximum principal stress theory : Rankines‟s theory Maximum principal strain theory: St. 332 Venant‟s theory 19 Maximum shear stress theory: guest‟s THEORIES OF theory ELASTIC Maximum strain energy theory or haigh‟s FAILURE theory Maximum shear strain energy (or distortion energy) theory Mises- Henky theory

Tests

20 Answer Key 352 TESTS

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INTRODUCTION

 There are three fundamental areas of engineering mechanics: i. Statics ii. Dynamics iii. Strength of materials or mechanics of materials  Statics and dynamics deals with the effect of forces on rigid bodies i.e. the bodies in which change in shape can be neglected.  Strength of material deals with the relation between externally applied loads and their internal effects on sold bodies.  “Strength of material is a branch of applied mechanics that deals with the behaviour of solid bodies subjected to various types of loading.”  The principal objective of strength of materials is to determine the stresses, strains, and displacements in structures and their components due to the loads acting on them.  An understanding of mechanical behaviour is essential for the safe design of all types of structures, whether airplanes and antennas, buildings and bridges, machines and motors, or ships and spacecraft.  In designing, engineer must consider both dimensions and material properties to satisfy the requirements of strength and rigidity.

Mechanics of Rigid Bodies

 The mechanics of rigid bodies is primarily concerned with the static and dynamic behavior under external forces of engineering components and systems which are treated as infinitely strong and undeformable. Primarily, we deal here with the forces and motions associated with particles and rigid bodies.

 A basic requirement for the study of the mechanics of deformable bodies and the mechanics of fluids is essential for the design and analysis of many types of structural members, mechanical components, electrical devices, etc, encountered in engineering.

 A rigid body does not deform under load

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Mechanics of deformable solids

Mechanics of Solids

 The mechanics of deformable solids is more concerned with the internal forces and associated changes in the geometry of the components involved. Of particular importance are the properties of the materials used, the strength of which will determine whether the components fail by breaking in service, and the stiffness of which will determine whether the amount of deformation they suffer is acceptable.

 Therefore, the subject of mechanics of materials or strength of materials is central to the whole activity of engineering design. Usually the objectives in analysis here will be the determination of the stresses, strains, and deflections produced by loads. Theoretical analyses and experimental results have an equal role in this field.

In short, Mechanics of Solids deals with the relation between the loads applied to a solid (non-rigid) body and the resulting internal forces and deformations induced in the body.

Principle Objective = determine the stresses, strains, and displacements in structures and their components due to loads acting on them.

Alternate Names = Strength of Materials or Mechanics of Deformable Bodies

These notes will provide a basis to determine:  The materials to be used in constructing a machine or structure to perform a given function.  The optimal sizes and proportions of various elements of a machine or structure.  If a given design is adequate and economical.  The actual load carrying capacity of a structure or machine. (structure may have been design for a purpose other than one being considered).

Guru Gyan

Mass is a property of matter that does not change from one location to another. Weight refers to the gravitational attraction of the earth on a body or quantity of mass. Its magnitude depends upon the elevation at which the mass is located Weight of a body is the gravitational force acting on it.

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CHAPTER 1 LOADS INTRODUCTION . Load may be defined as the external force or couple to which a component is subjected during its functioning. . Load is a vector quantity. . All the external forces acting on bodies are SURFACE forces. . Externally applied forces may be due to  Working environment  Service conditions  Contact with other members  Fluid pressure  Gravity or inertia forces . The forces acting on the body due to volume of the body is called BODY force. . Loads may be classified on following basis:  On the basis of time  On basis of direction of load  On the basis of area

ON THE BASIS OF TIME OF ACTION OF LOAD . On the basis of time of action of load, load may be classified as  Static load  Dynamic load STATIC LOAD may be  Dead load  Gradually applied load

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DEAD LOAD . Dead load includes loads that are relatively constant over time, including the weight of the structure itself. GRADUALLY APPLIED LOAD . Gradually applied load may be defined as the load whose magnitude increases gradually with the time. . Gradually applied loads become dead load after a certain period of time.

FIGURE 1.1

DYNAMIC LOAD may be

 Impact  Fatigue

IMPACT LOAD . The load which are acting for short interval time are said to be impact load. If t = time of application of load T = time period of vibration. Then 푇 푡 < 푖푚푝푎푐푡 푙표푎푑 2 푡 ≥ 3푇 푠푡푎푡푖푐 푙표푎푑 푡 = 2푇 푎푠푠푢푚푒푑 푖푚푝푎푐푡 푙표푎푑

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Since stress produced is greater than the static load.

휎푖푚푝푎푐푡 > 휎푠푡푎푡푖푐

휎푖푚푝푎푐푡 = 휎푠푡푎푡푖푐 × 푖푚푝푎푐푡 푓푎푐푡표푟

FIGURE 1.2

ퟐ풉 풊풎풑풂풄풕 풇풂풄풕풐풓 = ퟏ + ퟏ + 휹풔풕 Where

휎푖푚푝푎푐푡 = Stress due to impact loads

휎푠푡푎푡푖푐 = Stress due to static loads

훿푠푡 = Static deflection FATIGUE LOADS . Fatigue load are those load whose magnitude or direction or both magnitude and direction change with respective to time.

FIGURE 1.3

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Note:

ON THE BASIS OF DIRECTION OF LOAD . On the basis of direction of load, load may be classified as  Normal loads  Shear loads NORMAL LOADS . The loads which are acting perpendicular to the surface of the body. . Normal loads may be  Axial  Eccentric axial AXIAL LOADS . The loads which are perpendicular to the surface and passes through the longitudinal axis of the body are called axial loads.

Figure 1.4 ECCENTRIC AXIAL LOAD . The loads which are perpendicular to the surface and does not pass through the axis of the body.

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Figure 1.5

Note: Eccentric Load creates → Axial Load + Bending Moment

SHEAR LOADS . The loads which acts parallel to the surface of body. . Shear loads may be  Transverse shear load  Eccentric transverse shear load

TRANSVERSE SHEAR LOAD . The loads which are parallel to the surface and passes through the longitudinal axis of the body are called transverse shear loads.

Figure 1.6

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ECCENTRIC TRANSVERSE SHEAR LOAD . The loads which are parallel to the surface and do not pass through the longitudinal axis of the body are called transverse shear loads.

Figure 1.7

ON THE BASIS OF AREA OF ACTING THE LOAD . On the basis of area of acting the load, load may be classified as  Concentrated point load  Uniformly distributed load  Uniform variable load  Uniformly distributed moment  Uniform variable moment

CONCENTRATED POINT LOAD

Figure 1.8

. A point load is one which is concentrated at a point. . Self-weight of the body is assumed to be concentrated point load acting through the centre of gravity.

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UNIFORMLY DISTRIBUTED LOAD

Figure 1.9

. The load which is not acting through a point but is distributed uniformly over some area.

UNIFORM VARIABLE LOAD

Figure 1.10

. A uniformly varying load is one in which load intensity varies from one end to other. . They are also called linearly varying load.

UNIFORMLY DISTRIBUTED MOMENT . It is the moment which is not acting at a point but uniformly distributed over some area.

UNIFORM VARIABLE MOMENT . A uniformly varying moment is one in which moment intensity varies from one end to other.

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COUPLE . A couple consists of two parallel forces that are equal in magnitude and opposite in direction.

Figure 1.11

BENDING COUPLE . A couple is said to be bending couple if plane of couple is passing through the longitudinal axis of the body and perpendicular to the plane of cross-section.

Figure 1.12

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TWISTING COUPLE

. A couple is said to be twisting couple when plane of the couple is perpendicular to longitudinal axis of the member and parallel to the cross-section of plane .

Figure 1.13

OBLIQUE COUPLE . When member is subjected to bending and twisting couple when plane of couple acting in a plane inclined to the axis and plane is known as oblique plane.

푀푥 = 푀푐표푠훼 퐵푒푛푑푖푛푔

푀푦 = 푀푐표푠훽 퐵푒푛푑푖푛푔

푀푧 = 푀푐표푠훾 푡푤푖푠푡푖푛푔

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Figure 1.14 PURE BENDING . A member is said to be under pure bending when is subjected to two equal and opposite couple in a plane passing through the longitudinal axis of member and perpendicular to cross-section.

Figure 1.15

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PURE TORSION . A member is said to be in pure torsion couple in the plane perpendicular long axis of member and parallel to cross section of plane

Figure 1.16 FREE BODY DIAGRAM . A diagram showing all the forces acting directly or indirectly on the body that completely balance each other Steps involved for drawing a free body diagram . Step I. At the beginning, a clear decision is to be made by the analyst on the choice of the body to be considered for free body diagram. . Step II. Then that body is detached from all of its surrounding members including ground and only their forces on the free body are represented. . Step III The weight of the body and other external body forces like centrifugal, inertia, etc., should also be included in the diagram and they are assumed to act at the centre of gravity of the body. . Step IV When a structure involving many elements is considered for free body diagram, the forces acting in between the elements should not be brought into the diagram. . Step V The known forces acting on the body should be represented with proper magnitude and direction.

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. Step VI If the direction of unknown forces like reactions can be decided, they should be indicated clearly in the diagram.

Figure 1.17 SOLVED EXAMPLE 1.1 Consider the following figure. At point A, bar is rigidly fixed and at point D, bar is loaded in negative x,y and z- direction with loads P,Q and R respectively. Determine type and nature of load acting on each bar.

SOLUTION: On bar CD: Load P is acting perpendicular to cross section hence is a direct tensile load. Load Q and R is acting parallel to surface, hence are transverse shear loads.

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On bar BC: Load P is acting parallel to surface, hence are transverse shear loads. Load Q is acting perpendicular to surface but not passing through the axis, hence is eccentric axial tensile load. Load R is acting parallel to surface but not passing through the axis hence is eccentric transverse shear loads. On bar AB: Load P is acting perpendicular to cross section but not passing through the axis hence is a direct compressive load. Load Q is acting parallel to surface and passing through the axis hence is transverse shear load. Load R is acting parallel to surface but not passing through the axis hence is eccentric transverse shear loads

BAR AXIAL ECENTRIC TRANSVERSE ECCENTRIC AXIAL SHEAR TRANS V. SHEAR AB 0 -P Q R BC 0 +Q P R CD P 0 Q,R 0

Guru Gyan

Force: Magnitude (P), direction (arrow) and point of application (point A) is important. Change in any of the three specifications will alter the effect on any force. In case of rigid bodies, line of action of force is important (not its point of application if we are interested in only the resultant external effects of the force).

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CHAPTER 2 STRESSES INTRODUCTION

Analysis of Stress and Strain

 Concept of stress: Let us introduce the concept of stress as we know that the main problem of engineering mechanics of material is the investigation of the internal resistance of the body, i.e. the nature of forces set up within a body to balance the effect of the externally applied forces.

 As we know that in mechanics of deformable solids, externally applied forces acts on a body and body suffers a deformation. From equilibrium point of view, this action should be opposed or reacted by internal forces which are set up within the particles of material due to cohesion. These internal forces give rise to a concept of stress.

. Thus, we can say that when some external forces are applied on a body, the body get deformed. For the equilibrium, this action must be opposed or reacted by some internal forces which are set up in the body due to cohesion. These internal forces give rise to the concept of STRESS. . Thus, stress is the internal resistance offered by the body to deformation when it is acted upon by the body some external force or load. . A PRISMATIC BAR is a straight structural member having the same cross section throughout its length. . Examples of prismatic bar are the members of a bridge truss, connecting rod in automobile engines, spokes of bicycle wheels, columns in buildings, and wing struts in small airplanes. . A section cut perpendicular to longitudinal axis of bar is called cross section of the bar.

. Let us consider a rectangular bar of some cross sectional area and subjected to some load or force (in Newtons) Let us imagine that the same rectangular bar is assumed to be cut into two halves at section XX. The each portion of this rectangular bar is in equilibrium under the

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action of load P and the internal forces acting at the section XX has been shown.

. Now stress is defined as the force intensity or force per unit area. Here we use a symbol  to represent the stress. 푃 휎 = 퐴 Where 휎 represents the stress induced in the body. This expression is valid only when the force is uniformly distributed over the surface area. . Hence, Stress is also defined as the load applied per unit area of cross- section. . If the force is not uniformly distributed, then we choose a small area δA and a load 훿푃 acting on this area, then stress may be calculated as

훿퐹 휎 = lim 훿퐴→0 훿퐴

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Figure 2.1

푃 = 퐹1 + 퐹2 + 퐹3 + − − − + 퐹푛

If,

퐹1 = 퐹2 = 퐹3 = − − −= 퐹푛

푃 = 푛퐹1

푃 퐹 = 1 푛 퐹 휎 = 1 = 휎 = 휎 = − − − − −= 휎 = 휎 1 퐴 2 3 푛 푎푣푔

푃 휎 = 푎푣푔 퐴

UNITS OF STRESS

Stress has the unit of force per unit area.

퐹표푟푐푒 푁 푆푡푟푒푠푠 = = = 푃푎푠푐푎푙푠 (푃푎) 퐴푟푒푎 푚2

1 푁 푚푚2 = 106 푁 푚2 = 106푃푎 = 1푀푃푎

. In MKS system, unit of force is kgf.

1 푘푔푓 = 9.807푁 ≈ 10 푁

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1 푡표푛푛푒 = 9.807 푘푁 ≈ 10 푘푁 Hence unit of stress,

푘푔 105푁 1 = = 105푃푎 ≈ 0.1 푁/푚푚2 푐푚2 푚2

Saint Venant’s Principle – (used for gauge length in testing) . “It states that except in the region of extreme ends of a bar carrying direct loading, the stress distribution over the cross section is uniform.” . This is also called Principal of Rapid distribution of localised stresses.

DIFFERENCE BETWEEN PRESSURE AND STRESS

Sr. PRESSURE STRESS N o . 1 It is always due to normal It may be due to normal and force. shear force.

2 It is the force that has been It is developed internally. applied externally.

3 Pressure is a scalar Stress is a tensor or pointing property. property.

4 It can be measured using It cannot be measured. pressure gauge.

5 Pressure can‟t be developed Stresses may be developed due due to stresses. to pressure. 6. Example: Hydrostatic Example: Thermal Stresses Pressure

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Guru Gyan

 Stress provides a measure of the intensity of an internal force.

 Strain provides a measure of the intensity of a deformation.

STRENGTH . Strength is defined as the maximum or limiting value of stress at that a material can with stand any failure or fracture. . It has the same unit that of stress. . Strength is a constant property for a given material.

휎푖푛푑푢푐푒푑 ≤ 휎푦푖푒푙푑푖푛푔

푁표 푦푖푒푙푑푖푛푔 표푐푐푢푟푠 푖푛 푡푕푖푠 푐푎푠푒

휎푖푛푑푢푐푒푑 ≤ 휎푢푙푡푖푚푎푡푒

푁표 푓푎푖푙푢푟푒 푖푛 푡푕푖푠 푐푎푠푒

CLASSIFICATION OF STRESS Stresses may be classified as i. Normal stress or Direct stress  Tensile and compressive stresses  Axial and bending stresses ii. Shear stresses iii. Torsional stresses

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Stress

Normal Shear

Axial Bending Tensile Compressive Tensional Direct

NORMAL STRESS . When a stress acts in a direction perpendicular to the cross-section, it is called as normal or direct stress. . Normal stress occurs due to dimensional distortion. . Normal stress may be  Axial  Bending AXIAL STRESS . The stress acting along the longitudinal axis of the bar which tend to change the length of the body. 푃 휎 = 퐴

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. This is also known as uniaxial state of stress, because the stresses acts only in one direction however, such a state rarely exists, therefore we have biaxial and triaxial state of stresses where either the two mutually perpendicular normal stresses acts or three mutually perpendicular normal stresses acts as shown in the figures below :

. Axial stress may be  Tensile  Compressive  Bending stress . When a sign convention for normal stresses is required, it is customary to define tensile stresses as positive and compressive stresses as negative. TENSILE STRESS . When a body is stretched by a force, the resulting stress is known as tensile stress. . Tensile stress exists between two parts when two parts draws each other towards each other.

Figure 2.2

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COMPRESSIVE STRESS . When the forces are reversed and try to compress the body, the resulting stress is known as compressive stress. . Compressive stress exists between two parts when they try to push each other from it.

Figure 2.3

BENDING STRESS . Bending stresses are developed in beams due to bending of a member. . Loads perpendicular to the length of the beam causes bending moment. The stresses produced at any section to resist the bending moment are called bending stresses.

Figure 2.4

푀푥푥 휎푏 = 푍푥푥 (Discussed in detail in chapter of bending stresses in beams)

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SHEAR STRESS . When the force is acting parallel or tangential to the surface, the resulting stress is known as shear stress. . Shear stress exists between parts of a body when the two parts exerts equal and opposite forces on each other laterally in a direction tangential to their surface in contact. . Shear stress occurs due to shape deformation of body. . Shear stress may be  Direct Shear stress  Torsional shear stress

DIRECT SHEAR STRESS . Direct shear stress is developed in the body when the forces try to act through the material directly. . Let us consider now the situation, where the cross sectional area of a block of material is subject to a distribution of forces which are parallel, rather than normal, to the area concerned. Such forces are associated with a shearing of the material, and are referred to as shear forces.

. The resulting force intensities are known as shear stresses, the mean shear stress being equal to

The Greek symbol τ (tau) (suggesting tangential) is used to denote shear stress.

Where P is the total force and A the area over which it acts.

. As we know that the particular stress generally holds good only at a point therefore we can define shear stress at a point as

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Figure 2.5 . Shear stresses are also developed indirectly when a body is subjected to bending and torsion.

TORSIONAL SHEAR STRESS . Torsional shear stress is developed in the body when a torque is applied to body (due to twisting).

Figure 2.6

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Figure 2.7

(Discussed in detail in chapter of torsional stresses in beams)

However, it must be borne in mind that the stress (resultant stress) at any point in a body is basically resolved into two components σ and τ one acts perpendicular and other parallel to the area concerned, as it is clearly defined in the following figure.

The single shear takes place on the single plane and the shear area is the cross - sectional of the rivet, whereas the double shear takes place in the case of Butt joints of rivets and the shear area is the twice of the X - sectional area of the rivet.

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SIGN CONVENTION FOR SHEAR STRESS . Sign convention for shear stress a shear stress acing on positive face of an element is positive if it acts in the positive directing of one of the co- ordinate axes and is negative if it acts in the negative direction of an axis. . Shear stress acting on a negative face of an element is positive if it acts in the negative direction of an axis and negative if it acts in a positive direction. STRESS TENSOR . Stress tensor is used to define the state of stress (how many stress components acting at a point. . Stress at a point in 3D is given by 휎푥푥 휏푥푦 휏푥푧 휏 휎 휏 휎 3퐷 = 푦푥 푦푦 푦푧 휏푧푥 휏푧푦 휎푧푧 . The no. of stress component in a stress tensor for a point in 3D are 9 (3 normal and 6 shear). . Since complementary shear stress are equal and opposite

휏푥푦 = −휏푦푥

휏푥푧 = −휏푧푥

휏푦푧 = −휏푧푦 . The number of stress components in a stress tensor to define the state of stress at a point is 6 (3 normal and 3 shear )

PROB 1: How much quantity is sufficient to describe the stress at a point in a co-ordinate plane?

1] 3 2] 4 3] 6 4] 9

Correct Answer: 3

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The state of stress at a point

PROB 2: The steel and brass bars comprising the stepped shaft in figure are to suffer the same displacement under tensile force of 40 kN. The diameter of the brass bar is

(Take and )

1] 45.65 mm 2] 52.75 mm 3] 18.27 mm 4] 96.94 mm

Correct Answer: 4

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Figure 2.8 Plane Stress Problems . Stress tensor in 2D is given by 휎푥푥 휏푥푦 휎 2퐷 = 휏푦푥 휎푦푦 . The number of stress component in a stress tensor for a point in 2D are 4 (2 normal and 2 shear). . Since

휏푥푦 = −휏푦푥 . The number of stress components in a stress tensor to define the state of stress at a point is 3 (2 normal and 1 shear).

EFFECTS OF VARIOUS LOADS ACTING ON THE BODIES EFFECT DUE TO AXIAL LOAD Axial load causes elongation or compression of the member. Tensile stress . Tensile stress causes elongation of the bar. 푃 휎 = 퐴 . Change in length is given by 푃퐿 훿 = 퐴퐸

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Figure 2.9

Compressive Stress . Compressive stresses causes compression of the bar. 푃 휎 = − 퐴 . Change in length is given by 푃퐿 훿 = − 퐴퐸

Figure 2.10

EFFECT OF ECCENTRIC AXIAL LOAD . Eccentric axial load is equivalent to axial load and constant bending moment. . Shear force is zero in this case. . Hence resultant stress is given by

휎푟 = 휎푎푥푖푎푙 + 휎푏푒푛푑푖푛푔

휎푟 = 휎푎 + 휎푏

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휎푚푎푥 = 휎푎 + 휎푏 = 휎푡표푝

휎푚푖푛 = 휎푎 − 휎푏 = 휎푏표푡푡표푚

Figure 2.11

Figure 2.12

In case of tensile loads,

휎푚푎푥 = 휎푎 + 휎푏

Bending stress will be maximum in the top fibre i.e. ymax

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Hence, Bending Formula , M f E  IYR

푃 푀푦푚푎푥 휎푚푎푥 = + 퐴 퐼푥푥 푃 푃푒푦푚푎푥 휎푚푎푥 = + 퐴 퐼푥푥 Similarly,

푃 푃푒푦푚푎푥 휎푚푖푛 = − 퐴 퐼푥푥 In case of compressive loads,

푃 푃푒푦푚푎푥 휎푚푎푥 = − − 퐴 퐼푥푥

푃 푃푒푦푚푎푥 휎푚푖푛 = − + 퐴 퐼푥푥

EFFECT OF TRANSVERSE SHEAR LOAD . Transverse shear load is equivalent to a constant shear force and a variable bending moment. . Direct shear stress, 푃 휏 = 퐴

Figure 2.13

STRENGTH OF MATERIALS

. Bending stress,

At distance „a‟ from free end

푀푥푥 푃. 푎 휎푏 = = 푍푥푥 푍푥푥 At a distance „l’ from free end

푀푥푥 푃. 푙 휎푏 = = 푍푥푥 푍푥푥 In this case, bending moment will be varying along the length and shear force will be constant throughout the length.

EFFECT OF ECCENTRIC TRANSVERSE SHEAR LOAD . Eccentric transverse shear load is equivalent to a direct shear, a variable bending moment and a constant torque.

Figure 2.14 Direct shear stress, 푃 휏 = 퐴 Torsional Formula, T f N  IYL

Torsional stress,

푀푧푧 푃. 푒. 푦푚푎푥 휏푇 = = 푍푧푧 퐼푧푧

STRENGTH OF MATERIALS

Bending stresses,

푀푦푚푎푥 푃. 푎. 푦푚푎푥 휎푏 = = 퐼푥푥 퐼푥푥

EXAMPLE 2.1 Determine the shear force, bending moment, torsional moment in the following figure:-

Figure 2.15

On bar CD: P is acting parallel to cross-section, hence a shear force P is acting. A 푏푒푛푑푖푛푔 푚표푚푒푛푡 = 푃 × 퐶퐷 is induced at point C. On bar BC: Eccentric shear force P is acting on bar BC. Due to eccentric P, a 푡표푟푠푖표푛푎푙 푚표푚푒푛푡 = 푃 × 퐶퐷 is induced. Due to P, bending moment is induced at B, 퐵푀 = 푃 × 퐵퐶. On bar AB: Shear force P is acting on bar AB at E. Due to this P, 퐵. 푀. = 푃 × (퐴퐵 − 퐶퐷) is induced at A. 푇표푟푠푖표푛푎푙 푚표푚푒푛푡 = 푃 × 퐵퐶 is induced in bar.

BAR AXIAL SHEAR BENDING TORSIONAL LOAD FORCE MOMENT MOMENT CD 0 P P.(CD) 0 BC 0 P P.(BC) P.(CD) AB 0 P P.(AB-CD) P.(BC)

STRENGTH OF MATERIALS

EXAMPLE 2.2 Determine the stresses acting at points A, B, C and D in the following figure. The load P is acting out of plane and perpendicular to the plane.

Figure 2.16

SOLUTION: At point A,

푃 푀푦푦 푀푥푥 휎퐴 = − + 퐴 푍푦푦 푍푥푥 At point B,

푃 푀푦푦 푀푥푥 휎퐵 = − − 퐴 푍푦푦 푍푥푥 At point C,

푃 푀푦푦 푀푥푥 휎퐶 = + + 퐴 푍푦푦 푍푥푥 At point D,

푃 푀푦푦 푀푥푥 휎퐷 = + − 퐴 푍푦푦 푍푥푥

STRENGTH OF MATERIALS

Concept of True Stress and strain:

True Stress (σt): Original area (initial) P Stress()  ______(1) A0 As with elongation, ceoss section area varies Present area at that instant, P   t A

PPA0  t    AAA0

As volume remain same,

A0 L0 =A L

A L 0 AL0 PLL    t     ALL0 0   0 

LLL0   L t L0  L0

t 1    Engg Stress

STRENGTH OF MATERIALS

True Strain (εt) : L   Engg. strain L0 L dL   t  l L0 L Final Length

L0  Initial Length

L Ad00    ln or  ln   2ln   L0  A   d  L t  ln  L0 as L L0  L LL t ln  ln(1   ) L0

t ln(1   )  Engg . strain