--- title: "Stresses on plane at 60 deg to major principal" subject: "Strength of Materials" chapter: 4 key: "semester_3/strength_of_materials:principal-stresses-100t-40c-60deg-plane" source: https://www.bctengineer.com/q/semester_3/strength_of_materials/principal-stresses-100t-40c-60deg-plane site: BCT Engineering --- # Stresses on plane at 60 deg to major principal Strength of Materials · Chapter 4: Stress and Strain Analysis - Asked in: 2073 Shrawan, 8 marks - Video: [Principal stress and principal plane || Mohr's Circle || All formula || Numerical solution](https://www.youtube.com/watch?v=7fZPgk1NHVs) — Sudip Dhakal - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/principal-stresses-100t-40c-60deg-plane - Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-4 ## Question The principle stresses at a point in a bar are 100 MPa tensile and 40 MPa compressive. Find the normal stress shears and resultant stress on a plane inclined at $60^\circ$ to the axis of major principal stress. ## Short answer With major principal stress 100 MPa and minor -40 MPa, the plane inclined at 60° to the major axis has its normal at 30°. Normal stress = 65.00 MPa tensile, shear stress = 60.62 MPa. Resultant stress = 88.88 MPa, direction 43.00° from the normal. ## Answer When the two principal stresses are known, the plane is located by the angle $\theta$ between its normal and the major principal stress ($\sigma_1$) axis. The plane itself is inclined at $60^\circ$ to the axis of $\sigma_1$, so its normal makes $\theta = 90^\circ - 60^\circ = 30^\circ$ with that axis. Here $\sigma_1 = 100$ MPa and $\sigma_2 = -40$ MPa. #### Normal stress $$ \begin{aligned} \sigma_n &= \frac{\sigma_1+\sigma_2}{2} + \frac{\sigma_1-\sigma_2}{2}\cos 2\theta \\ &= \frac{100-40}{2} + \frac{100+40}{2}\cos 60^\circ = 30 + 70(0.5) = 65.00\ \text{MPa} \end{aligned} $$ (positive: tensile). #### Shear stress $$ \tau = \frac{\sigma_1-\sigma_2}{2}\sin 2\theta = 70\sin 60^\circ = 60.62\ \text{MPa} $$ #### Resultant stress and its direction $$ \sigma_R = \sqrt{\sigma_n^2+\tau^2} = \sqrt{(65.00)^2+(60.62)^2} = 88.88\ \text{MPa} $$ $$ \tan\phi = \frac{\tau}{\sigma_n} = \frac{60.62}{65.00} \Rightarrow \phi = 43.00^\circ $$ The resultant makes $43.00^\circ$ with the normal to the plane, i.e. $47.00^\circ$ with the plane itself. **Answer:** $\sigma_n = 65.00$ MPa (tensile), $\tau = 60.62$ MPa, resultant $= 88.88$ MPa at $43.00^\circ$ to the normal of the plane. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In this problem, how is the 40 MPa compressive principal stress entered in the formulas? A. sigma_2 = -40 MPa, so (s1+s2)/2 = 30 and (s1-s2)/2 = 70 B. sigma_2 = +40 MPa, so (s1+s2)/2 = 70 C. sigma_2 = -40 MPa, so (s1-s2)/2 = 30 D. sigma_2 = 40 MPa, so (s1-s2)/2 = 30 Answer: A. Tension is positive and compression negative, so sigma_2 = -40. This gives a mean of 30 and a radius of 70. (Key point: Sign of compressive principal stress) 2. Using sigma_n = 30 + 70 cos 2theta with theta = 60°, what is the normal stress and its nature? A. 65 MPa, tensile B. -5 MPa, compressive C. -5 MPa, tensile D. 35 MPa, tensile Answer: B. cos 120° = -0.5, so sigma_n = 30 - 35 = -5 MPa. The negative sign means compressive. (Key point: Normal stress on the plane) 3. After finding sigma_n = -5 MPa and tau = 60.62 MPa, how is the resultant stress obtained and how does it lie relative to the plane? A. sigma_n + tau = 55.62 MPa, almost along the normal B. sqrt(sigma_n^2 + tau^2) = 60.83 MPa, almost along the normal C. sqrt(sigma_n^2 + tau^2) = 60.83 MPa, almost along the plane because sigma_n is small D. tau - sigma_n = 65.62 MPa, along the plane Answer: C. The resultant is the vector sum of normal and shear stress. With tiny sigma_n it makes only 4.72° with the plane. 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