Report

6 Branches of Mechanics Engineering Mechanics Mechanics of Solids Rigid Bodies Statics Dynamics Mechanics of Fluids Deformable Bodies Strength of Materials Ideal Viscous Compressible Fluids Fluids Fluids Theory of Elasticity Theory of Plasticity 1 2 COURSE CONTENT IN BRIEF 1. Simple stress and strain 2. Stresses due to bending 3. Stresses due to shearing 4. Stresses due to Torsion in circular shaft CHAPTER – I Simple stress & strain • • • • • • • • Normal stress and strain Hooke’s law Modulus of elasticity Tension test on ductile and brittle materials Factor of safety and allowable stress Poisson's ratio Shear stress and shear strain Modulus of rigidity 4 CHAPTER – I 5 Introduction The subject strength of materials deals with the relations between externally applied loads and their internal effects on bodies. The bodies are no longer assumed to be rigid and the deformations, however small, are of major interest The subject, strength of materials or mechanics of materials involves analytical methods for determining the strength , stiffness (deformation characteristics), and stability of various load carrying members. Alternatively the subject may be called the mechanics of solids. 7 GENERAL CONCEPTS STRESS No engineering material is perfectly rigid and hence, when a material is subjected to external load, it undergoes deformation. While undergoing deformation, the particles of the material offer a resisting force (internal force). When this resisting force equals applied load the equilibrium condition exists and hence the deformation stops. These internal forces maintain the externally applied forces in equilibrium. 8 STRESS The internal force resisting the deformation per unit area is called as stress or intensity of stress. Stress = internal resisting force / resisting cross sectional area R A 9 STRESS SI unit for stress N/m2 also designated as a pascal (Pa) Pa = N/m2 kilopascal, 1kPa = 1000 N/m2 megapascal, 1 MPa = 1×106 N/m2 = 1×106 N/(106mm2) = 1N/mm2 1 MPa = 1 N/mm2 gigapascal, 1GPa = 1×109 N/m2 = 1×103 MPa = 1×103 N/mm2 10 AXIAL LOADING – NORMAL STRESS P Consider a uniform bar of cross sectional area A, subjected to a tensile force P. P R Consider a section AB normal to the direction of force P Let R is the total resisting force acting on the cross section AB. B A STRESS Then for equilibrium condition, R P R=P Then from the definition of stress, normal stress = σ = R/A = P/A P Symbol: σ = Normal Stress 11 AXIAL LOADING – NORMAL STRESS STRESS Direct or Normal Stress: Intensity of resisting force perpendicular to or normal to the section is called the normal stress. Normal stress may be tensile or compressive Tensile stress: stresses that cause pulling on the surface of the section, (particles of the materials tend to pull apart causing extension in the direction of force) Compressive stress: stresses that cause pushing on the surface of the section, (particles of the materials tend to push together causing shortening in the direction of force) 12 STRESS • The resultant of the internal forces for an axially loaded member is normal to a section cut perpendicular to the member axis. • The force intensity on that section is defined as the normal stress. F lim A0 A P ave A 13 Example 1 Illustrative Problem A composite bar consists of an aluminum section rigidly fastened between a bronze section and a steel section as shown in figure. Axial loads are applied at the positions indicated. Determine the stress in each section. Bronze A= 120 mm2 Aluminum A= 180 mm2 13kN 4kN 300mm 400mm Steel A= 160 mm2 2kN 7kN 500mm 14 Illustrative Problem To calculate the stresses, first determine the forces in each section. To find the Force in bronze section, consider a section bb1 as shown in the figure b 4kN 13kN 2kN 7kN Bronze b1 For equilibrium condition algebraic sum of forces on LHS of the section must be equal to that of RHS 15 Illustrative Problem b 4kN 13kN 2kN 7kN Bronze b1 4kN Bronze 4kN (= 13kN 2kN 7kN ) Force acting on Bronze section is 4kN, tensile Stress in Bronze section = Force in Bronze section Resisting c/s area of the Bronze section 4kN 4 1000 N 2 33 . 33 N / mm = = 33.33MPa 2 2 120mm 120mm Tensile stress 16 Force in Aluminum section 4kN Illustrative Problem 13kN 2kN 7kN Aluminum 4kN 13kN 9kN (= 2kN 7kN Aluminum Force acting on Aluminum section is 9kN, Compressive ) 17 Force in steel section 4kN 13kN Illustrative Problem 2kN 7kN steel 7kN 4kN 13kN 2kN steel Force acting on Steel section is 7kN, Compressive 7kN 18 Illustrative Problem Stress in Aluminum section = = Stress in Steel section = Force in Al section Resisting cross sectional area of the Al section 9kN 9 1000 N 2 50 N / mm = 50MPa 2 2 180mm 180mm Compressive stress Force in Steel section Resisting cross sectional area of the Steel section 7kN 7 1000 N 2 43 . 75 N / mm = 43.75MPa = 160mm2 160mm2 Compressive stress 19 STRAIN STRAIN : When a load acts on the material it will undergo deformation. Strain is a measure of deformation produced by the application of external forces. If a bar is subjected to a direct load, and hence a stress, the bar will change in length. If the bar has an original length L and change in length by an amount δL, the linear strain produced is defined as, Linear strain, L L Change in length = Original length Strain is a dimension less quantity. 20 Linear Strain P A 2 2L L P st ress A 2P P 2A A L normalst rain L 21 STRESS-STRAIN DIAGRAM In order to compare the strength of various materials it is necessary to carry out some standard form of test to establish their relative properties. One such test is the standard tensile test in which a circular bar of uniform cross section is subjected to a gradually increasing tensile load until failure occurs. Measurement of change in length over a selected gauge length of the bar are recorded throughout the loading operation by means of extensometers. A graph of load verses extension or stress against strain is drawn as shown in figure. 22 STRESS-STRAIN DIAGRAM Proportionality limit Typical tensile test curve for mild steel 23 STRESS-STRAIN DIAGRAM Typical tensile test curve for mild steel showing upper yield point and lower yield point and also the elastic range and plastic range 24 Stress-strain Diagram Limit of Proportionality : From the origin O to a point called proportionality limit the stress strain diagram is a straight line. That is stress is proportional to strain. Hence proportional limit is the maximum stress up to which the stress – strain relationship is a straight line and material behaves elastically. From this we deduce the well known relation, first postulated by Robert Hooke in 1678, that stress is proportional to strain. Beyond this point, the stress is no longer proportional to strain PP P = A Load at proportionality limit Original cross sectional area 25 Stress-strain Diagram Elastic limit: It is the stress beyond which the material will not return to its original shape when unloaded but will retain a permanent deformation called permanent set. For most practical purposes it can often be assumed that points corresponding proportional limit and elastic limit coincide. Beyond the elastic limit plastic deformation occurs and strains are not totally recoverable. There will be thus some permanent deformation when load is removed. PE E = A Load at elastic limit Original cross sectional area 26 Stress-strain Diagram Yield point: It is the point at which there is an appreciable elongation or yielding of the material without any corresponding increase of load. PY Y = A Load at yield point Original cross sectional area 27 Stress-strain Diagram Ultimate strength: It is the stress corresponding to maximum load recorded during the test. It is stress corresponding to maximum ordinate in the stressstrain graph. PU U = A Maximum load taken by the material Original cross sectional area 28 Stress-strain Diagram Rupture strength (Nominal Breaking stress): It is the stress at failure. For most ductile material including structural steel breaking stress is somewhat lower than ultimate strength because the rupture strength is computed by dividing the rupture load (Breaking load) by the original cross sectional area. PB B = A load at breaking (failure) Original cross sectional area True breaking stress, load at breaking (failure) = Actual cross sectional area 29 Stress-strain Diagram After yield point the graph becomes much more shallow and covers a much greater portion of the strain axis than the elastic range. The capacity of a material to allow these large plastic deformations is a measure of ductility of the material. Ductile Materials: The capacity of a material to allow large extension i.e. the ability to be drawn out plastically is termed as its ductility. Material with high ductility are termed ductile material. Example: Low carbon steel, mild steel, gold, silver, aluminum. 30 Stress-strain Diagram A measure of ductility is obtained by measurements of the percentage elongation or percentage reduction in area, defined as, Percentage elongation = increase in gauge length (up to fracture) ×100 original gauge length Reduction in cross sectional area of necked portion (at fracture) Percentage reduction in = area original area ×100 Brittle Materials : A brittle material is one which exhibits relatively small extensions before fracture so that plastic region of the tensile test graph is much reduced. Example: steel with higher carbon content, cast iron, concrete, brick 31 STRESS-STRAIN : DUCTILE MATERIAL 32 Stress-Strain Diagram: Ductile Materials pounds (lb ) squareinch (in 2 ) 1000 lb ksi in 2 psi Stress-Strain Diagram: Brittle Materials Stress-strain diagram for a typical brittle material 33 34 Stress-Strain Test Test specimen with tensile load Machine used to test tensile test specimen L = gauge length HOOKE”S LAW For all practical purposes, up to certain limit the relationship between normal stress and linear strain may be said to be linear for all materials stress (σ) α strain (ε) stress (σ) constant = strain (ε) Thomas Young in 1807 introduced a constant of proportionality that came to be known as Young’s modulus. stress (σ) E = strain (ε) = Young’s Modulus or Modulus of Elasticity 35 HOOKE”S LAW 36 Young’s Modulus is defined as the ratio of normal stress to linear strain within the proportionality limit. stress (σ) E = strain (ε) = P L PL A L AL The value of the Young’s modulus is a definite property of a material From the experiments, it is known that strain is always a very small quantity, hence E must be large. For Mild steel, E = 200GPa = 2×105MPa = 2×105N/mm2 37 Material Density Young's Modulus (kg/m3) 109 N/m2 Ultimate Strength Su 106 N/m2 Yield Strength Sy 106 N/m2 Steel 7860 200 400 250 Aluminum 2710 70 110 95 Glass 2190 65 50 ... Concrete 2320 30 40 ... Wood 525 13 50 ... Bone 1900 9 170 ... Polystyrene 1050 3 48 ... Hooke’s Law: Modulus of Elasticity 38 Below the yield stress E E Youngs Modulus or Modulus of Elasticity Strength is affected by alloying, heat treating, and manufacturing process but stiffness (Modulus of Elasticity) is not. Stress-strain diagram for Iron and different grades of steel Stress-strain diagram Hard drawn wire materials Various types of nylon and polycarbonate 39 Elastic vs. Plastic Behavior 40 • If the strain disappears when the stress is removed, the material is said to behave elastically. • The largest stress for which this occurs is called the elastic limit. • When the strain does not return to zero after the stress is removed, the material is said to behave plastically. Elastic vs. Plastic Behavior 41 For certain materials, for example, high carbon steel and non-ferrous metals, it is not possible to detect any difference between the upper and lower yield points and in some cases no yield point exists at all. In such cases a proof stress is used to indicate beginning of plastic strain. Proof stress is the stress corresponding to a fixed permanent strain in stress-strain diagram. For example: 0.1% proof stress indicates that stress which, when removed, produces a permanent strain or “set” of 0.1% of the original gauge length. Proof stress Determination of 0.1% proof stress Permanent deformation or “set” after straining beyond yield point 42 Deformations Under Axial Loading • From Hooke’s Law: E E P AE • From the definition of strain: L • Equating and solving for the deformation, PL AE • With variations in loading, crosssection or material properties, Pi Li i Ai Ei 43 Example 2 Illustrative Problem 44 A specimen of steel 20mm diameter with a gauge length of 200mm was tested to failure. It undergoes an extension of 0.20mm under a load of 60kN. Load at elastic limit is 120kN. The maximum load is 180kN. The breaking load is 160kN. Total extension is 50mm and the diameter at fracture is 16mm. Find: a) Stress at elastic limit b) Young’s modulus c) % elongation d) % reduction in area e) Ultimate strength f) Nominal breaking stress g) True breaking stress 45 Illustrative Problem Example 2 Solution: a) Stress at elastic limit, Load at elastic limit σE = Original c/s area PE 120kN 381.97 N 2 381.97 MPa 2 mm A 314.16mm b) Young’s Modulus, (consider a load which is within the elastic limit) P 60kN 2 190.98 314 . 16 mm A N E 190980 0.20mm mm2 L 110 3 L 190980 MPa 190.98GPa 200mm 46 Illustrative Problem Example 2 c) % elongation, % elongation = Final length at fracture – original length Original length 50 100 25% 200 Original c/s area -Final c/s area at fracture d) % reduction in area = Original c/s area 2 16 314.16 314.16 4 100 36% 47 Illustrative Problem Example 2 e) Ultimate strength, Maximum load Ultimate strength = Original c/s area 180kN 2 572 . 96 N / mm 314.16mm2 ( MPa) f) Nominal breaking Strength = Breaking load Original c/s area 160kN 509.29MPa 314.16 g) True breaking Strength = Breaking load c/s area at fracture 160kN 795.38MPa 2 201.06mm 48 Illustrative Problem Example 3 A composite bar consists of an aluminum section rigidly fastened between a bronze section and a steel section as shown in figure. Axial loads are applied at the positions indicated. Determine the change in each section and the change in total length. Given Ebr = 100GPa, Eal = 70GPa, Est = 200GPa 4kN Bronze A= 120 mm2 300mm Aluminum A= 180 mm2 13kN 400mm Steel A= 160 mm2 2kN 7kN 500mm 49 Illustrative Problem Example 3 From the Example 1, we know that, Pbr = +4kN (Tension) Pal = -9kN (Compression) Deformation due to compressive force is shortening in length, and is considered as -ve Pst = -7kN (Compression) PL stress (σ) = E = strain (ε) AL Change in length = PL L AE Change in length of bronze = Lbr 4000 N 300mm 120mm2 100 103 ( N / mm2 ) = 0.1mm 50 Illustrative Problem Example 3 Change in length of aluminum section = 9000 N 400mm Lal 180mm2 70 103 ( N / mm2 ) = -0.286mm Change in length of steel section = 7000 N 500mm Lst 160mm2 200 103 ( N / mm2 ) = -0.109mm Change in total length = Lbr Lal Lst +0.1 – 0.286 - 0.109 = -0.295mm 51 Illustrative Problem Example 4 An aluminum rod is fastened to a steel rod as shown. Axial loads are applied at the positions shown. The area of cross section of aluminum and steel rods are 600mm2 and 300mm2 respectively. Find maximum value of P that will satisfy the following conditions. a)σst ≤ 140 MPa Take Eal = 70GPa, Est = 200GPa b)σal ≤ 80 MPa c)Total elongation ≤ 1mm, 2P Aluminum 0.8m 4P 2P Steel 2.8m 52 Illustrative Problem solution To find P, based on the condition, σst ≤ 140 MPa Stress in steel must be less than or equal to 140MPa. Hence, σst = = 140MPa P 2P Pst 2 P 140 N / mm2 Ast Ast 140 Ast 21000 N 21kN 2 Aluminum 4P 2P 2P Steel 4P 2P 2P 2P Tensile 53 Illustrative Problem solution To find P, based on the condition, σal ≤ 80 MPa Stress in aluminum must be less than or equal to 80MPa. Hence, σal = 2P Pal 80 N / mm2 Aal Aal 80 Aal P 24000 N 24kN 2 = 80MPa 2P Aluminum 4P Steel 2P 4P 2P 2P 2P Compressive 2P 54 Illustrative Problem Solution To find P, based on the condition, total elongation ≤ 1mm Total elongation = elongation in aluminum + elongation in steel. 1mm PL PL AE AE al st 1mm 2 PLal 2 PLst Aal Eal Ast Est 1mm 2 P 800 2 P 2800 3 3 600 70 10 300 200 10 P = 18.1kN Ans: P = 18.1kN (minimum of the three values) Example: Shearing Stress 71 • Forces P and P‘ are applied transversely to the member AB. • Corresponding internal forces act in the plane of section C and are called shearing forces. • The resultant of the internal shear force distribution is defined as the shear of the section and is equal to the load P. • The corresponding average shear stress is, P ave A • The shear stress distribution cannot be assumed to be uniform. 72 Double shear Consider the simple riveted lap joint shown in the Fig.(a). When load is applied to the plates as shown in the figure the rivet is subjected to shear forces tending to shear it on one plane as indicated. But the joint with two cover plates, shown in Fig.(b), the rivet is subjected to possible shearing on two faces, which is called as double shear. In such cases twice the area of the rivet is resisting the applied forces so that the shear stress set up is given by Shear stress τ (in double shear) = P/2A P P P Fig. a P Fig. b 73 Examples Single Shear Double Shear P F ave A A P F ave A 2A 74 example: Shearing Stress Pin Shearing Stresses To find the shearing stress in pin. Rod BC • The cross-sectional area for pins at D 2 25 mm 6 2 A r 491 10 m 2 2 • The force on the pin at C is equal to the force exerted by the rod BC, C ,ave P 50 10 N 102 MPa 6 2 A 49110 m 3 75 Example: Double shear Rod AB • The cross-sectional area for pins at D, &E 2 25 mm 6 2 A r 49110 m 2 2 • The pin at A is in double shear with a total force equal to the force exerted by the boom AB, P 40 kN A,ave 40.7 MPa 6 2 2 A 2(49110 m ) 85 Poisson’s Ratio = µ For most engineering metals the value of µ lies between 0.25 and 0.33 y In general Lz Ly P P x Lx z l y Poisson’s Ratio = Lateral strain Strain in the direction of load applied l x ly lx l z OR l x lz lx 86 Poisson’s Ratio = µ In general y Ly Lz Px Px x Lx z Strain in X-direction = εx Strain in Y-direction = εy l x lx Strain in Z-direction = εz l z lz l x lx l y ly l x lx 87 Poisson’s Ratio Load applied in Y-direction y Py Lz Ly x Lx z Py l x Poisson’s Ratio = Lateral strain Strain in the direction of load applied Strain in X-direction = εx l x lx l y lx ly l y ly l z OR l y lz ly 88 Poisson’s Ratio Load applied in Z-direction y Pz Lz Ly x Lx z Pz l x Poisson’s Ratio = Lateral strain Strain in the direction of load applied Strain in X-direction = εx l x lx l z l y lx lz l z lz OR l z ly lz