Prof.Kodali Srinivas

Metallic engineering materials are classified as either ductile or brittle materials. Ductile materials  A ductile material is one having relatively large tensile strains up to the point of rupture. Ductile materials have a well-defined elastic region with linear stress- strain relationship, yield point, strain-harden and necking before the point of rupture in stress strain curve.  Example: structural steel, Copper, Brass and aluminum etc. Brittle materials A brittle material has a relatively small strain up to the point of rupture. A material is brittle if, when subjected to stress, it breaks with little elastic deformation and without significant plastic deformation. Brittle materials absorb relatively little energy prior to fracture, even those of high strength. Brittle materials do not have a well-defined yield point, and do not strain-harden. Therefore, the ultimate strength and breaking strength are the same. Example: Glass, Cast iron and Concrete etc. Malleability Mal...

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. 1.In case of a composite bar having two or more bars of different cross sections, the extensions or compression in each bar will be equal. The strain in the both materials will be same. Strain in bar 1 , ε 1  = σ 1 / E 1  = (P/AE) 1 Strain in bar 2 , ε 2  = σ 2 / E 2  = (P/AE) 2 Strain  ε 1  =  Strain   ε 2   = dl/L    ----(1) 2.The total load will be equal to the sum of the loads carried by each member.  P = P1 +P2 The Stress in bars are σ 1  and   σ 2 P= σ1 x A1 + σ2 x A2   -----(2) Solve the equation (1) and (2) for unknowns Assignment: 1.Find the stresses in the rods for the figure shown below. (Answer: 4.67, 1.39 N/mm2 ) 2. A 2m long bar of uniform section 50 mm2 extends 2mm under a limiting axial stress of 200N/mm2.W...

The Strain Energy in axially loaded bars can be expressed as U = 1/2 (Stress)x(Strain)x volume of the bar  U = 1/2 ( s ).( e ).V   = 1/2  ( s ). ( s /E).V   The stresses in Axial loaded bars due to different cases will be given as below 1)The stress induced in the bar due to gradually applied load  s  = P/A 2)The stress induced in the bar due to suddenly applied load  s  = 2P/A 3)The stress induced in the bar due to Impact load  s Example: 1.Calculate the strain energy stored in a bar 50mm diameter & 2.5 m long subjected to tensile load of 100 kN take E= 200 GPa if  a) load is gradually applied b) load is suddenly applied​ Solution: a) If load is gradually applied, the strain energy  U =  1/2 (Stress)x(Strain)x volume of the bar        = 1/2 ( s ).( e ).V   = 1/2  ( s ). ( s /E).V ---(1) The stress induced in the bar due to gradually applied load  s  = P/A  Load on the ...

ork is the product of the component of the force in the direction of the displacement and the magnitude of this displacement. Energy is defined as the capacity to do work. Strain energy Strain energy is the ability of a body to do work because of its deformation and its tendency to return to its original shape. This internal energy stored in a in the body due to its deformations. Strain energy (U) = Work done (W) Strain Energy is Always a positive quantity and is expressed in units of work (N-m,Joule). This may be calculated as the area under the stress-strain curve from the origin O to up to the elastic limit (the shaded area in the above figure).  Resilience The Strain Energy stored within elastic limit without creating a permanent distortion is called Resilience. The maximum energy stored at elastic limit is known as Proof Resilience .  Note : In Proof Resilience, stress is the value at the elastic limit or for non-ferrous materials,the stress is the Proof Stress. Proof Re...

The stresses induced in a body due to change in temperature are known as thermal stresses. The materials have a tendency to expand or contract when they exposed to temperature changes. If the length of the bar AB is L, the Coefficient of linear thermal expansion is  α  T = Rise or fall of temperature (t 2 - t 1 ) E = Young’s modulus The change in length dL due to temperature change is given by   ∆t =  α.T.L Thermal strain,  e  =  ∆t/L =  α . T  If the temperature deformations are not allowed freely, the stresses are induced in the body. These stresses are called thermal or temperature stresses. T hermal stress,    s tp =  e.E  =  α.T.E  Yielding of Supports Suppose if the support yield by an amount d, thermal stress   s tp  = ( D -  d ).E/L = (L a t -  d ).E/L Example:1 Q.1 A railway is laid so that there is no stress in rail at 10º C. If rails are 30 m long Calculate,  a. The stre...

Modulus of Elasticity (E)   It is the ratio between Normal stress to Normal strain within the elastic limit. Elastic Modulus E = Normal stress/Normal strain E =  s/e Modulus of Rigidity (G)  It is the ratio between Shear stress to Shear strain within the elastic limit. Rigidity Modulus G = Shear stress/ Shear strain G =    Ƭ / ø   Relation between Modulus of Elasticity and Modulus of Rigidity: Consider a solid cube  PQRS is  subjected to a shearing force F.  Let  Ƭ     be the shear stress produced in the faces PQ and RS due to this shear force. The complementary shear stress consequently produced in the vertical faces PS and RQ is also equal to same and shown in figure as Ƭ   Due to the pure shearing force, the cube is deformed PQRS to PQR'S' . The point   S moved to S' and point R moved to R' as shown in fig.  The shear strain = The angle of distortion  ø           ...