CASE STUDY AAssuming the role of a trainee on the verge of undertaking a structural integrity engineer position, you were presented with two technical problems involving beams (structural components). The first one is associated with the cantilever beam shown in figure 1a, subjected to a bending moment on its free-end (the magnitude of the force applied is negligible compared to the moment caused). The second one is associated with a similar cantilever beam, but, this time, subjected to a uniformly distributed load varying from 0 to 4 kN/m; as shown in figure 1.b, and with length L=6 m. [caption id="attachment_1735" align="aligncenter" width="600"] Figure 1 – Generic cantilever beam (a); beam under distributed load (b).[/caption] 1. Your task is to analyse and deduce the output values for the following:
20 marks2. Part of the role you are intending to assume consists in interpreting a physical situation, so as to describe it and model it properly. In order to show your understanding of different loading configurations and their causes, EXPLAIN what type of load is expected on the wing spar (main beam) of an aircraft wing, due to the elements highlighted in figure 2, and QUALITATIVELY draw a sketch of the imposed loads. The cross-section of the spar is wider at the connection with the fuselage and approximately constant at mid span. State your assumptions.
?200 words 10 marksNote: You are required to use your own words. [caption id="attachment_1736" align="aligncenter" width="600"] Figure 2 – Wing spar and some elements.[/caption]
-: Case Study A :-Question 1:- Part (a,b,c,d) for cantilever beam with given Moment at the free end. The shear force diagram and bemding moment diagram for the cantilever beam is calculated analytically and the diagrams are plotted using matlab as well as represental images are drawn in autocad software. For the first case the given cantilever beam is subjected to the end moment i.e the moment is subjected to the free end of the cantilever. Due to the only moment acting on the beam the net reaction force would be zero. Only reaction moment will be acting at the fixed end of the beam. The shear force diagram for the case when cantilever is subjected to moment only iszero line i.e no shear force is acting on the beam. The beam has only bending moment through out in the bending moment diagram. the bending moment acting is shown in the diagram. Calculations:- The above figure shows the cantilever beam with the end moment. The equations used to calculate the sfd and bms are as follows:-
-: Case Study D:-The above figure shows the chamber where laser scanner is kept which is maintained at 0oC. The ambient condition is 25oC.The The air chamber where the laser scanner is located needs to be maintained at 0ºC. The internal coefficient of convection heat transfer was determined to be . The lateral surfaces of the chamber are 35 mm thick, and made of a composite fabricated with expanded polystyrene foam and a plate of carbon fibre, with a coefficient of thermal conductivity of 0.021 W/(mK). The top surface of the chamber is made of the same material, but is 40 mm thick, and the bottom surface is well insulated. The temperature of the room housing the chamber is kept constant at 25ºC by an independent cooling system, and the coefficient of external convective heat transfer was estimated to be . The internal dimensions of the chamber were provided as 250 mm (length), 170 mm (width) and 152 mm (height). There are two red walls, two yellow walls and one roof (green top)from which the heat transfer takes place. The bottom is insulated properly so no heat transfer from the bottom. So in total there are five walls along which heat transfer takes place. The thermal conductivity of the walls is same. The internal and external heat transfer coefficient is given. Thickness of red wall and yellow wall is 35 mm. thickness of green wall is 40 mm. Heat transfer through red wall:- Let ambient temperature be T0 and internal temperature be Ti. Let h1 be the external heat transfer coefficient and h2 be internal heat transfer coefficient. Let k denote the thermal conductivity of the walls. Heat transfer through red wall (T0 – Ti) =QR((1/h1*AR) + ( LR/AR*k ) + ( 1/h2*AR )) Where AR = area of red wall = 250 * 152 = 38000 mm^2 = 0.038 m^2 LR = thickness of the red wall = 35 mm = 0.035 m. Substituting the values in the above equation , we get 25 – 0 = QR((1/3.1*0.038) + (0.035/0.021) + (1/12.5*0.038)) QR = 0.459W Hence total heat transfer through two red walls = 2 * 0.459 = 0.918 W. Heat transfer through yellow wall:- (T0 – Ti) =QY((1/h1*AY) + ( LY/AY*k ) + ( 1/h2*AY )) Where AY = area of yellow wall = 170 * 152 = 25840 mm^2 = 0.02584 m^2 LY = thickness of the yellow wall = 35 mm = 0.035 m. Substituting the values in the above equation , we get 25 – 0 = QY((1/3.1*0.02584) + (0.035/0.021) + (1/12.5*0.02584)) QY = 0.31 W Hence total heat transfer through two yellow walls = 2 * 0.31 = 0.62 W Heat transfer through green wall:- (T0 – Ti) =QG((1/h1*AG) + ( LG/AG*k ) + ( 1/h2*AG )) Where AG = area of green wall = 170 * 250 = 42500 mm^2 = 0.0425 m^2 LG = thickness of the green wall = 40 mm = 0.04 m. Substituting the values in the above equation , we get 25 – 0 = QG((1/3.1*0.0425) + (0.04/0.021) + (1/12.5*0.0425)) QY = 0.46 W Now the total heat transfer from all the walls = 2*QR + 2*QY + QG Qtot = 0.918 + 0.62 + 0.46 = 1.998 W -----Answer This is the amount of heat which will enter into the chamber from ambient . The total thermal resistance can be obtained by dividing total heat by temperature difference. Hence Rtot = 1.998/25 = 0.0799 W/K. --------Answer Since cooling system for the laser in the chamber is capable of converting 40% of its electric power in useful mechanical energy, and it is supplied with 2.52 W. Hence amount the amount of heat the cooling system injects in the chamber is 2.52-0.4*2.52 = 1.512 W Hence the total thermal load is = Qtot + 1.512 = 1.998 + 1.512 = 3.51 W. -----Answer The salient difference between natural convection and forced convection are as follows:-
-: Case Study C :-Part 1. In the present scenario the given things are that the barge is wrongly constructed and the cross section area of the barge is different from the original designed area. So in this question all the calculation will be done according to the present area not the one which was assumed to be kept. Now to solve this question we need to know Archimedes’ principle which says that weight of the body is equal to the buoyant force. Using this concept this problem can be solved. Given weight of the barge = 450000 N. Depth of the hull below free water surface = 1.5 m. Density of water = 998 kg/m^3. (A) :- weight of the body should be equal to the buoyant force. Buoyant force is equal to the volume submerged multiplied by the density of water multiplied by the acceleration due to gravity. Hence clubbing up the above idea we get equation as below:- WB = BF 450000 = h x A x 998 x 9.81 Here h = depth of hull below water surface, A is the area needs to be calculated. 450000 = 1.5 x A x 998 x 9.81 A = 30.64 m^2. (B) :- in this part we need to calculate the maximum weight of the barge such that the hull should not be below 2 m from the free water surface. It means we can use same equation as above but this time height h should be 2 m and the area which we need to use the one calculated above i.e 30.64 m^2. Therefore , Maximum weight is equal to the buoyancy force due the submerged volume of height 2 m. Wmax = 2 x 30.64 x 998 x 9.81 = 600000 N or 600 kN. (C) in this case they are asking to find the mass of the barge so that it floats on the surface of the water and no part of it is submerged in water. Here the length of the barge is three times the width. L = 3b Since this situation is only possible if the weight of the barge is less than equal to the surface tension of the water. Here the temperature of the water is given which is 20oC. at this temperature the surface tension of water is 0.0723 N, which is the standard value and can be checked from any standard data book. Weight of the barge = Surface tension force. To calculate surface tension force we need to get the length of the barge, which can be obtained from the area of the barge. A = L x b = 3b xb( because L = 3b given ) 30.64 = 3b^2 b = 3.2 m. Therefore L = 3 x 3.2 = 9.6 m. Hence Surface tension force = 0.0723 x 9.6 = 0.694 N This surface tension force is equal to the weight of the barge. WB = 0.694 Mg = 0.694 because weight is equal to mass times gravity Hence M= 0.07 kg. Part 2:-
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