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MECHANICS OF SOLIDS PRACTICE BOOK

ADITYA SILVER OAK INTITUTE OF TECHNOLOGY

MECHANICS OF SOLIDS-3130003 DEPARTMENT OF CIVIL ENGINEERING

Index Sr. No.

Content GTU Syllabus and Scheme Chapter wise Mark Distribution

1

Introduction

2

Fundamentals of Statics

3

Statically Determinant beams

4

Friction

5

Centroid and Moment of Inertia

6

Simple Stress and Strain

7

Stresses in beams

8

Torsion

9

Principle Stresses GTU Question Papers Objective type Questions

Remarks

GUJARAT TECHNOLOGICAL UNIVERSITY MECHANICS OF SOLIDS SUBJECT CODE: 3130003 B.E. 3RD SEMESTER Type of course: Applied Physics Prerequisite: System of units Laws of motion Basic idea of force Concept of centroid Fundamentals of stress, strain and their relationships

Rationale: Mechanics of Solids is conceptual applications of principles of mechanics in Engineering Teaching and Examination Scheme: Teaching Scheme Credits Examination Marks L T P C Theory Marks Practical Marks ESE PA (M) PA (V) PA (E) (I) PA ALA ESE OEP 4 0 2 6 70 20 10 20 10 20

Sr. No. 1

2

3

Topics Module 1 Introduction Definition of space, time, particle, rigid body, deformable body. Force, types of forces, Characteristics of a force, System of forces, Composition and resolution of forces. Fundamental Principles of mechanics: Principle of transmissibility, Principle of superposition, Law of gravitation, Law of parallelogram of forces. Fundamentals of Statics Coplanar concurrent and non-concurrent force system: Resultant, Equilibrant, Free body diagrams. Coplanar concurrent forces: Resultant of coplanar concurrent force system by analytical and graphical method, Law of triangle of forces, Law of polygon of forces, Equilibrium conditions for coplanar concurrent forces, Lami’s theorem. Application of statically determinate pin – jointed structures. Coplanar non-concurrent forces: Moments & couples, Characteristics of moment and couple, Equivalent couples, Force couple system, Varignon’s theorem, Resultant of non-concurrent forces by analytical method, Equilibrium conditions of coplanar non-concurrent force system, Application of these principles. Module 2 Applications of fundamentals of statics

Total Marks

150

Teaching Hrs.

Module Weightage

02

20

08

08

15

4

5

6

7

8

Statically determinate beams: Types of loads, Types of supports, Types of beams; Determination of support reactions, Relationship between loading, shear force & bending moment, Bending moment and shear force diagrams for beams subjected to only three types of loads :i) concentrated loads ii) uniformly distributed loads iii) couples and their combinations; Point of contraflexure, point & magnitude of maximum bending moment, maximum shear force. Module 3 Friction Theory of friction, Types of friction, Static and kinetic friction, Cone of friction, Angle of repose, Coefficient of friction, Laws of friction, Application of theory of friction: Friction on inclined plane, ladder friction, wedge friction, belt and rope friction. Centroid and moment of inertia Centroid: Centroid of lines, plane areas and volumes, Examples related to centroid of composite geometry, Pappus – Guldinus first and second theorems. Moment of inertia of planar cross-sections: Derivation of equation of moment of inertia of standard lamina using first principle, Parallel & perpendicular axes theorems, polar moment of inertia, radius of gyration of areas. Examples related to moment of inertia of composite geometry, Module 4 Simple stresses & strains Basics of stress and strain: 3-D state of stress (Concept only) Normal/axial stresses: Tensile & compressive Stresses :Shear and complementary shear Strains: Linear, shear, lateral, thermal and volumetric. Hooke’s law, Elastic Constants: Modulus of elasticity, Poisson’s ratio, Modulus of rigidity and bulk modulus and relations between them with derivation. Application of normal stress & strains: Homogeneous and composite bars having uniform & stepped sections subjected to axial loads and thermal loads, analysis of homogeneous prismatic bars under multidirectional stresses. Module 5 Stresses in Beams: Flexural stresses – Theory of simple bending, Assumptions, derivation of equation of bending, neutral axis, determination of bending stresses, section modulus of rectangular & circular (solid & hollow), I,T,Angle, channel sections Shear stresses – Derivation of formula, shear stress distribution across various beam sections like rectangular, circular, triangular, I, T, angle sections. Torsion: Derivation of equation of torsion, Assumptions, application of theory of torsion equation to solid & hollow circular shaft, torsional rigidity.

06

20

08

10

20

06

25

04

9

10

11

Principle stresses: Two dimensional system, stress at a point on a plane, principal stresses and principal planes, Mohr’s circle of stress, ellipse of stress and their applications Module –VI Physical & Mechanical properties of materials: (laboratory hours) Elastic, homogeneous, isotropic materials; Stress –Strain relationships for ductile and brittle materials, limits of elasticity and proportionality, yield limit, ultimate strength, strain hardening, proof stress, factor of safety, working stress, load factor, Properties related to axial, bending, and torsional & shear loading, Toughness, hardness, Ductility ,Brittleness Simple Machines: (laboratory hours) Basics of Machines, Definitions: Velocity ratio, mechanical advantage, efficiency, reversibility of machines. Law of Machines, Application of law of machine to simple machines such as levers, pulley and pulley blocks, wheel and differential axle, Single purchase, double purchase crab, screw jacks. Relevant problems.

04

05

50% (Practical) & 0% (Theory)

05

Course Outcome: After learning the course the students should be able to: 1. apply fundamental principles of mechanics & principles of equilibrium to simple and practical problems of engineering. 2. apply principles of statics to determine reactions & internal forces in statically determinate beams. 3. determine centroid and moment of inertia of a different geometrical shape and able to understand its importance. 4. know basics of friction and its importance through simple applications. 5. understand the different types of stresses and strains developed in the member subjected to axial, bending, shear, torsion & thermal loads. 6. know behaviour & properties of engineering materials. 7. know basics of simple machines and their working mechanism. List of Experiments: The students will have to solve atleast five examples and related theory from each topic as an assignment/tutorial. Students will have to perform following experiments in laboratory and prepare the laboratory manual. Mechanics of rigid body 1. 2. 3. 4. 5. 6. 7.

Equilibrium of coplanar concurrent forces Equilibrium of coplanar non-concurrent forces Equilibrium of coplanar parallel forces: Determination of reactions of simply supported beam Verification of principle of moment: Bell crank lever Determination of member force in a triangular truss Determination of coefficient of static friction using inclined plane Determination of parameters of machines (Any two)

(a) (b) (c) (d)

Wheel and differential axles Single purchase crab Double purchase crab System of pulleys

Mechanics of deformable body 8. Determination of hardness of metals: Brinell /Vicker/Rockwell hardness test 9. Determination of impact of metals: Izod/Charpy impact test 10. Determination of compression test on (a) Metals – mild steel and cast iron (b) Timber – along and parallel to the grains 11. Determination of tensile strength of metals 12. Determination of shear strength of metals Design based Problems (DP): (any two) 1. For a real industrial building having roof truss arrangement, (a) take photograph & identify type of truss, (b) draw sketch of truss with all geometrical dimension, cross sections details, type of joints, type of support conditions (c) prepare a model of truss (d) identify & determine types of load acts on it (d) determine support reactions & member forces due to dead load & live load only. 2. Take a case of the Mery-Go-Round used in the fun park. Draw its sketch showing radius of wheel, no of seats, capacity of each seats and other related information. Determine the amount of resultant produced at the centre of wheel during rest position, when (i) it is fully loaded (2) it is 30% loaded with symmetric arrangement. Draw support arrangement and determine support reactions. Also determine amount of torque required to start its operation. 3. Prepare working models for various types of beams with different shape of cross section, supporting conditions and study the effect of cross section on the deflection of beams. 4. Prepare working model of simple lifting machine using different types of pulley systems and calculate various parameters like load factor, velocity ratio, law of machine, efficiency of machine etc. Major Equipments: 1. 2. 3. 4. 5. 6. 7. 8. 9.

Force table Beam set up Truss set up Bell crank lever Friction set up Lifting machine Hardness testing machine Impact testing machine Universal testing machine with shear attachment

List of Open Source Software/learning website: www.nptel.iitm.ac.in/courses/ Active learning Assignments (AL) : Preparation of power-point slides, which include videos, animations, pictures, graphics for better understanding theory and practical work – The faculty will

allocate chapters/ parts of chapters to groups of students so that the entire syllabus to be covered. The power-point slides should be put up on the web-site of the College/ Institute, along with the names of the students of the group, the name of the faculty, Department and College on the first slide. The best three works should submit to GTU.

DEPARTMENT OF CIVIL ENGINEERING 3130003- Mechanics Of Solid

DEPARTMENT OF CIVIL ENGINEERING

CHAPTER 1: INTRODUCTION S.I. SYSTEM Fundamental units of S.I system Sr. No. 1 2 3 4

Physical quantities Length Mass Time Temperature

Unit Metre Kilogram Second Kelvin

symbol m Kg S K

Supplementary units of S.I. system Sr. No. Physical quantities 1 Plane angle

Unit

symbol Rad

Unit Newton Joule Watt Joule Square metre Cubic metre Pascal metre per second metre/second2 radian/second radian/second2 kilogram metre/second Newton metre Kilogram/metre3 Newton.metre Newton.metre

symbol N J, N.m W J, N.m m2 M3 Pa m/s m/s2 rad/s rad/s2 Kg.m/s N.m Kg/m3 N.m N.m

Radian Principal S.I. units

Sr. No. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Physical quantities Force Work Power Energy Area Volume Pressure Velocity/speed Acceleration Angular velocity Angular acceleration Momentum Torque Density Couple Moment

S.I. Prefixes Multiplication factor 1012 109 106 103 102 101 10-1 10-2 10-3 10-6 10-9 10-12

Prefix Tera Giga Mega kilo hecto deca deci centi milli micro nano pico

Symble T G M k h da d c m µ n p

DEPARTMENT OF CIVIL ENGINEERING

UNIT CONVERSION 1 m = 100 cm = 1000 mm 1 km = 1000 m 1 cm2 =100 mm2 1 m2 =106 mm2 1 kgf = 9.81 N = 10 N 1 kN = 103 N

1 Mpa = 1 N/mm2 1 Gpa =103 N/mm2 1 Pascal = 1 N/m2 1 degree =



180

radian

QUANTITY Scalar Quantity:

Vector Quantity:

“A Scalar Quantity is one which can be completely specified by its magnitude only”

“A vector Quantity is one which requires magnitude and direction both to completely specified it”

Examples: Length Density Volume Energy

Examples: Displacement Velocity

Mass Area Speed Work

Distance Temperature Time Moment of inertia

Angular velocity Moment

Force Angular displacement Momentum impulse

Weight Acceleration Angular acceleration

DEPARTMENT OF CIVIL ENGINEERING

CHAPTER-2: FUNDAMENTAL OF STATICS Force: “ An agent which produces or tends to produce, destroys or tends to destroy motion of body is called force” Unit: N, kN, Kg etc. Quantity : Vector Characteristics of Force:  Magnitude : Magnitude of force indicates the amount of force (expressed as N or kN) that will be exerted on another body  Direction: The direction in which it acts  Nature: The nature of force may be tensile or compressive  Point of Application : The point at which the force acts on the body is called point of application Types of Forces:  Contact Force  Body force  Point force and distributed force  External force and internal force  Action and Reaction  Friction force  Wind force  Hydrostatic force  Cohesion and Adhesion  Thermal force

System of Forces:  Coplanar Forces  Concurrent forces  Collinear forces  Coplanar concurrent forces  Coplanar non-concurrent forces  Non-coplanar concurrent forces  Non-coplanar non-concurrent forces  Like parallel forces  Unlike parallel forces  Spatial forces

Principle of Individual Forces 1. Principle of transmissibility: “If a force acts at a point on a rigid body, it may also be considered to act at any other point on its line of action, provided the point is rigidly connected with the body.” 2. Principle of Superposition of forces: “If two equal, opposite and collinear forces are added to or removed from the system of forces, there will be no change in the position of the body. This is known as principle of superposition of forces.”

COPLANAR CONCURRENT FORCES Resultant Force: If number of Forces acting simultaneously on a particle, it is possible to find out a single force which could replace them or produce the same effect as of all the given forces is called Resultant Force. Methods of Finding resultant :1. Parallelogram Law of Forces (For 2 Forces) 2. Triangle Law (For 2 Forces) 3. Lami’s theorem (For 3 forces) 4. Method of resolution (For more than 2 Forces)

DEPARTMENT OF CIVIL ENGINEERING

1

Parallelogram Law Of Forces “If two forces acting simultaneously on particle be represented in magnitude and direction by the two adjacent sides of a parallelogram, their resultant may be represented in magnitude and direction by the diagonal of the parallelogram, which passes through their point of intersection.”

R  P 2  Q 2  2PQ cos  Q sin  tan   P  Q cos  Where, R = Resultant force  = angle between P and Q

 = angle between P and R

2

Triangle law of forces “If two forces acting at a point presented in magnitude and direction by two sides of a triangle taken in order, then the third side of triangle will represent the resultant in magnitude and direction taken in opposite order.”

R  P 2  Q 2  2PQ cos  Where,   180   R =Resultant force

 = angle between P and Q  = angle between P and R Q  sin   R 

  sin 1 

3

Lami’s theorem “If three coplanar forces acting at a point be in equilibrium, then each force is proportional to the sine of angle between other two forces.”

P Q R   sin  sin  sin  Where, P, Q, R are given forces  = angle between Q and R

 = angle between P and R  = angle between P and Q

DEPARTMENT OF CIVIL ENGINEERING

4

Resolution of concurrent forces “The algebraic sum of the resolved parts of a number of forces in a given direction is equal to the resolved part of their resultant in the same direction.”

 H  P cos  P cos  P cos  P cos V  P sin   P sin   P sin   P sin  R    H    V  1

1

1

1

2

tan  

2

2

2

2

3

3

3

3

2

V H

Where, P1 , P2 , P3 , P4 are given forces

1 ,2 ,3 ,4 are angle of accordingly P1 , P2 , P3 , P4 forces from X-axes

R = Resultant of all forces  = angle of resultant with horizontal

4

4

4

4

DEPARTMENT OF CIVIL ENGINEERING

COPLANAR NON-CONCURRENT FORCES 1

2

3

Moment :M=PxX The moment of a force is equal to product of the force and Where, M= Moment the perpendicular distance of line of force from point about P = Force which the moment is required. X = Perpendicular distance between line of action of force and point about which Unit : KN.m, N.m moment is required Couple Two equal and Opposite parallel forces whose lines of action are not one form a couple Characteristic : Algebraic Sum of Forces forming a couple is zero  Couple will not give linear motion to the body, but Moment of Couple it will rotate the bod M Pd  Algebraic sum of moment of forces, forming a couple at any point is same and equal to the couple Where P = Force itself. d = arm of couple  A couple cannot be balanced b a single force, but can be balanced by couple of opposite nature. Varignon’s Theorem “The moment of a force about any point is equal to the sum of the moments of the components of the force about the same point.” Moment Mo = P1× d1+ P2× d2 Where, P = force P1= Component force 1 P2= Component force 2

DEPARTMENT OF CIVIL ENGINEERING Example - Law of Parallelogram Determine magnitude and direction of the resultant of the two forces shown in fig.

A boat is pulled along the river by two ropes with pulls P & Q inclined at 30° & 40°to the x-axis as shown in Fig. Find a) P and Q if their resultant R is 1000 N , parallelto x-axis b ) If P is inclined at 30° to x-axis find the minimum value of Q if R is same.

Two tensile forces of 20 kN and 30 kN are acting at a point with an angle of 60° between them. Find the magnitude & direction of the resultant force Two tensile forces acting at an angle 1200 between them. The bigger force is 50 kN. The resultant is perpendicular to the smaller force. Find the smaller force and the resultant force

Example - Law Of Triangle Of Force A system of forces is made of two forces of equal magnitude. Determine, using the triangle law of forces, the angle between two forces if magnitude of resultant force is equal to the magnitude of one of the forces.

Example - Resolution Of More Than Two Concurrent Forces Determine magnitude and direction of resultant force of the force system shown in fig.

Determine magnitude and direction of resultant force of the force system shown in fig

Find magnitude and direction of resultant for concurrent force system shown in fig

Find magnitude and inclination with +X axis of resultant of force system shown in fig. Identify type of force system

Find magnitude and direction of resultant of force system shown in fig

Determine the resultant of the forces 100 N, 200 N, 300 N, 400 N and 500 N are acting on one of the vertex of a regular hexagon, towards the other vertices, taken in order as shown in fig

DEPARTMENT OF CIVIL ENGINEERING Example - Lami’s Theorem Find the Tensions T1, T2 and T3 in respective strings as shown in Fig

A cylindrical roller weighing 1000 N is resting between two smooth surfaces inclined at 60º and 30º with horizontal as shown in fig. Draw free body diagram & determine reactions at contact points A and B.

Two traffic signals are temporarily suspended from a cable as shown in Fig- .Knowing that signal B weighs 300N, determine the weight of the signal at C

State Lami’s Theorem. Determine the force P required to keep the system as shown in Fig in equilibrium.

A chord supported at A and B carries a load of 20KN at point C and an unknown weight of W KN at D as shown in Fig. Find the value of unknown weight W so that CD remains horizontal.

An electric lamp in street as shown in fig is having 50N weight is suspended by two wires of 4m & 3m length. The horizontal distance between two fixed points are 5m from which two wires were suspended. Find out tension in both wires

Example - Coplanar Non-Concurrent Forces Find magnitude, direction and location of resultant w.r.t point ‘O’ of force system shown in Fig

Four forces are acting tangentially to a circle of radius 3 m as shown in fig. Determine the resultant in magnitude & it’s direction & location with respect to center or the circle

Determine the magnitude direction and position of resultant force of theforce system given in fig with reference to point A.

Determine the resultant and locate the same with respect to point ’A’ of a nonconcurrent force system shown in fig.

Some forces are acting on a rigid body as shown in fig. Find the resultant magnitude & direction with location WRT O.

Determine the magnitude direction & position of resultant force of the force system given in fig w.r.t A

DEPARTMENT OF CIVIL ENGINEERING Example - Miscellaneous Resultant force of a system of two forces is directed vertically downwards. The magnitude of resultant force R is 50 N One of the force of the system has magnitude of 30 N and is inclined at an angle of 600 with horizontal as shown in Fig.Determine the magnitude P & direction θ of the secondforce.

Two smooth sphere of weight 100 N each and radius 20 cm are in equilibrium in horizontal channel of width 72 cm as shown in fig. Find reactions at the contact surfaces A, B, and C. Assume sides of channel smooth

A uniform wheel of 80 cm diameter & 1500 N weight rests against a rigid rectangular block of thickness 30 cm as shown in fig. Considering all surfaces smooth, determine a) Least pull to be applied through the center of wheel to just turn it over the corner of the block, b) Reaction of the block

The line of action of the 2.6 kN force F runs through the points A & B shown in fig. Determine the x & y components of F

Replace the couple and force by a single force-couple applied at A for the lever shown in Fig. Also find the distance of a point C from A where only a single force can replace the force-couple system(July 11)

Find the Resultant force of a force system shown in FIG. Also sketch the Resultant force. (Jan-16)

DEPARTMENT OF CIVIL ENGINEERING

CHAPTER 3: STATICALLY DETERMINANT BEAMS Types of Load: 1) Point load

2) Uniformly distributed load

3) Uniformly varying load

1) Simply supported beam

2) Cantilever beam

3) Fixed beam

4) Continuous beam

5) Propped cantilever beam

Type of beam:

Type of support 1) Simple support

3)

Hinged support

2) Roller support

4)

Fixed support

DEPARTMENT OF CIVIL ENGINEERING Basic Relationship between the Rate of Loading, Shear Force and Bending Moment

Where, w = Load acting on beam F = Shear Force M = Bending Moment Sign Convention Bending Moment

+ve Positive BM Sagging

Shear Force

-ve Negative BM Hogging

+ve Positive SF

-ve Negative SF

Point of contra flexure:  It is a point where the bending moment diagram changes its sign from positive to negative or vice versa anywhere on the span of the beam is called the point of contra-flexure.

Shape of Shear Force and Bending Moment Diagram:

Type of Load Equation Shear Force

Shape Equation

Bending Moment

Shape

No Load Zone

Point Load

Moment

𝑦 = 𝑘𝑥 0 constant

Uniformly Distributed load

-

𝑦 = 𝑘𝑥 1

No effect

Horizontal Line

Vertical line

Inclined line

𝑦 = 𝑘𝑥 1

No effect

𝑦 = 𝑘𝑥 2

-

No effect

Square parabola

Vertical line

Inclined line

No effect

DEPARTMENT OF CIVIL ENGINEERING Example – Find out support reaction as shown in figure

Figure 1 Figure 2

Figure 3 Figure 4

Example – Find out support reaction & draw S.F.D & B.M.D as shown in figure

Figure 5

Figure 6

Figure 7

Figure 8

Figure 9

Figure 10

Figure 11 Figure 12

DEPARTMENT OF CIVIL ENGINEERING

CHAPTER 4: FRICTION Friction Force: “ When a body slide or tends to slide on a surface, a resisting force is developed against motion at the contact surface is called Friction force” Types of Friction:  Static Friction : Friction experienced when body is at rest is called static friction  Dynamic Friction : Friction experienced when body is in motion is called dynamic friction Coefficient of Friction “The ratio of Limiting friction to Normal reaction is called Coefficient of friction”



F N

W = weight of block N = normal reaction P=external force F = friction force Angle of Friction: “ The directional angle of Resultant R measured from Normal reaction is called Angle of Friction” angle of friction =  Angle of repose With increasing in angle of inclined surface, the maximum angle at which body starts sliding down is called angle of Repose

  tan    tan  Angle of repose = angle of friction =  Laws of Friction Static Friction  Friction Force acts always in direction opposite to that body tends to move  The magnitude of friction force is equal to the external force  The ratio of limiting friction and Normal reaction is constant and is called Static coefficient of friction  The friction force does not depends on area of contact between two surface  Friction force depends on roughness of surface Problem Types:   

Block and Inclined Plane Ladder Friction Wedge Friction

Dynamic Friction   

Friction Force acts always in direction opposite to that in which the body is moving The ratio of limiting friction and Normal reaction is constant and is called dynamic coefficient of friction For moderate speed, the friction force remains constant, But it decreases slightly with the increase in speed.

DEPARTMENT OF CIVIL ENGINEERING Example – Block and Inclined Plane Equilibrium of block is maintained by a pull P as shown in Fig.The co efficient of friction between block and surface is 0.2. Determine the values of P for which the block remains in equilibrium

Determine the horizontal force required to cause the motion of the block weighing 550N as shown in fig. Take μ= 0.55. a. To impend the motion downward b. To impend up the plane

Find the magnitude of the Horizontal force ‘P’ applied to the lower block to cause impending motion as shown in figure. Take μ = 0.3 at all contact surfaces. Weight of block ‘A’ is 300 N & weight of block ‘B’ is 1200 N

A block weighing 150 kN is placed on a rough inclined plane making angle 30° with horizontal. If coefficient of friction is 0.25, find out the force applied on the block parallel to the plane. So that the block is just on the point of moving up the plane. Also find angle of friction

What should be the value of _ in figurewhich will make the motion of 1000N block down the plane to impend? The coefficient of friction for all contact surfaces is 1/3.

Refer figure The coefficient of frictions between the block and the inclined plane is 0.2. Determine the least value of the force P required just to move the block up along the inclined plane

A weight 750 N just starts moving down a rough inclined plane supported by a force of 250 N acting parallel to the plane and it is at the point of moving up the plane when pulled by a force of 350N parallel to the plane. Find the inclination of the plane and the coefficient of friction between the inclined plane and the weight

Example – Ladder Friction A ladder 7 m long rests against a vertical wall with which it makes an angle of 45º and resting on a floor. If a man whose weight is one half of that the ladder, climbs it. At what distance along the ladder will he be when ladder is about to slip? μ= 1/3 at wall and 1/2 at floor A 4 m long ladder has to carry a person of 75 kg weight at 3.5 m distance from floor, along the length of ladder. The self weight of ladder is of 150 N. Find the maximum distance of lower end of ladder from vertical wall so that it does not slide. The coefficient of friction between floor and ladder is 0.3 and that between vertical wall and ladder is 0.2 A ladder of length 4 m, weighing 200 N is placed against a vertical wall making an angle of 60o with the floor. The coefficient of friction between the wall and the ladder is 0.2 and that between floor and the ladder is 0.3. The ladder, in addition to its own weight, has to support a man weighing 600 N at a distance of 3 m from foot of ladder. Calculate the minimum horizontal force to be applied at foot of ladder to prevent slipping

A ladder 6 m long, rests on horizontal ground and leans against a smooth vertical wall making an angle of 200 with the wall. Its weight is 1000 N and it is on the point of sliding when a man weighing 500 N stands on it at a distance of 2.2 m from the foot of the ladder. Calculate the coefficient of friction A uniform ladder AB weighing 230 N & 4m long, is supported by vertical wall at top end B & by horizontal floor at bottom end A. A man weighing 550 N stood at the top of the ladder. Determine minimum angle θ of ladder AB with floor for the stability of ladder. Take co efficient of friction between ladder & wall as 1/3 & between ladder & floor as 1/4 A uniform ladder of weight 250 N, and length 5 m is placed against a vertical wall in position where its inclination with vertical is 30°. A man weighing 800N climbs the ladder. At what position will he induced slipping. Take μ = 0.2 at all contact surface

DEPARTMENT OF CIVIL ENGINEERING

CHAPTER-5: CENTROID AND MOMENT OF INERTIA CENTRE OF GRAVITY “It is defined as an imaginary point on which entire weight of body is assumed to be concentrated” For basic Element

For Composite Element

𝑥=

𝑥 𝑑𝑤 𝑑𝑤

𝑥=

𝑥𝑖 𝑤𝑖 (𝑥1 𝑤1 ) + (𝑥2 𝑤2 ) + ⋯ + (𝑥𝑛 𝑤𝑛 ) = 𝑤𝑖 𝑤1 + 𝑤2 + ⋯ + 𝑤𝑛

𝑦=

𝑦 𝑑𝑤 𝑑𝑤

𝑦=

𝑦𝑖 𝑤𝑖 (𝑦1 𝑤1 ) + (𝑦2 𝑤2 ) + ⋯ + (𝑦𝑛 𝑤𝑛 ) = 𝑤𝑖 𝑤1 + 𝑤2 + ⋯ + 𝑤𝑛

CENTROID “ It is defined as an imaginary point on which entire length, area or volume of body is assumed to be concentrated” Element Type

Line

Area (Lamina)

Volume

For Basic Element

𝑥=

𝑥=

𝑥=

𝑥 𝑑𝑙 𝑦= 𝑑𝑙

𝑥 𝑑𝐴 𝑦= 𝑑𝐴

𝑥 𝑑𝑉 𝑦= 𝑑𝑉

𝑦 𝑑𝑙 𝑑𝑙

𝑦 𝑑𝐴 𝑑𝐴

𝑦 𝑑𝑉 𝑑𝑉

For Composite Element

𝑥=

𝑥𝑖 𝑙𝑖 (𝑥1 𝑙1 ) + (𝑥2 𝑙2 ) + ⋯ + (𝑥𝑛 𝑙𝑛 ) = 𝑙𝑖 𝑙1 + 𝑙2 + ⋯ + 𝑙𝑛

𝑦=

𝑦𝑖 𝑙𝑖 (𝑦1 𝑙1 ) + (𝑦2 𝑙2 ) + ⋯ + (𝑦𝑛 𝑙𝑛 ) = 𝑙𝑖 𝑙1 + 𝑙2 + ⋯ + 𝑙𝑛

𝑥=

𝑥𝑖 𝐴𝑖 (𝑥1 𝐴1 ) + (𝑥2 𝐴2 ) + ⋯ + (𝑥𝑛 𝐴𝑛 ) = 𝐴𝑖 𝐴1 + 𝐴2 + ⋯ + 𝐴𝑛

𝑦=

𝑦𝑖 𝐴𝑖 (𝑦1 𝐴1 ) + (𝑦2 𝐴2 ) + ⋯ + (𝑦𝑛 𝐴𝑛 ) = 𝐴𝑖 𝐴1 + 𝐴2 + ⋯ + 𝐴𝑛

𝑥=

𝑥𝑖 𝑉𝑖 (𝑥1 𝑉1 ) + (𝑥2 𝑉2 ) + ⋯ + (𝑥𝑛 𝑉𝑛 ) = 𝑉𝑖 𝑉1 + 𝑉2 + ⋯ + 𝑉𝑛

𝑦=

𝑦𝑖 𝑉𝑖 (𝑦1 𝑉1 ) + (𝑦2 𝑉2 ) + ⋯ + (𝑦𝑛 𝑉𝑛 ) = 𝑉𝑖 𝑉1 + 𝑉2 + ⋯ + 𝑉𝑛

DEPARTMENT OF CIVIL ENGINEERING Line Element Centroid – Basic Shape Element name

Geometrical Shape

Straight line

Length

𝒙

𝒚

L

𝐿 cos 𝜃 2

𝐿 sin 𝜃 2

𝐴 2

𝐵 2

𝐴2 + 𝐵2

Straight line

Circular wire

2𝜋𝑟

r

r

Semicircular

𝜋𝑟

r

2𝑟 𝜋

Quarter circular

𝜋𝑟 2

2𝑟 𝜋

2𝑟 𝜋

Circular arc

2𝑟𝛼 (𝛼 𝑖𝑛 𝑟𝑎𝑑𝑖𝑎𝑛)

𝑟 sin 𝛼 𝛼

On Axis of Symmetry

Area(Lamina) Element Centroid– Basic Shape Element name

Rectangle

Geometrical Shape

Area

𝒙

𝒚

bd

𝑏 2

𝑑 2

DEPARTMENT OF CIVIL ENGINEERING

Triangle

1 𝑏ℎ 2

𝑏 3

ℎ 3

Circle

𝜋𝑟 2

r

r

Semicircle

𝜋𝑟 2 2

r

4𝑟 3𝜋

Quarter circle

𝜋𝑟 2 4

4𝑟 3𝜋

4𝑟 3𝜋

Circular segment

𝛼𝑟 2 (𝛼 𝑖𝑛 𝑟𝑎𝑑𝑖𝑎𝑛)

2 𝑟 sin 𝛼 3 𝛼

On Axis of Symmetry

Solid Element Center of gravity– Basic Shape Element name

Geometrical Shape

Volume

𝒙

𝒚

𝒛

Cube

lbh

𝑙 2

ℎ 2

𝑏 2

Cylinder

𝜋𝑟 2 ℎ

𝑟

ℎ 2

r

DEPARTMENT OF CIVIL ENGINEERING

Cone

𝜋𝑟 2 ℎ 3

r

ℎ 4

r

Sphere

4𝜋𝑟 3 3

r

𝑟

r

Hemi-sphere

2𝜋𝑟 3 3

𝑟

𝑟

3𝑟 8

Pappus Guldinus theroems Theorem1 : For surface area “Surface area of revolution is equal to product of length of generating curve and the distance travelled by generating curve” Surface area 𝐴 = 𝜃 𝑙 Where, l = Length of generating curve and 𝜃 = Distance travelled by centroid of generating curve = 2𝜋𝑦 (For Full revolution) = 𝜋𝑦 (For Half revolution) Theorem 2 : For Volume “The volume of a body of revolution is equal to the product of the generating area and the distance travelled by centroid of the area” Surface volume 𝑉 = 𝜃 𝐴 Where, A = Area of generating curve and 𝜃 = Distance travelled by centroid of generating curve = 2𝜋𝑦 (For Full revolution) = 𝜋𝑦 (For Half revolution)

DEPARTMENT OF CIVIL ENGINEERING

Moment of Inertia Defination: ” It is defined as sum integral of product of the elemental areas and square of their distance from the reference axis” 𝐼𝑥𝑥 =

𝑦 2 𝑑𝐴 𝐼𝑦𝑦 =

𝑥 2 𝑑𝐴

Units: length4(mm4, m4, in4, etc.) Parallel Axis theorem “The Moment of Inertia of a planar element about a given reference axis is equal to the sum of moment of inertia about its centroidal axis and product of area of lamina and square of distance between the centroidal axis and reference axis” 𝐼𝑥𝑥 = 𝐼𝐺𝑋 + 𝐴ℎ𝑦2 𝐼𝑦𝑦 = 𝐼𝐺𝑌 + 𝐴ℎ𝑥2

Perpendicular Axis theorem “Moment of inertia of a liamina about an axis perpendicular to its plane is equal to the sum of moment of inertias of the planar element about other two orthogonal axis along the plane of lamina” 𝐼𝑧𝑧 = 𝐼𝑥𝑥 + 𝐼𝑦𝑦 Radius of Gyration (r) “ It is defined as a distance which when squared and multiplied with an area of planar element, gives the value of its moment of inertia” r = radius of gyration 𝐼 I = Moment of Inertia 𝑟= 𝐴 A= cross section area Area(Lamina) Element – Moment of Inertia (Basic Shape) Element name

Rectangle

Geometrical Shape

Area

𝑰𝒙𝒙

𝑰𝒚𝒚

bd

𝑏𝑑3 12

𝑑𝑏 3 12

DEPARTMENT OF CIVIL ENGINEERING

Triangle

1 𝑏ℎ 2

𝑏ℎ3 36

ℎ𝑏 3 36

Circle

𝜋𝑟 2

𝜋𝑑4 64

𝜋𝑑4 64

Semicircle

𝜋𝑟 2 2

0.11 𝑟 4

𝜋𝑑4 128

Quarter circle

𝜋𝑟 2 4

0.055 𝑟 4

0.055 𝑟 4

d= diameter

DEPARTMENT OF CIVIL ENGINEERING Example – Centroid For Line Element

Figure 3

` Figure 2 Figure 1

Figure 4

Figure 5 Figure 6

Example – Centroid For Area Element (Lamina)

Figure 9 Figure 8

Figure 7

Figure 10

Figure 11

DEPARTMENT OF CIVIL ENGINEERING Example – Area Of Revolution And Volume Of Revolution Determine surface area of revolution of the length ABCD revolved about x axis as shown in Fig

Figure 12

Determine the centroid of the shaded area shown in fig. Also calculate the volume of the article generated by revolving the area bout vertical axis “AB‟

Find surface area of the glass to manufacture an electric bulb shown in fig., using first theorem of Pappu- Guldinus

Figure 13

Determine volume of revolution generated by revolving plane lamina ABCDEA shown in fig. ,about y – y axis, to 2π rad. Write statement of theorem used for calculating volume

Figure 14

Figure 15

DEPARTMENT OF CIVIL ENGINEERING

Example – MOMENT OF INERTIA Determine moment of inertia of a section shown in Fig. about horizontal centroidal axis.

Determine the moment of inertia of the given lamina in fig about centroidal X axis

` Determine moment of inertia about horizontal centroidal axis for the section shown in fig

Find the moment of inertia of the area about x-x axis as shown in fig

Find the moment of inertia about the yaxis & x-axis for the area shown in Fig

Determine the location of centroid of plane lamina shown in fig with respect to point O

Determine the location of centroid, IXX & IYY of lamina shown in Fig

Find moment of inertia about horizontal centroidal axis of a shaded area shown in fig

DEPARTMENT OF CIVIL ENGINEERING

CHAPTER 6: SIMPLE STRESS AND STRAIN Stress “It is defined as Internal resisting force per unit area” A = Area

Force P R Stress    Area A A

P

P

P  A Unit:- N/mm2or kN/m2 (1MPa = 1 N/mm2, 1GPa = 103 N/mm2) Strain “ deformation or change in length per unit original length is called strain”



l

P

P

𝐶ℎ𝑎𝑛𝑔𝑒𝑖𝑛𝑙𝑒𝑛𝑔𝑡ℎ 𝑆𝑡𝑟𝑎𝑖𝑛 = 𝑂𝑟𝑖𝑔𝑖𝑛𝑎𝑙𝑙𝑒𝑛𝑔𝑡ℎ

l

l

Where,  l = change in length l = Original length

Hook’s law It states that “ Within Elastic limit , the stress is proportional to strain”

  E

Where,  = Stress  = Strain E = Modulus of elasticity of material or Young’s modulus Equation of deformations

l 

𝜎𝑙 Pl 𝛿𝑙 = 𝐸 AE or

Where, A= Cross-sectional Area

Area of uniformly Tapering circular section 𝜋 𝐴 = 𝑑1 𝑑2 4 Where,

d1 = Dia. of the bigger end of the bar

d 2 = Dia. of the smaller end of the bar

l

DEPARTMENT OF CIVIL ENGINEERING Types of Example- Bars in Series Type 1: Both ends or One end is free to expand

(tensile  positive, compressive  negative) Pl Pl Pl Pl l   1 1  2 2  3 3  4 4 A1 E1 A2 E2 A3 E3 A4 E4

Type 2: Both ends are Fixed

Equilibrium condition, P=P1+P2 Compatibility Condition, 𝛿𝑙11

=

𝛿𝑙 2

(tensile  positive, compressive  negative)

P

L1 Types of Example- Bars in parallel or Composite Bars When Elongation or Contraction of each bar is same

Equilibrium condition, P=P1+P2 Compatibility Condition, 𝛿𝑙11

=

𝛿𝑙 2

L2

DEPARTMENT OF CIVIL ENGINEERING

ELASTIC CONSTANTS Linear strain “ Deformation or Change in length per unit original dimension is called Strain”

𝐶𝑕𝑎𝑛𝑔𝑒 𝑖𝑛 𝑙𝑒𝑛𝑔𝑡𝑕 𝑂𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑙𝑒𝑛𝑔𝑡𝑕

𝑆𝑡𝑟𝑎𝑖𝑛 =



l

l = Original length

l

l Lateral strain

“Strain measured Lateral to the linear strain is called lateral Strain” OR “ it is defined as ratio of Change in lateral dimension to the original lateral dimension”

Poisson’s ratio

𝐿𝑎𝑡𝑒𝑟𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 =

' 

d

𝑐𝑕𝑎𝑛𝑔𝑒 𝑖𝑛 𝑙𝑎𝑡𝑒𝑟𝑎𝑙 𝑑𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛 𝑂𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑙𝑎𝑡𝑒𝑟𝑎𝑙 𝑑𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛

d = Dia. of bar ---- For Round bar

where

d b ' where b = width or thickness of bar---For   b rectangular section

𝑃𝑜𝑖𝑠𝑠𝑜𝑛′ 𝑠 𝑟𝑎𝑡𝑖𝑜 =

“Ratio of Lateral strain to Linear strain is called poisson’s ratio”

1 𝑚

“Ratio of Change in volume to the original volume is called volumetric strain” 𝜎𝑧 𝜎 𝑦

𝑉𝑜𝑙𝑢𝑚𝑒𝑡𝑟𝑖𝑐 𝑠𝑡𝑟𝑎𝑖𝑛 =

v  Where, 𝜀𝑥 =

𝜎𝑥

𝜎𝑥

𝜎𝑧

Tension : +ve Compression : -ve

𝜎𝑥 𝐸

𝑙𝑎𝑡𝑒𝑟𝑎𝑙 𝑠𝑡𝑟𝑎𝑖𝑛 𝑙𝑖𝑛𝑒𝑎𝑟 𝑠𝑡𝑟𝑎𝑖𝑛

𝑜𝑟 

Volumetric strain

𝜎𝑦

𝜎

V V

= 𝜀𝑥 + 𝜀𝑦 + 𝜀𝑧 𝜎

− 𝑚𝐸𝑦 − 𝑚𝐸𝑧

𝜎𝑦 𝜎𝑥 𝜎𝑧 − − 𝐸 𝑚𝐸 𝑚𝐸

𝜀𝑧 =

𝜎𝑦 𝜎𝑧 𝜎𝑥 − − 𝐸 𝑚𝐸 𝑚𝐸

𝐾=

K

K

' 

𝜀𝑦 =

“ The ratio of Direct stress to volumetric strain is called bulk modulus” 1-young’s modulus( E ) 2-Bulk modulus( K ) 1 3-poisson’s ratio((𝜇 𝑜𝑟 𝑚 )



𝐶𝑕𝑎𝑛𝑔𝑒 𝑖𝑛 𝑣𝑜𝑙𝑢𝑚𝑒 𝑂𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒

Bulk modulus

Relationship betn

 l = change in length

Where

l

mE 3(m  2)

𝐷𝑖𝑟𝑒𝑐𝑡 𝑠𝑡𝑟𝑒𝑠𝑠 𝑉𝑜𝑙𝑢𝑚𝑒𝑡𝑟𝑖𝑐 𝑠𝑡𝑟𝑎𝑖𝑛

 v

𝑂𝑅

𝐸 = 3𝐾 1 −

2 𝑚

DEPARTMENT OF CIVIL ENGINEERING

Shear stress (𝝉) “It is defined as the force acting tangentially to the cross-section per unit area” F

l

As  l

𝑠𝑕𝑒𝑎𝑟 𝑓𝑜𝑟𝑐𝑒 𝑠𝑕𝑒𝑎𝑟𝑖𝑛𝑔 𝑎𝑟𝑒𝑎

𝑆𝑕𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠 (𝝉) =



F As

l

Shear strain (∅) “ It is ratio of angular deformation and original dimension”

𝑑𝑒𝑓𝑜𝑟𝑚𝑎𝑡𝑖𝑜𝑛 𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑙𝑒𝑛𝑔𝑡𝑕

𝑆𝑕𝑒𝑎𝑟 𝑠𝑡𝑟𝑎𝑖𝑛 (∅) =



l l

Modulus of rigidity (shear modulus)-(G,N or C) “ it is defined as ratio of Shear stress and shear strain”

𝑀𝑜𝑑𝑢𝑙𝑢𝑠 𝑜𝑓 𝑟𝑖𝑔𝑖𝑑𝑖𝑡𝑦 =

G Relationship betn

Relationship betn

i-young’s modulus( E ) ii-modulus of rigidity( G ) iii-poisson’s ratio(  ) i-young’s modulus( E ) ii-Bulk modulus( K ) iii- modulus of rigidity( G )

G

E

𝑆𝑕𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠 𝑆𝑕𝑒𝑎𝑟 𝑠𝑡𝑟𝑎𝑖𝑛

 

mE 𝑂𝑅 2(m  1)

9GK G  3K

1

𝐸 = 2𝐺 1 + 𝑚

DEPARTMENT OF CIVIL ENGINEERING

TEMPERATURE STRESS AND STRAIN Change in length due to temp. effect (Free Expansion)

𝛿𝑙(𝑛𝑎𝑡) = 𝑙𝛼∆𝑡 Where, l = length of member  = coefficient of thermal expansion t = change in temperature

l

l

Force generated due to prevention of Free Expansion (Single bar or Parallel bar) Case 1: Free to Expand (No support on one end or both)

Case 3: Partially Restrained (Support on both end but, gap between support and member)

Case 2: Full restrained (Support on both)

𝛿

R

l

l

l

𝑁𝑜 𝐹𝑜𝑟𝑐𝑒 𝑑𝑒𝑣𝑒𝑙𝑜𝑝𝑒𝑑 𝑠𝑜, 𝑅 = 0 𝛿𝑙(𝑝𝑟𝑒𝑣) = No Stress

R

l

l

𝑅𝑙 𝐴𝐸

l

𝛿𝑙(𝑝𝑟𝑒𝑣) =

Where, 𝛿𝑙 𝑝𝑟𝑒𝑣 = 𝛿𝑙(𝑛𝑎𝑡) = 𝑙𝛼∆𝑡

𝑅𝑙 𝐴𝐸

Where, 𝛿𝑙 𝑝𝑟𝑒𝑣 = 𝛿𝑙(𝑛𝑎𝑡) − 𝛿 = 𝑙𝛼∆𝑡 − 𝛿

Temperature stress in Bars in Parallel (Composite bars) - Equilibrium Condition

1 A1   2 A2

Material 1

- Compatibility Condition

Material 2

1   2  t (1   2 )

Where,

[ 𝑇𝑎𝑘𝑒 𝛼1 𝐻𝑖𝑔ℎ𝑒𝑟]

𝜎 𝜀= 𝐸

Composite Section Material 1 Material 2 Free Expansion Material 1 Material 2 Composite Expansion

C T

DEPARTMENT OF CIVIL ENGINEERING Example – Simple Stress And Strain A circular rod of diameter 20 mm and 500 mm long is subjected to a tensile force 50kN. The modulus of elasticity for steel may be taken as200 kN/mm2. Find stress, strain and elongation of the bar due to applied load A 2.8 m long member is 60 mm deep and 40 mm wide. It is subjected to axial tensile force 210 kN. Determine change in dimension and in volume. Take E=200 Gpa and μ = 0.3

Example – Bars in series 5

2

A stepped bar carries loads as shown in Fig. Find total deformation in bar by taking modulus of elasticity 2x 10 N/mm . Length and cross sectional area are as under:

Parameter Length Area

AB 1.2m 30 mm x 30 mm

BC 1.35m 50mm dia

CD 1.1m 10mm x 50mm

Find the total deformation of a steel rod subjected to a force of 250kN, as shown in Fig. Length of rod is 1000mm and Modulus of Elasticity of steel is 200GPa

A stepped circular bar ABCD is axially loaded as shown in fig. is in equilibrium. Find unknown force P, and calculate tresses in each part and total change in length of the bar. Take Ecopper= 100 GPa, Ebrass= 80 GPa and E= 200GPa

For a bar shown in figure 8 find the diameter of the middle portion, if the stress at that location is to be limited to 140 N/mm2. Also find 5 2 the total change in the length of the bar. E = 2 X 10 N/mm

A steel member ABCD with three different circular crosssection and lengths as follows, is subjected to an axial pull of 150kN. Compute the net change in the length of the member if the modulus of elasticity (E)=200GPa.

• AB: diameter=40mm and length=750mm • BC: diameter=25mm and length=1000mm • CD: diameter=30mm and length=1200mm A stepped bar made of steel, copper and brass is under axial force as shown in figure 2and is in equilibrium. The diameter of steel is 12mm, diameter of copper is 16mm and the diameter of brass is 20mm.Determine (i) Magnitude of unknown force P (ii) stresses in each material and (iii) Total change in length of the bar. Take Esteel = 200GPa, Ecopper=100GPa and Ebrass = 80GPa

`

Example – Stresses In Composite Structure A short concrete column 300mm x 300mm in section is carrying axial load of 360 kN. The column is strengthened by four, 12mm diameter steel bars each one at corner. Calculate stresses in concrete and steel. Take Ec = 14 GPa and Es = 210 Gpa

A short concrete column 450mm x 450mm in section is axially loaded to 500 kN.The column is trengthened by four, 16mm diameter steel bars each one at corner.Calculate stresses in concrete and steel. Take Ec = 14 GPa and Es = 210 Gpa

A reinforced concrete column 500 mm x 500 mm in section is reinforced with four steel bars of 25 mm diameter, one in each corner. The column is carrying an axial load of 1000kN. Find the stresses in concrete and steel bars. Take E for steel = 210GPa and E for concrete = 14 GPa

DEPARTMENT OF CIVIL ENGINEERING Example – Thermal Stresses And Strain A bar 3 m long and 20 mm diameter is igidly fixed in two supports at certain temperature. If temperature is raised by 60° C, find thermal stress and strain of the bar. Also find thermal stress and strain if -6 support yields by 2 mm. Take α = 12 x 10 0 5 2 per C. E = 2.0 x 10 N/mm

A steel rod 25mm in diameter is inserted A steel tube of 2 m length is subjected to inside a brass tube of 25mm internal di500C rise in temperature. Determine (i) ameter & 35mm external diameter, the free natural expansion and (ii) stress ends are rigidly connected together. The developed in the tube, if expansion is 5 2 assembly is heated by 30o. Find value & prevented. Take Es = 2.0 x 10 N/mm nature of stress developed in both the andCo efficient of thermal expansion -6 0 materials. Take, E steel = 200GPa, E brass α=12 x10 per C -6 o =80 GPa, α steel =12 x 10 per C, α brass -6 o =18 x 10 per C. In an assembly of steel rod of 20 mm diameter passes centrally through a copper tube 40 mm external diameterand 30 mm internal diameter. The tube is closed at both ends by rigid plates of negligible thickness, is initially stress free. If the temperature of the 0 assembly is raised by 60 C, calculate stresses developed in copper and steel. Take Modulus of elasticity E for steel = 200 MPa and for copper =100 MPa, Co efficient of -6 0 -6 0 thermal expansion for steel = 12 x10 / C and for copper = 18 x 10 / C

A steel bar ABC having 25mm diameter and 500mm length of AB and 16mm diameter and 350mm length of BC is rigidly held between two supports at A and C. If 0 the temperature is raised by 30 Celsius determines the stresses developed in parts AB and BC. Take E= 200GPa and ά -6 o =12 x 10 / C An assembly made up from Aluminium and Steel bars as shown in the fig. is initially stress free at temperature 32° C .The assembly is heated to bring its temperature to 82° C. Find the tresses developed i n each bar. The coefficient of thermal expansions is -5 -5 1.25 x 10 / ° C & 2.25 x 10 / °C for steel and aluminium respectively. Take Es = 200 GPa & EAL= 75 GPa.

A steel circular bar of 16 mm diameter is placed inside a copper tube, having internal diameter of 20 mm and thickness of 2.5 mm as shown in fig. Both the ends are rigidly fixed and initially stress free. If the temperature of assembly is increased by 50ºC, compute magnitude& nature of stresses produced in each material. Take modulus of elasticity of steel and copper as 200 GPa and 100 GPa respectively. Take coefficient of thermal expansion -6 -6 (per ºC) for steel & copper as 12 x 10 and 18 x 10 respective

A steel rail is 10m long and is laid at a temperature of 20°C. The maximum temperature expected is 50°C, estimate the minimum gap -6 between two rails to be left so that temperature stresses do not developed. The coefficient of linear expansion a steel=12x10 per °C per unit length. A composite bar made up of steel and copper rods, connected in series. The ends of composite bars are fixed. Find the stress developed in steel and copper due to increase in temperature by 500C. Other relevant data are given below Copper Steel Length 1.2m 1.0m Diameter 30mm 30mm -6 0 -6 0 Coefficient of thermal 17 x 10 / C 12 x 10 / C expansion 5 5 Modulus of Elasticity 0.8 x 10 MPa 2 x 10 MPa

Example – ELASTIC CONSTANTS A rectangular block of material is 250 mm long, 100mm wide and 80mm thick. If it is subjected to a tensile load of 200kN, compressive load of 300kN and tensile load of 250kN along its length, width and thickness respectively. Find the change in volume of the block. Take E= 210GPa and Poisson’s ratio μ=0.25.

A cement concrete block having a shape of square cross section Of 250mm side and a uniform height of 350mm is tested in a compression testing machine by applying an axial compressive load of ‘P’. It was observed that the height decreased by 0.28mm and the side increased by 0.035mm. If the Modulus of Elasticity of concrete is 0.13x105 N/mm2, determine • Poison’s Ratio • The value of ‘P’  The volumetric strain of the block

DEPARTMENT OF CIVIL ENGINEERING A Steel bar is subjected to tensions as shown in fig. Determine change in volume of the bar, if Es = 200GPa and μs = 0.25 In order that there is no change in volume, what should be the revised value of load along X axis?

A rectangular block of 50mm X 50mm X 300mm is subjected to tensile stress of 200N/mm2 along the length in x direction and compressive stresses of 120 N/mm2 on the Remaining all faces in y and z directions. Find the strains produced along x, y and z Directions and calculate change in the volume. If 1/m = 0.25 & E = 200 KN/mm2

A rectangular block of size 350mm (l) x 50mm (b) x 150mm 5 (h) is subjected to forces shown in figure. E = 2 x 10 and Poisson’s ratio is 0.28, calculate the Change in volume of block.

A cube of 150mm x 150mm x 150mm is subjected to axial Tensile forces of 1000kN, 800kN and 600kN along X dir Y-dir and Z-dir respectively. Taking Poisson's ratio v = 0.25 and Modulus of Elasticity, E = 2 x 5 10 , Determine: 1) Change in each dimension 2) Change in Volume 3) Stress in each direction

Example – Miscellaneous Determine deformation in each part of the bar ABCD shown in 5 2 Figure . Take E = 2 x 10 N/mm

A reinforced concrete column of size 250 mm x 250 mm supports a load of 250 kN axially. The reinforcement consists of 4 steel roads each of 25 mm in diameter in each corner. Find how much load is carried by the rods and the concrete if Young’s modulus of steel is 15 times that of concrete

DEPARTMENT OF CIVIL ENGINEERING

CHAPTER 7 : STRESSES IN BEAMS BENDING STRESSES PURE BENDING : “When Beam length is subjected to only bending moment and zero shear force, such beam is said to be in Pure bending” Assumptions:  The material of beam is homogeneous and Isotropic  Bending stress are within elastic limit  A plane section before bending remains plane after bending  The section under consideration is under pure bending  The beam is in equilibrium i.e. bending tensile and compression force have same magnitude  The Modulus of Elasticity in tension and compression is same. EQUATION FOR PURE BENDING

M  E   I y R Where, M = moment of resistance I = M.I of the section about x-x axis  = bending stress y = distance of extreme fibre from N.A E = modulus of elasticity R = radius of curvature of the beam

BENDING STRESS DIAGREM B yt D

ZHollow Rectangular =

𝑏d2 6

𝐵𝐷3 − 𝑏𝑑 3 6𝐷

𝜎𝑏 =

ZCircle = (π/32)*d3 ZHollow Circular =

𝑀 𝑦𝑡 𝐼

Yb

SECTIONAL MODULUS (Z) = ZRectangular =

𝜎𝑡 =

(𝐷4 − 𝑑 4 ) 6𝐷

𝑀 𝑦𝑏 𝐼

𝑰 𝒀𝒎𝒂𝒙

ZTriangular =

𝑏ℎ2 24

ZSemicircular = 0.1911 r3

DEPARTMENT OF CIVIL ENGINEERING

SHEAR STRESSES  F

Ay Ib

A

Where,  = shear stress at section AB F = Shear Force A =area of section above AB I = moment of inertia b = width of beam at level AB y =distance between the centre of area A & neutral axes

B Y

b

SHEAR STRESS DIAGRAM Rectangular section  max  1.5 avg

Hollow Rectangular section

Solid circular section

Hollow Circular section

4 3

 max   avg

Triangular section I - section

4 3

At Neutral axis  na   avg

3 2

 max   avg ....at

h 2

τmax = 1.5 τavg 4

τNA = 3 τavg C - section

Plus - section

d

DEPARTMENT OF CIVIL ENGINEERING H - section

T- section

DEPARTMENT OF CIVIL ENGINEERING Example – Bending Stresses A simply supported beam of span 4.0 m has a cross-section 200 mm × 300 mm. If the permissible stress in the material of the beam is 20N/mm2, determine maximum udl it can carry

A circular pipe of 100 mm external Determine maximum bending stress in a cantidiameter and 80 mm internal diameter lever beam of length 2m. The beam carries a is used as a Simply Supported beam of udl of 8kN/m over the entire length of 2m and span 4 m. Find the safe concentrated a concentrated vertical downward load of25kN load that the beam can carry at the at the free end of cantilever. The cross-section midpoint, if the permissible stress in of the beam is a rectangle of size 350mm deep the beam is 120 N/mm2 and 250mm wide Find out maximum bending stresses at top and bottom of beam A simply supported beam has T-cross section as shown in FIG. It as shown in is subjected to Bending Moment of 50kN-m. Find Bending Stress at extreme fibers and draw bending stress distribution across the Section

A cast iron beam of T section (as per fig. 5), is loaded as shownin fig.If the tensile and compressive permissible stresses are 40MPa and 70MPa respectively, find the safe point load W. Neglect self weightof the beam.

Determine the maximum bending stress & draw bending stress distribution in a section as shown in Fig, if it is subjected to a bending moment of 20kN-m

Find out uniformly distributed load A beam having an I section with top flange 80 X which can be safely applied to a canti- 40 mm, web 120 X 20 mm and bottom flange lever beam having span 2m. The beam 160 X 40 mm, simply supported over a span of has rectangular cross section 6m, is subjected to uniformly distributed load 200x300mm. The allowable bending over entire span. If bending stress is limited to stress and allowable shear stress in 40 N/mm2 tensile and 120 N/mm2 compresbeam material is 15MPa and 10MPa sive, find max. value of U.D.L. the beam can respectively carry if the larger flange is in tension A beam of T shaped cross section shown A cast iron water pipe of 500 mm inside diin Fig. is subjected to bending about x-x ameter and 20 mm thick, is supported over a axis due to a moment of 20 kNm. Find span of 10 meters. Find the maximum bendthe bending stress at the top of the beam ing stress in the pipe metal, when the pipe is running full. Take density of cast iron = 70.6 kN/m and water =9.8 kN/m

`

The Rectangular block of size 300mm (b) x 450mm (d) is subjected to a uniform bending moment 120 kNm. Calculate the bending stresses at extreme fiber of the blocks. Also, find out total tensile and compressive forces due to bending stresses. Draw bending stress distribution diagram also

A mild steel simply supported beam of 3 m span has cross section 20 mm(width) x 50 mm (depth). Find the maximum uniformly distributed load that beam can carry in addition to its self weight, if maximum bending and shear stresses are limited to 150 N/mm2 and 10 N/mm2.Self weight of beam is 75N/m. A section of beam as shown in fig is subjected to a B.M of 10 kN.m about the major axis. Draw bending stress distribution across the section

DEPARTMENT OF CIVIL ENGINEERING Example – Shear Stresses A 50 mm X 100 mm in depth rectangular section of a beam is simply supported at the ends with 2m span. The beam is loaded with 20 kN point load at 0.5 m form R.H.S. Calculate the maximum shearing stress in the beam Determine the maximum shear stress and draw shear stress distribution A cross the section as shown in fig,if the section is subjected to a shear force of 40kN

A beam of rectangular section 100mm × 300mm is subjected to a shearforce of 10kN. Find shear stress at the top layer, at neutral layer and theaverage value of shear stress. Show the stress distribution diagram

A simply supported beam 6 m in span A simply supported beam of span 4.0 m carries udl of 18 kN/m. The croos- section has a cross-section 200mmX300mm. If the of beam is hollow rectangular section with permissible stress in the material of the 2 outer dimension 250 X 400 mm and 25 beam is 20N/mm , determine maximum mm thick. Determine shear stress at vari- udl it can carry ous locations. Fig shows a beam cross section subjected to A section of beam as shown in fig is shearing force of 200 KN. Determine the shear- subjected to a A S.F of 20KN. Draw ing stress at neutral axis and sketch the shear shear stress distribution across the stress distribution diagram across the section section

A T – shaped cross section of a beam has Determine maximum bending stress and maxiflange = 200 x 50 mm and web = 50 x 200 mum shear stress in a cantilever beam of length mm in size, is subjected to a vertical shear 2m. The beam carries a udl of 8kN/m overthe force of 100 kN. Calculate the shear entire length of 2m and a concentrated vertical stresses at important points and draw downward load of25kN at the free end of cantishear stress distribution diagram. Take- lever. The cross-section of the beam is arectangle moment of inertia about horizontal neutral of size 350mm deep and 250mm wide 6 axis = 113.4 x 10 mm 6 The Rectangular block of size 300mm (b) x A section of a beam shown in Fig. has moment of inertia about neutral axisis 11.6 x 10 450mm (d) is subjected to a shear force 80 mm4. The section is subjected to shear force of 14.5 kN. Determine value of maximum kN. Calculate the Shear stresses at neutral shear stress on the section axis and Junction of the blocks. Draw Shear stress distribution diagram also Draw shear stress distribution diagram across the cross section of a ‘T’ beam, having flange 150 x 20 mm and web 20 x 250 mm, carrying pure shear force of 50 kN at the section

DEPARTMENT OF CIVIL ENGINEERING

CHAPTER 8 : TORSION Theory of Torsion: “The moment of couple acting on shaft is called torque or torsional moment or twisting moment” Assumptions:  The material of Shaft is uniform throughout the length  The twist along the shaft is uniform  The shaft is of Uniform cross section throughout the length  Cross section of shaft, which are plane before twist remains plane after twist  All radii which are straight before twist remains straight after twist EQUATION FOR TORSION

DIAGREM

𝑇 𝜏 𝐶. 𝜃 = = 𝐽 𝑅 𝑙 Where, T = torque or twisting Moment J = Polar Moment of Inertia 𝜏 = Shear stress R= Radius of Shaft C = modulus of rigidity or Shear modulus 𝜃 = angular twist in radian l = Length of Shaft C 𝜃 = Torsional Rigidity

T

A’

θ

O

A

l

POLAR MOMENT OF INERTIA

For Solid Circular Shaft =

𝜋𝐷4 32

For Hollow Circular Shaft =

𝜋(𝐷4 − 𝑑 4 ) 32

DEPARTMENT OF CIVIL ENGINEERING Example – Torsion A solid shaft is 100 mm in diameter, transmits 120kW at 200 rpm. Find the maximum intensity of shear stress induced and angle of twist for a length of 6 4 meters. Take C = 8 X 10 N/mm2

A solid circular steel shaft of diameter 75 mm can resist maximum shear stress of 75 N/mm2. If shaft is rotating at 150 rpm, calculate the power transmitted by shaft. Also calculate the angle of twist for 1.4m long shaft if G=100 GPa.

A solid steel shaft is subjected to a torque of 45 kN m. If the angle of twist is 0.50 per meter length of shaft and shear stress is not to exceed 90 N/mm2. Find: (i) Suitable diameter of shaft (ii) Final maximum shear stress and angle of twist for diameter of shaft selected. Take G= 80 GPa

A solid steel shaft has to transmit 350 kW at 900 r.p.m. Find the diameter of the shaft if the shear stress is to be limited to 125 N/mm. Calculate the diameter of the shaft if hollow shaft is provided of internal diameter equals 0.75 times external diameter

Calculate the diameter of the shaft required to transmit 45 kW at 120 rpm. The maximum torque is likely to exceed the mean by 30% for a maximum permissible shear stress of 55 N/mm2. Calculate also the angle of twist for a length of 2 m. G = 3 80 X 10 N/mm2 A solid steel circular shaft is required to transmit a torque of 6.5 kNm. Determine minimum diameter of the shaft, if shear stress is limited to 40 N/mm and angle of twist should not exceed 0.5º per meter. Take Modulus of rigidity C = 85 GPa

DEPARTMENT OF CIVIL ENGINEERING

CHAPTER 9 : PRINCIPLE STRESSES Principal Plane  

The plane on which only direct stress is acting is called Principal plane. On the principle plane shear stress is zero

Principal Stresses 

Maximum and Minimum values of stresses are acting normal to the principal plane is known as Principal Stresses. Stresses On Inclined Plane Under Biaxial Stress And Shear Stress SIGN CONVENTIONS

σy σx

τ

+

σx

A

+ σy

C

O

σy

τ σx

τ

σx

+

+

θ τ D

B θ

τ

+ σy σx + σy

σn = (

2

σx − σy

σt =(

2

)+(

σx− σy 2

Where, ) cos2θ – τsin2θ

σn = Normal stress on inclined plane σt= Tengential stress on inclined plane σr = Resultant stress on inclined plane σx = Major direct stress σy = Minor direct stress θ = Angle of incined plane BC with the major plane τ = Shear stress

) sin2θ + τcos2θ

σr = σn2 + σt 2

Principal Stresses Principal Stresses

σmax/min = ( Angle of Principle Stress : tan 2𝜃 = Max Shear stress

:

τmax =

σx + σy 2

−2𝑞 𝜎𝑥 −𝜎𝑦

σmax − σmin 2

)+

σx− σy 2

(

2

) +

τ2

DEPARTMENT OF CIVIL ENGINEERING Example – Normal And Tangential Stresses A machine component is subjected to the stresses as shown in fig. Find the normal and shearing stresses on the section AB inclined at an angle of 60º with horizontal (x-x axis). Also find the resultant stress on the section.

The direct stresses at a point in the strained material are 150 N/mm2 compressive & 100 N/mm2 tensile as shown in fig. There is no shear stress. Find the normal & tangential stresses on the plane AC. Also find the resultant stress on AC

A point in a strained material is subjected to a tensile stress of 100 MPa and a compressive of 90 MPa acting on two mutually perpendicular planes and a shear stress of 25 MPa acts along these planes. Determine following stresses on a plane inclined at 35° with plane of compressive stress. (i) Normal Stress, (ii) Tangential Stress, (iii) Resultant Stress.

A point in a strained material is subjected to a tensile stress of 120N/mm2 and a compressive stress of 60N/mm2 acting at right angles to each other. Determine the Normal ,tangential and resultant stress on a plane inclined at 300 in anticlockwise direction with the direction of compressive stress The normal stress on plane AA is 20N/mm2 (tensile). If the principal stress in the material is limited to 60N/mm2 (compressive), determine the Allow able shear stress on plane AA. The normal stress on the planes perpendicular to plane AA is zero

Determine normal and tangential stress on plane AB, in a strained material shown in fig. Determine the stress by Mohr’s circle also.

Example – Principle Plane And Stresses For the state of stress as shown in figure determine location of principal planes, principal stresses and maximum shear stress

For an element shown in fig. (iii), find (i) principal stresses and location of corresponding principal planes (ii) Maximum shear stress and location of planes containing it

At a point in a strained material two mutually perpendicular tensile stresses of 420 N/mm2 and 280 N/mm2 are acting. There is also a clockwise shear stress of 200 N/mm2. Determine the values of principal stresses and location of principal plane

At a point in a strained material, stress conditions on two planes; making an angle of 60º between two, are as shown in Fig. Determine the principal planes and principal stresses through the point

At a point in a strained material the state of stress is as shown in fig 2. Determine (i) Location of Principal planes (ii) Principal stresses. (iii) Maximum shear stress and location of plane on which it acts.

Seat No.: ________

Enrolment No.___________

GUJARAT TECHNOLOGICAL UNIVERSITY BE - SEMESTER–III (NEW) - EXAMINATION – SUMMER 2018

Subject Code:2130003 Subject Name:Mechanics of Solids Time:10:30 AM to 01:00 PM

Date:18/05/2018 Total Marks: 70

Instructions: 1. Attempt all questions. 2. Make suitable assumptions wherever necessary. 3. Figures to the right indicate full marks.

Q.1 (a) Define (i) Couple (ii) Moment (iii) Equilibrant (b) State Lami’s Theorem. Determine tension in wire AB and BC shown in Fig 1. (c) Determine magnitude, direction and position of resultant for force system shown in fig 2. with respect to point O. Q.2 (a) State (i) Verignon’s theorem and (ii) Pappus-Guldinus Theorems (b) Determine C.G. of lamina shown in fig 3 (c) Derive equation for Ixx for triangular section with Base ‘B’ and Height ‘H’ OR (c) Determine Ixx and Iyy for section shown in fig 4 Q.3 (a) Define (i) Strain (ii) Poisson’s ratio (iii) Bulk Modulus (b) State Hook’s low. Draw stress strain curve for Mild Steel Specimen and explain each point in detail (c) A Reinforced concrete column is applied 700 kN load. Size of column is 300 mm X 400 mm, and it is reinforced with 6 bars of 16 mm dia. Determine load taken by concrete and steel. OR Q.3 (a) Define (i) Stress (ii) Young’s modulus (iii) Modulus of rigidity (b) Derive equation to find volumetric strain for cylindrical specimen. (c) A 2.8 m long member is 60 mm deep and 40 mm wide. It is subjected to axial tensile force 210 kN. Determine change in dimension and in volume. Take E=200 Gpa and µ = 0.3 Q.4 (a) State the assumption made in theory of bending. (b) A simply supported beam 5 m in span carries udl of 20 kN/m. The croos section of beam is I section. It is having flange dimension 200 X 20 mm. The thickness of web is 20 mm, depth 260 mm and overall depth of I section is 300 mm. Calculate maximum stresses. (c) A solid shaft is 100 mm in diameter, transmits 120kW at 200 rpm. Find the 7 maximum intensity of shear stress induced and angle of twist for a length of 6 meters. Take C = 8 X 104 N/mm2. OR Q.4 (a) Define (i) Shear Force (ii) Point of Contraflexture (iii) Neutral Axis (b) A simply supported beam 6 m in span carries udl of 18 kN/m. The croos- section of beam is hollow rectangular section with outer dimension 250 X 400 mm and 25 mm thick. Determine shear stress at various locations. (c) Draw shear force and bending moment for the beam shown in fig 5. Q.5 (a) Explain cone of friction with neat sketch. (b) State the laws of dry friction. (c) A uniform ladder of weight 250 N, and length 5 m is placed against a vertical wall in position where its inclination with vertical is 30°. A man weighing 800N climbs the ladder. At what position will he induced slipping. Take µ = 0.2 at all contact surface. OR

3 4 7 3 4 7 7 3 4 7

3 4 7

3 4

3 4

7 3 4 7

1

Prove that for rectangular section maximum shear stress is 1.5 time average stress (b) Find the reaction of beam shown in fig 6. (c) Determine normal and tangential stress on plane AB, in a strained material shown in fig 7. Determine the stress by Mohr’s circle also.

Q.5 (a)

3 4 7

C

A

B O

*************

2

Seat No.: ________

Enrolment No.___________

GUJARAT TECHNOLOGICAL UNIVERSITY BE - SEMESTER–III (NEW) EXAMINATION – WINTER 2017 Subject Code: 2130003 Date:09/11/2017

Subject Name: Mechanics of Solids Time: 10:30 AM to 01:00 PM

Total Marks: 70

Instructions: 1. Attempt all questions. 2. Make suitable assumptions wherever necessary. 3. Figures to the right indicate full marks.

Q.1

Q.2

Define: 1] Rigid body, 2] Newton’s second Law Define Force and classify the force system with neat sketch. Find magnitude and direction of resultant of force system shown in Fig.1 Define Moment & Couple giving two suitable examples Find magnitude, direction and location of resultant wrt point ‘O’ of force system shown in Fig.2 (c) Find the Tensions T1, T2 and T3 in respective strings as shown in Fig 3. OR A chord supported at A and B carries a load of 20KN at point C and an unknown (c)

(a) (b) (c) (a) (b)

03 04 07 03 04 07 07

weight of W KN at D as shown in Fig 4. Find the value of unknown weight W so that CD remains horizontal.

Q.3

(a) Derive with usual notations the theorem of perpendicular axis. (b) Define Centroid and With usual notations find the centroid of a triangle by method (c)

Q.3

Q.4

of integration Find the Centroid of the Lamina shown in Fig 5.

OR (a) Define: 1] Moment of Inertia, 2] Axis of symmetry (b) Explain the Pappu’s Guldinus Theorem I (c) Find the moment of Inertia of an I section shown in the Fig 6 about its centroidal axis. (a) Define: 1] Poison’s Ratio 2] composite bar (b) Derive relation between bulk modulus (K), Poisson’s ratio (1/m), and modulus of elasticity (E). (c) A stepped bar carries loads as shown in Fig 7. Find total deformation in bar by taking modulus of elasticity 2x 105 N/mm2. Length and cross sectional area are as under: Parameter AB BC CD Length 1.2 m 1.35 m 1.1 m Area 30 mm x 30 mm 50mm diameter 10 mm x 50 mm

03 04 07 03 04 07 03 04 07

OR

Q.4

Q.5

(a) Define: 1] Lateral stress, 2] Modulus of Rigidity (b) Derive the equation for deformation of a body due to self weight. (c) A steel rod 25mm in diameter is inserted inside a brass tube of 25mm internal diameter and 35mm external diameter, the ends are rigidly connected together. The assembly is heated by 30. Find value and nature of stress developed in both the materials. Take, E steel = 200GPa, E brass =80 GPa, α steel =12 x 10-6 per C, α brass =18 x 10-6 per C. (a) Define: 1]Point of Contra flexure, 2] Shear force (b) Explain with neat sketch types of beams, types of loads and types of supports (c) Draw the Shear force and Bending Moment diagram for the beam shown in the Fig 8

03 04 07

03 04 07 1

Q.5

OR (a) Define: Coefficient of Static Friction and state the Laws of Friction. (b) Derive with usual notation the relation between shear force and Bending moment. (c) At a point in a strained material the state of stress is as shown in Fig.9 Determine (i) location of principal planes (ii) principal stresses and (iii) maximum shear stress and location of plane on which it acts

03 04 07

2

Seat No.: ________

Enrolment No.___________

GUJARAT TECHNOLOGICAL UNIVERSITY BE - SEMESTER–III (NEW) - EXAMINATION – SUMMER 2017

Subject Code: 2130003 Subject Name: Mechanics of Solids Time: 10:30 AM to 01:00 PM

Date: 29/05/2017 Total Marks: 70

Instructions: 1. Attempt all questions. 2. Make suitable assumptions wherever necessary. 3. Figures to the right indicate full marks. MARKS

Q.1 1

2 3

4 5 6 7 8 9

10 11 12 13

14 Q.2

(a) (b) (c)

(c)

Short Questions As per ___________ law. Whenever a body exerts force on other body, the other body exerts similar force on the former body. (Newton’s, Pappu’s guldinus theorem, Lami’s theorem) Newton-meter is unit of ___________from following. (Pressure, Force, Angular Torque) ___________ is the property by virtue of which a body offers resistance to any change of its state of rest or motion. (Inertia, Matter, Mass, Motion) ___________ is the branch of dynamics, which deals with the bodies in motion due to the application of forces. Speed and distance are ___________ quantity. (Vector, Scalar) Moment is a ___________ vector, whereas couple is a ___________ vector. (Fixed, Free, Null) Define Modulus of rigidity. Define section modulus. One of the assumption in theory of pure bending is the value of ___________ is same in tension as well as compression. (Moment of Inertia, Modulus of Elasticity, Shear Stress, Bending Stress) Define principle of Superposition. Sketch qualitative shear stress distribution diagrams of ‘I’ section of the beams. Give mathematical expression of Lami’s theorem. The process of finding components of a force is called ___________ of forces. (Resolution, Splitting, Composition) Twisting of an object due to applied torques is known as ___________. (Bending, Shearing, Torsion, Rotation) State and prove Pappu’s guldinus theorem for surface area of bodies. Two tensile forces of 20 kN and 30 kN are acting at a point with an angle of 60° between them. Find the magnitude and direction of the resultant force. A point in a strained material is subjected to a tensile stress of 100 MPa and a compressive of 90 MPa acting on two mutually perpendicular planes and a shear stress of 25 MPa acts along these planes. (Figure 1) Determine following stresses on a plane inclined at 35° with plane of compressive stress. (i) Normal Stress, (ii) Tangential Stress, (iii) Resultant Stress. OR Four forces are acting tangentially to a circle of radius 3 m as shown in figure 2. Determine the resultant in magnitude and it’s direction and location

14

03 04 07

07 1

Q.3

(a) (b)

(c) Q.3

(a) (b) (c)

Q.4

(a) (b) (c)

with respect to center or the circle. Write assumption made in the theory of pure bending. An electric lamp in street as shown in figure 3 is having 50 N weight is suspended by two wires of 4 m and 3 m length. The horizontal distance between two fixed points are 5 m from which two wires were suspended. Find out tension in both wires. Find out centroid of thin homogeneous wire as shown in figure 4. OR Draw representative shear stress distribution diagrams for Hollow rectangle, b) I section, c) Hollow circle Find support reactions for beam shown in figure 5. For the beam shown in figure 6 calculate shear force and bending moments at salient points and draw shear force and bending moment diagrams. Explain various types beams and their support system. Calculate center of gravity of T-section having flange 20 X 2 cm and web 30 X 2 cm. also show position of C. G. on figure. A beam having an I section with top flange 80 X 40 mm, web 120 X 20 mm and bottom flange 160 X 40 mm, simply supported over a span of 6m, is subjected to uniformly distributed load over entire span. If bending stress is limited to 40 N/mm2 tensile and 120 N/mm2 compressive, find max. value of U.D.L. the beam can carry if the larger flange is in tension.

03 04

07 03 04 07 03 04 07

OR

Q.4

(a) (b)

(c)

Q.5

(a) (b)

(c)

Q.5

(a) (b)

(c)

Define: (i) coefficient of friction (ii) Angle of friction A block weighing 150 kN is placed on a rough inclined plane making angle 30° with horizontal. If coefficient of friction is 0.25, find out the force applied on the block parallel to the plane. So that the block is just on the point of moving up the plane. Also find angle of friction. Calculate the diameter of the shaft required to transmit 45 kW at 120 rpm. The maximum torque is likely to exceed the mean by 30% for a maximum permissible shear stress of 55 N/mm2. Calculate also the angle of twist for a length of 2 m. G = 80 X 103 N/mm2. Define: (i) Lateral strain, (ii) Poisson’s ratio, (iii) Modulus of rigidity. An M. S. bar of 20 mm diameter is acted upon by a tensile force of 60 kN. If the length of bar is 1.2 m and modulus of elasticity is 2.0 X 105 N/mm2. Find stress, strain and elongation of the bar. Determine moment of inertia of a plane area as shown in figure 7 about its base line a-a. OR Explain following terms: (i) rigid body, (ii) deformable body, (iii) Elastic body. A 50 mm X 100 mm in depth rectangular section of a beam is simply supported at the ends with 2m span. The beam is loaded with 20 kN point load at 0.5 m form R.H.S. Calculate the maximum shearing stress in the beam. For a bar shown in figure 8 find the diameter of the middle portion, if the stress at that location is to be limited to 140 N/mm2. Also find the total change in the length of the bar. E = 2 X 105 N/mm2.

03 04

07

03 04

07

03 04

07

2

Figure 2

Figure 1

Figure 3

Figure 4

Figure 6 Figure 5

Figure 8

Figure 7 *************

3

Seat No.: ________

Enrolment No.___________

GUJARAT TECHNOLOGICAL UNIVERSITY BE - SEMESTER–III(New) • EXAMINATION – WINTER 2016

Subject Code:2130003 Subject Name:Mechanics of Solids Time:10:30 AM to 01:00 PM

Date:31/12/2016 Total Marks: 70

Instructions: 1. Attempt all questions. 2. Make suitable assumptions wherever necessary. 3. Figures to the right indicate full marks. MARKS

Q.1 1 2

Short Questions: Two unlike parallel forces, will form a_______. (Couple, Bending Moment, Shear force). A particle is said to be in _____ when the resultant force acting on it is zero. (Equilibrium, Stable, Unstable)

14

The process of finding components of a force is called ______of forces. (Resolution, Splitting, Composition) 4 Define Law Of Transmissibility. 5 The Relation between Shear force and Bending moment is _________________. 3

6

A cylinder is a surface of revolution generated by revolving a ________line about a fixed axis. (Straight, circular)

7

Co-efficient of static friction will always be _____than the coefficient of kinetic friction. (greater, equal, lesser) The maximum value of Poisson’s ratio for most of the engineering material is____. (0.5, 1,1.5) Young’s modulus of elasticity for a perfectly rigid body is_____.(zero, infinity) The point where the Shear force is maximum, slope of the bending moment is_________. (maximum, minimum, zero) In a beam of I-section, the maximum shear stress is carried by the_________.(web, flange) A t the point of contraflexure _____changes it’s sign. (shear force, bending moment, axial force)

8 9 10 11 12 13 14 Q.2

(a) (b)

(c)

(c)

Shear stresses on principal planes are_________.(zero, maximum, minimum) For an element in pure shear, principal planes are oriented at_____ to the axis. (450, 900) State and explain Varignon’s theorem. Two tensile forces acting at an angle 1200 between them. The bigger force is 50 kN. The resultant is perpendicular to the smaller force. Find the smaller force and the resultant force. Two smooth sphere of weight 100 N each and radius 20 cm are in equilibrium in horizontal channel of width 72 cm as shown in figure 1. Find reactions at the contact surfaces A, B, and C. Assume sides of channel smooth. OR At a point in a strained material the state of stress is as shown in figure 2. Determine (i) Location of Principal planes (ii) Principal

03 04

07

07 1

stresses. (iii) Maximum shear stress and location of plane on which it acts. Q.3

Q.3

Q.4

Q.4

Q.5

Q.5

(a) For pure bending, prove that the neutral axis coincides with the centroid of the cross section. (b) A circular pipe of 100 mm external diameter and 80 mm internal diameter is used as a Simply Supported beam of span 4 m. Find the safe concentrated load that the beam can carry at the mid point, if the permissible stress in the beam is 120 N/mm2. (c) A solid steel shaft is subjected to a torque of 45 kN m. If the angle of twist is 0.50 per meter length of shaft and shear stress is not to exceed 90 N/mm2. Find: (i) Suitable diameter of shaft (ii) Final maximum shear stress and angle of twist for diameter of shaft selected. Take G= 80 GPa. OR (a) State assumptions made in theory of pure bending. (b) For a hollow circular section whose external diameter is twice the internal diameter, find the ratio of maximum shear stress to average shear stress. (c) What should be the value of ϴ in figure 3 which will make the motion of 1000N block down the plane to impend? The coefficient of friction for all contact surfaces is 1/3. (a) Define: (i) Lateral strain (ii) Poisson’s ratio (iii) Modulus of rigidity. (b) In a tension test, a bar of 20 mm diameter undergoes elongation of 14 mm in a gauge length of 150 mm and a decrease in diameter of 0.85 mm at a tensile load of 6 kN. Determine the two physical constants Poisson’s ratio and modulus of elasticity of the material. (c) Determine the centroid of the plane area in which a circular part of 40 mm radius, has been removed as shown in Figure 4. OR (a) Determine the surfaces area and volume of a right circular cone with radius of base R and height h using Pappus-Guldinus theorem. (b) Derive expression of moment of inertia of triangle by first principal. (c) A 6 m long steel rod having 20 mm diameter is connected to two grips and each end at a temperature of 1200 C. Find (i) pull exerted when temperature falls to 400 C and ends do not yield, (ii) pull exerted when temperature falls to 400 C and ends yield by 1.1 mm, (iii) the shortening allowed for no stress at 400 C and (iv) the minimum final temperature for shortening of 1.1 mm. Take Esteel= 205 GPa, αsteel =11 X 10-6/0C. (a) Define: (i) Coefficient of friction (ii) Angle of friction

03

(b) A solid circular steel shaft of diameter 75 mm can resist maximum shear stress of 75 N/mm2. If shaft is rotating at 150 rpm, calculate the power transmitted by shaft. Also calculate the angle of twist for 1.4m long shaft if G=100 GPa. (c) Draw Shear Force and Bending Moment diagram for the beam as shown in figure 5. OR (a) Enlist various type of loads and type of supports. (b) A steel bar of rectangular cross section is 60 mm wide and 50 mm thickness is subjected to an axial pull of 85 kN. Calculate Normal,

04

04

07

03 04

07

03 04

07

03

04 07

03

07

03 04 2

(c)

Tangential and Resultant stresses on an inclined plane at 300 to the cross section of bar. A weight 750 N just starts moving down a rough inclined plane supported by a force of 250 N acting parallel to the plane and it is at the point of moving up the plane when pulled by a force of 350N parallel to the plane. Find the inclination of the plane and the coefficient of friction between the inclined plane and the weight.

07

**************

3

Seat No.: ________

Enrolment No.___________

GUJARAT TECHNOLOGICAL UNIVERSITY BE - SEMESTER–III(New) EXAMINATION – SUMMER 2016

Subject Code:2130003 Subject Name:Mechanics of Solids Time:10:30 AM to 01:00 PM

Date:13/06/2016 Total Marks: 70

Instructions: 1. Attempt all questions. 2. Make suitable assumptions wherever necessary. 3. Figures to the right indicate full marks. MARKS

Q.1 1 2 3 4 5 6 7 8

9

10

11 12

Short Questions How will you differentiate scalar and vector quantities? Explain the term free body diagram What is the moment of inertia of a triangle about its base and horizontal centroidal axis? Define Modulus of rigidity and Modulus of elasticity. What is the relation between Load , Shear Force and Bending Moment. Define section modulus. What is the use of Mohr’s circle? _________ is equal and opposite to the resultant of several forces acting on a body. (Equilibrant, Resultant, Stress, Strain) The process of finding components of a force is called _____________ of forces. (Desolution, Resolution, Lami’s theorem, Composition) Angle of repose is equal to angle of static friction when _________. (motion is absent, system is in equilibrium, motion is impending, body is on a flat surface) Ratio of maximum to average shear stress in a rectangular section is ____________. (3/2, 1/2, 3/4, 5/2) The difference of angle between two principal plane is ___________ (180°, 90°,120°,45°)

One of the assumption in theory of pure bending is the value of _______________ is same in tension as well as compression. (Moment of Inertia, Modulus of Elasticity, Shear Stress, Bending Stress) 14 Twisting of an object due to applied torque is known as ____________ (Bending, Shearing, Torsion, Rotation) (a) What is Polygon law of forces? Explain graphical procedure to find resultant of the forces using polygon law of forces. (b) The line of action of the 2.6 kN force F runs through the points A and B shown in fig. 1. Determine the x and y components of F. (c) A uniform wheel of 80 cm diameter and 1500 N weight rests against a rigid rectangular block of thickness 30 cm

14

13

Q.2

03

04

07 1

(c)

Q.3

(a) (b)

(c)

Q.3

(a)

(b)

(c) Q.4

(a) (b)

(c)

as shown in fig. 2. Considering all surfaces smooth, determine a) Least pull to be applied through the center of wheel to just turn it over the corner of the block, b) Reaction of the block. OR A reinforced concrete column of size 250 mm x 250 mm supports a load of 250 kN axially. The reinforcement consists of 4 steel roads each of 25 mm in diameter in each corner. Find how much load is carried by the rods and the concrete if Young’s modulus of steel is 15 times that of concrete. Explain various types of statically determinant beams and their support system A bar 3m long and 20mm diameter is rigidly fixed in two supports at certain temperature. If temperature is raised by 60°C, find thermal stress and strain of the bar. Also find thermal stress and strain if support yields by 2mm. Take = 12 x 10-6 /°C and E = 2 x 105 N/mm2 Find centroid of the shaded area shown in fig.3 with reference to point ‘O’ OR A solid steel shaft is to transmit a torque of 1 kN.m. If the shearing stress is not to exceed 45 N/mm2. Find the minimum diameter of the shaft. Under what axial tensile load the diameter of a steel bar will be reduced from 50 mm to 49.899 cm? Take E = 2.0 x 105 N/mm2 and Poisson’s ratio = 0.33. Find moment of inertia about horizontal centroidal axis of a shaded area shown in fig. 4 Derive the formula for the elongation of a rectangular bar under the action of axial load. The shaded area shown in fig. 5 is revolved about Y axis. By means of a theorem of Pappus Gauldinus, determine the volume generated. Dimensions are in cm. Draw Shear Force and Bending Moment diagram for the beam shown in fig. 6

07

03 04

07

03

04

07 03 04

07

OR

Q.4

Q.5

(a) Draw representative shear stress distribution diagrams for a) hollow rectangle, b) I section, c) hollow circle (b) Determine reaction at supports for the Beam as shown in Fig. 7 (c) Determine the horizontal force required to cause the motion of the block weighing 550N as shown in fig.8. Take µ= 0.55. a. To impend the motion downward b. To impend up the plane (a) What is ellipse of stress? What is the use of it? (b) A material is subjected to tensile stresses 80 N/mm2 and 40 N/mm2 perpendiculars to each other. It is also subjected to shear stress of 60 N/mm2. Find normal, tangential and resultant stress on a plane inclined at 45° with smaller tensile stress. (c) A cast iron beam is of T section as shown in fig.9 The beam is simply supported on a span of 8m. The beam

03 04 07

03 04

07 2

Q.5

carries a UDL of 25 kN/m on the entire span. Determine the maximum tensile and maximum compressive stresses. Also draw bending stress distribution diagram. Take I = 3.14 x 106 mm4. OR (a) Explain theory of pure bending. (b) Prove that for a rectangular cross section of a beam the value of maximum shear stress is 1.5 times average shear stress. (c) A simply supported beam of length 5m and Its cross section as shown in fig.9. is uniformly loaded with 20kN/m. Find values of shear stress at critical points and draw shear stress distribution diagram across the c/s. Take I = 3.14 x 106 mm4.

03 04

07

*************

3

Fig. 2

Fig. 1

Fig. 3

Fig. 5 Fig. 4

Fig. 6

Fig. 7 100mm 20mm

80mm

Fig. 8 20mm

Fig. 9

4

Seat No.: ________

Enrolment No.___________

GUJARAT TECHNOLOGICAL UNIVERSITY BE - SEMESTER–III (New) EXAMINATION – WINTER 2015

Subject Code:2130003 Subject Name: Mechanics of Solids Time: 2:30pm to 5:00pm

Date:05/01/2016 Total Marks: 70

Instructions: 1. Attempt all questions. 2. Make suitable assumptions wherever necessary. 3. Figures to the right indicate full marks.

Marks

Q.1 a b c d e f g h i

j k l m

n Q.2

Q.3

(a) (b) (c)

(c) (a) (b)

Short Questions (each question carry one Mark) Define Pappu’s Guldinus theorem- I Define Principle of Superposition Define law of Transmissibility Define Couple Enlist types of supports. Sketch qualitative shear stress distribution diagrams of ‘I’ section of the beams. Give mathematical expression of Lami’s theorem, Fill in the blanks Lateral strains are ___________ longitudinal strains. (sometimes less than, always less than, never less than) The shape of shear force diagram for cantilever beam subjected to couple at free end is __________ [horizontal straight line, zero, parabola, incline straight line]. Moment is a ________ vector, whereas Couple is a ________ vector. (free, null, fixed) At the point of contraflexure ___________ changes its sign. (shear force, bending moment, axial force) The Relation between Shear force and Bending moment is _________________. The relation between, dynamic coefficient of friction (μd) is___________________ static coefficient of friction (μs). (less then/ greater then/Equal to) Relation between Modulus of Elasticity and Bulk Modulus is ________________________. Write assumption made in the theory of pure bending. State and Prove with usual notation ‘The law of Parallelogram’. The forces are acting on a rigid body as shown in Figure 1. Find the resultant of the given force system, in terms of magnitude and direction. Find the location of Resultant with respect to point A. OR Determine centroid of the section shown in Figure 2. At a point in a strained material, the state of stress is as shown in figure 3. Determine (i) principal stresses For problem above in Q.3 (a), figure 3 calculate (i) location of principal planes and (ii) maximum shear stress and its location.

14

03 04 07

07 03 04

1

(c)

Q.3

Q.4

Draw shear force diagram and bending moment diagram for beam shown in figure 4.

OR (a) A solid steel shaft has to transmit 350 kW at 900 r.p.m. Find the diameter of the shaft if the shear stress is to be limited to 125 N/mm 2. Calculate the diameter of the shaft. (b) A solid steel shaft has to transmit 350 kW at 900 r.p.m. Find the diameter of the shaft if the shear stress is to be limited to 125 N/mm2. Calculate the diameter of the shaft if hollow shaft is provided of internal diameter equals 0.75 times external diameter (c) A simply supported beam 10 m long carries three point loads at 50 kN, 60 kN and 80 kN at 3m, 5m and 8m from left support. Draw S.F. and B.M. diagram for the beam. (a) Define friction and State laws of Dry friction. (b) The T-section is manufactured by connecting two equal rectangular blocks having size 200 mm x 30 mm. Determine moment of inertia of the section about its horizontal axis. (c) Find the magnitude of the Horizontal force ‘P’ applied to the lower block to cause impending motion as shown in figure 5. Take µ = 0.3 at all contact surfaces. Weight of block ‘A’ is 300 N and weight of block ‘B’ is 1200 N.

07

03

04

07

03 04

07

OR

Q.4

Q.5

Q.5

(a) Determine reaction at supports for the Beam as shown in Figure 6. (b) Determine deformation in each part of the bar ABCD shown in Figure 7. Take E = 2 x 105 N/mm2. (c) A uniform ladder AB weighing 230 N and 4m long, is supported by vertical wall at top end B and by horizontal floor at bottom end A. A man weighing 550 N stood at the top of the ladder. Determine minimum angle θ of ladder AB with floor for the stability of ladder. Take co efficient of friction between ladder and wall as 1/3 and between ladder and floor as 1/4. (a) Derive formula for determine volumetric strain of circular bar of diameter ‘d’, length ‘L’, modulus of elasticity ‘E’ subjected to axial tensile force ‘P’ and Poisson’s ratio ‘µ’. (b) The Rectangular block of size 300mm (b) x 450mm (d) is subjected to a uniform bending moment 120 kNm. Calculate the bending stresses at extreme fiber of the blocks. Also, find out total tensile and compressive forces due to bending stresses. Draw bending stress distribution diagram also. (c) A rectangular block of size 350mm (l) x 50mm (b) x 150mm (h) is subjected to forces shown in figure 8. E = 2 x 105 N/mm2 and Poisson’s ratio is 0.28, calculate the Change in volume of block. OR (a) State the condition of equilibrium for Co-planner force system. (b) The Rectangular block of size 300mm (b) x 450mm (d) is subjected to a shear force 80 kN. Calculate the Shear stresses at neutral axis and Junction of the blocks. Draw Shear stress distribution diagram also. (c) A system of connected flexible cables shown in figure 9 is supporting two vertical forces 300 N and 400 N at points B and D. determine the forces in various segments of the cable.

03 04 07

03

04

07

03 04

07

*************

2

75 N 350 150 mm

175 mm 50 mm

450

A

50 N

20 N

Figure 1 60 N/mm2

30 N/mm2 90 N/mm2

90 N/mm2

30 N/mm2

60 N/mm2 Figure 3

Figure 5 300

E A

2m P

B

A

15 mm Ø

20 mm Ø

B

30 KN

C

10 mm Ø

D 80 KN

50 KN

1250 mm

1500 mm

1200 mm

Figure 7

Ø = diameter of bar

125 kN 75 kN

C

150 mm 150 kN

150 kN

300 A

figure 9 E

B 45

600

0

300 N 75 kN

350 mm 125 kN

50 mm Figure 8

D 400 N

3

Seat No.: ________

Enrolment No.___________

GUJARAT TECHNOLOGICAL UNIVERSITY B.E. SEMESTER– III (NEW) EXAMINATION – SUMMER 2015

Subject Code: 2130003 Subject Name: Mechanics of Solids Time:02.30pm-05.00pm

Date:11/06/2015 Total Marks: 70

Instructions: 1. Attempt all questions. 2. Make suitable assumptions wherever necessary. 3. Figures to the right indicate full marks.

Q.1

Q.2

Q.3

Q.3

Q.4

(a) Explain following terms (i) Rigid body, (ii) Deformable body (iii) Elastic body (b) State ‘Hooks Law’. Derive formula to determine change in length (δL) for the uniform, homogeneous axially loaded member of length (L), c/s area (A) and modulus of elasticity (E), subjected to axial tensile force (P). (c) A stepped circular bar ABC is axially loaded as shown in fig. (i), is in equilibrium. The diameter of part AB is 50 mm throughout its length, whereas diameter part BC is uniform decreasing from 40 mm at B to 30 mm at C. Determine (i) magnitude of unknown force ‘P’ (ii) stress in part AB and (iii) change in length of part BC. Take modulus of elasticity = 2 x 105 N/mm2.

03 04

Derive relation between Rate of loading, Shear force and Bending moment at a section in a beam subjected to general loading. Also derive condition for maximum BM in terms of SF. (b) Find magnitude and inclination with +X-axis of resultant of force system shown in fig (ii). Identify type of force system. OR (b) A cylindrical roller weighing 1000 N is resting between two smooth surfaces inclined at 60º and 30º with horizontal as shown in fig. (iii). Draw free body diagram and determine reactions at contact points A and B.

07

Calculate shear force and bending moment at salient points for the beam shown in fig. (iv). Also plot neat shear force and bending moment diagrams indicating values at above points. Locate point of contraflexure from support B.

07

(b) A reinforced concrete column 500 mm x 500 mm in section is reinforced with four steel bars of 25 mm diameter, one in each corner. The column is carrying an axial load of 1000 kN. Find the stresses in concrete and steel bars. Take E for steel = 210 GPa and E for concrete = 14 GPa. OR (a) Explain (i) Angle of limiting friction (ϕlim), (ii) Angle of repose (λ), and (iii) Coefficient of friction (μ) . Also prove that ϕlim and λ are numerically same. (b) Determine the location of centroid of plane lamina shown in fig. (v) with respect to point O.

07

Two blocks are connected by a horizontal link AB are supported on two rough planes as shown in fig. (vi). The coefficient of friction for block A is 0.4 and angle of friction for block B is ϕ = 15º What is the smallest weight WA of the block A for which the equilibrium of the system can exist? (b) State Parallel axis theorem and perpendicular axis theorem. Derive formula for moment of Inertia of a rectangular section about horizontal centroidal axis from first fundamental. Using parallel axis theorem derive formula for moment of inertia about base of the section. OR

07

(a)

(a)

(a)

1

07

07

07

07 07

07

Q.4

Q.5

Q.5

A steel circular bar of 16 mm diameter is placed inside a copper tube, having internal diameter of 20 mm and thickness of 2.5 mm as shown in fig. (vii). Both the ends are rigidly fixed and initially stress free. If the temperature of assembly is increased by 50ºC, compute magnitude and nature of stresses produced in each material. Take modulus of elasticity of steel and copper as 200 GPa and 100 GPa respectively. Take coefficient of thermal expansion (per ºC) for steel and copper as 12 x 10-6 and 18 x 10-6 respectively. (b) A rectangular block is 205 mm long, 100 mm wide and 80 mm thick. It is subjected to a tensile load of 200 kN, compressive load of 300 kN and tensile load of 250 kN along its length, width and thickness respectively. Find the change in volume of the block. Also calculate shear modulus for the block material. Take E = 210 GPa and poison’s ratio μ = 0.25

07

Draw bending stress distribution diagram across the cross section of a ‘T’ beam, having flange 150 x 20 mm and web 20 x 250 mm, carrying pure hogging bending moment of 30 kNm at the section. (b) Write the assumptions for finding out shear stress in a circular shaft, subjected to torsion. Prove that τ/R = Cθ/L with usual notations for circular shaft. OR (a) Draw shear stress distribution diagram across the cross section of a ‘T’ beam, having flange 150 x 20 mm and web 20 x 250 mm, carrying pure shear force of 50 kN at the section. (b) A machine component is subjected to the stresses as shown in fig. (viii). Find the normal and shearing stresses on the section AB inclined at an angle of 60º with horizontal (x-x axis). Also find the resultant stress on the section.

07

(a)

(a)

40kN

B P kN 400 mm

60kN -X

600 mm

O 60º

B

60°

+X

30°

100 kN

-Y Fig.(ii) Q.2(b)

16 kN

16 kN/m

WA = ?

Y

B

A

WB = 400 N

A

B

C 2m

1m 300mm

Fig.(iv) Q.3(a)

O

30º 400mm

150mm

X Fig.(vi) Q.4(a)

Fig.(v) Q.3(b) OR

Steel bar Copper tube 2.5 mm

50 MPa τ= 25 MPa A

16mm 20mm Longitudinal Section

07

Fig.(iii) Q.2(b) OR 25 kN

D

A

40° 30º

Fig. (i) Q.1(c)

4m

07

50 kN

20º

C

07

+Y

75 kN A

07

2.5 mm Cross Section

100 MPa

100 MPa

Fig. (vii) Q.4(a) OR

60º

τ= 25 MPa

B

50 MPa Fig. (viii) Q.5(b) OR

2

DEPARTMENT OF CIVIL ENGINEERING 3130003- MOS (SHORT QUESTIONS)

Short Questions Sr No 1

Question

Answer

Three coplanar non-parallel forces in equilibrium will _________ [always, never, sometimes] be concurrent

Sometimes

2

106 kg = _________ [103, 10-6,109, 10-3] Mg.

3

Coefficient of static fiction is ___________ [less than, more than, Equal to] coefficient of dynamics fiction.

More than

4

A mild steel bar under tension test shows property of ___________ [Malleability, ductility, tension ability].

Ductility

5

The shape of shear force diagram for cantilever beam subjected to couple at free end is ___________ [horizontal straight line, zero, Parabola, incline straight line].

zero

6

The ratio of the maximum shear stress to average shear stress is for 4/3, the cross section would be __________ [triangular, rectangular, Circular, hexagonal].

Circular

7

Which one of expressions is NOT true __________ [E = 2G (1+µ), E = 3K (1-2µ), E = 9KG/ (3G+K), M = s. I/y].

E = 9KG/ (3G+K)

8

Point of contra flexure is where ___________ [shear force is zero, shear force changes sign, bending moment changing sign, bending Moment is zero].

Bending moment changes its sign

9

Lateral strains are ___________ longitudinal strains. (always less than, sometimes less than, never less than)

Less than

10

Two forces under equilibrium must be _______( non rectilinear, rectilinear, parallel)

rectilinear

11

True relation between dynamic coefficient of friction ( µ d) and static coefficient of friction (µs) is __________ (µd < µs , µd = µs, µd > µs)

µd < µs

12

_____________ is a scalar quantity.( momentum , force , work)

Work

103

13

100 mm

-7

10-7

14

Beam extends beyond the support then that beam is known As________ beam.

Overhanging beam

15

Moment of inertia of any plane area is maximum about an axis Passing though_________.

Centre of gravity

16

At free end of a cantilever bending moment is always________ unless a concentrated moment is applied at the free end.

Zero

17

__________ of a force is the procedure of splitting a force into number of components

Resolution

18

_________ is equal and opposite to the resultant of several

________________

5

6

µm ( 10 , 10 , 10 )

Reaction

Forces, acting on a body.

19

Force of friction is ________ to the applied force, which tends to move the body.

Opposite

20

Splitting of force in given directions is known as __________ of force. (resolution, composition, division)

resolution

21

Poisson’s ratio is ratio of ___________ (longitudinal to lateral strain, lateral to longitudinal strain, shear stress to shear strain)

Lateral to longitudinal

22

Bending moment is ________ at a hinged support. (always maximum, always zero)

Always zero

23

Force can be defined in term of its_____,

24

Inertia is the resistance of body to any change in its _______[

25

Solid of mechanics comprises study of mechanics of ______ bodies and mechanics of deformable bodies

26

The process of replacing a set of forces by a single force is called_____ of force

27

A sketch showing a body in equilibrium along with all the forces acting on it, is called a __________ []

_______

, and point of

________

[.

magnitude, direction, point of application Velocity rigid Composition Free body diagram

DEPARTMENT OF CIVIL ENGINEERING 3130003- MOS (SHORT QUESTIONS)

28

Two unlike and parallel forces, will form a ________

29

Hinge support cannot prevent any ________ of the body that they support.

30

Centre of gravity is defined as the point on which the _______of the body acts.

31

The centroid is the _______ of a homogeneous object. [

32

A cylinder is a surface of revolution generated by revolving a _________about a fixed axis

33

Radius of gyration is given by the equation KX = __________

34

Friction ________the efficiency of a machine and _________ the work required to be done

reduces, increase

35

Dry friction occurs when unlubricated surfaces are in contact under condition of ______

Sliding

36

Fluid friction occurred between _______ of fluids at different _________[

37

At a point when motion is about to begin, the frictional force has reached its _________ value

limiting

38

Frictional force will always be in a direction ________ to that in which the body tends to move]

Opposite

39

The ration of shear stress to shear strain is known as _________. [

40

Poisson’s ratio for most of the materials is close to _________.

41

Young’s modulus of elasticity for a perfectly rigid body is _______.

42

A metal bar is heated without any restrain, the temperature stress will be _____.

Zero

43

Diagonal tensile and compressive stresses in the member are the result of ________ force.

Shear

44

Deformation of the statically indeterminate structures can be obtained by using _______ condition and ___________ condition.

equilibrium , capability

45

Lateral strains are ___________ longitudinal strains.

46

A __________ support is an example of a support with an unknown line of action.

Hinge

47

The stress in the beam is lesser if its section modulus is _______.

higher

48 49

Moment of resistance of the flanges of an I- section is _______ its web

higher than

IF arm of couple is doubled. Its moment will ________-

Be doubled

50

In a Cantilever, carrying a load whose intensity varies uniformly from zero at the free end to w per unit run at the fixed end, the SF diagram changes following a ______

51

A body is said to be in equilibrium if it has no linear motion [true, false]

52

If arm of couple is doubled its moment will

53

If a ladder is not in equilibrium against a smooth vertical wall, than it can be made in equilibrium by

54

Ductility is the ability of a material to be drawn in the form of a _________

Increasing angle of inclination with horizontal Wires

55

Statically determinate structures can be analyzed using the 3 equation of _______

Static equilibrium

56

The relationship between shear force and bending moment is given by the differential _______

Couple Rotation weight geometrical center . Straight line

layers, velocity

Modulus of rigidity 0.5 Infinity

always less than

Parabolic law False Be doubled

Equation (δM/δx)= -F