Elasticity Mech 24-25 Final PDF

Engineering Physics study materials by Prof. Praveen N. Vaidya, SDMCET Dharwad. 𝐶ℎ𝑎𝑛𝑔𝑒𝑖𝑛𝑑𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛 ∆𝐷 Syllabus: 𝑆𝑡𝑟𝑎𝑖𝑛 = =...

Engineering Physics study materials by Prof. Praveen N. Vaidya, SDMCET Dharwad. 𝐶ℎ𝑎𝑛𝑔𝑒𝑖𝑛𝑑𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛 ∆𝐷 Syllabus: 𝑆𝑡𝑟𝑎𝑖𝑛 = = 𝑂𝑟𝑖𝑔𝑖𝑛𝑎𝑙𝑑𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛 𝐷 Stress-Strain Curve, Stress hardening and softening. Elastic Moduli, Poisson’s ratio, Relation between Y, n and σ (with derivation), mention relation between K, Y Elastic Limit: This is the highest value of stress and σ, limiting values of Poisson’s ratio. Beams, for which body can maintain its elastic property. Bending moment and derivation of expression, Cantilever and I section girder and their Engineering For any magnitude of stress above this, object may Applications, Elastic materials (qualitative). Failures of engineering materials - Ductile fracture, Brittle fracture, lose its elastic property. Stress concentration, Fatigue and factors affecting Hooke’s Law: fatigue (only qualitative explanation), Numerical problems. Hook’s law states that, within the elastic limit, stress is directly proportional to strain; ELASTICITY Deforming force: 𝑆𝑡𝑟𝑒𝑠𝑠 Stress α Strain or 𝐸= 𝑆𝑡𝑟𝑎𝑖𝑛 The force by the application which, the shape and size of an object changes without displacement is E is constant of proportionality known as modulus called deforming force. of elasticity (Elastic constants). It is material Elasticity: dependent constant. The property of materials by the virtue of which, a There are three Moduli of Elasticity, namely body regain its original shape and size after the Young’s Modulus, Bulk Modulus, Rigidity removal of deforming force is called as elasticity. Modulus. Ex: Spider web, Steel, Graphene. Plasticity: Young’s Modulus of Elasticity: The property of materials by the virtue of which, It is defined as the ratio of Tensile stress to linear anobject continues to be in deformed condition (tensile) strain. even after removel of deforming force is called as 𝐿𝑎𝑛𝑔𝑖𝑡𝑢𝑑𝑖𝑛𝑎𝑙 (𝑡𝑒𝑛𝑠𝑖𝑙𝑒) 𝑠𝑡𝑟𝑒𝑠𝑠 𝑌𝑜𝑢𝑛𝑔′ 𝑠𝑀𝑜𝑑𝑢𝑙𝑢𝑠 = plasticity. Ex, Wet clay 𝑇𝑒𝑛𝑠𝑖𝑙𝑒 𝑠𝑡𝑟𝑎𝑖𝑛 𝐹/𝐴 𝐹 ×𝐿 Stress: 𝑌= = ∆𝐿/𝐿 𝐴 × ∆𝐿 The deforming force applied per unit area of an If a weight (m) suspended by an elastic wire of object is called stress, sometimes it is also called cross section area A = πr2, where r is radius and F pressure. Therefore, = mg,where g – acceleration due to gravity, then 𝑑𝑒𝑓𝑜𝑟𝑚𝑖𝑛𝑔𝑓𝑜𝑟𝑐𝑒 𝐹 𝑆𝑡𝑟𝑒𝑠𝑠 = = 𝑚𝑔𝐿 𝐴𝑟𝑒𝑎 𝐴 Young’s modulus𝑌 = 𝜋𝑟 2 ×∆𝐿 Strain: Longitudinal stress or tensile stress is applied along The Ratio of change in dimension of body by the the length and hence causes change in length. application of stress to original dimension Engineering Physics study materials by Prof. Praveen N. Vaidya, SDMCET Dharwad. Tensile strain is the ratio of change in length (Δl) to Distinction between Three Moduli of original length (l). elasticity Bulk Modulus of elasticity = It is the ratio of Young’s Bulk Modulus Modulus of volume stress to volume strain. Modulus Rigidity Bulk = volume stress / volume Strain Ratio of Tensil Ratio of Ratio of shear 𝐹/𝐴 𝐹×𝑉 stress to volume stress stress to shear 𝐵= = Tensile strain. to volume strain. ∆𝑉/𝑉 𝐴 × ∆𝑉 strain. 𝐹×𝑉 𝐵= deforming deforming deforming force 𝑡𝑜𝑡𝑎𝑙𝑠𝑢𝑟𝑓𝑎𝑐𝑒𝑎𝑟𝑒𝑎 × ∆𝑉 force acts at force acts at acts parallel to Application of normal (compressive) stress causes right angles to right angles to the free surface change in volume. Volume strain is the ratio of a cross a surface of the of an object, change in volume to original volume. sectional area object which is opposite of the wire. fixed one. Rigidity Modulus of Elasticity or Shear The object The object One of the under the under the surfaces of the Modulus (𝜂): deforming deforming object get This is given by the ratio of Tangential stress to force either force either displaced with gets gets increase or respect to another shearing strain. lengthened or decreased in surface. Shearing stress is applied tangential to a surface. shortened. volume. As a result, one surface is displaced with respec Strain = ratio Strain = Ratio Strain = Angle of change in of change in through which a length to volume to surface displaced original length original volume w.r.t opposite fixed surface. Y = FL/Cross B = FV/Total η = F/Area x tanθ sectional area surface area x x ΔL ΔV FACTOR OF SAFETY We know that stress is directly proportional to 𝜂 = Tangential stress / shear Strain strain under elastic limit. 𝑇𝑎𝑛𝑔𝑒𝑛𝑡𝑖𝑎𝑙𝑠𝑡𝑟𝑒𝑠𝑠 = 𝐹⁄𝐴 The maximum stress below which an elastic 𝑇𝑎𝑛𝑔𝑒𝑛𝑡𝑖𝑎𝑙𝑠𝑡𝑟𝑎𝑖𝑛 = ∆𝐿⁄𝐿 𝑭/𝑨 material restores its original shape and size is 𝛈= ⁄∆𝑳/𝑳 called working stress. ∆𝑳 𝒕𝒂𝒏𝜽 = , for small angle of shear tanθ = θ If the stress is above certain value, material fails to 𝑳 𝑭 Therefore𝛈 = restore its elastic nature and remain permanently in 𝑨×𝜽 deformed state is called breaking stress. The ratio of breaking stress to working stress is known as the factor of safety. Engineering Physics study materials by Prof. Praveen N. Vaidya, SDMCET Dharwad. 𝐁𝐫𝐞𝐚𝐤𝐢𝐧𝐠 𝐬𝐭𝐫𝐞𝐬𝐬 The stress strain curve for different material is 𝑭𝒂𝒄𝒕𝒐𝒓 𝒐𝒇 𝒔𝒂𝒇𝒆𝒕𝒚 = 𝐖𝐨𝐫𝐤𝐢𝐧𝐠 𝐬𝐭𝐫𝐞𝐬𝐬 different. It may vary due to the temperature and To avoid permanent deformation the engineering loading condition of the material tools are to be used within the elastic limit with a working stress Elastic Deformation: a) Proportional limit: Within this limit the Lateral strain -When Tensile stressappled, along hooks law if perfectly followed. So that with increase in length, there will be decrease in Stress α Strain. In this section the curve is the diameter of wire, then, ratio of decrease in straight. diameter (lateral contraction) to the original b) Elastic limit: there is always the limiting diameter is called as lateral strain. value for the stress below which strain totally Poisson’s Ratio: disappear on removal of stress. Material It is the ratio of longitudinal strain to lateral possesses elastic nature and properties till in strain. the elastic limit. ∆𝑹 𝑳𝒂𝒕𝒕𝒆𝒓𝒂𝒍𝒔𝒕𝒓𝒂𝒊𝒏 𝑷𝒐𝒊𝒔𝒔𝒐𝒏′ 𝒔𝒓𝒂𝒕𝒊𝒐 = = 𝑹 ∆𝑳 c) Yield point: The stress beyond which 𝒍𝒂𝒏𝒈𝒊𝒕𝒖𝒅𝒊𝒏𝒂𝒍𝒔𝒕𝒓𝒂𝒊𝒏 𝑳 material starts losing its elastic nature.If stress 𝑳 × ∆𝑹 = increased beyond this permanent deformation ∆𝑳 × 𝑹 It is unitless constant for a materials and its of material starts. limiting value will be vary from -1 to 0.5. Plastic Deformation: Stress-strain graph: Stress strain curve is the plot a) Ductile point: beyond this point neck of the graph of stress applied on given elastic forms. The structure of curve changes to material against the strain developed on it, taking give curvature material continues to gain stress along y- axis and corresponding strain on the plastic nature with partial elastic nature. x-axis. b) Ultimate point: The point up to which material can withstand extremely high stress and ultimate strength. Material shows maximum elongation. Above this point large deformation may possible before failure. c) Point of rupture: At this junction of graph either increase or decrease of stress, material shows complete failure and Engineering Physics study materials by Prof. Praveen N. Vaidya, SDMCET Dharwad. perfect plastic in nature or breaking of the A material that does not show further strain material. hardening is said to be perfectly plastic. The strain hardening coefficient is given by the Strain softening: expression σ = K εn. σ is the applied stress, ε is Strain softening is defined as the region in which the stress in the material is decreasing with an strain, n is the strain hardening coefficient K is the increase in strain. strength coefficient (elasticity). The value ‘n’ lies between 0.1 and 0.5 for most metals. e.g. Cold working can be easily demonstrated with a piece of metallic wire. On bending the straight section of wire back and forth several times,we notice that with increase in bending cycle we find Strain softening normally starts after yield point as hard to bend further. This is becausethe represented in the diagram. dislocations formed in bending point makes the The deterioration of material strength observed material hard and brittle, Continued bending will with increase in strain, eventually cause the wire to break at the bend due This phenomenon is typically observed in damaged to fatigue cracking. quasi brittle materials, including fiber reinforced composites and concrete. Relation between Y, η and σ: It is primarily a consequence of brittleness and x A’ P x A P’ heterogeneity of the material. Y X F Strain Hardening L θ θ When a material is strained beyond the yield point, more and more stress is required to produce D S additional plastic deformation, and the material becomes stronger and more difficult to deform this Consider the ray diagram of a cube with front face condition is known as Strain Hardening. APDS, by the application of stress ‘T’ on the side AD, The Face AP displace to A’P’ by an angle ‘θ’ The material is permanently deformed increasing with respect face DS. its resistance to further deformation. In this process, Diagonal PD expands by a length Under Strain hardening material looses its ductility increase in length gives a tensile strain = original length (PD) and acquire brittle nature. Therefore, Engineering Physics study materials by Prof. Praveen N. Vaidya, SDMCET Dharwad. Tensile strain 𝐘 = 𝟐𝛈(𝟏 + 𝛔) Tensile strain coefficient (α) = unit stress Increase in length/PD Relation between K, Y and σ: = 1 We know that increase in length = 1 PD = Y– Young′sModulus -------------- 1 α  Increase in length. = PD x α ---------- 1 β = σ − poison′s ratio --------------------- 2 α Similarly, Diagonal AS contracts by a length Y A, 1 Also, Bulkmodulus, K = 3(α−2β) -------- 3 Decrease in length gives a Lateral strain = Original length (AS) Rearrange the RHS of equation (3), 1⁄ α Lateral strain K= β ------------------------------ 4 Lateral strain coefficient (β) = 3(1−2 ) α unit stress Decrease in length/AS Decrease in length Substitute the value of equation (1) and (2) in (4). = 1 = AS Y  Decrease in length= AS x β =PDx β-----2 K= 3(1 − 2σ) Because From figure we have, PD= AS, Limiting value of poison’s ratio: From equations (1) and (2) we have, 𝑌 𝐾 = 3(1−2𝜎) ==>𝑌 = 3𝐾 (1 − 2𝜎) -------1 Total extension produced is given by, Also Y = 2η(1 + σ) ----------2 PX= PD x (α+ β) ------------------------- 3 Equating (1) and (2) But PD = √PS 2 + SD2 = √L2 + L2 3𝐾(1 − 2𝜎) = 2η(1 + σ) = √2L2 = √2L We must understand that the values of K, η, and σ x Also, P’X = 𝑃𝑃′ 𝑡𝑎𝑛45 = are all positive; therefore, RHS of the above √2 Substitute the value of PD and P’X in equation (3) equation is positive, x = √2L × (α + β) Along with this we have, (1 − 2𝜎) > 0 √2 Re-arrange the terms in above equation Or 1 >2𝜎 ==>𝜎 < 1⁄2 x β Also, the value of 𝜎 ≥ 0 = 2α(1 + ) L α x Therefore the 0 ≤ 𝜎 < 0.5 But, shear strain = θ = 𝐿 β 1 2 β θ = 2α(1 + α)or = θ (1 + α)------ 4 BEAM and its feature: α In the above equation, A beam is a long bar made of any metallic or any hard 1 𝛼 = 𝑌– 𝑌𝑜𝑢𝑛𝑔′ 𝑠𝑀𝑜𝑑𝑢𝑙𝑢𝑠, --------- 5 material, which has an uniform cross section and its For unit stress, T = 1 thickness is neglizible compared to its thickness. T 1 = = η − Regidity Modulus, ------- 6 θ θ Beam Neutral For unit stress, T=1 axis β And = σ − poison′s ratio ------ 7 α  Beams are used in supporting the load in the civil Put, eqns. 5, 6, 7 in 4 we get, structures and large machineries etc. For ex. Engineering Physics study materials by Prof. Praveen N. Vaidya, SDMCET Dharwad. support the load of multi floors in buildings, This type of beam does not allow for bending support the load of moving vehicle on bridges. moment production and will not have any vertical  The weight of the beam is negligible compared to movement or rotation. Fixed beams are most the load. frequently used in truss and similar structures.  There are negligible shearing forces over any Overhanging beams section of beams.  The cross section of the beams must be unaltered so that the geometrical moment of inertia of the beam remains same.  The curvature of the beam is small. An overhanging beam is one that is supported at Types of beams: two different areas, typically at one end and in the Simply supported beams middle of the beam, but does not have a support at Simply supported beams are those having support the other end of the beam, leaving it hanging. An at both end of the beam. overhanging beam is a combination of a simply supported beam and a cantilever beam. Cantilever beams A cantilever beam is afree hanging at one end of the beam and fixed at the other. This type of beam  These are most frequently utilized in general carries the load with bothbendingmoment construction.  They are very versatile in terms of the types of structures that they can be used with.  They do not have moment resistant at the and sheer stress. This type of beamis typically support area and is placed in a way that allows used during the construction of bridge trusses or for free rotation at the ends on columns or similar structures. walls. Bending moment of beam: Fixed beams Consider a beam with the axis XY and is called A fixed beam is one that is fixed on both ends of neutral axis, it is at the middle of the beam, and its the beam with supports. length remainssameeven after bending. Draw a line AB parallel to XY at a separation of distance ‘c’. Apply the force on both ends, so as the beam bends with AB at convex side. Engineering Physics study materials by Prof. Praveen N. Vaidya, SDMCET Dharwad. Beam shows tensile strain at convex side (above Moment of beam for a rod of cross section radius ‘r’ XY) and shear strain at concave side (below XY), Yπr4 is,𝜏 = 4R as shown in figure. Single cantilever: q The cantilever is any beam or bar supported at one AX Y B end or both ends.  θ R The beam which has support on both sides is called O as double cantilever; in this case the deforming Due to longitudinal strain, length of AB increases, force is acting at the center of the cantilever. hence, consider new length is AB The cantilever which fixed one end keeping other From the figure, we have end free is called as Single cantilever. Arc AB= (R+q) x θ The Elasticity (namely Young’s Modulus) of a (R+q) = OY+ YB– Radius of curvature material can be found by taking that in the form of θ – Angle between two radii drawn to center from two beam of suitable dimension fixed at its two ends or ends of the beam. one end keeping another end free. Original length of beam(XY) = R x θ Change in length on bending AB - XY = q θ, Change in length qθ q Tensil strain = = = original length Rθ R Tensil stress F⁄A F × R Young ′ s Modulus (Y) = = = Tensil strain q⁄ q×A Consider rectangular beam used as cantilever, whose R Y×q×A length is ‘l’, breadth ‘b’ and thickness ‘d’, then, the F= - for AB, similarly, contribution of whole R value of young’s modulus of cantilever is found by, beam is taken, 4Mgl 3 Y×q×A Y   F=∑ bd 3 R Now, moment of force or bending moment is given by Where M is mass loaded at the free end of single Y×q×A Y×q2×A τ=Fxq=∑ xq =∑ cantilever. R R Y 𝐈𝐠 ×𝐘 or𝜏 = ∑ q2 × A, or τ = ---------------- 1 R 𝐑 Failure of Engineering Materials. Equation (1) is bending moment of of the beam, Brittle Fracture: 2 Where,𝐼𝑔 = ∑ q × A, Moment of Inertia of beam,  Brittle fracture is the sudden and rapid metal Ybd3 For the rectangular beam, τ= failure in which the material shows little or no 12R plastic Deformation. Where, a and b are thickness and breadth respectively Engineering Physics study materials by Prof. Praveen N. Vaidya, SDMCET Dharwad.  This is characterized by quick failure without Factors affecting the fatigue strength of the any warning. The generated cracks propagate material rapidly and the material collapses suddenly. Fatigue is the declining in the elastic behaviour of a  Typically, this type of fracture produces a sharp given material after working for the number of snapping sound. cycles.  When the material breaks, little or no energy is absorbed as it is stressed. Fatigue life is the number of loading (stress or  When a brittle material is fractured, it is easy to strain) cycles a component can withstand before place the pieces back together and fit well due failure occurs. to the lack of deformation. For a given material there identified a given  The potential for material to become brittle number of cycles of loading cycle below which it depends on the type of material that is does suffer a fatigue failure is known as fatigue subjected to these low temperatures. strength.  Ex. The breaking of glass or rock is brittle Stress concentrations: fracture The existence of the notch, sharp bend, causes Ductile Fracture stress concentration, so that the maximum actual  Ductile fracture is the material failure that stress at the root of the notch is much greater than exhibits substantial plastic deformation prior to the nominal stress borne by the part, and the fracture. fatigue failure of the part often starts from here.  The ductile fracture process is slow and gives Residual stress enough warnings before final separation. In welding, residual stress is caused by the uneven Normally, a large amount of plastic flow is heating and cooling of the material causes material concentrated near the fracture faces. to contract, which intern produce internal stresses.  The material is stretches before fracture absorb The internal residual stress can significantly affect large amount of energy on application of stress. the fatigue performance of welded structures,  It is difficult place the broken pieces back and leading to premature failure of the structure due to fit well due to the lack of deformation. fatigue cracking.  The potential for material to become brittle depends on the type of material that is Influence of size factor subjected to high temperatures. Due to the in homogeneity of the material structure  For ex. Breaking of metallic wire and plastic and the existence of internal defects, the size will wire is ductile fracture. increase this on other hand increases the failure Engineering Physics study materials by Prof. Praveen N. Vaidya, SDMCET Dharwad. probability of the material, thus reducing the to operate under a certain number of weeks, will fatigue strength of the material. Normally material cause the metal material fatigue limit of the with same dimensions increase fatigue strength decline. Heat treatment and the effect of microstructure Effect of chemical composition Different heat treatment state will get different Under certain conditions, any alloy element that microstructure, therefore, the impact of heat can improve the tensile strength can improve the treatment on fatigue strength, in essence, is the fatigue strength of materials. Comparatively impact of microstructure. The same composition of speaking, carbon is the most important factor the metal material, due to different heat treatment, affecting the strength of materials. However, some although the same static strength can be obtained, impurity elements which form inclusions in steel but due to the different organization, fatigue have adverse effects on fatigue strength. strength can vary in a fairly large range. The influence of the surface processing state Numericals: Machined surfaces always have uneven machining Along with the numericals given here more marks, which are equivalent to tiny notches and number are solved in the class also considered. cause stress concentrations on the metal material 1. A mass of 2kg is hanging from steel wire of radius 0.5mm and length 3m. Compute the extension surface, thus reducing the fatigue strength of the produced. What should be the minimum radius of metal material. Tests have shown that for steel and the wire so that elastic limit should not exceed. (Elastic limit of steel = 2.4x108Nm-2, Young’s aluminum alloys, rough machining (rough turning) modulus of steel = 20x1010Nm-2) reduces the fatigue limit by 10%-20% or more m= 2Kg, r = 0.5mm = 0.5x10-3m, l =3m, Y = compared to longitudinal fine polishing. The higher 20x1010Nm-2 the strength of the metal material, the more 𝑚𝑔𝐿 1. Extension produce, ∆𝐿 = 𝜋𝑟 2×𝑌 sensitive it is to surface finish. 2×9.8×3 Effect of loading experience = 𝜋(0.5×10−3)2 ×20×1010 = 0.374x10-3 m In fact, no part is working under absolutely 2. The material of steel wire reaches to elastic limit constant stress amplitude conditions, the actual 2.4x108Nm-2 on application of force of 100Kg. Determine the longitudinal strain and work of the metal material overload and sub-load Elastic limit of steel = Stress = P = 2.4x108Nm-2, will have an impact on the fatigue limit of the Radius of wire =? F = mg = 2 x 9.8 = 19.6 kg 𝐹 𝐹 metal material, the test shows that the metal 𝑆𝑡𝑟𝑒𝑠𝑠, 𝑃 = = or 𝐴 𝜋𝑟2 material is commonly overload damage and sub- load exercise phenomenon. 𝐹 19.6 𝑟=√ =√ = 0.16 × 10−3 𝑚 𝜋×𝑃 3.142 × 2.4 × 108 The so-called overload damage refers to the metal material in higher than the fatigue limit of the load Engineering Physics study materials by Prof. Praveen N. Vaidya, SDMCET Dharwad. 3. On application of pressure of 107 Nm-2 on the 𝑳𝒂𝒕𝒕𝒆𝒓𝒂𝒍𝒔𝒕𝒓𝒂𝒊𝒏 𝑷𝒐𝒊𝒔𝒔𝒐𝒏′ 𝒔𝒓𝒂𝒕𝒊𝒐 = cylinders contain water and mercury, the volumes 𝒍𝒂𝒏𝒈𝒊𝒕𝒖𝒅𝒊𝒏𝒂𝒍𝒔𝒕𝒓𝒂𝒊𝒏 changed by the factor of 8.3ml and 0.35ml 5. A sphere contracts in volume by 0.02% when taken respectively. If the bulk modulus of water is to the bottom of sea 1km deep. Calculate the bulk 2.2x109Nm-2, find the bulk modulus of Mercury. modulus of the material of the sphere. You make Also determine the original volume of vessels if take density of sea water as 1030 kgm–3 and g = both are of same size. 9.8ms–2. 𝑃𝑋𝑉 𝑃𝑉 𝐵= 𝑜𝑟∆𝑉 = Solution: ∆𝑉 𝐵 h = 1km =1000m, D=1000kgm-3, g = 9.8ms-2 Bw= 2.2x109Nm-2 P = 107Nm-2ΔVw= 8.3ml Stress: P = hxDxg ΔVM= 0.35ml, BM = ? V = ? Change in volume produced: ∆𝑉 = 0.02, 𝑃𝑉 𝑂𝑟𝑖𝑔𝑖𝑛𝑎𝑙𝑣𝑜𝑙𝑢𝑚𝑒: 𝑉 = 100 For water, ∆𝑉𝑊 = -----1 𝐵 𝑃𝑉 For Mercury ∆𝑉𝐻𝑔 = --------2 𝑃 = ℎ𝐷𝑔 = 1000 × 1030 × 9.8 =10.1x106Nm-1 𝐵 ∆𝑉𝑊 × 𝐵𝑊 = 𝑃𝑉 -------3 or 𝑃𝑉 10.01 × 106 × 100 𝐵= = = 2.95 × 109 𝑁𝑚 −2 𝑑𝑉 0.02 ∆𝑉𝑀 × 𝐵𝑀 = 𝑃𝑉 -------- 4 Both enclosed in different cylinder of same size 6. A rubber ball is taken to a depth of 200m of sea water. The volume of ball is decreases by 0.1%. If and under application of same pressure, hence density of sea water is 1040kgm-3, determine the PV is same for both. Therefore from equation 3 Pressure on the ball and bulk modulus of elasticity of and equation 4 rubber. 𝐹 𝑚𝑔 𝑉𝐷𝑔 𝐴ℎ𝐷𝑔 𝑃 = 𝑆𝑡𝑟𝑒𝑠𝑠 = = = = = ℎ𝐷𝑔 ∆𝑉𝑊 × 𝐵𝑊 = ∆𝑉𝑀 × 𝐵𝑀 𝐴 𝐴 𝐴 𝐴 𝑃 = ℎ𝐷𝑔 = 200 × 1040 × 9.8 = 2.03 × 106 𝑁𝑚 −2 ∆𝑉𝑊 ×𝐵𝑊 𝐵𝑀 = , 𝑉 100 ∆𝑉𝑀 = = 103 𝑑𝑉 0.1 0.0083×2.2×109 𝐵𝑀 = = 5.21x109Nm-2. 0.00035 Therefore, Bulk modulus is given by, Volume of the cylinders is 𝑉 𝐵=𝑃× = 2.03 × 106 × 103 = 2.03 × 109 𝑁𝑚 −2 𝑑𝑉 𝑃𝑉 ∆𝑉𝑊 ×𝐵 ∆𝑉𝑊 = or = 𝑉. 𝐵 𝑃 7. A 0.6mm thick gold wire of length 1.12m 0.0083×2.2×109 elongates by 1mm, when stretched by a force of Therefore 𝑉 = = 1.826 lit. 107 160gm and twist 1rad when equal and opposite 4. A wire made of metallic alloy work under the torque of 10-4N as applied at its end. Find value of elastic limit for a load of 0.5kg. Determine the Poison’s ratio. Given η= 2.34x109Nm-2. length of wire required for elongation must not exceed more than 0.2mm. Diameter of wire is 8. A metal plate whose one of the opposite faces has 0.5mm. What is the should be the value of lateral fixed, has area 1mx1m and thickness of 1cm. Find the strain so that value poison’s should not exceed its force acting on free face, to produce the displacement limit of 0.005mm w.r.t. fixed face. Also find shearing stress, 𝑚𝑔 × ∆𝐿 shearing strain. (Modulus rigidity of material of plate = 𝐿= 𝜋𝑟 2 × 𝑌 4x1010Nm-2) Engineering Physics study materials by Prof. Praveen N. Vaidya, SDMCET Dharwad. Area of surface, A = 1m2, l = 1cm, Δl = 0.005, η=Modulus rigidity of material of plate = 4x1010Nm-2 𝐅×𝐥 C A’ 𝛈= C’ 𝐀 × ∆𝐥 𝛈𝐀∆𝐥 ==> 𝐹 = 𝐥 D D’ 𝟒 × 𝟏𝟎𝟏𝟎 × 𝟏 × 𝟎. 𝟎𝟎𝟓 × 𝟏𝟎−𝟑 = 𝟏A× 𝟏𝟎−𝟐 A’ F = 2x103N Shearing stress B A’ = F/A = 2x103/1 = 2x103Nm-2. -3 -2 Shearing strain = Δl/l = 0.005x10 /10 =0.0005