1. A concrete cube 0.65m on each side is to be held in equilibrium under water by attaching a light foam buoy to it. Specic weight of concrete and foam are 2.5! "#$cu.m. and 0.%& "#$cu.m. respecti'ely. Assume unit weight of water ( &.%& "#$cu.m. 1. )hat is the the minimum minimum 'olume 'olume of foam requir required* ed* 2. )hat is the the weight weight of of the foam* foam* . )hat is the the total weight weight of the the concrete concrete and foam* foam*
S+,-/+# 1. inimum inimum 'olume 'olume of foam requir required ed )1 )2 3 41 42 70.%&8 0.65 70.658 70.658 72.5!8 3 7&.%&8 0.65 70.658 70.658 7&.%&8 3 0.92 cu.m
2. )eight eight of of the the foam foam ) 3 0.%& 70.928 ) 3 0. "#
. otal weight weight of the concre concrete te and foam otal otal weight 3 0. 0.65 70.658 70.658 70.658 72.5!8 72.5!8 otal otal weight 3 6.!1 "#
2.A bloc: of wood 0.20 m thic: is ;oating in sea water. water. he specic gra'ity of wood is 0.65 while that of sea water is 1.0. ind the minimum area of a bloc: which will support a man weighing !0 :g. Solution:
<=' 3 0> 4 3 )man )wood ?wwood3 )man ?wood wood 71000 :g$m @ 1.08 wood3 !0 71000 :g$m @ 0.658 wood wood 3 0.2105 m 3 Area @ 0.2 Vwood= 1.05 square meter
.he section of a concrete gra'ity dam shown in the gure. he depth of water at the upstream side is 6 m. #eglect hydrostatic uplift uplift and use unit weight of concrete equal to 2.5 :#$m. oeBcient of friction between the base of the dam and the foundation is 0.6. Cetermine the following 7a8 factor of safety against slidingD 7b8 factor of safety
against o'erturning and 7c8 the o'erturning moment acting against the dam in :#Em.
3 ?whA 3 &.!1 :#$m 7876@18 3 1%6.5! :#
1
F 3
3
768
F 3 2 m.
)1 3 ?c 1 )1 3 2.5 <27!8718> )1 3 %6 :#
)2 3 ?c 1 1
)2 3 2.5 <
2
7287!8718>
)2 3 1!! :#
1
G1 3 9 E
2
728
G1 3 m
2
G2 3 7
3
8 728
G2 3 1. m
H@ 3 3 1%6.5! :#
Hy 3 )1 )2 H@ 3 %6 :# 1!! :# H@ 3 569 :#
Ss 3
μRy Rx
(
0.6 564
Ss 3
)
174.58
FSs = 1.916 (factor of safety against sliding
H 3 )1@1 )2@2 H 3 %6 78 1!! 71.8 H 3 1%!.601 :#Em
+ 3 @ y + 3 1%6.5! 728 !" = #5#.16 $%&m (o'erturning moment
So 3
RM OM 1378.601
So 3
353.16
FSo = #.90 ( factor of safety against o'erturning
9.
a. ind the appro@imate height of wa upstream of the dam of the headwater // metersD such that an air bubble. upon reaching e water surface has a a@le ones an rl had at the bottom* b. ompute the absolute pressure at the bottom of the darn c. ompute the gage reading at the bottom of the dam.
Solution a )y )oyle*s +aw. Iam 3101. :Ia I113I22 I1 101. &.!1J 7abs8 I23101. 7abs8 then 7101. &.!1J8 3 101.78 & !1J 3 101.78E 101. ,= -0.65m b /solute ressure: I 3 101. &.!1720.658 I 3 0.!! :Ia c age reading at t2e /ottom of t2e dam I30.!!E101. I 3 202.5! :Ia
5.
A 9! inch diams steel pipe 1$9 /nch thic:D carries oil of ofsp gr. 3 0.!22 under a head d 900 ft. of oil a. ompute the pressure inside the pipe /n psi. b. ompute the stress in the steel required to carry a pressure of 250 psiD with an allowable stress of 1!000 psi. c. ompute the thic:ness of steel required toK carry a pressure of 250 psiD with an allowable stress of 1!000 psi.
Solution a 3ressure inside t2e ie in si. 23 23 IC718 3 Ss 7t8718 27Ss8t 718 3IC718 Ss 3 IC 2t I 3 Fw h I 3 62.9 70.!22879008 I 3 2051% psf. 3= 1-.5 si. b Stress in t2e steel: Ss 3 IC 2t Ss 3 192.5 79!8
271$98 Ss =1#640 si. c: 2ic$ness of steel: Ss3 IC 2t 1!000 3 250 79!8 2t t= 0.### m.
6. A wooden buoy of sp.gr. 015 ;oats in a liquid with sp.grD of 0.!5 a. )hat is the percentage of the 'olume abo'e the liquid surface to the total 'olume of the buoy*
b. /f the 'olume abo'e the liquid surface is 0.0195mD what is the weight of the wooden buoy* c. )hat load that will cause the buoy to be fully submerged*
Solution: a: of 'ol. a/o'e liquid surface: ) 3 4 0.%5 7&.!183 &.!170.!58 2 3 1.1 2 2 3 0.!!2 1 3 1 (2 13 E 0.!!2 13 0.11! 1 30.11! V1 = 11.4 V /:7eig2t of wood: ) 3 7&.!1870.%58 13 0.11! 0.0195 3 0.11! 3 0.12 m ) 3 0.127&.!1870.%58 7 = 0.905
%$ c: +oad to cause t2e wood to su/merged: I3 1 7&.!1870.!58 I 3 0.01957&.!1870.!58 3 = 0.1-1
%$8. An /ceberg 7y.t.&E:#$rna8 ;oats in ocean water 7y3 10 :#$m8 with D000 m of the iceberg protruding abo'e the free surface. a. )hat is the 'olume of the iceberg below the free surface* b. )hat is the total 'olume of the iceberg* / what is the weight of the iceberg*
Solution a: Volume of t2e ice/erg /elow t2e free surface: w3 4 7 00087&8 3 7108 V= -8000 m# /: otal 'olume of t2e ice/erg: otal 'olume 3 2%000 000 otal 'olume = #0000 m# c: 7eig2t of t2e ice/erg: )eight of iceberg 30D0007&8 7eig2t of ice/erg = -80000
%$!.An open -Etube ha'ing a diameter of 10mm. contains mercury at depth of 120mm. Specic gra'ity of Jg is 1.6. /f 0.012 ,. of water is poured in the rightEhand leg. A.8 4.8
)hat is the height of water in the tube* )hat is the ultimate height of the liquid in the left side*
.8 )hat is the ultimate height of the liquid in the right side*
soln
gure
A.8 12ml3 0.012, 0.012 1000
3 0.000012 cu.m 3
0.000012 ( 1000 ) 3
12D000
cu.mm ol3Ah 12D0003
π
2
( )( 10) 4
h
h3152.!! mm. 4.8 − L
40
1.67&.%&8
1000
−
(
13.6 9.79 1000
) L
−
(
9.79 152.8 1000
1.67290E,8 ( 1.6 , ( 152.! 3 0 269 ( 1.6, ( 1.6, E 152.! 3 0 ,3 119.9 .8 290E, 3 126.6 7left side8 h , 3 152.! 119.9 3 26%.2mm 7right side8
)
3 0
&.An open tan: 1.!2m square weights 925 # and contains 0.&1m of water. /t is acted by an unbalanced force of 10900 # parallel to a pair of sides. )hat is the force acting in the side with the smallest depth*
Solution
I3 7-#/ )L/MJ87h bar87A8 Sol'e for a and y 3a310900 3)ALH A#" 31000771.!2871.!287.&188925$&.!1 36.92 :g 109003 6.927a8 a3 .0&2m$s 2 tan
∅
3 a$ g 3 y$.&1
I 3 &!10
7 .62$287.62871.!28 .0&2$&.!13y$.&1 7 A#S)LH8 y3 .2& m N .&1 m 7 o:8
I3 92 #
h 3 .&1Ey
10.A di'er and his suit weigh !&0 #. /t requires 10 # lead to sin: him in fresh water. /f the specic gra'ity of the lead is 11.D what is the 'olume of the di'er and his suit*
4 3 ?)C !&0 # 10 # 3 1 7&!10 #$m8 2 7&!10 #$m8
W 2
2 3
D L 130 N
2 3
(
11.3 9810
N
)
3
m
2 3 0.0011% m
!&0 # 10 # 3 1 7&!10 #$m8 0.0011% m 7&!10 #$m8 1 3 0.10 m
11.A rectangular tan: of internal width of 5 mD as shownD contains oil of sp. gr. 3 0.! and water. 7a8 ind the depth of oilD h. 7b8 /f a 1000E# bloc: of wood is ;oated in the oilD what is the rise in free surface of the water in contact with air*
Solution
7a8 Cepth of oil 7Hefer to igure a8 SumEup pressure head from oil surface 718 to water surface 728 in m of water. I1$F h 70.!8 E93 I 2$F 0 0.!h E 130 J3 1.-5 m 7b8
Hise of the water surface 7Hefer to igure b8 43 )
FoilC 3 w 7&!10 @ 0.!8 C 3 1000 C3 0.1-8 m# Since the 'olume of oil remain unchangedK oil7initial8 3 oil7nal8 70.58 758 71.258 3 70.58 758 7hO8 ( 0.12%9 hO3 1.01 m As shown in gure bD if the oilEwater interface drops by a distance of yD the free surface of water will rise by y$2D since the crossE sectional area of the right compartment is twice that of the left compartment.
SumEup pressure head from oil surface to water surface in m of water
0 1.01 70.!8 7Ey8 E9 E y$2 3 0 1.090! ( 1 Ey$2 3 0 y$2 3 0.090! y$2 3 0.016 m or 1.6 mm
hereforeK the free surface of water will rise 1#.6 mm.
12. A 'ertical triangular surface of height d and horiPontal base width b is submerged in a liquid with its 'erte@ at a liquid surface. Cetermine the total force acting on one side and its location from the liquid surface.
Solution
-sing the Iressure diagram
3 QRA diagram 2
R 3
3
1
d
K A3 2
3Q7 1
F=
3
3
1
d
Ɣbd
87
2
,ocation of
I g
e 3 A ȳ ȳ
2
3R3 bd
3
d
3
36
e3
d
( 1 bd )( 2 ) 2
e3
y p
2
3
d 12
3 R e
2
olume 1
bd
bd
3
3
3
(b x Ɣ d )( d )
1
8
F=
of
3
Ɣbd
2
the
pressure
y p
y p
2
3
3
=
d+
d 12
3d 4
1
F is located at t2e centroid of t2e diagram w2ic2 is
4
of t2e
altitude from t2e /ase. 1.A ft. diameter log 7sp.gr. 3 0.!28 di'ides two shallow ponds as shown. 1. ompute the net 'ertical reaction at point if the log is 12 ft. long. 2. ompute the horiPontal reaction at point . . ompute the direction of the resultant force with the horiPontal at point .
S+,-/+# 1. #et 'ertical reaction at point . 1 3 )
( π ) (1.5 )
2
1 3 62.9
2
7128
1 3 2696 lb.
( π ) (1.5 )
2
2 3 62.9
4
7128
2 3 12 lb. ) 3 T71.5827128762.9870.!28 ) 3 990 lb. H F 1 2 3 ) H F 2696 12 3 990 H F 3 %1 lb. 7up8 2. JoriPontal reaction at point . I1 3)hA I1 3 62.971.58787128 I2 3 )Ja
I2 3 62.970.%5871.587128 I2 3 !92 lb. HG 3 I1 ( I2 HG 3 %0 ( !92 HG 3 252! lb. 7to the right8
. Cirection of the resultant force with the horiPontal at point . an U 3
Ry Rx
371
an U 3
2528
V 3 !.5W
PROBLEM: 14 A dam is triangular in cross-section with the upstream face ertical! "ater is flushed with the top! #he dam is $m high % &m wide at the 'ase % weighs (!4 tons per cu'ic meter! #he coefficient of friction 'etween the 'ase % the foundation is )!$! *etermine the factors of safet+ against oerturning % against sliding!
,olution: ,p!gr! of conc ,conc. /conc / w ,p!gr! of conc ,conc. /conc / w ,p!gr! of conc ,conc. (!4 0onsider 1m length of dam " . /c " . 2/32(!43215(32&32$32136 " . 78!& / 9 . / h A . 2/32432$3213 9 . (/ R; . P . (/ R< . " . 78!&/ RM . "243 . 278!&/3243 RM . ()!4/
OM . P2$53 . 2(/32$53 OM . 2$7!/3 = . RM > OM R<
= . ()!4/ > $7!/ 78!&/ ; . (!71? m @ '5( 9,O . R< R; 9,O . 2)!$3278!&/3 (/ FSO = 1.44 9,, . RM OM 9,, . ()!4/ $7!/ FSS = 2.7
15. An object weighs 4 N in water and 5 N in alcohol having a sp. Gr. of 0.80. Assume
unit weight of water is 9.9 !N"cubic meter. #. $ind the volume of the object. %. $ind the densit& of the object. '. $ind the mass volume of the object. (olution)
Volume of the object
Note) * + * ,air- * ,fluid*+/$# *,air- 0.004 + 9.9 ,- 1111 Eqn. 1 *+/
%$ *,air- 0.005 + 9.9 ,.080- ,- 1111 Eqn. 2 23n. # to 23n. %) V = 0.0005107 cubic meter. Density of the object
*,air- + 9.9,- 0.004 *,air- + 9.9,0.0005#0- 0.004 *,air- + 9 N"9.8# m"s% ass + 0.9#4 !g 6 + ass"olume 6 + 0.9#4!g"0.0005#0m' ρ =1796.42 k!m"
#$ss Volume
s + #"6 s + #"#97.4% !g"m' Vs = 0.000557 m"!k
16.A 1.&0 m diameter closed cylinderD 2.%5 m high is completely lled with oil ha'ing sp. gr. +f .!0 under a pressure of 5 :g$ cm 2 at the top. 7A8 )hat angular speed can be imposed on the cylinder so that ma@imum pressure at the bottom of e tan: is 19 :g$cm2* 748 compute the pressure force e@erted by oil on the side of the tan: / :g.
Solution
-nit weight of oilD/. 1)))2!$3 3!00 :g$m 30.000! :g$cm 2a3 h . w(r( (g ,ole for h: p15/.75!)))$.&(7) cm h( .&(!7 m . p(5 X . 145)!)))$ . 187)) cm. 187 m h . h(- (!87 > p15 Y h . 187 > (!87 > &(!7 h . 1)?!87
1)?!87 . Y( 2)!?73( 5 (2 ?!$13 Y . 4$!$4 rad5sec = )5 pie Y . 4&&!44 rpm 9.XhA 9.$))218!&(73( π ( 0.95 )( 2.75 )¿ 9 . (!($=1)(
17. A vessel 'm in diameter containing %.4m of water is being raised. ,a-$ind the
pressure at the bottom of the vessel in !a when the velocit& is constant and ,b- find the pressure at the bottom of the vessel when it is accelerating 0.7m"s:% upwards. (;<=>?;N() $or vertical motion) p @+h,#a"gh+%.4m ,a- *hen the velocit& is constant a+0 then p @+h p @+9.8#,%.4p @+%'.544 !a ,pressure at the bottom,b- *hen a+0.7m"s:% ,use BC for upward motionp @+h,#a"gp @+ 9.8#,%.4-,#,#,0.7"9.8#--p @+ %4.9844 !a
1$!An open c+lindrical tan haing a radius of )) mm and a height of 1!( m is full of water! Cow fast should it 'e rotated a'out its own ertical a=is so that 87D of its olume will 'e spilled out ,olution
2
w r h. 2 g
2
since 87D of the total olume is spilled out the para'oloid will 'e formed a part outside the essel 2i!e! with its orte= 'elow the tan3 v spilled= v air =0.75 [ ԓ r v air = 0.9 ԓ r
But
2
( 1.2 ) ]
2
v air = v big paraboloid −v small paraboloid
)!? ԓ r 2
1
2
. 2
2
ԓ r
2
h
1
-
2
2
ԓ x y
2
1!$ r =r h − x y
EF! 1
B+ sFuared propert+ of para'ola:
2
2
2
x r r = ; x 2= y y h h
EF! (
Gn EF! 213 2
r 1!$ r =r h − h y ( y ) 2
2
2
1!$h. h − y But +.h-1!( 2
1!$h. h
2
−( h −1.2)
1!$h. h −( h −2.4 h + 1.44 ) 2
2
)!&h.1!44 C.(!4m 9inall+:
( 0.3) (!4. 2 ( 9.81 ) w
h
2
2
2
rad 30 x ".((!$8 sec ԓ
". 218.3rpm
multipl+ 'oth sides '+
r
2
1&.his 'ertical pipe line with attached gage and manometer contains oil and mercury as shown. he manometer is open to the atmosphere. here is no ;ow in the pipe. 7a8 )hat will be the gage pressure reading at A* 7b8 L@press the pressure reading at A in bar. 7c8 L@press the pressure reading at A in terms of head in meters.
7a8 I4 3 IA 7&.!1870.&08 I 3 0 0.%57&.!1871.5%8 I4 3 I IA 26.9!% 3 9&.&21 IA 3 2.9 "Ia
23.43
7b8 I 3
100
I 3 0.29 bar
7c8
P Ɣ 3
23.43
(
9.81 0.90
)
P Ɣ 3 2.65 m.
()! A ft. diameter log 7sp.gr. 3 0.!28 di'ides two shallow ponds as shown. 9. ompute the net 'ertical reaction at point if the log is 12 ft. long. 5. ompute the horiPontal reaction at point . 6. ompute the direction of the resultant force with the horiPontal at point .
S+,-/+# 9. #et 'ertical reaction at point . 1 3 )
( π ) (1.5 )
2
1 3 62.9
2
7128
1 3 2696 lb.
( π ) (1.5 )
2
2 3 62.9
4
7128
2 3 12 lb. ) 3 T71.5827128762.9870.!28 ) 3 990 lb. H F 1 2 3 ) H F 2696 12 3 990 H F 3 %1 lb. 7up8 5. JoriPontal reaction at point . I1 3)hA I1 3 62.971.58787128 I2 3 )Ja I2 3 62.970.%5871.587128 I2 3 !92 lb. HG 3 I1 ( I2 HG 3 %0 ( !92 HG 3 252! lb. 7to the right8 6. Cirection of the resultant force with the horiPontal at point . an U 3
Ry Rx
