Strength of Materials Problem Written and Solved for MET406 - Lab Report Example

Torque Required for an Athlete Diving Reflex Strength of Materials Problem Written and Solved for MET406 By On 28 November Instructor Course Date Strength of Materials Problem Written and Solved for MET406 Problem Statement Divers have an unusual ability to orient themselves when diving to ensure that they reach the water when they are they are straight or the desired shape. This is a skill that needs much accuracy, and one has to practice to ensure it is performed correctly. How much torque is required to ensure that a diver performs a stand, which requires him to rotate once before touching the water, from a height of 42ft? (Mott 2008). Given Height, h = 42 Find Torque, t Assumptions 1. The diver’s mass is 140 lbs. m= 140 lbs. 2. The diver will be modeled as five rectangles and one circle as shown in figure two. The diver is 72” tall (head to toe) and 22”long (shoulder to shoulder). 3. The head is 8” wide, the arms 5” wide; the legs 7” wide, and the body 10” wide. 4. The angular acceleration is uniform, and it takes place on the -axis. 5. The density is also uniform. Method 1. The initial action will be to calculate the length of time it will take for the diver to reach the water given the distance above. 2. Once we get the time, we use it to determine the angular acceleration required for the diver to turn 5400 before touching the water. 3. The Moment of inertia and the Centre of area of the diver is determined using the different values of the triangles and circles together with the distance of their centre of area from the -axis and the -axis. MET406 Topic(s) The topic(s) from MET406 that are illustrated in this problem are the center of moment of area and the time of inertia. Reference the list of topics in the syllabus. Figure 1 figure 2 Solution Step One When calculating the time it will take the diver to reach the water, the second equation of motion will be used. The second equation of motion is given by χ = χ0 + ѵ0t + at2. = 0 + ѵ0 + at2 42 = 0 + (0) + (32.2) t2 42 = (16.1) t2 t2 = 2.61S2 = 1.62S Step Two In step, two the speed at which the diver will be moving is calculated first, and then it is used it to get the angular velocity. The speed is given by and the acceleration is given by. = = = 5.82 α = = = 7.2 Step Three The locations of all the six parts are obtained from the model in figure 1. Xi is estimated from the distance of the centre of area with reference to the distance -axis and the Yi is estimated from the centre of area with reference to the distance of the -axis. Ai is the area of the parts. Measurements Part Length (in) Width (in) Breath (in) Head 7 8 Body 25 10 14 Legs 40 7 6 Legs 40 7 6 Arms 28 5 4 Arms 28 5 4 Area Body = 25” × 14” = 350 in2 Head = 7” × 3.142 = 22 in2 Legs = 40” × 6” = 240” in2 Arms = 28” × 4” = 112 in2 Legs = 40” × 7” = 240” in2 Arms = 28” × 4” = 112 in2 Part Ai (in2) (in) (in) (in3) (in3) 1 22 11 68.5 242 1507 2 350 11 52.5 3850 18375 3 240 7 51 1680 12240 4 240 13 51 3640 12240 5 112 2 20 224 2240 6 112 20 20 2240 2240 Total 1146 11876 48842 The composite centre of area of the diver is X = = = 10.36 in3 Y = = = 42.61 in3 Using the Y-axis as a reference point X is the distance to the composite centre of area ad using the X-axis as a reference point Y is the distance to the composite centre of area. Part (in4) (in) Aifi2(in4) + Aifi2(in4) 1 117.87 0.64 9.01 126.88 2 18229.12 0.64 143.36 18372.48 3 32000 2.64 1672.70 33672.70 4 32000 3.36 2709.04 34709.04 5 7317.33 8.36 7827.35 15144.68 6 7317.33 9.64 10408.12 17725.45 Total 119751.23 To get the centre of area that will be used to calculate the torque, we consider the centre of area of the individual member and subtract it from the composite centre of area that has been derived from the calculation above. As shown below. = X - Head = 11 − 10.36 = 0.64 in Leg = 10.36 − 7 = 3.36 in Arm = 10.36 − 2 = 8.36 in Body = 11 − 10.36 = 0.64 in Leg = 13 − 10.36 = 2.64 in Arm = 20 − 10.36 = 9.64 in To calculate the moment of inertia for each of the members, we use the formula for moment of inertia. The formula for calculating the moment of inertia for rectangles and circles are different rectangles, as shown below. The formula for a rectangle is, while that of a circle is = π d4 / 64. = and = π d4 / 64  Head = π 74 / 64 = 117.87 in4 Leg = = 32000 in4 Arm = = 7317.33 in4 Body = = 18229.12 in4 Leg = = 32000 in4 Arm = = 7317.33 in4 Part (in4) (in) AiDi2(in4) + AiDi2 (in4) 1 117.87 25.89 14746.42 14864.29 2 5716.67 9.89 34234.23 39950.90 3 720 8.39 16894.10 17614.1 4 720 8.39 16894.10 17614.1 5 149.33 22.61 57255.75 57405.08 6 149.33 22.61 57255.75 57405.08 Total 204853.55 To get the centre of area used to calculate the torque, we consider the centre of area of the individual member and subtract it from the composite centre of area that has been derived from the calculation above. As shown below. Di= Y− Head = 68.5 − 42.61 = 25.89 in Leg = 51 − 42.61 = 8.39 in Arm = 42.61 − 20 = 22.61 in Body = 52.5 − 42.61 = 9.89 in Leg = 51 − 42.61 = 8.39 in Arm = 42.61 − 20 = 22.61 in To calculate the moment of inertia for each of the members, we use the formula for moment of inertia. The formula for calculating the moment of inertia for rectangles and circles are different rectangles, as shown below. The formula for a rectangle is, while that of a circle is = π d4 / 64  = and = π d4 / 64 Head = π 74 / 64 = 117.87 in4 Leg = = 720 in4 Arm = = 149.33 in4 Body = = 5716.67 in4 Leg = = 720 in4 Arm = = 149.33 in4 Di (Di= Y−) is the distance of each part from the composite centre of area and it is along the –axis. The overall moment of inertia is IXT = IX1 + A1f12 + IX2 + A2f22 + …………. = 119751.23 IYT = Iy1 + A1d12 + Iy2 + A2d22 + ……….. = 204853.55 Step Four This is the last step of the method, where we use the values derived from the other steps to calculate the torque. The torque of the diver can be obtained by multiplying the mass moment of inertia the diver with the density of the diver. The density to the diver is obtained by dividing the mass of the athlete with volume of the diver. The mass is assumed, and the volume can be obtained by multiplying the areas of the different member parts with their width then adding them up V = (22 × 8) + (350 × 10) + (240 × 7) + (240 × 7) + (112 × 5) + (112 × 5) = 8156 in3 Once the volume is obtained, it is then used to get the density as shown below D, density = = 0.0172 Once the density is obtained, it is used together with the mass moment of inertia and the angular velocity to get the torque of the diver (Mott 2008). = (IYT ρ) = (204853.55 0.172 7.2) = 25369.1 Work Cited Mott, Robert L. Applied Strength of Materials. Upper Saddle River, N.J: Pearson Education, 2008. Print. Read More