--- title: "Shear centre of a channel section" subject: "Mechanics of Solids" chapter: 10 key: "semester_5/mechanics_of_solids:channel-shear-centre-numerical" source: https://www.bctengineer.com/q/semester_5/mechanics_of_solids/channel-shear-centre-numerical site: BCT Engineering --- # Shear centre of a channel section Mechanics of Solids · Chapter 10: Shear Centers for Thin- Wall Beam Cross Sections - Asked in: year not given, 8 marks - Video: [Shear Centre for Channel and Angle Section](https://www.youtube.com/watch?v=VkRXCyfN41E) — Being Civil - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/mechanics_of_solids/channel-shear-centre-numerical - Whole chapter: https://www.bctengineer.com/md/semester_5/mechanics_of_solids/chapter-10 ## Question A thin-walled channel section has a web of depth 160 mm and thickness 8 mm, and two flanges each 80 mm wide and 10 mm thick (the dimensions are measured on the centre-lines of the walls). A vertical shear force of 20 kN acts. (a) Find the shear flow distribution in the flange and web and its maximum values. (b) Locate the shear centre from the web centre-line. (c) Verify that the resultant force in each flange multiplied by the web depth equals $Ve$. ## Short answer Shear flow in flange is linear from 0 at tip to 98.7 N/mm at junction; in web parabolic from 98.7 to 138.2 N/mm at neutral axis (τ_max = 17.3 MPa). Shear centre is 31.6 mm from web centre-line on the side opposite flanges. Flange force couple equals V e. ## Answer #### Data $h = 160$ mm, $t_w = 8$ mm, $b = 80$ mm, $t_f = 10$ mm, $V = 20$ kN; the channel is open to the right and symmetric about the horizontal axis. ```text +-------- b = 80 -------+ t_f = 10 | ^ | web t_w = 8 | h = 160 | v +-----------------------+ S (shear centre at distance e to left of web) ``` #### Second moment of area (thin-wall approximation) $$ I_x = \frac{t_wh^3}{12} + 2\,b\,t_f\left(\frac h2\right)^2 = 2730667 + 10240000 = 12970667\ \text{mm}^4 $$ (The own inertia of the flanges, $bt_f^3/12$, is negligible.) #### (a) Shear flow **Flange** (distance $s$ from the free edge): $q = \dfrac{V\,(t_f s)(h/2)}{I_x}$, linear from zero at the tip to the maximum at the junction: $$ q_{junction} = \frac{20000\times(80\times10)\times80}{12970667} = 98.68\ \text{N/mm} $$ **Web** at the neutral axis (adds the half web): $$ q_{NA} = \frac{V}{I_x}\left[b\,t_f\frac h2 + t_w\frac h2\cdot\frac h4\right] = \frac{20000}{12970667}\left[64000 + 25600\right] = 138.16\ \text{N/mm} $$ The distribution in the web is parabolic between $98.7$ N/mm at the junctions and $138.2$ N/mm at mid-depth. Maximum shear stress in the web: $\tau = q_{NA}/t_w = 17.27$ MPa. #### (b) Position of the shear centre The resultant force in each flange is the area of the triangle of the shear flow: $$ F_f = \frac12\,q_{junction}\,b = \frac12\times98.68\times80 = 3947\ \text{N} $$ The web force is $\approx V = 20$ kN (vertical). Take moments about the web centre-line: the web force passes through it, and the two flange forces ($F_f$, in opposite directions) form a couple $F_f\,h$. The load $V$ at distance $e$ from the web must give the same moment: $$ V\,e = F_f\,h \ \Rightarrow\ e = \frac{3947\times160}{20000} = 31.58\ \text{mm} $$ General expression: $$ e = \frac{b^2h^2t_f}{4I_x} = \frac{3b^2t_f}{h\,t_w + 6\,b\,t_f} = \frac{3\times6400\times10}{160\times8 + 6\times80\times10} = \frac{192000}{6080} = 31.58\ \text{mm} $$ The shear centre lies **31.6 mm to the left of the web centre-line**, outside the section on the side opposite to the flanges. #### (c) Check $F_fh/V = 3947.4\times160/20000 = 31.58$ mm, equal to $e$. $\checkmark$ **Answer:** $q_{flange,max} = 98.7$ N/mm, $q_{web,max} = 138.2$ N/mm ($\tau_{max} = 17.3$ MPa); shear centre $e = 31.6$ mm from the web centre-line, on the side away from the flanges. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How does the shear flow vary along a flange of the channel, measured from the free tip to the web junction? A. Linearly from zero at the tip to a maximum at the junction B. Parabolically, maximum at the tip C. Constant along the whole flange D. Linearly from a maximum at the tip to zero at the junction Answer: A. Q grows linearly with s (Q = t_f s h/2) from the free edge, where it is zero, so q rises linearly to the junction. (Key point: Flange shear flow shape) 2. Why is the maximum shear flow in the web (138.2 N/mm) larger than the flange junction value (98.7 N/mm)? A. The web is thicker than the flange B. At the neutral axis Q includes the flange area plus the half-web area above it C. Because the load acts on the web only D. Because the flange inertia is ignored Answer: B. Q at the neutral axis adds the half-web contribution t_w(h/2)(h/4) to the flange term, giving 138.2 N/mm. (Key point: Web maximum shear flow) 3. How is the shear centre distance e found from the flange forces? A. Equate V e to the couple F_f x h formed by the two flange forces B. Equate V e to the flange force times the flange width C. Set e equal to half the flange width D. Set e so that the web force is zero Answer: A. Taking moments about the web centre-line, the web force has no moment, so V e = F_f h, giving e = 31.6 mm. (Key point: Shear centre by moments) 4. What is the resultant force in each flange, and how is it obtained? A. Peak flow times flange width, 7894 N B. Area of the triangular shear flow diagram, 1/2 x 98.68 x 80 = 3947 N C. V divided by 2 = 10 kN D. Peak flow times thickness, 987 N Answer: B. The flange flow is triangular, so the force is 1/2 q_junction b = 3947 N. (Key point: Flange force value) 5. Where does the shear centre of this channel lie? A. On the web centre-line B. At the centroid of the section C. Outside the section, about 31.6 mm from the web on the side away from the flanges D. Inside the flanges, 31.6 mm from the web towards the open side Answer: C. The flange couple requires the load line to be on the side away from the flanges, outside the web. 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