PHY
UNIT-1FRICTION AND MECHANICS OF SOLIDS Q1) A load of 4.0 Kg is suspended from a ceiling through a steel wire of radius 2.0 mm. Find the tensile stress developed in the wire when equilibrium is achieved .Take g=3.1 π m/s2. A1)
Q2) A copper wire of 3 m length and 1 mm diameter is subjected to a tension of 5N. Calculate the elongation produced in the wire if the young’s modulus of elasticity of copper is 120Gpa. A 2)
Q3) A cylindrical wire of length 1m and radius 5 mm is rigidly clamped at one end. Calculate the couple required to twist the free and through an angle 450. The rigidity modulus of the material of the wire is 200 X 109 Pa A 3)
Q4) State Hooke’s law?A4)
Q5) Define stress and strain. Write their units. A5) Deformation is the change in shape of the body when a force is applied on it. Even very small magnitude of forces is enough to cause some deformation. Two terms describe the forces on objects undergoing deformation: Stress Strain STRESS When an external force is acting on an elastic body, it causes deformation that means change in shape or in size or both. Stress = Force/Area = F/AThe stress is defined as the restoring force acting on unit area.Unit: The unit for stress is Newton Metre-2 or Nm-2 or Pascal written as Pa.Restoring Force: Due to elastic property, a force is developed within the material, which is equal and opposite to the applied force, this force is responsible to bring the body to its original shape and size. This force is Restoring Force.Since the applied force and the restoring force are equal in magnitude, the ‘stress’ is measured as the applied force acting per unit area.When the applied force tends to compress the body, the stress is compressive stress. When it tends to increase the length in the direction of the force, it is tensile stress. When it acts parallel to the surface of a body, the stress is tangential stress or Shear stress. When an object is being squeezed from all sides, like a submarine in the depths of an ocean, we call this kind of stress a bulk stress or volume stress. STRAIN Within the elastic limits, the strain produced in a body is directly proportional to the stress which causes it. i.e. strain ∝ stress or stress ∝ strain Strain: Change in dimensions to original dimensions is known as Strain. Unit: it has no unit. Q6) Name the types of strains and define it?A 6) Types of Strain (1) Linear Strain: When a wire or bar is subjected to two equal and opposite forces namely pulls at its ends there is an increase in the length. If the forces are tensile, the body is elongated. If the forces are compressive, the length is shortened in the direction of the forces. This is called the 'linear strain'.The linear strain is defined as the ratio of change in length to the original length. If the change (increase or decrease) in length is ' 5m ' in a wire or bar of original length L=20m. Linear strain = Change in length/ original length = 5m/20m = 0.25 As the linear strain is ratio of lengths, it has no unit. (2) Bulk (or) Volume Strain: When a force is applied uniformly and normally to the entire surface of the body, there is a change in volume of the body, without any change in its shape. This strain is called 'bulk or volume strain.Volume strain is defined as the ratio of change in volume to the original volume. If 'v' is the change in volume produced in a body of original volume ‘V’Bulk or volume strain = Change in volume v / original volume VIt has also no unit. (3) Shearing (or) Rigidity strain: When a force is applied parallel to one face of a body, the opposite side being fixed, there is a change in shape but not in size of the body. This strain is called the shearing strain. Solids alone can have a shearing strain. It is measured by the angle of the shear 'θ' in radian.Q7) Write various factors that affect elasticity?A7) It is found that bodies lose their elastic limit, due to elastic fatigue. Therefore, the manufacture should choose the material in such a way that it should regain its elastic property even when it is subjected to large number of cycles of stress. For example, substances like quartz, phosphor, bronze etc. May be employed in manufacturing of galvanometers, electrometers etc. after knowing their elastic properties. Apart from elastic fatigue some material will have change in their elastic property because of the following factors.Effect of stress Effect of annealing Hammering and Rolling temperature Effect of impurities Effect of nature of crystals Effect of Stress When a material is subjected to a large number of cycles of stresses, it loses its elastic property even within elastic limit, it results in the formation of large crystal grain, which reduces the elastic property of the material. Taking these into account, the working stress on an engineering piece is kept far below its ultimate tensile strength. Therefore, the working stress on the material should be kept lower than the ultimate tensile strengthen and the safety factor. 2. Effect of AnnealingAnnealing is a process by which the material is heated to a very high temperature and then it is slowly cooled. Usually this process is adopted for the material to increase the softness and ductility in the material. But if annealing is made to a material it results in the formation of large crystal grains, which ultimately reduces the elastic property of the material.3. Hammering and RollingCrystal grains break up into smaller units by Hammering and rolling. This results in increase in elasticity of material. 4. Effect of temperatureIntermolecular forces decrease with increase in temperature so The elastic property of the materials decreases with the rise in temperature. Normally the elasticity increases with the decrease in temperature and vice-versa. Examples1. Lead is not a good elastic material. But at low temperature it becomes a very good elastic material. The elastic property of lead increases when the temperature is decreased.2. A carbon filament which is highly elastic at normal temperature becomes plastic when it is at high temperature3. Invar is a special alloy used for making pendulums and its elasticity is not affected by the temperature change in the normal temperature range.5. Effect of impuritiesThe addition of impurities produces variation in the elastic property of the materials. The increase and decrease of elasticity depends upon the type of impurity added to it.Examples:1. When potassium is added to gold, the elastic property of gold increases.2. When carbon is added to molten iron, the elastic property of iron decreases provided the carbon content should be more than 1% in iron.6. Effect of Nature of crystalsThe elasticity also depends upon the types of the crystals, whether it is a single crystal or poly crystals. For a single crystal the elasticity is more and for a poly crystal the elasticity is less.Q 8) Write a note on elasticity?A 8) Unlike fluids, solids do not respond to outside force by flowing or easily compressing. The term elasticity refers to the manner in which solids respond to stress, or the application of force over a given unit area. An understanding of elasticity—a concept that helps to illuminate the properties of objects from steel bars to rubber bands to human bones.Elasticity, ability of a deformed material body to return to its original shape and size when the forces causing the deformation are removed. A body with this ability is said to behave (or respond) elastically.Most solid materials exhibit elastic behaviour to a greater or lesser extent, but there is a limit to the magnitude of the force and the accompanying deformation within which elastic recovery is possible for any given material. This limit, called the elastic limit.Elastic limit is the maximum stress or force per unit area within a solid material that can arise before the onset of permanent deformation. Stresses beyond the elastic limit cause a material to yield or flow. For these type of materials the elastic limit tells the end of elastic behaviour and the beginning of plastic behaviour. But for some brittle materials, stresses beyond the elastic limit result in fracture. No plastic deformation exists in these types of cases.Metals and ElasticityMetals, in fact, exhibit a number of interesting characteristics with regard to elasticity. With the notable exception of cast iron, metals tend to possess a high degree of ductility, or the ability to be deformed beyond their elastic limits without experiencing rupture. Up to a certain point, the ratio of tension to elongation for metals is high: in other words, a high amount of tension produces only a small amount of elongation. Beyond the elastic limit, however, the ratio is much lower: that is, a relatively small amount of tension produces a high degree of elongation.For example, a steel bar or wire can be extended elastically only about 1 percent of its original length, while for strips of certain rubber like materials, elastic extensions of up to 1,000 percent can be achieved. Thus elastic limit depends on the type of solid considered.Steel is much stronger than rubber, however, because the tensile force required to effect the maximum elastic extension in rubber is less (by a factor of about 0.01) than that required for steel. The elastic properties of many solids in tension lie between these two extremes.The different macroscopic elastic properties of steel and rubber are because of the different microscopic structures. Here is the point to note that the elasticity of steel and other metals arises from short-range interatomic forces that, when the material is unstressed, maintain the atoms in regular patterns. Under stress the atomic bonding can be broken at quite small deformations. Whereas at the microscopic level, rubber like materials and other polymers consist of long-chain molecules that uncoil as the material is extended and recoil in elastic recovery. The mathematical theory of elasticity and its application to engineering mechanics is concerned with the macroscopic response of the material and not with the underlying mechanism that causes it.The elastic limit is in principle different from the proportional limit, which marks the end of the kind of elastic behaviour that can be described by Hooke’s law, namely, that in which the stress is proportional to the strain (relative deformation) or equivalently that in which the load is proportional to the displacement. The elastic limit nearly coincides with the proportional limit for some elastic materials, so that at times the two are not distinguished; whereas for other materials a region of non-proportional elasticity exists between the two.Gases and liquids also possess elastic properties since their volume changes under the action of pressure. For small volume changes, the bulk modulus, κ, of a gas, liquid, or solid is defined by the equation P = −κ(V − V0)/V0, where P is the pressure that reduces the volume V0 of a fixed mass of material to V. Since gases can in general be compressed more easily than liquids or solids, the value of κ for a gas is very much less than that for a liquid or solid. By contrast with solids, fluids cannot support shearing stresses and have zero Young’s modulus. BonesThe tensile strength in bone fibers comes from the protein collagen, while the compressive strength is largely due to the presence of inorganic (non-living) salt crystals. It may be hard to believe, but bone actually has an ultimate strength, both in tension and compression, greater than that of concrete.The ultimate strength of most materials is rendered in factors of 108 N/m2—that is, 100,000,000 newtons (the metric unit of force) per square meter. For concrete under tensile stress, the ultimate strength is 0.02, whereas for bone, it is 1.3. Under compressive stress, the values are 0.2 and 1.7, respectively. In fact, the ultimate tensile strength of bone is close to that of cast iron (1.7), though the ultimate compressive strength of cast iron (5.5) is much higher than for bone.It may be hard to understand how bone can be stronger than concrete, but that is largely because the volume of concrete used in most situations is much greater than the volume of any bone in the body of a human being. By way of explanation, consider a piece of concrete no bigger than a typical bone: under relatively small amounts of stress, it would crumble.Q 9) Write a note on plasticity?A 9) The property of the material by virtue of which it does not regain its original shape when the external force acting on it is removed. It is called plasticity. Plasticity.It is also known as plastic deformation, is the ability of a solid material to undergo permanent deformation.Under the effect applied forces a non-reversible change of shape is occur. The transition from elastic behaviour to plastic behaviour is known as yielding.Paraffin wax, wet clay are the nearest approach to a perfectly plastic bodies.The solid acquires a permanent change in its structure if, on restoring the external environment (under which the deformation process has taken place) to the initial configuration, the solid fails to restore to the initial structure. The quantifiers for such structural changes can be the shape of the body (at a macroscopic level of observation) or a rearrangement of defects (at a microscopic level of observation).Unlike elasticity, plasticity is fundamentally a microscopic phenomenon.To illustrate the phenomenon of plasticity we consider a simple example of stretching a metal wire (with cylindrical cross-section) under an isothermal environment. A representative element of the wire is selected and the following two quantities are measured: The force per unit cross-sectional area (stress) and The change in length of the element with respect to some fixed reference state (strain). Below a certain critical value of stress and strain, the element under observation returns to its original state of stress and strain upon removal of the external mechanism for stretching. The collection of all such stress/strain values forms the elastic range associated with the material. The deformation remains elastic as long as the stress/strain values are restricted to the elastic range. For values beyond the elastic range, the element will undergo permanent structural changes. The boundary of the elastic range signifies the onset of plastic deformation and therefore can be appropriately termed as the yield limit for the transition from an elastic to a plastic deformation process. It should however be noted that for most materials an accurate measurement of the yield limit is difficult because of the gradual nature of the transition from elastic to plastic behaviour. It then becomes a matter of convention to choose an appropriate yield limit for such materials. Based on the nature of the yield limit, we distinguish between two models of plastic flow. The first model is known as perfect plasticity (or the ideal plastic flow), wherein the yield limit for stress is assumed to remain constant. The second model, however, allows the yield stress to vary with the structural changes in the body Note from that for a zero strain, the stress does not necessarily vanish. This can be attributed to the presence of microstructure in the form of defects in the unstrained state, which induces internal stresses so as to maintain an equilibrium configuration of defects in the element. Plasticity, ability of certain solids to flow or to change shape permanently when subjected to stresses of intermediate magnitude between those producing temporary deformation, or elastic behaviour, and those causing failure of the material, or rupture (see yield point). Plasticity enables a solid under the action of external forces to undergo permanent deformation without rupture. Elasticity, in comparison, enables a solid to return to its original shape after the load is removed. Plastic deformation occurs in many metal-forming processes (rolling, pressing, forging) and in geologic processes (rock folding and rock flow within the earth under extremely high pressures and at elevated temperatures).Plastic deformation is a property of ductile and malleable solids. Brittle materials, such as cast iron, cannot be plastically deformed, though at elevated temperatures some, such as glass, which is not a crystallized solid, do undergo plastic flow. Q 10) Write a note on strain hardening and strain softening?A 10) Strain hardening (also called work-hardening or cold-working) is the process of making a metal harder and stronger through plastic deformation.In the plastic region, the true stress increases continuously i.e when a metal is strained beyond the yield point, more and more stress is required to produce additional plastic deformation and the metal seems to have become more stronger and more difficult to deform. This implies that the metal is becoming stronger as the strain increases. Hence, it is called "Strain Hardening". The plastic portion of the true stress-strain curve (or flow stress curve) plotted on a log-log scale gives the n value as the slope and the K value as the value of true stress at true strain of one.
Figure 21 log(ø) = log(K)+ n x log(e)For materials following the power law, the true strain at the Ultimate Tensile Strength is equal to n. when you plot the log-log plot, use data points after the yield point (to avoid elastic points) and before instability (necking).A material that does not show any strain hardening (n=0) is classed as perfectly plastic. Such a material would show a constant flow stress irrespective of strain. K can be found by substituting n and a data point (from the plastic region) in the power law or from the y-intercept.Strain hardening reduces ductility and increases brittleness. e.g. Cold-working can be easily demonstrated with piece of wire or a paper clip. Bend a straight section back and forth several times. Notice that it is more difficult to bend the metal at the same place. In the strain hardened area dislocations have formed and become tangled, increasing the strength of the material. Continued bending will eventually cause the wire to break at the bend due to fatigue cracking.Stain Softening Strain softening is defined as the region in which the stress in the material is decreasing with an increase in strain. This can be observed in certain materials after yielding point. It causes deterioration of material strength with increasing strain, which is a phenomenon typically observed in damaged quasi brittle materials, including fibre reinforced composites and concrete. It is primarily a consequence of brittleness and heterogeneity of the material.Strain Softening Materials can also strain soften, for example soils. In this case, the stress-strain curve “turns down” as shown in figure. The yield surface for such a material will in general decrease in size with further straining.
Figure 2: uniaxial stress-strain curve for a strain-softening material Q 11) Define Friction A 11) Friction cannot be described by a simple formula, but macroscopic mechanics is hard to understand without some idea of the properties of friction. Friction arises when the surface of one body moves, or tries to move, along the surface of a second body. The magnitude of the force of friction varies in a complicated way with the nature of the surfaces and their relative velocity. In fact, the only thing we can always say about friction is that it opposes the motion which would occur in its absence. For instance, suppose that we try to push a book across a table. If we push gently, the book remains at rest; the force of friction assumes a value equal and opposite to the tangential force we apply. In this case, the force of friction assumes whatever value is needed to keep the book at rest. However, the friction force cannot increase indefinitely. If we push hard enough, the book starts to slide. For many surfaces the maximum value of the friction is found to be equal to μN, where N is the normal force and μ is the coefficient of friction. When a body slides across a surface, the friction force is directed opposite to the instantaneous velocity and has magnitude μN. For two given surfaces the force of sliding friction is essentially independent of the area of contact.Q 12) A block of mass m rests on a fixed wedge of angle θ. The coefficient of friction is μ. (For wooden blocks, μ is of the order of 0.2 to 0.5.) Find the value of θ at which the block starts to slide?A 12)
Figure 12In the absence of friction, the block would slide down the plane; hence the friction force f Points up the plane. With the coordinates shown, we have m
= Wsinθ – fand m
= N -Wcos
= 0. When sliding starts, f has its maximum value μN, and 
=0. The equations then giveWsinθmax = μNWcos
max = NHencetanθmax = μNotice that as the wedge angle is gradually increased from zero, the friction force grows in magnitude from zero toward its maximum value μN, since before the block begins to slide we havef = Wsinθ θ
θmax Q 13) A block of Mass M is moving with a velocity v on straight surface. What is the shortest distance and shortest time in which the block can be stopped if μ is coefficient of frictiona. v2/2μg,v/μg
b. v2/μg,v/μg
c. v2/2Mg,v/μg
d none of the above
A 13) Force of friction opposes the motion
Force of friction=μN=μmg
Therefore retardation =μmg/m=μg
From v2=u2+2as
or
s=v2/2μg
from v=u+at
or t=v/μg Q 14) Write a note on fracture?A 14) Fracture is the separation of an object or material into two or more pieces under the action of stress. The fracture of a solid usually occurs due to the development of certain displacement discontinuity surfaces within the solid. If a displacement develops perpendicular to the surface of displacement, it is called a normal tensile crack or simply a crack; if a displacement develops tangentially to the surface of displacement, it is called a shear crack. Brittle fractures occur with no apparent deformation before fracture; ductile fractures occur when visible deformation does occur before separation.Fracture strength, also known as breaking strength, is the stress at which a specimen fails via fracture. This is usually determined for a given specimen by a tensile test, which charts the stress-strain curve. The final recorded point is the fracture strength.Bones, on the whole, do not fracture due to tension or compression. Rather they generally fracture due to sideways impact or bending, resulting in the bone shearing or snapping. The behaviour of bones under tension and compression is important because it determines the load the bones can carry.
Figure 3: FractureBones are classified as weight-bearing structures such as columns in buildings and trees. Weight-bearing structures have special features; columns in building have steel-reinforcing rods while trees and bones are fibrous. The bones in different parts of the body serve different structural functions and are prone to different stresses. Thus, the bone in the top of the femur is arranged in thin sheets separated by marrow while, in other places, the bones can be cylindrical and filled with marrow or just solid. Overweight people have a tendency toward bone damage due to sustained compressions in bone joints and tendons. Q 15) Explain one dimensional stress strain curve?A 15) A stress-strain curve is a graphical representation of the behaviour of a material when it's subjected to a load or force.The two characteristics stress and strain are plotted. Stress along the y-axis and strain along the x-axis.We already discussed that Stress is the ratio of the load or force to the cross-sectional area of the material. The standard unit of measurement for stress is Nm2.Strain, on the other hand, is a measure of the deformation of the material as a result of the applied force. Deformation is a change in the shape of the material.The stress-strain graph provides engineers and designers a graphical measure of the strength and elasticity of a material. It allows them to predict the behaviour of materials used in a given application. To draw the graph, the material must first undergo a tensile test.
Figure 25: Stress Strain Curve When a ductile material consider a wire is put under stress, initially the resulting strain is proportional to the magnitude of the forces.This is represented by the straight-line OA. The straight-line implies that stress and strain share a linear or we can simply say that the material obeys Hooke’s law. Stress corresponding to point A is known as proportionality limit. In this region, the material behaves like elastic.Now when the stress is increased beyond A then for small stress there is large strain in the wire. Beyond the point A, stress and strain stop to share their linear relationship. The material in this region doesn’t obey Hooke’s law this fact can easily observe from its non-linear shape.Still, even though the material doesn’t obey Hooke’s Law, it manages to retain its elasticity.Point B is therefore known as the elastic point or elastic limit. Elastic limit represents the maximum stress on whose removal, the body regains its original shape or dimensions. This region OB is called the elasticity region.Point B is also known as yield point because if the stress is increased beyond point B, the strain increases and increased in strain is represented by BC part of curve. Now if the load is removed the wire does not regain its original shape. But the increase in length is permanent. This permanent strain in the wire is known as permanent set.Now further when the stress is increased beyond point C, there is large increase in the stain for small stress is represented by CD part of the curve.The material from C to D shows plastic behaviour or plastic deformation.The stress decreases until point C, which is known as the lower-yield point but as we can observe the material continues to elongate.When plasticity is induced in the material its internal molecular structure suffers constant rearrangements. The material tries to resist this change and tends to harden. This is known as strain hardening.However as the stress applied to it increases, it continues to elongate along its length, progressively growing longer and thinner until point D, which represents the material’s maximum strength. This is referred to as the material’s ultimate strength point.Stress greater than D is so unbearable, beyond this wire break. The wire break at point E is known as fracture point.Uses of stress-strain diagram 1. It is used to categorize the materials into ductile or brittle or plastic in nature. 2. It provides engineers and designers a graphical measure of the strength and elasticity of the material.Idealization of One dimensional Stress-Strain Curve:
Figure 26: Idealization of One dimensional Stress-Strain Curve Q 16) Discuss axial force, shear force, bending moment in details?A 16) FORCE ANALYSIS - AXIAL FORCE, SHEAR FORCE, BENDING MOMENT Shear force: The cutting force is the force on the radius that acts perpendicularly to its longitudinal axis (x). For design purposes, the ability of the beam to resist shear force is more important than its ability to resist axial force.Axial forceit is the force in the radius that acts parallel to the longitudinal axis.Below is a drawing of a simply supported beam of length L under uniform load, q:
Figure 27: Axial force This beam has the following support reactions:
Figure 28: Beam where Rl and R rare the responses at the left and right ends of the beam, correspondingly.The shear forces at the ends of the beam are equivalent to the vertical forces of the support responses. The shear force F(x) at some other point x on the beam can be establish by using the succeeding equation.
here x is the distance from the left side of the beam.Shear force diagrams are only plots of the shear force (on the y-axis) versus the position of various points along the beam (on the x-axis). Thus, the succeeding is the generalized shear force diagram for the beam shown above.
Figure 29: Shear force diagramsConcept of Shear Force and Bending moment in beams: When the beam is loaded in some arbitrarily manner, the internal forces and moments are developed and the terms shear force and bending moments come into pictures which are helpful to analyze the beams further. When a beam or frame is subjected to transverse loadings, the three possible internal forces that are developed are the normal or axial force, the shearing force, and the bending moment, as shown in section k of the cantilever of Figure 30. To predict the behaviour of structures, the magnitudes of these forces must be known.
Figure 30: The shearing force, and the bending moment, Normal Force The normal force at any section of a structure is defined as the algebraic sum of the axial forces acting on either side of the section.2. Shearing ForceThe shearing force (SF) is defined as the algebraic sum of all the transverse forces acting on either side of the section of a beam or a frame. The phrase “on either side” is important, as it implies that at any particular instance the shearing force can be obtained by summing up the transverse forces on the left side of the section or on the right side of the section.3. Bending MomentThe bending moment (BM) is defined as the algebraic sum of all the forces’ moments acting on either side of the section of a beam or a frame.4. Shearing Force DiagramThis is a graphical representation of the variation of the shearing force on a portion or the entire length of a beam or frame. As a convention, the shearing force diagram can be drawn above or below the x-centroidal axis of the structure, but it must be indicated if it is a positive or negative shear force.5. Bending Moment DiagramThis is a graphical representation of the variation of the bending moment on a segment or the entire length of a beam or frame. As a convention, the positive bending moments are drawn above the x-centroidal axis of the structure, while the negative bending moments are drawn below the axis.Sign ConventionAxial Force An axial force is regarded as positive if it tends to tier the member at the section under consideration. Such a force is regarded as tensile, while the member is said to be subjected to axial tension. On the other hand, an axial force is considered negative if it tends to crush the member at the section being considered. Such force is regarded as compressive, while the member is said to be in axial compression (see Figure 31(a) and Figure 31(b).2. Shear ForceA shear force that tends to move the left of the section upward or the right side of the section downward will be regarded as positive. Similarly, a shear force that has the tendency to move the left side of the section downward or the right side upward will be considered a negative shear force (see Figure 31c and Figure 31d).3. Bending MomentA bending moment is considered positive if it tends to cause concavity upward (sagging). If the bending moment tends to cause concavity downward (hogging), it will be considered a negative bending moment (see Figure 31e and Figure 31f).
Figure 31: Sign conventions for axial force, shearing force, and bending momentRelation Among Distributed Load, Shearing Force, and Bending MomentFor the derivation of the relations among w, V, and M, consider a simply supported beam subjected to a uniformly distributed load throughout its length, as shown in Figure 32. Let the shear force and bending moment at a section located at a distance of x from the left support be V and M, respectively, and at a section x + dx be V + dV and M + dM, respectively. The total load acting through the center of the infinitesimal length is wdx.
Figure 32: Simply supported beam
Conclusions: From the above relations, the following important conclusions may be drawn • From Equation (5), the area of the shear force diagram between any two points, from the basic calculus is the bending moment diagram The slope of bending moment diagram is the shear force, thus Thus, if F=0; the slope of the bending moment diagram is zero and the bending moment is therefore constant.' The maximum or minimum Bending moment occurs where 
=0. The slope of the shear force diagram is equal to the magnitude of the intensity of the distributed loading at any position along the beam. The –ve sign is as a consequence of our particular choice of sign conventions Q 17) Derive relation for twisting moment?A 17) Consider a cylindrical rod of length l radius r and coefficient of rigidity η. Its upper end is fixed and a couple is applied in a plane perpendicular to its length at lower end as shown in figure.
Figure 33: Twisting MomentConsider a cylinder is consisting a large number of co-axial hollow cylinders. Now, consider a one hollow cylinder of radius x and radial thickness dx as shown in fig.(b). Let θ is the twisting angle. The displacement is greatest at the rim and decreases as the center is approached where it becomes zero.
Q 18) Discuss the Motion on inclined planes?A 18) A body placed on an incline is subjected to downward pull due to gravity. The downward motion along the incline is, therefore, natural tendency of the body, which is opposed by friction acting in the opposite direction against the component of gravity along the incline. Depending on the relative strengths of two opposite forces, the body remains at rest or starts moving along the incline. The translational motion of a body on an incline is along its surface. There is no motion in the direction perpendicular to the incline surface. The force components in the direction perpendicular to the incline form a balanced force system. Let us examine the free body diagram as superimposed on the body systems and shown in the figure. In the direction of incline i.e. x – axis, we have :
Figure 14: Motion on inclined planes Motion on incline plane Motion of the block is opposed by static friction. ∑Fx =mgsinθ−FF=mawhere FF is the force of friction, which can be static friction, maximum static friction or kinetic friction. In the y-direction, ∑Fy = N−mgcosθ = 0⇒ N=mgcosθBased on the angle of incline” θ”, there are following possibilities: Case 1 : The friction is less than the maximum static friction and body is yet to start motion. Here, θ<tan−1(μs). In this case, friction is simply equal to component of gravity, parallel to the contact surface. Thus,
Figure 15: Motion on inclined planes Motion on incline plane Motion of the block is opposed by static friction. FF=fs=mgsinθThe net component of external forces in x-direction (including friction), therefore, is zero. As such, the block remains stationary. ∑Fx = mgsinθ − FF= mgsinθ−mgsinθ = 0Case 2: The friction is equal to the maximum static friction and body is yet to start motion. Here, θ=tan−1(μs). In this case also, the component of net external force in x-direction (inclusive of friction) is zero and the block is stationary. Here, ⇒tanθ = tanφ =μsCase 3: The friction is equal to the kinetic friction and body is in motion. Initially, θ>tan−1(μs). In this case, friction is given by: FF=Fk=μkNWe must understand here that friction can be equal to kinetic friction only when body has started moving. This is possible when component of gravity, parallel to contact surface is greater than the maximum static friction. However, once this condition is fulfilled, the friction between the surfaces reduces slightly to become equal to kinetic friction. Subsequently, the motion of the body can be either without acceleration (uniform velocity) or with acceleration. Motion with uniform velocity This is a case of balanced force system in x-direction. The component of net force in x-direction is zero and the block is moving with constant velocity. Here, ⇒tanθ=μkAccelerated motion This is a case of unbalanced force system in x-direction.
Figure 16: Accelerated motion Motion on incline plane Motion of the block is opposed by kinetic friction. In x – direction, ∑Fx= mgsinθ−FF = max ⇒ mgsinθ−μkN=maPutting value of normal force, N, we have: ⇒ mgsinθ−μkmgcosθ = ma ⇒a= gsinθ−μkgcosθFor the given value of “θ” and “μk”, acceleration is constant and the block, therefore, moves down with uniform acceleration. Q 19) A block of mass “m” slides down an incline, which makes an angle of 30° with the horizontal. If the acceleration of the block is g/3, then find the coefficient of kinetic friction. Consider g = 10 m/s2. A 19) The coefficient of kinetic friction is given by : μk=FkNIn order to evaluate this ratio, we need to find normal and friction forces. The free body diagram of the block as superimposed on the system is shown in the figure 16.
Figure 17: Motion on inclined planes
Q 20) Why do we need to reduce friction? Write down the methods of changing friction?A 20)Disadvantages of frictionUnwanted heating effects like forest fires Unwanted noise pollution like highway road sounds Wear and tear most commonly seen in shoes Reduced efficiency in machines Slows down many processes 1) Friction can be reduced by making the surface smooth by polishing The friction is due to the roughness of surface. If we make the surfaces smooth by polishing, then friction will be reduced. For example a slide in the park is polished to make its surface smooth and reduce friction. Due to reduced friction of a polished, smooth slide, children can slide down easily. 2) Friction can be reduced by applying lubricants (like oil or grease) to the rubbing surface When oil or grease is applied between the moving parts of a machine, a thin layer of oil or grease is formed between the two rubbing surfaces. This layer of oil separates the two rubbing surfaces a little bit due to which their interlocking is reduced to a large extent. This friction is reduced and movement becomes smooth. When oil or grease is applied to the moving parts of a machine, then their surfaces do not rub directly against each other, they rub through a layer of oil and grease. The substance which reduce friction are called lubricants Oil, grease, graphite and fine powder are lubricants. The applying of lubricants to a machine is called lubrication. Friction can be reduced by lubrication. A well lubricated machine runs more smoothly and lasts longer.A bicycle mechanic and motor mechanic uses grease between the moving parts of these machines to reduce friction and increase efficiency When a few drops of oil are poured on the hinges of a door, the friction is reduced and the door smoothly. We Sprinkle fine powder as dry lubricant on carrom board to reduce friction. In some machines an air cushion between the moving parts is used to reduce friction. For example the frictional drag from the sea on hovercraft is reduced by a cushion of compressed air. (A vehicle or craft which travels over land or water on a cushion of air provided by a downward blast is called hovercraft) 3) Friction can be reduced by using wheels to move objects It is quite difficult to move a heavy suitcase by dragging it on the ground because the sliding friction between the heavy suitcase and the ground is very large. If this heavy suitcase is fitted with small wheels then it can be pulled very easily. Because when we attach wheels then sliding friction is converted into rolling friction. And rolling friction between the wheels of suitcase and ground is much less. Friction can be reduced by attaching wheels to move the objects. 4) Friction can be reduced by using ball bearing between the moving parts of machineBall bearing is a device which consists of a ring of small metal balls. The small metal balls of a ball bearing can roll freely. They are designed to make the moving parts of a machine to roll over each other rather than slide. The ball bearing is introduced between the two surfaces which have to rotate over each other. The axle is fixed on the inner side of a ball bearing and wheel is fixed to the outer side of the ball bearing the ball bearing. They reduce friction by making the two surfaces (axle and wheel) to roll over each other. This happens due to the rolling action of small metal balls present inside the ball bearing. Even a wheel produces some friction where its central hole rubs with the axle. To reduce friction still further, wheels are mounted on ball bearing. The ball bearing is fixed between the hub of wheel and axle. When the wheel revolves the balls of ball bearing roll and reduce friction. The use of ball bearing makes the wheel roll smoothly over the axle. In most of the machine friction is reduced by using ball bearing. The Wheels of the bicycle turn on sets of ball bearing. These ball bearings reduce friction because they roll rather than slide. 5. Streamlined design of the sliding bodyA streamlined body helps in reducing friction especially in fluids. It enables the sliding body to easily pass through the medium with the least resistance. The streamlined body of fishes and birds are mimicked to ensure minimal friction. Streamlined body of bullet trains. 6. Reduce contact between surfacesReducing contact during sliding can reduce friction. we can reduce contact by magnetic levitation or by charging both the surfaces with same polarity. Maglev (Magnetic levitation) trains run on the principle of magnetic levitation to reduce friction in order to increase speed and travel experience. 7. Convert to rolling frictionSliding friction is always greater than rolling friction. Imagine a car with square wheels. Rolling friction is applied widely in our daily lives. Examples in our daily lives: Ball bearings use balls to separate two bearing races. Hence sliding friction is converted into rolling friction. The races won’t slide over each other now but transfer the load to the balls through rolling friction. 8. Reduce pressure or weight on the objectFriction is directly proportional to normal force applied on it. Since we are on earth, the weight of the body is the normal force on the object generally. In space there will be very less or no friction. So, choose materials wisely to reduce the friction. 9. Use fluid friction instead of the dry frictionFluids offer less friction compared to solid surfaces. That is why aircraft can travel at high speeds compared to cars. Also, when we are applying lubricants or oil onto surfaces, we are essentially converting dry friction into fluid friction.
The Tension of the wire is F= 4.0 X 3.1 π N. The area of cross-section is A= πr2 = π x (2.0 x 10-3 m)2 A = 4.0 π x 10-6 m2 Thus the tensile stress developed F/A = 4 X 3.1π / 4.0π x 10 N/m2 Stress = 3.1 x 106 N/m2
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Young’s Modulus Y = FL/Al l = FL/AY = (5 x 3) /π x(0.5 x 10-3)2 x 120 x109 = 15.9 x 10-3 m
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Given L = 1 m , r = 5 mm , ‖ = 450 = 45 x π /3600 η =200 X 109 Pa C = C= 54.056 Nm |
Hooke's Law In 17ᵗʰ century physicist Robert Hooke observed that the stress vs strain curve for many materials has a linear region.
Hooke's law: Within the elastic limits, the strain produced in a body is directly proportional to the stress which causes it.
i.e., strain ∝ stress or stress ∝ strain
So
This constant is called 'modulus of elasticity. Hooks law can be defined as
The force required to stretch an elastic object such as a metal spring is directly proportional to the extension of the spring. This is known as Hooke's law and commonly written:
F=−kx
Where F is the force, x is the length of extension/compression k is a constant of proportionality known as the spring constant which is usually given in N/m.
Negative sign signify that the restoring force due to the spring is in the opposite direction to the force which caused the displacement.
Dimensional Formula [ML-1T-2]
The total length L of a spring under extension is equal to the nominal length L0 plus the extension x, L=L0+x and during compression of spring, it is given by L=L0 – x.
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= Wsinθ – fand m = N -Wcos = 0. When sliding starts, f has its maximum value μN, and =0. The equations then giveWsinθmax = μNWcosmax = NHencetanθmax = μNotice that as the wedge angle is gradually increased from zero, the friction force grows in magnitude from zero toward its maximum value μN, since before the block begins to slide we havef = Wsinθ θ θmax Q 13) A block of Mass M is moving with a velocity v on straight surface. What is the shortest distance and shortest time in which the block can be stopped if μ is coefficient of frictiona. v2/2μg,v/μgb. v2/μg,v/μg
c. v2/2Mg,v/μg
d none of the above
A 13) Force of friction opposes the motion
Force of friction=μN=μmg
Therefore retardation =μmg/m=μg
From v2=u2+2as
or
s=v2/2μg
from v=u+at
or t=v/μg Q 14) Write a note on fracture?A 14) Fracture is the separation of an object or material into two or more pieces under the action of stress. The fracture of a solid usually occurs due to the development of certain displacement discontinuity surfaces within the solid. If a displacement develops perpendicular to the surface of displacement, it is called a normal tensile crack or simply a crack; if a displacement develops tangentially to the surface of displacement, it is called a shear crack. Brittle fractures occur with no apparent deformation before fracture; ductile fractures occur when visible deformation does occur before separation.Fracture strength, also known as breaking strength, is the stress at which a specimen fails via fracture. This is usually determined for a given specimen by a tensile test, which charts the stress-strain curve. The final recorded point is the fracture strength.Bones, on the whole, do not fracture due to tension or compression. Rather they generally fracture due to sideways impact or bending, resulting in the bone shearing or snapping. The behaviour of bones under tension and compression is important because it determines the load the bones can carry.
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To compute the bending moment at section x + dx, use the following:
Or Equation (1) implies that the first derivative of the bending moment with respect to the distance is equal to the shearing force. The equation also suggests that the slope of the moment diagram at a particular point is equal to the shear force at that same point. Equation (1) suggests the following expression:
Equation (2) states that the change in moment equals the area under the shear diagram. Similarly, the shearing force at section x + dx is as follows: Vx+dx = V – wdx V + dV = V – wdx or
Equation (3) implies that the first derivative of the shearing force with respect to the distance is equal to the intensity of the distributed load. Equation (3) suggests the following expression:
Equation (4) states that the change in the shear force is equal to the area under the load diagram. Equation (1) and (3) suggest the following:
Equation (5) implies that the second derivative of the bending moment with respect to the distance is equal to the intensity of the distributed load.
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As shown in fig.(a), Let AB be the line parallel to the axis OO’ before twist produced and on twisted B shifts to B’, then line AB become AB’. Before twisting if hollow cylinder cut along AB and flatted out, it will form the rectangular ABCD as shown in fig.(c). But if it will be cut after twisting it takes the shape of a parallelogram AB’C’D. The angle of shear ∠BAB’ = ɸ From fig.(c) BB’= Lɸ From fig.(b) BB’= xθ So Lɸ= xθ And ɸ = The modules of rigidity is η = F = η ɸ = η The surface area of this hollow cylinder = 2πxdx
Figure 34: Hollow cylinder ∴ Total shearing force on this area = η = The moment of this force= = Now, integrating between the limits x = 0 and x = r, We have, total twisting couple on the cylinder = = = ∴ Total twisting couple = Then, the twisting couple per unit twist (θ= 1) is C = This twisting couple per unit twist is also called the torsional rigidity of the cylinder or wire.
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Motion on incline plane Motion of the block is opposed by kinetic friction. In the x-direction, ∑Fx = mgsin300−Fk =max ⇒ mgsin300− Fk =m ⇒mg ⇒Fk = mg In the y-direction, there is no acceleration, ∑Fy = mgcos300−N = 0⇒ N=mgcos300 = Putting values in the expression of coefficient of friction, we have: μk =
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