Mechanical Engineering is one of the most elaborate disciplines of engineering. It plays a very important role in any organisation or industry. There are numerous sub-disciplines and topics that fall under the mechanical engineering umbrella. Universities all over the world pay special attention to the curriculum of mechanical engineering and to bring out the best from students, they device elaborative and complex mechanical engineering assignments. Most often, students do not understand how to approach their assignments in the right way to achieve the maximum marks. My Assignment Services provides
that encompasses all the different sub-disciplines and topics of mechanical engineering. Given below are some of the
that we provide assistance with.
Our mechanical engineering assignment help Australia provides error-free assignments on mechanical electronics, structural analysis, thermodynamics, and design and drafting etcetera. Students can take a look at the different mechanical engineering assignment samples online or in University databases.
Assignments are an important component in the entire study-cycle of a university. Professors take extra measures to devise assignments to bring out the comprehensive learning outcomes from students. However, students are often unable to perform well in their
. There are a lot of contributory factors that can bring down a student’s performance in assignment. Lack of time, multiple assignments, stringent deadlines, exam preparation and part-time jobs are some of the factors that affect the performance. In the case of Mechanical Engineering Assignment, there are a lot of terminologies, methodologies and mathematical theories and diagrams that the students have to be well-versed in order to write a high quality assignment. This is why students often get confused about where to start their assignment.
My Assignment Services has a team of dedicated subject matter experts who are not only well-versed in the different methodologies, topics and sub-topics of mechanical engineering but are also acquainted with university guidelines and marking rubric. They know exactly how much information to provide on which question and they always write assignments according to the marks allotted to each question in the marking rubric. Take a look at the below sample of mechanical engineering assignment to understand how our experts approach the question.
-: 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:-
- SFD = 0 --à as no load is acting
- BMD= -M ---àbecause only moment is acting.
Reactions at the end are calculated by equating the net force in the x-direction and y-directions
Therefore R
_{A} = 0 ----- > equation (1)
Now we can equate total moment at any point on the beam to get reaction.Only reaction moment acts at the fixed end of the beam which is –M. negative sign because taking clockwise as negative .
The sfd and bmd for the above figure is shown below.
The free end has no reaction.
Shear force at A left=0 Bending moment at A left=-M
Shear force at A right=0 Bending moment at A right =0
Shear force at B left=0 Bending moment at B left =0
Shear force at B right = 0 Bending moment at B right = -M
The shear force is zero so no maximum shear force. The maximum bending moment is M magnitude wise and its unit is kN-m. It acts throughout the beam but in negative direction.
Question 1part (a,b,c,d) for cantilever beam with given uniformly varying load.
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 second case the given cantilever beam is subjected to uniformly varying load. Intensity of load is highest at the fixed end and intensity of load is zero at the free end. To calculate the shear focre we can assume a concentrated load which is the area of the triangle formed due to loading. In thisquestion we are generating the general equation for the calculation of shear force for the cantilever beam subjected to uniformly varying load.
Let the uniformly varying load is w kN/m. let the length of the beam be L.
The load at any distance X from the left end is given by w
_{x}=w*X/L.
Hence the shear force F at any distance X is given by
F = -( w * X^2)/2*L
This is the parabolic variation of the shear force.
Bending moment at any distance X is given by
M = -(w*X^3)/3L
This shows the cubic variation of the bending moment diagram.
Solving the above situation with given load as 4 kN/m and given the length of the beam to be 6m.
Reaction at fixed end R
_{A} = 12kN ( total upward load = total download )
Now the sfd and bmd is solved using matlab and the code is attached below.
clearall;
clc
w=4;
L=6;
x=(0:0.01:6);
wx=w.*x/L;
F=-(w.*x.^2)/(2*L);
figure; plot(x,F)
xlabel('lenght of beam in meter');
ylabel('shear force in kilo newton');
M=-(w.*x.^3)/(6*L);
figure ;plot(x,M)
xlabel('lenght of beam in meter');
ylabel('bending moment in kilo newton meter');
The above code generates the sfd and bmd plots for this given situation. The representative images for sfd and bmd drawn in autocad are as follows:-
The matlab plots for the same sfd and bmd are shown below for unit length of the beam and for the given load.
The maximum shear is acting at the fixed end which can alsobe verified from the given general expression. The maximum bending moment is acting at the fixed end which can be verified from the given equation above.
Question part (b):-
Loads on Aircraft:-
When one thinks of aircraft loads, the air loads due to high-g maneuvering come immediately to mind. While important, maneuvering loads are only a part of the total loads that must be withstood by the aircraft structure. The various types of loads are as follows:-
- Air loads: - This includes maneuver , gust , control deflection, component interaction, buffet.
- Inertia loads: - This includes acceleration, rotation, dynamics, vibration, flutter.
- Power Plant: - This includes thrust, torque, gyroscopic, vibration, duct pressure.
- Landing: - This includes vertical load factor, spin-up, spring-back, crabbed, one wheel, arrested, and braking.
- Takeoff:- This includes catapult, aborted
- Other loads: - This includes towing, jacking, pressurization, bird brake, actuation, crash.
The largest load the aircraft is actually expected to encounter is called the “limit”, or “applied” load. The limit load on the wing occurs during an 8-g maneuver.
To provide a margin of safety he aircraft structure is always designed to withstand a higher load than the limit load. The highest load the structure is designed to withstand without breaking is the “design” or “ultimate” load.
Power Plant Loads:-
The engine mounts must be obviously be able to withstand the thrust of the engine as well as its drag when stopped or wind milling. The mounts must also vertically support the weight of the engine times the design load factor. The engine mounts are usually designed to support a lateral load equal to one-third of the vertical design load. The mount must withstand the gyroscopic loads caused by the rotating machinery (and propeller) at the maximum pitch and yaw rates.
For this case which is having propeller-powered aircraft, the engine mounts must withstand the torque of the engine times a safety factor based upon the number of cylinders. This reflects the greater jerkiness of an engine with few cylinders when one cylinder malfunctions.
For an engine with two cylinders, the safety factor is 4, with three cylinders 3, and with four cylinders 2. These safety factors are multiplied times the torque in normal operation to obtain the design torque for the engine mounts.
Landing-Gear Loads:-
The landing gear’s main purpose is to reduce the landing loads to a level that can be withstood by the aircraft. To analyze fully all the possible gear loads, a number of landing scenarios must be examined. These include a level landing, a tail down landing, a one wheel landing and crabbed landing. When the tires contact the ground they are not rotating. During the brief fraction of a second it takes for them to spin up, they exert a large rearward force by friction with the runway. This spin up force can be as much as half the vertical force due to landing.
When the tire is rotating at the correct speed the rearward force is relieved and the gear strut “spring back” forward, overshooting the original position and producing a “spring back” deflection load equal to or greater than the spin up load.
Fuel Tank loads:-
The fuel is uniformly distributed over the wing span through modules which can be disintegrated. The fuel tank can be assumed to be uniformly distributed load on the aircraft wing.The fuel tank near to wing act as load only with very less momentum.Due to tip tank wing experiences additional momentum. This situation is like a cantilever beam with the end moment. The situation when the fuel is along the wing is like uniform varying load. Larger aircrafts use wings to store the fuel this saves the lot of space. If the wings are tapered then the load due to fuel acts as uniformly varying load and if the wings are rectangular then the load due to fuel acts as a uniformly distributed load. For helicopters fuel tanks are kept overhead so that the fuel flows under the effect of gravity and this avoids the need of additional pump to pump the fuel.
The above figure shows the aircraft with various loads.
-: Case Study D:-
The above figure shows the chamber where laser scanner is kept which is maintained at 0
^{o}C. The ambient condition is 25
^{o}C.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) =Q
_{R}((1/h1*A
_{R}) + ( L
_{R}/A
_{R}*k ) + ( 1/h2*A
_{R} ))
Where A
_{R} = area of red wall = 250 * 152 = 38000 mm^2 = 0.038 m^2
L
_{R} = thickness of the red wall = 35 mm = 0.035 m.
Substituting the values in the above equation , we get
25 – 0 = Q
_{R}((1/3.1*0.038) + (0.035/0.021) + (1/12.5*0.038))
Q
_{R} = 0.459W
Hence total heat transfer through two red walls = 2 * 0.459 = 0.918 W.
Heat transfer through yellow wall:-
(T0 – Ti) =Q
_{Y}((1/h1*A
_{Y}) + ( L
_{Y}/A
_{Y}*k ) + ( 1/h2*A
_{Y} ))
Where A
_{Y} = area of yellow wall = 170 * 152 = 25840 mm^2 = 0.02584 m^2
L
_{Y} = thickness of the yellow wall = 35 mm = 0.035 m.
Substituting the values in the above equation , we get
25 – 0 = Q
_{Y}((1/3.1*0.02584) + (0.035/0.021) + (1/12.5*0.02584))
Q
_{Y} = 0.31 W
Hence total heat transfer through two yellow walls = 2 * 0.31 = 0.62 W
Heat transfer through green wall:-
(T0 – Ti) =Q
_{G}((1/h1*A
_{G}) + ( L
_{G}/A
_{G}*k ) + ( 1/h2*A
_{G} ))
Where A
_{G} = area of green wall = 170 * 250 = 42500 mm^2 = 0.0425 m^2
L
_{G} = thickness of the green wall = 40 mm = 0.04 m.
Substituting the values in the above equation , we get
25 – 0 = Q
_{G}((1/3.1*0.0425) + (0.04/0.021) + (1/12.5*0.0425))
Q
_{Y} = 0.46 W
Now the total heat transfer from all the walls = 2*Q
_{R} + 2*Q
_{Y} + Q
_{G}
Q_{tot} = 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
R_{tot} = 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 =
Q_{tot} + 1.512 = 1.998 + 1.512 = 3.51 W. -----Answer
The salient difference between natural convection and forced convection are as follows:-
- Natural convection is the result due to the density gradient when the wall or the surface is heated. As the temperature of the surface is increased the fluid near the wall tries to expand and its density decreases. Now this gradient in density gives rise to fluid motion i.e the lower density fluid near the wall goes up and the higher density fluid tries to come down which sets up the current for the fluid motion. There is no external forced is applied for the fluid motion, only temperature changes generates fluid flow.
- In case of forced convection the fluid is made to follow in presence of fan or blower or some driving mechanism or external force so it is called forced convection. In case of forced convection the equation of motion are similar to the fluid dynamics and are applicable with proper assumptions. Various non- dimensional number describes the forced convection for different situations.
-: 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:-
W
_{B} = B
_{F}
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.
W
_{max} = 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 20
^{o}C. 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.
W
_{B} = 0.694
Mg = 0.694 because weight is equal to mass times gravity
Hence M= 0.07 kg.
Part 2:-
- In this part of analyzing the situation we see that when the bottle is filled one third with water the center of gravity lies closer to the ground surface which means more stable condition. So the key concept is that we are trying to make center of gravity of the bottle with water close to the ground. If we fill the bottle completely or keep bottle empty then we are changing the center of gravity of the bottle there fore it becomes less stable and as a result of which the bottle doesn’t stand straight when flipped. But in the case when the bottle is filled with one third of the water it stand straight when flipped. The reason is in this physical problem we need to care about the angular momentum as well as the rotational force. When the bottle is flipped with optimum angular velocity it rotates and this rotation is faster and then slower and then again faster. This is due to the sloshing effect. When the bottle is flipped the water tries to spread through the bottle at this time the rotation is slower and when the rotation two third then the water rushes back to the bottom and giving more rotational effect which makes it faster and then it stands straight. This activity is very similar to the way the diver dives in a pool and also similar to pontoon.
- The trajectory of the critical point is sort of parabola. Considering the critical point to the c.g when it is at base, it follows a parabola.
The above shows the trajectories of the bottle flip. Now two important things first the angular momentum and rotational torque. The other thing is the fluid dynamics , so as the bottle flips liquid sloshes due to rotation changing c.g of the bottle, this makes bottle to slow down giving more to stand straight. Angular momentum is constant.
i.e L = m*v*r = constant.
Since mass is also constant hence this situation is controlled by the radius of flip r and spin given to bottle v. If v increases then r has t decrease to make situation stable and vice versa.
Angular momentum L = I * ?
Where I is moment of inertia and ? is the angular velocity ( v*r ).
Rotational torque T = I * ?
Where ?= angular acceleration. (=?/t)
Both the above equation can be equation as follows:-
T = L/t, t is time
Therefore T = (I * ?)/t
Or, I*? = (I * ?)/t
Or, v
_{min} = ?/r
Therefore v minimum is when radius of spin is large and angular velocity is less.
- The bottle filled one third doesn’t bounce back because as we flip the bottle the water inside sloshes and spread throughout the bottle distributing properly because of more space inside water bottle and force due to sloshing is very less which doesn’t make flip unstable and after flip it lands straight.
But in the case when is bottle is filled fully it can’t perform stable flip as the rotational energy inside bottle due to flip has no space to expand which makes bottle unstable.
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