Mohr’s Circles GLE/CEE 330 Lecture Notes Soil Mechanics William J. Likos, Ph.D. Department of Civil and Environmental Engineering University of Wisconsin-Madison

Sign Convention normal stress (s) shear stress (t) (+s) (-s) (+s) (-s) (USC) (-t) (+t) (+t) Element must have both pos. and neg. shear stress for equilibrium (-t)

Principal Stresses & Principal Planes Principal stress: normal stress on plane where shear stress is zero s1 = maximum (major) principal stress s3 = minimum (minor) principal stress s2 = intermediate principal stress (3D) If s1 = s2 = s3, then “isotropic stress” (thus no shear stress) (s1 - s3 ) = “deviator stress” (results in shear stress) Principal Plane: plane on which principal stress acts Two principal planes are perpendicular s1 s1 t = 0 Not necessarily horizontal And vertical! s3 s3

Derivation of Mohr’s Circle Consider a 2D element where s1 ≠ s3 Sub-element: s1 t s3 s sa s3 ds dy ta a dx s1 Substituting...

sa s3 Divide by ds… ds dy ta a dx Solve simultaneously… s1 (Note: if s1 = s3, ta = 0) Rewrite (2)… Square (1) and (3) and add together…

y Is an equation for a circle in the form r x x0 t ½(s1–s3) s s3 s1 ½(s1+s3)

t Mohr’s Circle: graphical representation of state of stress on every plane in a 2D element ½(s1–s3) s s3 s1 ½(s1+s3) Typically only plot top half (+t) in geotechnical practice.... Typically only have stresses (+s) in compressive regime… t +t -s +t +s s3 s1 s -t -s -t +s

Mohr’s Circle for effective stress (s’ = s – u) Circle for total stress (s) u s’ or s Shifts to left by pore pressure (u) We will see later that this is a less stable state of stress (soil is closer to failure conditions)

Finding principal stresses if given stresses on perpendicular planes… sy txy = -tyx tyx sx > sy txy sx t tmax (sx, txy) sy s3 sx s1 s (sy, tyx) 1/2(sx+sy)

Example: Find principal stresses and max shear stress (Analytical Solution) sy = 3000 psf txy = -300 psf sx = 2100 psf

Graphical Solution: draw to scale with x-scale = y-scale plot known points and construct circle “pick-off” principal stresses and max shear sy = 3000 psf txy = -300 psf sx = 2100 psf

Finding stresses on any plane… Graphical solution (Reference Plane Method) 1) Establish angle a from a reference plane (e.g,. horizontal). 2) Locate reference plane in Mohr’s circle (center of circle to stress on ref plane) 3) Measure 2a from ref plane in same direction (CW or CCW) 4) This intersects circle at state of stress on angled plane sy tyx txy sa sx ta a t Analytical solution: Reference Plane (sx, txy) sy s3 sx s1 s 2a (sa, ta) *See also “Pole Point” or “Origin of Planes” Methods (sy, tyx)

Example Graphical solution s1 = 2.5 kPa (C) Reference Plane (sa, ta) = (1.5, 1 kPa) 2a = 90 CCW sa s3 = 0.5 kPa (C) ta a = 45 Analytical solution:

Find orientation of principal planes: sy = 3000 psf Let horizontal be reference plane Find reference plane on circle Measure angle to principal plane (2a = 33.7 CCW) So major principal plane is 17 deg. CCW from horizontal tyx = -300 psf txy = 300 psf sx = 2100 psf s3 = 2009 psf 17 deg. s1 = 3091 psf

Example: Find stresses on plane 30˚ as shown 40 psi 30˚ 20 psi Let major principal plane be reference plane Find reference plane on circle Measure angle (2a = 60˚ CCW) 40 psi 30˚ 20 psi 35 8.7

Example: Find magnitude and orientation of principal stresses 20 psi 40 psi 10 psi -10 psi 30˚ Plot known points and draw circle. Find s1 = 44 psi, s3 = 16 psi Let plane 30˚ from horizontal be reference plane, (s,t) = (20,10) Find reference plane on circle Measure angle to minor principal plane (2a = 45˚ CCW) So minor principal plane is 22.5˚ CCW from reference plane, or 52.5˚ CCW from horizontal. 44 psi 52.5˚ 16 psi

Graphical solution (Pole Method) Draw line(s) on Mohr circle through known stress at known orientation. sy tyx Intersection of that line(s) with circle is the “Pole” Pole txy sa sx ta Draw a line from Pole at any angle of interest. This line intersects the circle at the state of stress on the plane defined by that angle a (sa, ta) a t (sx, txy) s1 s (sy, tyx)

So what happened to soil mechanics? s’h t uw s’v Is there a plane within the soil mass where the induced shear stresses exceed the shear strength of the soil (c’, f’) ? Will a “failure plane” develop? How does shear strength depend on effective stress?

Mohr-Coulomb Failure Criterion c’ = effective cohesion intercept f’ = effective friction angle s’ = effective stress on failure plane This is a straight line in s’-t space t failure M-C Failure Envelope no failure s’1 t s’ s’3 f’ Mohr’s Circle c’ s’3 s’1 s’

Bearing Capacity Failure Plane t s’v t s’h Stable f’ c’ s’

Active Earth Pressure Failure Retaining Wall Soil t s’v t s’h f’ c’ Stable s’