Solid Mechanics Chapter 1: Tension, Compression and Shear

Transcription

1 Solid Mechanics Chapter 1: Tension, Compression and Shear Dr. Imran Latif Department of Civil and Environmental Engineering College of Engineering University of Nizwa (UoN) 1

2 Why do we study Mechanics of Solids? 2 Anyone concerned with the strength and physical performance of natural/man made structures should study Mechanics of Solids

3 Introduction to Mechanics of Solids 3 Definition: Mechanics of solids is a branch of applied mechanics that deals with the behaviour of solid bodies subjected to various types of loading Compression Tension Bending Torsion Shearing (streteched) (twisted)

4 Introduction to Mechanics of Solids 4 Fundamental concepts stress and strain deformation and displacement elasticity and inelasticity load carrying capacity Design and analysis of mechanical and structural systems

5 Introduction to Mechanics of Solids 5 Examination of stresses and strains inside real bodies of finite dimensions that deform under loads In order to determine stresses and strains we use: Physical properties of materials Theoretical laws and concepts

6 Problem solving 6 Draw the free body diagram Check your diagram Calculate the unknowns Check your working Compute the problem Check your working Write the solution Check your working

7 Free Body Diagram 7 Free Body Diagram The unknowns: R A, R B, R C

8 Free Body Diagram 8 Free Body Diagram The unknowns: A x, A y, CB

9 Normal stress and strain 9 Most fundamental concepts in Mechanics of Materials are stress and strain Prismatic bar: Straight structural member with the same cross section throughout its length Axial force: Load directed along the axis of the member Axial force can be tensile or compressive Axial loads: Tension (+) and compression ( ) Type of loading for landing gear strut and for tow bar? Structural members subjected to axial loads

10 Normal stress and strain 10 A truss bridge is a type of beam bridge with a skeletal structure. The forces of tension, or pulling, are represented by red lines and the forces of compression, or squeezing, are represented by green lines.

11 Normal stress 11 Continuously distributed stresses acting over the entire cross section. Axial force P is the resultant of those stresses FBD of a segment of the bar Segment of the bar before loading Stress (σ) has units of force per unit area If stresses acting on cross section are uniformly distributed then: Segment of bar after loading Normal stresses in the bar Units of stress in USCS: pounds per square inch (psi) or kilopounds per square inch (ksi) SI units: newtons per square meter (N/m 2 ) which is equal to Pa

12 Limitations 12 The loads P are transmitted to the bar by pins that pass through the holes High localized stresses are produced around the holes!! Stress concentrations Steel eyebar subjected to tensile loads P

13 Normal strain A prismatic bar will change in length when under a uniaxial tensile force, and obviously it will become longer 13 Definition of elongation per unit length or strain (ε) FBD of a segment of the bar Segment of the bar before loading If bar is in tension, strain is tensile and if in compression the strain is compressive Strain is a dimensionless quantity (i.e. no units) Segment of bar after loading Normal stresses in the bar

14 Line of action of the axial forces for a uniform stress distribution 14 It can be demonstrated that in order to have uniform tension or compression in a prismatic bar, the axial force must act through the centroid of the crosssectional area.

15 Example 15 A short post constructed from a hollow circular tube of aluminum supports a compressive load of 26 kips (26000 lb). The inner and outer diameters of the tube are d 1 = 4 in. and d 2 = 4.5 in. respectively, and its length is 16 in. The shortening of the post due to the load is measured as in. Determine: (a) Compressive stress in the post. (b) Compressive strain in the post.

16 16

17 Problem 17 Problem A circular aluminum tube of length L = 400 mm is loaded in compression by forces P (see figure). The outside and inside diameters are 60 mm and 50 mm, respectively. A strain gage is placed on the outside of the bar to measure normal strains in the longitudinal direction. (a) If the measured strain is 550 x 10-6 bar?, what is the shortening of the (b) If the compressive stress in the bar is intended to be 40 MPa, what should be the load P?

18 18 Problem Two steel wires, AB and BC, support a lamp weighing 18 lb (see figure). Wire AB is at an angle α = 34 to the horizontal and wire BC is at an angle β = 48. Both wires have diameter 30 mils. (Wire diameters are often expressed in mils; one mil equals in.) Determine the tensile stresses AB and BC in the two wires.

19 Mechanical properties of materials 19 In order to understand the mechanical behaviour of materials we need to perform experimental testing in the lab A tensile test machine is a typical equipment of a mechanical testing lab ASTM (American Society for Testing and Materials)

20 Stress ( ) strain ( ) diagrams 20

21 diagrams 21 Nominal stress and strain (in the calculations we use the initial cross sectional area A) True stress (in the calculations we use the cross sectional area A when failure occurs) True strain if we use a strain gauge Stress strain diagrams contain important information about mechanical properties and behaviour Stress strain diagram for a typical structural steel in tension (not to scale)

22 diagrams 22 OA: Initial region which is linear and proportional (Slope of OA is called modulus of elasticity) BC: Considerable elongation occurs with no noticeable increase in stress (yielding) CD: Strain hardening changes in crystalline structure (increased resistance to further deformation) DE: Further stretching leads to reduction in the applied load and fracture OABCE : True stress strain curve

23 Tensile coupon test specimens showing elongation immediately prior to failure (G.J. Davies) 23

24 diagrams 24 The strains from zero to point A are so small as compared to the strains from point A to E and can not be seen (it is a vertical line ) Metals, such as structural steel, that undergo permanent large strains before failure are ductile Ductile materials absorb large amounts of strain energy Ductile materials: aluminium, copper, magnesium, lead, molybdenum, nickel, brass, nylon, teflon Stress strain diagram for a typical structural steel in tension (drawn to scale)

25 Aluminum alloys 25 Although ductile, aluminium alloys typically do not have a clearly definable yield point However, they have an initial linear region with a recognizable proportional limit Structural alloys have proportional limits in the range of MPa and ultimate stresses in the range of MPa Typical stress strain diagram for an aluminum alloy.

26 Offset method 26 When the yield point is not obvious, like in the previous case, and undergoes large strains, an arbitrary yield stress can be determined by the offset method (e.g or 0.2%) The intersection of the offset line and the stress strain curve (point A) defines the yield stress

27 Brittle materials 27 Brittle materials fail at relatively low strains and little elongation after the proportional limit Brittle materials: concrete, marble, glass, ceramics and metallic alloys The reduction in the crosssectional area until fracture (point B) is insignificant and the fracture stress (point B) is the same as the ultimate stress Typical stress strain diagram for a brittle material showing the proportional limit (point A) and fracture stress (point B)

28 28 Problem Three different materials, designated A, B,and C, are tested in tension using test specimens having diameters of in. and gage lengths of 2.0 in. (see figure). At failure, the distances between the gage marks are found to be 2.13, 2.48, and 2.78 in., respectively. Also, at the failure cross sections the diameters are found to be0.484, 0.398, and in., respectively. Determine the percent elongation and percent reduction in area of each specimen, and then, using your own judgment, classify each material as brittle or ductile.

29 29 Problem A specimen of a methacrylate plastic is tested in tension at room temperature (see figure), producing the stress-strain data listed in the accompanying table. Plot the stress-strain curve and determine the proportional limit, modulus of elasticity (i.e., the slope of the initial part of the stress-strain curve), and yield stress at 0.2% offset. Is the material ductile or brittle?

30 Solution to problem 30 Modulus of elasticity: 2.35 GPa Proportional limit: 47 MPa Yield stress: 52.5 MPa Material is brittle, because the strain after the proportional limit is exceeded is relatively small.

31 Elasticity 31 What happens when the load is removed (i.e. the material is unloaded)? Tensile load is applied from O to A (Fig 1) and when load is removed the material follows the same curve back. This property is called elasticity If we load the same material from O to B (Fig 2) and then unloading occurs, the material follows the line BC. Line OC represents the residual or permanent strain. Line CD represents the elastic recovery of the material. During unloading the material is partially elastic Fig. 1. Elastic behavior Fig. 2. Plastic behavior

32 Plasticity 32 Plasticity is the characteristic of a material which undergoes inelastic strains beyond the strain at the elastic limit When large deformations occur in a ductile material loaded in the plastic region, the material is undergoing plastic flow

33 Reloading of a material 33 If the material is in the elastic range, it can be loaded, unloaded and loaded again without significantly changing the behaviour When loaded in the plastic range, the internal structure of the material is altered and the properties change If the material is reloaded (fig 1 19), CB is a linearly elastic region with the same slope as the slope of the tangent to the original loading curve at origin O By stretching steel or aluminium into the plastic range, the properties of the material are changed

34 Creep 34 When loaded for periods of time, some materials develop additional strains and are said to creep Even though the load P remains constant after time t0, the bar gradually lengthens Relaxation is a process at which, after time t 0, the stress in the wire gradually diminishes and eventually is reaching a constant value Creep is more important at high temperatures and has to be considered in the design of engines and furnaces

35 Hooke s law 35 Many structural materials such as metals, wood, plastics and ceramics behave both elastically and linearly when first loaded and their stressstrain curve begin with a straight line passing through origin (line OA) Linear elastic materials are useful for designing structures and machines when permanent deformations, due to yielding, must be avoided

36 Hooke s law 36 The linear relationship between stress and strain for a bar in simple tension or compression is expressed by: σ is axial stress σ = E ε ε is axial strain E is modulus of elasticity Hooke s law Robert Hooke ( ) The above equation is a limited version of Hooke s Law relating only the longitudinal stresses and strains that are developed during the uniaxial loading of a prismatic bar Robert Hooke was an English inventor, microscopist, physicist, surveyor, astronomer, biologist and artist, who played an important role in the scientific revolution, through both theoretical and experimental work.

37 Modulus of Elasticity 37 E is called modulus of elasticity or Young s modulus and is a constant It is the slope of the stress strain curve in the linearly elastic region Units of E are the same as the units of stress (i.e. psi or Pa) Thomas Young was an English polymath, contributing to the scientific understanding of vision, light, solid mechanics, energy, physiology, and Egyptology.

38 Poisson s ratio 38 When a prismatic bar is loaded in tension the axial elongation is accompanied by lateral contraction longitudinal extension lateral contraction The lateral strain ε at any point in a bar is proportional to the axial strain ε at the same point if the material is linearly elastic The ratio of the above two strains is known as Poisson s ratio (ν) ν = (lateral strain / axial strain = (ε / ε )

39 Poisson s ratio The minus sign in the equation is because the lateral strain is negative (width of the bar decreases) and the axial tensile strain is positive. Therefore, the Poisson s ratio will have a positive value. When using the Poisson s ratio equation we need to know that it applies only to a prismatic bar in uni axial stress 39 Simeon Denis Poisson ( ) Poisson s value of concrete = Poisson s value of rubber = 0.5 Siméon Denis Poisson was a French mathematician, geometer, and physicist.

40 Limitations 40 Poisson s ratio is constant in the linearly elastic range Material must be homogeneous (same composition at every point) Materials having the same properties in all directions are called isotropic If the properties differ in various directions the materials called anisotropic

41 Shear stress and shear strain 41 Shear stress acts tangential to the surface of the material and not perpendicular perpendicular tangential Consider the bolted connection of Figure where A is a flat bar, C a clevis and B a bolt When load P is applied, the bar and clevis will press against the bolt and bearing stresses will be developed The bar and clevis tend to shear the bolt

42 Shear stress and shear strain 42 If we have a closer look from the side view (fig b) and draw a FBD (fig c) Bearing stresses exerted by the clevis against the bolt appear on the left hand side (1 and 3) Stresses from the bar are on the right hand side (2) Based on the assumption of uniform stress distribution we can calculate an average bearing stress σ b total bearing force bearing area

43 Shear stress and shear strain 43 The bearing area A b is defined as the projected area of the curved bearing surface. For example (for stresses labeled 1) the projected area on which the stresses act is a rectangle with height equal to the thickness of the clevis and width equal to the diameter of the bolt The bearing force F b (for stresses labeled 1) is equal to P/2 The same area and force apply for stresses labeled 3 For bearing stresses labeled 2 the bearing area is a rectangle with height equal to the thickness of the flat bar and width equal to the diameter of the bolt. The force is equal to P

44 Shear stress and shear strain 44 The FBD (fig. c) shows that there is a tendency to shear the bolt along the cross sections mn and pq From the FBD (fig. d) of the portion mnqp of the bolt we see that the shear forces V act over the cut surfaces of the bolt. There are two planes of shear (plane mn and plane pq). Therefore, the bolt is in double shear The shear stresses acting on the cross section mn are shown (fig. e) Shear stresses are denoted by τ

45 Single shear 45 The axial force P in the metal bar is transmitted to the flange of the steel column through a bolt A cross section of the column (fig b) shows more details Fig c shows the assumed distribution of the bearing stresses acting on the bolt Cutting through the bolt at section mn (fig d) we see the shear force V (equal to load P). V is the resultant of the shear stresses that act over the cross sectional area of the bolt

46 Single shear 46

47 Shear stress and strain 47 Discussing about bolted connections we disregard friction which is produced by tightening the bolts Average shear stress on the cross section of a bolt is obtained by dividing the total shear force V by the area A of the cross section on which it acts: Shear stresses have the same units as normal stresses The two previous examples (double and single shear) are examples of direct shear Direct shear arises in the design of bolts, pins, rivets, keys, welds and glued joints

48 48 Consider a small rectangular parallelepiped element Assume that a shear stress τ 1 is uniformly distributed over the right hand side area bc (τ 1bc ) For equilibrium in the y direction the τ 1bc must be balanced by an equal and of opposite direction shear force on the left hand side The forces τ 1bc acting on the right hand and left hand side faces form a couple having a moment about z axis equal to τ 1.abc (counterclockwise direction) Small element of material subjected to shear stresses

49 Equality of shear stresses on perpendicular planes 49 Similarly, in order to have equilibrium of the element, we have a shear force τ 2.ac and consequently a clockwise couple of moment τ 2.abc It is therefore evident that for moment equilibrium we have: τ 1 = τ 2 1. Shear stresses on opposite and parallel faces of an element are equal in magnitude and opposite in direction 2. Shear stresses on adjacent and perpendicular faces of an element are equal in magnitude and have directions such that both stresses point toward, or both point away from, the line of intersection of the faces

50 50 Shear stresses acting on an element of material (fig a) are accompanied by shear strains The lengths of the sides of the element do not change but, the shear stresses produce a change in the shape of the element Rectangular parallelepiped becomes oblique parallelepiped. Front and rear faces become rhomboids The angle γ (fig b) is a measure of distortion of the element and is called shear strain

51 Hooke s law in shear 51 We can plot shear stress strain diagrams Hooke s Law in shear shear stress τ = Gγ shear modulus of elasticity shear strain G has the same units as E (Young s modulus) G and E are also related by: G = E / (2(1+ν)) Poisson s ratio

52 Allowable stresses & allowable loads 52 The principal design interest is strength Strength is the capacity of the object to support or transmit loads The actual strength of a structure must exceed the required strength Factor of safety must be greater than 1 if failure is to be avoided Factors of safety from slightly above 1 to as much as 10 are used Factor of safety (n) = Actual strength / Required strength

53 Allowable stresses & allowable loads 53 Allowable stress = Yield strength / Factor of safety tension shear σ allow = σ Y / n 1 τ allow = τ Y / n 2 OR Allowable stress = Ultimate strength / Factor of safety tension shear σ allow = σ U / n 3 τ allow = τ U / n 4

54 Allowable loads 54 Allowable load is also called permissible load or the safe load Allowable load = (Allowable stress) (Area) For bars in tension or compression: P allow = σ allow. A For pins in direct shear: P allow = τ allow. A Permissible load based upon bearing: P allow = σ b. A