Application of Engineering Principles - Essay Example

?AEP One Question One a. Hooke’s Law. Hooke’s law s that the extension of a spring is directly proportional to the load applied to it. This is true as long as the material of the spring remains loaded within elastic limits. b. Based on the results obtained from Tensile Test Experiment in the laboratory: i. Obtain stress-strain curve for the given specimen. ii. If the same specimen is heated to 35oC and tested, what are the possible variations that could occur? Heating the sample would cause the length of the specimen to increase and this in turn would produce greater deflection under loading. However, this increase in deflection is going to be minimal. iii. Discuss the possibilities of errors occurring in the tensile test. Errors may be introduced into the tensile test due to various reasons such as: incorrect measurement of length of the sample due to incorrect reading from vernier calliper due to: zero error; parallax error. errors in dial calliper; zero error during initial calibration; parallax error while reading dial calliper. c. The figure shows a bar consisting of four lengths. i. Find the load carried by the bar at B, if the total extension is 0.063 mm. ii. Find the total extension of the bar, if loads applied at both ends are increased by 10%. If the loads at both ends are increased by 10% then inserting the new values into (1) gives: Take E = 2 x 105 N/mm2 for both the cases. d. A load of 500kN is applied on a short concrete column 400mmx500mm. The column is reinforced with 3 steel bars of 20mm diameter and 3 bars of 30mm diameter. If the modulus of elasticity for steel is 20 times that for concrete, find the stresses in both concrete and steel bars. Areas for concrete and steel bars are: AS = 400 x 500 = 2 x 105 mm2 AC = 3?(10)2 + 3?(15)2 = 3063.45 mm2 Solving simultaneously we get: Now: e. Discuss the load shared by composite cylinder will be more than the conventional cylinder. A composite cylinder can bear greater load than a conventional cylinder. This is due to the presence of materials that have greater tensile strength than the majority composing material. For example, in the case of steel reinforced steel bars, the steel has a greater tensile strength and being stronger can bear larger loads than what concrete could otherwise bear. Moreover, the steel in question tends to deform as much as the concrete which helps it to bear greater loads too. Another reason for composite cylinders being better at load bearing is because the constituent materials are distinct and are trying to move past each other within the overall structure. The presence of friction between the composing materials of a cylinder enable it to take larger loads too because the friction would need to be overtaken if the cylinder were to disintegrate. Question Two a. Based on the results obtained from Shear Force Measurement Experiment in the laboratory: i. Name the type of beam used. ????? ii. Critically analyse the variations of theoretical shear force and experimental shear force at the sensor for the given load condition in the laboratory. ????? iii. Usually measuring instruments absorbs some input energy to respond. Discuss how this affects the measurement. Measuring instruments absorb some of the energy that is produced in an experimental or other situation. The instruments need energy to generally overcome their own internal inertias. The absorption of energy tends to lower the amount of force actually available for measurement. However, in most situations the instruments are kept simple and light enough to ensure that the loss of energy is not too high to affect the actual measurement process. iv. Analyse the indication of a sudden change in shear force from negative to positive value. Positive shear tends to produce positive bending with geometrical minimas becoming apparent. On the other hand, negative shear tends to produce negative bending with geometrical maximas becoming apparent. As soon as the shear changes from negative to positive, the shape of the sag in the object changes as mentioned above indicating a change in shear. b. i. Define moment of inertia and discuss the moment of inertia as a function of the mass and radius of gyration. The moment of inertia for a rotating body is analogical to linear inertia for a linearly moving body. The inertia that a body experiences during rotation or simple a body’s resistance to rotation or changes in its rotation is better known as moment of inertia. Moment of inertia is generally computed around the axis of rotation of the concerned body. The radius of gyration (also gyradius) is the root mean square distance between an object’s parts from either the body’s centre of gravity or from an axis. The radius of gyration rg about any given axis for a body can be calculated in terms of the mass moment of inertia I as well as the total mass m as; ii. Discuss why the knowledge of moment of inertia is important in weight lifting machines. Weight lifting machines employ masses that are supposed to rotate around standard axis (both horizontal and vertical) as well as masses that are lifted. Determination of the moment of inertia helps to establish how much a human being would have to exert in order to rotate the mass such as at a butterfly machine (rotating weights using your arms). The moment of inertia is required merely for weight lifting machines that employ rotational techniques. If the moment of inertia is not determined in these machines, then there is little chance of developing ergonomically compliant machines. c. A timber beam with different loads is simply supported on a span of 20m. The beam carries both uniformly distributed loads and concentrated loads. The cross section of this timber beam is an I-section. i. Construct the SF and BM diagrams. Inclines are not noticeable as the loading gradients are too small so a representative table is attached below. Note that RA is assumed on the left side of the diagram while RB is assumed on the right side of the diagram of the timber beam. Length Along Beam Shear Force Equation Range 1 11362.44 RA 0 < x < 5 2 11362.44 3 11362.44 4 11362.44 5 11362.44 5 3362.438 RA – 8000 – 0.0078x 5 < x < 8 6 3362.43 7 3362.422 8 3362.415 9 3356.207 RA – 8000 – (0.0078+6.2)x 8 < x < 10 10 3349.999 10 -3150 RA – 8000 – (0.0078+6.2)x – 6500 10 < x < 12 11 -3156.21 12 -3162.42 13 -3162.42 RA – 8000 – (0.0078)x - 6500 12 < x < 14 14 -3162.43 14 -10162.4 RA – 8000 – (0.0078)x – 6500-7000 14 < x < 17 15 -10162.4 -10162.4 16 -10162.4 17 -10162.4 18 -10162.4 -10162.4 – 0.005x 17 < x < 20 19 -10162.4 20 -10162.4 Length Along Beam Bending Moment Equation Range 1 11362.44 RAx 0 < x < 5 2 22724.88 3 34087.31 4 45449.75 5 56812.19 6 56812.05 RAx – 8000(x - 5) - 0.0039x2 5 < x < 8 7 56812 8 56811.94 9 56560.77 RAx – 8000(x - 5) - 0.0039x2 – 3.1 x2 8 < x < 10 10 56501.8 11 56126.23 RAx – 8000(x - 5) - 0.0039x2 – 3.1 x2 – 6500(x – 10) 10 < x < 12 12 56054.84 13 56054.18 RAx – 8000(x - 5) - 0.0039x2 – 3.1 x2 – 6500(x – 10) – 0.0039 x2 12 < x < 14 14 56054.07 15 49054.07 RAx – 8000(x - 5) - 0.0039x2 – 3.1 x2 – 6500(x – 10) – 0.0039 x2 – 7000(x – 14) 14 < x < 17 16 35054.07 17 14054.07 18 14053.26 RAx – 8000(x - 5) - 0.0039x2 – 3.1 x2 – 6500(x – 10) – 0.0039 x2 – 7000(x – 14) – 0.0025 x2 17 < x < 20 19 14053.17 20 14053.07 ii. Calculate the maximum bending stress. The maximum bending stress occurs where the maximum bending moment occurs which is 56812.19 Nm. The moment of inertia for a rectangle is given as: For the I-beam in the question, the total inertia is: Using c = 45mm: iii. Discuss how the load diagram and bending moment diagram will be for cubic function of shear force diagram and write your inferences. A cubic function for the shear diagram means that the load is not changing linearly but rather quadratically. The shear force diagram will be cubic of course while the bending moment diagram will possess a greatest exponent of 4 and hence it will look more or less like a skewed quadratic plot. d. Prove that axial stress, principal stress (combination of bending stress and shear stress) and thermal stress are involved in drill bit during drilling. When drilling is being carried out, the drill bit is subject to a combination of stresses. Thermal stresses are common as the drill bit comes in contact with the material being drilled. The coefficient of friction between both surfaces ensures that both surfaces heat up as they move past each other. Other than thermal stress, the drill bit also experiences axial stress as it is being driven along its length into the hole being drilled. The drill bit also experiences bending stress as the drill bit encounters new material along the flukes as well as the contact of the drill bit with the sides of the holes produces bending stresses. Moreover, the drill bit is also subject to shear stresses as it encounters the situations listed above. This combination is also better known as the principal stress. Question Three a. Discuss the reasons why factor of safety is provided in designing any engineering components and elaborate how it helps a system to work without any damage even if the limit exceeds slightly. The factor of safety is simply a safe design practice according to which any component being designed is slightly overdesigned especially in respect to the principal stresses. The contention behind a factor of safety is that the material’s ultimate tensile strength (or other equivalent failure definition) should not be breached. Most components experience slightly higher than expected stresses in situations that the object was not designed for such as an automobile that is being used on off road tracks but was designed for movement on roads only. The factor of safety thus ensures that the component or system is adequately designed to resist slightly greater than normal working limits. b. A hollow shaft is to transmit 250kW at 120rpm. If the shear stress is not to exceed 60MN/m2 and diameter ratio is 4/5. Design the shaft, if the twist is 3o and length is 3m. Assuming maximum torque is 25% greater than mean. Take rigidity modulus as 75GN/m2. Justify on what basis the shaft is designed to transmit power safely. Maximum torque is 25% greater than mean so maximum torque is 38481262.5 Nmm. Given that the shaft is not going to exceed 60MN/m2, it can be seen that the FOS becomes 116/60 = 1.93 which is justifiable for an application of this kind. c. Elaborate the various mechanical power transmitting elements to be used in Well Engineering operations (excluding shafts). Well Engineering operations employ multiple power transmission elements such as pulleys, gears, hydraulic couplings, gear couplings, rigid couplings and the like. Pulleys are designed to operate by transmitting power and reducing or increasing the torque and speed. Gears are similarly designed to change the operating characteristics of one machine with respect to another. Gears could be spur, helical, herringbone, worm type etc. Couplings are also widely employed on various machines used for drilling wells and numerous kinds are available based on their specific application. Read More