Mechanical technology

    • An understanding of the fundamental concepts, laws and analytic methods for the solution of applied mechanics problems Skills • Proficiency in the use of mechanical technology, and the ability to evaluate and appraise the results of practical experiments. 1 Abstract • A self-contained summary of what you did and your main findings and conclusions. • Maximum length: 200 words. You should include a Word count at the end of the abstract. 10 2 Results • Full and complete table of experimental readings. • Well-presented and fully labelled graphs. 10 10 3 Analysis of results Discuss briefly: Experiment 1 • Show the relationship between deflection and beam material. • Comparison of experimental and theoretical values of Young’s modulus (E) for brass, aluminium and mild steel Experiment 2 • Estimate the inertia of the flywheel by treating it as a solid steel disc. Full working should be shown. • Critically analyse the tabulated data and comment on your graph and results and compare the value with the textbook value for the all three beam materials. 10 4 Discussion • A critical discussion of the accuracy of the results, which includes a description of the main sources of error/inaccuracy. 5 5 Quality of presentation of the report The use of an appropriate technical style for the presentation of the report. 5 Total marks for each report 50 Page 4 of 13 Part A: MOMENT OF INERTIA OF A FLYWHEEL During the course of the experiment a set of weights were hung in turn on a piece of cord from a pulley attached to the flywheel. The objects were allowed to fall under gravity from a height of 1 metre, as described in the laboratory sheet. This was performed three times and the data recorded is shown in the table below. The diameter of the spindle d (= 2r) was found to be 38 mm (0.038 m). Total weight (including hanger) (N) 1 st descent time (s) 2 nd descent time (s) 3 rd descent time (s) 3.2 29.69 34.60 34.60 6.0 24.25 24.28 25.22 8.5 19.60 19.70 19.95 11.0 17.60 17.06 17.29 13.5 15.12 15.65 15.12 Complete the following tasks: • The flywheel dimensions are as follows: diameter = 390mm, mean thickness = 30mm. Also, it is made of steel, with density 7850 kg/m3 . Use this information to estimate the inertia, by considering the formula for the inertia of a solid disc. This value should be used as an approximate validation of the inertia derived. • Critically analyse the tabulated data and use it to determine the inertia of the flywheel and the frictional torque Tf. You should use Excel to calculate and tabulate relevant values, and to plot Pxr versus α, as per the laboratory sheet. NB: do not confuse weight with mass (a common mistake!) Part B: STIFFNESS OF DIFFERENT MATERIALS During the experiment, three beams of different materials (Aluminium, Brass and Steel), but of the same dimensions (9.5 mm x 3 mm), were tested to find their material stiffness values. All steps to perform the complete experiment are given in the laboratory instruction sheet. Tables 1 to 3 were obtained after completing the experiment. Complete the following tasks: • Complete tables 1 to 3 and obtain the values for equation 48δI / L3 . Page 5 of 13 • Plot your load and equation 48δI / L3 results on the chart paper (as shown in Figure 1). • Find the gradient of the chart. • Critically analyse the tabulated data and comment on your graph and results and compare the value with the textbook value for the all three beam materials. Beam Modulus Load (m) (g) Load (W) (N) Deflection (mm) Deflection (δ) (m) 48δI / L3 0 0 0 100 0.98 0.65 200 1.96 1.16 300 2.94 1.73 400 3.92 2.23 500 4.9 2.74 Material: Aluminium Beam Size: b= 9.5 mm d= 3 mm I = 2.1375 x 10-11 m4 Table 1: Results table for Aluminium. Beam Modulus Load (m) (g) Load (W) (N) Deflection (mm) Deflection (δ) (m) 48δI / L3 0 0 0 100 0.98 0.36 200 1.96 0.80 300 2.94 1.06 400 3.92 1.41 500 4.9 1.75 Material: Brass Beam Size: b= 9.5 mm d= 3 mm I = 2.1375 x 10-11 m4 Table 2: Results table for Brass. Beam Modulus Page 6 of 13 Load (m) (g) Load (W) (N) Deflection (mm) Deflection (δ) (m) 48δI / L3 0 0 0 100 0.98 0.19 200 1.96 0.42 300 2.94 0.58 400 3.92 0.78 500 4.9 0.98 Material: Steel Beam Size: b= 9.5 mm d= 3 mm I = 2.1375 x 10-11 m4 Table 3: Results table for Steel. Figure 1: Blank results chart. Laboratory instructions Page 7 of 13 TITLE EAT103: MOMENT OF INERTIA OF A FLYWHEEL OBJECT To determine the moment of inertia of a flywheel by accelerating it with a falling mass. THEORY APPARATUS m d a P P mg ,  Page 8 of 13 Flywheel, variety of masses m attached to length of cord. m = falling mass, kg h = distance fallen by mass, m t = time of fall of mass, s d = diameter of spindle, m r = radius of spindle, m (= d/2) a = acceleration of mass, m/s2  = angular velocity of flywheel when mass reaches ground, rad/s  = angular acceleration of flywheel with mass falling, rad/s2 P = tension in cord, N g = gravitational acceleration = 9.81 m/s2 METHOD 1. For a particular mass, time the rate of descent and measure the distance fallen. 2. Use the formula given below in conjunction with Excel to determine the moment of inertia of the flywheel, by plotting the straight line described. 3. Repeat steps 1-3 for a number of different masses (try at least five values of m). 4. Estimate the inertia using the formula for the flywheel of a solid disc, to enable you to check that your experimental results are of the correct order of magnitude. Considering acceleration of the flywheel:  T I d T = I  P − f = 2 ------------------------------------- (1) Where, Tf = friction torque due to bearings etc. I = moment of inertia of flywheel and spindle Newton’s 2nd law applied to falling mass: mg − P = ma ------- (2) Page 9 of 13 From (s = ut + 0.5at2 ), acceleration: 2 2 t h a = ------------------------ (3) Thus: (3) into (2)  2 2 t mh P = mg − ---------------------------------- (4) From (v = u + at), and using the expression for acceleration a given by equation (3), the speed of mass m when it hits ground is: t h v 2 = ---------------------------- (5) Since  = v/r, from (5) the angular velocity  of the flywheel when the mass hits ground is: rt 2h  == --------------------------- (6) Also, the angular acceleration of the flywheel as mass falls is: t   = ---------------------------- (7) Thus, (6) into (7)  t r h 2 2  = --------------------------- (8) Equation (1) can be re-written as follows: y mx c P r I Tf  = +  =  + --------------------- (9) In other words, if we plot P r (y-axis) versus  (x-axis), we should get a straight line, where the gradient is I, the inertia of the flywheel, and the intercept with the y-axis is Tf, the friction torque. P is calculated using equation (4), and  is calculated using equation (8). TITLE Page 10 of 13 EAT103: Stiffness of materials OBJECT To achieve an understanding of the stiffness of different materials under load. THEORY Stiffness is an important property for many applications of materials. The stiffness of a structure depends on both the properties of the material from which it is made and on the geometry of the structure. Theory: Deflection of a simply supported beam. This formula applies to beams which are subjected to a central load acting at a right angle to their length. The equation for the deflection of a beam supported between two points is given by: Δ = 𝑊𝐿 3 48𝐾 Where: 𝑊 = Load (N) Δ = deflection (m) L = Length (m) K = Flexural rigidity constant K may be calculated due to the second moment of area of the material I, and the value of Young’s modulus E Page 11 of 13 𝐾=𝐸.𝐼 Some standard values of (E) for the following materials: Brass 105 Gpa Aluminium 69 GPa Mild Steel 200 GPa Second moment of area (I) for solid and hollow beams: For a solid beam of rectangular cross section: 𝐼 = 𝑏𝑑 3 12 The units of I are m4 APPARATUS The ES4 apparatus uses an example of a very simple structure, a beam in three point bending. The apparatus explores that beams of different material but identical dimensions (and I value) deflect by different amounts for the same load, providing that their material property (Young’s Modulus) affects deflection. The apparatus is very simple to use, the operator places the specimen into the apparatus and applies a load. A dial indicator measures the deflection of the specimen at the point of loading. The apparatus is simple and demonstrated in Figure 1. Page 12 of 13 Figure 1: Experiment Setup METHOD Page 13 of 13 Experimental Procedure – Beam Modulus 1. Find the three beams of different materials (Aluminium, Brass and Steel) but of the same dimensions (9.5 mm x 3 mm). The aluminium beam weight less than the other two and the brass beam has a golden colour. 2. For each beam, use the Dial Calliper Gauge (supplied with the kid) to accurately measure the beam dimensions (b and d). Calculate the I values for the beam and note this and the other beam properties into your results table 1-3. 3. Fit the aluminium beam to the supports. 4. Adjust the Dial indicator and Wire ‘stirrup’ along to the middle of the beam (180 mm). 5. Set and zero the Dial indicator. 6. In increments of 100 g, add weights up to a maximum of 500 g. Each time you add a weight. Tap the work panel to reduce the effects of friction. Record the deflection at each weight increment. 7. Calculate the load in Newton and calculate the results of the equation shown in the table 1. 8. Rearranging the equation in the form y = mx+c means that the gradient of a chart of W against 48 δI 𝐿 3 will give the value of E. 9. Plot your load and equation results on the chart paper. 10.Find the gradient of the chart. 11.Compare the value with the textbook value for the beam material. 12. Repeat for the other two beams.          
Franklin Delano Roosevelt, known to many as FDR, was not only a leader for his time, but a leader for all times. Not only did FDR successfully lead his nation through possibly the two most dramatic, and consequential world events of the 20th century, the Great Depression and WW2; he is also responsible for spearheading the creation of the United Nations, as well as the series of programs, projects, policies and laws that came to be known as the “New Deal”. Through his experimentation and willingness to listen and learn from others, FDR demonstrated that he was the man to lead America forward. FDR’s actions throughout his presidency shaped not only the way Americans lived their lives, but how the world perceived America for decades to come. Throughout his presidency, FDR was able to display his brilliance in leadership. However, what was the greatest achievement in each of his four terms? And perhaps most interesting of all, what was his greatest failure? For his brilliance in leadership, as well as a lack of competent competitors, the American people rewarded him with an unprecedented four terms in office. In his first term he successfully tiptoed around the edge of the precipice, leading the American people from the turmoil caused by the Great Depression. In his second term, perhaps his most undervalued achievement was the packing of the supreme court, for it was here where you can see the long lasting impact. His third term was dominated by World War Two and the successes enjoyed by the allies. In his fourth term, despite the brevity of it, he was able to establish the foundations of the modern United Nations. Truly a leader for all times. From the moment FDR won the governorship of New York State, to the moment he became president of the United States of America, you see an unparalleled level of political victories. In his first presidential campaign, the elites of the democratic party thought that they could control him. In the end, their underestimation of FDRs coolness proved costly, with him ultimately using it to his advantage. History tells us that the world was overall, a better place because of his actions, but what exactly did he do to transform America in specific, and

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