LECTURE #3 1.060 ENGINEERING MECHANICS II HYDROSTATICS (from Recitation #1) In a fluid at rest there can be no shear stress only normal stress. In fluids at rest normal stress is isotropic (same in all directions) and positive if compression. It is referred to as "the pressure". [Diagram: A small cube is shown with arrows representing normal stresses acting on each face.] Hydrostatic Pressure Distribution [Diagram: A coordinate system is shown with the z-axis pointing upwards. A pressure gradient is indicated as grad p = -ρg ê_z.] dp/dz = -ρg => p = constant p(z) = p(0) - ρgz = p_0 + ρgz_0 - ρgz = (constant) within a constant density fluid at rest. To apply this: Need pressure P_0 at specified elevation z_0 For liquids with a free surface: We choose z = 0 at the free surface. P_o = P as z → 0⁻ = pressure above surface (except in cases when surface tension counts) So: P + ρgz = P_o - Constant If "fluid" above the free surface is air at atmospheric pressure, then P_o = P_atm = 101.3 kPa = 1.013 bar = 1013 mbar (Standard) and p - P_atm + ρgz = 0 We make a deal: From now on p' = P - P_atm = PRESSURE (gauge pressure) P_atm = 0 (gauge) by definition P_atm = 101.3 kPa (absolute) P_absolute = P + P_atm P_o = P_atm = 0 gauge [Diagram: A vertical column of fluid is shown. The top surface is exposed to the atmosphere, with pressure P_o. The bottom of the column is at a depth h, with pressure P.] -P_o -ρgh P = ρgh Pressure at depth h simply balances the weight per unit horizontal area of the fluid above. (Plus pressure at the free surface, if this is not atmospheric, i.e. 0 gauge). In a fluid at rest, the pressure varies linearly in the vertical direction so long as the fluid density is constant. P₂ - P₁ = -ρg(z₂ - z₁) = -ρg Δz If Δz > 0, i.e., going up from ① to ②, then Δp = P₂ - P₁ < 0 (less fluid on top at ②) and if Δz < 0 (down from ① to ②), then Δz < 0 and Δp > 0 (more fluid on top at ①). In many cases involving a gas, e.g., air, ρ is so small that ρgΔz ≈ 0 and P₂ ≈ P₁ (constant). Same rules" apply if the plane surface is inclined to vertical, so long as it is recalled that the pressure force is ⊥ to the surface upon which it acts. Total force on inclined surface (of length L = h/cosθ from vertical) F_1 = (1/2 ρgh)L = (1/2 ρgh)(h/cosθ) = (1/2 ρgh²) (1/cosθ) = P_c * A Area of pressure prism = (1/2 h)L = (1/2 h)(h/cosθ) = (1/2 h²) (1/cosθ) = A Pressure at center of gravity of Area x Area: F_x = F_1 cosθ = (1/2 ρgh²) = Pressure force on area's projection on a vertical plane F_y = F_1 (-sinθ) = - (1/2 ρgh²) tanθ = -(weight of fluid above the area) = -(1/2 h)(h tanθ)ρg = -(1/2)ρgh² tanθ This makes sense, since the triangle of fluid above the surface must be in equilibrium, i.e., Net force in x = 0 Net force in y must balance gravity (weight). pressure from surface on fluid

Hydrostatics, Forces on Plane and Curved Surfaces

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