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Scenario You are working as a Mechanical Engineer for Scania AB, which is a heavy trucks manufacturer. Your job is to inspect the new model of Scania LBS141 cab-over truck with Scania’s 14.2-litre turbocharged V8 DS14 engine. You will also have to research on cases where the truck passes over the bridge, where the beams of the bridge have to be considered while solving numerical problems.

You have to perform the following tasks: –

Task 1 Physical Properties of Engineering Materials

Section 1: Debate the relationship between the elastic constants. Demonstrate the effects of two-dimensional and three dimensional loading on the dimensions of a given material.

Relation between elastic constants are shown below: –

E = 2G(1 + μ)

E = 2K(1 – 2μ)

μ ,E ,G ,K are elastic constant

μ = Poission ratio

E = Young modulus

G = Shear modulus

K = Bulk modulus

The effects of two-dimensional and three dimensional loading on the dimensions of a given material is shown below: –

For 2D

Total elastic constant = 4 (E,K,G,μ)

Total independent elastic constant =2 (E,μ)

For 3D

Total elastic constant = 12(Ex,Kx,Gx,μx,Ey,Ky,Gy,μy,Ez,Kz,Gz,μz)

Total independent elastic constant =9 (Ex,Kx,μx,Ey,Ky,μy,Ez,Kz,μz)

Section 2: Determine the volumetric strain and change in volume due to three dimensional loading.

A material has stresses of 2 MgPa in the x direction and 3 MgPa in the z direction. Given the elastic constants given the elastic constant is 205GPa and v is 0,27 calculate the strains in both directions.

Given data shown below:

σ_x= 2 MPa

σ_z = 3 MPa

μ = 0.27

E = 205 GPa

Formula for volumetric strain

ϵ_v = 1.12 * 10^-5

Change in volume

Formula for Strain in x direction

Formula for Strain in z direction

Section 3: Assess the effects of volumetric thermal expansion and contraction on isotropic materials

Isotropic materials are those materials which have same properties in every direction .Volumetric thermal expansion coefficient shows the effect on any material .Basically this coefficient is depend on the properties like temperature, pressure and different points of material .So change in any one of property may change the value of coefficient of volumetric thermal expansion .But for isotropic material , as per definition there is no change in the property as in all directions the property of material remain same so volumetric thermal expansion coefficient is same at all points therefore no effect  on isotropic materials.

A beam has a hollow circular cross section and 40 mm outer diameter and 30 mm inner diameter. It is made from a metal with a modulus of elasticity of 205 GPa. The maximum allowable tensile strength is 305GPa.

Given data

Young modulus E = 205 GPa

Maximum strength σ = 305 GPa

Shaft outer diameter D = 40 mm

Shaft inner diameter d = 30 mm

Calculate i) the maximum allowable bending moment

Maximum allowable bending moment is given by

I = 85.9 * 10³ mm⁴

ii) the radius of curvature(R).

Section 4: Analyze the behavioral characteristics of materials subjected to complex loading.

Ductility of this material is very high. The material having high value of young modulus (E) therefore this material is elastic in nature. This material can easily take up large tensile and compressive stresses and likely to fail in shear stresses due to ductile in nature. Due to application of complex stress, load act on the material along x direction, y direction which are tensile or compressive in nature and also shear stress act clockwise or anticlockwise direction.

Task 2: Forces and Macaulay’s Method

Section 1: Appraise the variation of slope and deflection along a simply supported beam P5

Solution:As the truck passes through the bridge then load acts on the beams of bridge. Consider the case of simply supported beam having uniform distributed load of W N/m act on beam.

Although the applied load is uniform but slope and deflection vary at all point.

Deflection is zero at the support A and support B. For uniform distributed load, deflection is maximum at the center of beam. Maximum value of deflection (y) for above case is given below:

Location at which deflection is maximum, the slope will be zero. Slope is maximum at the supports. Slope of beam at support A is equal to that of at support B but are opposite in nature. Maximum value of slope () for above case is given below:

      

Section 2: Determine the principal stresses that occur in a thin walled cylindrical pressure vessel and a pressurized thick-walled cylinder.

Principal stresses act on cylinder are given below: –

p = pressure act on cylinder

t = thickness of cylinder

d = diameter of thin cylinder

R₁ = Internal radius of thick cylinder

R₂ = External radius of thick cylinder

Section 3: Evaluation a suitable size universal beam from appropriate data tables, which conforms to, given design specifications for slope and deflection. A beam must withstand a bending moment of 360 NM. If the maximum stress must not exceed 260 MPa determine the elastic modulus z and determine a suitable beam.

Given data is shown below

Bending moment M =360 Nm or 360*1000 N mm

Maximum stress =260 MPa

Calculate the section modulus z

Determine suitable beam

I/y = z

To determine the beam dimension, design criteria should consider.

For safe design

I/y > 1384.6

Consider the combination provided below:

Beam width = 18 mm

Beam depth = 22 mm

As the conditions are satisfied therefore the design is safe.

Suitable beam size is width 18 mm and depth 22 mm

Section 4: Critique and justify your choice of suitable size universal beam using appropriate computer software to model the application by explaining any assumptions that could affect the selection.

As the selected beam having size of 18 mm x 22 mm is tested on required loading conditions.

As Bending Moment M =360 Nm

Consider the case of simply supported beam having load at centre. Maximum bending moment of this beam is shown below.

Consider the length of beam is 600 mm

Therefore, beam should be test on center load of 2400 N for safe design.

Software analysis of beam shown below.

• Initially the beam of 600 mm length is prepared in software as shown below.

 

• After creating a beam, supports are applied at its ends followed by load of 2400 N is applied at the center, this process made the beam simply supported with loading at the center. The supports and loading on beam shown below.

 

• After meshing, run this study. After that the results obtained from the input data shown below. The below results shows that maximum value of stress generated due to applied loading in the beam is 171 MPa but the maximum permissible value of stress is 260 MPa therefore the design is sufficiently safe.

Task 3: Differential Gears

Section 1: Discuss the initial tension requirements for the operation of a v-belt drive.

V Belt drive are power transmission system in which belts are used for power transmission. V belts are having trapezoidal cross-section and manufactured for certain specified length. When the v-belt is installed over the pulley then the belt retains initial tension due to its smaller length of belt as compared to the center distances between two pulleys. Initial tension need to be exist so that firm grip maintains between pulley and v-belt throughout motion. This initial tension is same throughout the belt length and when there is no motion. Initial tension in belt given by: –

During motion of v belt drive, tight side tension should be higher than the initial tension and slack side tension should be lower than the initial tension.

Section 2: Examine the force requirements to engage a friction clutch in a mechanical system. Study the holding torque and power transmitted through epicyclical gear trains.

Clutches are generally used for power transmission of machines and shafts which are either fully or partially loaded. Some of main requirements to engage a friction clutch in a mechanical system given below.

Material used for friction surface should have high coefficient of friction and uniform throughout. Material of friction surface should not affect by oil/ moisture, withstand high temperature, high thermal conductivity and high wearing resistant. Clutch moving parts should have low weight so that inertia force should be reduced.

Holding/Braking torque and power in epicyclical gear train

There are 3 torque on epicyclical gear train, input torque (T1) on driving shaft, output torque (T2) on driven shaft and holding/braking torque (T3) on the fixed member.

• Holding/braking torque (T3) is given by:

• Power transmitted (P) is given by

 Section 3: Critically analyze both the uniform wear and uniform pressure theories of friction clutches for their effectiveness in theoretical calculations.

Uniform wear theory

Under this theory pressure is inversely proportional to the r distance from the axis of clutch. Pressure intensity is given by

For maximum pressure r = r2
For minimum pressure r = r1

Total frictional torque acting on clutch plate given by

                    

Uniform wear theory generally effective for old clutches because in old clutches pressure is not uniform throughout due to wear on the clutch surface.

Uniform pressure theory

• Under this theory pressure is uniformly distributed over frictional area as shown above. Pressure intensity is given by

• Total frictional torque acting on clutch plate given by

• Uniform pressure theory generally effective for new clutches because in new clutches pressure is uniform throughout.

Section 4: Evaluate the conditions needed for an epicyclical gear train to become a differential, and show how a differential works in this application

As the bevel gears are used in epicyclic gear train then this system becomes more compact and used as differential gear system.

Working of differential gears

The differential gears used in rear part of automobile to rotate the rear wheels at different speed while taking a turn to avoid skidding. In the round path while taking a turn, outer wheel has to travel more distance than inner wheel, so inner wheel should have lesser speed than outer wheel, this relative speed reflects on gear c and gear D. Due to this gear E and gear F start rotating about is own axis. Due to motion of bevel gear E, F speed of inner rear wheel decrease and outer rear wheel increases in the same amount.

Task 4: Cam Profiles

Section 1: Explore the profiles of both radial plate and cylindrical cams that will achieve a specified motion. Show the mass of a flywheel needed to keep a machine speed within specified limits. A cam must produce a lift of 30mm.This must produce 60 degrees of rotation. The follower must stay at this level for a further 30 degrees and then fall over the next 6 degrees. Then there is 210 degrees to complete the cycle. Design a cam cycle so the velocity is constant when rising and falling.

Radial cam

In the radial cam, follower oscillates or reciprocated in the direction perpendicular to the cam axis. Radial cam is widely used as compared to cylindrical cam. Profile of radial cam follower shown below.

Cylindrical cam

In the cylindrical cam, follower oscillates or reciprocated in the direction parallel to the cam axis. Follower moves in the groove present in the cam surface. Profile of cylindrical cam follower shown below.

Mass (m)of a flywheel needed to keep a machine speed within specified limits.

∆E = Maximum fluctuation of energy

R = Radius of flywheel

C_S = coefficient of fluctuation of speed

ω = mean speed in rad/sec

Cam cycle design

Cam profile

Displacement profile

Section 2: Investigate the balancing masses required to obtain dynamic equilibrium in a rotating system. Four masses are 200kg, 300 kg, 240 kg and 250 kg respectively. The corresponding radii of rotation are 0.2 m, 9.15 m, 0.25 m and 0.3 m respectively and the angles between successive masses are 45, 75 and 135 degrees. Find the position and magnitude of the balance mass required if its radius of rotation is 0.2m.

Given data

Mass m₁ = 200 Kg, m₂ = 300 Kg,m₃ = 240 Kg, m₄ = 250 Kg

Radii of rotation r₁ = 0.2 m, r₂ = 0.5 m, r₁ = 0.25 m, r₂ 2= 0.3 m

Angle of rotation θ₁ = 0°, θ₂ = 45°, θ₃ = 45 + 75 = 120°, θ₄ = 120 + 135 = 255°

The position and magnitude of the balance mass required if its radius of rotation is 0.2m.

r = 0.2 m

Centrifugal force in each case

m₁ * r₁ = 200 * 0.2 = 40 Kg m

m₂ * r₂ = 300 * 0.5 = 45 Kg m

m₃ * r₃ = 240 * 0.25 = 60 Kg m

m₄ * r₄ = 250 * 0.3 = 75 Kg m

Now resolve the component horizontally

∑H = m₁ * r₁ cos⁡ θ₁ + m₂ * r ₂ cosθ₂ + m₃ * r₃ cos⁡θ₃ + m₄ * r₄ cosθ₄

∑H = 40 cos 0 + 45 cos 45 + 60 cos 120 + 75 cos 255

∑H = 22.38 Kg m

Now resolve the component Vertically

∑V = m₁ * r₁ cos⁡ θ₁ + m₂ * r ₂ cosθ₂ + m₃ * r₃ cos⁡θ₃ + m₄ * r₄ cosθ₄

∑V = 40 sin 0 + 45 sin 45 + 60 sin 120 + 75 sin 255

∑V = 11.35 Kg m

Resultant value

R = √(∑ H² +∑ V²)

R = √22.38² + 11.35²

R = 25.1 Nm

Magnitude of the balance mass

R = m * r

25.1 = m * 0.2

m = 125.5 Kg

Position of the balance mass

θ^* = 26.89

Position of balanced mass from first load is 180 + 26.89 = 206.9

Section 3: Evaluate the effects of misalignment of shafts and the measures that are taken to prevent problems from occurring.

Effects of misalignment of shafts

Shaft misalignment can cause excessive vibration in the system. Due to shaft misalignment, noise can be produces while machine in operation due to striking of parts of machines. Equipment life is reduced as frequent maintenance is required. Product quality highly reduced due to ineffective operation due to shaft misalignment.

Prevention from misalignment of shafts

Regular inspection is necessary by visual inspection, dial indicators and laser inspection. Maintenance is required for shaft, bearing and other rotating mass. Balancing if rotating masses are necessary to prevent shaft from misalignment.

Section 4: Critically evaluate and justify the different choices of cam follower that could be selected to achieve a specified motion, explaining the advantages and disadvantages of each application.

Under cam with knife edge follower, follower edge is sharp as knife. Motion of follower is sliding and this system is generally used because of least contact between surfaces. But as the edge is so sharp therefore there is chances of wear on the cam surface. Under cam with roller follower, follower is round in shape so no wear on cam surface, this system is widely used in aircraft engines. This system required large space and high side thrust. Under cam with mushroom follower, follower surface is completely flat and side thrust is reduced. In this system the wear can be further reduced by offsetting the follower axis.

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