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Table of Contents

• Task 1: Static mechanical system
• Task 2: Dynamic mechanical system
• Task 3: Mechanical power transmission systems
• Task 4: Damped vibrations
• Reference

Task 1: Static Mechanical System

Section 1: For shear force, bending moment and stress due to bending calculate the distribution into simply supported beams.

The axle rod of a car can be thought of as an enchanter beam fixed at both ends. The car weighs 5.56 KN. Assuming a uniform continuous load .Treat the ends of the contact points as pinned draw the shear and moment diagrams if the distance of the axle is 0.9m.

The company has a carport with a cantilever beam. The length of the main beam in the carport is 4m. Calculate bending moment for this beam at points A, BC and D, which are 1 m apart.

• Answer: Given data in the above asked question is mentioned in the below section:
  • The weighs of the car is = 5.5 KN
  • Distance between the axel is 0.09m
  • Length of the Car port is equals to 4m
  • And distance between the points such A, BC & D is 1m
• Here in the first section we have to draw the shear as well as the moment diagram of the given issue.
• Additionally, in the next section the query is about to calculate the bending moment of the above-mentioned points.

Here in the below section, the two diagrams of shear, moreover, moment diagram has been mentioned:

Fig: Shear and Moment Diagram
(Source: Created through the author)

Here in the above-mentioned diagrams the W is standing in order to the weighs of the car, and L is standing in order to the total length. Hereby we solved the first section, which is about the moment and shear diagram of the given issues of static mechanical system.

In the below mentioned part, the query of the second part has been solved:

Here supposing that the bending moment at the points such A, BC, D is M. so that M= Wl = 5.56 KN* 0.09m = 0.5004 KNm.

Section 2: Justify the selection of standard rolled steel sections for beams and columns used in the carport.

• Answer:
  Beam Section:

  

  (Fig: Beam Section)
  (Source: Created through the author)

Columns Section:

(Fig: Columns Section)
(Source: Created through the author)

Section 3: Determine the distribution of shear stress and the angular deflection due to torsion in solid and hollow circular shafts. See section 1.

Stress concerning the shears within beams: in which point a beam is addressed toward bending that is no uniform. In this case, both, the moment that is recognized as the bending moment I s mentioned here as M and the V that is standing in order to shear forces, applicable upon the section that is horizontal. Therefore, there is a stress concerning shears that is in cross section among layers that is considered as horizontal layer concerning the particular and significant beam, consequently, stresses concerning the shear that is crosswise in manner upon the transverse section that is recognized as vertical.   The formula in order to calculate the shear stress is mentioned in the below section moreover the underneath section will also include the amount of shear stress:

Supposing that distribution of shear stress = S= Force (F)/ Area (A) = 5.56 KN/ 16m^2= 0.3475KNm^2

The angular deflection in order to the torsion into the hard and empty shafts concerning the particular circle is 45 . The below mentioned will fig. will convey the angular deflection:

(Fig: Angular Deflection)
(Source: Created through the author)

Section 4: Determine the material of a circular bar from experimental data of angle of twist obtained from a torsion test. Research experimental data on this test and produce your findings Calculate them magnitude of shear force and bending moment in cantilever and enchanter beams for a variety of applications. see section 1.

There are several data has been gathered from the circular bar of twist of angle obtained from a test that is related to torsion. The experimental data has been mentioned in the below section:

The angular deflection is 45 as per the previous solution. Additionally, the distribution of shear stress is equal to 0.3475KNm^ 2 as per the aforementioned solution.

The magnitude of share stress and the bending moment is equal to 0.3475KNm^2*0.5004 KNm = 0.173889 KN^2m^3.

Task 2: Dynamic Mechanical System

Section 1: Explain the effects of energy transfer in mechanical systems with uniform acceleration present.

The Director`s Aston Martin car can accelerate from 0 to 60 km/h in 4.3 s. Calculate its acceleration. The car has a mass of 1695kg calculate its momentum at the final speed.  Calculate the kinetic energy at the final speed.

The effect of energy transfer in mechanical system:

The exchange concerning the power among a pair of specific molecules those are place as like as nanometers except perform a vital responsibility within several manufacturing, involving solar panel, system in order to gather information about the quantum, displaying moreover the metal detector (Zinoviev, 2018), likewise within physics that is related to biology named as biophysics for measuring length concerning the nanometer moreover within the vital process that is performed within the tree named as photosynthesis. As we, all knew that energy could not be destroyed. It only can be transfer from one place to another place (Johnson, 2017). There several effects in the above mentioned section has been accomplished through the transfer of energy.

The car can accelerate = 60/4.3= 13.9534km/s^2

Monument at the final speeds= mass * velocity= 1695*(0+13.9534*4.3) = 101699.3559kgkm^3

Kinetic energy=½ m*v^2= ½ * 1695*(0+13.9534*4.3) = 50849.67795W

Section 2: Identify the magnitude and effect of gyroscopic reaction torque. Construct diagrams of the vector solutions of velocities and accelerations within planar mechanisms.

The company motorcycle travels at 80 km/h around a left bend of radius 30 m. The wheels have a mass of 2.8kg and an outer diameter of 0.5 m and a radius of gyration of 240 mm .Calculate the following:

1. The angular velocity of the wheels:

Given data in the above query:

Travelled speed = 80km/hr

Wheel’s mass= 2.8 KG

Radius of wheel= 30 m

Outer diameter of 0.5 m

Radius of gyration of 240 mm

Suppose that the angular velocity is equals to W=?

We know that W= v/r =1/2 *(2.8* 80)/ 30 =224/30= 7.46ω

2. The moment of inertia of each wheel

Supposing that the moment of inertia of each wheel is equal to I

Then the I = mr^2 = 2.8*30^2= 2,688 Kg m

3. The magnitude of gyroscopic torque produced.

Suppose that the moment of inertia of each wheel:

T= dL/L

= f (d240) = d240/240

=d/dd240 [f (d/240)] =f′ (d/240) = 1/240

=d^2/dd^2240 [f (d/240)] =f′′ (d/240) = 0

Section 3: Construct diagrams of the vector solutions of velocities and accelerations within planar mechanisms.

Research planar mechanisms and construct solutions from applications in mechanical engineering.

(Fig: vector solutions of velocities and accelerations within planar mechanisms)
(Source: Created by Author)

Example:

We will assume the θ12 and the utilized derivatives:

∞12= θ12=, ∞12= θ12=  are given.

Then the closure of the loop and the conjunction that is difficult is equal to

S14 + ic =+ a3e13 = a212 …………. (I)

S14 – ic =+ a3e13 = a212 …………. (I)

Section 4: Calculate solutions of velocities and accelerations within planar mechanisms

Using trigonometric methodology Identify, the magnitude and effect of gyroscopic reaction torque

The solution of the velocity is denoted as V=

The solution of acceleration is denoted through the equation: a=

Trigonometric methodology:

(Fig: Trigonometric methodology)
(Source: Cnx.org. 2019)

The changed angular momentum is

 netτ = ΔL/Δt

=f (d240) = d30/d3600

=d/dd240 [f (d240)] =f′ (d240) = 0

Task 3: Mechanical Power Transmission Systems

Section 1: Determine the velocity ratio for compound gear systems and the holding torque required to mount securely a gearbox. A compound gear chain in a machine in the company has a gear with 40 teeth engaging with a gear having 160 teeth .Attached to the same shaft is another gear that has 48 teeth and meshes with another gear having 96 teeth. Determine the velocity ratio of the system.

• Suppose that,In order to case 1,N’= speed of gear 1And N’’= speed of gear 2Given data,T’= amount of teeth in the gear 1 = 40 teethT’’= amount of the teeth in the gear 2=160 teethSuppose that,In order to case 2,N’= speed of gear 1And N’’= speed of gear 2Given data,T’’’= amount of teeth in the gear 1= 48 teethT’’’’’= amount of the teeth in the gear 2= 96 teeth

  Velocity ratio in the case 1,

  N’/N’’ = T’/T’’= 40/160= 1/40

  Velocity ratio in the case 2,

  N’/N’’=T’’’/T’’’’= 48/96 = ½

  Such that, the total velocity ration is equals to N’/N’’= T’/T’’* T’’’/T’’’’=1/40*½ = 1/80= 0.0125 m/sec (Ans.).

Section 2: Calculate the operating efficiency of lead screws and screw jacks.

In the company garage, there is a screw jack. It is being used to support a car axle  the load on it being 4.8 N .the screw jack has an effort effective radius of 200 mm and a single square start thread, having a lead of 5mm.Determine the efficiency of the jack if an effort of 60 N is required to raise the car axle.

• Given data,Load in initial case = 4.8NDistance moved in initial case = 200mmIn the second case the load or force is = 60NAnd, the second case the distance moved is = (200+5) mm= 205mmWe have to find the overall efficiency,Ƞ= (output of the work/ input of the work)*100%=?We know that the output of the work = (Further Load * Further Distance Covered)

  = 60 N * 205 mm =12300 J

  And the input of the work= (Initial Load * Initial Distance Covered)

  = 4.8 N * 200mm =960 J

  Such that Ƞ= (output of the work/ input of the work)*100 %

  = (12300/960)*100%= 12.8125%

Section 3: Explain the conditions required for a constant velocity ratio between two joined shafts

The proportion of speeds is equivalent to the proportion of boundaries of shafts or sprockets utilized for moving force (Darby and Chhabra, 2016). As it were, the speed proportion is not consistent just when cone like sprockets are utilized for moving force, for example, VCT overdrives.

Section 4: Examine the cause of a documented case of mechanical power transmission failure and the steps taken to correct the problem and rectify any design faults.

Cause for failure:

Spilling transmission liquid:

The least demanding approach to foresee transmission issues is to search for indications of a physical hole. The liquid is typically red in shading. Holes are commonly found close to the channel fitting or seal between the transmission and the motor. Transmission liquid is basic for the task of your vehicle or truck (Scholz, 2019). It is critical to fix the break before any lasting harm is finished.

Low transmission liquid level moreover quality: 

Clearly, releasing liquid can prompt a low liquid level. You ought to ensure likewise that the liquid is still in respectable enough condition to play out its activity. As the liquid moves through the transmission, it greases up and cools the segments. Within the event, there is insufficient liquid or the liquid is beginning to corrupt, the transmission will perform inadequately. At one point, the transmission will quit working out and out.

Obstructed transmission liquid channel:

Many people do not have a clue about there is a channel for the transmission liquid. This channel helps keep the liquid clean and of high caliber. In the event that the channel ends up stopped up, it will not let the appropriate measure of liquid through (Patankar, 2018). Either the absence of greasing up and cooling transmission liquid will make the transmission come to a standstill, or it will overheat. Neither of these results is useful for your vehicle and both will result in a weighty fix bill.

Solution:

By following below mentioned step, we can overcome the problems of mechanical power transmission failure:

• A better investigation of indication concerning physical hole has been needed.
• Checking of the working capability of the used fluid has been required.
• Checking the clarity and high efficiency of the fluid is much needed.

Fig: Mechanical Power Transmission
(Source: Chegg.com.2019)

Task 4: Damped Vibrations

Section 1: Explain the natural frequency of vibration in a mass-spring system in a car suspension system how cans the cars suspension systems are viewed as an ideal mass spring system.

Tremble recognizes as the phenomenon that is mechanical in manner in which the process named as oscillation has been demonstrated upon a specific point that is named as the equilibrium point.

The discussion might consider that there are three separate deviations concerning the power that is engaged toward the method related to the spring-mass, is mentioned in the below section:

• Rotor Excitation models the impact of a turning machine mounted on an adaptable floor (Gieras Piech and Tomczuk, 2018). The wrenches with little mass pivots at steady rakish speed, making the mass m vibrate.
• Outer Forcing models the conduct of a framework, which has a period fluctuating power following up on it. A model may be a seaward structure exposed to wave stacking.
• Base Excitation models the conduct of a vibration segregation framework. The base of the spring is given a recommended movement, making the mass vibrate (Worden, 2019). This framework can be utilized to show a vehicle suspension framework or the tremor reaction of a structure.

Section 2: Determine the amplitude and phase angle of the transient response within a mass-spring damper system.

This part of this report includes each formula that the problem requires to solve.

Motion’s formula:

(Source: Brown.edu. 2019)

 

(Source: Brown.edu. 2019)

Solution concerning Steady state:

(Source: Brown.edu. 2019)

 

Fig: amplitude and phase angle
(Source: Brown.edu. 2019)

 

Fig: Solution concerning outer Forcing
(Source: Brown.edu. 2019)

Section 3:  Unnecessary vibrations need to be prevented in the suspension .Identify the conditions needed for mechanical resonance and measures that are taken to prevent this from occurring.

Several conditions are present there mechanical resonance. In this part of the discussion the several major conditions has been mentioned:

1. The goal should hold minimum one number of general Tremble rather frequency (Bloomberg and Volpe, 2018).
2. The goal should be carried on through an outer frequency power.
3. The Tremble concerning the outer Tremble power need to overview the general Tremble concerning the object.
4. While the Tremble of the features of powering links the general Tremble concerning specific goal, this will start for resonating (Mohanty, 2018).
5. A powering feature in a likewise Tremble such as the general Tremble: in the method that is considered as the mechanical process the powering feature remains as a power that is recognized as the variable power.

There are three major things are presented in order to measures, which are mentioned in, the below section of the discussion:

1. Stimulation
2. Scattering
3. Momentum

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