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PRINCIPAL STRESSES

Principal Plane is the Plane in which  no Shear Stresses are acting. The normal stresses which are acting on Principal Plane are called as Principal Stress.
Stresses on Oblique Plane 
The normal stresses (sx and sy) and the shear stress (txy) are acting vary smoothly on a body, the normal and tangential stresses or shear stress acting on a Oblique plane making a rotation of an angle q to the vertical face are given by,

Normal stress on inclined plane  
sq = (sx+sy) / 2 + (sx-sy)/2cos2q + txysin2q  -- (1)

Shear stress in the inclined plane 
tq =(sx-sy)/2sin2q -txycos2q - (2)

In above equations, there exist a couple of particular angles where the stresses take on special values.
First, there exists an angle qp where the shear stress txy becomes zero. That angle is found by setting txy to zero in the above shear transformation equation and solving for q (set equal to qp). The result is, the principal palenes are located at
tanqp = 2 xy / (x - y)  ---(3)
The principal stresses are 
--(4) The angl…
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PROBLEMS IN PRINCIPAL STRESSES

Problem no 1 For a given loading conditions the state of stress in the wall of a cylinder is subjected to normal stresses 85 MN/m2 tensile,in x- direction and  25 MN/m2 tensile in y- direction,along with  Shear stresses of 60 MN/m2 on the planes on which the stresses are act.  the sheer couple acting on planes carrying the 25 MN/m2 stress is clockwise. a)Calculate the principal stresses and the planes on which they act.  b)What would be the effect on these results if owing to a change of loading if the 85MN/m2 becomes compressive while other stresses are remain unchanged Solution:

a)The principle stresses are given by the formula


For finding out the planes on which the principle stresses act us the equation
Tan ø = 2 xy / x - y The solution of this equation will yeild two values qi.e they q1 and q2 giving, q1= 31071' and q2= 121071' (b) In this case only the loading (a) is changed i.e. its direction had been changed. While the other stresses remains unchanged hence now the block diag…

Mohr's Circle of stress

The transformation equations for plane stress can be represented in graphical form by a plot known as Mohr’s Circle. 

To establish Mohr's Circle, we first recall the stress  transformation formulas for plane stress at a given location,
sq = (sx+sy) / 2 + (sx-sy)/2cos2q +txysin2q - - (1) tq = (sx-sy)/2sin2q -txycos2q  -- (2)
Using a basic trigonometric relation (cos22q + sin22q = 1) to combine the two above equations 1 and 2 after squring the equations we have, (sq- save)2  + t2q= R2where
save = (sx + sy)/2  This is the equation of a circle, plotted on a graph where the abscissa is the normal stress and the ordinate is the shear stress.This circle is with radius of 'R' and centered at  C

Failure theories

When a component is subject to increasing loads it eventually fails.  It is comparatively  easy to determine the point of failure of a component subject to a single tensile force.  The strength data on the material identifies this strength.   However when the material is subject to a number of loads in different directions some of which are tensile and some of which are shear, then the determination of the point of failure is more complicated. Metals can be broadly separated into 'Ductile' metals and 'Brittle' metals. Examples of ductile metals include mild steel, copper etc . Cast iron is a typical brittle metal.Ductile metals under high stress levels initially deform plastically at a definite yield point or progressively yield.   In the latter case a artificial value of yielding past the elastic limit is selected in lieu of the yield point e.g 2%proof stress.  At failure a ductile metal will have experienced a significant degree of elongation. Brittle metals experienc…

PROBLEMS IN PRINCIPAL STRESSES -2

 Question Bank
Unit -1 Short Answer questions
1Define the term obliquity and explain about Normal and tangential stresses           on an inclined plane2Define the terms principal planes and principal stresses 3Normal stresses in X and Y axis and  shear stress  act at a point. Find the             principal stresses and the principal planes. 4Explain graphical method for locating principal axes. 5Define and explain maximum strain energy theory. 6Discuss briefly the maximum principal stress theory. 7Discuss in brief various prominent theories of failure
Long Answer questions 1. For a given loading conditions the state of stress in the wall of a cylinder is subjected to normal stresses 85 MN/m2 tensile,in x- direction and  25 MN/m2 tensile in y- direction,along with  Shear stresses of 60 MN/m2 on the planes on which the stresses are act a)Calculate the principal stresses and the planes on which they act.  b)What would be the effect on these results if owing to a change of loading if the 85MN/m2 be…