Mechanics of Solids [3 1 0 4]
CIE 101 / 102
First Year B.E. Degree
Mechanics of Solids
PART- IIPART- I
Mechanics of Deformable Bodies
Mechanics of Rigid Bodies
COURSE CONTENT IN BRIEF
1. Resultant of concurrent and non-concurrent coplanar forces.
2. Equilibrium of concurrent and non-concurrent coplanar forces.
3. Centroid of plane areas
4. Moment of Inertia of plane areas
5. Kinetics: Newton’s second law, D’Alembert’s principle, Work- Energy, and Impulse- Momentum principle.
PART I Mechanics of Rigid Bodies
PART II Mechanics of Deformable bodies6. Simple stresses and strains
7. Statically indeterminate problems and thermal stresses
8. Stresses on inclined planes
9. Stresses due to fluid pressure in thin cylinders
Books for Reference1.Engineering Mechanics, by Meriam & Craige, John Wiley & Sons.
2.Engineering Mechanics, by Irwing Shames, Prentice Hall of India.
3.Mechanics for Engineers, by Beer and Johnston, McGraw Hills Edition
4.Engineering Mechanics, by K.L. Kumar, Tata McGraw Hills Co.5. Machanics of Materials, by E.P.Popov
6. Machanics of Materials, by E J Hearn
7. Strength of materials, by Beer and Johnston
8. Strength of materials, by F L Singer & Andrew Pytel
9. Strength of Materials, by B.S. Basavarajaiah & P. Mahadevappa
10. Strength of Materials, by Ramamruthum
11. Strength of Materials, by S S Bhavikatti
PART - I
MECHANICS OF RIGID BODIES
Definition of Mechanics :In its broadest sense the term ‘Mechanics’ may be defined as the ‘Science which describes and predicts the conditions of rest or motion of bodies under the action of forces’.
INTRODUCTION
PART - I Mechanics of Rigid Bodies
This Course on Engineering Mechanics comprises of Mechanics of Rigid bodies and the sub-divisions that come under it.
Engineering Mechanics
Mechanics of Solids Mechanics of Fluids
Rigid Bodies DeformableBodies
Statics Dynamics
Kinematics Kinetics
Strength of Materials
Theory of Elasticity
Theory of Plasticity
Ideal Fluids
ViscousFluids
CompresFluids
Branches of Mechanics
Concept of Rigid Body :
It is defined as a definite amount of matter the parts of which are fixed in position relative to one another under the application of load.
Actually solid bodies are never rigid; they deform under the action of applied forces. In those cases where this deformationis negligible compared to the size of the body, the body may be considered to be rigid.
Particle
A body whose dimensions are negligible when compared to the distances involved in the discussion of its motion is called a ‘Particle’.
For example, while studying the motion of sun and earth, they are considered as particles since their dimensions are small when compared with the distance between them.
Force
It is that agent which causes or tends to cause, changes or
tends to change the state of rest or of motion of a mass.
A force is fully defined only when the following four
characteristics are known:
(i) Magnitude
(ii) Direction
(iii) Point of application
(iv) Sense.
Force:
characteristics of the force 100 kN are :
(i) Magnitude = 100 kN
(ii) Direction = at an inclination of 300 to the x-axis
(iii) Point of application = at point A shown
(iv) Sense = towards point A
300
100 kN
A
Scalars and Vectors
A quantity is said to be a ‘scalar’ if it is completely defined by
its magnitude alone.
Example : Length, Area, and Time.
A quantity is said to be a ‘vector’ if it is completely defined only
when its magnitude and direction are specified.
Example : Force, Velocity, and Acceleration.
Principle of Transmissibility : It is stated as follows : ‘The external effect of a force on a rigid body is the same for all points of application along its line of action’.
P PA B
For example, consider the above figure. The motion of the block will be the same if a force of magnitude P is applied as a push at A or as a pull at B.
The same is true when the force is applied at a point O.
P PO
1. RESULTANT OF COPLANAR FORCES
Resultant, R : It is defined as that single force which can replace a set of forces, in a force system, and cause the same external effect.
R
θ
=
same isA particle,on effect 321
externalFFFR ++=F3
F1
F2
A A
Resultant of two forces acting at a point
Parallelogram law of forces : ‘If two forces acting at a point are represented in magnitude and direction by the two adjacent sides of a parallelogram, then the resultant of these two forcesis represented in magnitude and direction by the diagonal of the parallelogram passing through the same point.’
B C
AO
P2
P1
Rα
θ
Contd..
B C
AO
P2
P1
Rα
θ
In the above figure, P1 and P2, represented by the sides OA and OB haveR as their resultant represented by the diagonal OC of the parallelogram OACB.
It can be shown that the magnitude of the resultant is given by:R = √P1
2 + P22 + 2P1P2Cos α
Inclination of the resultant w.r.t. the force P1 is given by:
θ = tan-1 [( P2 Sin α) / ( P1 + P2 Cos α )]
Resultant of two forces acting at a point at right angle
B C
AO
P2
P1
Rα
θ
If α = 900 , (two forces acting at a point are at right angle)
222
1 PPR +=
1
2tanPP
=θθ
B C
AO
P2
P1
R
Triangle law of forces
‘If two forces acting at a point can be represented both in magnitude and direction, by the two sides of a triangle taken in tip to tail order, the third side of the triangle represents both in magnitude and direction the resultant force F, the sense of the same is defined by its tail at the tail of the first force and its tip at the tip of the second force’.
Triangle law of forcesLet F1 and F2 be the two forces acting at a point A and θ is the included angle.
θA
F1
F2
θF1
F2
R=
‘Arrange the two forces as two sides of a triangle taken in tip to tail order, the third side of the triangle represents both in magnitude and direction the resultant force R.
the sense of the resultant force is defined by its tail at the tail of the first force and its tip at the tip of the second force’.
Triangle law of forces
θF1
F2
R
θA
F1
F2
=
β
α
R
F1
F2 )180sin(sinsin21
βααβ −−==
RFF
(180 - α - β) = θ
where α and β are the angles made by the resultant force with the force F1 and F2 respectively.
Component of a force :
Component of a force, in simple terms, is the effect of aforce in a certain direction. A force can be split into infinite number of components along infinite directions.
Usually, a force is split into two mutually perpendicularcomponents, one along the x-direction and the other along y-direction (generally horizontal and vertical, respectively).
Such components that are mutually perpendicular are called‘Rectangular Components’.
The process of obtaining the components of a force is called ‘Resolution of a force’.
Rectangular component of a force
θx
F
θx
F
θx
F
Fx
Fy = Fy
Fx
Consider a force F making an angle θx with x-axis.
Then the resolved part of the force F along x-axis is given by Fx = F cos θx
The resolved part of the force F along y-axis is given by
Fy = F sin θx
Oblique component of a force
Let F1 and F2 be the oblique components of a force F. The components F1 and F2 can be found using the ‘triangle law of forces’.
β
α
F
F1
F2
The resolved part of the force F along OM and ON can obtained by using the equation of a triangle.
βα
F
F1
F2 M
O
N
F1 / Sin β = F2 / Sin α = F / Sin(180 - α - β)
Sign Convention for force components:
+ve
+vex
xyy
The adjacent diagram gives the sign convention for force components, i.e., force components that are directed along positive x-direction are taken +ve for summation along the x-direction.
Also force components that are directed along +ve y-direction are taken +ve for summation along the y-direction.
Classification of force system
Force system
Coplanar Forces Non-Coplanar Forces
Concurrent Non-concurrentConcurrent Non-concurrent
Like parallel Unlike parallelLike parallel Unlike parallel
A force that can replace a set of forces, in a force system,and cause the same ‘external effect’ is called the Resultant.
RESULTANT OF COPLANAR NON CONCURRENT FORCE SYSTEM
Coplanar Non-concurrent Force System:This is the force system in which lines of action of
individual forces lie in the same plane but act at different points of applications.
Fig. 1
F2F1
F3
Fig. 2
F1 F2
F5F4
F3
1. Parallel Force System – Lines of action of individual forces are parallel to each other.
2. Non-Parallel Force System – Lines of action of the forces are not parallel to each other.
MOMENT OF A FORCE ABOUT AN AXIS
The applied force can also tend to rotate the body about an axis in addition to motion. This rotational tendency is known as moment.
Definition: Moment is the tendency of a force to make a rigid body to rotate about an axis.
This is a vector quantity having both magnitude and direction.
MOMENT OF A FORCE ABOUT AN AXIS
Moment Axis: This is the axis about which rotational tendency is determined. It is perpendicular to the plane comprising moment arm and line of action of the force (axis 0-0 in the figure)
Moment Center: This is the position of axis on co-planar system. (A).
Moment Arm:Perpendicular distance from the line of action of the force to moment center. Distance AB = d.
Magnitude of moment:
It is computed as the product of the of the force and the perpendicular distance from the line of actionto the point about which moment is computed. (Moment center).
MA = F×d
= Rotation effect because of the force F, about the point A (about an axis 0-0)
Unit – kN-m, N-mm etc.
The sense is obtained by ‘Right Hand Thumb’ rule.
‘If the fingers of the right hand are curled in the direction of rotational tendency of the body, the extended thumb represents the sense of moment vector’.
Sense of moment:
For the purpose of additions, the moment direction may be considered by using a suitable sign convention such as +vefor counterclockwise and –vefor clockwise rotations or vice-versa.
MM
VARIGNON’S THEOREM (PRINCIPLE OF MOMENTS)
Statement: The moment of a force about a moment center or axis is equal to the algebraic sum of the moments of its component forces about the same moment center (axis).
θ
P
A
dθ
P
A
P sinθ
P cosθd1
d2
Moment of Force P about the point A,
P x d
Algebraic sum of Moments of components of the Force P about the point A,
P cosθ x d1 + P sinθ x d2
=
VARIGNON’S THEOREM (PRINCIPLE OF MOMENTS)
Proof (by Scalar Formulation):Proof (by Scalar Formulation):
Let ‘R’ be the given force.
‘P’ & ‘Q’ are component forces of ‘R’. ‘O’ is the moment center.
QR
αAO
β p
qr P
γ
Y
X
p, r and q are moment arms from ‘O’of P, R and Q respectively.
α, β and γ are the inclinations of ‘P’, ‘R’ and ‘Q’ respectively w.r.to X –axis.
We have,Ry = Py + QyR Sinβ = P Sinα + Q Sin γ ----(1)From ∆le AOB, p/AO = Sin αFrom ∆le AOC, r/AO = Sin βFrom ∆le AOD, q/AO = Sin γFrom (1),∴ R ×(r/AO) = P ×(p/AO) + Q ×(q/AO)
i.e., R × r = P × p + Q × q
Moment of resultant R about O = algebraic sum of moments of component forces P & Q about same moment center ‘O’.
Ry
Py
R
A O
Q
P
p
q r
α
βγ
Y
X
B
CD
Qy
COUPLE
Two parallel, non collinear (separated by certain
distance) forces that are equal in magnitude and opposite
in direction form ‘couple’.
d
F
F
The algebraic summation of the
two forces forming couple is zero.
=Hence, couple does not produce any
translation and produces only rotation. M = F x d
RESOLUTION OF A FORCE INTO A FORCE-COUPLE SYSTEM
Replace the force F acting at the point A to the point B F
AB
Apply two equal and opposite forces of same magnitude & direction as Force F at point B, so that external effect is unchanged
F
AB
F
Fd
Of these three forces, two forces i.e., one at A and the other oppositely directed at B form a couple. Moment of this couple, M = F × d.
Thus, the force F acting at a point such as A in a rigid body can be moved to any other given point B, by adding a couple M. The moment of the couple is equal to moment of the force in its original position about B.
Third force at B is acting in the same direction as that at P.
F
AB
F
F
AB
F
=M = F x dd
TYPES OF LOADS ON BEAMS
1. Concentrated Loads – This is the load acting for very small length of the beam.
(also known as point load, Total load W is acting at one point )
2. Uniformly distributed load – This is the load acting for a considerable length of the beam with same intensity of w kN/m throughout its spread.
Total intensity, W = w × L
(acts at L/2 from one end of the spread)
W kN
w kN/m
L
W = (w x L) kN
L
L/2
3. Uniformly varying load – This load acts for a considerable length of the beam with intensity varying linearly from ‘0’ at one end to w kN/m to the other representing a triangular distribution.
Total intensity of load = area of triangular spread of the load
W = 1/2× w × L.
(acts at 2×L/3 from ‘Zero’ load end)
w kN/m
L
L
W = ½ × L × w
2/3 ×L 1/3 ×L
EXERCISE PROBLEMS1. Resultant of force system
Q1 A body of negligible weight, subjected to two forces F1= 1200N, and F2=400N acting along the vertical, and the horizontal respectively, is shown in figure. Find the component of each force parallel, and perpendicular to the plane.
= 1200 N
X
34
YF2
F1
= 400 N
Ans : F1X = -720 N, F1Y = -960N, F2X = 320N, F2Y = -240N
EXERCISE PROBLEMS1. Resultant of force system
Q2. Determine the X and Y components of each of the forces shown in the figure.
30º40º
12
5
300 N
390 N
400 N
X
Y
F1 =
F2 =
F3 =
(Ans : F1X = 259.81 N, F1Y= -150 N, F2X= -150N, F2Y= 360 N,
F3X = -306.42 N, F3Y= -257.12N )
EXERCISE PROBLEMS1. Resultant of force system
Q3. Obtain the resultant of the concurrent coplanar forces shown in the figure
600N
200N
800N
20º 40º
30º
(Ans: R = 522.67 N, θ = 68.43º)
EXERCISE PROBLEMS1. Resultant of force system
Q4. A disabled ship is pulled by means of two tug boats as shown in FIG. 4. If the resultant of the two forces T1 and T2 exerted by the ropes is a 300 N force acting parallel to the X – direction, find :
(a) Force exerted by each of the tug boats knowing α = 30º.(b) The value of α such that the force of tugboat 2 is minimum,
while that of 1 acts in the same direction.Find the corresponding force to be exerted by tugboat 2.
T2
R = 300 N
T1
α
20º
FIG. 4
X - direction
( Ans: a. T1= 195.81 N, T2 = 133.94 Nb. α = 70º, T1 = 281.91 N, T2(min) = 102.61 N )
EXERCISE PROBLEMS1. Resultant of force system
Q5. An automobile which is disabled is pulled by two ropes as shown in the figure. Find the force P and resultant R, such that R is directed as shown in the figure.
P
Q = 5 kN
R20º
40º
(Ans: P = 9.4 kN , R = 12.66 kN)
EXERCISE PROBLEMS1. Resultant of force system
Q6. A collar, which may slide on a vertical rod, is subjected to three forces as shown in figure. The direction of the force F may be varied Determine the direction of the force F, so that resultant of the three forces is horizontal, knowing that the magnitude of F is equal to(a) 2400 N, (b)1400N
1200 N
800 N60º
θF
RODCOLLAR
( Ans: a. θ = 41.81º ; b. The resultant cannot be horizontal.)
Q7. Determine the angle α and the magnitude of the force Q such that the resultant of the three forces on the pole is vertically downwards and of magnitude 12 kN. Refer figure
EXERCISE PROBLEMS1. Resultant of force system
8kN5kN
Q30º
α
Fig. 7(Ans: α = 10.7 º, Q = 9.479 kN )
Q8. Determine the resultant of the parallel coplanarforce system shown in figure.
400 N
1000 No
60º
60º 30º10º
(Ans. R=800N towards left, d=627.5mm)
EXERCISE PROBLEMS1. Resultant of force system
600 N
2000 N
Q9. Four forces of magnitudes 10N, 20N, 30N and 40Nacting respectively along the four sides of a squareABCD as shown in the figure. Determine the magnitude, direction and position of resultant w.r.t. A.
20N
10N
40N
a
a
30N
AB
C
(Ans:R=28.28N, θ=45º, x=1.77a)
EXERCISE PROBLEMS1. Resultant of force system
D
Q10. Four parallel forces of magnitudes 100N, 150N, 25N and 200N acting at left end, 0.9m, 2.1m and 2.85m respectively from the left end of a horizontalbar of 2.85m. Determine the magnitude of resultantand also the distance of the resultant from the leftend.
(Ans: R = 125 N, x = 3.06 m)
EXERCISE PROBLEMS1. Resultant of force system
Q11. Reduce the given forces into a single force and a couple at A.
70.7 kN200 kN
1.5m
A1m
100 N80 N
30º45º
30º
(Ans:F=320kN, θ=14.48º, M=284.8kNm)
EXERCISE PROBLEMS1. Resultant of force system
Q12. Determine the resultant w.r.t. point A.
(Ans: R = 450 kN, X = 7.5 kNm)
EXERCISE PROBLEMS1. Resultant of force system
A
150 N
1.5m3m1.5m
150 Nm
500 N100 N
2. EQUILIBRIUM OF FORCE SYSTEMS
2. EQUILIBRIUM OF FORCE SYSTEMS
EQUILIBRIUM OF - CONCURRENT COPLANAR FORCE SYSTEMS
Definition:-If a system of forces acting on a body, keeps the body in a state of rest or in a state of uniform motion along a straight line, then the system of forces is said to be in equilibrium.ALTERNATIVELY, if the resultant of the force system is zero, then, the force system is said to be in equilibrium.
EQUILIBRIUM OF - CONCURRENT COPLANAR FORCE SYSTEMS
Conditions for Equilibrium :
A coplanar concurrent force system will be in equilibrium if itsatisfies the following two conditions:
i) ∑ Fx = 0; and ii) ∑ Fy = 0
i.e. Algebraic sum of components of all the forces of the system, along two mutually perpendicular directions, is ZERO.
XY
Graphical conditions for Equilibrium
Triangle Law: If three forces are in equilibrium, then, they form a closed triangle when represented in a Tip to Tail arrangement, as shown in Fig 2.1
Fig 2.1
F 3 F2
F1
F2
F3
F1
Polygonal Law: If more than three forces are in equilibrium, then, they form a closed polygon when represented in a Tip to Tail arrangement, as shown in Fig. 2.2.
Fig 2.2 F5
F 4 F3
F2
F1
F5
F4
F3 F2
F1
LAMI’S THEOREM
If a system of three forces is in equilibrium, then, each force of the system is proportional to sine of the angle between the other two forces (and constant of proportionality is the same for all the forces). Thus, with reference to Fig.2.3, we have,
F2
β
α
γ
Fig. 2.3
F1
F3
γβα SinF
SinF
SinF 321 ==
Note: While using Lami’s theorem, all the three forces should be either directed away or all directed towards the point of concurrence.
EQUILIBRIUM OF NON-CONCURRENT COPLANAR FORCE SYSTEM
When a body is in equilibrium, it has neither translatory nor rotatory motion in any direction.
Thus the resultant force R and the resultant couple M are both zero, and we have the equilibrium equations for two dimensional force system
∑ Fx = 0; ∑ Fy = 0 ∑M = 0
These requirements are both necessary and sufficient conditions for equilibrium.
SPACE DIAGRAMS & FREE BODY DIAGRAMS
Space Diagram (SPD) : The sketch showing the physical conditions of the problem, like, the nature of supports provided; size, shape and location of various bodies; forces applied on the bodies, etc., is known as space diagram.
eg, Fig 2.4 is a space diagramWeight of sphere = 0.5 kN,Radius = 1m
Cable
P = 2kN30°
Fig. 2.4 SPDwall
3 m θ
Sphere
Free Body Diagram (FBD) :
It is an isolated diagram of the body being analyzed (called free body), in which, the body is shown freed from all its supports and contacting bodies/surfaces. Instead of the supports and contacting bodies/surfaces, the reactive forces exerted by them on the free body is shown, along with all other applied forces.
A Few Guidelines for Drawing FBD
1) Tensile Force: It is a force trying to pull or extend the body. It is represented by a vector directed away from the body.
2) Compressive Force: It is force trying to push or contract the body. It is represented by a vector directed towards the body.
3) Reactions at smooth surfaces: The reactions of smooth surfaces, like walls, floors, Inclined planes, etc. will be normal to the surface and pointing towards the body.
4)Forces in Link rods/connecting rods: These forces will be acting along the axis of the rod, either towards or away from the body.(They are either compressive or tensile in nature).
5) Forces in Cables (Strings or Chords): These can only be tensile forces. Thus, these forces will be along the cable and directed away from the body.
T = Tension in the cable
Rw = Reaction of the wall
W = self weight of the sphere
P = external load acting on the sphere
Free Body Diagrams of the sphere shown in Fig. 2.4
Fig. 2.5 F B D of Sphere
P = 2kN
30°
θ
Sphere
T
W=0.5kN
Rw
Detach the sphere from all contacts and replace that with forces like:Cable contact is replaced by the force tension = TContact with the smooth wall is replaced by the reaction Rw.
Supports: A structure is subjected to external forces and transfers these forces through the supports on to the foundation. Therefore the support reactions and the external forces together keep the structure in equilibrium.
There are different types of supports.a) Roller Support b) Hinged or pinned support c) Fixed or built in support
Supports
Types of Supports Action on body
(a) Flexible cable ,belt ,chain, ropeBODY
TBODYForce exerted by cable is always a tension away from the body in the direction of cable
(b) Smooth surfaces
Contact forces are normal to the surfaces
F F
900
900
Supports
(c) Roller support
A
A
Contact force is normal to the surface on which the roller moves. The reaction will always be perpendicular to the plane of the roller . Roller support will offer only one independent reaction component. (Whose direction is known.)
Supports
(d) pinned Support / hinged support
RvR
Rh
θ
AA
This support does not allow any translatory movement of the rigid body. There will be two independent reaction components at the support. The resultant reaction can be resolved into two mutually perpendicular components. Or it can be shown as resultant reaction inclined at an angle with respect to a reference direction.
Supports
(e) Fixed or Built-in Support
RAV
MRAH
A A
This type of support not only prevents the translatorymovement of the rigid body, but also the rotation of the rigid body. Hence there will be 3 independent reaction components of forces. Hence there will be 3 unknown components of forces, two mutually perpendicular reactive force component and a reactive moment as shown in the figure.
TYPES OF BEAMS
A member which is subjected to predominantly transverse loads and supported in such a way that rigid body motion is prevented is known as beam. It is classified based on the support conditions. A beam generally supported by a hinge or roller at the ends having one span (distance between the support) is called as simply supported beam. A beam which is fixed at one end and free at another end is called as a cantilever beam.
span
A B
span
A B
(a) Simply supported beam
TYPES OF BEAMS
B
span
A
Rv
MRH A B
(b) Cantilever beam
TYPES OF BEAMS
If one end or both ends of the beam project beyond the support it is known as overhanging beam.
A B
A B
(c) Overhanging beam (right overhang)
Statically determinate beam
Using the equations of equilibrium given below, if all the reaction components can be found out, then the beam is a statically determinate beam
the equations of equilibrium∑ Fx = 0; ∑ Fy = 0 ∑M = 0
FRICTION
Friction is defined as the contact resistance exerted by one body upon another body when one body moves or tends to move past another body. This force which opposes the movement or tendency of movement is known as frictional resistance or friction. Friction is due to the resistance offered by minute projections at the contact surfaces. Hence friction is the retarding force, always opposite to the direction of motion. Friction has both advantages & disadvantages.
Disadvantages ---- Power loss, wear and tear etc.
Advantages ---- Brakes, traction for vehicles etc.
FRICTION
NF (Friction)
P
W
Hills & Vales Magnified Surface
Frictional resistance is dependent on the amount of wedging action between the hills and vales of contact surfaces. The wedging action is dependent on the normal reaction N.
FRICTION
Frictional resistance has the remarkable property of adjusting itself in magnitude of force producing or tending to produce the motion so that the motion is prevented.
When P = 0, F = 0 block under equilibrium
When P increases, F also increases proportionately to maintain equilibrium. However there is a limit beyond which the magnitude of this friction cannot increase.
FRICTION
When the block is on the verge of motion(motion of the block is impending) F attains maximum possible value, which is termed as Limiting Friction. When the applied force is less than the limiting friction, the body remains at rest andsuch frictional resistance is called the static friction.
Further if P is increased, the value of F decreases rapidly and then remains fairly a constant thereafter. However at high speeds it tends to decrease. This frictional resistance experienced by the body while in motion is known as Dynamic friction OR Kinetic Friction.
FRICTION
Sliding friction friction experienced when a body slides over another surface.
Dynamic FrictionRolling friction friction experienced by a body when it rolls over a surface.
FRICTION
Where Fmax = Limiting Friction
N= Normal Reaction between the contact surfaces
µ =Coefficient of friction
Fmax
N
Fmax
P
W
φ
N R
F α N
Fmax = µN
µ =
Note : Static friction varies from zero to a maximum value. Dynamic friction is fairly a constant.
FRICTIONAngle of Friction
The angle between N & R depends on the value of F.
This angle θ, between the resultant R and the normal reaction N is termed as angle of friction.
Fmax
P
W
φ
N RAs F increases, θ also increases and will reach to a maximum value of φ when F is Fmax (limiting friction)
i.e. tanφ = (Fmax )/N = µ
Angle φ is known as Angle of limiting Friction.
Angle of limiting friction is defined as the angle between the resultant reaction (of limiting friction and normal reaction) and the normal to the plane on which the motion of the body is impending.
Angle of reposeWhen granular material is heaped, there exists a limit for the inclination of the surface. Beyond that angle, the grains start rolling down. This limiting angle upto which the grains repose (sleep) is called the angle of repose of the granular material.
FRICTION
FRICTION
Significance of Angle of repose:
The angle that an inclined plane makes with the horizontal, when the body supported on the plane is on the verge of motion due to its self -weight is equal to the angle of repose.
Angle of repose is numerically equal to Angle of limiting friction
FRICTIONLaws of dry friction
1. The magnitude of limiting friction bears a constant ratioto the normal reaction between the two surfaces. (Experimentally proved)
2. The force of friction is independent of the area of contactbetween the two surfaces.
3. For low velocities the total amount of friction that canbe developed is practically independent of velocity.It is less than the frictional force correspondingto impending motion.
Q1. A 10kN roller rests on a smooth horizontal floor and is held by the bar AC as shown in Fig(1). Determine the magnitude and nature of the force in the bar AC and reaction from the floor under the action of the forces applied on the roller. [Ans:FAC=0.058 kN(T),R=14.98 kN]
C
7kN
5kN
Fig(1)
A
450
300
2. EQUILIBRIUM OF FORCE SYSTEMS
EXERCISE PROBLEMS
Q2. A 10 kN weight is suspended from a rope as shown infigure. Determine the magnitude and direction of the least force P required to pull the rope, so that, the weight is shifted horizontally by 0.5m. Also, determine, tension in the rope in its new position. [Ans: P= 2.43 kN, θ = 14.480 ; T= 9.7kN.]
2m
10kN
Pθ
2. EQUILIBRIUM OF FORCE SYSTEMSEXERCISE PROBLEMS
Q3. Determine the value of P and the nature of the forces in thebars for equilibrium of the system shown in figure.[Ans: P = 3.04 kN, Forces in bars are Compressive.]
60
7545 45
P2kN
2. EQUILIBRIUM OF FORCE SYSTEMSEXERCISE PROBLEMS
Q4. A cable fixed as shown in Fig. supports three loads. Determine the value of the load W and the inclination of the segment BC. [Ans: W=25kN, θ = 54.780]
Loads are in kNW
22.520
B
C
DA
θ 60
30
2. EQUILIBRIUM OF FORCE SYSTEMSEXERCISE PROBLEMS
2. EQUILIBRIUM OF FORCE SYSTEMSEXERCISE PROBLEMS
Q5. Find the reactions at A,B,C and D for the beam loaded as shown in the figure. (Ans.RA=RB =34kN;RC=28.84kN;
MC=-140kNm ; θC=-33.69 ˚ )
12kN/m
4kN/m
12kN/m
4kN/m
20 kN
30kN
1m 2m 1m 1m 2m 1m 1m 2m
A BC
34
40kNm
2. EQUILIBRIUM OF FORCE SYSTEMSEXERCISE PROBLEMS
Q6. A uniform bar AB of weight 50N shown in the figure supports a load of 200N at its end. Determine the tension developed in the string and the force supported by the pin at B. (Ans. T=529.12N;RB=807.15N, θB=64.6˚)
2.5m 2.5m200N
2.5m
A
B60˚
string
2. EQUILIBRIUM OF FORCE SYSTEMSEXERCISE PROBLEMS
Q7. Find the position of the hinged support (x),such that the reactions developed at the supports of the beam are equal..
(Ans.x=2m.)
2.0m 1.4m1.0m 3.0m0.6
15kN18kN/m
10kN/m
x
Q8. A right angled bar ABC hinged at A as shown in fig carries two loads W and 2W applied at B &C .Neglecting self weight of the bar find the angle made by AB with vertical (Ans:θ =18.44˚)
2. EQUILIBRIUM OF FORCE SYSTEMSEXERCISE PROBLEMS
0.5L 2W
θ
B
L m
WC
A
2. EQUILIBRIUM OF FORCE SYSTEMSEXERCISE PROBLEMS
Q9. For the block shown in fig., determine the smallest force P required
a) to start the block up the plane b) to prevent the block moving down the plane.
Take µ = 0.20
P
25°
θ100N
[Ans.: (a) Pmin = 59.2N
(b) Pmin = 23.7N (b) θ = 11.3o]
2. EQUILIBRIUM OF FORCE SYSTEMSEXERCISE PROBLEMS
Q10. A block of weight 2000 N is attached to a cord passing over a frictionless pulley and supporting a weight of 800N as shown in fig. If µ between the block and the plane is 0.35, determine the unknown force P for impending motion(a) to the right (b) to the left
[Ans.: (a) P = 132.8N (b) P = 1252N]
2000N P800N30°
2. EQUILIBRIUM OF FORCE SYSTEMSEXERCISE PROBLEMS
Q11. Determine value of angle θ to cause the motion of 500N block to impend down the plane, if µ for all contact surfaces is 0.30.
500N
200N
θ = ?
[Ans.: θ = 28.4°]
2. EQUILIBRIUM OF FORCE SYSTEMSEXERCISE PROBLEMS
Q12. A horizontal bar 10m long and of negligible weight rests on rough inclines as shown in fig. If angle of friction is 15o, how close to B may the 200N force be applied before the motion impends.
100N 200N
2 m X = ?A B
30° 60°
[Ans.: x = 3.5m]
2. EQUILIBRIUM OF FORCE SYSTEMSEXERCISE PROBLEMS
Q13. Determine the vertical force P required to drive the wedge B downwards in the arrangements shown in fig. Angle of friction for all contact surfaces is 12o.Weight of block A= 1600 N.
AB
P
20°
[Ans.: P = 328.42N]
2. EQUILIBRIUM OF FORCE SYSTEMSEXERCISE PROBLEMS
Q14. Determine the force P which is necessary to start the wedge to raise the block A weighing 1000N. Self weight of the wedge may be ignored. Take angle of friction, φ = 15o for all contact surfaces.
20°
A
Pwedge
[Ans.: P = 1192N]
2. EQUILIBRIUM OF FORCE SYSTEMSEXERCISE PROBLEMS
Q15. A ladder of weight 200N, 6m long is supported as shown in fig. If µ between the floor and the ladder is 0.5 & between the wall and the ladder is 0.25 and it supports a vertical load of 1000N, determine a) the least value of α at which the ladder may be placed without slippingb) the reactions at A & B
[Ans.: (a) α = 56.3o (b) RA = 1193 N, RB = 550N]
α
1000N
5m
A
B
2. EQUILIBRIUM OF FORCE SYSTEMSEXERCISE PROBLEMS
Q16. An uniform ladder of weight 250N is placed against a smooth vertical wall with its lower end 5m from the wall. µbetween the ladder and the floor is 0.3. Show that the ladder remains in equilibrium in this position. What is the frictional resistance on the ladder at the point of contact between the ladder and the floor?
Smooth wall
12m
5mA
B
[Ans.: FA = 52 N]
2. EQUILIBRIUM OF FORCE SYSTEMSEXERCISE PROBLEMS
Q17. A ladder of length 5m weighing 500N is placed at 45o
against a vertical wall. µ between the ladder and the wall is 0.20 & between ladder and ground is 0.50. If a man weighing 600N ascends the ladder, how high will he be when the ladder just slips. If a boy now stands on the bottom rung of the ladder, what must be his least weight so that the man can go to the top of the ladder.[Ans.: (a) x = 2.92m (b) Wboy = 458N]
3. CENTROID OF PLANE AREA
3. CENTROID
Centre of gravity : of a body is the point at which the whole
weight of the body may be assumed to be concentrated.
A body is having only one center of gravity for all
positions of the body.
It is represented by CG. or simply G or C.
Contd.
CENTRE OF GRAVITY
Consider a block of uniform thickness and having a uniform mass m.
It is possible to support (hold) the block in stable position by a rod as shown in the figure provided rod must be positioned exactly at the point of intersection of the diagonals.
Contd.
Or the rod must be supported exactly below where the total weight of the block act.
W
R
CENTRE OF GRAVITY
The block can be supported from any position provided the support rod and the line of action of weight are in same line.
W
This indicates that the whole weight of the block act through one point. This point is called as centre of gravity.
Centre of gravity is that point about which the summation of the first moments of the weights of the elements of the body is zero.
R
Contd.
To determine mathematically the
location of the “centre of gravity”
of any body, we apply the
“principle of moments” to the
parallel system of gravitational
forces
X1
CENTRE OF GRAVITYW
W1
W2
W3W4
x
X2
Contd.
The moment of the resultant gravitational force W, about any axis
=the algebraic sum of the moments about the same axisof the gravitational forces dWacting on all infinitesimal elements of the body.
X1
CENTRE OF GRAVITYW
dW1
dW2
dW3dW4
x
X2
∫ ×=⋅ dWxWx
nn xdWxdWxdWxWx ×+×+×+×=⋅ ................dW 332211
∫ dWWhere W = Contd.
x
CENTRE OF GRAVITY
where = x- coordinate of centre of gravity
x
x
W
dWxx ∫ ⋅=
Similarly, y and z coordinates of the centre of gravity are
W
dWzz ∫ ⋅=
W
dWyy ∫ ⋅= and ----(1)
x
CENTRE OF MASS
With the substitution of W= m g and dW = g dm
(if ‘g’ is assumed constant for all particles, then )
m
dmxx ∫ ⋅=
m
dmyy ∫ ⋅=
m
dmzz ∫ ⋅=,, ----(2)
Equation 2 is independent of g and therefore define a unique point in the body which is a function solely of the distribution of mass. This point is called the centre of mass and clearly coincides with the centre of gravity as long as the gravity field is treated as uniform and parallel
Contd.
When speaking of an actual physical body, we use the term
“centre of mass”.
The term centroid is used when the calculation concerns a
geometrical shape only.
Calculation of centroid falls within three distinct categories,
depending on whether we can model the shape of the body
involved as a line, an area or a volume.
Contd.
The centroid “C” of the Volume segment,
V
dVxx ∫ ⋅=
V
dVyy ∫ ⋅=
V
dVzz ∫ ⋅=, ,
The centroid “C” of the line segment,
L
dLxx ∫ ⋅=
L
dLyy ∫ ⋅=
L
dLzz ∫ ⋅=, ,
The centroid “C” of the Area segment,
AREA: when the density ρ, is constant and the body has a small constant thickness t, the body can be modeled as a surface area. The mass of an element becomes dm = ρ t dA.
If ρ and t are constant over entire area, the coordinates of the ‘centre of mass’ also becomes the coordinates of the centroid, C of the surface area and which may be written as
A
dAyy ∫ ⋅=
A
dAzz ∫ ⋅=
A
dAxx ∫ ⋅= ,,
Contd.
Centroid of Simple figures: using method of moment ( First moment of area)
Centroid of an area may or may not lie on the area in question.
It is a unique point for a given area regardless of the choice of the origin and the orientation of the axes about which we take the moment.
The coordinates of the centroid of the surface area about any axis can be calculated by using the equn.
(A) x = (a1) x1 + (a2) x2 + (a3) x3 + ……….+(an) xn
= First moment of area
Algebraic Sum of moment of elemental ‘dA’ about the same axis
Moment of Total area ‘A’ about y-axis =
where (A = a1 + a2 + a3 + a4 + ……..+ an)
AXIS of SYMMETRY:
It is an axis w.r.t. which for an elementary area on one side of the axis , there is a corresponding elementary area on the other side of the axis (the first moment of these elementary areas about the axis balance each other)
If an area has an axis of symmetry, then the centroidmust lie on that axis.
If an area has two axes of symmetry, then the centroidmust lie at the point of intersection of these axes.
Contd.
For example:
The rectangular shown in the figure has two axis of symmetry, X-X and Y-Y. Therefore intersection of these two axes gives the centroid of the rectangle.
B
DD/2
D/2
B/2 B/2
X X
Y
Y
xx
dada
da × x = da × x
Moment of areas,daabout y-axis cancel each other
da × x + da × x = 0 Contd.
AXIS of SYMMETYRY
‘C’ must lie at the intersectionof the axes of symmetry
‘C’ must lie on the axis
of symmetry
‘C’ must lie on the axis of symmetry
To locate the centroid of simple rectangular area from first principles
D
B
To locate the centroid w.r.t. the base line x-x
X Xy
D
Let the distance of centroidfrom the base line x-x be y
Then from the Principle of Moments
∫ ⋅=⋅ dayyASum of moment of elemental area dAabout the same axis
Moment of Total area Aabout x-axis
=
Contd.
Consider a elemental area dA at a distance y from the base line (x-x)
Let the thickness of the element be ‘dy’
Area of small element = dA = B .dy
x xy
y
dyB
dAMoment of this elemental area about x-x axis
= (area) x (distance)
= (B.dy) . (y)
Contd.
Sum of Moment of all such elemental areas comprising the total area =
∫ ⋅= day
∫ ⋅⋅= ydyBD
By
0
2
2 ⎥⎦
⎤⎢⎣
⎡= ⎥
⎦
⎤⎢⎣
⎡=
2
2BD
2
2BDyA =Then from the Principle of Moments
ABDy2
2
=BD
BDy2
2
=2Dy =,,
Contd.
Similarly we can show
x
Y
Y
B
xdx B
Dx
0
2
2 ⎥⎦
⎤⎢⎣
⎡=∫ ⋅=⋅ daxxA ( )∫ ⋅⋅= dxDx
x = B/2
To locate the centroid of simple right angle triangular area from first principles
B
H
To locate the centroid w.r.t. the base line x-x.
Let the distance of centroid from the base line x-x be y
H
x xy
Then from the Principle of Moments
∫ ⋅=⋅ dayyAMoment of Total area A aboutx-axis
Algebraic Sum of moment of elementary area ‘dA’ about the same axis.
=
Contd.
Consider a small elemental area ‘dA’ at a distance ‘y’from the base line (x-x)
Let the thickness of the element be ‘dy’
Area of small element = dA = b .dy
x xy
y
dy
b
dAMoment of this small elemental area about x-x axis
= (area) x (distance)
= (b.dy) . (y)Contd.
Then from the Principle of Moments
( ) dyH
yHBdybda ×−
=×=∫ ⋅=⋅ dayyA
( )H
yHBb −=∫ ⋅⋅=⋅ ydybyA
( )∫ ⋅⋅
−=⋅ ydy
HyHByA
y
y = H/3
X X
HH/3
BContd.
Similarly we can prove that x = B/3( ) dx
BxBHdxhda ⋅
−=⋅=∫ ⋅=⋅ daxxA
( )∫ ⋅⋅
−=⋅ xdx
BxBHxA
∫ ⋅⋅=⋅ xdxbxA( )
BxBHb −
=
x = B/3
Y
xH
YB/3 B
dx
xb
Contd.
The centroid of simple right angled triangle area from the base
B
H
B/3
Centroid
H/3
To locate the centroid of Semi Circular Area w.r.t. the diameter AB from first principles
Consider a semicircle of radius R,
A = area = 8
2Dπ D = 2R
Let ‘G’ be the centroid of the Semicircle, and y is its distance from the diameter AB.
Contd.
drdrdA ××= θ
Consider a small elemental area da, located at distance yfrom the diameter AB,
Let ‘r’ =radial distance of area ‘da’ from centre of the semi circle.
drdrda ××= θ
θsin×= ry
Moment of this elemental area about the diameter AB =
θθ ddrr ⋅⋅= sin2
Contd.
Moment of all such elemental area ‘da’ about the diameter AB
∫ ∫ ⋅⋅=π
θθO
R
O
ddrr .sin2
∫ ⋅⋅⎥⎦
⎤⎢⎣
⎡=
π
θθO
R
O
dr .sin3
3
[ ] )cos(3
3πθ O
R−=
3.2 3R
= Contd.
Then from the Principle of Moments
∫ ⋅=⋅ dayyA
∫ ∫ ⋅⋅π
θθO
R
O
ddrr .sin2
=⋅ yA
=⋅ yA
3.2 3R
π34Ry =
xx
y
y
Centroid
R
π34Ry =
Because of symmetry x = 0
R
To locate the centroid of Quarter Circular Area w.r.t. the boundary radial line AB from first principle
r
dr
dθθ
R
∫ ⋅=⋅ dayyA
y
∫ ∫ ⋅⋅2
2 .sin
π
θθO
R
O
ddrr=⋅ yA
π34Ry =
A B
Contd.
π34Ry =
π34Ry =
π34Rx =
Centroid of Quarter Circular Area w.r.t. the boundary radial lines
Centroid
To locate the centroid of Circular Sector w.r.t. the y-axis shown from first principle
Consider a triangle of differential area = da =
heightbase××=21
RdR ×××= θ21
Distance of the differential area ‘da’, from y-axis = xc =
θcos32
××= R
y
daXc=(2/3)Rcosθ
Contd.
To locate the centroid of Circular Sector w.r.t. the y-axis shown from first principle
Consider a triangle of differential area = da =
heightbase××=21
RdR ×××= θ21
Distance of the differential area ‘da’, from y-axis = xc =
θcos32
××= RContd.
Then from the Principle of Moments
∫ ×=× cxdaxAαθ
α
α×=××= ∫−
221 RRdRA
( ) θθαα
α
dRRxR 22
21cos
32
×⋅=× ∫−
( ) αα sin32 32 ××=× RxR
ααsin
32 ×
=Rx
ααsin
32 ×
=Rx
For a semicircular area 2α = π, if we use this value in the above formula we get
π34 Ry =
To locate the centroid of area under the curve x = k y3
from x = 0 to x = a from first principle
Consider a vertical element of area da = y dx at a distance x from the y-axis.
xx
da
To find x- coordinate,
∫ ×=× xdaxAX
At x = a, y = b,
i.e. a = k b3, k = a/b3∫ ⋅=a
dxyA0
Contd.
Substituting y = (x/k)1/3 and k = a/b3
∫∫ =aa
xydxydxx00
dxkxxdx
kxx
aa
∫∫ ⎟⎠⎞
⎜⎝⎛=⎟
⎠⎞
⎜⎝⎛
0
31
0
31
73
43 2baxab
=⋅
ax74
= Contd.
To find y, Coordinate of centroid of the rectangular element is yc = y/2
dAyyA c ×=× ∫
( )∫∫ ⎟⎠⎞
⎜⎝⎛=
aa
ydxyydxy00 2
Substituting, y = b( x/a)1/3
103
43 2abyab
=⋅
by52
=
EXERCISE PROBLEMS3. Centroid of plane area
Problem No.1:
Locate the centroid of the shaded area shown
50
40
10
10
Ans: x=12.5, y=17.5
EXERCISE PROBLEMS3. Centroid of plane area
Problem No.2:
Locate the centroid of the shaded area shown
300
300
500
1000 m
m
1000 mm
500
r=600D=600
Ans: x=474mm, y=474mm
EXERCISE PROBLEMS3. Centroid of plane area
Problem No.3:Locate the centroid of the shaded area w.r.t. to the axes shown
x-axis
y-axis
90
20
20
60
120
r=40
Ans: x=34.4, y=40.3
EXERCISE PROBLEMS3. Centroid of plane area
Problem No.4:
Locate the centroid of the shaded area w.r.t. to the axes shown
250 mm2010
380
10
200 mm x-axis
y-axis
10
Ans: x= -5mm, y=282mm
EXERCISE PROBLEMS3. Centroid of plane area
Problem No.5
Locate the centroid of the shaded area w.r.t. to the axes shown
3050
40
20 20
40
x
y
r=20
30
Ans:x =38.94, y=31.46
EXERCISE PROBLEMS3. Centroid of plane area
Problem No.6
Locate the centroid of the shaded area w.r.t. to the axes shown
2.4
m
1.0
1.5
1.5
1.0 1.0
r=0.6x
y
Ans: x=0.817, y=0.24
EXERCISE PROBLEMS3. Centroid of plane area
Problem No.7
Locate the centroid of the shaded area w.r.t. to the axes shown
Ans: x= -30.43, y= +9.58
EXERCISE PROBLEMS3. Centroid of plane area
Problem No.8
Locate the centroid of the shaded area.
20
Ans: x= 0, y= 67.22(about base)
EXERCISE PROBLEMS3. Centroid of plane area
Problem No.9
Locate the centroid of the shaded area w.r.t. to the base line.
Ans: x=5.9, y= 8.17
2
EXERCISE PROBLEMS3. Centroid of plane area
Problem No.10
Locate the centroid of the shaded area w.r.t. to the axes shown
Ans: x=21.11, y= 21.11
EXERCISE PROBLEMS3. Centroid of plane area
Problem No.11
Locate the centroid of the shaded area w.r.t. to the axes shown
Ans: x= y= 22.22
EXERCISE PROBLEMS3. Centroid of plane area
Problem No.12
Locate the centroid of the shaded area w.r.t. to the axes shown
R=25
X
Y
R=25
75
50
50
80
Ans: x= 24.33 y= 4.723
4. MOMENT OF INERTIA OF PLANE AREA
4. MOMENT OF INERTIA
Moment of Inertia( Second moment area)
The product of the elemental area and square of the perpendicular distance between the centroid of area and the axis of reference is the “Moment of Inertia” about the reference axis
y
x
ydA
o
Iox= da1 y12 + da2 y2
2+ da3 y32+ --
= ∑ da y2
Ioy = da1 x12 + da2 x2
2 + da3 x32+ ----
= ∑ da x2
Radius of Gyration
kk
kk
kk
Elemental areaElemental area
rr11
rr22
rr33
AA AA
BB BBRadius of gyration is defined as a constant distance of all Radius of gyration is defined as a constant distance of all elemental areas which have been rearranged with out elemental areas which have been rearranged with out altering the total moment of inertia.altering the total moment of inertia.
IIABAB= da k2 + da k2 + ----------- IIABAB= A k2
IIAB AB = ∑ da k2 k=k=√√IIABAB/A/A
Polar moment of Inertia (Perpendicular Axes theorem)
The moment of inertia of an area about an axis perpendicular to the plane of the area is called “Polar Moment of Inertia” and it is denoted by symbol Izz or J or Ip. The moment of inertia of an area in xy plane w.r.to z. axis is Izz = Ip = J = ∫r2dA = ∫(x2 + y2) dA = ∫x2dA + ∫y2dA = Ixx +Iyy
O
y
x
r
z
Y
x
Polar moment of Inertia (Perpendicular Axes theorem)
Hence polar M.I. for an area w.r.t. an axis perpendicular to its plane of area is equal to the sum of the M.I. about any two mutually perpendicular axes in its plane, passing through the point of intersection of the polar axis and the area.
Parallel Axis Theorem
x x
dA
y*G
d
A BMoment of inertia of any area about an axis AB is equal to the M.I. about parallel centroidal axis plus the product of the total area and square of the distance between the two axes.
IAB =∑dA (d +y)2
= ∑dA (d2 + y2 + 2 × d × y)
=∑dA. d2 +∑dA y2 + ∑ 2×d×dA y
= ∑dA. d2 +∑dA y2 + 2×d. ∑ y dA
In the above term (2×d) is constant & ∑ y dA = 0
IAB = Ixx + A.d2
MOMENT OF INERTIA BY DIRECT INTEGRATION
Moment of inertia of rectangular area about centroidalhorizontal axis by direct integration
M.I. about its horizontal centroidal axis :
12
)(
3
2/
2/
2
22/
2/
BD
ydyB
ydAID
D
D
Dxx
=
∫ ××=
×∫=
+
−
+
−
G
B
xx
dy
yD
D/2
.
Moment of Inertia of rectangular area about its base-
(about the line AB) using Parallel Axis Theorem
IAB = IXX + A(d)2
Where d = D/2, the distance between axes xx and AB
BA
G
B
xx
dy
yD
D/2
=BD3/12+(BD)(D/2)2
=BD3/12+BD3/4
=BD3/3
Moment of inertia of Triangular area about the base by direct integration
hx
Bh/3
x
b
xy
(h-y)dy
AA
From similar triangles b/h = x/(h-y)
∴ x = b . (h-y)/h
hx
Bh/3
x
b
xy
(h-y)dy
AA
IAB = ∫ dA.y2 = ∫ (x.dy)y2
hIxx = ∫ (b . (h-y) y2.dy) /h
0
= b[ h (y3/3) – y4/4 ]/h= bh3/12
Moment of inertia of Triangular area about the centroidal horizontal axis
Using Parallel axis theorem .MI about any line(AB) = MI about cenroidal parallel axis + Ad2
hx
Bh/3
x
b
xy
(h-y)dy
AA
CentroidalCentroidal horizontal axishorizontal axis
IAB = Ixx + Ad2
Ixx = MI about centroidal axis x xIAB= MI about the Base line AB
Ixx = IAB – Ad2
= bh3/12 – bh/2 . (h/3)2
= bh3/36
Moment of inertia of Circular area about the centroidalhorizontal axis
Ixx = ∫ dA . y2
R 2π= ∫ ∫ (x.dθ.dr) r2Sin2θ
0 0
R 2π=∫ ∫ r3.dr Sin2θ dθ
0 0
R 2π=∫ r3 dr ∫ {(1- Cos2θ)/2} dθ
0 R
0 2π
=[r4/4] [θ/2 – Sin2θ/4] 0 0
= R4/4[π - 0] = πR4/4
IXX = π R4/4 = πD4/64
BA
xxRθ
dθ
y=rSinθr
Moment of inertia of Semi-circular area about the Base & centroidal horizontal axis
IAB = ∫ dA . y2
R π= ∫ ∫ (r.dθ.dr) r2Sin2θ
0R
0π
=∫ r3.dr ∫ Sin2θ dθ0 0
R π=∫ ∫ r3 dr (1- Cos2θ)/2) dθ
0 0 π
=[R4/4] [θ/2 – Sin2θ/4]0
= R4/4[π/2 - 0] = πR4/8 = (πD4/32)
4R/3π
y0
y0
BAxx R
Moment of inertia of Semi-circular area about the centriodal horizontal axis
using parallel axis theorem:
IAB = Ixx + A(d)2
Ixx = IAB – A(d)2
= π R4/8 πR2/2 . (4R/3π)2
Ixx = 0.11R4
IAB = ICD
R π/2IAB = ∫ ∫ (r.dθ.dr). r2Sin2θ
0 0
R π/2=∫ r3.dr ∫ Sin2θ dθ
0 0
R π/2 =∫ r3 dr ∫ (1- Cos2θ)/2) dθ
0 0 π/2
=[R4/4] [θ/2 – (Sin2 θ)/4] 0
= R4 (π/16 – 0) = πR4/16
A B
x x
C
D y
y
4R/3π
Moment of inertia of Quarter-circular area about the base & centroidal horizontal axis
4R/3π
Moment of inertia about Centroidal axis,Ixx = IAB - Ad2
= πR4/16 - πR2. (0. 424R)2
= 0.055R4
Sl.No Figure I x0-x0I y0-y0
I xx I yy
1bd3/12 - bd3/3 -
2bh3/36 - bh3/12 -
3πR4/4 πR4/4 - -
40.11R4 πR4/8 πR4/8 -
50.055R4 0.055R4 πR4/16 πR4/16
b
dx0
x
x0
xd/2
b
h
xxx0
x0h/3
x0x0
y0
y0
OR
y0
y0xxx0
x0
4R/3π
x0
y y0
4R/3π
4R/3π
Y
Y Xo
EXERCISE PROBLEMS
Q1. Determine the moment of inertia about the centroidal axes.
20
30mm
30mm
30mm
100mm
[Ans: Y = 27.69mm Ixx = 1.801 x 106mm4
Iyy = 1.855 x 106mm4]
EXERCISE PROBLEMS
Q2. Determine second moment of area about the centroidalhorizontal and vertical axes. 300mm
[Ans: X = 99.7mm from A, Y = 265 mm
Ixx = 10.29 x 109mm4, Iyy = 16.97 x 109mm4]
200mm
200
300mm
900mm
EXERCISE PROBLEMS
Q3. Determine M.I. Of the built up section about the horizontal and vertical centroidal axes and the radii of gyration.
200mm20
60
100mm
140mm
20
[Ans: Ixx = 45.54 x 106mm4, Iyy = 24.15 x 106mm4
rxx = 62.66mm, ryy = 45.63mm]
EXERCISE PROBLEMS
Q4. Determine the horizontal and vertical centroidal M.I. Of the shaded portion of the figure.
60X X2020
60 60
[Ans: X = 83.1mm
Ixx = 2228.94 x 104mm4, Iyy = 4789.61 x 104mm4]
EXERCISE PROBLEMS
Q5. Determine the spacing of the symmetrically placed vertical blocks such that Ixx = Iyy for the shaded area.
[Ans: d/2 = 223.9mm d=447.8mm]
200mm
600mm
d400mm
200mm
200mm
200mm
EXERCISE PROBLEMS
Q6. Find the horizontal and vertical centroidal moment of inertia of the section shown in Fig. built up with R.S.J. (I-Section) 250 x 250 and two plates 400 x 16 mm each attached one to each.
Properties of I section are
Ixx = 7983.9 x 104mm4
Iyy = 2011.7 x 104mm4
[Ans: Ixx = 30.653 x 107mm4, Iyy = 19.078 x 107mm4]
4000mm
2500mm
160mm
160mmCross sectional area=6971mm2
EXERCISE PROBLEMS
Q7. Find the horizontal and vertical centroidal moment of inertia of built up section shown in Figure. The section consists of 4 symmetrically placed ISA 60 x 60 with two plates 300 x 20 mm2.
[Ans: Ixx = 111.078 x 107mm4, Iyy = 39.574 x 107mm4]300mm
Properties of ISA
Cross sectional area = 4400mm2
Ixx = Iyy ;Cxx = Cyy =18.5mm
18.5mm18.5mm
20mm
200mm
Q8. The R.S. Channel section ISAIC 300 are placed back to back with required to keep them in place. Determine the clear distance d between them so that Ixx = Iyy for the composite section.
[Ans: d = 183.1mm]
Properties of ISMC300
C/S Area = 4564mm2
Ixx = 6362.6 x 104mm4
Iyy = 310.8 x 104mm4
Cyy = 23.6mm X X
Y
Y
Lacing
d
380mm
23.6mm
EXERCISE PROBLEMS
EXERCISE PROBLEMS
Q9. Determine horizontal and vertical centroidal M.I. for the section shown in figure.
[Ans: Ixx = 2870.43 x 104mm4, Iyy = 521.64 x 104mm4]90mm
160mm
40mm
40mm
40mm
5. Kinetics of rectilinear motion
5. Kinetics of rectilinear motion
In this chapter we will be studying the relationship between forces on a body/particle and the accompanying motion
Newton’s Second law of motion:
Newton’s first and third law of motion were used extensively in the study of statics (the bodies at rest) whereas Newton’s second law of motion is used extensively in the study of the kinetics.
Work done by force:
F
F
s
α αA B
F cosθ F cosθ
F sinθF sinθ
Work done by a force is the product of the force and the distance moved by the point of application in the direction of the force. It is a scalar quantity.
Work done = (F cosα ) × s
X- component of force F moves through distance a S,
S = displacement of force from A to B Unit: Nm ( Joule )
POWER:-
It is defined as the time rate of doing work.
Power = work done /Time= (force × distance) /Time
= force × velocity
Unit: (Nm)/s = [watt] (kN m)/s = [kilo watt]
1 metric H.P=735.75 watts.
Energy:-
It is defined as the capacity to do work. It is a scalar quantity.
Unit :- N m (Joule)
Momentum:-Quantity of motion possessed by a body is called momentum. It is the product of mass and velocity. It is a vector quantity.
Unit:- N s.
Impulse of a Force:-It is defined as the product of force and the time over which itacts. It is a vector quantity.
Unit:- N s.
Newton’s second law of motion.
“If the resultant force acting on a particle is not zero , the particle will have an acceleration proportional to the magnitude of the resultant force and its direction is along that of the resultant force.”
F α aF =Resultant of forcesa = Acceleration of the particle.
F = mam= mass of the particle.
The constant value obtained for the ratio of the magnitude of the force and acceleration is characteristic of the particle and is denoted by ‘m’. Where ‘m’ is mass of the particle
Since ‘m’ is a +ve scalar, the vectors of force ‘F’and acceleration ‘a’ have the same direction.
UnitsForce in Newtons (N) N = 1 Kgm/s2
Acceleration in m/s2
F1
F2
R = m a
Body will m
ove in th
e
directio
n of resultant
F3
R = Resultant of forces F1,F2 and F3
Using the rectangular coordinate system we have components along axes as,
ΣFx = max
ΣFy = may
ΣFz = maz
where Fx ,Fy Fz and ax , ay ,az are rectangular components of resultant forces and accelerations respectively.
Newton’s second law may also be expressed by considering a force vector of magnitude ‘ma’ but of sense opposite to that of the acceleration. This vector is denoted by (ma)rev. The subscript indicates that the sense of acceleration has been reversed and is called the inertia force vector.
F1
F2
R = m a
F3
R = Resultant of forces F1,F2 and F3
Body w
ill mov
e in th
e dire
ction
of
resulta
nt
F1
F2
R = m a
F3
Inert
ia For
ce
F1
F2
F = R = m
a
F3
Inertia Forc
e If the inertia force vector is added to the forces acting on the particle we obtain a system of forces whose resultant is zero.
Resultant of forces F1,F2, F3 and Inertia force = 0
0321 =+++ maFFF
The particle may thus be considered to be in equilibrium. (THIS IS DYNAMIC EQUILIBRIUM)
It was pointed out by D’Alembert (Alembert, Jean le Rond d’ (1717-1783), French mathematician and philosopher) that problems of kinetics can be solved by using the principles of statics only (the equations of equilibrium) by considering an inertia force in a direction directly opposite to the acceleration in addition to the real forces acting on the system
D’Alembert’s principle states thatWhen different forces act on a system such that it is in
motion with an acceleration in a particular direction, the vectorial sum of all the forces acting on the system including the inertia force (‘ma’ taken in the opposite direction to the direction of the acceleration) is zero.
The problem under consideration may be solved by using the method developed earlier in statics. The particle is said to be in dynamic equilibrium.
If
ΣFx = 0
ΣFy= 0 including inertia force vector
ΣFz = 0
This principle is known as D’Alembert’s principle
From Newton’s second law of motion∑ F = m × a = -----------(1)
Also a = dv/dt =( dv/dt) ×( ds/ds) = v × dv/dssubstituting in (1)
∑ F = m × v × dv/ds
ΣF × ds = m × v × dv ------------------(2)
Let the initial velocity be u and the final velocity after it moves through a distance ‘ s’ be v
Work-Energy relation for translation
Integrating both sides, we get
Therefore work done by a system of forces acting on a body while causing a displacement is equal to the change in kinetic energy of the body during the displacement.
( )∑ −=×
∑ =×
∫=∑ ∫
22
2
v
u0
21
)2
(
dv
uvmsF
vmsF
vmdsFs
v
u
Impulse-momentum relationship
F = m × aF = m × (v - u)/t = (mv – mu)/t
Force = Rate of change of momentum
F × t = mv – mu
Impulse = final momentum – Initial momentum
The component of the resultant linear impulse along any direction is equal to change in the component of momentum in that direction.
EXERCISE PROBLEMS5. Kinetics
Q1. Blocks A and B of mass 10 kg and 30 kg respectively are connected by an inextensible cord passing over a smooth pulley as shown in Fig. Determine the velocity of the system 4 sec. after starting from rest. Assume coefficient of friction =0.3 for all surfaces in contact.
B A
60o 30o
Ans: v=13.6m/s
EXERCISE PROBLEMS5. Kinetics
Q2. A tram car weighs 150kN. The tractive resistance being 1% of the weight of car. What power will be required to move the car at uniform speed of 20 kmph
(i) Up an incline 1 in 300
(ii) (ii) Down an incline 1 in 250. Take efficiency 75%.
Ans: Pull = 2 kN
a) Output power=11.12kW
EXERCISE PROBLEMS5. Kinetics
Q3. Two masses of 5 kg and 3 kg rest on two smooth inclined plane, each of inclination 30º and are connected by a string passing over a common apex. Find the velocity of 3 kg mass after 2 sec when released from rest. Find the distance it will cover before changing direction of motion, if 5kg mass is cut off after two sec of its release from rest.
30º 30º
5kg 3kg
V = 4.45 m/s
s = 0.61 m
EXERCISE PROBLEMS5. Kinetics
Q4. A locomotive weighing 900 kN pulls a train of 10 coaches each weighing 300 kN at 72 Kmph on a level track against a resistance of 7 N/kN. If the rear 4 coaches get snapped from the train, find the speed of the engine and the remaining coaches after 120 secs. Assume no change in resistance and draw bar pull. Find also distance traveled by detached coaches before coming to rest.
4x300=1200kN 6x300=1800kN 900kN
P
V = 23.66 m/s
s = 2.9 km
EXERCISE PROBLEMS5. Kinetics
Q5. Find the tension in the cord supporting body C in Fig. below. The pulleys are frictionless and of negligible weight.
150 kN
450 kN
A
B
C
Assume all blocks moving either downward or upward and accordingly draw FBD
0=aA +2aB +cC
300 kN
Ans : T=211.72 kN
EXERCISE PROBLEMS5. Kinetics
Q6. Two blocks A and B are released from rest on a 30o
inclined plane with horizontal, when they are 20m apart. The coefficient of friction under the upper block is 0.2 and that under lower block is 0.4. compute the time elapsed until the block touch. After they touch and move as a unit what will be the constant forces between them.
(Ans : t = 4.85 s, contact force=8.65 N)
EXERCISE PROBLEMS5. Kinetics
Q7. An elevator cage of a mine shaft weighing 8kN when empty is lifted or lowered by means of rope. Once a man weighing 600N entered it and lowered at uniform acceleratin such that when a distance of 187.5 m was covered, the velocity of the cage was 25m/s. Determine the tension in the cable and force exerted by man on the floor of the cage.
(Ans: T=7139 N and R=498 N)
EXERCISE PROBLEMS5. Kinetics
Q8. A small block starts from rest at point a and slides down the inclined plane. At what distance along the horizontal will it travel
before coming to rest . Take µk=0.3 [Ans :s=6m ]
A3
4
5m
B Cs
EXERCISE PROBLEMS5. Kinetics
Q9. The system starts from rest in the position shown . How much further will block ‘A’ move up the incline after block B hits the ground . assume the pulley to be frictionless and massless and µ is 0.2 .WA=1000N, WB=2000N. [ Answer s =1.27m]
3m
A
B34
EXERCISE PROBLEMS5. Kinetics
Q10. A 1500Kg automobile travels at a uniform rate of 50kmph to 75kmph . During the entire motion, the automobile is traveling on a level horizontal road and rolling resistance is 2 % of weight of automobile . Find (i) maximum power developed (ii) power required to maintain a constant speed of 75kmph.
[ ANSWER: power developed = 6.131KN]
EXERCISE PROBLEMS5. Kinetics
Q11. Two bodies A and B weighing 2000N and 5200N are connected as shown in the figure . find the further distance moved by block a after the block B hits Wall. µ=0.2 .[ Answer s=1.34m]
512
A
B 3m
EXERCISE PROBLEMS5. Kinetics
Q12. A spring is used to stop 60kg package which is sliding on a horizontal surface . the spring has a constant k = 20kN/m and is held by cable such that it is initially compressed at 120mm. knowing that the package has a velocity of 2.5m/s in position shown and maximum additional displacement of spring is 40mm . Determine the coefficient of kinetic friction between package and surface. (Answer µk=0.2)
600m
60kg
2.5 m/s
EXERCISE PROBLEMS5. Kinetics
Q13. The system shown in figure has a rightward velocity of 4m/s, just before force P is applied. Determine the value of P that will give a leftward velocity of 6m/s in a time interval of 20sec. Take µ = 0.2 & assume ideal pulley. [Answer P=645.41N]
1000N
400N
P
EXERCISE PROBLEMS5. Kinetics
Q14. A locomotive of weight 500kN pulls a train of weight of 2500kN. The tractive resistance, due to friction is 10N/kN. The train can go with a maximum speed of 27kmph on a grade of 1in100. Determine (a) Power of the locomotive. (b) Maximum speed it can attain on a straight level track with the tractive resistance remaining same. [Answer (a) Power= 450kN (b) v=15m/s]
Q15. A wagon weighing 400kN starts from rest, runs 30m down a 1% grade & strikes a post. If the rolling resistance of the track is 5N/kN, find the velocity of the wagon when it strikes the post.
If the impact is to be cushioned by means of one bumper string, which compresses 1mm per 20 kN weight, determine how much the bumper spring will be compressed. [Answer v=1.716m/s, x=77.5mm]
Q16. A train whose weight is 20kN moves at the rate of 60kmph. After brakes are applied, it is brought to rest in 500m. Find the force exerted, assuming it to be uniform
a) Use work-energy relation
b) Use D’Alemberts equation.
Ans: F = 5.663 kN
EXERCISE PROBLEMS5. Kinetics
EXERCISE PROBLEMS5. Kinetics
Q17. The blocks A and B having weights 100 N and 300 N start from rest. The horizontal plane and the pulleys are frictionless. Determine the acceleration and the tension in the string.
A
B
string
string
Frictionless pulley
Frictionless pulley
Ans: aA=8.403m/s2
aB=4.201 m/s2
T= 85.71 N
EXERCISE PROBLEMS5. Kinetics
Q18. The blocks A and B having weights 100 N and 300 N start from rest. The horizontal plane and the pulleys are frictionless. Determine the velocity of block B after 0.5 seconds and the tension in the string. Use impulse-momentum relation.
A
B
string
string
Frictionless pulley
Frictionless pulley
EXERCISE PROBLEMS5. Kinetics
Q19. The blocks A and B having weights 100 N and 300 N start from rest when a load of 100 N is applied on the block A as shown in the figure. The horizontal plane and the pulleys are frictionless. Determine the acceleration and the tension in the string.
A
B
string
string
Frictionless pulley
Frictionless pulley
100 N
EXERCISE PROBLEMS5. Kinetics
Q20. Two blocks A and B are connected as shown in the figure. At the instant of their release if the block A, which is on smooth horizontal plane has a left ward velocity of 2 m/s, what would be its velocity 5 seconds after their release. The blocks A and B weigh 100 N and 300 N respectively.
A
B
string
string
Frictionless pulley
Frictionless pulley
EXERCISE PROBLEMS5. Kinetics
Q21. An engine weighing 500 kN drags carriages weighing 1500kN up an incline of 1 in 100 against a resistance of 5N/kN starting from rest. It attains a velocity of 36 kmph (10m/s) in 1 km distance with a constant draw bar pull supplied by the engine. What is the power required for the same ? What is the tension developed in the link connecting the engine and carriages?
1500N500N
1001
P
Ans:
Pull=40.19kN, power=401kW, T=30.15kN
Q22. what velocity the block A will attain after 2 seconds starting from rest? Take µ = 0.2. WA = 1500N, WB = 2000N. Use impulse-momentum relation.
4
3
34
AB
EXERCISE PROBLEMS5. Kinetics
Mechanics of Deformable Bodies
PART - II
COURSE CONTENT IN BRIEF
6. Simple stresses and strains
7. Statically indeterminate problems and thermal stresses
8. Stresses on inclined planes
9. Stresses due to fluid pressure in thin cylinders
6. Simple stresses and strains
The subject strength of materials deals with the relations between externally applied loads and their internal effects on bodies. The bodies are no longer assumed to be rigid and the deformations, however small, are of major interest
Alternatively the subject may be called the mechanics of solids.
The subject, strength of materials or mechanics of materials involves analytical methods for determining the strength , stiffness (deformation characteristics), and stability of various load carrying members.
GENERAL CONCEPTS
STRESS
No engineering material is perfectly rigid and hence, when a material is subjected to external load, it undergoes deformation.
While undergoing deformation, the particles of the material offer a resisting force (internal force). When this resisting force equals applied load the equilibrium condition exists and hence the deformation stops.
These internal forces maintain the externally applied forces in equilibrium.
STRESS
The internal force resisting the deformation per unit area is
called as stress or intensity of stress.
Stress = internal resisting force / resisting cross sectional area
AR
=
STRESS
SI unit for stress
N/m2 also designated as a pascal (Pa)
Pa = N/m2
kilopascal, 1kPa = 1000 N/m2
megapascal, 1 MPa = 1×106 N/m2
= 1×106 N/(106mm2) = 1N/mm2
1 MPa = 1 N/mm2
gigapascal, 1GPa = 1×109 N/m2
= 1×103 MPa
= 1×103 N/mm2
STRESSAXIAL LOADING – NORMAL STRESS
Consider a uniform bar of cross sectional area A, subjected to a tensile force P.
Consider a section AB normal to the direction of force P
P
P
P
R
BA
R
P
Let R is the total resisting force acting on the cross section AB.
Then for equilibrium condition,
R = P
Then from the definition of stress, normal stress = σ = R/A = P/A
σ = Normal StressSymbol:
STRESSAXIAL LOADING – NORMAL STRESS
Direct or Normal Stress:
Intensity of resisting force perpendicular to or normal to the section is called the normal stress.
Normal stress may be tensile or compressive
Tensile stress: stresses that cause pulling on the surface of the section, (particles of the materials tend to pull apart causing extension in the direction of force)
Compressive stress: stresses that cause pushing on the surface of the section, (particles of the materials tend to push together causing shortening in the direction of force)
STRESS
• The resultant of the internal forces for an axially loaded member is normalto a section cut perpendicular to the member axis.
• The force intensity on that section is defined as the normal stress.
AP
AF
aveA
=∆∆
=→∆
σσ0
lim
STRAIN
STRAIN :
when a load acts on the material it will undergo deformation. Strain is a measure of deformation produced by the application of external forces.
If a bar is subjected to a direct load, and hence a stress, the bar will changes in length. If the bar has an original length L and change in length by an amount δL, the linear strain produced is defined as,
LLδε =
Original lengthChange in length
=Linear strain,
Strain is a dimensionless quantity.
Linear Strain
LL
AP
δδε
σ
==
=
22
L
AP
AP
δε
σ
=
==22
strain normal
stress
==
==
L
AP
δε
σ
STRESS-STRAIN DIAGRAM
In order to compare the strength of various materials it is necessary to carry out some standard form of test to establish their relative properties.
One such test is the standard tensile test in which a circular bar of uniform cross section is subjected to a gradually increasing tensile load until failure occurs.
Measurement of change in length over a selected gauge length of the bar are recorded throughout the loading operation by means of extensometers.
A graph of load verses extension or stress against strain is drawn as shown in figure.
STRESS-STRAIN DIAGRAM
Proportionality limit
Typical tensile test curve for mild steel
STRESS-STRAIN DIAGRAM
Typical tensile test curve for mild steel showing upper yield point and lower yield point and also the elastic range and plastic range
Stress-strain Diagram
Limit of Proportionality :
From the origin O to a point called proportionality limit the stress strain diagram is a straight line. That is stress is proportional to strain. Hence proportional limit is the maximum stress up to which the stress – strain relationship is a straight line and material behaves elastically.
From this we deduce the well known relation, first postulated by Robert Hooke, that stress is proportional to strain.
Beyond this point, the stress is no longer proportional to strain
APP
P =σ Load at proportionality limitOriginal cross sectional area
=
Stress-strain Diagram
Elastic limit:
It is the stress beyond which the material will not return to its original shape when unloaded but will retain a permanent deformation called permanent set. For most practical purposes it can often be assumed that points corresponding proportional limit and elastic limit coincide. Beyond the elastic limit plastic deformation occurs and strains are not totally recoverable. There will be thus some permanent deformation when load is removed.
APE
E =σ Load at proportional limitOriginal cross sectional area=
Stress-strain Diagram
Yield point:
It is the point at which there is an appreciable elongation or yielding of the material without any corresponding increase of load.
APY
Y =σ Load at yield pointOriginal cross sectional area=
Ultimate strength:
It is the stress corresponding to maximum load recorded during the test. It is stress corresponding to maximum ordinate in the stress-strain graph.
APU
U =σ Maximum load taken by the materialOriginal cross sectional area=
Stress-strain Diagram
Rupture strength (Nominal Breaking stress):
It is the stress at failure.
For most ductile material including structural steel breaking stress is somewhat lower than ultimate strength because the rupture strength is computed by dividing the rupture load (Breaking load) by the original cross sectional area.
APB
B =σ load at breaking (failure)Original cross sectional area=
True breaking stress = load at breaking (failure)Actual cross sectional area
Stress-strain Diagram
After yield point the graph becomes much more shallow and covers a much greater portion of the strain axis than the elastic range.
The capacity of a material to allow these large plastic deformations is a measure of ductility of the material
Ductile Materials:
The capacity of a material to allow large extension i.e. the ability to be drawn out plastically is termed as its ductility. Material with high ductility are termed ductile material.
Example: Low carbon steel, mild steel, gold, silver, aluminum
Stress-strain Diagram
Percentage elongation
A measure of ductility is obtained by measurements of the percentage elongation or percentage reduction in area, defined as, increase in gauge length (up to fracture)
original gauge length×100=
Percentage reduction in area original area
×100Reduction in cross sectional area of necked portion (at fracture)
=
Cup and cone fracture for a Ductile Material
Stress-strain Diagram
Brittle Materials :A brittle material is one which exhibits relatively small extensions before fracture so that plastic region of the tensiletest graph is much reduced.
Example: steel with higher carbon content, cast iron, concrete, brick
Stress-strain diagram for a typical brittle material
HOOKE’S LAW
Hooke’s Law
For all practical purposes, up to certain limit the relationship between normal stress and linear strain may be said to be linear for all materials
stress (σ) α strain (ε)stress (σ) strain (ε) = constant
Thomas Young introduced a constant of proportionality that came to be known as Young’s modulus.
stress (σ) strain (ε) =
Young’s ModulusE =Modulus of Elasticity
or
HOOKE’S LAW
Young’s Modulus is defined as the ratio of normal stress to linear strain within the proportionality limit.
stress (σ) strain (ε) LA
PLLL
AP
δδ
=÷==E
The value of the Young’s modulus is a definite property of a material
From the experiments, it is known that strain is always a very small quantity, hence E must be large.
For Mild steel, E = 200GPa = 2×105MPa = 2×105N/mm2
Deformations Under Axial Loading
AEP
EE ===
σεεσ
• From Hooke’s Law:
• From the definition of strain:
Lδε =
• Equating and solving for the deformation,
AEPL
=δ
• With variations in loading, cross-section or material properties,
∑=i ii
iiEALPδ
Consider an element of length, δx at a distance x from A
B
WW
Axd1 d2dx
( ) xL
ddd ×−
+= 121 c/s area at x, ( )21
21
44 kxdd +==ππ
xkd ×+= 1
Diameter at x,
Change in length over a length dx is ( ) ⎟⎟
⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛
×+=⎟
⎠⎞
⎜⎝⎛=
Ekxd
WdxAEPL
dx 214
π
Change in length over a length L is ( )
∫⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛
×+=
L
Ekxd
Wdx0 2
14π
Consider an element of length, δx at a distance x from A
( )∫
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛
×+=
L
Ekxd
Wdx0 2
14π
( )∫
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛
×=
L
Et
kdtW
0 2
4π
Change in length over a length L is
Put d1+kx = t,
Then k dx = dt
LLL
kxdEkW
tEkWt
EkW
0100
12
)(1414
14
⎥⎦
⎤⎢⎣
⎡+
−=⎥⎦
⎤⎢⎣⎡−=⎥
⎦
⎤⎢⎣
⎡−
=+−
πππ
EddWL
dEdWL
×==
4
42121
ππ
Derive an expression for the total extension of the tapered bar AB of rectangular cross section and uniform thickness, as shown in the figure, when subjected to an axial tensile load ,W.
WW
A B
L
d1d2
bb
W W
A B
x
d1d2
bb
dxConsider an element of length, δx at a distance x from A
( ) xL
ddd ×−
+= 121 c/s area at x, ( )bkxd += 1
xkd ×+= 1
depth at x,
( ) ⎟⎟⎠
⎞⎜⎜⎝
⎛×+
=⎟⎠⎞
⎜⎝⎛=
EbkxdWdx
AEPL
dx 1
Change in length over a length dx is
Change in length over a length L is ( )∫ ⎟⎟
⎠
⎞⎜⎜⎝
⎛×+
=L
EbkxdWdx
01
( )12 loglog ddkEb
Pee −
××=
( ) ( )1212
loglog302.2 ddddEbLP
−−××××
=
Derive an expression for the total extension produced by self weight of a uniform bar, when the bar is suspended vertically.
L
Diameter d
P1x
dx dx
elementP1 = weight of the bar below
the section,= volume × specific weight= (π d2/4)× x × γ= A× x ×γ
Diameter d Extension of
the element due to weight of the bar below that,
AEdxxA
AEdxP
AEPL
dx
)(1 ρ××==⎥⎦
⎤⎢⎣⎡=
Hence the total extension entire bar
EL
Ex
AEdxxA
LL
22)( 2
0
2
0
γγγ=⎥
⎦
⎤⎢⎣
⎡=
××= ∫
AEPL
AELAL
AA
EL
×=×
=×=21
2)(
2
2 γγThe above expression can also be written as
Where, P = (AL)×γ= total weight of the bar
SHEAR STRESS
Consider a block or portion of a material shown in Fig.(a) subjected to a set of equal and opposite forces P. then there isa tendency for one layer of the material to slide over another to produce the form failure as shown in Fig.(b)
PP
R RP P
Fig. cFig. a Fig. b
The resisting force developed by any plane ( or section) of the block will be parallel to the surface as shown in Fig.(c).
The resisting forces acting parallel to the surface per unit area is called as shear stress.
Shear stress (τ)
=Shear resistanceArea resisting shear
τ
If block ABCD subjected to shearing stress as shown in Fig.(d), then it undergoes deformation. The shape will not remain rectangular, it changes into the form shown in Fig.(e), as AB'C'D.
B
Shear strain
AP
=
This shear stress will always be tangential to the area on whichit acts
τD
C
A
τB'
D
C'
Aτ
B C
Fig. d Fig. e
τ Shear strain is defined as the change in angle between two line element which are originally right angles to one another.
The angle of deformation is measured in radians and hence is non-dimensional.
D
B' C'B
A
C
φ
τFig. e
φφ ≈=′
= tanstrain shear ABBB
The angle of deformation is then termed as shear strain φ
SHEAR MODULUS
For materials within the proportionality limit the shear strain is proportional to the shear stress. Hence the ratio of shear stress to shear strain is a constant within the proportionality limit.
Shear Modulus or
Modulus of Rigidity
Shear stress (τ) Shear strain (φ) G= == constant
The value of the modulus of rigidity is a definite property of a material
For Mild steel, G= 80GPa = 80,000MPa = 80,000N/mm2
example: Shearing Stress
• Forces P and P‘ are applied transversely to the member AB.
AP
=aveτ
• The corresponding average shear stress is,
• The resultant of the internal shear force distribution is defined as the shear of the section and is equal to the load P.
• Corresponding internal forces act in the plane of section C and are called shearing forces.
• The shear stress distribution cannot be assumed to be uniform.
State of simple shear
Consider an element ABCD in a strained material subjected to shear stress, τ as shown in the figure
τBA
D Cτ
Force on the face AB = P = τ × AB × t
Where, t is the thickness of the element.
Force on the face DC is also equal to P
State of simple shear
Now consider the equilibrium of the element.(i.e., ΣFx = 0, ΣFy = 0, ΣM = 0.)
P
The element is subjected to a clockwise moment
P × AD = (τ × AB × t) × AD
BAFor the force diagram shown,ΣFx = 0, & ΣFy = 0, But ΣM = 0 D C
Pforce
But, as the element is actually in equilibrium, there must be another pair of forces say P' acting on faces AD and BC, such that they produce a anticlockwise moment equal to ( P × AD )
State of simple shear
P BAP ' × AB = P × AD= (τ × AB × t)× AD ----- (1) P ' P '
CDP
If τ1 is the intensity of the shear stress on the faces AD and BC, then P ' can be written as, P ' = τ ' × AD × t
τ BA
τ ' τ ' CD
τEqun.(1) can be written as
(τ ' × AD× t ) × AB = (τ × AB × t) × AD ----- (1)
τ ' = τ
State of simple shear
Thus in a strained material a shear stress is always accompanied by a balancing shear of same intensity at right angles to itself. This balancing shear is called “complementary shear”.
τ
The shear and the complementary shear together constitute a state of simple shear
BA
τ'= τ τ'= τ
CDτ
Direct stress due to pure shear
Consider a square element of side ‘a’ subjected to shear stress as shown in the Fig.(a). Let the thickness of the square be unity.
a
A B
CD
τ
τ
τ
τa
τAB
Fig.(b) shows the deformed shape of the element. The length of diagonal DB increases, indicating that it is subjected to tensile stress. Similarly the length of diagonal AC decreases indicatingthat compressive stress.
C
τaa
τ
Dτ
Fig.(a). Fig.(b).
Direct stress due to pure shear
Now consider the section, ADC of the element, Fig.(c).
Resolving the forces in σn direction, i.e., in the X-direction shown
a
Fig.(c).
aa
A
CD
( )a2
For equilibrium
A σn
C
a
X
τ
Dτ
( ) ( )τσ
τσ=
××−×××=
=∑
n
n aa
Fx
45cos212
0
Direct stress due to pure shear
Therefore the intensity of normal tensile stress developed on plane BD is numerically equal to the intensity of shear stress.
Similarly it can be proved that the intensity of compressive stress developed on plane AC is numerically equal to the intensity of shear stress.
POISSON’S RATIO
Poisson’s Ratio:
Consider the rectangular bar shown in Fig.(a) subjected to a tensile load. Under the action of this load the bar will increase in length by an amount δL giving a longitudinal strain in the bar of
ll
lδε =
Fig.(a)
The associated lateral strains will be equal and are of opposite sense to the longitudinal strain.
POISSON’S RATIO
The bar will also exhibit, reduction in dimension laterally, i.e. its breadth and depth will both reduce. These change in lateral dimension is measured as strains in the lateral direction as given below.
dd
bb
latδδε −=−=
Provided the load on the material is retained within the elasticrange the ratio of the lateral and longitudinal strains will always be constant. This ratio is termed Poisson’s ratio (µ)
POISSON’S RATIOLateral strainLongitudinal strain
=l
ld
d
δ
δ )(−=
ll
bb
δ
δ )(−OR
Poisson’s Ratio = µ
For most engineering metals the value of µ lies between 0.25 and 0.33
In general y
Lx
LyLz
PP x
z
x
x
y
y
ll
ll
δ
δ−
=
x
x
z
z
ll
ll
δ
δ−
=Lateral strain
Poisson’s Ratio
OR= Strain in the direction of load applied
Poisson’s Ratio = µ
In general
Strain in X-direction = εx
z
y
xPxPx
Lx
LyLz
x
x
llδ
=
Strain in Y-direction = εy
x
x
y
y
ll
ll δµδ
==
x
x
z
z
ll
ll δµδ==Strain in Z-direction = εz
Load applied in Y-direction
z
y
x
Py
Lx
LyLz
Py
y
y
x
x
ll
ll
δ
δ−
=
y
y
z
z
ll
ll
δ
δ−
=Lateral strain
Poisson’s Ratio
OR= Strain in the direction of load applied
y
y
x
x
ll
ll δ
µδ==Strain in X-direction = εx
Load applied in Z-direction
y
z
x
Pz
Lx
LyLz
Pz
z
z
y
y
ll
ll
δ
δ−
=
z
z
x
x
ll
ll
δ
δ−
=Lateral strain
Poisson’s Ratio
OR= Strain in the direction of load applied
z
z
x
x
ll
ll δµδ==Strain in X-direction = εx
Load applied in X & Y direction
Strain in X-direction = εx
z
y
xPxPx
Lx
LyLz
Py
EEyx σ
µσ−=
Py
EExy σµ
σ−=Strain in Y-direction = εy
EExy σµ
σµ −−=Strain in Z-direction = εz
General case:
Strain in X-direction = εx
Py
Strain in Y-direction = εy
Strain in Z-direction = εz
z
y
xPxPx
Pz
PyPzEEE
zyxx
σµσ
µσε −−=
σx
σzσy
σx
σz σy
EEEzxy
yσµσµ
σε −−=
EEExyz
zσµ
σµσε −−=
Bulk Modulus
Bulk Modulus
A body subjected to three mutually perpendicular equal direct stresses undergoes volumetric change without distortion of shape. If V is the original volume and dV is the change in volume, then dV/V is called volumetric strain.
Bulk modulus, K
A body subjected to three mutually perpendicular equal direct stresses then the ratio of stress to volumetric strain is calledBulk Modulus.
⎟⎠⎞
⎜⎝⎛
=
VdVσ
Relationship between volumetric strain and linear strain
Consider a cube of side 1unit, subjected to three mutually perpendicular direct stresses as shown in the figure.
Relative to the unstressed state, the change in volume per unit volume is
( )( )( )[ ] [ ]
eunit volumper in volume change
1111111
=
++=
+++−=+++−=
zyx
zyxzyxdV
εεε
εεεεεε
Relationship between volumetric strain and linear strain
Volumetric strain
⎟⎟⎠
⎞⎜⎜⎝
⎛−−=
EEEzyx σµ
σµσ
⎟⎟⎠
⎞⎜⎜⎝
⎛−−+
EEEzxy σµσµ
σ⎟⎟⎠
⎞⎜⎜⎝
⎛−−+
EEExyz σµ
σµσ
zyxVdV εεε ++=
( )zyxEσσσµ
++−
=21
For element subjected to uniform hydrostatic pressure,
σσσσ === zyx
( )
( ) 321
21
σµ
σσσµ
EVdV
EVdV
zyx
−=
++−
=
⎟⎠⎞
⎜⎝⎛
=
VdV
K σ
( )
( )µ
µ
2-13KEor
modulusbulk 213
=
=−
=EK
Relationship between young’s modulus of elasticity (E) and modulus of rigidity (G) :-
A
D τ
B
τ
aa
45˚A1
φφ
B1
C
H
Consider a square element ABCD of side ‘a’ subjected to pure shear ‘τ’. DA'B'C is the deformed shape due to shear τ. Drop a perpendicular AH to diagonal A'C.Strain in the diagonal AC = τ /E – µ (- τ /E) [ σn= τ ]
= τ /E [ 1 + µ ] -----------(1)Strain along the diagonal AC=(A'C–AC)/AC=(A'C–CH)/AC=A'H/AC
In ∆le AA'HCos 45˚ = A'H/AA'A'H= AA' × 1/√2AC = √2 × AD ( AC = √ AD2 +AD2)Strain along the diagonal AC = AA'/ (√2 × √2 × AD)=φ/2 ----(2)Modulus of rigidity = G = τ /φ
φ = τ /GSubstituting in (2) Strain along the diagonal AC = τ /2G -----------(3)Equating (1) & (3)τ /2G = τ /E[1+µ]
E=2G(1+ µ)
Relationship between E, G, and K:-
We have
E = 2G( 1+ µ) -----------(1)
E = 3K( 1-2µ) -----------(2)
Equating (1) & (2)
2G( 1+ µ) =3K( 1- 2µ)
2G + 2Gµ=3K- 6Kµ
µ= (3K- 2G) /(2G +6K)
Substituting in (1)
E = 2G[ 1+(3K – 2G)/ (2G+6K)]
E = 18GK/( 2G+6K)
E = 9GK/(G+3K)
Working stress: It is obvious that one cannot take risk of loading a member to its ultimate strength, in practice. The maximum stress to which the material of a member is subjected to in practice is called working stress.
This value should be well within the elastic limit in elastic design method.
Factor of safety: Because of uncertainty of loading conditions, design procedure, production methods, etc., designers generally introduce a factor of safety into their design, defined as follows
Factor of safety =Allowable working stress
Maximum stress Allowable working stress
Yield stress (or proof stress)or
Malleability: A property closely related to ductility, which defines a material’s ability to be hammered out in to thin sheets
Homogeneous: A material which has a uniform structure throughout without any flaws or discontinuities.
Isotropic: If a material exhibits uniform properties throughout in all directions ,it is said to be isotropic.
Anisotropic: If a material does not exhibit uniform properties throughout in all directions ,it is said to be anisotropic or nonisotropic.
Exercise Problems
Q1. An aluminum tube is rigidly fastened between a brass rod and steel rod. Axial loads are applied as indicated in the figure. Determine the stresses in each material and total deformation. Take Ea=70GPa, Eb=100GPa, Es=200GPa
500mm 700mm600mm
steelaluminum
brass20kN 15kN 15kN 10kN
Ab=700mm2Aa=1000mm2
As=800mm2
Ans: σb=28.57MPa, σa=5MPa, σs=12.5MPa, δl = - 0.142mm
Q2. A 2.4m long steel bar has uniform diameter of 40mm for a length of 1.2m and in the next 0.6m of its length its diameter gradually reduces to ‘D’ mm and for remaining 0.6m of its length diameter remains the same as shown in the figure. When a load of 200kN is applied to this bar extension observed is equal to 2.59mm. Determine the diameter ‘D’ of the bar. Take E =200GPa
Ф = 40mm
Ф = D mm
200kN200kN
500mm500mm1000mm
Q3. The diameter of a specimen is found to reduce by 0.004mm when it is subjected to a tensile force of 19kN. The initial diameter of the specimen was 20mm. Taking modulus of rigidity as 40GPa determine the value of E and µ
Ans: E=110GPa, µ=0.36
Q.4 A circular bar of brass is to be loaded by a shear load of 30kN. Determine the necessary diameter of the bars (a) in single shear (b) in double shear, if the shear stress in material must not exceed 50MPa.
Ans: 27.6, 19.5mm
Q.5 Determine the largest weight W that can be supported by the two wires shown. Stresses in wires AB and AC are not to exceed 100MPa and 150MPa respectively. The cross sectional areas of the two wires are 400mm2 for AB and 200mm2 for AC.
Ans: 33.4kN
WA
CB300 450
Q.6 A homogeneous rigid bar of weight 1500N carries a 2000N load as shown. The bar is supported by a pin at B and a 10mm diameter cable CD. Determine the stress in the cable
Ans: 87.53MPa
3m
A CB
2000 N
3m
D
Q.7. A stepped bar with three different cross-sectional areas, is fixed at one end and loaded as shown in the figure. Determine the stress and deformation in each portions. Also find the net change in the length of the bar. Take E = 200GPa
250mm 270mm320mm
300mm2 450mm2
250mm2
10kN40kN20kN
Ans: -33.33, -120, 22.2MPa, -0.042, -0.192, 0.03mm, -0.204mm
Q.8 The coupling shown in figure is constructed from steelof rectangular cross-section and is designed to transmit a tensile force of 50kN. If the bolt is of 15mm diameter calculate:
a) The shear stress in the bolt;b) The direct stress in the plate;c) The direct stress in the forked end of the coupling.
Ans: a)141.5MPa, b)166.7MPa, c)83.3MPa
Q.9 The maximum safe compressive stress in a hardened steel punch is limited to 1000MPa, and the punch is used to pierce circular holes in mild steel plate 20mm thick. If the ultimate shearing stress is 312.5MPa, calculate the smallest diameter of hole that can be pierced.
Ans: 25mm
Q.10 A rectangular bar of 250mm long is 75mm wide and 25mm thick. It is loaded with an axial tensile load of 200kN, together with a normal compressive force of 2000kN on face 75mm×250mm and a tensile force 400kN on face 25mm×250mm. Calculate the change in length, breadth, thickness and volume. Take E = 200GPa & µ=0.3
Ans: 0.15,0.024,0.0197mm, 60mm3
Q.11 A piece of 180mm long by 30mm square is in compression under a load of 90kN as shown in the figure. If the modulus of elasticity of the material is 120GPa and Poisson’s ratio is 0.25, find the change in the length if all lateral strain is prevented by the application of uniform lateral external pressure of suitable intensity.
180
3030
90kN
Ans: 0.125mm
Q.12 Define the terms: stress, strain, elastic limit, proportionality limit, yield stress, ultimate stress, proof stress, true stress, factor of safety, Young’s modulus, modulus of rigidity, bulk modulus, Poisson's ratio,
Q.13 Draw a typical stress-strain diagram for mild steel rod under tension and mark the salient points.
Q.14 Diameter of a bar of length ‘L’ varies from D1 at one end to D2 at the other end. Find the extension of the bar under the axial load P
Q.15 Derive the relationship between Young’s modulus and modulus of rigidity.
Q.16 Derive the relationship between Young’s modulus and Bulk modulus.
Q.17 A flat plate of thickness ‘t’ tapers uniformly from a width b1at one end to b2 at the other end, in a length of L units. Determine the extension of the plate due to a pull P.
Q.18 Find the extension of a conical rod due to its own weight when suspended vertically with its base at the top.
Q.19 Prove that a material subjected to pure shear in two perpendicular planes has a diagonal tension and compression of same magnitude at 45o to the planes of shear.
Q.20 For a given material E=1.1×105N/mm2& G=0.43×105N/mm2 .Find bulk modulus & lateral contraction of round bar of 40mm diameter & 2.5m length when stretched by 2.5mm.
ANS: K=83.33Gpa, Lateral contraction=0.011mm
Q.21 The modulus of rigidity of a material is 0.8×105N/mm2 , when 6mm×6mm bar of this material subjected to an axial pull of 3600N.It was found that the lateral dimension of the bar is changed to 5.9991mm×5.9991mm. Find µ & E. ANS: µ=0.31, E= 210Gpa.
7. STATICALLY INDETERMINATE MEMBERS
&
THERMAL STRESSES
STATICALLY INDETERMINATE MEMBERS
Structure for which equilibrium equations are sufficient to obtain the solution are classified as statically determinate. But for some combination of members subjected to axial loads, the solution cannot be obtained by merely using equilibrium equations. The structural problems with number of unknowns greater than the number independent equilibrium equations are called statically indeterminate.
The following equations are required to solve the problems on statically indeterminate structure.
1) Equilibrium equations based on free body diagram of the structure or part of the structure.
2) Equations based on geometric relations regarding elastic deformations, produced by the loads.
COMPOUND BAR
L2
Material(2)
L1Material(1)
A compound bar is one which is made of two or more than two materials rigidly fixed, so that they sustain together an externally applied load. In such cases
(i) Change in length in all the materials are same.
(ii) Applied load is equal to sum of the loads carried by hb
W
(dL)1 = (dL)2
(σ1/ E1)L1 = (σ2 /E2)L2
σ1 = σ2 ×( E1/E2)(L1/L2) (1)
E1/E2 is called modular ratio
Total load = load carried by material (1) + load carried by material(2)
W = σ1 A1 + σ2 A2 (2)
From Equation (1) & (2) σ1 and σ2 can be calculated
Temperature Stress
L
A B
L
AB
L
A B
B´P
αTL
Any material is capable of expanding or contracting freely due to rise or fall in temperature. If it is subjected to rise in temperature of T˚C, it expands freely by an amount ‘αTL’ as shown in figure. Where α is the coefficient of linear expansion, T˚C = rise in temperature and L = original length.
From the above figure it is seen that ‘B’ shifts to B' by an amount ‘αTL’. If this expansion is to be prevented a compressive force is required at B'.
Temperature strain = αTL/(L + αTL) ≈ αTL/L= αTTemperature stress = αTE
Hence the temperature strain is the ratio of expansion or contraction prevented to its original length.
If a gap δ is provided for expansion then Temperature strain = (αTL – δ) / LTemperature stress = [(αTL – δ)/L] E
Temperature stress in compound bars:-
Material(2)
Material(1)
α2TL
∆
α1TL
(dL)1
P1
(dL)2P2
x
xWhen a compound bar is subjected to change in temperature, both the materials will experience stresses of opposite nature.
Compressive force on material (1) = tensile force on material (2)
σ1A1 = σ2A2 (there is no external load) σ1=( σ2A2)/A1 (1)
As the two bars are connected together, the actual position of the bars will be at XX.
Actual expansion in material (1) = actual expansion in material (2)
α1TL – (dL)1 = α2TL + (dL)2
α1TL – (σ1 / E1) L =α2TL + (σ2 / E2) L
αT – (σ1 / E1) = α2T + σ2 / E2 --------------------------(2)
From (1) and (2) magnitude of σ1 and σ2 can be found out.
Exercise problems
Q.1 A circular concrete pillar consists of six steel rods of 22mm diameter each reinforced into it. Determine the diameter of pillar required when it has to carry a load of 1000kN. Take allowable stresses for steel & concrete as 140Mpa & 8Mpa respectively. The modular ratio is 15 ANS: D=344.3mm
Q.2 Determine the stresses & deformation induced in Bronze & steel as shown in figure. Given As=1000mm2, Ab=600mm2, Es= 200Gpa, Eb= 83Gpa ANS: ( σb=55Mpa, σs=93.5Mpa, dLs=dLb=0.093mm)
160kN
Bronze BronzeSteel
Q..3 A cart wheel of 1.2m diameter is to be provided with steel tyre. Assume the wheel to be rigid. If the stress in steel does not exceed 140MPa, calculate minimum diameter of steel tyre & minimum temperature to which it should be heated before on to the wheel.
ANS: d=1199.16mm T=58.330C Q.4 A brass rod 20mm diameter enclosed in a steel tube of 25mm internal diameter & 10mm thick. The bar & the tube are initially 2m long & rigidly fastened at both the ends. The temperature is raised from 200C to 800C. Find the stresses in both the materials.
If the composite bar is then subjected to an axial pull of 50kN, findthe total stress. Es=200GPa, Eb=80GPa, αs=12×10-6/0C, αb=19×10-6/0C.
ANS: σb=8.81N/mm2 ( C ) , σs=47.99N/mm2( T )
8. STRESSES ON INCLINED PLANES
INTRODUCTION
The state of stress on any plane in a strained body is said to be ‘Compound Stress’, if, both Normal and Shear stresses are acting on that plane. For, example, the state of stress on any vertical plane of a beam subjected to transverse loads will, in general, be a Compound Stress. In actual practice the state of Compound Stress is of more common occurrence than Simple state of stress.
In a compound state of stress, the normal and shear stress may have a greater magnitude on some planes which are inclined (or, Oblique) to the given stress plane.
Hence in compound state of stresses it is necessary to find the following
(i) The normal and shear stress on a plane which is inclined (Oblique) to the given stress plane;
(ii) The inclination of max. and min. normal stress planes and values of the normal stress (max. / min.) on them;
(iii) The inclination of max. shear stress planes and the values of the shear stress (max.) on them.
EQUATIONS, NOTATIONS & SIGN CONVENTIONS :
(i) Normal & Shear stress on plane inclined (Oblique) to given stress plane:
Consider an element ABCD subjected to a state of compound stress as shown in the Fig. Let:σx Normal Stress in x- directionσy Normal Stress in y- directionτ Shear Stresses in x & y – directionsθ Angle made by inclined plane AE wrt verticalσθ Normal Stress on inclined planeτθ Shear Stress on inclined plane
A
E
D
C B
σy
τ
τ
σx σxθ
σy
σθ
τθ
Normal Stress, σθ, and Shear stress, τθ, on inclined plane are given by:
)1(2sin2cos22
−−−+⎟⎟⎠
⎞⎜⎜⎝
⎛ −+⎟⎟
⎠
⎞⎜⎜⎝
⎛ += θτθ
σσσσσθ
yxyx
)2(2cos2sin2
−−−−⎟⎟⎠
⎞⎜⎜⎝
⎛ −= θτθ
σστ θ
yx
(ii) The inclination of max. and min. normal stress planes and the values of the normal stress (max. / min.) on them
Let, θP be the inclination of the plane of max. or min. normal stress and σP be the value of the max. or min. normal stress on that plane, then, from Eqn. (1):
(1)--- 2sin2cos22 PP
yxyxP θτθ
σσσσσ +⎟⎟
⎠
⎞⎜⎜⎝
⎛ −+⎟⎟
⎠
⎞⎜⎜⎝
⎛ +=
( ) ( )
( ) ( )
0
02cos2sin2
02cos22sin22
0dθ
dσ min.,or max. be toσFor PP
=⇒
=−⎟⎟⎠
⎞⎜⎜⎝
⎛ −⇒
=+−⎟⎟⎠
⎞⎜⎜⎝
⎛ −⇒
=
P
PPyx
PPyx
θτ
θτθσσ
θτθσσ
Thus, the condition for max. or min. normal stress to occur on a plane is, shear stress on that plane should be zero.These planes on which shear stress is zero and the normal stress on them being either the max. or the min. are called ‘PRINCIPAL PLANES’.
( ) ( ) (3) ---
2
2tan02cos2sin2
have, we,0,
⎟⎟⎠
⎞⎜⎜⎝
⎛ −=⇒=−⎟⎟
⎠
⎞⎜⎜⎝
⎛ −
=
yxPPP
yx
PFrom
σστθθτθ
σσ
τθ
The above Eqn. (3), gives two values for θP, which differ by 900. Thus, there are two mutually perpendicular Principal planes, on which there are only normal stresses, shear stress being zero onthem.
On one of them, the value of the normal stress is the max.; it is called the ‘Major Principal plane’, the max. normal stress on it is called the ‘Major Principal Stress’.
On the other principal plane, the value of the normal stress is the min.; it is called the ‘Minor Principal plane’, the min. normal stress on it is called the ‘Minor Principal Stress’.
( )
2
2
2
2
2)(
2/2cos
2)(
2sin
get, we,
2
2tan From
τσσ
σσθ
τσσ
τθ
σστθ
+⎥⎦⎤
⎢⎣⎡ −
−±=
+⎥⎦⎤
⎢⎣⎡ −
±=
⎟⎟⎠
⎞⎜⎜⎝
⎛ −=
YX
YXP
YX
P
yxP
(σx-σy)/ 2
2θP
τ
±√[(σ
x - σ
y/2]2
+ τ2
Substituting for sin 2θP and cos 2θP in Eqn. (1), and simplifying, we get the equation for principal stresses as:
( ))4(
222
2
−−−+⎟⎟⎠
⎞⎜⎜⎝
⎛ −±
+= τ
σσσσσ yxyx
P
The above equation (4) gives two values for principal stresses. The numerically max. of the two values (+ ve or − ve) is the Major Principal Stress, (σMajor or σMax); The numerically min. (+ ve or − ve) is the Minor Principal Stress (σMinoror σMin).
(iii) Inclination of max. shear stress planes, Max. shear stress Equation. Let, θS be the inclination of the plane of max. or min. shear stress and τS be the value of the max. or min. shear stress on that plane, then, from Eqn. (2):
)2( 2cos2sin2
−−−−⎟⎟⎠
⎞⎜⎜⎝
⎛ −= SS
yxS θτθ
σστ
( ) ( )
12tan 2 thave We:
(5) --- 2
2tan
02sin22cos22
0dθd min.,or max. be toFor
S
SS
−=×
⎟⎟⎠
⎞⎜⎜⎝
⎛ −
−=⇒
=−−⎟⎟⎠
⎞⎜⎜⎝
⎛ −⇒
=
SP
yx
S
SSyx
anNOTE θθτ
σσ
θ
θτθσσ
ττEqn. (5) gives two values for θS, which differ by 900. Thus, there are two mutually perpendicular planes, on which shear stress are max.; numerically equal but opposite in sense.The planes of Max. Shear stresses are inclined at 450 to the Principal planes.
( )2
2
2
2
2)(
2/2sin
2)(
2cos
get, we,2
2tan From
τσσ
σσθ
τσσ
τθ
τ
σσ
θ
+⎥⎦
⎤⎢⎣
⎡ −
−±=
+⎥⎦
⎤⎢⎣
⎡ −±=
⎟⎟⎠
⎞⎜⎜⎝
⎛ −
−=
yx
yxS
yx
s
yx
S
(σx-σy)/ 2
2θS
τ
±√[(σ
x − σ
y)/2]
2 + τ2
Substituting for sin 2θS and cos 2θS in Eqn. (2), we get the equation for Max. shear stresses as:
)6(
2
2
.max
22
.max
−−−
⎥⎥⎥⎥⎥
⎦
⎤
⎟⎟⎠
⎞⎜⎜⎝
⎛ −±=⇒
+⎟⎟⎠
⎞⎜⎜⎝
⎛ −±=
MinorMajor
yx
σστ
τσσ
τ
The above equation (6) gives two values for Max. shear stresses,which are numerically equal but opposite in sense.
NOTATIONS & SIGN CONVENTIONS
σx Normal Stress in x- directionσy Normal Stress in y- directionτ Shear Stresses in x & y – directionsθ Angle made by inclined plane wrt verticalσθ Normal Stress on inclined plane AEτθ Shear Stress on inclined plane AE
A
E
D
C B
σy
τ
τ
σx σxθ
σy
σθ
τθ
θP Inclination of Principal planesσP Principal stresses θS Inclination of Max. shear stress planes [θS = θP + 450].
All the parameters are shown in their +ve sense in the Fig.
Normal stresses, σ Tensile stresses +ve.Shear Stresses, τ, in x – direction & Inclined Plane Clockwise +ve.Shear Stresses, τ, in y – direction Anti-Clockwise +ve.Angle, θ measured w r t vertical, Anti-Clockwise +ve.
Exercise Problems:
Q.1 The principal stresses at a point in a strained material are 80 MPa(C) and 40 MPa(T). Find the normal, tangential and resultant stress on a plane inclined at 50o to the major principal plane. [Ans: - 9.58MPa, -59MPa].
Q.2 The stresses in a strained material is as shown in Fig. Find the normal and shear stresses on plane inclined at 30o to the horizontal. Also determine the intensity and position of the plane upon which there is only shear stress. Sketch the plane.
Ans:σ = − 22.5MPa, τ = − 64.95MPa. Pure shear plane σ = 0, Ө = 39.23o w.r.t. horizontal.τpure = − 73.48MPa.
60MPa
30o
90 MPa
Q.3 A plane element is subjected to the system of stresses as shown in Fig. Determine (i) the principal stresses and inclination of their planes (ii) maximum shearing stresses and inclination of their planes. Represent your answers in neat sketches.
160 MPa
200 MPa
80 MPa
Ans:σ1 = 262.46MPa, σ2 = 97.54MPa. Өp = 37.98 or 127.98 with horizontalτtmax = 82.46, Өs = 82.98 or 172.98 with horizontal
Q.4 At a point in a structural member subjected to stresses asshown in fig. determine the principal stresses and the maximum shear stress. Also determine and sketch planes on which these stressesact.
100
80
40
[Ans:131.23, 48.77, − 37.98o, − 127.98o
41.23, 7o, 970 , Angles w.r.t. horizontal]
Q.5 At a point in a material under stress, the intensity of resultant stress on a certain plane is 60 MPa, directed outwards and inclined at 30o to the normal of that plane. The stress on the plane at right angles to this has a normal stress component of 40 Mpa (T). Find (i) the principal stresses and inclination of their planes, (ii) the maximum shear stresses and inclination of their planes .
[Ans: 76.57MPa, 15.39MPa, 39.36o , 129.36o and 30.59 MPa, 84.36o , 174.36o, Angles w.r.t. horizontal]
9. Stresses due to fluid pressure in thin cylinders
9 -THIN CYLINDERS
They are,
1. Hoop or Circumferential Stress (σC) – This is directed along the tangent to the circumference and tensile in nature. Thus, there will be increase in diameter.
In many engineering applications, cylinders are frequently used for transporting or storing of liquids, gases or fluids.
Eg: Pipes, Boilers, storage tanks etc. These cylinders are subjected to fluid pressures. When
a cylinder is subjected to internal pressure, at any point on the cylinder wall, three types of stresses are induced on three mutually perpendicular planes.
INTRODUCTION:
2. Longitudinal Stress (σL) – This stress is directed along the length of the cylinder. This is also tensile in nature and tends to increase the length.
3. Radial pressure (σ r) – It is compressive in nature. Its magnitude is equal to fluid pressure on the inside wall and zero on the outer wall if it is open to atmosphere.
σC σL
1. Hoop/Circumferential
Stress (σC)
2. Longitudinal Stress (σL) 3. Radial Stress (σr)
Element on the cylinder wall subjected to these three stresses
σL
σL
σL
p pσr
σCσC
σC
p
σLσL
σCσr
σrσC
A cylinder or spherical shell is considered to be thin when the metal thickness is small compared to internal diameter.
i. e., when the wall thickness, ‘t’ is equal to or less than ‘d/20’, where ‘d’ is the internal diameter of the cylinder or shell, we consider the cylinder or shell to be thin, otherwise thick.
Magnitude of radial pressure is very small compared to other two stresses in case of thin cylinders and hence neglected.
)1....(....................t2dpσ stress, ntialCircumfere c ×
×=
)2 .........(..........t4dpσ stress, alLongitudin L ×
×=
Wherep = internal fluid pressured= internal diameter, t = thickness of the wall
Maximum Shear stress, τmax = (σc- σL) / 2
τmax = ( p x d) / (8 x t)
EVALUATION OF STRAINS
σ L=(pd)/(4t)
σ C=(pd)/(2t) σ C=(pd)/(2t)
σ L=(pd)/(4t)
A point on the surface of thin cylinder is subjected to biaxial stress system, (Hoop stress and Longitudinal stress) mutually perpendicular to each other, as shown in the figure. The strains due to these stresses i.e., circumferential and longitudinal are obtained by applying Hooke’s law and Poisson’s theory for elastic materials.
µ)2(Eσ
Eσµ
Eσ2
Eσµ
Eσε
:ε strain, ntialCircumfere
L
LLLCC
C
−×=
×−×=×−=
µ)21(Eσ
E)σ2(µ
Eσ
Eσµ
Eσε
:ε strain, alLongitudin
LLLCLL
L
×−×=×
×−=×−=
)3..(..............................µ)2(Et4
dp dδdε i.e., C −×
×××
==
C=(pd)/(2t)
L=(pd)/(4t)
L=(pd)/(4t)
C=(pd)/(2t)
)4..(..............................µ)21(Et4
dpLδl ε i.e., L ×−×
×××
==
µ)2(Et4
dp2µ)21(Et4
dp
dd2
LdL
Ld4π
dd24πLdLd
4π
Vdv
dd24πLdLd
4πdV
L.d4πV volume,have We
2
2
2
2
−×××
×+×−×××
=
×+=××
××××+××=
××××+××=
××=
dd
d
= εL+2× εCVdV
)5.......(..........µ)45(Et4
dpVdv i.e., ×−
×××
=
σC=(pd)/(2t)σC=(pd)/(2t)
σL=(pd)/(4t)
σL=(pd)/(4t)
Vdv STRAIN, VOLUMETRIC
JOINT EFFICIENCY
The cylindrical shells like boilers are having two types of joints namely Longitudinal and Circumferential joints. Due to the holes for rivets, the net area of cross section decreases and hence the stresses increase. If the efficiencies of these joints are known, the stresses can be calculated as follows.
Let ηL=Efficiency of Longitudinal jointand ηC=Efficiency of Circumferential joint.
...(1).......... ηt2
dpσ L
C ×××
=
Circumferential stress is given by,
Longitudinal stress is given by,
...(2).......... ηt4dpσ
CL ××
×=
Note: In longitudinal joint, the circumferential stress is developed and in circumferential joint, longitudinal stress is developed.
Exercise Problems
Q.1Calculate the circumferential and longitudinal strains for a boiler of 1000mm diameter when it is subjected to an internal pressure of 1MPa. The wall thickness is such that the safe maximum tensile stress in the boiler material is 35 MPa. Take E=200GPa and µ= 0.25.
(Ans: ε C=0.0001531, ε L=0.00004375)
Q.2A water main 1m in diameter contains water at a pressure head of120m. Find the thickness of the metal if the working stress in the pipe metal is 30 MPa. Take unit weight of water = 10 kN/m3.
(Ans: t=20mm)
Q.3A gravity main 2m in diameter and 15mm in thickness. It is subjected to an internal fluid pressure of 1.5 MPa. Calculate the hoop and longitudinal stresses induced in the pipe material. If a factor of safety 4 was used in the design, what is the ultimate tensile stress in the pipe material?
(Ans: σC=100 MPa, σL=50 MPa, σU=400 MPa)Q.4At a point in a thin cylinder subjected to internal fluid pressure, the value of hoop strain is 600×10-4 (tensile). Compute hoop and longitudinal stresses. How much is the percentage change in the volume of the cylinder? Take E=200GPa and µ= 0.2857. (Ans: σC=140 MPa, σL=70 MPa, %age change=0.135%.)
Q.5A cylindrical tank of 750mm internal diameter and 1.5m long is to be filled with an oil of specific weight 7.85 kN/m3 under a pressure head of 365 m. If the longitudinal joint efficiency is 75% and circumferential joint efficiency is 40%, find the thickness of the tank required. Also calculate the error of calculation in the quantity of oil in the tank if the volumetric strain of the tank is neglected. Take permissible tensile stress as 120 MPa, E=200GPa and µ= 0.3 for the tank material.
(Ans: t=12 mm, error=0.085%.)