1 Contents CHAPTER 1 FORCE ON MATERIALS.................................................................................................... 3 1.0 TYPES OF LOADS FORCE ON MATERIALS ............................................................................... 3 1.1 TYPES OF FORCES ........................................................................................................................ 3 1.3 EFFECT OF FORCE.................................................................................................................. 4 1.4 STRESS AND STRAIN.................................................................................................................... 4 1.5 YOUNG’S MODULUS.............................................................................................................. 5 1.5.1 Example 1.0.................................................................................................................. 5 1.5.2 Example 1.2................................................................................................................... 6 1.5.3 Example 1.3.......................................................................................................................... 6 1.5.4 Example 1.4.......................................................................................................................... 6 1.5.5 Example 1.5.......................................................................................................................... 7 1.6 TENSILE TEST............................................................................................................................... 7 1.7 OUTCOME OF TENSILE TEST........................................................................................................ 9 1.7.1 Example 1.6.......................................................................................................................... 9 1.8 HOOKE’S LAW ........................................................................................................................... 10 1.9 SAFETY FACTOR......................................................................................................................... 11 1.10 STRAIN ENERGY....................................................................................................................... 12 1.10.1 Example 1.7...................................................................................................................... 12 1.11 POISSON RATIO....................................................................................................................... 13 1.11.1 Example 1.8...................................................................................................................... 13 1.11.2 Example 1.9...................................................................................................................... 14 1.12 SHEAR STRESS AND SHEAR STRAIN ........................................................................................ 15 1.13 MODULUS OF RIGIDITY......................................................................................................... 16 1.13.1 Example 1.10.................................................................................................................... 16 1.13.2 Example 1.11.................................................................................................................... 17 1.13.3 Example 1.12.................................................................................................................... 17 1.13.4 Example 1.13 .................................................................................................................... 18 1.4 EXERCISES................................................................................................................................ 19 CHAPTER 2 THERMAL STRESS AND COMPOSITE BAR......................................................................22 2.0 COMPOSITE BAR ...................................................................................................................... 22 2.1 SERIES COMPOSITE BAR............................................................................................................ 22 2.2 PARALLEL COMPOSITE BAR ....................................................................................................... 23 2.2.1 Example 2.0........................................................................................................................ 23 2.2.3 Example 2.1........................................................................................................................ 25
2 2.2.4 Example 2.2........................................................................................................................ 25 2.2.5 Example 2.3........................................................................................................................ 27 2.2.6 Example 2.4........................................................................................................................ 28 2.3 THERMAL STRESS ..................................................................................................................... 29 2.4 TEMPERATURE COEFFICIENT OF LINEAR EXPANSION............................................................... 29 2.5 THERMAL STRESS ON COMPOSITE BAR..................................................................................... 30 2.5.1 Example 2.5 ........................................................................................................................ 31 2.5.2 Example 2.6 ........................................................................................................................ 32 2.5.3 Example 2.7 ........................................................................................................................ 34 2.5.4 Example 2.8 ........................................................................................................................ 36 2.6 EXERCISES ................................................................................................................................. 38 CHAPTER 3 SHEAR FORCE AND BENDING MOMENT .......................................................................44 3.0 TYPES OF BEAMS....................................................................................................................... 44 3.1 TYPES OF LOADS........................................................................................................................ 45 3.2 REACTION FORCE ...................................................................................................................... 46 3.3 SHEAR FORCE ............................................................................................................................ 46 3.4 BENDING MOMENT .................................................................................................................. 46 3.5 SHEAR FORCE AND BENDING MOMENT DIAGRAMS................................................................. 47 3.5.1 Example 3.0........................................................................................................................ 47 3.5.2 Example 3.1........................................................................................................................ 50 3.5.3 Example 3.2........................................................................................................................ 53 3.6 POINT OF CONTRA-FLEXURE ..................................................................................................... 55 3.7 MAXIMUM BENDING MOMENT................................................................................................ 55 3.7.1 Example 3.3........................................................................................................................ 56 3.7.2 Example 3.3........................................................................................................................ 58 3.7.4 Example 3.4........................................................................................................................ 61 3.7.5 Example 3.5........................................................................................................................ 63 3.7.6 Example 3.6........................................................................................................................ 65 3.8 EXERCISES ................................................................................................................................. 68
3 CHAPTER 1 FORCE ON MATERIALS 1.0TYPES OF LOADS FORCE ON MATERIALS a) Static Load – known as a fixed load or unchanged load. i.e.: buildings, mountains and hills b) Dynamic Load – known as the load is constantly changing i.e.: the vehicle crossing a bridge, people ride on the elevator c) Impact Load – known as the load acting on an object immediately i.e.: a process using a hammer to drive a nail, a piling process d) Fatigue and Alternating Load – known as the load acting on a certain time only i.e.: windmill (fatigue load) and piston in the engine of a vehicle (alternating load) 1.1 TYPES OF FORCES a) Direct Force – a situation where the reaction on a cross-sectional of the material which a force is applied always perpendicular to the external force. – If the direction of the force applied is tensile force will result in elongation on materials, while the compressive force will result in a shortening of the material. b) Shear Force – a situation where reaction occurs on a part of material which is parallel to the applied shear force causes slippage on that part. P P P P Tensile Force Compressive Force P P Shear Force Shearing occurs on this part
4 1.3EFFECT OF FORCE The effect of force on a material will cause deformation of the material. Among the effects is: a) Shortening of the material b) Elongation of the material c) Bending of the material d) Twisting on the material e) Shearing of the materia 1.4 STRESS AND STRAIN • If the tensile or compressive force is applied to a material, there will be internal forces always act in the opposite direction of the applied load to the crosssectional area of the material. • This internal force is called stress which is the reaction to the applied load. Its purposed is to ensure a stability in the material system. Stress, σ = Applied Load, P Area of cross sectional, A • The reaction of force that causing elongation or shortening is known as deformation of the material. This deformation is called strain which is defined as the elongation or shortening of the size of the material per unit of length. Strain, ε = Elongation of Material,∆L Length,L = Final Length,Lf −Original Length,L Original Length,L P P Internal force act in opposite direction of the applied load, σ σ σ L 1/2∆L 1/2∆L P P
5 1.5 YOUNG’S MODULUS • Young's modulus (E) or the modulus of elasticity is a measure of a solid’s stiffness or resistance to elastic deformation under load. It relates stress (force per unit area) to strain (proportional deformation) along an axis or line Young′ s Modulus = Stress, σ Strain, ε 1.5.1 Example 1.0 A bar with a diameter of 30 mm and a length of 80 mm is subjected to tensile force of 100 kN causes the bar elongated to 0.0585 mm. Determine the stress and strain on the bar. Solution Area of cross sectional, A A = π 4 D2 = π 4 (0.03)2 = 7.069 x 10-4 m2 Stress, σ = P A = 100 x 103 7.069 x 10-4 = 141.47 x 106 N/m2 Strain, ε = ∆L L = 0.0585 80 = 7.31 x 10-4 100 kN 100 kN φ30 mm 80 mm Figure 1.1
6 1.5.2 Example 1.2 A metal bar is applied compression load 120kN as shown Figure 1.2 causes elongation of the bar to 0.075 mm. Calculate the stress and strain in the bar. Solution Area of cross sectional, A = 0.085 x 0.04 = 3.4 x 10-3 m2 Stress, σ = P A = 120 x 103 3.4 x 10-3 = 32.29 x 106 N/m2 Strain, ε = ∆L L = 0.075 350 = 2.14 x 10-4 1.5.3 Example 1.3 A circular wire has a tensile force of 66N applied to it and this force produces a stress of 3.08 MPa in the wire. Determine the diameter of wire. Solution Cross section, A = P σ = 66 3.08 x 106 = 2.143 x 10-5 m2 The diameter of wire, A = π 4 D2 2.143 x 10-5 = π 4 D2 D2 = 4 π(2.143 x 10-5) = D = 5.22 × 10−3 m 1.5.4 Example 1.4 A copper wire of 4 m long carrying a load of 100 kN. If the stress on the wire 60 MN/m2, calculate: Given, Ecopper = 112 GN/m2 a) Strain in copper wire b) Elongation of copper wire 120 kN 120 kN 40 mm 350 mm 85 mm Figure 1.2 100 kN 100 kN 4 m
7 Solution a) Strain, ε = Stress, σ Young's Modulus, E = 60 x 106 112 x 109 = 5.357 x 10-4 b) Elongation, ∆L = ε.L = 5.357 x 10-4(4) = 2.143 x 10-3m 1.5.5 Example 1.5 A 5.6 m rod with a cross sectional area of 1150 mm2 elongate by 7.56 mm when a 70kN tensile force is applied on both sides. Determine: a) The tensile stress b) The tensile strain c) Young’s Modulus of the rod Solution a) The tensile stress Stress, σ = P A = 120 x 103 3.4 x 10-3 = 32.29 x 106 N/m2 b) The tensile strain Strain, ε = ∆L L = 7.56 x 10-3 5.6 = 1.35 x 10-3 c) Young’s Modulus of the rod Young's Modulus, E = Stress, σ Strain, ε = 60.87 x 106 1.35 x 10-3 = 45.04 x 109 N mm 2 1.6 TENSILE TEST • The tests carried out on a round bar using 'Universal Tensile Machine' which aims to measure the elongation of the bar due to gradually load increasing. • The test is conducted by measuring the elongation of the bar is proportional to the increase of the load until the bar is break apart. • From this test graphs can be plotted, load versus elongation that can determine the level of elasticity of the materials used during the test.
8 1.6.1 TEST SAMPLE 1.6.2 Graph Load Versus Elongation Elasticity Limit : A point where the bar will return to its original shape after the load is released. Yield Point : A point where elongation occurs without an increase in load. This situation will create the upper yield point and lower yield point. Maximum Load : A point where the highest loads imposed on the bar. It also called for the ultimate tensile strength. Break Point : A point where necking occurs on bar which causes decreasing in cross-sectional area of the bar until the bar is broken into two parts. Elasticity Region : Bar in this region will return to the original form in which stress is directly proportional to the strain. Plasticity Limit : Bar in this region did not return to its original shape after the load is released that causes permanent elongation occurs to the bar. Load, P Elasticity limit Elongation, ∆L Upper yield point Maximum Load Lower yield point Break Point Elasticity region Plasticity region Figure 1.3
9 1.7 OUTCOME OF TENSILE TEST There are several outcomes from tensile test measurements: - a) Yield Stress, σ = Load at upper yield, Py Area of cross sectional, A b) Maximum Tensile Stress, σmax = Maximum Load, Pmax Area of cross sectional, A c) Percentage of Elongation = Increase in Length, ∆L Original Length, L x 100% d) Percentage of Size Reduction = Original Size, A - Final Area, Af Original Size, A x 100% 1.7.1 Example 1.6 The tensile tests conducted on a mild steel using specimens having a diameter of 14.28 mm and gauge length 50 mm. In the early stage of the test results is record. Load, kN 0 20 40 60 80 Elongation, mm 0 0.030 0.062 0.091 0.121 Other data collected are: – Final gauge length after fracture = 64.45 mm – The diameter of the rod in the neck after breaking = 11:51 mm a) Using the appropriate scale, draw a graph of load versus elongation up to 80kN load. Determine the slope of the graph. b) Calculate: i) Young's modulus of the material ii) Percent elongation iii) Percentage depletion of area
10 Solution a) Slope of the graph Slope = ∆P ∆L = 72 - 26 0.11 - 0.04 = 657 kN/mm b) i) Young’s modulus A = π 4 D2 = π 4 �14.282 � = 160.16 mm2 E = PL A∆L = slope x L A = 657 x 50 160.16 = 205 kN/mm2 ii) Percent elongation % ∆L = � Lf - L L � x 100% = � 64.65 - 50 50 � x 100% = 28.9% iii) Percentage depletion of area % ∆A = � π 4 D2 - π 4 Df 2 π 4 D2 �x 100% = � D2 - Df 2 D2 � x 100% = 14.282 -11.512 14.282 = 35% 1.8 HOOKE’S LAW • Elastic material is a material that deforms easily when the application of load. When the load is no longer subject to the material will return to its original length. • Most of the mechanical properties of the metal showed a range of stress is called the elastic range. If we draw a graph of stress versus strain, we find the straight line OA in Figure 1.4 • Hooke's law states that within the elastic limit of a material, the stress is directly proportional to the strain
11 • Hooke's Law constants in the equation is called the modulus of Elasticity or Young's modulus, E. Therefore Modulus Young, E = Stress, σ Strain, ε • By changing σ = P A and ε = ∆L L into the above equation will forms E = PL A∆L ∆L = PL AE unit for E is N/m2 • The following is the value of E for several materials Material Modulus Young (GN/ m2) Steel 200 – 220 Aluminum 60 – 80 Copper 90 – 110 Wood 10 1.9 SAFETY FACTOR • The safety factor is defined as the ratio between the maximum load with the workload or the ultimate stress with work stress Safety Factor (SF) = Maximum Load, Pmax Working Load, Pwork = Ultimate Stress, σult Working Stress, σwork Load, P Elongation, ∆L ∆P ∆L O Figure 1.4 Stress (σ) α strain (ε) or σ ε = constant
12 • The safety factor usage depends on the following factors: - i) the possibility of overload ii) the type of load used - static, impact, dynamic or alternating iii) the possibility of a defect in the materials use iv) for special purposes or due to failure 1.10 STRAIN ENERGY • Strain energy is defined as the work done by the load on a component to produce a strain in the component Strain Energy, U = 1 2 P.∆L unit : Joule (J) • Strain energy also can be defined by calculating the area under the graph, OAB of load versus elongation. whereas AL = V U = σ2V 2E 1.10.1 Example 1.7 A bar bears a tensile force such as in the diagram below. Calculate the strain energy stored in the bar. Given that E = 200 GN/m2. Solution U = 1 2P.∆L whereas ∆L = PL AE A = π 4 D2 = π 4 (0.03)2 = 7.069 x 10-4 m2 Load, P Elongation, ∆L O A B Strain U = 1 2P.∆L where P = σA and ∆L= εL U = 1 2P.∆L = 1 2 σA.εL = σA 2 . σL E = σ2AL 2E d = 30 mm 600 mm 150 kN 150 kN Figure 1.5
13 ∆L = PL AE = 150x103 (0.6) 7.069 x 10-4(200x2009 ) = 6.37 x 10-4 m U = 1 2P.∆L = 1 2 (150x103 )(6.37 x 10-4) = 47.74 J OR U = P2 L 2AE = �150x103 � 2 �600x10-3� 2 � π 4 �30x10-3� 2 � (200x2009 ) = 47.74 J 1.11 POISSON RATIO • Poisson's ratio is the ratio between the lateral strain and longitudinal strain produced by a single stress. Lateral Strain, εy = Change in diameter, ∆d Original diameter, D Longitudinal Strain, εx = Change in length, ∆L Original length, L • In engineering practice, the value of Poisson ratio between 0.25 to 0.35 1.11.1 Example 1.8 A steel bar has length of 420 mm and diameter of 45 mm was loaded with tensile load 65 kN and the elongation of the bar is 0.032 mm and the diameter reduction of 0.01 mm. From the test, determine: i) Stress in the bar ii) Strain relatives to x-axis iii) Strain relatives to y-axis iv) Poisson ratio v) Safety factor if the ultimate stress was 200 MN/m2 P P Poisson Ratio, = Lateral Strain, εy Longitudinal Strain, εx
14 Solution i) Stress in the bar A = π 4 D2 = π 4 (0.045)2 = 1.59 x 10-3 m2 σ = P A = 65 x 103 1.59 x 10-3 = 40.9 x 106 N/m2 ii) Longitudinal strain εx = ∆L L = 0.032 420 = 7.62 x 10-5 iii) Lateral strain εy = ∆D D = 0.01 45 = 2.22 x 10-4 vi) Poisson ratio ν = εy εx = 2.22 x 10-4 7.62 x 10-5 = 2.9 iv) Safety factor, SF S.F = σult σworking = 200 x 106 40.9 x 106 = 4.89 1.11.2 Example 1.9 A hollow steel column has an internal diameter of 200 mm, calculate minimum external diameter which carries a load of 1.2 MN if the factor of safety is 5. Ultimate stress for steel is 470 MN/m2. Solution σworking= σult Safety Factor = 470 x 106 5 = 94 x 106 N/m2 Area, A = P σ = 1.2 x 106 94 x 106 = 0.0128 m2 If D is the external diameter, A = π 4 (D2 − d2 ) D2 − 0.22 = 4 π(0.0128) = 0.0163
15 D2 = 0.0163 + 0.22 =………. D = 0.237 m2 1.12 SHEAR STRESS AND SHEAR STRAIN • Shear stress is defined as the component of stress coplanar with a material cross section. Shear stress arises from the force vector component parallel to the cross section. • For double shear as three plates are riveting each other, the shear stress • For three-piece rectangular plate affixed below, the shear stress is obtained • Shear strain is defined as the tangent of that angle, and is equal to the length of deformation at its maximum divided by the perpendicular length in the plane of force application which sometimes makes it easier to calculate Shear Stress, = Shear force, V Area of cross section, A Shear Stress, τ = V A = P/2 A Shear Stress, τ = V A = P/2 Lw P P P L w
16 1.13 MODULUS OF RIGIDITY • Modulus of Rigidity is defined as the ratio of shear stress to shear strain 1.13.1 Example 1.10 Figure shows a punch with a diameter of 19 mm was used to produce a hole in the plate thickness of 6 mm. A force of 116 kN applied to the punch, determine the shear stress occurring in the plates. Solution Given D = 19 mm; t = 6 mm; V = 116 kN A = πDt = π(0.019)(0.006) = 3.58 x 10-4 m2 Shear Stress, τ = Shear Force, V Area, A = 116 x 103 3.58 x 10-4 = 324.0 x 106 N/m2 Shear strain, ∅ = Length of demormation, x Perpendicular length, L Modulus of Rigidity, G= Shear stress, τ Shear strain, ∅ 6 mm 19 mm
17 1.13.2 Example 1.11 Three plates are connected by means of two rivets as in the figure below. If the shear stress τ = 40 N/mm2, calculate the appropriate diameter of the rivet. Solution Cross sectional of two rivet, A = 2π 4 D2 Shear Stress, τ = Shear Force, V Area, A 2π 4 D2 = 50 x 103 2(4 x 106 ) D2 = 4(6.25 x 10−3) 2π D = 0.063 m 1.13.3 Example 1.12 Two steel plates each of thickness 10mm are joined together by two rivet shown as Figure 1.4. Diameter of rivet is 14mm. If the plates are pulled apart by a force of 30kN, find the shear stress in the rivet. Solution Cross sectional of two rivet, A = 2π 4 D2 = 2π 4 (0.014)2 = 3.08 × 10-4 m2 Shear Stress, τ = V A = 50 x 103 3.08 × 10-4 = 97.4 x 106 N/m2 30 kN 30 kN Figure 1.4
18 1.13.4 Example 1.13 Figure 1.5 shown that a 20mm diameter rivet joins the plates that are each 110mm wide. The allowable stresses are 120MPa for bearing in the plate material and 60 MPa for shearing of rivet. Determine: a) The minimum thickness of each plate b) The largest average tensile stress in the plates Solution From shearing of rivet: Cross sectional of the rivet = π 4 D2 = π 4 (0.02)2 = 3.14 × 10-4 m2 Shear Force, V = τA = 60 x 106 (3.14 × 10-4) = 18.85 x 103 N From bearing of plate material: Allowable Stress, σ = V A 120 x 106 = 18.85 x 103 0.02 t t = 18.85 x 103 120 x 106 (0.02) t = 0.00785 m Largest average tensile stress in the plate: Allowable Stress, σ = V A = 18.85 x 103 0.00785(0.11 − 0.02) = 26.68 ×106 N/m2 P 110 mm P P P ∅ 20 mm t t Figure 1.5
19 1.4 EXERCISES 1. a) Define the following: i) Stresses ii) Strain iii) Young's Modulus b) Give THREE (3) types of loads and give an example. c) Graph stress versus strain of tensile tests performed on mild steel is shown in Figure 1 below. Label the points A to D and range of E and F 2. A rod with 2.5 m long and cross-sectional area of 1290 mm2 have been extended to 1.5 mm when 142 kN force is applied. Find: a) Stress in rod b) Strain of rod c) Young’s Modulus d) Factor of safety when maximum stress is 432MN/m2 (Ans: 110.08 MN/m2, 6 x 10-4, 183.47 GN/m2, 3.92) 3. A bar with 30 mm in diameter and 80 mm in length is subjected to a tensile load of 100kN. The ultimate stress in the bar is 230MN/m2. Due to applied load, the bar elongates to 0.00585 mm and the diameter become 29.994 mm. Calculate: a) Stress occurs in the bar b) Young’s Modulus c) Factor of safety d) Poisson ratio (Ans: 141.47 MN/m2, 193.45 GN/m2, 1.63, 0.27) 4. A aluminium bar has length of 420 mm and diameter of 45 mm was loaded with tensile load 65kN and the elongation of the bar is 0.032mm and the diameter reduction of 0.01 mm after the test. Determine: a) Stress in the bar b) Strain relatives to x axis c) Poisson ratio (Ans: 40.88 MN/m2, 7.62 x 10-5, 2.914) E F A B C D Stress, σ Strain, ε Figure 1
20 5. A copper wire of 4 meters in length is applied with a force of 10 kN. If the stress in the wire is 60 MPa. Given Ecopper = 112 GN/m2, calculate: a) The strain in the wire. b) The elongation of the wire. c) The factor of safety, if the ultimate stress is 230 MPa. d) The diameter of the wire. (Ans: 5.36 x 10-4, 2.14 x 10-3 m, 3.8, 0.015 m) 6. A bar of thickness 21mm and having a rectangular cross section caries a load of 150kN. Determine the minimum width of the bar if the limit of maximum stress is 300 MPa. The bar, which is 1.5 m long, extends by 3.2 mm when carrying the load. Determine the modulus of Elasticity of the material. (Ans: 0.024m, 140.9GPa) 7. In a tension test on a mild steel rod of 13mm diameter using a 202 mm gauge length extensometer, the following observations were made: Extension under load of 17kN = 0.15mm Load at yield point = 28.5kN Ultimate load = 63KN Breaking load = 43kN Diameter at the neck = 7.66mm Calculate the: a) Young’s modulus b) Yield stress c) Ultimate stress d) Actual stress at breaking load (Ans: 172.38 GN/m2, 214.72 MN/m2, 474.6 MN/m2, 933.08 MN/m2) 8. a) Define the following in engineering terms: i) Young’s Modulus ii) Lateral Strain iii) Longitudinal Strain b) An aluminium tube of length, L = 500 mm is loaded with tensile forces, P as shown in Figure 2). The outside and inside diameters are 90 mm and 40 mm respectively. A strain gauge is placed on the outside of the bar to measure the normal strain in the longitudinal direction. i) If the measurement on the strain gauge, ε = 650 x 10-6, calculate the elongation, ∆L of the tube. ii) If If the stress applied to the bar is 30 MPa, calculate the load P is applied. (Ans: 3.25 x 10-4 m, 153.2 kN) P P Figure 2 500 mm
21 9. A hollow metal rod with external diameter of 200 mm and internal diameter of 180 mm elongates 2.0 mm when subjected to a tension of F kN. If stress is 15 MN/m2, calculate: i) tensile force, F ii) strain if the length of the rod is 3.0 m iii) Young's modulus of the metal rod iv) Strain energy v) The safety factor if the ultimate stress is 40 MN/m2 (Ans: 89.5 kN, 6.67 x 10-4, 22.5 GN/m2, 89.5 J, 2.7) 10. a) List FOUR (4) types of loads with examples b) Identify the formulas and units of the following terms; i) Poisson ratio ii) Safety factor iii) Young’s Modulus c) An 1130 mm2 cross sectional area with 50 mm length of steel bar is subjected to a tensile force of 20 kN and cause the elongation of the bar to 0.0435 mm. Determine: i) The stress in bar ii) Normal strain iii) Young’s Modulus (Ans: 17.7 MN/m2, 8.7 x 10-4 , 20.34 MN/m2) 11. A bracket is connected to a tie bar by means of four bolts. The permissible shear stress in the bolts is 90MN/m2. If the maximum load in the bar is 50 kN, calculate the minimum diameter of the bolts. (Ans: 0.0133m) 12. a) Calculate the shear force needed to pinch a hole of 30mm diameter in a sheet of metal 3 mm thick given that the ultimate shear stress is 60 MPa b) Calculate the force to shear a pin 8 mm diameter given that the ultimate shear stress is 60 MPa (Ans: 16.97 kN, 3.02kN) 13. As in Figure 3, a hole is to be punched out of a plate having a shearing stress of 40 kPa. The compressive stress in the punch is limited to 50 kPa. a) Compute the maximum thickness of plate in which a hole 63 mm in diameter can be punched. b) If the plate is 6.35 mm thick, determine the diameter of the smallest hole that can be punched (Answer: 0.0197m, 0.02m) Figure 3
22 CHAPTER 2 THERMAL STRESS AND COMPOSITE BAR 2.0 COMPOSITE BAR • Composite is defined as a combination of two or more bars that have the same or different materials, the same or different length and the same or different diameter. • There are two types of composite bar: a) Series composite bar b) Parallel composite bar 2.1 SERIES COMPOSITE BAR • Elongation or shortening occurs in series composite bar is considering different in each bar while external load occurs at each bar is same. P L1 L2 φD1 φD2 Bar 1 Bar 2 a) P = P1 = P2 b) ∆L = ∆L1 + ∆L2 P P P P
23 2.2 PARALLEL COMPOSITE BAR • Elongation or shortening occurs in parallel composite bar is considering same for each bar while external load occurs at each bar is different. 2.2.1 Example 2.0 Figure shows a series of composite bar which are made of the same material are subjected to loads of P1 = 50 kN, P3 = 480 kN and P4 =160 kN. Calculate: i) P2 to balance the system ii) Total elongation if E = 210 GN/m2 Solution i) Using the law of equilibrium of forces, The total force acting on the left = The total force acting on the right P1 + P3 = P2 + P4 50 + 480 = P2 + 160 P2 = 370 kN a) P = P1 + P2 b) ∆L = ∆L1 = ∆L2 P Steel wire Copper wire P P Bar 1 Bar 2 D1 d D2 Bar 2 1 Bar 1
24 ii) Total elongation is obtained by determining the forces on each side of the bar Force on bar A PA = 50 kN (Tensile) Force on bar B PB = − 320 kN (Compressive) Force on bar C PC = 160 kN (Tensile) ii) Total elongation of the composite bar AA = π 4 DA 2 = π 4 (0.015)2 = 1.767 x 10-4 m2 AB = π 4 DB 2 = π 4 (0.04)2 = 1.257 x 10-3 m2 AC = π 4 DC 2 = π 4 (0.025)2 = 4.909 x 10-4 m2 ∆L = ∆LA+∆LB+∆LC = � PL AE� A + � PL AE� B + � PL AE� C = 1 E � PALA AA + PBLB AB + PCLC AC � = 1 210 x 109 � 50 x 103(1.5) 1.767 x 10-4 + −320 x 103(0.8) 1.257 x 10-3 + 160 x103(1) 4.909 x 10-4� = 2.6 x 10−3 m 50 kN 50 kN PB = P3 − P4 = 480 − 160 = 320 kN PB = P2 − P1 = 370 − 50 = 320 kN 160 kN 160 kN
25 2.2.3 Example 2.1 A series composite bar consists of a copper bar with a diameter of 30 mm and a length of 0.25 m and steel bars with a diameter of 25 mm and length of 0.5 m. If both ends are rigid connected and subjected to a tensile force of 15 kN, determine: i) Stresses in copper and steel bars ii) Elongation of the two bars Given, ES= 200 GN/m2; EC = 100 GN/m2 Solution Given DS = 25 mm; LS = 0.5 m; ES = 200 GN/m2 DC = 30 mm; LC = 0.25 m; EC = 100 GN/m2 Get area for copper and steel bars AS = π 4 DS 2 = π 4 (0.025)2 = 4.91 x 10-4 m2 AC = π 4 DC 2 = π 4 (0.03)2 = 7.07 x 10-4 m2 i) Stresses in copper and steel bars σS = PS AS = 15 x 103 4.91 x 10-4 = 30.55 x 106 N/m2 σC = PC AC = 15 x 103 7.07 x 10-4 = 21.22 x 106 N/m2 ii) Elongation of the two bars ∆L = ∆LS + ∆LC = � σL E � S + � σL E � C = 30.55 x 106 (0.5) 200 x 109 + 21.22 x 106 (0.25) 100 x 109 = 1.294 x 10-4 m 2.2.4 Example 2.2 a) For parallel and serial composite bar, explain. i) extension of the composite bar ii) shortening of the composite bar iii) external load of the composite bar b) A composite bar made of steel bar with a square cross-section and copper rod with diameter of 10 mm. The length of steel bar is 2 m connected in series with the copper rod which is 1 m in length. The end of the copper rod welded to a steel beam and the end of steel bar is left hanging free. Determine: i) Maximum load that can be hung on the free end ii) Cross sectional area of the steel bar
26 iii) Total elongation of the composite bar due to action of the load Take Ecopper = 100kN/mm2 σcopper = 56N/mm2 Esteel = 200kN/mm2 σsteel = 140N/mm2 Solution a) i) extension of the composite bar - For parallel composite bar, extension occurs at each bar is same, while for series composite bar, extension occurs at each bar is different. ii) shortening of the composite bar - For parallel composite bar, retardation occurs at each bar is same, while for series composite bar, shortening occurs at each bar is different. iii) external load of the composite bar - For parallel composite bar, external load occurs at each bar is different, while for series composite bar, external load occurs at each bar is same. b) Given DC = 10 mm; LC = 1 m; LS = 2 m; EC = 100 GN/m2; ES = 200 GN/m2; σC = 56 MN/m2; σS = 140 MN/m2 i) Maximum load that can be hung on the free end AC = π 4 DC 2 = π 4 (0.01)2 = 7.854 x 10-5 m2 PC = σC.AC = 56 x 106 (7.854 x 10-5) = 4.398 x 103 N ∴ PC = P = 4.398 x 103 N ii) Cross sectional area of the steel bar PC = PS = 4.398 x 103 N AS = PS σS = 4.398 x 103 140 x 106 = 3.14 x 10-5 m2 iii) Total elongation of the composite bar due to action of the load ∆L = ∆LS + ∆LC = � σL E � S + � σL E � C = 140 x 106 (2) 200 x 109 + 56 x 106 (1) 100 x 109 = 1.96 x 10-3 m
27 2.2.5 Example 2.3 A steel rod with diameter of 40 mm and 150 mm length was placed into the copper tube with outer diameter of 60 mm and inner diameter of 40 mm. The length of copper tube is the same as steel rod. This composite bar was subjected to compression force 200 kN. Calculate: i) The stress occurred in steel rod. ii) The stress occurred in copper rod. iii) Change of length for the composite bar. Given Esteel = 207 GN/m2; Ecopper = 107 GN/m2 Solution Given DS = 40 mm; DC = 60 mm; dC = 40 mm; ES = 207 GN/m2; EC = 107 GN/m2 i) The stress occurred in steel rod. AS = π 4 DS 2 = π 4 (0.04)2 = 1.257 x 10-3 m2 AC = π 4 (DC 2 - dC 2 ) = π 4 (0.062 - 0.042 ) = 1.571 x 10-3 m2 PS + PC = 200 σSAS + σCAC = 200 σS�1.257 x 10-3� + σC�1.571 x 10-3�= 200k − − − − − − − − − (1) ∆LS = ∆LC → � σ E � S = � σ E � C σS 207 x 109 = σC 107 x 109 σC = 0.517 σS − − − − − − − − − (2) By substituting equation (2) into equation (1) yields, σS�1.257 x 10-3�+ 0.517σS�1.571 x 10-3� = 200k 2.069 x 10-3 σS = 200k σS = 96.66 x 106 N/m2 (compressive) ii) The stress occurred in copper rod. σC = 0.517(96.66 x 106 ) = 49.97 x 106 N/m2 (compressive) We can also solve the above problem using the formula, σS = PES ASES+ ACEC = 200 x 103 ( 207 x 109 ) 1.257 x 10-3�207 x 109 �+1.571 x 10-3(107 x 109 ) = 96.66 x 106 N/m2 (compressive)
28 σC = PEC ASES+ ACEC = 200 x 103 ( 107 x 109 ) 1.257 x 10-3�207 x 109 �+1.571 x 10-3(107 x 109 ) = 49.97 x 106 N/m2 (compressive) iii) Change of length for the composite bar ∆L = ∆LS = � σL E � S = 96.66 x 106 (0.15) 207 x 109 = 7.005 x 10-5 m 2.2.6 Example 2.4 A composite bar has a radius of 50 mm constructed by inserting a brass tube to a solid steel rod with a diameter of 60 mm. If the composite bar subjected with a compressive force of 200 kN, determine the forces and stresses occur. Given, ES = 210 GN/m2 and EB = 100 GN/m2 Solution Given DS = 60 mm; DB = 100 mm; dB = 60 mm; ES = 210 GN/m2; EB = 100 GN/m2 The force and stress occur in steel and brass AS = π 4 DS 2 = π 4 (0.06)2 = 2.83 x 10-3 m2 AB = π 4 (DB 2 - dB 2 ) = π 4 (0.12 - 0.062 ) = 5.03 x 10-3 m2 PS + PB = 200k − − − − − − − − − −(1) ∆LS = ∆LB → � P AE� S = � P AE� B whereas LS=LB PS 2.83 x 10-3(210 x 109 ) = PB 5.03 x 10-3(107 x 109 ) PS = 1.181 PB − − − − − − − − − −(2) By substituting equation (2) into equation (1) yields, 1.181PB + PB = 200k 2.181PB = 200k PB = 91.68 x 103 N (compressive) And PS = 0.517(91.68 x 103 ) = 108.27 x 103 N (compressive)
29 We can also solve the above problem using the formula, PS = PASES ASES+ ABEB = 200 x 103 (2.83 x 10-3)( 207 x 109 ) 2.83 x 10-3�207 x 109 �+5.03 x 10-3(107 x 109 ) = 108.27 x 103 N (compressive) PB = PABES ASES+ ABEB = 200 x 103 (5.03 x 10-3)( 207 x 109 ) 2.83 x 10-3�207 x 109 �+5.03 x 10-3(107 x 109 ) = 91.68 x 103 N (compressive) The stresses occur in the composite bars, σS = PS AS = 108.27 x 103 1.257 x 10-3 = 38.26 x 106 N/m2 (compressive) σC = PC AC = 91.68 x 103 5.03 x 10-3 = 18.23 x 106 N/m2 (compressive) 2.3 THERMAL STRESS • When the temperature in material is raised or lowered, there will be expansion or contraction of the material. Any restrictions on the reaction of temperature of the materials, its will result thermal stress on the material. • The most obvious example is internal combustion engine in which components such as connecting rod or piston that generate high temperatures which produce thermal stress that must be considered when designing the engine. 2.4 TEMPERATURE COEFFICIENT OF LINEAR EXPANSION • Define as changes in the length of material for a unit of length when the temperature changes by 1o. Unit for linear expansion coefficient, α is per oC (C-1) or per oK (K-1) ∆L = αL∆t ∆L = amount of expansion ∆t = changes in temperature (increase or decrease) L = length of bar
30 • The following is linear expansion coefficient for such materials: αsteel = 11.5 x 10-6/ oC to 13.0 x 10-6/ oC αaluminum = 23.0 x 10-6/ oC to 24.0 x 10-6/ oC αcopper = 17.0 x 10-6/ oC to 18.0 x 10-6/ oC αcast iron = 11.0 x 10-6/ oC to 12.0 x 10-6/ oC 2.5 THERMAL STRESS ON COMPOSITE BAR a) Series Composite Bar b) Parallel Composite Bar i) P1 = P2 ii) � PL AE� 1 + � PL AE� 2 = (α1L1+ α2L2)∆t i) P1 = P2 ii) � σ E � 1 + � σ E � 2 = (α2 - α1)∆t
31 2.5.1 Example 2.5 Figure shows a series of composite bar mounted on a rigid wall at both ends. Calculate the stress that occurs in every bar if the temperature is raised by 50 °C. Material Young Modulus Linear Coefficient of Expansion Area Steel ES = 200 GN/m2 αS = 12 x 10-6/ oC AS = 150 mm2 Aluminum EA = 69 GN/m2 αA = 23 x 10-6/ oC AA = 300 mm2 Solution Equation of a series composite bar under a temperature change � σL E � S + � σL E � A = (αSLS+αALA)∆t � σS(0.1) 200 x 109� + � σA(0.2) 69 x 109� = �12 x 10-6 (0.2) + 23 x 10-6(0.1)�(50) � σS(0.1) 200 x 109� + � σA(0.2) 69 x 109� = 2.35 x 10-4 -------------------------- (1) Internal load for both bar is same, Ps = PA σSAS = σAAA σS = 300 150 σA = 2σA --------------------------- (2) Substitute equation (2) into equation (1) � 2σA(0.1) 200 x 109� + � σA(0.2) 69 x 109� = 2.35 x 10-4 3.898 x 10-12σA = 2.35 x 10-4 σA = 60.28 x 106 N/m2 and σS = 2(60.28 x 106 ) = 120.56 x 106 N/m2
32 This problem can be solved by using the following formula σS = (αsAS+αALA)∆t AS � LS ASES + LA AAEA � = �12 x 10-6 (0.2) + 23 x 10-6(0.1)�(50) 1.5 x 10-4 � 0.1 1.5 x 10-4(200 x 109 ) + 0.2 3.0 x 10-4(69 x 1009� = 2.35 x 10-4 1.5 x 10-4(1.2995 x 10-8) = 120.56 x 106 N/m2 σA = (αsAS+αALA)∆t AA � LS ASES + LA AAEA � = �12 x 10-6 (0.2) + 23 x 10-6(0.1)�(50) 3.0 x 10-4 � 0.1 1.5 x 10-4(200 x 109 ) + 0.2 3.0 x 10-4(69 x 1009� = 2.35 x 10-4 3.0 x 10-4(1.2995 x 10-8) = 60.28 x 106 N/m2 2.5.2 Example 2.6 A serial composite bar consisting of aluminum and copper as shown in figure below is expand by 0.5 mm when heated. Calculate: i) Stress for aluminum and copper ii) Change of temperature for this composite bar Given: EA = 69GN/m2; αA = 23 x 10-6/ oC EC = 107GN/m2; αC = 17.5 x 10-6/ oC 80 mm φ30 mm φ15 mm 60 mm A B C Aluminium Copper
33 Solution i) Stress for aluminum and copper AA = π 4 DA 2 = π 4 (0.03)2 = 7.069 x 10-4 m2 AC = π 4 DC 2 = π 4 (0.15)2 = 1.767 x 10-4 m2 Equation of a series composite bar under a temperature change � σL E � C + � σL E � A = (αCLC+αALA)∆t � σC(0.06) 107 x 109� + � σA(0.08) 69 x 109� = 0.5 x 10-3 -------------------------------- (1) Internal load for both bar is same, PC = PA σCAC = σAAA σC = 7.069 x 10-4 1.767 x 10-4 σA= 4.0σA ---------------------------------- (2) Substitute equation (2) into equation (1) � 4.0σA(0.06) 107 x 109 � + � σA(0.08) 69 x 109� = 0.5 x 10-3 3.402 x 10-12σA = 0.5 x 10-3 σA = 146.97 x 106 N/m2 and σC = 4.0(146.97 x 106 ) = 587.89 x 106 N/m2 Using the formula σC = (αCAC+αALA)∆t AC � LC ACEC + LA AAEA � = 0.5 x 10-3 1.767 x 10-4 � 0.06 1.767 x 10-4(107 x 109 ) + 0.08 7.069 x 10-4(69 x 1009� = 0.5 x 10-3 1.767 x 10-4(4.814 x 10-9) = 587.8 x 106 N/m2
34 σA = (αCAC+αALA)∆t AA � LC ACEC + LA AAEA � = 0.5 x 10-3 7.069 x 10-4 � 0.06 1.767 x 10-4(107 x 109 ) + 0.08 7.069 x 10-4(69 x 1009� = 0.5 x 10-3 7.069 x 10-4(4.814 x 10-9) = 146.93 x 106 N/m2 ii) Change of temperature for this compound bar ∆L = (αCLC+αALA)∆t 0.5 x 10-3 = �23 x 10-6 (0.08) + 17.5 x 10-6(0.06)�(∆t) 0.5 x 10-3 = 2.89 x 10-6 ∆t ∆t = 173 oC 2.5.3 Example 2.7 Refer to figure below, an alloy tube outer diameter of 40 mm and a thickness of 5 mm. Inside of alloy tubes containing a copper bar. Both of the materials are rigidly tied at the end so that when subjected to temperature, it will always be horizontal. Calculate the stress of copper and alloy when the change of temperature is 100oC. Take Ealloy = 180GN/m2 αalloy = 11 x 10-6/ oC Ecopper = 100GN/m2 αcopper = 18 x 10-6/ oC Solution Given: DA = 40 mm; dA = 30 mm; DC = 30 mm; ∆t = 100oC; EA = 180GN/m2; EC = 100 GN/m2; αA = 11 x 10-6/ oC; αC = 18 x 10-6/ oC The stress of copper and alloy Alloy tube Copper bar
35 AC = π 4 DC 2 = π 4 (0.03)2 = 7.069 × 10-4 m2 AA = π 4 (DA 2 − dA 2 ) = π 4 (0.042 − 0.032 ) = 5.498 × 10-4 m2 Equation of a parallel composite bar under a temperature change � σ E � C + � σ E � A = (αC - αA)∆t � σC 100 × 109� + � σA 180 × 109� = [(18 - 11) × 10-6](100) � σC 100 × 109� + � σA 180 × 109� = 7 × 10-4------------------------------- (1) Internal load for both bar is same, PC = PA σCAC = σAAA σC= 5.498 × 10-4 7.069 × 10-4 σA = 0.778σA ---------------------------- (2) Substitute equation (2) into equation (1) � 0.778σA 100 × 109� + � σA 180 × 109� = 7 × 10-4 1.335 × 10-11σA = 7 × 10-4 σA = 52.49 × 106 N/m2 and σC = 0.778(52.49 × 106 ) = 40.84 × 106 N/m2 This problem can be solved by using the formula σC = (αC − αA)∆t AC � 1 ACEC + 1 AAEA � = [(18 – 11) × 10-6](100) 7.069 × 10-4 � 1 7.069 × 10-4(100 × 109 ) + 1 5.498 × 10-4(180 × 1009� = 7 × 10-4 7.069 × 10-4(2.425 × 10-8) = 40.83 × 106 N/m2 σA = (αC − αA)∆t AA � 1 ACEC + 1 AAEA �
36 = [(18 - 11) × 10-6](100) 5.498 × 10-4 � 1 7.069 × 10-4(100 × 109 ) + 1 5.498 × 10-4(180 × 1009� = 7 × 10-4 5.498 × 10-4(2.425 × 10-8) = 52.49 × 106 N/m2 2.5.4 Example 2.8 The steel tube external diameter of 2.4 cm and internal diameter of 1.8 cm is surrounding the copper rod of 1.5 cm and mounted rigidly at both ends. If there is no stress at 10oC, calculate the stresses in the rod and tube if the temperature was increased by 200 oC. Given: ES = 200GN/m2 αS = 12 x 10-6/ oC EC = 107GN/m2 αC = 17.5 x 10-6/ oC Solution Given: DS = 2.4 cm; dS = 1.8 cm; DC = 1.5 cm; ∆t0 = 10 oC; ∆t1 = 200 oC ES = 210 GN/m2; αS = 12 x 10-6/ oC; EC = 107 MN/m2; αC = 17.5 x 10-6/ oC The stress of copper and steel AC = π 4 DC 2 = π 4 (0.015)2 = 1.767 × 10−4 m2 AS = π 4 (DS 2 − dS 2 ) = π 4 (0.0242 − 0.0182 ) = 1.979 × 10−4 m2 Equation of a parallel composite bar under a temperature change � σ E � S + � σ E � C = [αC - αS](∆t1 − ∆t0) � σS 210 × 109� + � σC 107 × 109� = [17.5 × 10-6 - 12 × 10-6](100 − 10) � σS 210 × 109� + � σC 107 × 109� = 1.045 × 10-3------------------------------- (1) Internal load for both bar is same, PS = PC σSAS = σCAC
37 σS= 1.767 × 10-4 1.979 × 10-4 σC = 0.893 σC ------------------------------ (2) Substitute equation (2) into equation (1) � 0.893σC 210 × 109� + � σC 107 × 109� = 1.045 × 10-3 1.36 × 10-11σC = 1.045 × 10-3 σC = 76.85 × 106 N/m2 and σS = 0.893(76.85 × 106 ) = 68.63 × 106 N/m2 The problem also can be solved by using the formula σS = [αC - αS](∆t1 − ∆t0) AS � 1 ASES + 1 ACEC � = [17.5 × 10-6 - 12 × 10-6](100 − 10) 1.979 × 10-4 � 1 1.979 × 10-4(100 × 109 ) + 1 1.767 × 10-4(180 × 1009� = 1.045 × 10-3 1.523 × 10-11 = 68.62 × 106 N/m2 σC = [αC - αS](∆t1 − ∆t0) AC � 1 ASES + 1 ACEC � = [17.5 × 10-6 - 12 × 10-6](100 − 10) 1.767 × 10−4 � 1 1.979 × 10-4(100 × 109 ) + 1 1.767 × 10-4(180 × 1009� = 1.045 × 10-3 1.360 × 10-11 = 76.85 × 106 N/m2
38 2.6 EXERCISES 1. A composite bar made of steel bar with a square cross-section and copper rod with diameter of 10 mm. The length of steel bar is 2 m connected in series with the copper rod which is 1 m in length. The end of the copper rod welded to a steel beam and the end of steel bar is left hanging free. Determine: 4.4kN, i) Maximum load that can be hung on the free end ii) Cross sectional area of the steel bar iii) Total elongation of the composite bar due to action of the load Take Ecopper = 100kN/mm2 ; σcopper = 56N/mm2 Esteel = 200kN/mm2 ; σsteel = 140N/mm2 2. Figure 1 shows a composite bar consisting of 3 parts, ABCD that subjected to tensile force P, AB and CD consists of steel having the same length of 100 cm and diameter of 15 mm. While BC is aluminium bar which has a length of 200 cm and a diameter of 10 mm. When a tensile force P is applied, the BC experience elongation of 0.85 mm. Determine: (98 mm) i) Force P ii) Total elongation of the composite bar Given Esteel = 200GPa and AAluminium = 80 GPa (Ans: 2.67kN, 1 mm) 3. a) Define the following; i) composite bar ii) temperature coefficient of linear expansion b) A steel rod with diameter of 40 mm and 150 mm length was placed into the copper tube with outer diameter of 60 mm and inner diameter of 40 mm. The length of copper tube is the same as steel rod. This composite bar was subjected to compression force 200 kN. Calculate: i) The stress occurred in steel rod. ii) The stress occurred in copper rod. iii) Change of length for the composite bar. Given Esteel = 207 GN/m2; Ecopper = 107 GN/m2 (Ans: 96.7MN/m2, 50MN/m2, 0.07mm) 100 cm 200 cm 100 cm A B C D P Figure 1
39 4. A compound parallel bar made of two bars that is steel and copper. The crosssectional of area of steel is 600 mm2 and copper rod is 1000 mm2. Both ends have rigidly mounted each other. Calculate the stress in both bars if 60 kN load is applied. Given ESteel = 206 GN/m2 and ECopper = 107 GN/m2. (Ans: 53.59MN/m2, 27.84MN/m2) 5. A series composite bar consists of a copper bar with a diameter of 30 mm and a length of 0.25 m and steel bars with a diameter of 25 mm and length of 0.5 m. If both ends are rigid connected and subjected to a tensile force of 15 kN, determine: a) Stresses in copper and steel bars b) Elongation of the two bars Given, ES= 200 GN/m2 ; EC = 100 GN/m2 (Ans: 30.55MN/m2, 21.22MN/m2, 0.13mm) 6. A compound bar consisting of 1 m long of copper rod with a diameter of 50 mm is mounted into a steel tube of 50 mm with internal diameter of 50 mm and external diameter of 60 mm. The bars are subjected to a tensile force of 150 kN. Calculate: a) The stress in steel tube b) The stress in copper rod Given ES = 200 GN/m2 and EC = 100 GN/m2. (Ans: 81.34MN/m2, 40.67MN/m2) 7. Figure 2 shows two rod A and B, have the same diameter, and carry a load of 500 N through a rigid length of the horizontal bar. If the stress in the rod A is 1.76 MN/m2 , calculate: a) Stress in the bar B b) The diameter of the rod Given, EA = 200 GN/m2; EB = 90 GN/m2 (Ans: 0.79MN/m2, 15.8mm) 8. The temperature of a 0.5 m long rod was raised from 30oC to 80oC. Calculate the extension of this rod (Ans: 0.3 mm) 500 N Rod A Rod B
40 9. A composite bar is rigidly mounted on both of each end. Calculate the stresses occurs in every bar if the temperature is lowered to 50oC. Given: ES = 200 GN/m2; αS = 12 x 10-6/ oC; AS = 100 mm2 EA = 69 GN/m2; αA = 23 x 10-6/ oC; AA = 200 mm2 (Ans: 139MN/m2, 69.5MN/m2) 10. A steel rod as shown in Figure below is subjected to tensile force of 18 kN. Calculate the change of temperature that is required to reduce the tensile force to 10 kN. Given ES = 200GN/m2 and αS = 12 x 10-6/o C. (Ans: 2.97oC) 11. The specimens set in Figure A and Figure B are subjected to the same temperature that raised at 80oC. Given: ES = 200 GPa; αS = 11.7 x 10-6/ oC; AS = 1000 mm2; LS = 4200 mm EC = 120 GPa; αC = 17.0 x 10-6/ oC; AC = 900 mm2; LC = 4200 mm Calculate: a) The stress in bar of Figure A b) The stresses in both bars of Figure B (Ans: 187.2MN/m2, 29.74MN/m2, 33.04MN/m2) Steel Figure A Steel Copper Figure B Steel Aluminum 150 mm 100 mm φ50 mm φ25 mm 18 kN 18 kN 450 mm 150 mm
41 12. A composite bar consisting of a series of aluminium and steel bar placed between two rigid walls at a temperature of 60o C as shown in Figure below. If the temperature is lowered to 40o C, calculate: i) Stresses in both bar if the walls in not elastic. ii) Stresses in both bar if the distance between the two walls reduced by 0.1 mm Given ES = 21MN/cm2; αS = 11.7 x 10-6/ oC; AS = 2 cm2 EA = 7MN/cm2 ; αA = 23.4 x 10-6/ oC ; AA = 3 cm2 (Ans: 49.1MN/m2, 32.8MN/m2, 31.6MN/m2, 21.1MN/m2) 13. A hollow mild steel has an outer diameter of 30 mm, thickness of 5 mm and 1.5m length. A solid of copper rod is fixed inside the mild steel. The composite bar is rigidly fixed at both ends. Calculate the stress developed in each bar when the temperature is increased to 70oC. Given: ES = 206 GN/m2; αS = 12 x 10-6/ oC EC = 109 GN/m2; αA = 18.5 x 10-6/ oC (Ans: 27.88MN/m2, 34.84MN/m2) 14. A series composite bar consists of 900 mm2 copper rod and 600 mm2 steel rod bounded rigidly with initial temperature of 10oC. Calculate the stress in each rod when temperature raised to 22oC. If the stress in steel rod is increased by 35%, determine the new stresses in each bar and the new temperature change. Assuming each rod is 1m length. Given: ES = 200 GPa; αS = 12 x 10-6/ oC EC = 100 GPa; αA = 17 x 10-6/ oC (Ans: 19.89MN/m2, 29.83MN/m2, 40.27MN/m2, 26.85MN/m2, 16.2oC) 60 cm 30 cm Steel Aluminum
42 15. A composite bar ABC which connected in series is fixed at both ends. AB is made of steel and BC is made of aluminium. If the temperature is increase by 60 ˚C, calculate the stress in each part. Given: Configuration Steel Aluminium Modulus Young 210 GN/m2 80 GN/m2 Cross Sectional Area (A) 200 mm2 50 mm2 Coeficient of Linear Expansion (oC) 12 x 10-6/˚C 18 x 10-6/˚C Length (L) 100 mm 50 mm (Ans: 42.3MN/m2, 169.4MN/m2) 16. Three bars are used to support a load of 100 kN as in Figure below. Each bar has the diameter of 20 mm and 1 m length. If the temperature is increased by 100oC from the initial, calculate the stress occurred due to the force and temperature change in each individual bars. Assume the bar is remains in horizontal. Given: ES = 206 GN/m2; αS = 12 x 10-6/ oC EA = 70 GN/m2; αA = 23 x 10-6/ oC (Ans: 169.04MN/m2, -19.58MN/m2) 100 kN Steel Aluminium Steel
43 17. A composite parallel bar is made of two bars that is steel and copper with an initial temperature of 30oC as shown in Figure below. The steel cross-sectional of area is 700 mm2 and 1200 mm2 for copper. The bar is rigidly fixed at both ends. Calculate the stress in each bar when the temperature is risen to 80oC. Given: ESteel = 206 GN/m2; αSteel = 12 x 10-6/ oC ECopper = 107 GN/m2; αCopper = 17.5 x 10-6/ oC (Ans: 26.69MN/m2, 15.56MN/m2) 18. A copper tube has an internal diameter of 27 mm, thickness of 5 mm and length of 0.9 m. Inside the copper tube is a steel bar with a diameter of 15 mm. The bar is rigidly fixed at both ends. Given: ES = 210 GN/m2; αS = 12 x 10-6/ oC EC = 109 GN/m2; αC = 18.5 x 10-6/ oC Calculate: a) The stress in both bar if a compressive force is subjected until the composite bars has experience elongation of 0.35 mm. b) The stress developed in both bar when temperature is increased to 65oC. (Ans: 81.67MN/m2, 42.39MN/m2, 52.95MN/m2 , 18.53MN/ Steel Copper