150 likes | 178 Views
College Of Engineering Electrical Engineering Department. Engineering Mechanics-Static Centroid Lecture-3 By Dr. Salah M. Swadi 2018-2019. 9.4 Resultant of a General Distributed Loading. Pressure Distribution over a Surface
E N D
College Of Engineering Electrical Engineering Department Engineering Mechanics-Static Centroid Lecture-3 By Dr. Salah M. Swadi 2018-2019
9.4 Resultant of a General Distributed Loading Pressure Distribution over a Surface • Consider the flat plate subjected to the loading function ρ = ρ(x, y) Pa • Determine the force dF acting on the differential area dA m2 of the plate, located at the differential point (x, y) dF = [ρ(x, y) N/m2](d A m2) = [ρ(x, y) d A]N • Entire loading represented as infinite parallel forces acting on separate differential area dA
9.4 Resultant of a General Distributed Loading Pressure Distribution over a Surface • This system will be simplified to a single resultant force FR acting through a unique point on the plate
9.4 Resultant of a General Distributed Loading Magnitude of Resultant Force • To determine magnitude of FR, sum the differential forces dF acting over the plate’s entire surface area dA • Magnitude of resultant force = total volume under the distributed loading diagram • Location of Resultant Force is
9.5 Fluid Pressure • According to Pascal’s law, a fluid at rest creates a pressure ρ at a point that is the same in all directions • Magnitude of ρ depends on the specific weight γ or mass density ρ of the fluid and the depth z of the point from the fluid surface ρ = γz = ρgz • Valid for incompressible fluids • Gas are compressible fluids and the above equation cannot be used
9.5 Fluid Pressure Flat Plate of Constant Width • Consider flat rectangular plate of constant width submerged in a liquid having a specific weight γ • Plane of the plate makes an angle with the horizontal as shown
9.5 Fluid Pressure Flat Plate of Constant Width • As pressure varies linearly with depth, the distribution of pressure over the plate’s surface is represented by a trapezoidal volume having an intensity of ρ1= γz1 at depth z1 and ρ2 = γz2 at depth z2 • Magnitude of the resultant force FR= volume of this loading diagram
9.5 Fluid Pressure Curved Plate of Constant Width • When the submerged plate is curved, the pressure acting normal to the plate continuously changes direction • Integration can be used to determine FR and location of center of centroid C or pressure P
9.5 Fluid Pressure Flat Plate of Variable Width • Consider the pressure distribution acting on the surface of a submerged plate having a variable width • Since uniform pressure ρ = γz (force/area) acts on dA, the magnitude of the differential force dF dF = dV = ρ dA = γz(xdy’)
9.5 Fluid Pressure Flat Plate of Variable Width • Centroid V defines the point which FR acts • The center of pressure which lies on the surface of the plate just below C has the coordinates P defined by the equations • This point should not be mistaken for centroid of the plate’s area
Example 9.14 Determine the magnitude and location of the resultant hydrostatic force acting on the submerged rectangular plate AB. The plate has a width of 1.5m; ρw = 1000kg/m3.
Solution The water pressures at depth A and B are For intensities of the load at A and B,
Solution For magnitude of the resultant force FR created by the distributed load. This force acts through the centroid of the area, measured upwards from B
Solution Same results can be obtained by considering two components of FR defined by the triangle and rectangle. Each force acts through its associated centroid and has a magnitude of Hence
Solution Location of FR is determined by summing moments about B