Writing Assignment Answer All Questions On Engineering - 1650 Words

Writing Assignment Answer All Questions On Engineering (Math Problem Sample) Content: NameInstitutionTutorDateAssignment 1TASK 1 1 Figure 1 represents a light beam ABCD of length 10m which is simply supported at the points A and C. The beam carries a uniformly distributed load of 30kN/m between A and B and point loads of 100kN and 70kN at points B D, as shown.Show that the reactions at A C are 120kN and 230kN respectively8Rav +270 = 1805+ 10028Rav = 900+200-1408Rav =960Rav is therefore 120 kN8Rc =1803 + (1006) + (7010)8Rc = 1840Rc = 230 Kn 29199273297002919937855300259720316307825892631707623088597155394SFD(KN) 6-X117565753191299676968673380974455118902083265700187441316896900 X1766725171914002604887433800BMD(KN)3991702675910039957129159300 (-)(+)X/120 = (6-X)/160X= 18Max shear= x 960 x 18=8.640KNTask 2 1 Suitable I-beam from tableThe suitable beam will be of 200.9 kg/m 2 v1 = 120v2 = 120-180 = -60v3 = -60-100 =-160v4 = -160= 230 = 70therefore max mom ent = 120KNM at point Ai = 3 x 160/(2 ) = 240/1002.4MPaWhich is more than there it will fail.3. The standard beams such as the I-beams resists shear forces, while the flanges resist most of the bending moment experienced by the beam. Beam theory shows that the I-shaped section is a very efficient form for carrying both bending and shear loads in the plane of the web. On the other hand, the cross-section has a reduced capacity in the transverse direction, and is also inefficient in carrying torsion, for which hollow structural sections are often preferred. Therefore, the design and structure of the I beam makes it uniquely capable of handling a variety of loads.Task 3 1 A hollow circular shaft has an external diameter of 80mm and an internal diameter of 50mm. The shaft is designed to transmit 450kW of power at 1200rpm.Taking G = 75GPa, answer the following questions: * Find the maximum and minimum shear stress acting on the shaft.I=32D4-d4external diameter = 0.08 minternal diameter = 0.05 m= 36 (0.084-x0.054) =363.47110-5=3.4110-6Nxm4TorqueT = p62N = 45010006021200=3579.55 NmShear maximum = TRJ = 3575.550.043.4110-6= 4.1954107 Pa change to mPa= 4.1954101 mPaMinimum shear=3575.550.0253.4110-6=26.21 mPa(b) angle of twist= á ¼ ° LGR =41.951.6106751090.04 =0.022 radiansConverting to degrees = 1800.0223 = 1.28Angle of twist for minimum=26.211061.6751090.025 = 0.0223 radianConverting to degrees = 180 0.0223=1.28Hence the angles are the sameangle of twist = 1.28.(c) Find the diameter of a solid circular shaft that would perform the same operation as above. * Polar momentTorque =3575.55Power =450 KWRPm =1200G =75 GpaL =1.6 MJ = D432 hence3.4110-6 = 32D4D =0.07676 m =0.077 m=77 mmTask 3Question two2) A hollow steel shaft with a diameter ratio of 0.8 and a length of 2.5 m is required to transmit 1250 kW at 200rpm. The maximum shear stress i s not to exceed 60 MPa. Determine the following: 1 The diameters of the shaftDiameter ratio = 0.8P = 1.25 x 106 wattsPolar moments =J = pi/32(D4 -d4)Pi/32 (D4 0.8D4)Pi/32 x 0.5904= 0.05799D4Torsional loads= (p x 60)/2 x pi x N(1.250 X 106 X 60)/( 2 X 3.142 X 200)=59659.091 NmTJ = á ¼ ° =59659.091Nm0.05799D2 = 260106D=2601060.05799D459659.09D Nm =116.64D4DD3 =1116.64D3 =8.573210-3D =∛8.573210-3 = 0.2047 metersLarger diameter= 204.7 mmBut d= 0.8D4 = (0.8204.7) mmThe smaller diameter = 163.73 mm 2 The maximum angle of twist=TLJ = 59659.0912.5GJGiven that J=0.057990.20474G=75109=1.017510-4 = 0.01954 radianConverting to degrees = 0.01954180=1.12(c) Hollow shafts are much better to take torsional loads compared to solid shafts. shear stress in a shaft subjected to torsion varies linearly from zero at the center to the maximum at the boundary. Inside a solid shaft, most of the material carries a shear stress whose value is much below the maximum shear stress which is the Interior portion of the shaft. But at the same they are adding to the weight, without contributing much to the capability of the shaft to carry torsional load. Similarly, a Hollow shaft has a greater Strength to weight ratio.Assignment 2Task 1. 1 An Airbus A 380-800, with the specification given in table 1, shown below, sits on a runway ready for take-off. Assuming it has undercarriage wheels of diameter 1.4m, is operating at maximum take-off weight (MTOW) and accelerates uniformly from rest to a take-off speed of 300 km/h in the distance given in table. Determine: * The take-off speed in ms-1300 x 1000 =83.33 m/s3600 * The linear acceleration of the aircraft from rest to take-off speed311 x 4 x 1000560 000=2.221m/s2 * The time taken to reach take-off speedV= u + 2at83.33 = 2 x 2.221x tt= 18.76 * The angular velocity of the undercarriage wheels at take-off speed=V/r= 83.33/ 0.7=119.04 rad/s * The angular acceleration of the undercarriage w heelsAngular velocity/ time=119.04/18.76=6.346rad/s2 * The number of revolutions made by each wheel during the take-off run( angular accelaration)time= x 119.04 x 18.76=1116.59.52radsTherefore revolution per wheel=1116.5952/(2 x 3.142)=177.64 revs * The force required to produce the linear accelerationT= Mr2=560000 x 2.23 x 0.7= 874.160NTask 2 1 In a manufacturing plant as part of a certain process, a cylindrical drum of diameter 0.7 m and mass 15kg is rolled down an incline from rest. The radius of gyration of each drum is 0.3m. If the drum rolls down the slope without slipping and descends a height of 0.5m, calculate the following:a.The moment of inertia, I, of the drumI = Mk2 where M is mass and K is the radius of gyrationI = 15 x 0.32=1.35kg m2b. The angular velocity of the drumW2 = (2 x 9.81 x 0.5)/ (0.352 x 0.32)W = 29.83 rad/sc. The linear velocity of the drumv = wr= 29.83 x 0.35=10.44m/s * The potential energy lost by the drumP.E = mgz=15 x 9.81 x 5= 735.75 J * The kinetic energy gained by the drumK.E = mv2/215 x 10.442/2= 817.45J 2 Derive the equation for = MGZ = MV2/2 + Iw2/2W2 = 2gz/r2 + k2Task 3A piston performs reciprocal motion in a straight line which can be modeled as s.h.m. Its velocity is 8 ms-1 when the displacement is 90mm from the mid-position, and 2ms-1 when the displacement is 140mm from the mid-position.Determine: * The frequency and amplitude of the motion.Frequency = 8/0.09