EDEXCEL NATIONAL CERTIFICATE/DIPLOMA MECHANICAL PRINCIPLES AND APPLICATIONS NQF LEVEL 3 OUTCOME 1 - LOADING SYSTEMS TUTORIAL 3 LOADED COMPONENTS

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1 EDEXCEL NATIONAL CERTIICATE/DIPLOMA MECHANICAL PRINCIPLES AND APPLICATIONS NQ LEVEL 3 OUTCOME 1 - LOADING SYSTEMS TUTORIAL 3 LOADED COMPONENTS 1. Be able to determine the effects of loading in static engineering systems Non-concurrent coplanar force systems: graphical representation e.g. space and free body diagrams; resolution of forces in perpendicular directions e.g. x Cosθ, y Sinθ, vector addition of forces, conditions for static equilibrium (Σ x 0, Σ y 0, Σ M 0), resultant, equilibrant, line of action Simply supported beams: conditions for static equilibrium; loading (concentrated loads, uniformly distributed loads, support reactions) Loaded components: elastic constants (modulus of elasticity, shear modulus); loading (uniaxial loading, shear loading); effects e.g. direct stress and strain including dimensional change, shear stress and strain, factor of safety INTRODUCTION This tutorial is about stress and strain, not only in structures like beams partly covered in the last tutorial but also in many other engineering components and structures. Typical values for the various material properties mentioned in this tutorial may be found in tables on the home page for D.J.Dunn 1

2 1. STATIC ORCES and STRUCTURES STATIC ORCES will deform a body in one or more of the following manners. a. The force may stretch the body, in which case it is called a TENSILE ORCE. b. The force may squeeze the body in which case it is called a COMPRESSIVE ORCE. A rod or rope used in a frame to take a tensile load is called a TIE. If it takes a compressive force it is called a STRUT. A strut that is thick compared to its length is called a COLUMN. c. The force may bend the body in which case both tensile and compressive forces may occur. A structure used to support a bending load is called a BEAM or JOIST. d. The force may try to shear the body in which case the force is called a SHEAR ORCE. A scissors or guillotine produces shear forces. e. The force may twist the body in which case SHEAR ORCES occur. A structure that transmits rotation is called a SHAT and it experiences TORSION. D.J.Dunn

3 RAMES Struts and ties make up the members of lattice frames such as the simple one shown. The two side members are compressed and so are struts but the bottom one is stretched and could be a chain so it is a tie. BEAMS A beam is a structure, which is loaded transversely (sideways). The loads may be point loads or uniformly distributed loads (udl). The diagrams show the way that point loads and uniform loads are illustrated. Transverse loading causes bending and bending is a very severe form of stressing a structure. The bent beam goes into tension (stretched) on one side and compression on the other. In this unit you are not required to calculate the stress in a beam due to bending. D.J.Dunn 3

4 . SAETY ACTORS In the following sections you will learn how to calculate the stress in some basic engineering structures and components. If we want to be confident that the structure or component does not fail by being overstressed, we design it so that working stress is smaller than the stress at which it fails. If we want to define what failure means, we would need to study stress and strain and material science to a much greater depth than covered by this unit. ailure might mean that the material has yielded or that it has broken. Also calculation of maximum stress in a component is much more complicated than the cases studies here. If a material has direct stress and shear stress at the same time, the maximum stress may well exceed either of them on their own. Also the stress may be raised by things like sharp corners and undercuts. Other factors such as fatigue and creep also affect the stress. Don't think that this tutorial is anything but the start of your studies in that area. If we decide what the fail stress is for a given material, then we design the component so that the working stress is less. We define the safety factor as the ratio such that: 3. DIRECT STRESS σ Safety actor Stress at ailure Maximum working stress When a force is applied to an elastic body, the body deforms. The way in which the body deforms depends upon the type of force applied to it. A compression force makes the body shorter. A tensile force makes the body longer. Tensile and compressive forces are called DIRECT ORCES. Stress is the force per unit area upon which it acts. Stress σ orce/area N/m or Pascals. The symbol σ is called SIGMA NOTE ON UNITS The fundamental unit of stress is 1 N/m and this is called a Pascal. This is a small quantity in most fields of engineering so we use the multiples kpa, MPa and GPa. Areas may be calculated in mm and units of stress in N/mm are quite acceptable. Since 1 N/mm converts to N/m then it follows that the N/mm is the same as a MPa D.J.Dunn 4

5 4. DIRECT STRAIN ε In each case, a force produces a deformation x. In engineering we usually change this force into stress and the deformation into strain and we define these as follows. Strain is the deformation per unit of the original length Strain ε x/l The symbol ε is called EPSILON Strain has no units since it is a ratio of length to length. Most engineering materials do not stretch very much before they become damaged so strain values are very small figures. It is quite normal to change small numbers in to the exponent for of Engineers use the abbreviation µε (micro strain) to denote this multiple. or example a strain of could be written as 68 x 10-6 but engineers would write 68 µε. Note that when conducting a British Standard tensile test the symbols for original area are S o and for Length is L o. WORKED EXAMPLE No.1 A metal wire is.5 mm diameter and m long. A force of 100 N is applied to it and it stretches 0.3 mm. Assume the material is elastic. Determine the following. i. The stress in the wire σ. ii. The strain in the wire ε. If the stress that produces failure is 300 MPa, calculate the safety factor. πd π x A mm σ 44 N/mm 4 4 A Answer (i) is hence 44 MPa x 0.3 mm ε or 150 µε L 000 Safety actor 300/ SEL ASSESSMENT EXERCISE No.1 1. A steel bar is 10 mm diameter and m long. It is stretched with a force of 0 kn and extends by 0. mm. Calculate the stress and strain. If the stress at failure is 300 MPa calculate the safety factor. (Answers 54.6 MPa, 100 µε and 1.18). A rod is 0.5 m long and 5 mm diameter. It is stretched 0.06 mm by a force of 3 kn. Calculate the stress and strain. If the stress at failure is 50 MPa calculate the safety factor. (Answers 15.8 MPa, 10µε and 1.64) D.J.Dunn 5

6 5. MODULUS O ELASTICITY E Elastic materials always spring back into shape when released. They also obey HOOKE'S LAW. This is the law of a spring which states that deformation is directly proportional to the force. /x stiffness k N/m The stiffness is different for different materials and different sizes of the material. We may eliminate the size by using stress and strain instead of force and deformation as follows. If and x refer to direct stress and strain then σa x εl hence x σa εl L and Ax σ ε The stiffness is now in terms of stress and strain only and this constant is called the MODULUS of ELASTICITY and it has a symbol E. L σ E Ax ε A graph of stress against strain will be a straight line with a gradient of E. The units of E are the same as the units of stress. 6. ULTIMATE TENSILE STRESS If a material is stretched until it breaks, the tensile stress has reached the absolute limit and this stress level is called the ultimate tensile stress. WORKED EXAMPLE No. A steel tensile test specimen has a cross sectional area of 100 mm and a gauge length of 50 mm. In a tensile test the graph of force - extension produces a gradient in the elastic section of 410 x 10 3 N/mm. Determine the modulus of elasticity. The specimen breaks when the force is 35 kn. What is the ultimate tensile stress based on the original area? The gradient gives the ratio /A and this may be used to find E. σ L 3 50 E x 410 x 10 x N/mm or MPa or 05 GPa ε x A 100 Breaking orce σu 350N/mm or 350MPa Area 100 D.J.Dunn 6

7 WORKED EXAMPLE No. 3 A Steel column is 3 m long and 0.4 m diameter. It carries a load of 50 MN. Given that the modulus of elasticity is 00 GPa, calculate the compressive stress and strain and determine how much the column is compressed. πd A 4 σ E ε x ε L π x so ε σ E 0.16 m x x10 σ x so x ε L x 3000 mm 5.97 mm 6 A x10 6 Pa SEL ASSESSMENT EXERCISE No A bar is 500 mm long and is stretched to mm with a force of 15 kn. The bar is 10 mm diameter. Calculate the stress and strain. Assuming that the material has remained within the elastic limit, determine the modulus of elasticity. If the fail stress is 50 MPa calculate the safety factor. (Answers 191 MPa, 900µε, 1. GPa and 1.31.). The maximum safe stress in a steel bar is 300 MPa and the modulus of elasticity is 05 GPa. The bar is 80 mm diameter and 40 mm long. If a factor of safety of is to be used, determine the following. i. The maximum allowable stress. (150 MPa) ii. The maximum tensile force allowable. (754 kn) iii. The corresponding strain at this force. (731.7 µε) iv. The change in length. (0.176 mm) 3. A circular metal column is to support a load of 500 Tonne and it must not compress more than 0.1 mm. The modulus of elasticity is 10 GPa. The column is m long. Calculate the following. i. The cross sectional area(0.467 m ) ii. The diameter. (0.771 m) Note 1 Tonne is 1000 kg. D.J.Dunn 7

8 7. SHEAR STRESS τ Shear force is a force applied sideways on to the material (transversely loaded). This occurs typically : Shearing Punching Transverse orce Shear Pin Shear stress is the force per unit area carrying the load. This means the cross sectional area of the material being cut, the beam and pin respectively. Shear stress τ /A The symbol τ is called Tau. Note that A is the area being sheared. As with direct stress, we may define a safety factor as the ratio of the fail shear stress to the actual working stress. The sign convention for shear force and stress is based on how it shears the materials and this is shown below. 8. SHEAR STRAIN γ In order to understand the basic theory of shearing, consider a block of material being deformed sideways as shown. (Note in reality the deformation is very small and the diagram is much exaggerated) The shear force causes the material to deform as shown. The shear strain is defined as the ratio of the distance deformed to the height x/l. The end face rotates through an angle γ. Since this is a very small angle, it is accurate to say the distance x is the length of an arc of radius L and angle γ so that γ x/l It follows that γ is the shear strain. The symbol γ is called Gamma. D.J.Dunn 8

9 9. MODULUS O RIGIDITY G If we were to conduct an experiment and measure x for various values of, we would find that if the material is elastic, it behave like a spring and so long as we do not damage the material by using too big a force, the graph of and x is a straight line as shown. The gradient of the graph is constant so /x constant and this is the shear spring stiffness of the block in N/m. If we divide by the area A and x by the height L, the relationship is still a constant and we get x L constant A L Ax x L τ But /A τ and x/l γ so constant A L Ax γ This constant will have a special value for each elastic material and is called the Modulus of Rigidity with symbol G. τ G γ 10. ULTIMATE SHEAR STRESS If a material is sheared beyond a certain limit it becomes permanently distorted and does not spring all the way back to its original shape. The elastic limit has been exceeded. If the material is stressed to the limit so that it parts into two (e.g. a guillotine or punch), the ultimate limit has been reached. The ultimate shear stress is τ u and this value is used to calculate the force needed by shears and punches. WORKED EXAMPLE No. 4 Calculate the force needed to guillotine a sheet of metal 5 mm thick and 0.8 m wide given that the ultimate shear stress is 50 MPa. The area to be cut is a rectangle800 mm x 5 mm A 800 x mm τ A so The ultimate shear stress is 50 N/mm τ x A 50 x N or 00 kn D.J.Dunn 9

10 WORKED EXAMPLE No. 5 Calculate the force needed to punch a hole 30 mm diameter in a sheet of metal 3 mm thick given that the ultimate shear stress is 60 MPa. The area to be cut is the circumference x thickness πd x t A π x 30 x mm τ A so The ultimate shear stress is 60 N/mm τ x A 60 x N or kn WORKED EXAMPLE No. 6 Calculate the force needed to shear a pin 8 mm diameter given that the ultimate shear stress is 60 MPa. The area to be sheared is the circular area A π x 8 A 4 τ so A 50.6mm πd 4 The ultimate shear stress is 60 N/mm τ x A 60 x N or kn SEL ASSESSMENT EXERCISE No A guillotine must shear a sheet of metal 0.6 m wide and 3 mm thick. The ultimate shear stress is 45 MPa. Calculate the force required. (Answer 81 kn). A punch must cut a hole 30 mm diameter in a sheet of steel mm thick. The ultimate shear stress is 55 MPa. Calculate the force required. (Answer10.37 kn) 3. Two strips of metal are pinned together as shown with a rod 10 mm diameter. The ultimate shear stress for the rod is 60 MPa. Determine the maximum force required to break the pin. (Answer 4.71 kn) D.J.Dunn 10

11 11. DOUBLE SHEAR Consider a pin joint with a support on both ends as shown. This is called a CLEVIS and CLEVIS PIN. If the pin shears it will do so as shown. By balance of forces, the force in the two supports is / each. The area sheared is twice the cross section of the pin so it takes twice as much force to break the pin as for a case of single shear. Double shear arrangements doubles the maximum force allowed in the pin. WORKED EXAMPLE No. 7 A pin is used to attach a clevis to a rope. The force in the rope will be a maximum of 60 kn. The ultimate shear stress allowed in the pin is 10 MPa. Calculate the diameter of a suitable pin if the safety factor must be 3.0 based on the ultimate value. The working stress 10/3 40 MPa The pin is in double shear so the shear sress isτ A τ x 6 πd 4 x m A 750 mm d 30.9 mm 6 x 40x10 4 π A SEL ASSESSMENT EXERCISE No A clevis pin joint as shown is used to joint two steel rods in a structure. The pin is 8 mm diameter. The ultimate shear stress for the pin material is 80 MPa. Determine the maximum force that can be exerted using a safety factor of.0. (Answer 4.0 kn). The steel tie rods used in question 1 are 0 mm diameter. The steel yield stress is 160 MPa. Based on the answer to question 1, determine the safety factor for the rods based on yield stress. (Answer 7) D.J.Dunn 11