--- title: "Shear flow in a closed box beam" subject: "Mechanics of Solids" chapter: 10 key: "semester_5/mechanics_of_solids:box-beam-shear-flow-numerical" source: https://www.bctengineer.com/q/semester_5/mechanics_of_solids/box-beam-shear-flow-numerical site: BCT Engineering --- # Shear flow in a closed box beam Mechanics of Solids · Chapter 10: Shear Centers for Thin- Wall Beam Cross Sections - Asked in: year not given, 6 marks - Video: [Topic # 7.5 - Shear Flow in Thin-Walled Sections (PROBLEM EXAMPLE)](https://www.youtube.com/watch?v=AT7-a3l20SI) — Solid Mechanics - Prof. Jay Ghosh (IIT Bombay) - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/mechanics_of_solids/box-beam-shear-flow-numerical - Whole chapter: https://www.bctengineer.com/md/semester_5/mechanics_of_solids/chapter-10 ## Question A symmetrical box beam has a rectangular thin-wall section of mean width 120 mm and mean depth 200 mm with uniform wall thickness 5 mm. A vertical shear force of 30 kN acts through the vertical axis of symmetry. (a) Find the position of the shear centre. (b) Draw the shear flow variation and find the shear flow at the corners and at the neutral axis. (c) Find the maximum shear stress and the share of the shear force carried by the webs. ## Short answer Shear centre is at the centroid. Shear flow: 0 at mid-flange, 48.2 N/mm at corners, 88.4 N/mm at neutral axis. Maximum shear stress 17.7 MPa. The two webs carry the entire 30 kN (15 kN each). ## Answer #### Data $b = 120$ mm, $h = 200$ mm, $t = 5$ mm, $V = 30$ kN. #### (a) Shear centre The section is symmetric about both the vertical and horizontal axes, so the shear centre coincides with the **centroid** (centre of the box). A load through the vertical axis of symmetry therefore causes no twist. #### Second moment of area $$ I_x = 2(b\,t)\left(\frac h2\right)^2 + 2\,\frac{t\,h^3}{12} = 1.200\times10^{7} + 6.667\times10^{6} = 1.8667\times10^{7}\ \text{mm}^4 $$ #### (b) Shear flow By symmetry the shear flow is zero at the middle of the top (and bottom) flange, so we make the imaginary cut there. Then $q = VQ/I_x$, where $Q$ is the first moment of the area measured from the cut. **Top flange**, $s$ measured from the middle: $Q = t\,s\,\dfrac h2$ $$ q_{corner} = \frac{30000}{1.8667\times10^{7}}\times(5\times60\times100) = 48.21\ \text{N/mm} $$ **Web**, $y$ measured down from the corner: $Q = (5\times60\times100) + 5\,y\,(100 - y/2)$. At the neutral axis ($y = 100$) this gives $30000 + 25000 = 55000$ mm$^3$: $$ q_{NA} = \frac{30000\times55000}{1.8667\times10^{7}} = 88.39\ \text{N/mm} $$ ```text 0 ----- 48.2 -> (top flange: linear) | | 88.4 | | 88.4 (web: parabolic, max at NA) | | 0 ----- 48.2 -> ``` | Location | Shear flow (N/mm) | |---|---| | Middle of flange | 0 | | Corner | 48.2 | | Neutral axis (web) | 88.4 | #### (c) Maximum shear stress and force in webs $$ \tau_{max} = \frac{q_{NA}}{t} = \frac{88.39}{5} = 17.68\ \text{MPa} $$ Vertical force in one web: $$ F_w = \frac{V}{I_x}\int_0^h \left[30000 + 5\,y(100 - y/2)\right]dy = 1.607\times10^{-3}\times9.333\times10^6 = 15000\ \text{N} $$ The two webs carry $30$ kN, which is **100 percent** of $V$. The flanges carry equal and opposite horizontal forces that cancel, so they give no net vertical force. **Answer:** The shear centre is at the centroid. The shear flow is $q = 0$ at mid-flange, $48.2$ N/mm at the corners and $88.4$ N/mm at the neutral axis. $\tau_{max} = 17.7$ MPa, and the two webs carry the whole 30 kN (15 kN each). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why does the shear centre of this box coincide with the centroid? A. Because the walls are thin B. Because the thickness is uniform only C. Because the section is symmetric about both axes D. Because the shear force is vertical Answer: C. With two axes of symmetry, a load through the centre produces no twist, so the shear centre is at the centroid. (Key point: Shear centre of symmetric box) 2. Where is the imaginary cut made, and why? A. At the corner, where the flow is maximum B. At the middle of the top flange, where symmetry makes q zero C. At the neutral axis, where stress is zero D. At the bottom of the web, to make q maximum Answer: B. By symmetry the shear flow is zero at the flange mid-point, so q = VQ/I can be measured from there. (Key point: Location of imaginary cut) 3. How does shear flow vary in the web, and what is its peak value? A. Constant at 48.2 N/mm B. Linear, maximum 48.2 N/mm at the corner C. Parabolic, maximum 88.4 N/mm at the neutral axis D. Parabolic, maximum 88.4 N/mm at the corners Answer: C. Q increases down the web to 55000 mm^3 at the neutral axis, giving q = 30000 x 55000 / 1.8667e7 = 88.4 N/mm. (Key point: Where web flow peaks) 4. What fraction of the vertical shear force is carried by the two webs? A. 100 percent; flange forces cancel horizontally B. 50 percent, the flanges carry the rest C. About 75 percent D. Only the neutral-axis fibres carry it Answer: A. The flanges' horizontal forces are equal and opposite, so the two webs carry all 30 kN, 15 kN each. 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