--- title: "Principal stresses at web-flange junction" subject: "Strength of Materials" chapter: 9 key: "semester_4/strength_of_materials_mechanical:principal-stresses-web-flange-junction" source: https://www.bctengineer.com/q/semester_4/strength_of_materials_mechanical/principal-stresses-web-flange-junction site: BCT Engineering --- # Principal stresses at web-flange junction Strength of Materials · Chapter 9: Design of Beams and shafts - Asked in: year not given, 6 marks - Video: [Problem 8.2 | Principal Stresses under Given Loading || MOM by Beer & Johnston || Solved Problem](https://www.youtube.com/watch?v=C1xdE-HOUT8) — Engr. Adnan Rasheed Mechanical - 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/principal-stresses-web-flange-junction - Whole chapter: https://www.bctengineer.com/md/semester_4/strength_of_materials_mechanical/chapter-9 ## Question An I-section has flanges 150 mm x 20 mm and a web 20 mm x 200 mm (overall depth 240 mm, I = 86.13 x 10^6 mm^4). At a section the bending moment is 60 kN m and the shear force is 100 kN. For a point in the web just below the top flange, find the bending stress, the shear stress, the principal stresses and the maximum shear stress, and compare the principal stress with the maximum bending stress at the extreme fibre. ## Short answer Bending stress 69.7 MPa (compression), shear stress 19.2 MPa. Mohr's circle gives principal stresses +4.9 MPa and -74.6 MPa, maximum shear 39.8 MPa. Extreme fibre bending stress 83.6 MPa exceeds the principal compression, so the junction does not govern. ## Answer The point is at a distance $y = 100$ mm from the neutral axis (the junction of web and flange). #### Bending stress $$ \sigma = \frac{My}{I} = \frac{60\times10^6\times100}{86.13\times10^6} = 69.7\ \text{MPa (compression at the top, for a sagging moment)} $$ #### Shear stress $Q$ of the flange about the NA: $150\times20\times110 = 330\,000\ \text{mm}^3$. Width at this level (web) $b = 20$ mm. $$ \tau = \frac{VQ}{Ib} = \frac{100\times10^3\times330\,000}{86.13\times10^6\times20} = 19.2\ \text{MPa} $$ #### Principal stresses Take $\sigma_x = -69.7$ MPa (axial direction of the beam), $\sigma_y = 0$, $\tau_{xy} = 19.2$ MPa. $$ \frac{\sigma_x}{2} = -34.8\ \text{MPa}, \qquad R = \sqrt{34.8^2 + 19.2^2} = 39.75\ \text{MPa} $$ $$ \sigma_1 = -34.8 + 39.75 = +4.9\ \text{MPa (tension)} $$ $$ \sigma_2 = -34.8 - 39.75 = -74.6\ \text{MPa (compression)} $$ Maximum shear stress $\tau_{max} = R = 39.75\ \text{MPa}$. Direction: $\tan2\theta = \dfrac{2\tau}{|\sigma_x|} = 0.551$, so $2\theta = 28.8^\circ$ and the compressive principal stress acts at $\theta = 14.4^\circ$ to the beam axis. #### Comparison with the extreme fibre At the extreme fibre ($y = 120$ mm) there is no shear stress and $$ \sigma_{max} = \frac{60\times10^6\times120}{86.13\times10^6} = 83.6\ \text{MPa} $$ The principal stress at the junction ($74.6$ MPa) is smaller than the extreme-fibre stress ($83.6$ MPa), so for this loading the extreme fibre governs. For deep plate girders, or where a large shear force and large moment act together, the junction can govern, so it must be checked. **Answer: $\sigma = 69.7$ MPa, $\tau = 19.2$ MPa; $\sigma_1 = 4.9$ MPa, $\sigma_2 = -74.6$ MPa; $\tau_{max} = 39.8$ MPa; $\sigma_{max,fibre} = 83.6$ MPa.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What are Q and b used for the shear stress at the web just below the flange? A. Q = 150×20×100 mm³, b = 150 mm B. Q = 330 000 mm³, b = 150 mm C. Q = 150×20×110 = 330 000 mm³, b = 20 mm D. Q = 150×20×120 mm³, b = 20 mm Answer: C. Flange centroid is 110 mm from NA, and the width in the web is 20 mm. (Key point: Shear stress at junction) 2. What stresses act at the point? A. σx = 83.6 MPa, τ = 0 B. σx = 69.7 MPa tension, τ = 0 C. σx = 34.8 MPa, τ = 39.75 MPa D. σx = 69.7 MPa compression, τ = 19.2 MPa, σy = 0 Answer: D. y = 100 mm gives σ = My/I = 69.7 MPa and τ = VQ/Ib = 19.2 MPa. (Key point: Stress state) 3. What are the principal stresses? A. +4.9 MPa and -74.6 MPa B. +34.8 MPa and -34.8 MPa C. +39.75 MPa and -39.75 MPa D. -69.7 MPa and 0 Answer: A. -34.8 ± 39.75. (Key point: Principal stresses) 4. How does the junction compare with the extreme fibre? A. Junction governs since it has shear B. The junction principal stress (74.6) is less than 83.6 at the extreme fibre, which has no shear C. They are equal D. The junction stress is greater than 83.6 Answer: B. Extreme fibre 83.6 MPa governs here, but the junction must be checked for deep girders. 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