ENGINEERING COMPONENTS EDEXCEL NATIONAL CERTIFICATE/DIPLOMA MECHANICAL PRINCIPLES OUTCOME ENGINEERING COMPONENTS TUTORIAL 1 STRUCTURAL MEMBERS Structural members: struts and ties; direct stress and strain, dimensional changes; combined effects of direct and thermal loading, factor of safety. Compound members: series and parallel connected compound bars made up of two materials; direct stress and strain in each material, dimensional changes. Fastenings: shear stress in fastenings e.g. riveted joints, bolted joints and hinge pins subjected to single and double shearing forces You should judge your progress by completing the self assessment exercises. These may be sent for marking at a cost (see home page). On completion of this tutorial you should be able to do the following. Define structural members. Calculate direct stress and strain. Calculate changes in dimensions. Solve basic problems involving stress, strain and modulus. Explain and calculate Safety Factor. Explain and calculate stresses due to temperature changes. It is assumed that the student is already familiar with the concepts of FORCE. D.J.DUNN FREESTUDY.CO.UK 1
1. TYPES OF STRUCTURAL MEMBERS Engineering structures come in many forms. Here are some. STRUTS AND TIES A strut is a long thin member that is compressed and usually fails by buckling. A tie is a member that is stretched so it cannot buckle. A tie could be a rope or chain as well as a rigid length of material. Figure 1 FRAMES 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. Figure COLUMNS A column is a thick compression member. Struts fail due to bending but columns fail in compression. Columns are usually made of brittle material which is strong in compression such as cast iron, stone and concrete. These materials are weak in tension so it is important to ensure that bending does not produce tensile stresses in them. If the compressive stress is too big, they fail by crumbling and cracking Figure 3 D.J.DUNN FREESTUDY.CO.UK
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. Figure 4 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.. DIRECT STRESS σ Figure 5 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. Figure 6 Tensile and compressive forces are called DIRECT FORCES. Stress is the force per unit area upon which it acts. Stress = σ = Force/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 1 000 000 N/m then it follows that the N/mm is the same as a MPa D.J.DUNN FREESTUDY.CO.UK 3
3. DIRECT STRAIN ε In each case, a force F 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 10-6. Engineers use the abbreviation µε (micro strain) to denote this multiple. For example a strain of 0.000068 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 1 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. πd π x.5 A = = = 4.909 mm 4 4 F 1 σ = = =.44 N/mm A 4.909 Answer (i) is hence.44 MPa x 0.3 mm ε = = = 0.00015 or 150 µε L 000 SELF 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. (Answers 54.6 MPa and 100 µε). 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. (Answers 15.8 MPa and 10µε) D.J.DUNN FREESTUDY.CO.UK 4
4. MODULUS OF 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. F/x = stiffness = k N/m Figure 7 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 F and x refer to direct stress and strain then F = σa x = εl hence F x = σa εl FL 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. FL 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. Values for different materials may be found in various sources such as the web site Matweb. WORKED EXAMPLE No. A steel tensile test specimen has a cross sectional area of 100 mm and a gauge length of 50 mm, the gradient of the elastic section is 410 x 10 3 N/mm. Determine the modulus of elasticity. The gradient gives the ratio F/A = and this may be used to find E. σ F L 3 50 E = = x = 410 x 10 x = 05 000 N/mm or 05 000 MPa or 05 GPa ε x A 100 D.J.DUNN FREESTUDY.CO.UK 5
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 π x 0.4 A = = = 0.16 m 4 4 6 F 50 x10 6 σ = = = 397.9 x10 Pa A 0.16 6 σ σ 397.9 x10 E = so ε = = = 0.001989 9 ε E 00 x10 x ε = so x = ε L = 0.001989 x 3000 mm = 5.97 mm L 5. SAFETY FACTOR The stress at which a material is deemed to fail might be the ultimate stress or the yield stress. It might also be some other value based on some other criterion such as fatigue and creep. We should also bear in mind that the working stress is often higher than that predicted in the theory covered so far because of local factors such as grooves and sharp corners that raise the stress level. We will not be studying this here. The safety factor is the ratio of the maximum stress allowed and the actual stress. SF = Maximum Allowable Stress/Working Stress WORKED EXAMPLE No.4 A Steel tie rod is 0 mm diameter. It carries a load of 4 MN. Given that the maximum allowable stress is 460 MPa, calculate the safety factor. πd A = 4 π x 0. = 4 F 4 x10 σ = = A 31.4 x 10 460 SF = = 3.61 17.3 6 3 = 31.4 x 10 3 = 17.3 x10 6 m Pa D.J.DUNN FREESTUDY.CO.UK 6
SELF ASSESSMENT EXERCISE No. 1. A bar is 500 mm long and is stretched to 505 mm with a force of 50 kn. The bar is 10 mm diameter. Calculate the stress and strain. The material has remained within the elastic limit. Determine the modulus of elasticity. (Answers 636.6 MPa, 0.01 and 63.66 GPa.). A steel bar is stressed to 80 MPa. The modulus of elasticity is 05 GPa. The bar is 80 mm diameter and 40 mm long. Determine the following. i. The strain. (0.00136) ii. The force. (1.407 MN) 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 cross sectional area and the diameter. (0.467 m and 0.771 m) Note 1 Tonne is 1000 kg. 4. A Steel tie rod is 10 mm diameter. It carries a load of 3 MN. Given that the maximum allowable stress is 500 MPa, calculate the safety factor. (Answer 1.31) D.J.DUNN FREESTUDY.CO.UK 7
6. TEMPERATURE STRESSES It is not clear in the syllabus how much attention should be paid to thermal stresses and perhaps a description only would suffice. This work should be studied if you think it is required to calculate thermally induced stresses. Metals expand when heated. This can be put to good use. For example a ring may be expanded by warming it and then fitted onto a shaft and on cooling grips the shaft very tightly. Thermal expansion can also produce unwanted stresses in structures. For example, suddenly allowing hot fluid into a badly designed pipe could cause it to fracture as it tries to get longer but is prevented from doing so. COEFFICIENT OF LINEAR EXPANSION All engineering materials expand when heated and this expansion is usually equal in all directions. If a bar of material of length L has its temperature increased by θ degrees, the increase of length L is directly proportional to the original length L and to the temperature change θ. Hence L =constant x L θ The constant of proportionality is called the coefficient of linear expansion (α). L = α L θ INDUCED STRESS IN A CONSTRAINED BAR When a material is heated and not allowed to expand freely, stresses are induced which are known as "temperature stresses." Suppose the bar was allowed to expand freely by distance L and then changed back to its original length. The strain is then ε = L /L = α L θ/l = α θ Since stress/strain = modulus of elasticity (E) then the induced stress is σ = E ε = E α θ This may be a tensile stress or a compressive stress depending whether the bar was pulled back to its original length or pushed back. D.J.DUNN FREESTUDY.CO.UK 8
WORKED EXAMPLE No.5 A thin steel band 850 mm diameter must be expanded to fit around a disc 851 mm diameter. Calculate the temperature change needed and the stress produced in the ring. The coefficient of linear expansion is 15 x 10-6 per o C and the modulus of elasticity E is 00 GPa. Initial circumference of ring = πd = π x 850 = 670.35 mm Required circumference = π x 851 = 673.50 mm L = 673.50-670.35 = 3.15 mm L = α L θ 3.15 = 15 x 10-6 x 670.35 x θ θ = 3.15/(15 x 10-6 x 670.35) = 78.6 Kelvin σ = Eα θ = 00 x 10 9 x 15 x 10-6 x 78.6 = 35.8 MPa. Alternatively strain ε = L/ L = 3.15/670.35 = 0.011796 stress σ = Eε = 00 x 10 9 x 0.011796 = 35.8 MPa WORKED EXAMPLE No.6 A brass bar is 600 mm long and it is turned on a centre lathe to 100 mm diameter. It is held between the chuck jaws and a running tail stock so that it is not free to expand. During the turning process it has become heated from 0 o C to 95 o C. Calculate the thermal stress induced in the bar and the resulting thrust on the chuck and tail stock. E for brass is 90 GPa and α is 18 x 10-6 per o C. Stress induced = σ = Eα θ = 90 x 10 9 x 18 x 10-6 x (95 0) = 11.5 MPa Force = stress x cross sectional area = 11.5 x 10 6 x π x 0.1 /4 = 954.3 kn WORKED EXAMPLE No.7 Determine the induced stress and thrust if the centre lathe flexed so that the bar changed length. Check your solution here. The free expansion of the bar is L = α L θ = 18 x 10-6 x 600 x (95 0) = 0.81 mm Actual change in length is 0.6 mm. Strain induced = change in length/original length = (0.81 0.6)/600 = 0.00035 Stress induced σ = E ε = 90 x 10 9 x 0.00035 = 31.5 MPa Force = stress x cross sectional area = 31.5 x 10 6 x π x 0.01 /4 = 47.4 kn D.J.DUNN FREESTUDY.CO.UK 9
SELF ASSESSMENT EXERCISE No.3 1. A steel ring is 50 mm diameter and mm thick. It must be fitted onto a shaft 50.1 mm diameter. Calculate the temperature to which it must be heated in order to fit on the shaft. The initial temperature is 0 o C and the coefficient of linear expansion is 15 x 10-6 per o C. (Answer 133.3 K). A stub shaft 85. mm diameter must be shrunk to 85 mm diameter in order to insert it into a housing. By how much must the temperature be reduced? Take the coefficient of linear expansion is 1 x 10-6 per o C. (Answer -195.6 K) 3. A steam pipe is 10 mm outer diameter and 100 mm inner diameter. It has a length of 30 m and passes through a wall at both ends where it is rigidly constrained. Steam at 00 o C is suddenly released into the pipe. The initial temperature of the pipe is 15 o C and the coefficient of linear expansion is 15 x 10-6 per o C. E is 00 GPa. Calculate i. the thermal stress produced. (555 MPa) ii. the force exerted by the pipe against the walls. (1.918 MN) D.J.DUNN FREESTUDY.CO.UK 10