--- title: "Pressure on isosceles triangular plate" subject: "Hydro Power" chapter: 1 key: "semester_6/hydro_power:triangular-plate-pressure" source: https://www.bctengineer.com/q/semester_6/hydro_power/triangular-plate-pressure site: BCT Engineering --- # Pressure on isosceles triangular plate Hydro Power · Chapter 1: Introduction to Basic Fluid Mechanics and Hydraulics - Asked in: 2070 Magh, 7 marks - Video: [Determine the total pressure and centre of pressure on an isoscles triangular plate of base 4m .....](https://www.youtube.com/watch?v=rT-tE1_dax4) — Yogi Mechanics - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/hydro_power/triangular-plate-pressure - Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-1 ## Question An isosceles triangle plate of base 4 m and altitude 6 m is immersed vertically in water. Its axis of symmetry is parallel to the free surface and at a depth of 6 m from the free water surface. Calculate the magnitude and location of the total pressure force. ## Short answer Vertical isosceles triangle (base 4 m, altitude 6 m) with horizontal axis at 6 m depth gives force 706.32 kN from F = ρgA h̄. Center of pressure at 6.111 m depth, 2 m from base, using h* = h̄ + IG/(A h̄) with IG = 8 m⁴. ## Answer The plate is vertical and its axis of symmetry (the altitude) is horizontal, at 6 m depth. So the 4 m base is vertical, and the centroid lies on the axis at depth 6 m. ```text free surface ========================== : 6 m |\ : 4 m | \ v base |----G-----> axis (horizontal) (vert.)| / 6 m altitude |/ ``` **Given:** base $b = 4$ m, altitude $h = 6$ m, water. #### Area and centroid $$ A = \frac{1}{2}\times 4 \times 6 = 12\ \text{m}^2, \qquad \bar h = 6\ \text{m} $$ The centroid lies on the axis, at $h/3 = 2$ m from the base (4 m from the apex). #### Total pressure $$ F = \rho g A\bar h = 1000 \times 9.81 \times 12 \times 6 = 706\,320\ \text{N} = 706.32\ \text{kN} $$ #### Second moment about the horizontal centroidal axis (the axis of symmetry) The triangle is two right triangles, each of base 6 m (along the axis) and height 2 m, on either side of the axis: $$ I_G = 2 \times \frac{6 \times 2^3}{12} = \frac{h b^3}{48} = \frac{6 \times 4^3}{48} = 8\ \text{m}^4 $$ #### Centre of pressure (depth) $$ \begin{aligned} h^* &= \bar h + \frac{I_G}{A\bar h} = 6 + \frac{8}{12 \times 6} \\ &= 6 + 0.111 = 6.111\ \text{m} \end{aligned} $$ Horizontally, the plate is symmetric about the axis (product of inertia about the centroidal axes is zero), so the centre of pressure lies at the same horizontal position as the centroid: 2 m from the base. **Answer:** $F = 706.32$ kN, acting at 6.111 m depth (0.111 m below the axis), 2 m from the vertical base (4 m from the apex). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The axis of symmetry (altitude 6 m) of the triangle is horizontal at 6 m depth. What does this imply? A. The base is horizontal B. The apex is at the surface C. The 4 m base is vertical and the centroid lies on the axis at 6 m depth D. The plate is inclined Answer: C. The altitude is perpendicular to the base, so a horizontal altitude means the base is vertical. The centroid lies on the axis at depth 6 m. (Key point: Orientation) 2. What is the total pressure force on the triangle? A. 706.32 kN B. 353.16 kN C. 1412.64 kN D. 235.44 kN Answer: A. A = (1/2)(4)(6) = 12 m^2, h_bar = 6 m, F = 9810 x 12 x 6 = 706.32 kN. (Key point: Total force) 3. Why is I_G = h b^3 / 48 = 8 m^4 used rather than b h^3/36? A. I_G is always h b^3/48 B. b h^3/36 is for the wrong shape C. The triangle is rotated by 45 deg D. The axis for depth variation is the horizontal axis of symmetry, about which the triangle's vertical extent is the 4 m base Answer: D. The moment must be about the horizontal centroidal axis, which is the symmetry axis; the vertical dimension is b = 4 m, giving h b^3/48 = 8 m^4. (Key point: Second moment about the axis) 4. Where is the centre of pressure? A. 6.5 m depth B. Exactly at the centroid, 6 m C. 5.889 m depth D. 6.111 m depth, on the axis Answer: D. h* = 6 + 8/(12 x 6) = 6.111 m. Symmetry puts it on the axis horizontally, with the same horizontal position as the centroid. 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