Friday 3 September 2010

HOW TO FIND THE RESULTANT OF A FORCE SYSTEM?

For a force system i.e. a system of several forces acting on an object, it is possible to get the same effect on the object by the force system replacing it by a single force, that will be equivalent to the summation of the component forces acting on the object. The single force that will produce exactly the same effect on the object in stead of the force system is called Resultant of the force system.


We know that two forces acting on an object lying on a plane can be added together by
  • (i) Triangle's Law or
  • (ii) Parallelogram Law.
    For more than two vectors we use
  • (iii) Polygon Law of Force Addition.
  • (iv) Force Resolution Method.

The resultant of a force system is the Force which produces same effect as the combined forces of the force system would do. So if we replace all components of the force by the resultant force, then there will be no change in effect.

The Resultant of a force system is a vector addition of all the components of the force system. The magnitude as well as direction of a resultant can be measured through analytical method.

THE STEPS TO FIND A RESULTANT OF A CON-CURRENT FORCE SYSTEM:


STEP 1:

RESOLVE ALL THE COMPONENT FORCES ALONG X-AXIS AND Y-AXIS.


If a force F acts on an object at an angle ß with the positive X-axis, then its component along X-axis is F cosß, and that along Y-axis is F sinß.


STEP 2:

ADD ALL THE X-COMPONENTS OR HORIZONTAL COMPONENTS AND IT IS DENOTED BY ΣFx AND

ADD ALL THE Y-COMPONENTS OR VERTICAL COMPONENTS AND IT IS DENOTED BY ΣFy.


STEP 3:

FIND THE MAGNITUDE OF THE RESULTANT R


We know from Geometry that

R = √{(ΣFx)2 + (ΣFy)2}


STEP 4:

FIND THE DIRECTION (α) OF THE RESULTANT FORCE (R)


We know that

tan α = (ΣFy/ΣFx)

hence,

α = tan-1(ΣFy/ΣFx)

Thursday 26 August 2010

TWO DIMENSIONAL FORCE SYSTEM

Q: WHAT DO YOU UNDERSTAND BY THE TERM "FORCE"? WHAT IS THE EFFECT OF FORCE ON A PARTICLE AND A RIGID BODY? EXPLAIN WITH SUITABLE EXAMPLES.

Answer:

FORMAL DEFINITION:

A FORCE is that which can cause an object with mass to ACCELERATE. Force has both MAGNITUDE and DIRECTION, making it a vector quantity. According to Newton's second law, an object with constant mass will accelerate in proportion to the net force acting upon it and in INVERSE PROPORTION TO ITS MASS (M). An equivalent formulation is that the net force on an object is equal to the RATE OF CHANGE OF MOMENTUM it experiences. Forces acting on three-dimensional objects may also cause them to rotate or deform, or result in a change in pressure. The tendency of a force to cause angular acceleration about an axis is called TORQUE. Deformation and pressure are the result of stress forces within an object.


EXPLANATION OF MECHANICAL FORCE AND IT'S EFFECT ON A PARTICLE:

CHANGE IN POSITION:

To know force well, first we have to understand what do we mean by Change. What does it mean when we say the position of the body has been changed? Whenever we find the state of object becomes different than that of the same object before some time say Δt, then we say that there exists a change in the state of the object. Suppose the change occurs in the position of the body. But to find the initial position of a body, we need a co-ordinate system.

THE CAUSE OF CHANGE:

It has been seen that to induced a change or to make a change in the position of an object we must have to change the energy possess by the body. To transfer energy into the object we shall have to apply FORCE on the body. Therefore Force is the agency that makes a change in position of a body.

THE CONCLUSION: GALILEO'S LAW OF INERTIA OR NEWTON'S FIRST LAW OF MOTION.

So, if there is no force on an object the position of the object won't change with respect to time. It means if a body at rest would remain at rest and a body at uniform motion would remain in a steady motion. This law is known as Galileo's Law of Inertia or Newton's first law of motion.

  • 2 DIMENSIONAL FORCE
In physics, force is a vector quantity that is used to describe the interaction between two objects. In a two-dimensional system, forces can act in two different directions, which are typically labeled as the x-axis and the y-axis.










When dealing with two-dimensional force, it is essential to use vector addition to determine the net force acting on an object. The net force is the vector sum of all the forces acting on the object. The direction of the net force is determined by the angle of the resultant force vector.








To calculate the net force in two dimensions, we must first break down each force into its x and y components. The x-component of a force is the amount of force acting in the x-direction, and the y-component is the amount of force acting in the y-direction. Once we have the x and y components for each force, we can add them together to find the net force.









The magnitude of the net force can be calculated using the Pythagorean theorem, which states that the square of the hypotenuse of a right triangle is equal to the sum of the squares of the other two sides. In this case, the hypotenuse is the magnitude of the net force, and the other two sides are the x and y components of the net force.

In summary, when dealing with two-dimensional force, it is essential to use vector addition to determine the net force acting on an object. To calculate the net force, we must first break down each force into its x and y components and then add them together. The magnitude and direction of the net force can be determined using trigonometry.








  • ORTHOGONAL RESOLUTION OF A FORCE
Orthogonal resolution of a force is a technique used in physics to break down a force vector into its components along two orthogonal axes, typically the x and y axes. This technique is useful in analyzing the motion of an object under the influence of a force and can help determine the net force acting on an object.


To perform orthogonal resolution of a force, we first need to identify the angle that the force vector makes with respect to one of the axes, usually the x-axis. We can then use trigonometry to determine the components of the force vector along the x and y axes.

If the angle between the force vector and the x-axis is θ, the x-component of the force can be found using the equation Fx = F cos(θ), where F is the magnitude of the force. Similarly, the y-component of the force can be found using the equation Fy = F sin(θ).

Once we have the x and y components of the force, we can use vector addition to determine the net force acting on an object. The net force is the vector sum of all the forces acting on the object and can be found by adding the x and y components of each force separately.

Orthogonal resolution of a force is a powerful technique that is used in many areas of physics, including mechanics, electromagnetism, and fluid dynamics. By breaking down a force vector into its components, we can better understand the forces acting on an object and predict its motion under different conditions.

WHAT IS A FORCE SYSTEM? CAN WE CLASSIFY FORCE SYSTEMS?


ANSWER:
                         
A force system may be defined as a system where more than one force act on the body. It means that whenever multiple forces act on a body, we term the forces as a force system. We can further classify force system into different sub-categories depending upon the nature of forces and the point of application of the forces. Almost any system of known forces can be resolved into a single force called a resultant force or simply a Resultant. The resultant is a representative force which has the same effect on the body as the group of forces it replaces. (A couple is an exception to this) It, as one single force, can represent any number of forces and is very useful when resolving multiple groups of forces. It is important to note that for any given system of forces, there is only one resultant.


Different types of force system:

  • (i) COPLANAR FORCES:
If two or more forces rest on a plane, then they are called coplanar forces. There are many ways in which forces can be manipulated. It is often easier to work with a large, complicated system of forces by reducing it an ever decreasing number of smaller problems. This is called the "resolution" of forces or force systems. This is one way to simplify what may otherwise seem to be an impossible system of forces acting on a body. Certain systems of forces are easier to resolve than others. Coplanar force systems have all the forces acting in in one plane. They may be concurrent, parallel, non-concurrent or non-parallel. All of these systems can be resolved by using graphic statics or algebra.
  • (ii) CONCURRENT FORCES:
A concurrent coplanar force system is a system of two or more forces whose lines of action ALL intersect at a common point. However, all of the individual vectors might not actually be in contact with the common point. These are the most simple force systems to resolve with any one of many graphical or algebraic options. If the line of actions of two or more forces passes through a certain point simultaneously then they are called concurrent forces. concurrent forces may or may not be coplanar.
  • (iii) LIKE FORCES:
A parallel coplanar force system consists of two or more forces whose lines of action are ALL parallel. This is commonly the situation when simple beams are analyzed under gravity loads. These can be solved graphically, but are combined most easily using algebraic methods. If the lines of action of two or more forces are parallel to each other, they are called parallel forces and if their directions are same, then they are called LIKE FORCES.
  • (iv) UNLIKE FORCES:
If the parallel forces are such that their directions are opposite to each other, then they are termed as "UNLIKE FORCE".
  • (v) NON COPLANAR FORCES:
The last illustration is of a "non-concurrent and non-parallel system". This consists of a number of vectors that do not meet at a single point and none of them are parallel. These systems are essentially a jumble of forces and take considerable care to resolve.

on 20th November, 2010: ©subhankar

CENTROID OF COMPLEX GEOMETRIC FIGURES:




So in this articles, we are going to discuss the concepts of centroid for one dimensional as well as two dimensional objects. 

Let's first discuss about 1D and 2D objects, one by one, an 1D object is a line, practically a metallic rod will be considered as a linear, 1D object. Where as any thin plate of negligible thickness can be considered a 2D body. Suppose we have a thin metallic rectangular plate. If it is considered as a 2D rectangular area of b X h.

The concept of centroid has been developed on the basis of resultant of several areas. We know that an area can be represented as the cross product of two vectors, hence it is also an vector. Suppose we have an area A, in a cartesian 2D coordinate system. We just divide the area into n parts, and represent them as a1, a2, a3, ..... an.

Let the elemental areas are at a distance x1, x2, x3, ..... xn, from Y axis and y1, y2, y3, ...yn from X axis.

The total moments produced about Y axis will be equal to the summation of all the individual moments produced by n elemental areas. 


Now moment is a vector quantity and we know vectors of same kind can be added together, therefore, all the n moment vectors can be added to get a single value of Resultant Moment. 

We also know this resultant moment's position vector. Let the resultant moment passes through a point G. The point through which resultant moment passes through is called Center of the Area or Centroid.

How can we find out the point G, whose coordinates are (Xg,Yg)?

As moment of an area also obeys VARIGNON'S THEOREM OF MOMENT, then sum of all the moments produced by individual elemental areas will exactly be equal to the moment produced by the total area, i.e. the resultant of all those elemental areas. Now if all the areas are added to have the resultant area which will pass through the centroid G such that it produces a moment of XgA about Y axis and YgA about X axis.

But Varignon's theorem states us that, for a vector system, resultant vector produces the moment about a point, is exactly equal to the sum of all the moments produced by all elemental areas about the same point and in the same plane. Hence, we can write now that,

Sum(a1x1+ a2x2+ + +anxn) = AXg
we can use summation sign ∑ to represent these equations,
∑aixi = (∑ai)Xg
=> Xg = (∑aixi)/((∑ai)


Sum(a1y1+ a2y2+ + +anyn) = AYg
∑aiyi = (∑ai)Yg
=> Yg = (∑aiyi)/((∑ai)

Algorithm to find out the Centroid G(Xg, Yg) of a Complex Geometric Figure.


Step1:
Take a complex 2D figure like an Area or Lamina.


Step2:
Try to identify the basic figures whose algebraic combination produces our problem figure, whose centroid we shall find out.


Step3:
Choose a coordinate system, and make it as our frame of reference. All the distances and coordinate must be define with respect to our frame of reference.


Step4:
Compute the area (ai), coordinates of their own centroid Gi (xi, yi) for each and every elemental areas. While measuring the centroids, all the measurements will be based on according to our chosen Axes.


Step5:
If any particular area has to subtracted to get the complex figure, the area will be negative, where as any area addition will be positive area.


Step6:
If the Centroid of the complex figure be G(Xg,Yg)then,

=> Xg = (∑aixi)/((∑ai)

=> Yg = (∑aiyi)/((∑ai)


Here G1 is the centroid of the part one where G2 is the centroid of the circular area that has to be removed where as G3 is the centroid of the triangular area that has to be removed also.

If we are asked to find moment of inertia of an area, which is nothing but the "second moment of area" then we shall have to find the centroidal moment of inertia first. Then we shall transfer the Moment of Inertia to another axis ie we shall apply parallel axis theorem to transfer moment of inertia from one axis (here centroidal axis) to another parallel axis.

Wednesday 25 August 2010

INTELLIGENT OBJECTIVE QUESTIONS IN MECHANICS

1) A cantilever beam of square cross-section (100 mm X 100 mm) and length 2 m carries a concentrated load of 5 kN at its free end. What is the maximum normal bending stress at its mid-length cross-section?

(a) 10 N/mm²
(b) 20 N/mm²
(c) 30 N/mm²
(d) 40 N/mm²

2) A hollow shaft of outside diameter 40 mm and inside diameter 20 mm is to replaced by a solid shaft of 30 mm diameter. If the maximum shear stresses induced in the two shafts are to be equal, what is the ratio of the maximum resistible torque in the hollow to that of solid shaft?

(a) 10/9
(b) 20/9
(c) 30/9
(d) 40/9

(3) A cannonball is fired from a tower 80 m above the ground with a horizontal velocity of 100 m/s. Determine the horizontal distance at which the ball will hit the ground. (take g=10 m/s²)

(a) 400 m,
(b) 280 m,
(c) 200 m,
(d) 100 m.

(4) Water drops from a tap at the rate of four droplets per second. Determine the vertical separation between two consecutive drops after the lower drop attained a velocity of 4 m/s. Take g=10 m/s².

(a) 0.49 m
(b) 0.31 m
(c) 0.50 m
(d) 0.30 m

Wednesday 11 August 2010

NEED OF QUALITY ASSURANCE

NEED OF QUALITY ASSURANCE

Quality assurance can be a confusing realm for those who don't have any prior experience in this field. Many commonly asked questions by first timers include wanting to know exactly what quality assurance is and why they require such a service. Read on to find out the answers.

How Can We Define Quality Assurance?

Quality assurance is the process of ascertaining, through a systematic set of procedures, whether or not a product or service satisfies the customers' requirements. This is the simplest and most basic definition for quality assurance.

Why Do We Need Quality Assurance?

If your company has manufactured a certain product, it is necessary to get that product checked to verify that it conforms to the expectations and requirements of the customer. This process of checking and verifying a product's quality is known as quality assurance.

On the other hand, if you are the customer, you would definitely want to ascertain that the product that you are purchasing satisfies your requirements. If the product fails in some way to meet your expectations, you can provide feedback to the company who will then try to improve their quality standards for that particular product in order to improve their product performance.

Thus quality assurance works both ways, ensuring satisfied manufacturers as well as customers.

The mass industrialization period saw the widespread introduction of mass production and piecework, which created problems as workmen could now earn more money by the production of extra products, which in turn led to bad workmanship being passed on to the assembly lines. To counter bad workmanship, full time inspectors were introduced into the factory to identify, quarantine and ideally correct product quality failures. Quality control by inspection in the 1920s and 1930s led to the growth of quality inspection functions, separately organised from production and big enough to be headed by superintendents.

The systematic approach to quality started in industrial manufacture during the 1930s, mostly in the USA, when some attention was given to the cost of scrap and rework. With the impact of mass production, which was required during the Second World War, it became necessary to introduce a more appropriate form of quality control which can be identified as Statistical Quality Control, or SQC. Some of the initial work for SQC is credited to Walter A. Shewhart of Bell Labs, starting with his famous one-page memorandum of 1924.

SQC came about with the realization that quality cannot be fully inspected into an important batch of items. By extending the inspection phase and making inspection organizations more efficient, it provides inspectors with control tools such as sampling and control charts, even where 100 per cent inspection is not practicable. Standard statistical techniques allow the producer to sample and test a certain proportion of the products for quality to achieve the desired level of confidence in the quality of the entire batch or production run.


At Total Quality Assurance, all your quality needs are taken care of.
Visit click for more to find out more.

short url for this entry: http://goo.gl/SF07

WHAT IS QUALITY?

QUALITY SYSTEM & MANAGEMENT:
 
Definitions: QUALITY, QUALITY ASSURANCE, QUALITY CONTROL
 
QUALITY SYSTEM & MANAGEMENT: - quality

INTRODUCTION:

Quality of product signifies the "degree of its excellence and fitness for the purpose. Although some of the quality characteristics can be specified in quantitative terms, but no single characteristics can be used to measure quality of a product on an "absolute scale".

So, quality of a product means all those activities which are directed to maintaining and improving such as

(i) Setting of Quality Targets,
(ii) Appraisal of Conformance,
(iii) Adopting Corrective Action

where any deviation is noticed, analysed and planning for improvements in Quality.



DEFINITIONS OF QUALITY

1. General: Measure of excellence or state of being free from defects, deficiencies, and significant variations. ISO 8402-1986 standard defines quality as "the totality of features and characteristics of a product or service that bears its ability to satisfy stated or implied needs."

2. Manufacturing: Strict and consistent adherence to measurable and verifiable standards to achieve uniformity of output that satisfies specific customer or user requirements.

3. Objective: Measurable and verifiable aspect of a thing or phenomenon, expressed in numbers or quantities, such as lightness or heaviness, thickness or thinness, softness or hardness.

4. Subjective: Attribute, characteristic, or property of a thing or phenomenon that can be observed and interpreted, and may be approximated (quantified) but cannot be measured, such as beauty, feel, flavor, taste.



CHARACTERISTICS OF QUALITY:


(i) Quality in business, engineering and manufacturing has a pragmatic interpretation as the non-inferiority or superiority of something.

(ii) Quality is a perceptual, conditional and somewhat subjective attribute and may be understood differently by different people.

(iii) Consumers may focus on the specification quality of a product/service, or how it compares to competitors in the marketplace.

(iv) Producers might measure the conformance quality, or degree to which the product/service was produced correctly.

Numerous definitions and methodologies have been created to assist in managing the quality-affecting aspects of business operations.

Many different techniques and concepts have evolved to improve product or service quality.



QUALITY RELATED FUNCTIONS:

There are two common quality-related functions within a business.

(a) One is quality assurance which is the prevention of defects, such as by the deployment of a quality management system and preventative activities like FMEA.

(b) The other is quality control which is the detection of defects, most commonly associated with testing which takes place within a quality management system typically referred to as verification and validation.




BUSINESS DEFINITION OF QUALITY:


(a) The common element of the business definitions is that the quality of a product or service refers to the perception of the degree to which the product or service meets the customer's expectations.

(b) Quality has no specific meaning unless related to a specific function and/or object. Quality is a perceptual, conditional and somewhat subjective attribute.



QUALITY & PRODUCTIVITY:


In the manufacturing industry it is commonly stated that “Quality drives productivity.” Improved productivity is a source of greater revenues, employment opportunities and technological advances.






CHARACTERISTICS OF MODERN QUALITY MANAGEMENT SYSTEM


However, there is one characteristic of modern quality that is universal. In the past, when we tried to improve quality, typically defined as producing fewer defective parts, we did so at the expense of increased cost, increased task time, longer cycle time, etc. We could not get fewer defective parts and lower cost and shorter cycle times, and so on.

However, when modern quality techniques are applied correctly to business, engineering, manufacturing or assembly processes, all aspects of quality - customer satisfaction and fewer defects/errors and cycle time and task time/productivity and total cost, etc.- must all improve or, if one of these aspects does not improve, it must at least stay stable and not decline. So modern quality has the characteristic that it creates AND-based benefits, not OR-based benefits.

The most progressive view of quality is that it is defined entirely by the customer or end user and is based upon that person's evaluation of his or her entire customer experience. The customer experience is the aggregate of all the touch points that customers have with the company's product and services, and is by definition a combination of these. For example, any time one buys a product one forms an impression based on how it was sold, how it was delivered, how it performed, how well it was supported etc.

Quality Management Techniques:
_____________________________________
* Quality Management Systems
* Total Quality Management (TQM)
* Design of experiments
* Continuous improvement
* Six Sigma
* Statistical Process Control (SPC)
* Quality circles
* Requirements analysis
* Verification and Validation
* Zero Defects
* Theory of Constraints (TOC)
* Business Process Management (BPM)
* Business process re-engineering
* Capability Maturity Models

Quality Awards:

* Malcolm Baldrige National Quality Award
* EFQM
* Deming Prize

Monday 7 December 2009

ME-101 MOCK QUESTION PAPER; ENGINEERING MECHANICS

Vivekanand Institute of Technology & Science; Ghaziabad
PRE-SEMESTER EXAMINATION (odd SEMESTER 2009-10)
B.Tech…first Semester

Sub Name: Engineering Mechanics Max. Marks: 100
Sub Code: EME-102 Max. Time: 3: 00 Hr

(i) This paper is in three sections, section A carries 20 marks, section B carries 30 marks and section C carries 50 marks.
(ii) Attempt all the questions. Marks are indicated against each question
(iii) Assume missing data suitably if any.

Group A

Q.1 Answer the following questions as per the instructions 2x20=20
Choose the correct answer of the following questions:

(i) The magnitudes of the force of friction between two bodies, one lying above the another depends upon the roughness of the
(a) Upper body;                 (b) Lower body
(c) Both the bodies               (d) The body having more roughness

(ii)The moment of inertia of a circular section of diameter D about its centroidal axis is given by the expression
(a) π(D)4/16               (b) π(D)4/32
(c) π(D)4/64               (d) π(D)4/4

Fill in the blanks in the following questions:

(iii)The distance of the centroid of an equilateral triangle with each side(a) is …………. From any of the three sides.
(iv)Poisson’s ratio is defined as the ratio between ……………………. and
………………………… .
(v)If two forces of equal magnitudes P having an angle 2Ө between them,
then their resultant force will be equal to ________ .

Match the following columns for the following two parts:

(vi) Match the column I to an entry from the column II:
COLUMN – I COLUMN - II
(i) BMD of an UDL(a) stored strain energy per unit volume
(ii) Resilience is(b) brittle materials
(iii) Bulk Modulus (c) parabolic in nature
(iv) Yield Point (d) volumetric stress & strain
(e) Ductile materials
(f) Shear stress

(vii) Match the Following columns:
COLUMN – I COLUMN - II
(i) Square of side (b) (p) π b4 / 64
(ii) Equilateral Triangle of side (b)(q) b4 / 12
(iii) Circle of diameter (b)(r) b4/ 36
(iv) Isosceles right angle triangle of base (b) (s) b4/(32√3)
(e) Ductile materials
(f) Shear stress

Column II gives the value of Moment of Inertia Ixx about a centroidal axis.

Choose correct answer for the following parts:

(viii) Statement 1:
In the yielding zone, strain increases even if stress is decreased.

Statement 2:
At yielding point, the deformation becomes plastic by nature.

(i) Statement 1 is true, Statement 2 is true.
(ii) Statement 1 is true, Statement 2 is true and they are unrelated with
each other
(iii) Statement 1 is true, statement 2 is false.
(iv) Statement 1 is false, Statement 2 is false.

(ix) Statement 1:
For two identical mass, one of which lies on a horizontal plane and another is kept at an inclined plane having same co-efficient of friction, still frictional forces differs from each other.

Statement 2:
Frictional force always depends upon the magnitude of the normal force.

(i) Statement 1 is true, Statement 2 is true.
(ii) Statement 1 is true, Statement 2 is true and they are unrelated with
each other
(iii) Statement 1 is true, statement 2 is false.
(iv) Statement 1 is false, Statement 2 is false.

Choose the correct word/s.
(x) In a truss if the no.of joints (j) is related with no. of links (m) by the equation m > 2j- 3, then it is an example of (redundant/ deficient/ perfect) Truss.

SECTION-B

Q.2: Answer any three parts of the followings: 10X3=30
(a) Find the shear force and moment equation for the beam as shown in the figure. Also sketch SFD (shear force diagram) and BMD (bending moment diagram)

(b) Explain and prove “the parallel axis theorem of moment of inertia.”
Also find the centroid of the following composite area.


(c) Find the centroidal Moment of Inertia of the following shaded area.
(d) Two cylinders P and Q rest in a channel as shown in fig. below. The cylinder P has a diameter of 100 mm and weighs 200 kN, where as the cylinder Q has a diameter of 180 mm and weighs 500 kN. Find the support reactions at all the point of contact.

(e)
Two blocks A and B of weights 1 kN and 2 kN respectively are in equilibrium as Shown in the figure.
If the co-efficient of friction everywhere is 0.3, find the force P required to move the block B.

SECTION C:

(3) Answer any two parts of the following 5X2=10

(a) A simply supported beam of circular cross section of radius 4 cm having a length 2 m long in concentrated load of 5 kN (acting perpendicular to the axis of beam) at a point 0.75 m from one of the supports. Determine
(i) the maximum fiber stress (σb)max); (ii) the stress in a fiber located at a distance of 1 cm from the top of the beam at mid-span.

(b) Describe the procedure of Truss Analysis by Section method.

(c)A flywheel is making 180 rpm and after 20 second it is running at 120 rpm.
How many revolutions will it make and what time will elapse before it stops, if the retardation is constant.


(4) Answer any one part of the following: 1X10=10

(a) Explain the terms polar moment of inertia and radius of gyration.
Derive the area moment of inertias for a quarter circle of radius R.

(b) Analyse the following truss:



(5) Answer any three question;                           3X10=30

(a) Compare the stress – strain diagrams of a ductile material to that of a brittle Material. Also explain the term poisson's ratio and modulus of rigidity.

(b) Explain and prove the “Torsion Equation” (T/J)=(τ/r)=(Gθ/L). What is
called as Section Modulus?

(c) A cylinder of radius R, length L and total mass M is suspended vertically
from the floor, if the modulus of elasticity of the cylinder be E, find the total
deflection and maximum stress induced due to self weight.

(d) A cylinder of 200 mm diameter is subjected to a twisting moment of
250 kN-m, the length of the cylinders is 1 m, if the modulus of rigidity of the
cylinder material be 150 GPa, find the maximum shear stress induced in the
cylinder. Also find the total angular deformation.