--- title: "Mohr's circle for a plane stress element" subject: "Strength of Materials" chapter: 6 key: "semester_4/strength_of_materials_mechanical:mohr-circle-plane-stress-construction" source: https://www.bctengineer.com/q/semester_4/strength_of_materials_mechanical/mohr-circle-plane-stress-construction site: BCT Engineering --- # Mohr's circle for a plane stress element Strength of Materials · Chapter 6: Transformation of stress and strain - Asked in: year not given, 8 marks - Video: [HOW TO PLOT THE STRESSES ON AN IN PLANE STRESS ELEMENT ON THE MOHR'S CIRCLE | STRENGTH OF MATERIALS](https://www.youtube.com/watch?v=4koyXom4PlE) — Engr. Jobert S. De La Cruz - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/strength_of_materials_mechanical/mohr-circle-plane-stress-construction - Whole chapter: https://www.bctengineer.com/md/semester_4/strength_of_materials_mechanical/chapter-6 ## Question The stresses at a point are sigma_x = 100 MPa (tensile), sigma_y = 20 MPa (tensile) and a shear stress of 40 MPa on the x faces tending to rotate the element clockwise (the shear on the y faces is therefore anticlockwise). Construct Mohr's circle and use it to find the principal stresses, the maximum in-plane shear stress, the orientation of the principal planes, and the stresses on a plane whose normal is rotated 20 degrees anticlockwise from the x axis. ## Short answer Mohr's circle centre 60 MPa, radius 56.6 MPa. Principal stresses: 116.6 MPa and 3.4 MPa. Maximum in-plane shear: 56.6 MPa. Principal planes at 22.5° clockwise from x-axis. On a plane 20° anticlockwise from x, normal stress 64.9 MPa, shear 56.4 MPa clockwise. ## Answer #### Mohr's circle construction Convention: tension is positive to the right on the $\sigma$ axis; a shear stress that rotates the element **clockwise** is plotted **upward**. With this convention, an angle $2\theta$ on the circle is measured in the same sense as the rotation $\theta$ of the plane. 1. Plot X $(\sigma_x,\ \tau) = (100,\ +40)$ for the x face. 2. Plot Y $(\sigma_y,\ \tau) = (20,\ -40)$ for the y face. 3. Join XY; it cuts the $\sigma$ axis at the centre $C$. 4. Draw the circle with centre $C$ and diameter XY. $$ \sigma_{avg} = \frac{100+20}{2} = 60\ \text{MPa}, \qquad R = \sqrt{40^2 + 40^2} = 56.57\ \text{MPa} $$ ```text tau (cw +) | X(100,40) | .-*-. | .' | '. ---+---*----C----*---- sigma 0 B 60 A | '. | .' | '-*-' | Y(20,-40) ``` #### Principal stresses (points A and B) $$ \sigma_1 = 60 + 56.57 = 116.6\ \text{MPa}, \qquad \sigma_2 = 60 - 56.57 = 3.43\ \text{MPa} $$ #### Maximum in-plane shear stress At the top and bottom points of the circle: $$ \tau_{max,\,in\text{-}plane} = R = 56.6\ \text{MPa}, \quad \text{with normal stress } 60\ \text{MPa} $$ Because both principal stresses are tensile, the absolute maximum shear stress (including the third principal stress, $\sigma_3=0$) is $\sigma_1/2 = 58.3$ MPa. #### Principal planes On the circle, the angle XCA is $2\theta_p$ with $\tan2\theta_p = 40/40 = 1$, so $2\theta_p = 45^\circ$, measured **clockwise** from CX to CA. (By formula, a clockwise shear on the x face is $\tau_{xy} = -40$ MPa, so $\tan2\theta_p = \dfrac{2(-40)}{100-20} = -1$ and $\theta_p = -22.5^\circ$.) So the plane carrying $\sigma_1$ is **$22.5^\circ$ clockwise** from the x axis; $\sigma_2$ acts on the plane $90^\circ$ from it. #### Stresses on the plane at 20 degrees anticlockwise Rotate the radius CX by $2\theta = 40^\circ$ anticlockwise. CX starts at $45^\circ$ above the $\sigma$ axis, so the new radius makes $45^\circ + 40^\circ = 85^\circ$ with the $\sigma$ axis: $$ \sigma = 60 + 56.57\cos85^\circ = 64.9\ \text{MPa} $$ $$ \tau = 56.57\sin85^\circ = 56.4\ \text{MPa (clockwise sense)} $$ Check by formula with $\tau_{xy} = -40$ MPa: $\sigma = 60 + 40\cos40^\circ + (-40)\sin40^\circ = 64.9$ MPa. **Answer: $\sigma_1 = 116.6$ MPa, $\sigma_2 = 3.43$ MPa; $\tau_{max} = 56.6$ MPa; principal plane $22.5^\circ$ clockwise from x; on the 20 degree plane $\sigma = 64.9$ MPa, $\tau = 56.4$ MPa.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With a clockwise shear on the x faces taken as upward, which points are plotted for the x and y faces? A. X(100, +40) and Y(20, +40) B. X(100, -40) and Y(20, -40) C. X(100, +40) and Y(20, -40) D. X(20, +40) and Y(100, -40) Answer: C. The x face has clockwise shear (plotted up) and the y face has the opposite shear (plotted down). (Key point: Plotting X and Y) 2. What are the centre and radius of the circle? A. Centre 60 MPa, radius 40 MPa B. Centre 80 MPa, radius 56.57 MPa C. Centre 60 MPa, radius 80 MPa D. Centre 60 MPa, radius 56.57 MPa Answer: D. Centre = (100+20)/2 = 60 and R = sqrt(40^2 + 40^2) = 56.57 MPa, since half the difference is 40. (Key point: Centre and radius) 3. Why is the absolute maximum shear stress 58.3 MPa and not the in-plane 56.6 MPa? A. Both principal stresses are tensile, so the third principal stress (zero) gives σ1/2 as the largest circle B. Because the circle radius changes with the plane chosen C. Because shear on the x faces adds 40 MPa to the radius D. Because the average stress is added to the radius Answer: A. With σ3 = 0 and σ1 = 116.6, the circle through σ1 and σ3 has radius σ1/2 = 58.3 MPa. (Key point: Absolute maximum shear) 4. How is the stress on the plane 20° anticlockwise found from the circle? A. Rotate CX by 20° so the radius is at 65° B. Rotate radius CX by 40° (double the angle) in the same sense, so the radius is at 85° to the σ axis C. Rotate CX by 40° the opposite sense, giving 5° D. Rotate CX by 80°, giving 125° Answer: B. Angles on Mohr's circle are doubled; 45° + 40° = 85° gives σ = 60 + 56.57cos85° = 64.9 MPa. 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