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FLAMMANT’S PROBLEM
By
Karamcheti Anantha Venkateswara Prasad
203621011
 POINT LOAD
Point load have a finite force acting on a surface of zero area.
 LINE LOAD
Line load acts on an infinitely long line rather than a point.
 Consider a line load acting on a surface of a halfspace.
 Flammant used Boussinesq solution along with principle of
superposition to solve for stress field in halfspace with line load.
 Consider point A in the figure.
 Since the load is acting on the Y axis from infinity to infinity, we can
have origin at any point on Y axis.
 We will try to find out the σzz component of stress at point A due to
line load.
 Consider an elemental length dy of the y axis.
 For any point inside the halfspace, including A, the line load ρ
acting on the element dy looks like a point load.
 There will be increment of stress dσzz caused by this point load ρdy.
 dσzz is given by Boussinesq solution.
 Now to find stress σzz, we have to integrate both sides
As the line load varies from -∞ to ∞
 From the figure
b = (x2+z2)
1/2
y = (b tanØ)
dy = b sec2Ø
We can rewrite as
 We can find other components of stress similarly
 The line load is effectively a sequence of point loads side by side and we are using
superposition to derive σzz.
 This is possible only when we consider linear elastic theory.
 Flamants solution is one of the applications of Boussinesq solution.
 It is an example of plain strain problem.
 Plain strain problems have 1 spatial direction in which only rigid motions occur.
 As a result certain strain will be identically zero.
 Non zero strain functions are not functions of y.
 A particle, that initially has coordinate y0 in reference configuration will always have
coordinate y0 in any deformed plane; unless and otherwise a rigid translation in the
y direction occurs.
 Now let us consider a cylindrical surface of radius b alligned with line load.
 We could carry out an analysis to find tractions that act on this surface by using
above equations.
 After doing analysis we find traction vector
where n is the unit normal vector to the cylindrical surface
 What this means is cylindrical surface itself is a principal surface.
 The major principal stress
 The intermediate principal surface is defined by vector n = [0,1,0]
 The intermediate principal stress
 The minor principal surface is perpendicular to the cylindrical surface and to the
intermediate principal surface.
 Minor principal stress = 0
 When the vertical component of the traction is integrated along the cylindrical
surface, it equilibriates the applied load ρ.
 Another interesting characteristic of Flammant’s problem is the distribution of the
principal stress in space.
 Consider the locus of points on which major principal stress σ1 is constant.
 It is an equation of a circle with centre on z axis at a depth c beyond the origin and
radius as c.
 At each point on this circle σ1 is constant.
 It points directly at origin.
 If we consider larger value for c, circle will be larger as c ∞ (1/ σ1).
 This result gives the pressure bulb in the soil beneath the
foundation.
 It is helpfull in visualizing stress fields.
THE END

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flammants problem by line load and point load

  • 1. FLAMMANT’S PROBLEM By Karamcheti Anantha Venkateswara Prasad 203621011
  • 2.  POINT LOAD Point load have a finite force acting on a surface of zero area.  LINE LOAD Line load acts on an infinitely long line rather than a point.
  • 3.  Consider a line load acting on a surface of a halfspace.
  • 4.  Flammant used Boussinesq solution along with principle of superposition to solve for stress field in halfspace with line load.  Consider point A in the figure.  Since the load is acting on the Y axis from infinity to infinity, we can have origin at any point on Y axis.  We will try to find out the σzz component of stress at point A due to line load.
  • 5.  Consider an elemental length dy of the y axis.
  • 6.  For any point inside the halfspace, including A, the line load ρ acting on the element dy looks like a point load.  There will be increment of stress dσzz caused by this point load ρdy.  dσzz is given by Boussinesq solution.
  • 7.  Now to find stress σzz, we have to integrate both sides As the line load varies from -∞ to ∞
  • 8.  From the figure b = (x2+z2) 1/2 y = (b tanØ) dy = b sec2Ø We can rewrite as
  • 9.  We can find other components of stress similarly
  • 10.  The line load is effectively a sequence of point loads side by side and we are using superposition to derive σzz.  This is possible only when we consider linear elastic theory.  Flamants solution is one of the applications of Boussinesq solution.  It is an example of plain strain problem.  Plain strain problems have 1 spatial direction in which only rigid motions occur.  As a result certain strain will be identically zero.  Non zero strain functions are not functions of y.  A particle, that initially has coordinate y0 in reference configuration will always have coordinate y0 in any deformed plane; unless and otherwise a rigid translation in the y direction occurs.
  • 11.  Now let us consider a cylindrical surface of radius b alligned with line load.  We could carry out an analysis to find tractions that act on this surface by using above equations.  After doing analysis we find traction vector where n is the unit normal vector to the cylindrical surface
  • 12.  What this means is cylindrical surface itself is a principal surface.  The major principal stress  The intermediate principal surface is defined by vector n = [0,1,0]  The intermediate principal stress  The minor principal surface is perpendicular to the cylindrical surface and to the intermediate principal surface.  Minor principal stress = 0
  • 13.  When the vertical component of the traction is integrated along the cylindrical surface, it equilibriates the applied load ρ.  Another interesting characteristic of Flammant’s problem is the distribution of the principal stress in space.  Consider the locus of points on which major principal stress σ1 is constant.  It is an equation of a circle with centre on z axis at a depth c beyond the origin and radius as c.
  • 14.
  • 15.  At each point on this circle σ1 is constant.  It points directly at origin.  If we consider larger value for c, circle will be larger as c ∞ (1/ σ1).  This result gives the pressure bulb in the soil beneath the foundation.  It is helpfull in visualizing stress fields.