--- title: "Normal and resultant stress on 60 degree plane" subject: "Strength of Materials" chapter: 4 key: "semester_3/strength_of_materials:normal-resultant-stress-60deg-plane-100" source: https://www.bctengineer.com/q/semester_3/strength_of_materials/normal-resultant-stress-60deg-plane-100 site: BCT Engineering --- # Normal and resultant stress on 60 degree plane Strength of Materials · Chapter 4: Stress and Strain Analysis - Asked in: 2068 Chaitra, 8 marks - Video: [Normal and tangential stress (direct stress in one plane)](https://www.youtube.com/watch?v=wWF-0bDxnIo) — TECHNICAL CLASSES - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/normal-resultant-stress-60deg-plane-100 - Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-4 ## Question For stresses shown in figure below, find the normal and resultant stresses on the plane shown. Find the direction of resultant stresses. Show the results diagrammatically. [Figure: element with $150\ \text{MN/m}^2$ tensile on the left and right faces, $100\ \text{MN/m}^2$ on the top and bottom faces and shear stress $60\ \text{MN/m}^2$; the plane is inclined at $60^\circ$ to the horizontal.] ## Short answer From σx=150, σy=100, τxy=60 MN/m², plane normal at -30° gives normal stress 85.54 MN/m², shear 51.65 MN/m². Resultant stress is 99.92 MN/m², inclined at 31.12° to the normal. ## Answer **Reading of the figure:** $\sigma_x = 150$ MN/m² (left/right faces), $\sigma_y = 100$ MN/m² (top/bottom faces), $\tau_{xy} = 60$ MN/m². The plane makes $60^\circ$ with the horizontal, so its normal makes $60^\circ$ with the vertical. Measured anticlockwise from the $x$-axis, the plane is at $60^\circ$ and its normal is at $\theta = 150^\circ$ (equivalent to $-30^\circ$). #### Normal stress $$ \begin{aligned} \sigma_n &= \frac{150+100}{2} + \frac{150-100}{2}\cos 300^\circ + 60\sin 300^\circ \\ &= 125 + 25(0.5) + 60(-0.866) = 85.54\ \text{MN/m}^2 \end{aligned} $$ #### Shear stress $$ \tau = -\frac{150-100}{2}\sin 300^\circ + 60\cos 300^\circ = -25(-0.866) + 60(0.5) = 51.65\ \text{MN/m}^2 $$ #### Resultant stress and direction $$ \sigma_R = \sqrt{85.54^2 + 51.65^2} = 99.92\ \text{MN/m}^2 $$ $$ \tan\phi = \frac{\tau}{\sigma_n} = \frac{51.65}{85.54} \Rightarrow \phi = 31.12^\circ $$ The resultant makes $31.12^\circ$ with the normal to the plane (toward the shear direction), i.e. $58.88^\circ$ with the plane itself. ```text resultant 99.92 ^ / \ /31.1\ normal sn = 85.54 / deg \ ---> --/--------\-- plane (60 deg to horizontal) shear tau = 51.65 acts along the plane ``` **Answer:** $\sigma_n = 85.54$ MN/m², $\tau = 51.65$ MN/m², $\sigma_R = 99.92$ MN/m² at $31.12^\circ$ to the normal. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The plane makes 60° with the horizontal. What angle theta is used in the formulas? A. The angle of its normal from the x-axis, theta = 150° (or -30°) B. 60°, the angle of the plane itself C. 30° measured clockwise only as 30° D. 120° Answer: A. The formulas need the angle of the normal measured from the x-axis; the plane at 60° has a normal at 150°. (Key point: Angle of the normal) 2. With 2theta = 300°, what is the normal stress? A. 125 + 25 cos300° - 60 sin300° = 188.5 MN/m2 B. 125 + 25 cos300° + 60 sin300° = 85.54 MN/m2 C. 125 + 25 sin300° + 60 cos300° = 133.4 MN/m2 D. 125 MN/m2 (the average) Answer: B. cos300° = 0.5 and sin300° = -0.866, giving 125 + 12.5 - 51.96 = 85.54. (Key point: Normal stress value) 3. How is the resultant stress on the plane obtained? A. 85.54 + 51.65 = 137.2 MN/m2 B. 85.54 - 51.65 = 33.9 MN/m2 C. sqrt(85.54^2 + 51.65^2) = 99.92 MN/m2 D. sqrt(150^2 + 100^2) = 180 MN/m2 Answer: C. Normal and shear stress are perpendicular so the resultant is their vector sum. (Key point: Resultant stress) 4. How is the direction of the resultant given in the answer? A. tan(phi) = tau/sigma_n, phi = 31.12° from the normal B. tan(phi) = sigma_n/tau, phi = 31.12° from the plane C. tan(phi) = tau/sigma_n, phi = 31.12° from the x-axis D. tan(phi) = tau/sigma_n, phi = 58.88° from the normal Answer: A. The angle with the normal has tangent tau/sigma_n = 0.604; so 31.12° from the normal and 58.88° from the plane. 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