Calculations in Hydraulic Engineering: Fluid pressure, and the calculations of its effects in engineering structuresLongmans, Green, 1898 |
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Página 14
... resultant " of the fluid pressure . In the same way a pressure of uniform intensity acting normally upon any one of the plane surfaces , illustrated in the last article , may be directly opposed by an equal and opposite force or ...
... resultant " of the fluid pressure . In the same way a pressure of uniform intensity acting normally upon any one of the plane surfaces , illustrated in the last article , may be directly opposed by an equal and opposite force or ...
Página 23
... resultant forces which are worthy of careful study , although they seem to have escaped the notice of most hydraulicians . In ordinary practice , it is true that their effects have not often been noticed ; but it is impossible to doubt ...
... resultant forces which are worthy of careful study , although they seem to have escaped the notice of most hydraulicians . In ordinary practice , it is true that their effects have not often been noticed ; but it is impossible to doubt ...
Página 25
... resultant has the direction OD , and has exactly the magnitude already given by the above formula . Thus , regarding the fluid contents of the pipe EFGH in Fig . 15 as forming the voussoir of a fluid arch , the planes EG and FH ...
... resultant has the direction OD , and has exactly the magnitude already given by the above formula . Thus , regarding the fluid contents of the pipe EFGH in Fig . 15 as forming the voussoir of a fluid arch , the planes EG and FH ...
Página 26
Thomas Claxton Fidler. upon the familiar principles of the parallelogram gives for their resultant the same value- 0 S = 2P tan 2 It follows , then , that one method of equilibrating the dis- placing forces upon a polygonal bend would be ...
Thomas Claxton Fidler. upon the familiar principles of the parallelogram gives for their resultant the same value- 0 S = 2P tan 2 It follows , then , that one method of equilibrating the dis- placing forces upon a polygonal bend would be ...
Página 29
... resultant of the unbalanced pressures on the internal surfaces of the bend AB , will therefore have the magnitude- 0 N = 2pd3 . sin 4 2 ( 5 ) 0 = 2P sin ( 5a ) 2 chord AB = P X radius OA ( 56 ) If we proceed to subdivide the segmental ...
... resultant of the unbalanced pressures on the internal surfaces of the bend AB , will therefore have the magnitude- 0 N = 2pd3 . sin 4 2 ( 5 ) 0 = 2P sin ( 5a ) 2 chord AB = P X radius OA ( 56 ) If we proceed to subdivide the segmental ...
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Términos y frases comunes
annular axial fluid stress axial stress axis B₁ bending moment bending stresses bolts buoyant pressure calculated centre of action centre of buoyancy centre of gravity centre-line CHEMISTRY compressive stress Crown 8vo curvature curve of buoyancy cylinder denotes depth diagram of moments direction displacing force distance elastic engineering equal equilibrium example expansion joint external load feet flange floating fluid arch fluid pressure foot lineal girder horizontal hydraulic pressure hydrostatic hydrostatic pressure Illustrations inclined internal diameter length lifting force longitudinal stress mean head measured metacentre metacentric height metal moment of inertia ordinates P₁ P₂ parabolic pipe plane pontoon proportional quantity radius represent resistance right angles segmental side sketched in Fig socket square inch straight line stress Q surface tensile strength tensile stress thrust tons transverse triangle tube uniform uniformly distributed vertical component vertical force vessel water-line wedge weight
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