Prof.Kodali Srinivas

  Poisson’s ratio   It is the ratio of lateral strain to longitudinal strain and it is constant with in the elastic limit of the material.  Poisson's Ratio is generally denoted by 'μ ' or  u The ratio is named after the French mathematician and physicist Siméon Poisson . Note: The tensile longitudinal stress produces compressive lateral strains or the compressive longitudinal stress produces tensile lateral strains. (+ve Longitudinal strains produce -ve lateral strains or vice versa)     Let us consider a rectangular bar of cross section (bxd) and length l and it is subjected to a axial load acts in the direction of length of the bar. The longitudinal tensile strain in the bar  e  longitudinal   = δl / l Lateral compressive strains in the directions of breadth b and  depth d of bar e lateral  = δb/ b and  e lateral    =   δd /d where δl = Change in length δb = Change in width =b' - b δd = Change in depth = d' - d The poisson's Ratio                             

1. Elongation of Uniformly tapering Circular rod Let us consider the uniformly tapering circular rod as shown in figure, length of the uniformly tapering circular rod of length L and larger diameter of the rod is D1 at one end and the diameter of other end is D2. The bar is  subjected with an axial tensile load P Let us consider a section at distance of x form lager end and take one infinitesimal smaller element of length dx.  The diameter of infinitesimal smaller element is Dx Dx = D1-[(D1-D2)/L] X Dx = D1- KX Where we have assumed that K= (D1-D2)/L The area of cross section of circular bar at a distance x from its larger diameter end is Ax  Ax = (П/4) (Dx)2 Ax = (П/4) (D1- KX) 2 Stress The stress induced in circular bar at a distance x from its larger diameter end is σx and we will be given by σx = P/ Ax σx = P/ [(П/4) (D1- KX) 2] σx = 4P/ [П (D1- KX) 2] Strain The strain induced in circular bar at a distance x from its larger diameter end is Ԑx and we will be given by  Strain = Stre

Determination of stress, strain and deformations in the bars subjected to axial loads. Principle of Superposition The resultant deformation of the body is equal to the algebraic sum of the deformation of the individual section. Such principle is called as principle of superposition In the analysis of axial loaded bars we use the  principle of superposition. By  draw the free body diagrams of each section of the bar for finding out the required stresses and strains or deformations. we can understand the concept with the following example. Example: Q.1. Determine the stress in each section of the bar shown in the figure.The PQ and RS segments are having a length of 1 meter and QR segment is 1.5 m.  The cross section of bar is 10 mm square. What will be the total extension of the bar? The Young's modulus of bar material E = 210000 MPa. Solution : Check  the equilibrium of the bar, the sum of total forces in X- axis is  ∑ Fx = 0                           - 63 + 35 + 49 - 21 = 0 Write t

The behavior of any material under loading can be analyzed using stress strain curve of the material.  To draw the stress - strain diagram, the testing metal specimen be placed in U.T.M (universal testing machine) and applied an axial load. 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. 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. Engineering stress-strain curve The curve based on the original cross-section and gauge length is called the engineering stress-strain curve, while the curve based on the instantaneous cross-section area and length is called the true stress-strain curve . Unless stated otherwise, engineering stress-

Types of Stress Generally the internal resisting force per unit area is called stress.  If this stress is normal (perpendicular) to the resisting cross section is called Normal stress or Axial stress and corresponding Strain is called Normal Strain . If the stress is tangential (parallel) to the resisting cross section is called Tangential Stress or Shear stress and corresponding strain is called Shear strain. Shear Stress The internal resistance of material per unit area in the tangential direction of the cross section is called  Tangential Stress or Shear stress. Generally the internal tangential resistance is called Shear force. So the resistance of  shear force  per unit area is called Shear stress. The tangential force or Shear force acting on surface  = S The cross sectional area of face  = A Therefore the Shear stress       Ƭ = Shear force /Area of cross   section = S/A The unit of Shear stress is expressed as 1 N/m 2   , kN /m 2  ,    MN/m 2  Note: Unlike normal stress, the d

Elasticity It is the property by virtue of which certain materials return back to their original position after the removal of the external force. Plasticity It is the property of the solid material by virtue of which it tends to retain its deformed shape even after the removal of external load.     Normal Stress or Direct Stress The internal resistance per unit area, offered by a body against deformation is known as Normal stress or direct stress or axial stress. where  P = External force or load applied normal to the cross section.  R = Internal resistance developed at the cross section as shown in fig, A = Cross-sectional area of bar But to satisfy the equilibrium of the bar,  ∑ Fx = 0 R - P = 0 R = P ( The internal resistance R is always equal to normal force P) The internal resistance per unit area is called normal stress, and is given by s     =R/A =P/A.  The unit of Stress is expressed as 1 N/m 2 , kN /m 2  ,  MN/m 2  ,  MPa = Megapascal  =  10 6   N/m 2  = 1   N/mm 2 Normal