Prof.Kodali Srinivas

A tank or pipe carrying a fluid or gas under a pressure is subjected to tensile forces, which resist bursting, developed across longitudinal and transverse sections. If the thickness of the wall of the cylindrical vessel is less than 1/15 to 1/20 of its internal diameter,the cylindrical vessel is known as Thin Cylinder.  Tangential Stress (Circumferential Stress) or Hoop stress Consider the tank shown being subjected to an internal pressure p. The length of the tank is L and the wall thickness is t . Considering the right half of the tank:  The forces acting on this right half of the vessel are the total pressures F caused by the internal pressure p, and the total tension T in the walls. The projected area subjected to internal pressure = A = DL F = pxA = p.DL The tangential stress in the walls =  s t T =    s t  A wall =  s t .  tL But F = 2T  p.DL  = 2( s t .  tL)  The tangential stress  s t = pD/2t If there exist an external pressure po and an internal pressure pi, the the Tangent

In case of unsymmetrical sections, neutral axis will not pass through the geometrical centre of the section. In unsymmetrical beam sections the distance of  outermost layers i.e. for topmost layer and bottom layer of the section from neutral axis will not be same. In order to calculate the bending stress for unsymmetrical sections, we must have to find the value of centre of gravity (centroid) of the given unsymmetrical section. In order to calculate the maximum bending stress for unsymmetrical sections, we will use the bigger value of y. In this section, the following notation will be use:  fbt = flexure stress of fiber in tension fbc = flexure stress of fiber in compression N.A. = neutral axis I = Moment of Inertia about N.A yt = distance of fiber in tension from N.A. yc = distance of fiber in compression from N.A. M = resisting moment Mc = resisting moment in compression Mt = resisting moment in tension  Max.flexure stress in tension  f bt = (M/I).y t Max.flexure stress in compress