water displaced.
(14' - 13') X .7854 X 27 = volume of the cylinder walls.
13' X .7854 X J X 2 = volume of the cylinder ends.
.261 lb. = weight of a cubic inch of cast iron, then,
[(14' - 13') X .7854 X 27 + 13' X .7854 X ^ X 2] X .261 =
167 lb., nearly, = weight of cylinder. Since weight of
cylinder is greater than the weight of the water displaced,
it will sink. Ans.
(487) 2 lb. ā 1 lb. 5 oz. = 11 oz., weight of water.
1 lb. 15.34 oz. ā 1 lb. 5 oz. = 10.34 oz., weight of oil.
10.34 ^ 11 = .94 = Sp. Gr. of oil. Ans.
(488) Head = 41 h- . 434 = 94. 47 ft. Using formula 36,
Ā«;=.98i/2^/! =.98i/2 X 32.16 X 94.47 = 76.39 ft. per sec. Ans.
This is not the mean velocity, vā.
(489) (a) Use formula 39.
(2a = .815^|/27^, or
1.5' X .7854
(2a = .815 X ,,'. X |/2X 32.16X94.47 X 60 =
144
46.77 cu. ft. per min. Ans.
196 HYDROMECHANICS.
(d) See Art. 1005.
The theoretical velocity of discharge is t^ = ^%gTi, and as
// = 41 -^.434 = 94. 47 ft., we have 7/ = -^/2 X 32716 X 1)4.47
= 77.95 ft. per sec.
Using formula 31, Q ā A i\ and multiplying by 60 to
reduce the discharge from cu. ft. per sec. to cu. ft. per min. ,
we have
^ ^ā '^ f^/ X 77.95 X 60 = 57.39 cu. ft. per min. Ans.
144
(490) {b) Use formulas 31 and 38.
1 KĀ» v" 7854
Qz=Av= ^': X 77.95 = .9568 cu. ft. per sec. Ans.
144
{a) (2a= ā¢ 615 (2 = . 615 X . 9568 = . 5884 cu. ft. per sec. Ans.
a_.5884_
^'^ e"- 79568-'^^^- ^'''ā¢
(491) {a) 9 X 5 X .7854 = 35.343 sq. in. = area of base.
2' X .7854 = 3.1416 sq. in. = area of hole.
77ā tt-t; = rrr-r-r^ J hencc, the area of the base is less than 20
o5.o4o 11.25
times the area of the orifice, and formula 35 must be used.
v-^/J^ - 4 / ^x 32.16 x-e"- _
^-^ , a'~^ , (2' X .7854)^ " ^'^- ^"^^ "*
^-A^ ^- (9X5X.7854)- P^^ ^^^^^^
(<^) .434 X 6 = 2.604, or say 2.6 lb. per sq. in. Ans.
(492) 6 X 4 X .7854 = 18.85 sq. in. = area of upper
surface.
15' X .7854 = 176.715 sq. in. = area of base.
132
- ^-5- X 176.715 = 1,237.5 lb., pressure due to weight on
lo. o5
upper surface.
HYDROMECHANICS.
197
.03617 X 24 X 176.715 = 153.4 lb., pressure due to water
in vessel.
1,237.5 + 153.4 = 1,390.9 lb., total pressure. Ans.
(493) Use formula 31 or 32.
Divide by 60 X GO to get the discharge in gallons per
second, and by 7.48 to get the discharge in cubic feet per
second.
e X .7854
144
= 5.106 ft. per sec.
.. = Ā« =
Area in sq. ft. =
12,000 X 144
A 60 X 60 X 7.48 X 4^ X .7854
Ans.
(494) A sketch of the arrangement is shown in Fig. 45.
40,000 lb.
Pig. 45.
(a) Area of pump piston = (-\ x .7854 = .19635 sq.
Area of plunger = 10" X .7854 = 78.54 sq. in.
m.
198 HYDROMECHANICS.
Pressure per square inch exerted by piston = lb.
.19635
Hence, according to Pascal's law, the pressure on the
plunger is ^^ x 78.54 = 40,000 lb. Ans.
(d) Velocity ratio = li : .00375 = 400 : 1. Ans.
{c) According to the principle given in Art. 981, Px
1 ā inches = Wx distance moved by plunger, or 100 X
1 5- = 40,000 X required distance; hence, the required dis-
tance = ^^^ ^}}^ = .00375 in. Ans.
40, 000
(495) {a) Use formula 44, and multiply by 7.48 and
60 to reduce the discharge from cu. ft. per sec. to gal. per
min.
^ā = .41 ^/^[/J* - 4/^'] X 60 X 7.48 =
.41 X 12 X 4/64:32 rWo + ^V- 4/9^1 X 60 X 7.48 =
(d) In the second case,
(2a = .41 X^ X 4^64:32 rW9 + i|y- 4/9^1x60x7. 48 =
13,323 gal. Ans.
(496) {a) Area of weir = 14 X 20 ^ 144 sq. ft. Use
formula 32, and divide by^O X L^8 to reduce gal. per min.
to cu. ft. per sec.
Q 13,502X144 ,-.Ā«., A
'^m = -^ = TTx s-77; Ti 777: = I0.47 ft. per sec. Ans.
A 60 X 7.48 X 14 X 20 ^
... 13,323X144 ..^āār. A
(^> ^"= 60X7.48X14X20 = ^^-^^^^P^^^"- ^^ā¢
(497) (a) See Art. 997. .
J^=2 lb. 8|oz. =40.5oz.
w = 12 oz.
W = 1 lb. 11 oz. = 27 oz.
HYDROMECHANICS. 199
By formula 30,
W-ta 40.5-12 28.5 ,^ .
SP-Gr.=-^^,.;-^ = -^^^-^ = ā =1.9. Ans.
(^) 15 oz. = ^Ib. = .9375 lb. .9375^.03617 = 25.92 cu.
Id
in. = volume of water = volume of slate. Therefore, the
volume of the slate = 25.92 cu. in. Ans.
(498) In Art. 1019 it is stated that the theoretical
o 2
mean velocity is -^^ "Igh, Hence, ^;ā,= -|/2 X 32.16 X 3 =
o o
9.26 ft. per sec. Ans.
(499) {a) 4 ft. 9 in. = 4.75 ft. 19 - 4.75 = 14.25.
Range = 4/4 liy = |/4 x 4.75 X 14.25= 16.454 ft. Ans.
ip) 19 - 4.75 = 14.25 ft. Ans.
{c) 19 -T- 2 = 9.5. Greatest range = 4/4 X 9.5" = 19 ft.
Ans. (See Art. 1009.)
(500) Use formulas 46 and 50.
From the table, /= .0230 for z'ā = 4 and .0214 for v^ = 6.
.0230 -.0214 =.0016. 6-4 = 2.
4. 5397 -4 = .5397. Then, 2 : .5397::.0016 : jr, or x =
.0004. Hence, .0230 - .0004 = .0226 =/for vā, = 4.5397.
e = 60 X 60 X .09445 X 6' X / .0,^6 x 1,300 + i X 5 =
17,350 gal. per hr. Ans.
. (501 ) Obtain the values by approximating to those given
in Art. 1033. Thus, for vā, = 2, /= .0265; for vā, = 3,/ =
.0243; .0205 - .0243 = .0022. 2.37 - 2 = .37. Hence, 1 : 37
:: .0022 : ;r, or .r = .0008. Then, .0265 - .0008 = .0257 =/
for vā.= 2.37. Ans.
For z;,ā = 3,/= .0243; for z;ā = 4,/= .0230; .0243 - .02.30
= .0013. 3.19 -3 = .19. Hence, 1 : .19:: .0013 : jr, or 4r =
.0002. Then, . 0243 - . 0002 = . 024 1 = / for v^ = 3. 19. Ans.
aOO HYDROMECHANICS.
For z;ā = 4,/=.0230; forz/ā = 6,/= .0214; .0230 -.0214
= .0016. 5.8-4=1.8. 6-4=2. Hence, 2 : 1.8 :: .0016 :
;r, or;ir = .0014. Then, .0230 - .0014 = .0216 =/for v,^ = 5.8.
Ans.
Forz/,ā= 6,/= .0214; for z/^ = 8, /= .0205; .0214 -.0205
= .0009. 7.4 - 6 = 1.4. 8-6 = 2. Hence, 2 : 1.4 =.0009 : x,
or;i:=.0006. Then, .0214 - .0006 = .0208 =/ for ?/ā. =
7.4. Ans.
Forz/ā. = 8,/=.0205; forz/ā= 12,/= .0193; .0205 -.0193
= .0012. 9.83-8 = 1.83. 12-8 = 4. Hence, 4 : 1.83::
.0012 : ;r, or ;r = .0005. Then, .0205 ā .0005 = .02 =/ for
z/ā=9.83. Ans.
For v^ - 8, /= .0205; for v^ = 12, /= .0193; .0205 -
.0193 = .0012. 11.5-8=3.5. 12-8 = 4. Hence, 4 : 3.5
::.0012 : .r, or;r = .0011. .0205 ā .0011 = .0194 = /for z;,ā =
11.5. Ans.
(502) Specific gravity of sea-water is 1.026. Total area
of cube = 10.5' X 6 = 661.5 sq. in. 1 mile = 5,280 ft. Hence,
total pressure on the cube = 661.5 X 5,280 X 3.5 X .434 X
1.026 = 5,443,383 lb. Ans.
(503) 19' X .7854 X 80 = 22,682 lb. Ans.
PNEUMATICS.
(QUESTIONS 504-553.)
(504) The force with which a confined gas presses
against the walls of the vessel which contains it.
(505) (a) 4 X 12 X .49 = 23.52 lb. per sq. in. Ans.
(d) 23.52 -f- 14.7 = 1.6 atmospheres. Ans.
(506) {a) A column of water 1 foot high exerts a pres-
sure of .434 lb. per sq, in. Hence, .434 X 19 = 8.246 lb. per
sq. in., the required tension. A column of mercury 1 in. high
exerts a pressure of ,49 lb. per sq. in. Hence, 8.246 -^- .49 =
16.828 in. = height of the mercury column. Ans.
(d) Pressure above the mercury = 14,7 ā 8.246 = 6.454
lb. per sq. in. Ans.
(507) Using formula 63, A = ~= ^^^'^ ^ j^^ ^ ^ =
17.64 lb. Ans.
(508) {c) Using formula 61 ,
.3705^ IV T^ .37052 X 7. 14 X 535 ^ ^^^^^ ^^ ^^ ^^^
/ 22.05
{T= 460Ā° + 76Ā°=;535Ā°, and/ = 14.7 X 1.5 = 22.05 lb.
per sq. in.)
(a) 7.14 -^ .08 = 89.25 cu. ft., the original volume. Ans.
{d) If 1 cu. ft. weighs .08 lb., 1 lb. contains 1^.08 =
12.5 cu. ft. Hence, using formula 60, />V= .37052 T, or
T = ^ ^ = ^'^-^^ X 1'^-^ = 743.887Ā°. 743.887 - 460 =
.37052 .37052
283.887Ā°. Ans.
202 PNEUMATICS.
(509) Substituting in formula 69, / = 40, / = 120, and
/ = 55
' ' ^/4G0+55\ 40 X 515
= 35.517 Ibr Ans.
580
(5 1 0) Using formula 6 1 , / F = . 37052 W T, or
W = ,J^!^ ^ . T = 460Ā° + 60Ā° = 520Ā°.
Therefore, IV= Jtl ^ Ln = -076296 lb. Ans.
' .370o2 X 520
(511) 175,000 ^ 144 = pounds per sq. in.
(175,000 -^ 144)-=- 14.7 = 82.672 = atmospheres.
Ans.
(512) Extending formula 63 to include 3 gases, we have
PV = J>,v^-\-/>^v^-{-p^v^, or4=0xP= 1 X12+ 2x10 -f 3x8.
Hence, P=^ = IA atmos. = 1.4 X 14.7 = 20.58 lb. per
40
sq. in. Ans.
(513) In the last example, PV=5Q. In the present
23 56 56
case, P= ā -r; atmos. Therefore, V = -p = -^tt- = 35.79
cu. ft. Ans. 14?r
(514) For / = 280Ā°, r=740Ā°; for ^ = 77Ā°, ^=537''.
p r=. 37052 WT, or W=-^^^^. (Formula 61.)
w ā¢ u. fu ^ ā 14.7X10,000 ;,on-,Qiu
Weight of hot air = ā^^^- ā ā = 536.13 lb.
^ .37052 X 740
yjT ' U^ t ā¢ A- ^ A 14.7X10,000 ^oo 01 IK
Weight of air displaced = 3^052 x 537 ~ ^^^-^^ ^^ā¢
738. 81 - 536. 13 = 202. 68. 202. 68 - 100 = 102. 68 lb. Ans.
(515) According to formula 64,
PK=(^+^)7-,or
-x- = (^^li^ + ^-)ā ā ā¢ .
PNEUMATICS. 203
Therefore, 7"== āāāāā - = 439.35Ā°. Since this is less
I.411I00
than 460Ā°, the temperature is 460 ā 439.35 = 20.65Ā° below
zero, or ā 20.65Ā°. Ans.
(516) A hollow space from which all air or other gas
(or gaseous vapor) has been removed. An example would
be the space above the mercury in a barometer.
(517) One inch of mercury corresponds to a pressure of
.49 lb. per sq. in.
1 . .49
-ā inch of mercury corresponds to a pressure of 'āā lb. per
40 40
49
sq. in. '-ā X 144 = 1.764 lb. per sq. ft. Ans.
40
(518) (a) 325 X .14= 45.5 lb. = force necessary to
overcome the friction. 6 X 12 = 72" = length of cylinder.
72 ā 40 = 32 = distance which the piston must move. Since
the area of the cylinder remains the same, any variation in
the volume will be proportional to the variation in the
length between the head and piston. By formula 53,
rru c ^ A^, 14.7X40 _,2 ,,
pv=pv^. Therefore, / =:'^-^ā^= ā = 8.1- lb. per
sq. in. = pressure when piston is at the end of the cylinder.
Since there is the atmospheric pressure of 14. 7 lb. on one
2
side of the piston and only 8.1 -lb. on the other side, the
o
2
force required to pull it out of the cylinder is 14.7 ā 8.1- =
O
6.5- lb. per sq. in. Area of piston = 40' X .7854 = 1,256.64
o
sq. in. Total force = 1,256.64 x 6.5^- = 8,210.05. Adding
o
the friction, 8,210.05 + 45.5 = 8,255.55 lb. Ans.
{Jti) Proceeding likewise in the second case, pv =. p^ z/ā or
/ ='^J-^ = i^t^iA^ ^ ^g ^^ ,^g - 14.7 = 83.3 lb. per sq. in.
1,266.64 X 83.3 -f 45.5 = 104,723.612 lb. Ans.
204 PNEUMATICS.
(519) 8.47 = original volume = i-,. 8.47 ā 4.5 = 3.97
cu. ft. = new volume = v. By formula 53,
^ pv 3.97x38 ,^oioiu ā¢ A
p=<-~ = ā ā ā ā = 17.812 lb. per sq. m. An&
t'j 8. 47
(520) Original weight = \V= .5 lb. = 8oz. ; new weight
= ]V^ = 1 lb. 6 oz. = 22 oz. According to formula 56,
^iir ^ ijir J. ^^, 14.7x22 ,. .^. ,,
/ IF. =A W, or/, = -^ = = 40.420 lb. per
sq. in. Ans.
(521) Applying formula 58,
/460 + t\ 4,516 / 460 + 80 \ , ^^ ,, .
t' =v I ā *- 1 = I ā ^ ' ^^^ I = 1.96 cu. ft. Ans.
\460 -H / / 1,728 V460 + 260/
(522) According to formula 61,/ F=. 37052 W T, or
TJ7 /^ 14.7X1.25X55 . ^-^ i. .
^=:mE2T= .37052X548 = ^'^'^ ''^- ^ns.
(523) Using formula 63, PV = pv +/, z'ā or P x 7. 5
= 14.7 X 2 X 7.5 + 40 X 7.5, or P= 69.4 lb. persq. in. Ans.
(524) 48', 36', and 24' = 4', 3', and 2', respectively.
Hence, 4 X 3 X 2 = 24 cu. ft. = the volume of the block.
The block will weigh as much more in a vacuum as the
weight of the air it displaces. In example 510, it was found
that 1 cu. ft. of air at a temperature of 60Ā° weighed
.076296 lb. .076296 X 24 -f- 1,200 = 1,201.83 lb. Ans.
(525) (a) (See Art. 1088.) 127 + 16 = 143.
{^y X .7854 X 1 X 125 X 62.5 X 143 ^ ^^ ^^^^ ^ ^
00,000
14.9563 -^ .75 = 19.942 H.P. Ans.
{d) Discharge in gallons per hour = volume of cylinder
in cu. ft. X number of strokes per minute X 7.48 X 60 =
ivi) ^ '^^^^ ^ ^^^ X 7.48 X 60 = 24,784.3 gal. per hr. Ans.
(526) In this example, the number of times that the
pump delivers water in 1 minute is 100 -^ 2 = 50; in the last
example, 125. Hence, the number of gallons discharged
50
per hour in this case will be 24,786 X -tk^ = 9,914.4 gal. Ans.
125
PNEUMATICS. 205
(527) See Art. 1043.
30 23
Pressure in condenser = ā ā x 14.7 = 3.43 lb. per sq.
in. Ans.
(528) 144 X 14.7 = 2,116.8 lb. per sq. ft. Ans.
(529) .27 -7- 3 = .09 = weight of 1 cu. ft. Using formula
56,
/ W, =/, W, or 30 PT, = 65 X .09. IV^ = .195 lb. Ans.
(530) Using formula 61,
p V= .37052 fT r, or 30 X 1 = .37052 X .09 X T.
^ = o^A-o nn = 899. 6Ā°. 899. 6Ā° - 460Ā° = 439. 6Ā°. Ans.
.370o2x.09
(531 ) 460Ā°+ 32Ā° = 492Ā° ; 460Ā°+ 212Ā° = 672Ā° ; 460Ā°+62Ā°
= 522Ā°, and 460Ā° + ( - 40Ā°) = 420Ā°.
(532) Using formula 61,/ F=. 37052 W T, and
substituting,
(14.7X10) X4=. 37052X3.5 X r, or r=imf|^* =
453.417Ā°. 453.417 - 460Ā° = - 6.583Ā°. Ans.
(533) Using formula 63, V P=vp + v^p^, we find
o 15X63+19X14.7X3 ^, oi.. lu a
P= ār- =71.316 lb. Ans.
25
(534) Using formula 60,/F=. 37052 T, or P =
.3705 2X540 . . ,, . , .
-r = 20 lb. per sq. m., nearly, Ans.
(535) One inch of mercury represents a pressure of
.49 lb. Therefore, the height of the mercury column is
12.5 -i- .49 = 25.51 in. Ans.
(536) Thirty inches of mercury corresponds to 34 ft.
of water. (See Art. 1043.) Therefore,
30' : 34 ft. ::27' : x ft., or x = 30.6 ft. Ans.
A more accurate way is (27 X .49) -^ .434 = 30.5 ft.
206 PNEUMATICS.
(537) (a) 30 ā 17.5 = 12.5 in. = original tension of gas
in inches of mercury. 30 ā 5 = 25 in. = new tension in
inches of mercury.
VP=vp-\-v^p^ (formula 63), or 6. 7 X 25= 6. 7 X 12. 5+z', X 30.
6.7X25-6.7X12.5 ^ā_1 ,^ .
v^ = jr-r = 2.79 - cu. ft. Ans.
oO D
{d) To produce a vacuum of inches,
6.7X30-6.7X12.5 _ ^__ ,. .
v^ = āJ = 3. 908 cu. ft. Ans.
o\)
(538) 11 + 25 = 36, final volume of gas. 2.4-^36 =
=- lb. Ans.
lo
(539) Using formula 59,
,/460 + /A ,ā /460 + 300\ ,ā ^, ,,
m.
Ans.
(540) r= 460 + 212 = 672Ā°. UsingformulaGl,/ F =
.37052 WT, we have 14.7 X 1 = .37052 X H^X 672, or IV=
14.7
.37052 X 672
= .059039 lb. Ans.
(541) (a) ^^'^-"^^^^^^^ = 5.8178 cu. ft. = volume
of cylinder.
32 ā 26 = 6 in., length of stroke unfinished.
5.8178 X ;^ = 1.0908 cu. ft. Ans.
{b) By formula 61, taking the values of /, V, and T at
the beginning of stroke,
^rz Q^r^^o uz-r u^ P^ 14.7x5.8178
/r=. 37052 WT, or ^ - g^O-" .37052 X 535 =
.43143 lb. Ans.
{c) Now, substituting in formula 61 the values of F,
W, and T at time of discharge,
^ .37052 I^r .37052 X .43143 X 585 oi. ^..^ ,u
/= V = 170908 =^^-^^^ ^^- P^'
sq. in, Ans.
I>NEtJMATlC^. 20'}'
<542) Using formula 63, VP= vp-\-vJ^, or 30 X 35
"2r
= 19 X 12 + 21A. or A == 30X35-19X12 ^ ^^^^ ^^ ^^
sq. in. Ans.
(543) Use formula 64. PV=U^J^^^^T.
r= 460 + 72 = 532.
/ 13X45 nxCO X
V^20~ + ~540~r^^
Therefore, P=-^ ā ^ =26.723 lb. persq.
in. Ans.
(544) One inch corresponds to a pressure of .49 lb.
Therefore, the gauge will show 4.5 -^.49 = 9.18 + in.
See Art. 1043.
(545) Sp. Gr. of alcohol =.8. Therefore, 16 X. 434 X. 8=
pressure exerted by the column of alcohol. '-jr 'ā =
11.337 in. = height of a column of mercury that will give
the same pressure as 16 ft. of alcohol = number of inches
shown by the gauge. Ans.
(546) {a) 14.7 + 9 = 23.7 lb. per sq. in. Using
formula 53,
pv 14.7X80 ,. ^^.
distance between piston and end of stroke. Since the area
of the piston remains constant, the volume at any point of
the stroke is proportional to the distance passed over by the
piston. Hence, we may use the latter for the former in the
formula. 80 ā 49.62 = 30.38 in. Ans.
{b) Area of piston = 80' X .7854. The volume of air at
point of discharge is 80' X .7854 X 49.62 cu. in. =
80* X. 7854X49. 62 _, _ . -^ .
1;728 = 144.34 cu. ft. Ans.
(547) Using formula 56,/ W=p^ fF, or 3.5x14.7x2=
A X 13; hence,/. = 14-7 X 3.5 X 2 ^ ^ ^^^ ^ ^^ ^^^ ^ .^^
^^ Am.
208 PNE-aMATlCS.
(548) CO' ā 50' = 10'. Since the volumes are proper
tional to the lengths of the spaces between the piston and
the end of the stroke, we may apply formula 62,
iV^AZi, 14.7 X GO ^ /. X 10
T r, ' ^^ 460 + 60 460 + 130*
_,, - ^ 14.7 X COX 590 ,^^^^1, . .
Therefore, p, = -ā r r^r = 100. 07 lb. per sq. m. Ans.
''^' 520 X 10 r M
(549) T = 127Ā° + 460Ā° = 587Ā°. Using formula 60,
p V= .37052 r, or p^^ -37052 X 587 ^ ^^^^ ^^ ^^ ^^^^
A/ I
(550) T - 100Ā° + 4C0Ā° = 5C0Ā°.
Substituting in formula ei,/>V= .37052 W T, or
rr .37052 W T .37052 X .5 X 560
p 4,000
144
= 3.735 cu. ft. Ans.
(551) Use formula 64. PV=U^+^^T.
r= 110Ā° + 460Ā° = 570Ā°; T^ = 100Ā° + 460Ā° = 560Ā°; T, =
130Ā° + 460Ā° = 590Ā°.
/ 90 X 40 80 X 57 \^yQ
Therefore, V= \-^ ^^^ / = 67.248 cu. ft.
^^^ Ans.
(552) The pressure exerted by squeezing the bulb may
be found from formula 53, in which / is 14.7, v, the orig-
inal volume = 20 cu. in., and v^, the new volume, = 5 cu. in.
ā pV 14.7X20 r,o o ^u n.u a ^ *u
p,-=- ā = = = 58.8 lb. The pressure due to the
atmosphere must be deducted, since there is an equal pres-
sure on the outside which balances it. 58.8 ā 14.7 =44.1
lb. per sq. in. = pressure due to squeezing the bulb.
3'X .7854= 7.0686 sq. in. = area of bottom. 7-0686 X 44.1 =
311.725 lb. 7.0686 X .434= 3.068 lb. = pressure due to
weight of water. 311.725 + 3.068 = 314.793 lb. Ans.
(553) Use formula 58.
'4G0_+
,4C0 +
Z//4C0 + /A ,/460+115\ ,^ . .
STRENGTH OF MATERIALS.
(QUESTIONS 554-613.)
(554) See Arts. 1094, 1097, and 1096.
(555) See Arts. 1 1 02, 1 1 03, 1 1 1 0, and 1 1 1 2.
(556) See Art. 11 05.
(557) Use formula 67.
E = -Tā ; therefore, e =
A =.7854 X 2'; /=10X 12; /*= 40 X 2,000; i5"= 25,000,000.
Therefore, . ^ ,^^'7 ^A", n ,1 = -l^^'- ^ns.
' .7854 X 4 X 25,000,000
(558) Using formula 67,
^ PI 7,000 X 7i _- ^^_ o^_ - ,,
E ā -r- = ^^^/ ā ,.,., ^ ^,,,, = 29,708,853.2 lb. per sq. in.
A e .78o4 X (i) X .009 ' ' ,
' Ans.
(559) Using formula 67,
r, PI ā AeE 1^X2X .000 X 15,000,000
^ = :?7'^'^=ā r- = 93o2 =
2,500 lb. Ans.
(560) By formula 67,
P /Ā»/ , AeE .7854 X3'X.05X 1,500,000 __ ^^,
^=^.or/ = -^= Poo = 265.07.
' Ans.
(561) Using a factor of safety of 4 (see Table 24),
formula 65 becomes
210 STRENGTH OF MATERIALS.
p-^^. ā, 4-^^- ^X^X^-QQQ - 8727272 so in
P- -^, or A- ~^ - ^^^ .8727272 sq. in.
(562) From Table 19, the weight of a piece of cast iron
1' square and 1 ft. long is 3.125 lb.; hence, each foot of
length of the bar makes a load of 3.125 lb. per sq. in. The
breaking load- ā that is, the ultimate tensile strength ā is
20,000 lb. per sq. in. Hence, the length required to break
the bar is ^^ = 6,400 ft. Ans.
(563) Let / = the thickness of the bolt head;
d = diameter of bolt.
Area subject to shear = t: d f. ' '
Area subjected to tension = āizd'^.
4
S^ = 55,000. 5, = 50,000.
Then, in order that the bolt shall be equally strong in both
tension and shear, t: d t S^ = āTt d"^ S^,
o,, = ^ = ^.=|2qM00^.200'. Ans.
4:T:dS^ 4.S, 4 X 50,000
(564) Using a factor of safety of 15 for brick, formula
65 gives
15 ā
A = {2^X 3^) sq. ft. = 30 X 42 = l.,260 sq. in. ; S^ = 2,500.
Therefore, P= 1>^^0 X ^^^00 ^ ^10,000 lb. = 105 tons. Ans.
(665) The horizontal component of the force Pis Pcos
30Ā° = 3,500 X .866 = 3,031 lb. The area A is i a, the ulti-
mate shearing strength, S^, 600 lb. , and the factor of safety, 8.
STRENGTH OF MATERIALS.
211
Hence, from formula 65,
AS. 4:aS. aS.
a =
%P 2X3,031
8 8 2 ā¢ " 5,
(566) See Art. 1124.
600
10.1'. Ans.
ScaU of forces 1^160016.
Scale of distance 1^32^
Fig. 46.
(567) Using formula 68, with the factor of safety of 4,
^, 2/5, /5, , ^pd 2X120X48
^^=^ = -2"'"^^ = -^= 55,000 ā¢
212 STRENGTH OF MATERIALS.
Since 40^ of the plate is removed by the rivet holes, 60%
remains, cmd the actual thickness required is
_^^ 2X120X4 8^
.60 .00X55,000
(568) Using a factor of safety of G, in formula 68,
^ 6 3
Hence . _ 3/^ _ 3 X 6 X 200 _
Hence, ^ - 5 - ^0,000 " '^^ ' ^"'ā¢
(569) Using formula 71, with a factor of safety of 10,
9,600,000/'-"' nr-AAAA^''"
^= 10/^ =960,000^
'ā 'Ā«/ //^ _ ^.18/130 X 12 X 12 X 3
Hence, /= f -Ā£ - = f -
= .272".
960,000 ^ 960,000
Ans.
(570) From formula 70,
._ S^ or. - ^^ - ^>OOOXt _4,000_
/*-;-4_/'Ā°^^- ^^-^-2,800-2,000- 800-^- ^''^ā¢
(571) See Fig. 46. (a) Upon the load line, the loads
0-1^ 1-2, and 2-3 are laid off equal, respectively, to
F^, i%, and F^ ; the pole P is chosen, and the rays drawn in
the usual manner; the pole distance //= 2,000 lb. The
equilibrium polygon is constructed by drawing a c, c d, d e,
and e f parallel to PO, Pi, P 2, and PS, respectively, and
finally drawing the closing line/" a to the starting point n.
P in is drawn parallel to the latter line, dividing the load
line into the reactions in = P^, and 3 vi = R^. The shear
axis in n is drawn through in, and the shear diagram
h I . . . . s' n m \s constructed in the usual manner. To
the scale of forces in = 1,440 lb., and 3 in = 2,160 lb. To
the scale of distances the maximum vertical intercept j' =
^'^=31.2 ft., which, multiplied by N, = 21.2 x 2,000==
62,400 ft. -lb. = 748,800 in. -lb. Ans.
{d) The shear at a point 30 ft. from the left support =
in = l,UO\h. Ans.
{c) The maximum shear = ns' ā ā 2,160 lb. Ans.
STRENGTH OF MATERIALS.
213
(572) See Fig. 47. Draw the force polygon 0-1-2-3-U-5-0
in the usual manner, 0-1 being equal to and parallel to
/^,, 1-2 equal to and parallel to F^, etc. 0-5 is the resultant.
Scale of forces 2'='40lb^
,K Scale of distance 1=2^'
Fig. 47.
Choose the pole P, and draw the rays PO^ P 1^ P2, etc.
Choose any point, a on 7^ā and draw through it a line
parallel to the ray P 1. From the intersection b of this line
with F^, draw a line parallel to P 2; from the intersection
c of the latter line with /% produced, draw a parallel to
P S, intersecting F^ produced in </. Finally, through d,
draw a line parallel to P ^, intersecting F^ produced in (\
Now, through a draw a line parallel to P 0, and through
e a line parallel to P5; their intersection/ is a point on the
resultant. Through / draw the resultant P parallel to
0-5. It will be found by measurement that P = 05 lb., that
it makes an angle of 22^^" with ;// w, and intersects it at a
distance of 1;^' from the point of intersection of F^ and m ti.
(573) See Fig. 48. The construction is entirely similar
to those given in the text. 0-1^ 1-2, and 2-S are laid off to
represent /^ā F^, and F^ ; the pole P is chosen and the rays
214
STRENGTH OF MATERIALS.
drawn. Parallel to the rays are drawn the lines of the
equilibrium polygon a b c d g a. The closing line g a'\%
found to be parallel to Pi. Consequently, 0-1 is the left
reaction and 1-3 the right reaction, the former being 6 tons
2
Scale of forces 1=5 torts.
Scale of distance 1^5^
Fig. 48.
and the latter 3 tons. The shear diagram is drawn in the
usual manner ; it has the peculiarity of being zero between
F^ and F^.
(574) The maximum moment occurs when the shear
line crosses the shear axis. In the present case the shear
line and shear axis coincide with s t, between F and F ;
hence, the bending moment is the same (and maximum) at
F^ and i%, and at all points between. This is seen to be true
from the diagram, since k h and b c are parallel. Ans.
{Jb) By measurement, the moment is found to be 24 X 12 =
288 inch-tons. Ans.
(r) 288 X 2,000 = 576,000 inch-pounds. Ans.
STRENGTH OF MATERIALS.
(575) See Arts. 1 1 33 to 1 1 37.
215
(576) See Fig. 49. The force polygon 0-1-2-3-^-0 is
drawn as in Fig. 47, 0-4 being the resultant. The equilibrium
polygon ab c d g a g^^i^ ^f f^^^^^ JZ50 lb.
is then drawn, the Scale of distance I'^S"
point g" lying on the
resultant. The re-
sultant R is drawn
through g, parallel to
and equal to O-4. A
line is drawn through
C, parallel to R.
Through g the lines
ge and g/ are drawn
parallel, respectively,
to PO and P4, and
intersecting the par-
allel to R, through C
in e and / ; then, e/
is the intercept, and
Pu^ perpendicular to
0-4, is the pole dis-
tance. Pii = d3lh.;
^/=1.32'. Hence,
the resultant mo-
FiG. 49.
ment is 33 X 1.32 = 43.6 in. -lb. Ans.
,/. /.
(577) The maximum bending moment, M= W-^ (see
Fig. 6 of table of Bending Moments) = 4 X 2,000 X "^^ ^ ^ =
40,727tV ft.-lb. =488,727 in.-lb. Then, according to for-
mula 74,
ST
fc
= 488,727.
/ 488,727/ 488,727X8 ,.,,,ā,
7 = S, = 9,000 = ^'^^ā¢^^*-
21G
STRENGTH OF MATERIALS.
c \d
ā b d*^ and, according to the conditions
of the problem, b ^= ā d.
Therefore, ~=\bd^ = ^d' = 434.424.
d' = 5,213.088
d = nv. I
b = 81". f
Ans.
(578) The beam, with the moment and shear diagrams,
is shown in Fig. 50. On the line, through the left reaction,