静水总压力
第2章
2.3 解:
'p,p,p,227,95,132kPa(1) a
132 Pat,,1.3469at98
'p,p,p,95,70,25kPa(2) va
p25vh,,,2.55m 水柱高 v,g9.8
AA h1 h1ρgh 1h ρgh 21h h 32ρgh 2
ρgh 3BB
(a)(b)ρgh 22.4 解:
2.8 解:
p,,gh,p,,gh,,gh AABBp
,,p,p,,gh,h,,gh ABABp
,,,g,,,h,1,,gh, p
AAh 1ρgh h1
ρghh
Rh 2ρg(h-h2) ρg(h+R)BBρg(h-h2) (c)(d)
,,,,9.8,0.36,1,133,28,0.36,
,34.6528 kPa
2.12 解: T 静水总压力:
1,60?,,P,,g2L,Lsin60,L,B 1ρgL1sin60.12LAP
L
ρg(L1+L)sin60.
1,,,1,9.8,2,2,2.5,sin60,2.5,1.5,, 2
,103.4329kN
L,,,h,L,sin60,2.8146m 或: ,,C12,,
P,,ghA,1,9.8,2.8146,1.5,2.5,103.4359kN C
合力作用点距A点的距离:压力中心至闸门底边的距离:
,,,,LLgsin60,2,L,Lgsin60,,,,11 L,L,A,,,,,,3L,gsin60,L,L,gsin6011
,,2.52.0,22.0,2.5,,,2.5,,,1.4103m ,,32.0,2.5,2.0
ICx,,yy 或:压力中心的位置: DCyAC
13,1.5,2.52.5,,12,2.0,,,,2.52,, ,,2.0,,1.5,2.5,,2,,
,,,3.4103m
,PL,TLcos60,M,0 : AA
PL103.4359,1.4103A,,T,,,116.6969kN ,,Lcos602.5,cos60
2.15 解:
受力示意图:
ρgh1
h1
zh2
Oρg(h1+h2)
x0
bb12
(1)力
1122 P,,gh,,9.8,10,490kNx1122
P,01z
112222 ,,,, P,,gh,h,h,h,b,,9.8,2,10,40,10,1521121122
,5300.16kN
,,,,b15,,1,,,, P,Pcosarctan,5300.16,cosarctan,4646.72kN,,22x,,,,h40,,2,,,,
,,,,b15,,1,,,, P,Psinarctan,5300.16,sinarctan,1861.01kN ,,22z,,,,h40,,2,,,,
P,P,P,490,4646.72,5136.72kN x1x2x
P,P,P,0,1861.01,1861.01kN z1z2z
22P,P,P,5463.44kN xz
(2)对o点的矩
P的矩: 1
21,,,,M,P,h,h,h,490,,10,40,21233.33kN,m ,,,,P1x121133,,,,
P至坝趾的斜距: 2
2222h,b,,2h,h,,10,152,10,402112 L,,,7.2111m P2,,,,3h,h310,4012
P的矩: 2
,,,,15,,,,,,sinarctan ,,MPbbL,,,,P212P2,,240,,,,,,
,,15,, ,,,5300.16,15,40,sinarctan,7.2111,64135.59kN,m,,,,40,,,,
,M,M,M,85368.92kN,m PP12
2.17 解:
PzPzPzPz
(d)(a)(b)(c)
2.19 解:
解法一:
水平分力:
2211,, ,,,,P,,grsin45,b,,9.8,2,sin45,4,39.2kNx22
铅直分力:
224514512,2,2 ,,,,A,,r,r,sin45,,,,2,2,sin45,0.5708mP236023602
P,A,b,,g,0.5708,4,9.8,22.3752kN zP2
22P,P,P,45.1364kN xz
P22.3752,,2,,arctan,arctan,29.7175,2943'3'' P39.2x
解法二:
水平分力:
11, h,rsin,,,2,sin45,0.7071c22
,2h,brsin,,4,2,sin45,5.6568m c
P,,ghA,9.8,0.7071,5.6568,39.2kN xcx
铅直分力:
11,,2P,,gV,9.8,,r,rcos,,rsin,b z,,82,,
111,,22,9.8,,,,2,,2,,4,22.3752kN ,,822,,
其余同解法一。