Question

2 A plane rectangular surface is submerged vertically in a liquid with its upper edge below the surface and parallel to it. Develop from first principles an expression for the depth to the centre of pressure A vertical bulkhead has a door 2m high and 1 m wide fastened by two hinges situated 15cm below the top and 15 cm above the bottom of one vertical edge, and by a bolt in the centre of the other vertical edge. Calculate the forces on each hinge and the bolt when one face of the bulkhead is subjected to water pres- sure, the water surface being 1m above the top of the door

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Answer 1

WIKI - - fх - "B,D هو ما ملل TUTTI I TOT Front view side view Take small element of thickness doe & width W at distance & from top edge of Rectangular plate. plate dimentions - HXW distance of upper edge from free surface =a ressure at element = sg69+x) ostatic force on element= Pressure . X Area dF= sgcatx) w dze H so total hydrostatic force on element, F= Sdf = seg(atxdz w F = [ęgnat sgt²] W

F= sg HW [a+ ] How centre of pressure is a point on surface through which This resultant tore F due to pressure passes. This can be found out ash =s@+9df ñ= S sgcarxWdx IgHW [a+H/

ñ = S (az+ 2ax +32) doc H [a+ H/2] h = 92H+ 294² - 4 H[atti ū= a?+ 2H+633 [a+H12] - H312 + [a+H/2] [a+m)* Mow see the numerical. H= 2m Weim a=im a 0-15m -- F→ Force on Istninge Fato force on and Hinge 0.15m 2 2 e Il F3+ Force on For Force on bolt ñ = 12+1x 2+ 233 e = 13 m 1+212 -1 W I - Fi h H F2 o torce on 0.15m Mow Fi+Fz+F3 = F for equilibrium of gate. F=sg Hw[a+H4/23 = 39.240KN

- 4 - -Naim 0.154 N 1.7m - AⓇFI om F-- - - BLOOF OF F3 Im in 0.5m 0.5m Now Take moment about vertical axis pop, Empezo for equilibrium . FX0.5 – F3X 150 F3 = 72 = 39,240 53= 19.62 KN Take moment about Horizontal axis NAN EMNN=0 for. - equilibrium, Fx x It Fix 0.15+ F2x (0.15+1.7) - FX716-0 19.62+0.156, +1-85 F2 = 45.78 - A FitF2 +F3 = F

From ean ③ kg we get Fi= 5.963kN F2= 13.657 KN
numerical values of F1 and F2 are correct in above solution.