Tuesday, 12 September 2017

Deflection of Beams

Strength of Materials -1, Unit -5
The deformation of a beam is usually expressed in terms of its deflection from its original unloaded position. The deflection is measured from the original neutral surface of the beam to the neutral surface of the deformed beam. The configuration assumed by the deformed neutral surface is known as the elastic curve of the beam.

The angle through which the cross-section rotates with respect to the original position is called the angular rotation of the section.

Methods of Determining Beam Deflections:

1. Double integration method

The double integration method is a powerful tool in solving deflection and slope of a beam at any point because we will be able to get the equation of the elastic curve.

In Cartesian co-ordinates, the radius of curvature(R) of a curve y = f(x) is given by 

  In the derivation of flexure formula, the radius of curvature of a beam is given as

 1/R = M /EI

ThusThe deferential equitation of elastic curve is EI dy2 / dx2  = MX
The product EI is called the flexural rigidity of the beam, If EI is constant,the General equation may be written as:

 EI dy2 / dx2 =  MX     ---  (1)

Note :The down ward deflection will consider here as -ve ,
where x and y are the coordinates shown in the figure of the elastic curve of the beam under load, y is the deflection of the beam at any distance x. E is the modulus of elasticity of the beam, I represent the moment of inertia about the neutral axis, and M represents the bending moment at a distance x from the end of the beam. 
Direct Integration Method or Double Integration Method
The first integration of equation of (1) yields the slope of the elastic curve and the second integration equation (1) gives the deflection of the beam at any distance x.
The resulting solution must contain two constants of integration since EI y" = M is of second order.
These two Integral constants must be evaluated from known conditions concerning the slope deflection at certain points of the beam.
1.A simply supported beam with rigid supports, 
at x = 0 and x = L, the deflection y = 0, and in locating the point of maximum deflection, we simply set the slope of the elastic curve y' to zero.
2.A Fixed support,
at fixed end, the deflection y is zero and slope dy/dx ( ø)is zero.
Macaulay's Method For Beam Deflections
For complex loading, specially where the span is partially loaded or loaded with number of concentrated loads,this Macaulay's method is more useful.
Note : 
1. While using the Macaulay's method, the section 'x' is to be taken in the last portion of the beam.
2. the quantity with in brackets ( ), should be integrated as a whole.
3.The expression for Bending Moment (Mx) equation can be used at any point, provided the term within the brackets becomes negative is omitted. 

Friday, 4 August 2017


The stresses produced due to constant Bending Moment (with zero Shear Force or pure bending) are know as Bending stresses.

Assumptions in Theory of Bending :
In deriving the relations between the bending moments and flexure (bending)stresses and between the Shear forces and Sharing stresses the following assumptions are made.
1.Transverse sections of the beam that were plane before bending remain so even after bending.
2.The material of the beam is isotropic and homogeneous and follows Hooke's law and has the same value of Young's Modulus in tension and compression.
3.The beam is subjected to Pure bending and therefore bends in an arc of a circle.
4.The radius of curvature is large compared to the dimension of the cross-section.
5.Each layer is independent to enlarge or contract.
6.The stresses are purely longitudinal and local effects of point loads are neglected.

Flexure Formula :

Stresses caused by the bending moment are known as flexural or bending stresses. Consider a beam to be loaded as shown.


where R is the radius of curvature of the beam in mm, M is the bending moment in N·mm, fb is the flexural stress in MPa., I is the centroidal moment of inertia in mm4, and y is the distance from the neutral axis to the  fiber in mm.


M/I = f/y = E/R 
M = Bending Moment
I = Moment of inertia about Neutral axis (N.A.)
f = Bending stress
y = Distance of the fiber from N.A.
R = Radius of Curvature
E = Young's Modulus
This equation may be remember as 

" May I flow you Every Rule "

M/I = f/y = E/R

Sunday, 30 July 2017


Definition of a Beam

A beam is a bar subject to forces or couples that lie in a plane containing the longitudinal section of the bar. 
According to determinacy, a beam may be determinate or indeterminate.

Statically Determinate Beams

Statically determinate beams are those beams in which the reactions of the supports may be determined by the use of the three static equilibrium equations

Statically Indeterminate Beams
If the number of reactions exerted upon a beam exceeds the number of equations in static equilibrium, the beam is said to be statically indeterminate. In order to solve the reactions of the beam, the static equations must be supplemented by equations based upon the elastic deformations of the beam.
The degree of indeterminacy is taken as the difference between the umber of reactions to the number of equations in static equilibrium that can be applied. 

Types of Supports

The supports of the beam may consists of 
a) Simply support or Roller Support
b) Hinge Support or Pinned Support
c) Fixed Support or Built in support

Types of Loading

Loads applied to the beam may consist of 
a) concentrated load (load applied at a point), 
b)uniform distributed load (u.d.l.), 
c) uniformly varying load, 
d) an applied couple or moment. 

Shear Force
The internal vertical resistance is called Shear Force.
 The Shear Force at a section may be defined as "the algebraic sum of the total vertical forces on either side of the cross section".
If a left portion of a section is considered ,upward forces will be positive and right side portion is considered down ward forces will be positive.

Bending moment
The moment witch bends the beam is called Bending moment.
The Bending moment at a section through a structural element may be defined as "the algebraic sum of the moments about that section of all external forces acting to one side of that section". 
The forces and moments on either side of the section must be equal in order to counteract each other and maintain a state of equilibrium so the same bending moment will result from summing the moments, regardless of which side of the section is selected.

If clockwise bending moments are taken as negative, then a negative bending moment within an element will cause "sagging", and a positive moment will cause "hogging"
It is more common to use the convention that a clockwise bending moment to the left of the point under consideration is taken as positive. 


1.Shear force at a section ,such that the portion to the right of the section slides up wards with respect to the left of the section are +ve and vice versa.

2.Bending moment at a section is +ve if it is sagging and -ve if it is hogging.

Shear Force and Bending moment diagrams 
Critical values within the beam are most commonly annotated using as a Shear force and bending moment diagrams, where negative values are plotted to scale below a horizontal line and positive values are plotted above the line.
A Shear Force diagram is one which shows the variation of the shear force along the length of the beam and a bending moment diagram is one which shows the variation of the bending moment along the length of the beam.
Bending moment  varies linearly over unloaded sections, and parabolically over uniformly loaded sections.
Point of contraflexure
A point of zero bending moment within a beam is called as point of contraflexure—that is the point of transition from hogging to sagging or vice versa.
1.Bending Moment is maximum at the point where the Shear Force is zero or where it changes direction from +ve to -ve or vice versa.
2.The point of contra-flexure or point of inflexion is the point where the Bending moment changes its direction from +ve to -ve or vice versa. Where the magnitude of B.M. is zero.
3. Relationship between intensity of load 'w' ,Shear Force 'F',and Bending Moment 'M'  :
    a) Rate of change of Shear Force is equal to Intensity of loading.
                          dF/dx = w
    b) Rate of change of Bending Moment is equal to Shear force.
                          dM/dx = F


 Elasticity is the property by virtue of which certain materials return back to their original position after the removal of the external force.
     Normal Stress or Direct Stress

The internal resistance per unit area, offered by a body against deformation is 
       known as   Normal stress.  
      The stress is given by   =R/A  =P/A.                                                               
      where P = External force or load;
                  A = Cross-sectional area.
   Stress is expressed as kgf/m2, kgf/cm2, N/m2 and N/mm2, or MPa

  1. The stress induced in a body, which is subjected to two equal and opposite pulls,      is known as  tensile stress
  2. The stress induced in a body, which is subjected to two equal and opposite pushes, is known as compressive stress.

Normal Strain
The deformation for unit length is called Strain.
   The ratio of change of dimension of the body to the original dimension is
       known as strain.
ROBERT HOOKE (1655-1705) 
Hooke’s law  : It states that within elastic limit,  the stress is proportional to the corresponding strain.

THOMAS YOUNG (1773-1829) 
 Young’s modulus: 
   The ratio of tensile stress (or compressive stress) to the corresponding strain is known as Young’s modulus or modulus of elasticity and is denoted by E.
The Normal Stress required to produce one unit normal strain with in elastic limit is also called Young's Modulus.

Rigidity Modulus:
The ratio of Shear Stress to the corresponding Shear Shear Strain with in the elastic limit is known as Modulus of Rigidity or Shear Modulus.
     It is denoted by 'C' or 'G'

11. The Total Change in length of a bar, when it subjected to an axial load of 'P' is
      dL = PL / AE
Factor of Safety.
12.The ratio between Ultimate stress to Working stress is known as Factor of Safety.

13. A composite bar is made up of two or more bars of equal lengths but of different materials rigidly fixed with each other and behaving as one unit for extension or compression.
14.The stresses induced in a body due to change in temperature are known as thermal stresses.
15. Thermal strain and thermal stress is given by
      thermal strain, e = α . T and
      thermal stress, p = α.T.E
      where       α = Co-efficient of linear expansion
                        T = Rise or fall of temperature
                        E = Young’s modulus

16. In case of a composite bar having two or more bars of different lengths, the extensions or compression in each bar will be equal. And the total load will be equal to the sum of the loads carried by each member.
17. In case of nut and bolt used on a tube with washers, the tensile load on the bolt is equal to the compressive load on the tube.
18. Elongation of a bar due to its own weight is given by
                   dL = W L / 2E
      where       ω = Weight per unit volume of the bar material (W/L)
                         L = Length of bar.

Example :

For you to feel the situation, position yourself in pull-up exercise with your hands on the bar and your body hang freely above the ground. 
Notice that your arms suffer all your weight and your lower body fells no stress (center of weight is approximately just below the chest). 
If your body is the bar, the elongation will occur at the upper half of it.




01. A 2m long bar of uniform section 50 mm2 extends 2mm under a limiting axial stress of 200N/mm2.What is the modulus of resilience for the bar.. 

02. Derive the expression for temperature stresses of a bar.

03. A material has modulus of rigidity equal to 0.4 x 105 N/mm2 and bulk modulus equal to 0.75 x 105 N/mm2. Find the Young’s Modulus and Poisson’s Ratio.

04. A steel rod 30 mm diameter and 300mm long is subjected to tensile force P acting axially.The temperature of the rod is then raised through 600C and total extension measured as 0.30mm. Calculate the value of tensile force P. Take ES for of steel =200 GN/m2 and thermal coefficient of steel =12 x 10-6 /oC

05. A hollow cast-iron cylinder 4 m long, 300 mm outer diameter, and thickness of metal 50 mm is subjected to a central load on the top when standing straight. The stress produced is 75000 kN/m2. Assume Young's modulus of cast iron as 1.5 x 108 KN/m2, find

i) Magnitude of the load,ii) Longitudinal strain produced and iii) Total decrease in length

06. Define Resilience and derive the equation of stresses for a body subjected to sudden and Impact loading

07. A circul rod 0.2m long, tapers form 20mm diameter at one end to 10mm at other end. On applying an axial pull of 6KN it was found to extend by 0.068mm. Determine the Young’s Modulus of the rod material.

08. A vertical circular bar20mm diameter, 3m long carries a tensile load of 150kN. Calculate a) Elongation b) Decrease in diameter and c) Volumetric strain.

09.As shown in Fig., a rigid bar with negligible mass is pinned at O and attached to two vertical rods. Assuming that the rods were initially stress-free, what maximum load P can be applied without exceeding stresses of 150 MPa in the steel rod and 70 MPa in the bronze rod.

10.Shown in Fig.is a section through a balcony. The total uniform load of 600 kN is supported by three rods of the same area and material. Compute the load in each rod. Assume the floor to be rigid, but note that it does not necessarily remain horizontal.


stress-strain diagram

Suppose that a metal specimen be placed in U.T.M (universal testing machine) As the axial load is gradually increased in increments, the total elongation over the gauge length is measured at each increment of the load and this is continued until failure of the specimen takes place. Knowing the original cross-sectional area and length of the specimen, the normal stress and the strain can be obtained.
Stress-strain diagram of a medium-carbon structural steel
The graph of these quantities with the stress along the y-axis and the strain along the x-axis is called the stress-strain diagram. The stress-strain diagram differs in form for various materials. The diagram shown below is that for a medium-carbon structural steel. Metallic engineering materials are classified as either ductile or brittle materials. A ductile material is one having relatively large tensile strains up to the point of rupture like structural steel and aluminum, whereas brittle materials has a relatively small strain up to the point of rupture like cast iron and concrete. An arbitrary strain of 0.05 mm/mm is frequently taken as the dividing line between these two classes. 
Proportional Limit (Hooke's Law)
From the origin O to the point called proportional limit, the stress-strain curve is a straight line. This linear relation between elongation and the axial force causing was first noticed by Sir Robert Hooke in 1678 and is called Hooke’s Law that within the proportional limit, the stress is directly proportional to strain or

The constant of proportionality 'E' is called the Modulus of Elasticity or Young’s Modulus and is equal to the slope of the stress-strain diagram from O to P. Then
Elastic Limit
The elastic limit is the limit beyond which the material will no longer go back to its original shape when the load is removed, or it is the maximum stress that may e developed such that there is no permanent or residual deformation when the load is entirely removed.

Elastic and Plastic Ranges

The region in stress-strain diagram from O to P is called the elastic range. The region from P to R is called the plastic range.

Yield Point

Yield point is the point at which the material will have an appreciable elongation or yielding without any increase in load.

Ultimate Strength

The maximum ordinate in the stress-strain diagram is the ultimate strength or tensile strength.

Rapture Strength

Rapture strength is the strength of the material at rupture. This is also known as the breaking strength.

Modulus of Resilience

Modulus of resilience is the work done on a unit volume of material as the force is gradually increased from O to P, in N⋅m/m3. This may be calculated as the area under the stress-strain curve from the origin O to up to the elastic limit E (the shaded area in the figure). The resilience of the material is its ability to absorb energy without creating a permanent distortion.

Modulus of Toughness

Modulus of toughness is the work done on a unit volume of material as the force is gradually increased from O to R, in N⋅m/m3. This may be calculated as the area under the entire stress-strain curve (from O to R). The toughness of a material is its ability to absorb energy without causing it to break.

Working Stress, Allowable Stress, and Factor of Safety

Working stress is defined as the actual stress of a material under a given loading. The maximum safe stress that a material can carry is termed as the allowable stress. The allowable stress should be limited to values not exceeding the proportional limit. However, since proportional limit is difficult to determine accurately, the allowable tress is taken as either the yield point or ultimate strength divided by a factor of safety. The ratio of this strength (ultimate or yield strength) to allowable strength is called the factor of safety.

Elastic Constants - S.M.

Poisson’s ratio 
1. It is the ratio of lateral strain to longitudinal strain.  It is generally denoted by  'μ '.

2.   The tensile longitudinal stress produces compressive lateral strains.

3.   If a load acts in the direction of length of rectangular bar, 
      then longitudinal strain = δl / l
      and Lateral strain =  δb/ b, δd /d

      where δl = Change in length
                  δb = Change in width
                  δd = Change in depth

4.   The ratio of change in volume to original volume is known as volumetric strain.

5.   Volumetric strain (v') for a rectangular bar subjected to an axial load P, is given by .

                   v' = δl / l(1-2μ)
complementary shear stresses

6.  Principle of complementary shear stresses states that a set of shear stresses across a plane is always accompanied by a set of balancing shear stresses (i.e., of the same intensity) across the plane and normal to it.

7. Volumetric strain of a cylindrical rod, subjected to an axial tensile load is given by                  longitudinal strain – 2 x strain of diameter 
Bulk modulus :

8. Bulk modulus ( K ) is the ratio of direct stress to the corresponding volumetric strain.

9. The relation between Young’s modulus and bulk modulus is given by,
                       E = 3K(1-2μ) 

10. When an element is subjected to simple shear stresses then:
i)    The planes of maximum normal stresses are perpendicular to each other.
ii)   The planes of maximum normal stresses are inclined at an angle of 450 to the plane of pure shear.
iii)  One of the maximum normal stress is tensile while the other maximum normal stress is compressive.
iv)  The maximum normal stresses are of the same magnitude and are equal to the shear stress on the plane of pure shear.
11. Modulus of Rigidity is the ratio between Shear stress to Shear strain with in elastic 

12.The relation between modulus of elasticity (E) and modulus of rigidity(C)  is given by
          E =  2C (1 + μ )   or   C = E/2(1 + μ )
13. The relation between modulus of elasticity (E) and Bulk modulus (K)   is given by
          E = 3K (1 - 2 μ )
14.The relation between modulus of elasticity (E), modulus of rigidity(C) and Bulk modulus (K)   is given by
          E = 9KC/ (3K+C)

Deflection of Beams

Strength of Materials -1, Unit -5 The deformation of a beam is usually expressed in terms of its deflection from its original unloaded ...