Electrical Engineering

3. Wooden joists are used to support a floor load of 6.95kPa exclusive of its own weight. The joists will have an effective span of 4.25m and be placed at 0.40m on centers. Weight of wood is 7.5kN/m2 A. Design the wooden joists so as not to exceed the allowable bending stress of 10.35MPa. B. Design the wooden joists so as not to exceed the allowable shearing stress of 0.85MPa. C. Design the wooden joists so as not to exceed the allowable deflection of 10mm. E=12135 MPa Hint: In designing joists (not beam) please follow next page for market size of joist. Also, in the design you have read the problem carefully. If the problem states to design economically then you have to consider the stress if it did not mention economically then go for conservative design.

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The figure shows a side view of a tilting table. The left support controls the tilt angle of the table by means of a threaded shaft between pins C and D that raises and lowers the scissor mechanism. The table is pinned on the right to two vertical support posts. The scissor mechanism is located along the centerline of the table which lies midway between the right-side support posts. If the tabletop is horizontal and a uniform 55-kg crate is placed along the centerline at the position shown, determine the magnitudes of the forces induced in pin E and in the threaded shaft between pins C and D. The length b = 140 mm and θ = 21°.

The figure shows a side view of a tilting table. The left support controls the tilt angle of the table by means of a threaded shaft between pins C and D that raises and lowers the scissor mechanism. The table is pinned on the right to two vertical support posts. The scissor mechanism is located …

The figure shows a side view of a tilting table. The left support controls the tilt angle of the table by means of a threaded shaft between pins C and D that raises and lowers the scissor mechanism. The table is pinned on the right to two vertical support posts. The scissor mechanism is located along the centerline of the table which lies midway between the right-side support posts. If the tabletop is horizontal and a uniform 55-kg crate is placed along the centerline at the position shown, determine the magnitudes of the forces induced in pin E and in the threaded shaft between pins C and D. The length b = 140 mm and θ = 21°. Read More »

Consider the problem of locating a new machine to an existing layout considering the position of four machines. These machines are located at the following coordinates of a plane: (3,0), (0,-3), (2,1), and (1,4). Let the coordinates of the new machine be (x1,x2). Formulate the problem of finding an optimal location for the new machine as a linear program given that the objective is to minimize the sum of the distances from the new machine to the four existing machines. Use the “street” distance (rectilinear norm); for example, the distance from (x1,x2) to the first machine located at (3,0) is absolute (x1-3) + absolute (x2-0). In your formation, clearly define the variables, and state the objective function and constraints with proper justification.

Consider the problem of locating a new machine to an existing layout considering the position of four machines. These machines are located at the following coordinates of a plane: (3,0), (0,-3), (2,1), and (1,4). Let the coordinates of the new machine be (x1,x2). Formulate the problem of finding an optimal location for the new machine …

Consider the problem of locating a new machine to an existing layout considering the position of four machines. These machines are located at the following coordinates of a plane: (3,0), (0,-3), (2,1), and (1,4). Let the coordinates of the new machine be (x1,x2). Formulate the problem of finding an optimal location for the new machine as a linear program given that the objective is to minimize the sum of the distances from the new machine to the four existing machines. Use the “street” distance (rectilinear norm); for example, the distance from (x1,x2) to the first machine located at (3,0) is absolute (x1-3) + absolute (x2-0). In your formation, clearly define the variables, and state the objective function and constraints with proper justification. Read More »

Substantiate how a particular forecasting model can be applied in this situation 10 marks

Substantiate how a particular forecasting model can be applied in this situation 10 marks EXPERT ANSWER ANSWER There are 3 basic types—qualitative techniques, statistic analysis and projection, and causative models. The primary uses qualitative knowledge (expert opinion, for example) and data regarding special events of the type already mentioned, and will or might not take …

Substantiate how a particular forecasting model can be applied in this situation 10 marks Read More »

Ventilating air enters a fan through a duct 1.6 m2 in area. The inlet static pressure is 2.54 cm of water less than atmospheric pressure. The air leaves the fan through a duct 0.87 m2 in area and the discharge static pressure is 7.68 cm water above atmospheric pressure. If the specific weight of the air is 1.2 kg/m3, and the fan delivers 9.5m3/s of air, determine: a.The power input on the air; b.The total dynamic head in water c.The fan efficiency if the power input to the fan is 12.5 Kw.

EXPERT ANSWER

The frame is subjected to the load of 4 kN which acts on member ABD at D. The following shows two ways of calculating the force acting along the member BC. Obviously, the two solutions differ from each other. Please comment on which method is incorrect/correct. 4 KN 1 m -1.5 m 45° D 1.5 m B 1.5 m Method 1: Using the entire structure as FBD, Fx = 0 4000(cos 45) = Fax FAX = 2828 N ΣMg = 0 4000(sin 45)(2.5) – Fay (2.5) + Fax (3) = 0 Fay = 6222 N Using the method of joint on Point D, 4000 N FCD 45 dearee D FBD Σε, = 0 -4000 sin 45 + Fbd = 0 Fbp = 2828 N Using the method of joint on Point B, FBD FCB 45 dearee Fay = FAB Σκαι, – 0 Fco sin 45+ FBD – FAB = 0 Fcb sin 45 + 2828 – 6222 = 0 FCB = 4800 N Method 2: Using the entire structure as FBD, ΣΜ 0 MA = 0 4000(cos 45)(3) – Fe (2.5) = 0 Fe = 3394 N Then, using the member ECD as FBD, im 15m Dx ° 45 F Dy Foc E Mp = 0 0 Fe(2.5) – FBC sin 45 (1.5) = 0 3394(2.5) – Fbc sin 45 (1.5) = 0 FBC = 8000 N O Both methods are correct. The values of force (BC) calculated are different because of different free body diagrams used. Both methods are wrongs. For the method 1, the force (CB) in the free body diagram should act away from point B. For Method 2, the 4 kN should be included in the free body diagram. O Method 1 is correct. It is the method taught in the Mechanics of Structures (MOS) for analyzing truss system. Furthermore, the principle of static equilibrium is applied correctly. O Method 2 is correct. The structure is treated as a frame rather than a truss system. Method 1 assumes that the member DBA is subjected to axial load only, and each member in the structure is treated as a two-force member. In fact, DBA is never a two-force member. The load at Point B will also cause bending moment and shear force to be developed in the member DBA. Thus, the loading condition of that member deviates from pure axial loading.

EXPERT ANSWER 1– option A is incorrect because values of forces will be same whether you solve by method 1 or method 2. 2– option B is incorrect because in method 2 , 4kN is included in the free body diagram. 3– option C is incorrect because principle of static equilibrium is not applied correctly …

The frame is subjected to the load of 4 kN which acts on member ABD at D. The following shows two ways of calculating the force acting along the member BC. Obviously, the two solutions differ from each other. Please comment on which method is incorrect/correct. 4 KN 1 m -1.5 m 45° D 1.5 m B 1.5 m Method 1: Using the entire structure as FBD, Fx = 0 4000(cos 45) = Fax FAX = 2828 N ΣMg = 0 4000(sin 45)(2.5) – Fay (2.5) + Fax (3) = 0 Fay = 6222 N Using the method of joint on Point D, 4000 N FCD 45 dearee D FBD Σε, = 0 -4000 sin 45 + Fbd = 0 Fbp = 2828 N Using the method of joint on Point B, FBD FCB 45 dearee Fay = FAB Σκαι, – 0 Fco sin 45+ FBD – FAB = 0 Fcb sin 45 + 2828 – 6222 = 0 FCB = 4800 N Method 2: Using the entire structure as FBD, ΣΜ 0 MA = 0 4000(cos 45)(3) – Fe (2.5) = 0 Fe = 3394 N Then, using the member ECD as FBD, im 15m Dx ° 45 F Dy Foc E Mp = 0 0 Fe(2.5) – FBC sin 45 (1.5) = 0 3394(2.5) – Fbc sin 45 (1.5) = 0 FBC = 8000 N O Both methods are correct. The values of force (BC) calculated are different because of different free body diagrams used. Both methods are wrongs. For the method 1, the force (CB) in the free body diagram should act away from point B. For Method 2, the 4 kN should be included in the free body diagram. O Method 1 is correct. It is the method taught in the Mechanics of Structures (MOS) for analyzing truss system. Furthermore, the principle of static equilibrium is applied correctly. O Method 2 is correct. The structure is treated as a frame rather than a truss system. Method 1 assumes that the member DBA is subjected to axial load only, and each member in the structure is treated as a two-force member. In fact, DBA is never a two-force member. The load at Point B will also cause bending moment and shear force to be developed in the member DBA. Thus, the loading condition of that member deviates from pure axial loading. Read More »