1. BENDING MOMENT ANS SHEARING FORCE
STRENGHT OF MATERIALS
BENDING MOMENT ANS SHEARING FORCE
Concept of beams and type of loading.
Concept of end supports-Roller,Hinged and fixed.
Concept of bending moment and shearing force .
B.M and S.F. Diagram for cantilever and simply supported beams with and without overhang subjected to concentrated and U.D.L.

2. BEAM:- A beam is a structural member which can take loads acting at right angles to its longitudinal axis. Generally, a beam is a horizontal member of moderate size and is made up of one piece.

3. TYPES OF END SUPPORTS OF BEAMS The following are the important types of support of beams:- Free support:- when the beam rest freely on the support, the support is known as free support or simply support.

4. Hinge Supports
The hinge support is capable of resisting forces acting in any direction of the plane. This support does not provide any resistance to rotation. The horizontal and vertical component of reaction can be determined using equation of equilibrium. Hinge support may also be used in three hinged arched bridges at the banks supports while at the center internal hinge is introduced. It is also used in doors to produce only rotation in a door. Hinge support reduces sensitivity to earthquake.

5.   Roller Supports Roller supports are free to rotate and translate along the surface upon which the roller rests. The surface may be horizontal, vertical or slopped at any angle. Roller supports are commonly located at one end of long bridges in the form of bearing pads. This support allows bridge structure to expand and contract with temperature changes and without this expansion the forces can fracture the supports at the banks. This support cannot provide resistance to lateral forces. Roller support is also used in frame cranes in heavy industries as shown in figure, the support can move towards left, right and rotate by resisting vertical loads thus a heavy load can be shifted from one place to another horizontally.

6. Fixed Supports Fixed support can resist vertical and horizontal forces as well as moment since they restrain both rotation and translation. They are also known as rigid support For the stability of a structure there should be one fixed support. A flagpole at concrete base is common example of fixed support In RCC structures the steel reinforcement of a beam is embedded in a column to produce a fixed support as shown in above image. Similarly all the riveted and welded joints in steel structure are the examples of fixed supports Riveted connection are not very much common now a days due to the introduction of bolted joints.

7. CLASSIFICATION OF BEAMS   The beam may be classified in several ways, but the commonly used classification is based on end conditions .On this basis, the beams can be divided into six types: 1.Cantilever beams Simply supported beam 3.Overhanging beams Propped cantilever beam 5.Fixed beams Continuous beam

8. CLASSIFICATION OF BEAMS : Depending upon the type of supports, beams are classified as follows : 1) Cantilever 2) Simply (or freely) supported beam 3) Overhanging beam 4) Fixed beam 5) Continuous beam   CANTILEVER : A cantilever is a beam whose one end is fixed and the other end free. In other words, a cantilever is a beam anchored at only one end. The beam carries the load to the support where it is forced against by a moment and shear stress. Cantilevers can also be constructed with

9.  SIMPLY SUPPORTED BEAM : A  simply supported beam is one whose ends freely rest on walls or columns or knife edges. In all such cases, the reaction is upwards. It is the one of the simplest structural elements in existence.   It is the type of beam that has pinned support at one end and roller support at the other end. Depending on the load applied, it undergoes shearing and bending.

10.   OVERHANGING BEAM :  An overhanging beam is one in which the supports are not situated at the ends i.e. one or both the ends project beyond the supports.

11. FIXED BEAM : A fixed beam is one whose both ends are rigidly fixed or built in into its supporting walls or columns

12. CONTINUOUS BEAM : A continuous beam is one which has more than two supports. The supports at the extreme left and right are called the end supports and all other supports except the extreme are called intermediate supports.   It may be noted that the first three types of beams (i.e. cantilevers, simply supported beams and overhanging beam) are known as Statically Determinate Beams as the reactions of these beams at their supports can be determined by the use of equations of static equilibrium and the reactions are independent of the deformation of beams. The last two types of beams ( i.e. fixed beams and continuous beams) are known as Statically Indeterminate Beams as their reactions cannot be determined by the use of equations of static equilibrium.

13. Types of Load There are three types of load
Types of Load There are three types of load. These are: Point load that is also called as concentrated load. Distributed load Coupled load   POINT LOAD Point load is that load which acts over a small distance. Because of concentration over small distance this load can may be considered as acting on a point. Point load is denoted by P and symbol of point load is arrow heading downward (↓).

14. TYPES OF DISTRIBUTED LOAD
Distributed load is that acts over a considerable length or you can say “over a length which is measurable. Distributed load is measured as per unit length.
EXAMPLE
If a 10k/ft load is acting on a beam having length 10′. Then it can be read as “ten kips of load is acting per foot”. If it is 10′then total point load acting is 100Kips over the length.
TYPES OF DISTRIBUTED LOAD
Distributed load is further divided into two types.
Uniformly Distributed load (UDL)
Uniformly Varying load (Non-uniformly distributed load).
UNIFORMLY DISTRIBUTED LOAD (UDL)
Uniformly distributed load is that whose magnitude remains uniform throughout the length. For Example: If 10k/ft load is acting on a beam whose length is 15ft. Then 10k/ft is actingthroughout the length of 15ft.
Uniformly distributed load is usually represented by W and is pronounced as intensity of udl over the beam, slab etc.

15. What is shear force?] Below a force of 10N is exerted at point A on a beam. This is an external force. However because the beam is a rigid structure,the force will be internally transferred all along the beam. This internal force is known as shear force. The shear force between point A and B is usually plotted on a shear force diagram. As the shear force is 10N all along the beam, the plot is just a straight line, in this example.

17. Basic bending moment diagram:- Bending moment refers to the internal moment that causes something to bend. When you bend a ruler, even though apply the forces/moments at the ends of the ruler, bending occurs all along the ruler, which indicates that there is a bending moment acting all along the ruler. Hence bending moment is shown on a bending moment diagram. The same case from before will be used here:

18.   Shear Force & Bending Moment Diagram of Cantilever Beam Shear force on cantilever beam is the sum of vertical forces acting on a particular section of a beam. While bending moment is the algebraic sum of moments about the centroidal axis of any selected section of all the loads acting up to the section. Example: Draw shear force and bending moment diagrams of the cantilever beam carrying point loads. As shown in figure;  Home > SFD & BMD > Shear Force & Bending Moment Diagram of Cantilever Beam Shear Force & Bending Moment Diagram of Cantilever Beam  Jalal Afsar  October 15, 2014  SFD & BMD  No Comments Shear force on cantilever beam is the sum of vertical forces acting on a particular section of a beam. While bending moment is the algebraic sum of moments about the centroidal axis of any selected section of all the loads acting up to the section. Example: Draw shear force and bending moment diagrams of the cantilever beam carrying point loads. As shown in figure;

19. SOLUTION Shear Force To draw a shear force diagram
SOLUTION Shear Force To draw a shear force diagram. First find value of shear force between varying loads. Let start from left side. Shear force Between point D and C S.F (D-C) = -100 kg. Shear force value increases gradually as we move towards fixed end. Shear force Between Point C and B S.F (C-B) = -( ) = -300 kg. Now one can see, shear force between point C and B is the sum of point loads acting up to that point. Shear force Between Point B and A S.F (B – A) = -( ) = – 600 kg. One can see shear force between B and A is the sum of all point loads acting on it. This shows shear force is maximum at fixed end and minimum at free end of cantilever beam. Bending Moment Bending moment at point D = B.M (D) = 0 Bending moment at point C = B.M (C) = -(100×1) = -100 kg.m Bending moment at point B = B.M (B) = – (100×2 +200×1) B.M (B) = -400 kg.m

21. Calculate Cantilever subjected to U. D. L
  Calculate Cantilever subjected to U.D.L. over whole length   Consider a cantilever beam AB carrying a UDL of ‘w’ kN/m over the entire length of the cantilever, as shown in figure. Let the length of the beam be ‘L’. Then at any section (X – X’), at a distance x from end B.

22. Simply supported beam with point load at center   Consider a simply supported beam AB, with span ‘L’, and subjected to point load (W) at the centre, as shown in figure.   To draw SFD and BMD, we need RA and RB.   RB.L – W. L/2 = 0   ⇒ RB = W/2   Also, from condition of static equilibrium Σ Fy = 0 i.e.,   RA + RB – W = 0   ⇒ RA = W – RB = W – W/2   RA = W/2

23. Chapter 5.1 BENDING STRESSES
The resistance induced to resist bending is known as bending stress or longitudinal stress and the resistance induced to resist the shearing is known as shearing stress.
5.2 PURE BENDING OR SIMPLE BENDING
The bending of the beam not accompanied by any shear force is known
as pure bending or simple bending.
Fig. 5.1 : Pure bending
5.3 ASSUMPTIONS IN THE THEORY OF SIMPLE BENDING
The following assumptions are made in the theory of simple bending :
1. The transverse section of the beam, which was plane before bending , will remain plane after bending.

24. 2. Every cross-section of the beam is symmetrical about the plane of bending.
3. The material of the beam is homogeneous (i.e. same type throughout) and isotropic (i.e. of same elastic properties in all directions).
4. The material of beam is stressed within its elastic limit and Hook's low.
5. The value of Young's modulus of elasticity is the same in tension and compressive.
5.4 NEUTRAL AXIS
The intersection of the neutral surface with the crosssection of the beam perpendicular to its longitudinal axis is called neutral axis (N.A.).
5.5 DERIVATOON OF BENDING EQUATIONM
Let us consider a small portion ABCD of the beam which is subjected to bending moment M unaccompanied by any shear force between two parallel section AB and CD . Let this portion of the beam bends an arc circle with o as centre.

25. 0 = Angle subtended by the arc at the centre,
R = Radius of curvature i.e. radius of neutral layer.
Fibres above neutral axis will be under compression and fibres neutral axis will be under tension. Let us consider a layer EF at a distance y from neutral axis. After deformation, EF will become EF.

26. change in lemgth of EF original length of EF
Changer in length of layer EF = EF - EF
= EF - GH (.-.EF=GH)
= (R+Y) 0 -R0 (Length of are x Angle subtended by the arc the centre)
= R0 + y0 —r0
= Y0
Striain in layer EF,E =
change in lemgth of EF original length of EF
Y0 _y R0 =R
strain, E =-
But
When
0= Intending of stress in layer GH,
E= YOUNG'S MODUPUS.
From equation (1) and (2), we get
L-0
R =E Or

27. We can write the above relation as below :
---Y ~R
We can write the above relation as below : E
S = - Y R
Here E a and R are constant fir given beam. Hence, stress in any fibre is proportional to the distance of the fibre from the maximum bending stress of opposite nature will be at extreme on both sides of the neutral axis.
Let up consider elementary areas a1,a2, etc , at a
^distance of y1, y2, etc. respectively from neutral axis.
Let a 1, o be the stresses on elementary areas a1,a2 ....etc .
Force on elementary area a1= alal
Force on elementary area a2= &2a2... Moment of all these forces about neutral-axis,
M= alalyl +— ... (3)
a_E
y R

28. Putting the values of al, o2f.... in equation (3), we get
but
o2 = E
a = -y
R J
al = —yl Ry
= —yl and os on. Ry
Putting the values of al, o2f.... in equation (3), we get
M=-ylalyl + -y2a2y2 ...
=-a1y12 + -a2y
r J r J
(a1yl2 +~a2y )
Wher E
= R!
M E ~~R
I=Moment of inertia of beam section
about neutral axis
Hence
M a E z “ y ~R
The above equation is known as bending equation.

29. section about neutral axis,
The sum of moments of the internal forces about neutral axis is known as the moment of resistance or flexural strength.
Form bending equation, we have
M _ o
i y
a I
Or M= x/ = jx ■
y y
Or M=aZ
Where Z is known as section modulus . Thus moment of resistance of a beam is directly proportional to its section modulus for a given material of the beam.
5.7 SECTION MODULUS
The ratio of moment of inertia of a section about the neutral axis to the distance of the extreme fibre from the neutral axis .IT is denoted by the symbol Z.
Z=- y
Where I= Moment of inertia of the beam
section about neutral axis,
Y=Distance of the extreme fibre from the neutral axis.

30. HR 5.9 MOMENT OF RUPTURE l-.f-d4 d1 32 6b l„. -^(D4-d4)
Section modulus
Type of section
Moment of Inertia
Jmii
RMMflll
paralleogram
Hoik>w rectangular section
bd3 bid?
lw* * - -
-(bd3- b,cf} )
db3 di&i
La _•
Zyy = l(db3 - d,b>, ) 6b
Circular socl on
l-.f-d4 d1 32
Hollow curcular section
Zyy-Z
l„. -^(D4-d4)
— (D4 - d4)
320
section
bd3 b,d?
Z» - —(bd3 - b,d? )
l«w ■ - -
db3
Zyy - —(db3 - d,t?, ) 6b HR
l„ » 1 (bd3 - (b - l) df? )
Mangto
5.9 MOMENT OF RUPTURE

32. Chapter:-6
COLUMNS AND STRUTS

33. Strut:- It is an inclined member which carries an axial compressive
Column:- It is a vertical member of structure which carries an axial compressive load.
Strut:- It is an inclined member which carries an axial compressive
load

34. Properties of an ideal colums
The column should be perfectly straight.
The material of the column should be homogeneous & isotropic i.e the elastic properties should be same in all the direction.
The cross-section of column must be uniform
The compressive load must be perfectly axial.

35. Various terms used in analysis of column
Actual length of the column is the distance b/w centres of effective lateral reshaints.It is represented by L.
The equivalent or effective length of the column is the distance b/w adjacent point of inflexion.

36. End Condition:- These can be three condition a the end of column.
Fixed End :-In this case end is fixed both in position and direction for such end,the deflection and slope is zero.
y=0 & dy/dx=0
Pinned End :-In the case,end is fixed in position only i.e
y=0
Free End :-In the case,the column end is neither fixed in position non in direction.
Thus,the colums are of form types are as follows ;

37. Types of column are:-

38. MOMENT OF INTERTIA

39. SECOND MOMENT OF AREA:-
MOMENT OF INERTIA:-
Moment of inertia of a mass about any reference axis is the second moment of mass about that axis. Moment of inertia, also termed as second moment of mass, is denoted by I.
I = Mass * Square of the perpendicular distance between the C.G. of the mass and reference axis.
SECOND MOMENT OF AREA:-
The product of the area and the square of the distance of the centroid of the area from reference axis is known as second moment of area. It is called second moment of area because we are taking moment of area about reference axis twice. The S.I. unit of second moment area is m^4.
RADIUS OF GYRATION:-
The distance of a point where the whole area of a figure is assumed to be concentrated from a given axis is called radius of gyration.

40. THEOREM OF PARALLEL AXIS
It states that the moment of Inertia of a laminar about any axis in the plane of the lamina is equal to the sum of the moment of inertia of that lamina about its centroidal axis paralllel to the given axis and the product of the area of lamina and square of the perpendicular distance between the two axes.

42. CENTROID OF COMMON PLANE GEOMETRY FIGURES

43. MOMENT O INERTIA OF PLANE GEOMETRICAL FIGURES

44. THEOREM OF PERPENDICULAR AXIS
It states that the moment of inertia of a lamina about an axis perpendicular to the lamina and passing through the intersection of other two mutually perpendicular axes in the plane of lamina is equal to the sum of moment of inertia of the given lamina about two mutually perpendicular axes in the plane of lamina.

46. Chapter-2
Resilience

47. Strain Energy:- The work done in straining the body within elastic limit is known as strain energy.
Resilience:- The capacity of a strained body for doing work (when it springs back) on the removal of the straining force.
Proof Resilience:- The maximum strain energy which can be stored in a body upto the elastic limit is called proof resilience.
Modulus of Resilience:- Proof resilience per unit volume the body is known as modulus of resilience.
Modulus of resilience= proof resilience .
volume of the body

48. STRAIN ENERGY STORED IN A BODY DUE TO GRADUALLY APPLIED LOAD
TYPES OF LOADING
A load may act on the body in the following ways:
Gradually
Suddenly
With impact
STRAIN ENERGY STORED IN A BODY DUE TO GRADUALLY APPLIED LOAD
A Gradually applied load is that which is applied
gradually on the body i.e. loading begins from zero
And increases gradually till the body is fully loaded.

49. The below derivation is of STRAIN ENERGY STORED IN A BODY DUE TO GRADUALLY APPLIED LOAD

50. STRAIN ENERGY IN A BODY DUE TO SUDDENLY APPLIED LOAD
A load applied suddenly on a body is called suddenly applied load. strain energy stored in the bar(U)=Workdone on the bar by of the load =Load*Extension =P*𝛿l

51. STRAIN ENERGY STORED IN A BODY DUE TO IMPACT LOAD
A load applied with some velocity is called impact load.

52. STRAIN ENERGY STORED IN A BODY DUE TO SHEAR STRESS

53. SPRINGS Detailed Content:- Introduction Stress deformation
Closed coil helical springs subjected to axial load and calculation of:
Stress deformation
Stiffness and angle of twist and strain energy
Strain energy and proof resilience.
Determination of number of plates of laminated spring (semi elliptical type only)

54. Springs:- Spring is an elastic member which distorts under load and regain its original shape when distorting load are removed.
Function of Springs:-
To absorb or control energy due to shock or vibrations.
To apply force and to control motion.
To store energy.
To measure force.

55. Types of Springs:- 1. Helical Springs:- (a) Close-Coiled
(b) Open-Coiled
2. Laminated Springs (Leaf springs):-
(a) Full-Elliptical
(b) Semi-Elliptical
(c) Cantilever
3. Spiral Springs
Helical Springs: A length of wire when wound into a helix form a helical springs.
(a) Close-Coiled:- It is a coil in such a way that the coils are in contact with each other and slope of helix is so small that bending effect can be neglected.
(b) Open-Coiled:- In open coil helical springs, there is a large gap between two consecutive coil.

56. Term Relating To Helical Springs
Solid Length:- The solid length of the spring is the product of total number of coils
and the diameter of the wire from which the spring is formed.
i.e. Solid Length = Number of coils*Diameter of wire = n*d
Free Length:- The length of the spring in the free or unloaded condition is called
free length.
Spring Index:- The ratio of mean diameter of the coil and diameter
of the wire is called spring index.
i.e. Spring Index , C= D/d
where D=mean diameter of the coil; d=diameter of the wire

57. Pitch:- The axial distance between adjacent coils in unloaded state is pitch.
Pitch of coil= Free length/(n’-1)
where n’ is the total number of coils.
Helix Angle:- It is the angle which the axis of the spring wire makes with a horizontal
line perpendicular to the axis of the spring.
Stiffness of Spring:- It is defined as the load required per unit deflection of the
spring.
i.e. S= W/ẟ 
where W= Axial Load on spring
ẟ= Deflection of spring due to axial load 

58. Angle of twist:- It is the angle through which the cross section of bar is twisted
due to the twisting moment or torque.
Proof load:- It is the maximum load which a spring can carry without undergoing
any permanent distortion.
Proof Stress:- The maximum stress developed in a spring due to the proof Load is
called proof stress.
Proof Resilience:- It is the maximum resilience of the spring without the occur-
hence of permanent deformation in the spring.
Resilience:- When a body is stressed within the elastic limit. The amount of
internal energy stored is called resilience.

59. CLOSED COIL HELICAL SPRING SUBJECTED TO AXIAL LOAD
According to figure:-
A closed coil helical spring subjected to axial compressive load .
The upper end of the spring is fixed. An axial load W is acting at free end
of spring.
Due to this-
(a) Twisting moment(T=W*R) will act on the section of spring
(b) Due to shear load W, bending stress (fb) will induce in wire of the spring
d = Diameter of spring wire or coil
p = Pitch of the helical spring
D = Mean diameter of spring

60. Where:- n = Number of spring coils W = Load applied on spring axially
G= Modulus of rigidity
τ = Maximum shear stress developed in the spring wire
θ = Angle of twist in wire of spring
L = Length of the spring
δ = Deflection of spring under axial load
R = Mean radius of spring

61. Minimum Shear Stress:-
Twisting moment, T=W*R
Also Twisting moment, T=π /16fsd³ or
W*R= π /16fsd³
fs= 16WR/ πd³
Angle of twisting:- From Torsion Equation
T/Ip=Gθ/l
where
T= Twisting moment acting on spring
Ip= M.O.I. of cross section of wire about ZZ axis
Izz=Ixx+Iyy= π/64d4+ π/64d4= π/32d4
l= length of wire = 2 πR*n
we have, W.R./ π32d4=Gθ/l
θ=WR.l/G πd4*32= 32WR/ πGd4*2 πRn
i.e. Θ= 64WR2n/Gd4

62. Deflection of Spring:-
When there is no load on spring, AB will remain horizontal. By the application of to Axial load W, AB is displaced to position AB’. The BB’ will show the deflection delta(ẟ ) of spring.
From △AAB’ , tanθ = BB’/AB
BB’= ABtanθ
Since is very small so tanθ= θ
BB’=AB*θ
ẟ =R*θ
ẟ= R*64WR2n/Gd4 or
ẟ= 64WR³n/Gd4

63. Strain Energy Stored in the Spring:-
Strain energy is the energy stored in the spring due to deflection-delta(ẟ ).
i.e. Strain Energy, U=1/2W. ẟ = 1/2W*64WR³n/Gd4
U=32W²R³n/Gd4
Stiffness of Coil:- It is the amount of the load required per unit deflection.
i.e S=W/ẟ=W/64WR³/Gd4
S= Gd4/64R³n

64. LAMINATED SPRINGS [ Semi Elliptical Springs] : -These springs are widely used to absorb shocks in the automobiles. These springs consists of number of parallel strips or plates of Metal having different length, but same width .these are placed one over another. All the plates are initially bent to the same radius. Plates are secured together at the centre with the help of centre bolt. U –Clamps are provided to secure compactness.

65. In the slides we will cover the following Topics:-
STRENGTH OF MATERIALS
In the slides we will cover the following Topics:-
Concept of load ,Stress ,Strain
Tensile ,Compressive & Shear Stress
Linear strain ,Lateral strain ,Shear Strain ,Volumetric Strain
Concept of Elasticity ,Elastic Limit ,& Limit of proportionality
Hooke’s Law & Elastic Constant
Stress-strain curve for ductile and brittle materials
Nominal stress
Yield point , Plastic stage
Ultimate stress & Breaking stress
Percentage elongation
Proof stress and working stress
Factors of Saftey
Poisson’s Ratio
Thermal stress & strain
Longitudinal and circumferential stress in seamless thin walled cylindrical shell

66. DIRECT OR NORMAL STRESS
 When a force is transmitted through a body, the
body tends to change its shape or deform. The
body is said to be strained.
 Direct Stress = Applied Force (F)
Cross Sectional Area (A)
 Units: Usually N/m2 (Pa), N/mm2, MN/m2, GN/m2 or
N/cm2
 Note: 1 N/mm2 = 1 MN/m2 = 1 MPa

67. Hooke’s Law • States that providing the limit of proportionality of
a material is not exceeded, the stress is directly
proportional to the strain produced.
• If a graph of stress and strain is plotted as load is
gradually applied, the first portion of the graph will
be a straight line.
• The slope of this line is the constant of
proportionality called modulus of Elasticity, E or
Young’s Modulus.
• It is a measure of the stiffness of a material.

68. Linear strain :- Shear strain :- Volumetric Strain :-
Linear strain of a deformed body is defined as the ratio of the change in length of the body due to the deformation to its original length in the direction of the force.
Shear strain :-
Shear strain is defined as the tangent of the angle, and is equal to the length of deformation at its maximum divided by the perpendicular length in the plane of force application, which sometimes makes it easier to calculate.
Volumetric Strain :-
The volumetric strain is the unit change in volume, i.e. the change in volume divided by the original volume .

69. Elastic limit & Limit of Proportionality

70. Hooke’s Law
• States that providing the limit of proportionality of a material is not exceeded, the stress is directly proportional to the strain produced. • If a graph of stress and strain is plotted as load is gradually applied, the first portion of the graph will be a straight line. • The slope of this line is the constant of proportionality called modulus of Elasticity, E or Young’s Modulus. • It is a measure of the stiffness of a material.

71. Elastic Constant :-
An elastic modulus (also known as modulus of elasticity) is a quantity that measures an object or substance's resistance to being deformed elastically (i.e., non-permanently) when a stress is applied to it. The elastic modulus of an object is defined as the slope of its stress–strain curve in the elastic deformation region:
Specifying how stress and strain are to be measured, including directions, allows for many types of elastic moduli to be defined. The three primary ones are:
Young's modulus (E) describes tensile elasticity, or the tendency of an object to deform along an axis when opposing forces are applied along that axis; it is defined as the ratio of tensile stress tensile strain. It is often referred to simply as the elastic modulus.
The shear modulus or modulus of rigidity (G or μμ) describes an object's tendency to shear (the deformation of shape at constant volume) when acted upon by opposing forces; it is defined as shear stresss over shear strain. The shear modulus is part of the derivation of viscosity.
The bulk modulus (K) describes volumetric elasticity, or the tendency of an object to deform in all directions when uniformly loaded in all directions; it is defined as volumetric stress over volumetric strain, and is the inverse of compressibility. The bulk modulus is an extension of Young's modulus to three dimensions.

72. Stress Strain Curve

73. Stress-Strain Curve for Brittle Materials
Brittle materials such as concrete or carbon fiber do not have a yield point, and do not strain-harden. Therefore, the ultimate strength and breaking strength are the same. A typical stress–strain curve is A typical stress–strain curve for a brittle material will be linear.

74. Stress-Strain Curve for Ductile Materials
From the diagram one can see the different mark points on the curve. It is because, when a ductile material like mild steel is subjected to tensile test, then it passes various stages before fracture.

75. Nominal Stress:-
Nominal stress: the force on the object divided athe original area. True stress: the force on the object divided by the actual area. E.g., when necking occurs the true stress is the force applied divided by the area of the neck. The unit for stress is Nm-2, or Pa.

76. Yield Point :- The stress point beyond which a material becomes plastic.
Plastic stage :- A permanent deformation or change in shape of a solid body without fracture under the action of a sustained force

77. Ultimate stress &Breaking Stress
Ultimate stress :- is the maximum value of stress it is the point at which the size (cross section area) decreases and a neck is formed Breaking stress is the stress at which the material actually breaks down, the value of ultimate stress is always more than the breaking stress as formation of neck

78. Percentage Elongation

79. Proof Stress & Working Stress
Proof stress is the amount of stress that a material, usually metal or plastic, undergoes a level of deformation. Proof stress is important to monitor if the material being used in manufacturing is undergoing any deformation or what's commonly referred to as plastic deformation.
Working stress :-
Safe working stress is known as the maximum allowable stressthat a material or object will be subjected to when in service. This stress is always lower than the Yield stress and the Ultimate Tensile Stress (UTS)

80. Factor of safety
The ratio of a structure's absolute strength (structural capability) to actual applied load; this is a measure of the reliability of a particular design. This a calculated value, and is sometimes referred to, for the sake of clarity, as a realized factor of safety.

81. Poissons Ratio:-
The ratio of the proportional decrease in a lateral measurement to the proportional increase in length in a sample of material that is elastically stretched.

82. Thermal stress & Strain
If temperature deformation is permitted to occur freely, no load or stress will be induced in the structure. In some cases where temperature deformation is not permitted, an internal stress is created. The internal stress created is termed as Thermal stress. The corresponding strain due to temperature stress is known as Temperature Strain

83. Longitudinal and circumferential stress in samless thin walled circumferential
Hoop (Circumferential) Stress. The hoop stress is acting circumferential and perpendicular to the axis and the radius of the cylinder wall. ...
Longitudinal (Axial) Stress. For a cylinder closed closed in both ends the internal pressure creates a force along the axis of the cylinder. ...
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 bethin, otherwise thick  Magnitude of radial pressure is very small compared to other two stresses in case of thin cylinders and hence neglected.

84. Thin Cylindrical formula

85. Introduction to principle stresses
Principal Stresses. It is defined as the normal stress calculated at an angle when shear stress is considered as zero. The normal stress can be obtained for maximum and minimum values.

86. Chapter:-7
Torsion

87. Concept of torsion:-
The product of the tangential circumfrence of the shaft and radius of shaft is known is Torque.
Thus a shaft is said to be in pure torsion is it is subjected to equal and opposite end torques whose axes coincide with the axis of the shaft.Ex- Axles of automobiles & sterring rods.

88. The SI unit of Torque is Nm
THE DIFFERENCE B/W TORSION & TORQUE
Torque is a measureable concept, whereas torsion is a concept, which is mathematically projected by the shear stress or the twist angle.
Torque requires at least one force and torsion requires at least two forces to happen.
Torque depends only on the magnitude, directions and the separation of the forces applied, while torsion depends on the torque, the type of material and the shape of the object.

89. Assumption in the theory of of pure Torsion
The material of the shaft is homogeneous through out the length of shaft and obey Hooks law.
The twist is uniform throughout the length of the shaft
Maximum shear stress induced in the shaft due to application of torque doesnot exceed its elastic limit.
The shaft is of uniform cross section through out its length

90. Dimension of Torque:- M L2 T−2
Torsion Equation for solid shaft

91. Difference b/w Hollow & Solid Shaft
Hallow Shafts are stronger then solid shaft having the same height
The stifness of the hallow shaft is more than the solid shaft with the same weight
The Hollow shaft is costilier than a solid shaft
The strength of hallow shaft is more than the solid shaft with the same weight
Solid shaft when subjected to bending to bending are stronger than that of a hollow shaft