Example 2.A Width is 1 m. Find: F and the position that the fluid pressure exerted. ΣF=0=Fy-Fy - W FH -2 m Fv tw Fx F 4m y Fy=pA = 9810*4*2*1 = 78.5 kN π 2 W = WV = y = y₁² 4 = 9810*0.25 **4*1 = 30.8 kN F₁ = 30.8+78.5 y F₁ = 109.3 KN *1 ΣF=0=FH-Fx Fx = PA X = 5*9810*2*1 Fx = 98.1kN

Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Y cp
FH
W
LIX
Fy
F
Example 2.A
4 m
x cp
x cp
Ycp = y +
xw
I
JA
XcpFy = Fy *1+W* xw
ср у
=
4r
3π
5+0.067 5.067m
=
= 0.957 m
0.849 m
78.5*1+30.8*0.849
109.3
Transcribed Image Text:Y cp FH W LIX Fy F Example 2.A 4 m x cp x cp Ycp = y + xw I JA XcpFy = Fy *1+W* xw ср у = 4r 3π 5+0.067 5.067m = = 0.957 m 0.849 m 78.5*1+30.8*0.849 109.3
Example 2.A
Width is 1 m. Find: F and the position that the fluid pressure exerted.
ΣF=0=Fy-Fy – W
FH
Fx
W
F
4m
y
Fy = PA
= 9810*4*2*1
= 78.5 kN
T
W = y = y ²¹ r² *1
4
= 9810*0.25*π *4*1
= 30.8 kN
Fy = 30.8+78.5
F₁, = 109.3 kN
ΣF=0= FH - Fx
X
Fx = pA
= 5*9810*2*1
F₂ = 98.1kN
X
Transcribed Image Text:Example 2.A Width is 1 m. Find: F and the position that the fluid pressure exerted. ΣF=0=Fy-Fy – W FH Fx W F 4m y Fy = PA = 9810*4*2*1 = 78.5 kN T W = y = y ²¹ r² *1 4 = 9810*0.25*π *4*1 = 30.8 kN Fy = 30.8+78.5 F₁, = 109.3 kN ΣF=0= FH - Fx X Fx = pA = 5*9810*2*1 F₂ = 98.1kN X
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