Prof.Kodali Srinivas

Kani's method was introduced by Gasper Kani in 1940’s. It involves distributing the unknown fixed end moments of structural members to adjacent joints, in order to satisfy the conditions of continuity of slopes and displacements. K ani's method is also known as Rotation contribution method. Advantages of Kani's Method : 1. Hardy Cross method distributed only the unbalanced moments at joints, whereas Kani’s method distributes the total joint moment at any stage of iteration. 2. The more significant feature of Kani’s method is that the process is self corrective. Any error at any stage of iteration is corrected in subsequent steps. Framed structures are rarely symmetric and subjected to side sway, hence Kani’s method is best and much simpler than other methods like moment distribution method and slope displacement method. Procedure  1. Rotation stiffness at each end of all members of a structure is determined depending upon the end conditions. The stiffness factors   a) If Bo...

Short Answer Questions: 1.What do you mean by direct stress and bending stress?  2 Draw stress distribution across the section due to Bi axial bending stress and direct stress.  3 Find core diameter of a solid circular section, if diameter is ‘d’  4 Explain about the term kernel and determine the size of kernel for a rectangular section 200 mm x 300mm  5 Explain the conditions for stability of dam.  6 Find core diameter of a hollow section, if external and internal diameter are ‘D’ and ‘d’. Assignment : 1 A hollow rectangular column of external depth of 1 m and external width 1 m is 10 cm thick.. Calculate the maximum and minimum stress in the section of the column, if vertical load of 200 kN is acting with an eccentricity of 20 cm. 2 A short column of external diameter 40 cm and internal diameter 20 cm carries an eccentric load of 80 kN. Find the greatest eccentricity which the load can have without producing tension on the cr...

 Assignment : 1. the stresses and deflection for the mid section of the I beam by unsymmetrical method Also identify the position of the neutral axis  2. A 240 mm × 120 mm steel beam of I-section is simply supported over a span of 6m and carries two equal concentrated loads at points 2 m from each end. The properties of the section are Ixx = 6012.32 × 104mm4, Iyy = 452.48 × 104 mm4. a) Determine the magnitude of the loads when the plane of the loads is vertical through YY. The permissible stress is 150 N/mm2 in compression and tension. b) Determine the degree of inclination of the plane of these loads to the vertical principal plane YY that will result in 20 percent greater bending stress than permitted under (A)  3. A T-Section of dimensions 150 wide x 200 mm deep, with 10 mm thickness of flange and web, is used as simply supported a beam on a span of 6 m. Find the maximum value of ‘w’ in kN/m, the permissible stress in the material is 120 MPa. The plane of loading is in...

Show that the shearing stress developed at the neutral axis of a beam with circular cross section is  τ max = (4/3)(F/πr2). Assume that the shearing stress is uniformly distributed across the neutral axis.  Solution : Let us consider the circular section of a beam as displayed in following figure. We have assumed one layer EF at a distance y1 from the neutral axis of the circular section of the beam Shear stress at a section will be given by following formula as mentioned here Where, F = Shear force (N) τ = Shear stress (N/mm2) A = Area of section, where shear stress is to be determined (mm2) ȳ = Distance of C.G of the area, where shear stress is to be determined, from neutral axis of the beam section (m) Q = A. ȳ = Moment of the whole shaded area about the neutral axis I = Moment of inertia of the given section about the neutral axis (mm4) For circular cross-section, Moment of inertia, I = ПR4/4 b = Width of the given section where shear stress is to be determined. Let us co...

1. Shearing Stress Distribution in Rectangular cross Section  Let us consider the rectangular section ABCD of a beam as shown in figure.  We have assumed one layer EF at a distance y from the neutral axis of the beam section. The shear force acting on section is = F b= Width of the rectangular section d= Depth of the rectangular section N.A: Neutral axis of the beam section EF: Layer of the beam at a distance y from the neutral axis of the beam section A- Area of section CDEF, where shear stress is to be determined  A = b (d/2 - y) ȳ = Distance of C.G of the area CDEF from neutral axis of the beam section ȳ = y + (d/2 - y)/2 = (d/2 +y)/2 I = Moment of inertia of the given section about the neutral axis (mm4)  I = bd3/12 Shear stress at a section will be given by following formula  substitute the values in the above shear stress equation, the shear stress in the layer EF is, when y= d/2 , the shear stress is zero y = 0 , the shear stress  at neutral axis is...

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...

Shear stress distribution for Tee section. T - section is not symmetrical over neutral axis, shear stress distribution also will not be symmetrical. The method of finding shear stress distribution in t section is explained in the following  example.  Example 1. A T-section contains a flange of size 150 mm × 50 mm and web of size 50 mm × 150 mm. This section is subjected to a vertical shear force of 100 kN. Determine the maximum shear stress and draw the shear stress distribution diagram along with values at vital points. Centroid of section is 125 mm from the bottom face and moment of inertia around the centroidal axis is I = 5312.5 × 104 mm4. Solution The Shear stress at a layer is  τ y  = V x ay'/ Ib =  V x Q/ Ib Where  a = Area between the extreme face of beam and the plane at which the shear stress is  τ y   y' = Distance of the centroid of area ‘a’ from N.A Q = a.y' = the first moment of an area of the shaded section about N.A. V x ...

When a beam is loaded, bending moment and shear force are developed at all section of the beam. In Unit 2, we have already learnt the methods of determining shear force and bending moment in a beam section under the given loading  conditions.  Previously, we have seen the bending stress distribution at any cross section of the beam. Now, we will study the shear stress distribution at any section of the beam. Shear Stress Distribution in Beams The vertical shear force at any section of a beam produces shear stress at that section that varies along the depth of the section. This vertical shear stress is accompanied by a horizontal shear stress of equal magnitude, known as complementary shear stress. So at any point in a cross section of the beam, there is a  vertical shear stress and a horizontal shear stress of equal magnitude of  τ .  These two shear stresses cause the diagonal tension and compression inclined at 45 degrees to the horizontal. Let us conside...

When a beam is subjected to external loads, shear forces and bending moments develop in the beam. Therefore, a beam must resist these external shear forces and bending moments. The beam itself must develop internal resistance to resist shear forces and bending moments. The stresses caused by the bending moments are called bending stresses. For beam design purposes, it is very important to calculate the shear stresses and bending stresses at various locations of a beam.  The bending stress varies from zero at the neutral axis to a maximum at the tensile and compressive side of the beam. Procedures for determining bending stresses   Stress at a Given Point:  1.Use the method of sections to determine the bending moment M at the cross section containing the given point.  2. Determine the location of the neutral axis. 3. Compute the moment of inertia I of the cross- sectional area about the neutral axis. 4. Determine the y-coordinate of the given point. Note that y is pos...