Prof.Kodali Srinivas

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...

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...

According to Theory of bending,w ithin the elastic limit, the bending stress (f) is directly proportional to the distance y from the neutral axis. The maximum value of bending stress ( f max ) will be the outermost fiber ie,y =  y max                                       f max = [M/I]. y max W e can write this equation as, f max  = [M/Z] Where Z = I/ y max and Z is called as  Elastic Section Modulus. The stress in the outermost section of beam is computed with the help of section modulus (Z). Section Modulus It is a ratio between second moment of area (area Moment of Inertia) and distance from Neutral axis to the extreme fiber.  Section modulus,  Z = I/ y max Where, I    = Moment of inertia about Neutral axis (N.A.)  y max  =The maximum distance of the fiber from N.A Note: 1.S ection modulus is also expressed as the r...

Bending stress or Flexural stress can be define  as the internal resistance per unit area developed by material of the beam against the external bending moment. According to Theory of bending, the Bending formula is given by, M/I = f/y = E/R Where M  = Bending Moment at a section  I    = Moment of inertia about Neutral axis (N.A.)  f    = Bending stress in the fiber at distance of y                 form the N.A y   = Distance of the fiber from N.A.  R  = Radius of Curvature  E  = Young's Modulus  Within the elastic limit, the bending stress (f) is directly proportional to the distance (y) from the neutral axis. The bending stress f = [M/I].y The maximum value of bending stress will be the outermost fiber ie,y =  y max                        f max = [M/I]. y max W e can write this...