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Stepwise procedure of Mohr's circle

Short answer

Draw σ-τ axes; plot X(σx, τxy) and Y(σy, −τxy) taking clockwise shear as positive. Centre C = ((σx+σy)/2, 0); radius R = √[((σx−σy)/2)² + τxy²]. Draw the circle. For an inclined plane, rotate radius CX by 2θ to read σn and τθ. Principal stresses: OC±R; τmax = R.

Write down the stepwise procedure (with sketch) for the determination of stresses in inclined plane for General stresses conditions (σx\sigma_x, σy\sigma_y and τxy\tau_{xy} are given) using the Mohr's circle method.

Mohr's circle is a graphical method that represents the state of plane stress at a point; every point on the circle gives the normal and shear stress on one plane.

Given: σx\sigma_x, σy\sigma_y and τxy\tau_{xy} on an element, and a plane whose normal is at angle θ\theta to the xx-axis.

Procedure

  1. Draw the σ\sigma-axis horizontally (tension positive, to the right of O) and the τ\tau-axis vertically through O. Select a suitable scale.
  2. Sign rule for plotting shear: a shear stress that tends to turn the element clockwise is plotted above the σ\sigma-axis, and one that turns it anticlockwise is plotted below. With this rule, a rotation of 2θ2\theta on the circle is in the same sense as the rotation θ\theta of the plane. Plot point X(σx,τxy)X(\sigma_x, \tau_{xy}) for the face on which σx\sigma_x acts.
  3. Plot point Y(σy,−τxy)Y(\sigma_y, -\tau_{xy}) for the perpendicular face (its shear turns the element the opposite way).
  4. Join X and Y. The line cuts the σ\sigma-axis at CC, the centre, where OC=σx+σy2OC = \dfrac{\sigma_x+\sigma_y}{2}.
  5. With C as centre and CXCX as radius, draw the circle. Radius R=(σx−σy2)2+τxy2R = \sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2 + \tau_{xy}^2}.
  6. The circle cuts the σ\sigma-axis at P1P_1 and P2P_2: σ1=OC+R\sigma_1 = OC + R, σ2=OC−R\sigma_2 = OC - R. The angle XCP1=2θpXCP_1 = 2\theta_p gives the principal plane.
  7. For the stresses on the given inclined plane, rotate the radius CX through 2θ2\theta about C in the same sense as the plane is rotated, reaching point QQ. The abscissa of Q is the normal stress σn\sigma_n and the ordinate is the shear stress τθ\tau_\theta. The diametrically opposite point Q′Q' gives the stresses on the perpendicular plane.
  8. The resultant stress is OQOQ, and its obliquity is ∠QOC\angle QOC measured from the σ\sigma-axis.
  9. The top and bottom points of the circle give the maximum shear stress τmax=R\tau_{max} = R.

Sketch

        tau
         |        T (tau max = R)
         |     .--*--.
         |   /    |    \
         |  *Q    |     * X(sx, txy)
 --------+-P2-----C------P1------- sigma
         O  \     |    /
         |   *Y   |   /
         |     '--*--'
         |        T'
 P2 = sigma_2, P1 = sigma_1, angle XCP1 = 2*theta_p

✓ Answer checked by Claude Opus 5.5 (high effort)

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