TASK 1 –

Mechanics of Materials Mechanics of materials provide us with valuable information regarding the behaviour of solid bodies under the influence of loads. You are required to define and explain the following concepts that relate to mechanics of materials. Use equations and diagrams to supplement your answer:

a) Strain and how to measure it experimentally

b) Young’s Modulus of Elasticity

c) The Stress resulting from different types of loading such as Tension/Compression, Shear, Bending and Torsion

a) Strain and how to measure it experimentally:

Strain:

Strain typically denotes toward the distortion of a substantial under strain. It's a quantity of how much an entity is stretched or compressed compared to its original size. There are different types of strain, including linear strain, shear strain, and volumetric strain, each describing different ways in which a material can deform under stress.

We have up till now concentrated on the tension inside structural components. An object deforms when stress is applied to it. When you tug on a rubber band, it stretches and becomes longer. An object's displacement is a measure of its stretching, and its stress is the ratio of it stretching to its initial length. Seen as a percentage of elongation, strain represents the amount that an object's grows or contracts as it is loaded.

A system may encounter two different kinds of strain in addition to stress:

1. Normal Strain

2. Shear Strain.

A power crops a standard pressure once it turns perpendicular—called as "normal"—towards an object's shallow. A power products a fleece strain after it acts similar to an item's external.

Let's think about a rod that is in uniaxial tension. The usual rinsing is the proportion of this slight distortion to the bar's initial span. Under this stress, the rod extends to a higher length.

a. Rod

b. Uniaxially Loaded Rod

Strain

ε = δ / L

The sum that an article expands or contracts in answer to an practical weight is unhurrid untilllessly as draining. The Greek symbol epsilon represents normal stress, which is the lengthening of a substance in retort to a regular tension (i.e., erect to a external). A compression value is represented by a negative value, and a tensile strain by an optimistic value. Shear strain, represented by the Greek letter gamma, happens when an item deforms in response to a shear stress, or when it is perpendicular to its exterior.

Mechanical Behavior of Materials

It's obvious that strain and stress are related. By measuring the amount of stress needed to stretch an object, we may objectively ascertain the correlation between stress and strain, which is governed by a constitutive rule. Tensile testing might be used for this measurement. In the most basic

scenario, a component expands in direct proportion to the amount of pull applied to it; for low

strains values, this connection is linear in nature. Hooke's Law describes the elastic, linear

relationship between stresses and strains. When stress and strain are plotted, the graph will be

straight for small varieties, and the gradient of the line represents Young's elasticity modulus of

the substance being studied. Since 1 kPa for Jello to 100 GPa, its value might fluctuate

significantly. for steel. The line section of the stress-strain illustration lone appears for actual

minor stresses (<0.1%) for the majority of engineered materials. Only linearly flexible (i.e.,

according to Hooke's law) and non-isotropic (behaving the similar regardless of pulling way)

materials will be covered in this course.

A straightforward connection for the displacement of a material may be obtained with ease using

Hooke's law variable our descriptions of tension and straining.

σ = Eε

P

A

= E δ

L

δ = PL / EA

δ_total = ∑_i P_i L_i / (E_i A_i)

This test seems to have sense automatically: apply greater load to get a greater distortion; smear

the exact similar force to a stronger before stiffer physical to become a smaller distortion. We can

divide up these many loadings using the notion of superposition if a structure varies material, form,

or loading at various sites.

Generalized Hooke's Law

The previous session introduced us to Hooke's law, which explains the association amid pressure and straining. However, we have only discussed stress or strain in a single direction at this opinion, and we have lone examined a actual basic interpretation of Hooke's law. For consistent, isotropic, or and elastic substances subjected to stresses on several axes, we will discuss a broader Hooke's law in this lesson.

To begin with, many substances tend to deform in each of the three orthogonal dimensions even when pulled (or pushed) in a single direction. Now let's return to the initial strain example. The fact that an object compresses horizontally in the longitudinal directions when pulled axially will be taken into consideration this time:

As a result, it shrinks in the y and z dimensions when pulled in the x direction. The Poisson's relation, which is represented by the Grecian communication nu and has the following definition,

ν = - ε_y / ε_x = - ε_z / ε_x

is a characteristic of a substance:

picture above:

Being a ratio of two varieties, Poisson's ratio is devoid of units since strain has no dimensions. A

material feature is Poisson's ratio. A value of 0.5 to -1 can be assigned to Poisson's relation. A

Poisson's proportion of about 0.3 is found in most engineering supplies, such as steel or aluminum,

while 0.5 is found in rubber, which is why rubber is called "indestructible". Its volume won't vary

since it is incompressible, which basically implies any degree of compression in one dimension

will cause it to expand equally in both dimensions.

A fascinating body of work was done in the past ten years to construct structural substances that

use geometry and elasticity variabilities—a subject we'll touch on quickly in a later talk—to

produce auxetic materials—materials having an adverse Poisson's proportion. In terms of physics,

this indicates that the material extends in all directions when pulled in one direction, and vice

versa:

We now have a formula that connects stress in the y or z direction with draining in the z way thanks to Poisson's ratio. By Hooke's law, we may then connect this to stress. It's crucial to

remember that tugging on an item in a single motion only results in stress in that direction; pulling

in every single direction generates strain. Hence, sigma_x = 0 = sigma_y? In the context of the strain

in the x track, lease's inscribe available the strain that occur in the y and z directions.

εx = σx / E

εy = εz = -ν σx / E

Consider that we have only discussed uniaxial deformation up to this point. In actuality, buildings

have the capacity to be loaded concurrently in several directions, leading to stress across these

directions. To have a better understanding of this, picture a little "cube" of substantial inside of an

article. The stresses that are typical for every surface of that cubes could look like this:

As with normal straining in the y and z directions, placing a weight in the x track results in standard pressure in this way as well. Furthermore, tension in ace course results in draining in all 3 guidelines, as we nowadays understand. Therefore, we now apply this concept to Hooke's law and record the reckonings aimed at the rinsing in all directions as

ε_x = σ_x/E - ν σ_y/E - ν σ_z/E

ε_y = -ν σ_x/E + σ_y/E - ν σ_z/E

ε_z = -ν σ_x/E - ν σ_y/E + σ_z/E

The straining in apiece way (or apiece constituent of tension) be contingent on the regular pressure in that way and the Poisson's proportion areas the straining in the other two guidelines. These comparisons appear more complicated than they actually are. We now possess the formulas for the three orthogonal directions in which an item will undergo shape change. A substance will change its volume if it shifts form in all three directions. You can find a straightforward measurement of this volume change by summing the three typical components of strain:

e = e_x + e_y + e_z

With the volume change, or expansion, equation now expressed in footings of standard straining, we container now recast it in values of usual pressures.

e = (σ_x + σ_y + σ_z) / E - 2ν(σ_x + σ_y + σ_z) / E

e = (1 - 2ν) / E (σ_x + σ_y + σ_z)

Hydrostatic stress is a relatively shared kind of pressure that results in dilatation. All that is happening to the material is an equal pressure. Given that it is operating similarly, this implies:

p = σ_x = σ_y = σ_z

Hooke's Rule in Shave

The correlations amid normal strain then normal stress were developed in the section before. We need to discuss shear now. Let us return to that hypothetical cube of matter. Not only can external forces produce stresses parallel to each cube face, but they can also produce stresses normal to each cube surface. Furthermore, tensions similar to a irritated unit are known to be fleece pressures.

Tauxy = tauxy is the equation that the cube must satisfy to be in equilibrium; otherwise, it would spin.

Six stresses (σ_x, σ_y, σ_z, τ_xy, τ_yz, and τ_zx) therefore define the stress condition inside a same, isotropous, stretchy substance.

Consider the cube's right face as an example. Ordinary stresses in the x-axis correspond to the stresses on this face. This surface has two parallel stresses: one that points in the direction of y (denoted τ_xy) and the other that points in the direction of the z (denoted τ_xz). Tauxy = tauxy is the equation that the cube must satisfy to be in equilibrium; otherwise, it would spin. Six stresses (σ_x, σ_y, σ_z, τ_xy, τ_yz, and τ_zx) therefore define the stress condition inside a same, isotropous, stretchy substance.

In what way then fix these shear straining and strains tell toward one other? Hooke's law in fleece bears a striking resemblance to the formula we observed for typical stress and strain:

τ = Gγ

The fleece modulus of a physical is defined as this equation, which represents the association amid shear pressure besides shear rinsing. The equation container remain rewritten more clearly by means of its machineries, x, y, and z, but that is its general form. We shall then have the

generalised Hooke's law for elastic, homogeneous, isotropic resources.

ε_x = σ_x / E - ν σ_y / E - ν σ_z / E

ε_y = -ν σ_x / E + σ_y / E - ν σ_z / E

ε_z = -ν σ_x / E - ν σ_y / E + σ_z / E

γ_xy = T_xy / G γ_yz = T_yz / G γ_zx = T_zx / G

Six elements of tension and strain and three physical characteristics make up our generalised

Hooke's law. An obvious question to ask is: What is the relationship between these three materials

properties? The following formula provides this connection.

G = E / 2(1+ν)

b) Young's Modulus of Elasticity:

Alternately referred to as the coefficient of elasticity, the modulus of Young's is the structural

ability of a material to tolerate extension or compression relative to its length.

E or Y are used to denote it.

A measure of the physical properties of elastic linear solids, such as wires, rods, and the like, is

called Young's modulus (also called elastic modulus or tension modulus). Offspring's modulus is

the greatest often rummage-sale number to quantify a fabric's flexible capabilities; extra values

that are used include its bulk modulus and shearing modulus. The rationale for this is that it

provides insight into a material's tensile elasticity, or its capacity to undergo deformation along a

plane.

The link between strain (in proportion deformation) and stresses (due to the per unit area) is

described by Young's modulus. in a thing). The British physicist Thomas Young is honored by the

name Young's modulus. An application of a specific load causes a solid object to break down. If

an elastic object is released from pressure, the body returns to its initial shape. Yonder a sure grade

of misrepresentation, numerous substances remain non uniform and flexible. Lone linear y flexible

resources are topic to the continuous Young's modulus.

Young’s Modulus chMethod:

E = σ/ε

Young's Modulus Calculation Using Different Quantities:

E ≡ σ/ε = F/A ÷ ΔL/L = FL/(AΔL)

The Young's Modulus Formulary's Notations

Young's modulus in Pa is E.

In Helper, C remains the uniaxial stress.

ε is the proportionate deformation or strain.

The force that an object under strain exerts is denoted by F.

The true cross-section part is type A.

Units And Dimension:

SI unit Pa

Imperial unit psi

Dimension ML^-1T^-2

For such a material, the ratio of stress to strain—which represents the material's stress—gives the

Young's modulus. Below is the relation.

Young's modulus = Stress/Strain

where E, expressed in N/m2, is the material's Young's modulus.

σ signifies the pressure used on the physical.

ε represents the strain that corresponds to applied stress.

The stiffness of the body can be calculated using a material's Young's modulus value. This is so

that we may learn more about the body's resistance to deformation when force is applied.

Below are the values of Young's modulus for various materials:

Glass: 65 Steel: 200

Wood: thirteen

Polystyrene (plastic): three

Young’s Modulus Factors

Steel has a lower tendency to buckle under pressure than wood or polypropylene, which leads us

to conclude that steel has a higher modulus of elasticity than these materials. A material's potential

for deformation in response to an applied force can also be ascertained using Young's modulus.

It's also important to remember that the body experiences more distortion in materials with lower

Young's Modulus values; in the case of clay and wood, for example, this deformation can vary

within the sample. A steel bar will deform uniformly throughout, whereas a clay sample will

deform more in one area than the other.

Consider figuring out how long a steel beam will alter. With an applied tension of 1.5N/^2, whose

original length was 200 m. The table above contains the steel's modulus of elasticity.

a) The amount of stress brought on by various loading scenarios, including bending,

torsion, shear, and tension/compression

1. Tightness/Density Stress:

Tightness stress is the stress that happens once a physical is strained or dragged apart, while

compression pressure happens once a physical is beaten or squeezed.

Equation: σ = F / A

anywhere σ is the tension or compression strain, F is the power, and A is the cross-section area.

2. Shave Pressure:

Shave pressure is the pressure that happens once a physical is exposed toward a force that

reasons the situation to distort through descending lengthways a plane.

Equation: τ = F / A

anywhere τ is the shave pressure, F is the power, and A is the cross-section part.

3. Bending Pressure:

Bending stress is the stress that happens once a material is exposed to a force that causes the situation to curve or curve.

Equation: σ = (M × y) / (I × W)

wherever σ is the winding pressure, M is the winding instant, y is the coldness after the neutral alliance, I is the second moment of area, and W is the width.

4. Torsion Stress:

Definition: Torsion stress is the pressure that happens once a physical is exposed to a winding power.

Equation: τ = (T × r) / (J × W)

where τ is the torsion pressure, T is the rotation, r is the range, J is the glacial instant of apathy, and W is the width.

These definitions and equations assume a homogeneous and isotropic material, and the stresses are calculated at a specific point or section.

Task 2

Your next task is to find the following required values a) Axial Loads are applied at indicated locations on a bronze bar fastened between a steel and an aluminum bar as shown in the below diagram. What should be the highest value of P that will:

a. Limit the total elongation to 3 mm

b. Limit the stress in the steel to 140 MPa

c. Limit the stress in the aluminum to 80 MPa

d. Limit the stress in the bronze to 120 MPa

The assembly is assumed to be suitably braced so buckling will not occur. Consider Est = 200 GPa, Eal = 70 GPa, and Ebr = 83 GPa.

b) The below diagram represents:

a. A 2.5 m diameter driving wheel at 80°C

b. A steel tire (90 mm wide and 12 mm thick) that just fits over the wheel at 25°C

The temperature difference will make the steel tire shrunk onto the wheel. If the deformation of the wheel caused by the pressure of the tire is neglected and the assembly cools to 25°C.; Find the thermal strain of the tire and the contact pressure between the tire and wheel. Consider α = 13 μm/(m·°C) and E = 250 GPa.

Answer:

a)

Considering permissible stress levels:

a) For steel, Pst=σstAstPst=σstAst

P = 140(480) = 67200N

P = 140(480) = 67.2kN

b) Copper: Pbr=σbrAbrPbr=σbrAbr

2P=120(650)=78000N

P = 39000N = 39kN

P = 39000N = 39kN

c) Aluminium: P=galAalPal=σalMfl

2P=80(320)=25600N

P=12800N=12.8kN

d) In accordance with permitted deformation:

bronze shortens, steel and aluminium lengthen.

δ=δst-δbr+δalδ=δst-δbr+δal P(1000)480/(200000)

= 3.-2P(2000)/650(83000) plus

2P(1500)/320(70000) P3=(196000-226975+322400) = 3

=(1/196000-2/226975+3/322400)P P=42733.52N=42.73kN

P = 12.8 kN is the least value of P to use.

b)

we can calculate the thermal strain and contact pressure as follows:

Thermal Strain:

ΔT = 80°C - 25°C = 55°C

α = 13 μm/(m·°C)

Thermal strain = α × ΔT = 13 × 10^(-6) × 55 = 715 × 10^(-6) (or 0.000715%)

There is compressive thermal strain because the tyre is shrinking onto the wheel.

Contact Pressure: Assume that at 25°C, the steel tire's inner diameter is the same as the wheel's diameter (2.5 m). The tire's inner diameter will drop as a result of the temperature differential.

New inner diameter = 2.5 m - (2.5 m × 0.000715) = 2.498225 m

The change in diameter is:

Δd = 2.5 m - 2.498225 m = 0.001775 m

The contact pressure can be calculated using the formula:

P = (E × Δd) / (2 × width)

where width = 90 mm = 0.09 m

P = (200 × 10^9 × 0.001775) / (2 × 0.09) = 197.22 MPa

As a result, the tyre has a compressive thermal strain of 0.0715% and a contact pressure of roughly 197.22 MPa with the wheel.

TASK 3

Beams & Theory of Bending Aircraft beam structures can be represented using free-body

diagrams in order to carry out certain calculations of the loads exerted on them.

a) Define Centroid, Neutral Axis and second moment of Area. Explain the Parallel Axis

Theorem and its iapplications

b) For the shape below find the second moment of area about the axis AA

a) Using the shear diagram in the below graph

a. Construct the bending moment diagram

a. Construct the load diagram

c. Provide the values at each change of Load points

d. Provide the values at each point of Zero Shear

e. Given that the section modulus of the beam is 3250 cm^3, determine the allowable bending

stress, while ignoring the weight of the beam

b) The below diagram represents a steel bandsaw plate wrapped around a pulley (R=160mm). The dimensions of the plate are: width = 20 mm and thickness = 2 mm.

a. Calculate the maximum flexural stress developed in the bandsaw

b. Find the minimum diameter pulley if the flexural stress is limited to 200 MPa Consider E = 180 GPa.

Answer:

a) Definitions:

- Centroid: A shape's or object's geometric center, or center of mass, is its centroid.

A key idea in geometry is the centroid, which has to do with a triangle. A triangle is a delimited figure consisting of three sides and three angles inside. A triangle can be divided into several categories according to its sides and angles, including

• scalene triangles.

• isosceles triangle

• equilateral triangle

• acute-angled triangle

• Irregularly shaped triangle

• A triangular shape

One crucial aspect of a triangle is its centroid. Let's go into further detail about the

definition, attributes, formula, and centroid of various geometric shapes.

Centroid Alternative Definition:

The object's center point is known as the centroid.

The point where the three triangle medians converge is known as the centroid of a triangle. It may also refer to the point of convergence of the three medians.

The median is the line that joins the midpoint of one side to the opposite vertex of the triangle. A 2:1 ratio is obtained by dividing the median by the centroid of the triangle. The average of all the triangle's vertices' x- and y-coordinate points can be used to determine it.

The Centroid Hypothesis:

The centroid of the triangle is situated two thirds of the way between the vertex and side midpoints, according to the centroid theorem.

Features of the centroid

As the center of gravity, the centroid has the following characteristics:

• it is the place where the medians concur

• it is the center of the object

• it should always reside inside the object.

- Neutral Axis: The neutral axis is a hypothetical line that runs through the centroid and is perpendicular to the bending plane, the location of the zero bending stress.

Neutral axis of a beam:

The neutral axis of a beam is the line that passes through its centeroidal depth and through which there is no longitudinal stress (compressive or tensile) or strain. The neutral axis is represented by a dotted line. The beam's upper half is under compressive stress from the neutral axis, while the lower neutral axis is under tensile stress. This means that given a beam exposed to the same basic bending moment from the neutral axis or centroidal axis, the moment of area of every construction with respect to the neutral line is always zero.

A Beam Neutral Axis: What Is It?

Think of a beam that is initially unstressed. Now let's look at direct tension across the surface of the beam. This direct tension results in some bending in the beam because it places compressive stress on the top fiber and tensile stress on the bottom fiber.

Neutral Axis for Various Cross Section Types

We are aware that the centroid axis, often known as the center of gravity, is where the neutral axis always passes. As a result, the neutral axis has zero total tension. Tell us about the various cross sectional varieties and their neutral axis.

Key Information about the Neutral Axis:

Let's review some important information regarding the neutral axis. These points are useful for identifying or locating the neutral axis of different forms.

- The neutral axis of a symmetrical construction is identical to the centroidal axis in all directions. This suggests that every applied load-generated stress has a zero point on the

neutral axis. But the unsymmetry that results from integrating steel into an RCC beam leads the neutral axis to deviate from the centroidal axis, or the center of gravity.

• In the unsymmetrical section, the neutral axis is calculated as per individual symmetrical member about th of e single centroid axis that we assume.

• That line, like in a complete member where the neutral axis is passing, creates a plane known as the neutral axis plane at which the position of the particle is unaffected by the app lied load or stresses.

- Second Moment of Area (I):

Second Moment of Area (I): Also known as the moment inertia, the second moment of ar ea describes how resistant a shape is to bending.

It is calculated by taking the square root of the distance from the neutral axis of each small area.

The area moment of inertia of a two-dimensional plane shape, sometimes called its second area moment or second moment of area, is a property that shows the distribution of its points along a given axis in the cross-sectional plane.

This property basically characterises the deflection of the planar shape under a load.

The area moment of inertia for an axis in a plane is usually denoted by the letter I.

When the axis is perpendicular to the plane, it can also be symbolised by the letter J.

Length to the power of four, or L4, is the dimension unit for the second area moment. A meter to the power of four, or m4, is the unit of measurement used by the International System of Units. Inches to the fourth power, or in4, can be obtained using the Imperial System of Units.

It is common to encounter this concept in the field of structural engineering.

In this instance, the area moment of inertia of the beam is stated to be the indicator of its flexural stiffness.

It's an important property that's used to measure how resistant a beam is to bending and how much deflection it would encounter.

In this situation, we have two cases to look at.

First, a simple method for determining or characterising a beam's resistance to bending is the planar second moment of area, where the force is perpendicular to the neutral axis at its location.

Secondly, the resistance of the beam can be determined using the polar second moment of area when the applied moment is parallel to the cross-section of the beam.

Area Moment of Inertia Determination:

The product of the body's mass and the square of the distance between the two lines determines the body's moment of inertia along any axis. It also determines the body's moment of inertia about a parallel axis that passes through its centre of mass.

D is the perpendicular distance between the two lines, with I = Ig + Md2 and M = Body Mass.

The sum of the moments of inertia around two perpendicular axes that coincide with a planar body’s perpendicular axis represents the moment of inertia for the body along that perpendicular axis which is located on the plane of the body.

It can be found in the formula IZ = Ix + Iy.

Area Moments of Inertia for a Few Typical Shapes

The area moments of inertia for various forms are listed below.

Shape/ Figure Description of the Axis Area Moments of Inertia

Rectangle Centroid along the Cartesian Axis [unreadable]

Square Centroid along the Cartesian Axis [unreadable]

Disk Centroid [unreadable]

Ellipse Centroid [unreadable]

Annulus Centroid [unreadable]

-Parallel Axis Theorem:

This theorem states that the second moment of area (Io) about the neutral axis plus the second moment of area (I) about any axis parallel to the neutral axis is equal to the product of the area (A) and the square of the distance between the two axes. In other words, I is Io plus Ad^2.

Parallel Axis Theorem:

What is the Parallel Axis Theorem?

The parallel axis theorem states that the body's moment of inertia about an axis that runs parallel to it and passes through its center is equal to the product of the body's mass times the square of the distance between the two axes.

- M is the body's mass

- I is the figure's second of inactivity

- Ic is the second of inactivity around the midpoint

- h2 is the square of the separation between the two axes

Origin of the Parallel Axis Theorem:

Let Ic be the moment of inertia of an axis passing through the center of mass (AB in the illustration), and let me represent the moment of inertia about axis A'B' at a distance of h. Consider a particle with mass m that is situated r from the center of gravity of the body. Hence, the distance from A'B' equals r + h. I = ∑ m (r + h)∑m (r2 + h2 + 2rh) = 2 I ∑mr2 + ∑mh2 + ∑2rh = I

I = 2h∑mr + Ic + h2∑m

Ic + Mh2 + 0 is my equal.

I am the same as Ic + Mh2.

Thus, the parallel axis theorem method is as follows.

Uses:

By disassembling complex shapes into their smaller parts and One can calculate those shapes' second moment of area by applying the Parallel Axis Theorem.

b) Second moment of area about axis AA:

We must compute the moment of inertia in order to determine the second moment of area (I) around axis AA. of each rectangular section and sum them up. Let's assume the dimensions are in meters:

I = (1/12) * b * h3 + (1/12) * (w - b) * (H - h)3

where b = 0.2 m, h = 0.3 m, w = 0.6 m, and H = 0.8 m

I ≈ 0.01333 m4

a) Shear diagram:

The provided shear diagram displays the shear force (V) fluctuation over the beam's length.

b) Bending moment diagram:

The shear diagram can be integrated to create the bending moment diagram. The area under the shear diagram up to that point is the bending moment (M) at any given position.

c) Load diagram:

The load diagram displays point loads as well as distributed loads. We can determine the locations where the load changes based on the shear diagram:

- Point A: 40 kN (beam start)

- Point B: 16 meters from the beginning, 24 kN

- Point C: 20 meters from the beginning, 36 kN

- Point D: the beam's terminus, 0 kN

d) Values at every load point change:

40 kN at Point A; 24 kN at Point B; and 36 kN at Point C

- Point D: 0 kN

e) Values at every zero shear point:

The locations where the shear force is 0 can be found on the shear diagram:

- Point B: 16 meters from the beginning, 24 kN

- Point D: the beam's terminus, 0 kN

Permissible bending stress: Using the following formula, we can determine the permitted bending stress (σ) given the section modulus (Z) of 3250 cm^3.

Where M is the maximal bending moment, σ = M / Z.

The greatest bending moment is found at Point C, as shown by the bending moment diagram:

M is an approximate value of 72 kN/m.

σ = (3250 cm^3) / (72 kN*m) ≈ 22.15 MPa

Please take note that the calculation makes simplified assumptions and does not account for the beam's weight. In actually, more variables would need to be taken into account, such as loading circumstances, safety considerations, and material qualities.

b)

a) Maximum flexural stress:

The bandsaw plate can be thought of as a curved beam with a pulley radius (R = 160 mm)-sized radius of curvature. The formula to compute the maximum flexural stress (σ) is σ = (E * t) / (2 * R).

Where M is the range (160 mm), t is the thickness (2 mm), and E is the modulus of elasticity (180 GPa).

σ = (2 * 160 mm) / (180 GPa * 2 mm) ≈ 225 MPa

b) Minimum pulley diameter:

We can reorganize the formula to solve for R in order to restrict flexural stress to 200 MPa:

(E * t) / (2 * σ) = R

Changing the values:

R = (2 * 200 MPa) / (180 GPa * 2 mm) ≈ 180 mm

The pulley with the smallest diameter would be D = 2 * R = 2 * 180 mm = 360 millimeter.

TASK 4

Torsion Drive shafts and drive couplings are all subjected to twisting and torsional loads.

An important use of shafts is to transmit power between parallel planes, as in aircraft engines or helicopter rotors.

a) Explain the Engineer's Theory of Torsion and state the assumptions made in this theory

b) A steel propeller shaft is to transmit 5.5 MW at 180 rpm without exceeding a shearing stress of 60 MPa or twisting through more than 1° in a length of 25 diameters. Calculate the proper diameter if G = 83 GPa.

c) A solid shaft with built-in ends is subject to a torque T as shown in below figure.

a) Prove that the resisting torques at the walls are T1 = Tb/L and T2 = Ta/L.

b. How would these values be changed if the shaft were hollow?

d) The below diagram represents 2 shafts, one is hollow, and one is solid. The hollow is made of bronze with D = 7 mm and d = 5mm. the solid shaft goes inside the hollow one with diameter equal to the inner diameter of the hollow shaft (d=5mm). both shafts are rigidly fastened together at their ends.

a. Calculate the torque that can be applied to the composite shaft while limiting the shear stresses in the bronze to 55 MPa and in the steel to 83 MPa

Consider G = 40 GPa for the bronze and G = 83 GPa for the steel

Answer:

a) Explain the Engineer’s Theory of Torsion and state the assumptions made in this theory:

Torsion is the twisting of an item caused by a torque applied to it in the study of solid mechanics. Pascal (Pa), or newtons per square meter, is used to express torsion. The shear stress that results is perpendicular to the radius because certain parts of the object are perpendicular to the torque axis.

Warping occurs in non-circular cross-sections as a result of twisting. In this Physics article, we will learn the derivation of torsion equation and its assumptions.

What is Torsion?

Torsion refers to the twisting or rotation of an object around its axis. After a power is practical to one end of an thing while the other end remains fixed, it creates a twisting motion. This twisting force is called torsion. It's similar to when you twist a towel to wring out water.

Torsion Units:

Torsion is measured in pounds per square inch (psi) or newtons per square meter (N/m2).

Assumptions of Torsion Equation:

Some of the assumptions are as follows:

• The substance is uniform/ elastic.

• Hooke's law should apply throughout the entire text.

• There should be a linear relationship in the material between shear stress and strain.

• A plane cross-sectional area is what the cross-sectional area should have.

• For the circular segment to work, it must be round.

• The material's diameters should all rotate at the same angle.

• The elastic limit of the material must not be exceeded by stress.

An essential idea in mechanical engineering is the Engineer's Theory of Torsion, or Torsion

Theory.

that describes the twisting of shafts and beams under torsional loads. The theory assumes:

1. The shaft is straight and circular in cross-section.

2. The material is homogeneous and isotropic.

3. The twist is uniform along the length of the shaft.

4The linear distribution of stress is directly proportional to the distance from the axis of rotation.

5. The shaft is free from stress concentrations and irregularities.

According to this theory, the torsional shear stress (τ) and angular twist (θ) are related by the formula:

τ = (G * φ) / L

where G is the modulus of inflexibility, φ is the angular rotation in rads, and L is the distance of the choke.

Rearranging the formula to solve for diameter (D = 2r), we get:

D = √(16T / (πτ))

where T is the rotation, r is the range of the tube, and J is the glacial instant of inertia.

D ≈ 0.355 m or 355 mm

c)

a) To prove that the resisting torques at the walls are T1 = Tb/L and T2 = Ta/L.

The idea of torque equilibrium can be applied. There can be no torque acting on the shaft if it is in static equilibrium.

Let's consider a section of the shaft of length L, with torques Ta and Tb applied at the ends. The resisting torques at the walls can be represented by T1 and T2.

Since the shaft is in equilibrium, we can write:

Ta + Tb = T1 + T2

Since the shaft is solid and has built-in ends, the resisting torques are proportional to the length of the section. Let's assume the resisting torque per unit length is k.

T1 = k(L-b)

T2 = kb

Substituting these expressions into the equilibrium equation, we get:

Ta + Tb = k(L-b) + kb

Simplifying and rearranging, we get:

T1 = Tb/L

T2 = Ta/L

b) How would these values be changed if the shaft were hollow?

The altered moment of inertia and polar moment of inertia would have an impact on the resisting torques if the shaft were hollow. The formulae for T1 and T2 would remain the same, but the values would be different due to the changed geometry.

For a hollow shaft, the polar moment of inertia (J) would be reduced, which would increase the torsional shear stress (τ) for a given torque (T). To maintain the same stress, the diameter of the hollow shaft would need to be increased. The exact calculation would depend on the specific dimensions and geometry of the hollow shaft.

1. First, we essential to discover the polar moment of inertia (J) of the composite shaft.

d) The two shafts shown in the diagram below are solid and hollow, respectively. Bronze makes up the hollow, which has D = 7 mm and d = 5 mm. The hollow shaft's inner diameter (d = 5 mm) is the same as the diameter of the solid shaft as it enters the hollow.

At their ends, the two shafts are firmly secured to one another.

a. Determine the maximum torque that may be given to the composite shaft (a) while keeping the shear stresses in the steel and bronze at 83 MPa and 55 MPa, respectively.

Think about G = 83 GPa for the steel and G = 40 GPa for the bronze.

For the hollow bronze shaft:

J_b = (π/32) * (D^4 - d^4) = (π/32) * (7^4 - 5^4) = 0.000246 m^4

For the solid steel shaft:

J_s = (π/32) * d^4 = (π/32) * 5^4 = 0.000096 m^4

The total polar instant of apathy (J) of the compound shaft is the amount of the polar instants of apathy of the individual shafts:

J = J_b + J_s = 0.000246 + 0.000096 = 0.000342 m^4

2. Next, we need to find the rotation (T) that container remain applied to the compound trough.

The torque remains incomplete by the fleece pressure in the figure (τ_b) and the steel (τ_s). We can calculate the torque using the following formula:

T = (J * τ) / r

anywhere r is the range of the shaft.

For the bronze:

T_b = (J * τ_b) / (D/2) = (0.000342 * 55) / (7/2) = 1.33 Nm

For the steel:

T_s = (J * τ_s) / (d/2) = (0.000342 * 83) / (5/2) = 2.33 Nm

The smaller of the two figures represents the torque that can be applied to the composite shaft:

Min(T_b, T_s) = 1.33 Nm for T.

As a result, 1.33 Nm of torque may be given to the composite shaft while keeping the shear stresses in the steel and bronze at 83 MPa and 55 MPa, respectively.

Note: The units of the torque are in Nm (Newton-meters).

TASK 5

Dynamics

The laws of motion define the relationship between forces and how they act on certain bodies that result in uniformly accelerated motion. Since the wing can be regarded as a static structure, we will look into the more dynamic components and systems on the aircraft.

a) Define and explain the following concepts that relate Newton’s Law of motion. Use diagrams to support your answer.

i.Moment of Inertia and its calculation

ii.Radius of Gyration and its calcuation

b) A Large train accelerates from rest slowly. The angular acceleration of the wheels is 0.25 rad/s2 and the radius is 0.35 m. The wheels complete 200 revolutions (assume no slippage):

a Calculate how far the train has moved down the track

b Find the final angular velocity of the wheels and the linear velocity of the train

c) Two boxes are pulled by a force F as shown in below figure. If the friction assumed to be zero and the box X has an acceleration of 4m/s2, Calculate the acceleration of the box Y.

y=5kg

x=2kg

F=30N

d) While stopping at the side of the road, a police car spots a speeding motorist travelling at 160 /h. An immediate chase start, with the police car accelerating at a constant

2.5 m/s2.

i. Find the time required for the police car to intercept the speeder

ii. Calculate the distance travelled by the police car before the interception

iii. What was the velocity of the police car at the interception

e) The time taken for an object to move from A to D along the red path in 30 minutes and

50 seconds.

i. Calculate the average speed of the object in m/s

ii. Calculate the average velocity of the object in m/s

iii. What was the velocity of the object at the interception

Answer:

Newton's Laws of motion:

The cornerstone of classical mechanics are Newton's laws of motion, three propositions initially put out by English physicist and mathematician Isaac Newton that characterize the relationships between the forces acting on a thing and its motion.

Newton's first law, known as the inertia law,

Example:

Newton's laws of motion and basketball

The ball always travels in an arc when a basketball player makes a jump shot. Because of Isaac

Newton's principles of motion, the ball travels along this route.

Galileo Galilei first proposed the law of inertia for horizontal motion on Earth, and René Descartes later generalized it. Despite being the foundation of classical mechanics and its starting point, the idea of inertia is not immediately apparent to the untrained eye. In both common experience and Aristotelian mechanics, non-pushed objects have a tendency to come to rest. Galileo used his experiments with balls rolling down inclined planes to derive the law of inertia.

Galileo's primary scientific objective was to explain how it is possible that, even if Earth is actually rotating on its axis and orbiting the Sun, we are not aware of this motion. To do this, he needed to understand the concept of inertia. The solution lies in the principle of inertia, which states that even though we are moving with Earth and have a natural propensity to continue moving, Earth appears to be at rest to us. As a result, the principle of inertia was previously a major source of disagreement among scientists rather than a statement of the obvious.

The fact that the motion of the Earth's surface is not uniform motion in a straight line allowed for the correct accounting of the minor deviations from this image by the time Newton had worked out all the specifics (the effects of rotating motion are explained below). The frequent

observation that non-pushed bodies tend to come to rest is explained by the Newtonian formulation as the result of unbalanced forces acting on them, such as air resistance and friction.

F = ma is Newton's second law:

F = ma is Newton's second law, which provides a quantitative explanation of the modifications that a force can bring about in a body's motion. It asserts that the force applied to a body has an equal and opposite effect on the temporal rate at which its momentum changes. A body's momentum is equal to the product of its mass and velocity. Similar to velocity, momentum has a magnitude and a direction. When a force is applied to a body, the momentum's direction, magnitude, or both can be altered. Among the most significant laws in all of physics is Newton's second law.

A body with a constant mass m can be expressed as follows: F = ma, where an is the acceleration and F is the force. A body accelerates according to the equation if there is a net force acting on it. On the other hand, there is no net force applied on a body if it is not propelled.

Newton's third law, known as the action-reaction law:

According to Newton's third law, forces applied by two bodies in contact with one another must be equal in magnitude and directed in the opposite direction. The law of action and reaction is another name for the third law. This law applies to bodies in uniform or rapid motion and is useful in the analysis of situations involving static equilibrium, in which all forces are balanced. The forces it talks about are not just accounting tricks; they are actual forces. A book laying on a table, for instance, exerts downward force equal to the weight of the book on the table.

The third law states that the book is subject to an equal and opposing force from the table. This force is produced when the book's weight slightly distorts the table, causing it to push back on the book like a coil spring.

According to the second law, a body experiences accelerated motion if there is a net force acting on it. A body is considered to be in equilibrium if there is no net force acting on it, either because there are no forces at all or because all forces are perfectly balanced by opposing forces. In this scenario, the body does not accelerate. On the other hand, it can be assumed that a body experiencing no acceleration has no net force acting upon it.

Impact of Newton's laws:

Known as Principia, this 1687 masterwork Philosophiae Naturalis Principia Mathematica is where Newton originally introduced his laws. Nicolaus Copernicus proposed in 1543 that the Sun, not Earth, may be the center of the universe. In the years that followed, Galileo, Johannes Kepler, and Descartes established the groundwork for a brand-new science that would explain the mechanisms of a heliocentric cosmos and supersede the Aristotelian worldview that the ancient Greeks had left behind. Newton invented that new science in the Principia. Though he was successful in explaining why planet orbits are ellipses rather than circles, it turned out that he explained much more when he created his three laws. The term "scientific revolution" refers to the sequence of events from Copernicus to Newton.

The most fundamental rules of physics were superseded by quantum mechanics and relativity in the 20th century, replacing Newton's laws. Newton's laws still accurately describe the natural world, with the exception of extremely small objects like electrons and objects travelling very close to the speed of light. For larger bodies or slower-moving bodies, Newton's rules apply, as does relativity and quantum mechanics.

What is the significance of Newton's laws of motion?

Since they provide the basis of one of the primary fields of physics, classical mechanics, Newton's laws of motion are significant. The science of mechanics examines how forces cause objects to move or remain still.

a)

The concept of Moment of Inertia (I):

It pertains to the ability of an item to withstand modifications in its rotating motion. The calculation is based on the mass distribution inside the item and is oriented around a certain rotational axis.

Formula:

The formula is I = Σ (m_i * r_i^2),

where m_i is the particle's mass and r_i is its distance from the axis of rotation. The formula takes into account all of the particles in the object.

Diagram:

- Imagine a wheel with mass distributed evenly along its circumference.

-Summing the product of each mass element (m_i) and its squared distance from the axis of rotation (r_i) yields the moment of inertia (I).

ii. Radius of Gyration (k):

This quantity, which is connected to the moment of inertia, indicates how mass is distributed throughout an object. It shows the separation between the object's center of mass and rotational axis at which the object's moment of inertia remains constant.

Calculation:

The calculation is k = √(I / m),

where m is the object's total mass and I is its moment of inertia.

Diagram:

- Using the same wheel example, one would find the radius of gyration (k) by dividing the total mass (m) by the square root of the moment of inertia (I).

- This is the separation between the wheel's center of mass and rotational axis at which the wheel's moment of inertia remains constant.

Figure - A

Figure - B

b)

a) We may use the calculation s = θ * r to determine how far the train has traveled down the track.

where s is the distance traveled, r is the wheel's radius, and θ is the angular displacement (in radians).

Assumed:

θ = 200 revolutions * 2π radians/revolution = 400π radians

r = 0.35 m

s = 400π * 0.35 = 440π m ≈ 1385.4 m

b) The formula ω = α * t can be used to get the wheels' final angular velocity (ω).

anywhere α is the angular acceleration (0.25 rad/s^2) and t is the time. Since the train accelerates from rest, the initial angular velocity is 0. We can calculate the time (t) using the formula:

t = Δθ / α

where Δθ is the angular displacement (400π radians).

t = 400π / 0.25 = 1600π s ≈ 5026.5 s

ω = 0.25 * 5026.5 ≈ 1256.6 rad/s

The following formula can be used to determine the train's final linear velocity (v):

v = ω * r

v ≈ 1256.6 * 0.35 ≈ 439.3 m/s

c)

Since the friction is assumed to be zero, the force (F) applied to the two boxes remains equivalent toward the mass of box X increased through the situation acceleration (4 m/s^2).

F is equal to m * a * X.

The formula for calculating the acceleration of box Y (a_Y) is a_Y = F / m_Y, where m_Y is the

mass of box Y. Because there is no difference in the force (F) between the two boxes, box Y's acceleration will be inversely proportional to its mass.

a_Y = (m_X / m_Y) * a_X

Note: The exact value of a_Y depends on the ratio of the masses of the two boxes.

d]

i.We can use the following calculation to determine how long it will take the police car to catch the speeder:

t = (v_speeder - v_police) / a_police

where v_speeder is the initial velocity of the speeder (160 km/h = 44.44 m/s), v_police is the initial velocity of the forces carriage (0 m/s, since it's starting from rest), and a_police is the acceleration of the police car (2.5 m/s^2).

t = (44.44 - 0) / 2.5 = 17.776 s ≈ 17.8 s

ii. Using the formula

s = ut + (1/2)at^2,

we can get the police car's pre-interception coldness: s is the detachment, u is the early speed (0 m/s), t is the duration (17.8 s), and an is the acceleration of the police car (2.5 m/s^2).

s = 0(17.8) + (1/2)(2.5)(17.8)^2 ≈ 445.5 m

iii. We can use the following formula to determine the police car's velocity at the intersection:

v = u + at

where an is the acceleration (2.5 m/s^2), t is the time, v is the end velocity, and u is the beginning velocity (0 m/s).

v = 0 + 2.5(17.8) ≈ 44.5 m/s

e]

i. The formula average speed = total distance / total time can be used to determine the object's average speed.

The total distance is the sum of the distances along each segment of the red path:

total distance = 100 + 200 + 150 + 50 = 500 m

The total time is given as 30 minutes and 50 seconds, which is equivalent to:

total time = 1850 s

average speed = 500 / 1850 ≈ 0.27 m/s

ii. To calculate the average velocity, we need to consider the displacement (the shortest distance

between the starting and ending points) rather than the total distance. The displacement is:

displacement = 100 + 150 - 200 = 50 m

average velocity = displacement / total time

= 50 / 1850

≈ 0.027 m/s

Note: The regular speed is much lower than the regular haste because the object is moving in a non-straight path, resulting in a smaller displacement compared to the total distance traveled.