CITY AND GUILDS 9210 Unit 130 MECHANICS OF MACHINES AND STRENGTH OF MATERIALS OUTCOME 1 TUTORIAL 1 - BASIC STRESS AND STRAIN

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1 CITY AND GUILDS 910 Unit 130 MECHANICS O MACHINES AND STRENGTH O MATERIALS OUTCOME 1 TUTORIAL 1 - BASIC STRESS AND STRAIN Outcome 1 Explain static equilibrium, Newton's laws, and calculation of reaction and internal forces and the basic concepts of different stresses and strains and determine stresses and strains of components under loading conditions. The learner can: 1. Explain the concept of stresses and strains: Compressive and shear stresses and linear, lateral, shear, thermal and volumetric strains.. Explain static equilibrium, Newton s Laws and Calculation of reaction and internal forces. 3. Explain Hooke s law, Poisson s ratio, Modulus of Elasticity and Modulus of Rigidity, Bulk Modulus. 4. Explain stress-strain diagrams for ductile and brittle materials and for various strengths of materialyield strength, ultimate tensile strength etc. 5. Explain the interrelation between elastic constants, proof stress and true stress and strain, axial force diagrams, stresses and strains in determinate, homogeneous and composite bars under concentrated loads and self weight. 6. Explain temperature stresses in simple and composite members. 7. Explain strain energy due to axial load (gradual, sudden and impact) and strain energy due to self weight. This tutorial should be seen as providing basic knowledge of stress and strain that many degree students should already know. In this case use it as revision or skip it. On completion of this tutorial you should be able to do the following. Define direct stress and strain. Define shear stress and strain. Define the modulus of elasticity and rigidity. Solve basic problems involving stress, strain and modulus. It is assumed that the student is already familiar with the concepts of ORCE. 1

2 1. STATIC ORCES and STRUCTURES STATIC ORCES will deform a body in one or more of the following manners. The force may stretch the body, in which case it is called a TENSILE ORCE. 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. 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. 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. 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.

3 RAMES Struts and ties make up the members of lattice frames such as the simple one shown. In this example the sloping 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 shows examples. Transverse loading causes bending and bending is a very severe form of stressing a structure. You might note how easy it is to snap a stick by bending but not so easy by stretching or twisting. The bent beam goes into tension (stretched) on one side and compression on the other. ailure or fracture will always occur on the tension side and spread across the section. 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 studied 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 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 and the symbol is (called SIGMA). The units are N/m or Pascals. 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 4

5 4. DIRECT STRAIN 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. The symbol 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 into the exponent form of Engineers use the abbreviation (micro strain) to denote this multiple. or example a strain of could be written as 68 x 10-6 but engineers would write 68. 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. ii. The strain in the wire. If the stress that produces failure is 300 MPa, calculate the safety factor. πd π x.5 A mm 4 4 Answer (i) is hence 44 MPa σ A N/mm x 0.3 mm or 150 L 000 Safety actor = 300/44 = 1.3 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 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, 10 and 1.64) 5

6 5. STRESS AND STRAIN RELATIONSHIPS OR MATERIALS The relationship between direct stress and strain is determined with a tensile test. A typical testing machine is shown. These measure and plot the force against extension of the test piece and converts them into stress and strain so that the relationship between them may be determined. The details of the test are not needed here. The tensile test is conducted in order to find the following properties of a material. The yield stress. The ultimate tensile stress. The elastic range. The ductile range. The modulus of elasticity. Ductility or lack of ductility is indicated by two results from the test as follows. The % elongation. The % area reduction. Pictures courtesy of Lloyd Instruments The following applies to materials tested in tension (direct stress). Similar tests and descriptions apply to materials in torsion (the torsion test). The diagram below shows typical orce extension graphs for some materials. Graph A is typical for a brittle material. The force is directly proportional to extension right up to the point of fracture and hence obeys Hooke's Law. The material fractures without warning. Cast Iron is a typical example. Graph B shoes the result for a material that has some ductility such as steel. The material obeys Hooke's Law up to a point where the structure of the material starts to give and slip; a phenomenon known as yielding. After yielding the material continues to extend with little or no increase in the force and this is known as plasticity. Graph C is typical for a very ductile material such as copper with a very large plastic range. In the elastic range the material will spring back when the force is removed. In the plastic range the material is permanently damaged and remains stretched when the force is removed. The permanent damage starts at the yield point. Usually this is the same point as the limit of proportionality. At the necking point the material starts to narrow at the point where it is going to break and a pronounced neck forms. Because the cross sectional area is reduced, the force needed to stretch it further becomes smaller but the stress in the material remains the same. Eventually at some point the specimen will break. 6

7 DIRECT STRESS AND STRAIN Stress Strain Yield stress Ultimate Tensile Stress = orce/area = Extension/Gauge length y = Yield Load/Original Area. u = Maximum Load/Original Area Normally the ultimate tensile stress is based on the maximum load as the reduction in area causes the load (but not the stress) to fall. 6. MODULUS of ELASTICITY Elastic materials always spring back into shape when released. They usually obey HOOKE'S LAW. This is the law of a spring which states that deformation is directly proportional to the force. As long as the material is within the elastic range and the graph is a straight line, the ratio of /x and hence / is constant. The ratio / is called the modulus of Elasticity and has a symbol E. This property determines how elastic the material is. The units of E are the same as the units of stress. WORKED EXAMPLE No. An aluminium tensile test specimen is 5 mm diameter with a gauge length of 50 mm. The force measure at the yield point was 98 N and the maximum force was 1.6 kn. Calculate the yield stress and the ultimate tensile stress. Cross sectional area A = π d /4 = π x 5 /4 = mm Yield Stress = 98/19.63 = 50 N/mm or 50 MPa (Note 1 MPa = 1 N/mm ) Ultimate Tensile Stress = 1600/19.63 = 81.5 N/mm or 81.5 MPa At a point on the proportional section the extension was 0.03 mm and the force 800 N. Calculate the modulus of Elasticity (Young's Modulus) Lo 800 x 50 E N/mm or 67.9 GPa Ax x 0.03 SEL ASSESSMENT EXERCISE No. A tensile test specimen has a cross sectional area of 100 mm. The force measure at the yield point was 41 kn and the maximum force was 4 kn. Calculate the following. The yield stress (410 MPa) The tensile strength. (40 MPa) The same tensile test showed that the force at the limit of proportionality was 40 KN and the extension was mm over a gauge length of 50 mm. Calculate the modulus of elasticity. (10 GPa) 7

8 WORKED EXAMPLE No. 3 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 MPaor 05 GPa ε x A 100 Breaking orce σu 350N/mm or 350MPa Area 100 WORKED EXAMPLE No. 4 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 8

9 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. 9

10 7. SHEAR STRESS Shear force is a force applied sideways on to the material (transversely loaded). Here are some examples: Shear stress is the force per unit area carrying the load. This means the cross sectional area of the material being cut and of the beam and pin. Shear stress = /A The symbol 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 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 (x) to the height (L). The end face rotates through an angle. 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 so that = x/l It follows that is the shear strain. The symbol is called Gamma. 10

11 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 (obeying Hooke's Law). 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: But /A = and x/l = so: This constant will have a special value for each elastic material and is called the Modulus of Rigidity with symbol G. 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 u and this value is used to calculate the force needed by shears and punches. WORKED EXAMPLE No. 5 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 rectangle 800 mm x 5 mm A 800 x mm A so The ultimate shear stressis 50 N/mm x A 50 x N or 00 kn 11

12 WORKED EXAMPLE No. 6 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 A so The ultimate shear stressis 60 N/mm x A 60 x N or kn WORKED EXAMPLE No. 7 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 x 8 A 4 so A 50.6mm πd 4 The ultimate shear stressis 60 N/mm 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) 1

13 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. 8 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 is A τ A d 4 x x 10 m A 750 mm d 30.9 mm 6 x 40x10 4 π 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 1.5) 13