Fig. bo.
are laid off the loads in order. Thus, 0-i = 40 X 8 = 320 lb.,
is the uniform load between the left support and /', ; l-'2 is
STRENGTH OF MATERIALS. 217
F^ = 2,000 lb. ; ^-3 =40 Xl2 = 480 lb., is the uniform load
between F^ and F^\ 3-4- = 2,000 X 1.3 = 2, GOO lb., is F„ and
j^.5 = 40 X 10 = 400 lb., is the uniform load between F^ and
the right support. The pole P is chosen and the rays drawn.
Since the uniform load is very small compared with F^ and /%,
it will be sufficiently accurate to consider the three portions
of it concentrated at their respective centers of gravity
X, j\ and ^. Drawing the equilibrium polygon parallel to
the rays, we obtain the moment diagram a Ji b k c I d a.
From /*, drawing P in parallel to the closing line a d, we
obtain the reactions^ m andrnS equal, respectively, to 2,930
and 2,870 lb. Ans. The shear axis ;;/ x, and the shear dia-
gram r s t u V tt ;«, are drawn in the usual manner. The
greatest shear is m, 2,930 lb. The shear line cuts the shear
axis under F^. Hence, the maximum moment is under /'"",. By
measurement, £'^is64', andP^r is 5,000 lb. ; hence, the maxi-
mum bending moment is 64 X 5,000 = 320,000 in. -lb. Ans.
(579) From the table of Bending Moments, the great-
est bending moment of such a beam is — ^— , or, in this case,
o
w X 240'
8
By formula 74,
_ wx_240]_ _ 5/ _ 45^000 280
8 fc 4 12 ^ 2
^, f 45,000 X 280 X 8 ^^ no lu • u r
Therefore, zu = — -f — = 72.92 lb. per mch of
' 240 X 240 X 4 X ^
length = 72.92 X 12 = 875 lb. per foot of length. Ans.
(580) From the table of Bending Moments, the maxi-
mum bending moment is
From formula 74,
/=,^(^*-<*)=5G.945;r=:i-^=^=3i;5=38,000;/=G.
218 STRENGTH OF MATERIALS.
n o. I// 38,000 X 56.945
Hence, 2AIV= ^ ^ ^^^^ .
38,000 X 56.945 _
'^ 24X6X3.25 -*'^^*^^- ^^s.
(581) {^) From the table of Bending Moments,
From formula 74,
«;Xl92' _5y
^'^- ' 8 -A'
5, = 7,200;/=8; 7=1^^^ = ^; ^ = 1-^=5.
Then ^^ X 192' ^ 7,200 ^^ 2,000
jf; =
12 X 5
7,200 X 2,000 X 8
8 X 12 X 5 X 192 X 192
6.51 lb. per in. = 6.51 X 12 = 78.12 lb. per ft. Ans.
/A\ r 1 A ^3 10 X 2" 80 ,^
w X 192' _ 7,200 80
8 8 12 X 1
7,200 X 80 X 8 . ^ ,,
w = - — ^— -— — - = 1.3 lb. per m.
8 X 12 X 192 X 192 ^
= 1.3 X 12 = 15.6 lb. per ft. Ans.
(582) {(t) From the table of Bending Moments, the
K T///S
deflection of a beam uniformly loaded is — — ^ , . In
384 Ji I
Example 579, W= 874 X 20 = 17,480 lb.; /= 240', E =
25,000,000, and 7=280.
XX A a ^' 5X 17,480 X 240' ,^ . .
Hence, deflection s = 77-7^ — ^- \,.r. r.,^r. — ^t7T7: = -45 m. Ans.
' 384 X 20,000,000 X 280
1 IfV
{b) From the table of Bending Moments, s = — ^ , .
STRENGTH OF MATERIALS. 219
In Example 580, IF = 4,624 lb.; /=96 in.; ^=15,000,000,
and /= 50.945.
„ 4,624 X 96' -,» , A
"^"^"' ' = 48X15,000,000X56.945 = "^ ' ""^'"^>^- ^"^-
5 W/*
(f) .y = gg^ -^j-. In Example 581 (a), W= 78.12 X 16;
/=192; i5"= 1,500,000, and 7=^1^.
12
„ 5X78.12X16X192* ,,.,, .
^""^^' ^ ^ 384 X 1,500,000 XHF ^-''"' '^'''•
(583) Area of piston = ^7r^» = ^rx 14\
1
IV=- pressure on piston = - z x 14' X 80.
From the table of Bending Moments, the maximum
bending moment for a cantilever uniformly loaded is
^ ^ i^ ^ i-'X 14-^X80x4 ^ ^ / g^^ j^^_^^,^ ^^
&= 15^ = 4,500. L^lklf^^ a-,
f 10 c ^a 32
„ iJtXU'x80x4 4,500ira"
Hence, ^ = -^5—,
^^^.^ 14-x 80X4X32 ^
4 X 2 X 4,500
^ = ^^55775 = 3. 82'. Ans.
(584) Substituting in formula 76, 5, = 90,000; A =
6' X .7854; /==6; /=14x 12=168; ^=5,000; /=^X
6\ we obtain
4+77) t+^;^^^;^xi4p«:) ^ -
64
5
(585) For timber, 5, = 8,000 and /=8; hence, -^ =
8,000 , ^^^
-^-3— = 1,000.
220 STRENGTH OF MATERIALS.
Substituting in formula 65,
P= A -^ = 1,000 A.
. P 7 X 2,000 , , . ,.
^ = i;ooa = 1,000 = '^ '^- ^"•' "^^^«^^^y ^'^^ «^ ^
short column to support the given load. Since the column
is quite long, assume it to be 6" square. Thea ./i = 3d
Formula 76 gives
/ = 30 X 12 = 360, and ^ = 3,000.
5, _ 14,000 / 36 X 360- \ _
7 - 36 v^ + 3,000 X 108 ; -^'^-^O' "^^^ly-
Since this value is much too large, the column must be
made larger. Trying 9" square, A =Sl, 1= 546f.
Then -^\ - 14,000 / 81 X 360 X 360 \ _
^ ^^"' / - TT^V + XOOO X 546f ) - ^'^^^•
5
This value of -^ is much nearer the required value, 1,000.
Trying 10" square, A = 100, /= ^^^ = 833^.
1/u
5, 14,000 /, , 100 X 360 X 360\ -„. ,
/ = -Too- V + 3,000 X 833^ ) == «"•'' ""^''y-
Since this value of -^ is less than 1,000, the column is a
little too large'; hence, it is between 9 and 10 inches square.
9f' will give 997.4 lb. as the value of -^; hence, the column
should be 9|' square.
This problem may be more readily solved by formula
77, which gives
_a/ TX 2000 X 8 / 7x 2000x8 / 7x2000x8 12x360* \
^~^ 2x800 â– '"r 8000 V 4 X 8000 "*" liOOO J ^
V^ + /I473.5 + 518.4) = V 92.479 = 9.61' = 9f, nearly.
STRENGTH OF MATERIALS. 221
(586) Here IF = 21,000; /= 10 ; 5, = 150,000; ^ =
6,250; / = 7.5 X 12 = 90*. For using formula 78, we have
.3183 IVf _ .3183 X 21,000 X 10 _
S, ~ 150,000 - 4455.
11^^11X81(^^20.7360.
g- 6250
Therefore,
d= 1.41421^.4456 + i^.U56 (.4456 + 20.7300) =
1.4142-/. 4456 + 3.0722 = 2.65', or say 2f'.
(587) For this case, A = 3.1416 sq. in. ; /= 4 X 12 =
48'; 5, = 55,000; /= 10; /= .7854; ^=20,250.
Substituting these values in formula 76,
S,A 55,000X3.1416
W=
A^'\ .J. . 3.1416 X 48'
/(x.f) 4,
20,250 X .7854/
5,500 X 3.1416
1.4551
Steam pressure = 60 lb. per sq. in.
rru c • . r,c.. ^. ^ 5,500 X 3.1416
Then, area of piston = .7854 d* = — = — .
f loo^a gQ 1.4551X60
Hence d^ - 5>500X 3.1416 _
^^"""^'"^ - .7854 X 1.4551 X 60 " ^^^' "^^'^^y*
and d= |/252 = 15|-", nearly. Ans.
(588) {a) The strength of a beam varies directly as the
width and square of the depth and inversely as the length.
Hence, the ratio between the loads is
6X 8'.4x 12' ^^ ,^ -, , A
—Jq— • — iQ— = 16 : 15, or Ijiy. Ans.
{d) The deflections vary directly as the cube of the
lengths, and inversely as the breadths and cubes of the
depths.
Hence, the ratio between the deflections is
222 STRENGTH OF MATERIALS.
(589) Substituting the value of <r, , from Table 27. in
formula 80, we obtain
(^) ^=^/-^=4.92f^ = 3.739'. Ans.
(^)
d=cy^=L92^=L65\ Ans.
N '^ 100
(590) Using formula 80,
^4,000
d = 5. 59 y -^^ = 14. 06'.
oO
Since this result is greater than 13.0", formula 81 must
be used, in which
a = k^y -y-= S.Sy =14. 22 . Ans.
(591) From formula 80,
N
.*/77 d* N
dz=c^y^, or H=^-^. ^, = 4.11. (Table 27.)
Hence, H = ,^ ° = 71. 775 H. P. Ans.
4.11
(592) Using formula 83,
^=^^iV^/i^^\=.0212xlOo(t?l^^)=717.7H.P. Ans
(.0212 is the value of ^, from Table 28.)
(593) (a) Using formula 84,
P= 100 C = 100 X 8' = 6,400 lb. Ans.
(^) Using formula 85,
^=eooc..o.c- = 4 = 5^ = io.
C= /lb = 3.162'.
d=^C= 1.05^'. Ans.
o
(c) Using formula 86,
(7=yT3|^=3,65r. Ans.
/^-lOOOr'-T'- ^ - 6tX 2,000 _
STRENGTH OP MATERIALS. 223
(594) (a) Using formula 87,
P= 12,000 fl^' = 12,000 X (^) = 9,187.5 lb. Ans.
(d) Formula 88 gives P= 18,000^*.
Therefore, ^=VCj:=4/W^/4^^^^, An.
(595) The deflection is, by formula 75,
Wl' 1 W/"
s = a ,, ■■= — — „ , , the coefficient being found from the
is / 192 -c / *
table of Bending Moments.
Transposing, IV =-^, ; /= 120; ^= 30,000,000; /=
.7854; s = ^.
_,, „. 192 X 30,000,000 X .7854 oo-v or lu a
Then, IV = g ^ ^^q, = 327.25 lb. Ans.
(596) (a) The maximum bending moment is, accord-
4. ^u ^ u^ c Ti A- Tvj . ^^ 6,000X60
mg to the table of Bendmg Moments, —r— = — j =
90,000 inch-pounds.
By formula 74,
5 /
J/ =90,000 = -^ -.
5, = 120,000;/= 10. ^= '^^ '^^^
Hence, 90,000 =
c ^d 32
120,000 Ttd'
10 32 '
, .V90,000 X 10 X 32 , - .,, ,,, ,
^^ ^=^ 120,000X3.1416 = ^-'^^ = ^ ' "^^^^-^^
(^) Using formula 80,
*/~H */75
<i=<^,y j^ = ^-'iV â– ^Q = H\ne2ir\Y. Ans.
224
STRENGTH OF MATERIALS.
(597) {a) The graphic solution is shown in Fig. 51.
On the vertical through the support 0-1 is laid off equal to
the uniform load between the support and F^ ; 1-S is laid off
-g^^ - -^
lb.
Scale of forces 1^800 Ih.
Scale of distance 1^4^
Fig. 51.
to represent F^. 2-3 represents to the same scale the re-
mainder of the uniform load, and 3 in represents F^. The
pole P is chosen and the rays drawn. The polygon a b c c fJi
is then drawn, the sides being parallel, respectively, to the
corresponding rays. If the uniform load between F^ and /^,
be considered as concentrated at its center of gravity, the
polygon will follow the broken line c e f. It will be better
in this case to divide the uniform load into several parts,
^-.4> -4-^, ^-^, etc., thus obtaining the line of the polygon
STRENGTH OF MATERIALS. 2-25
c d f. To draw the shear diagram, project the point 1
across the vertical through F^^ and draw O s. Next project
the point 2 across to /, and 3 across to u, and draw / ?/.
O s t 71 ft 111 is the shear diagram. The maximum moment
is seen to be at the support, and is equal to a hx P w/. To
the scale of distances, a /; = 58.8 in., while Pm = H =
1,400 lb. to the scale of forces. Hence, the maximum bend-
ing moment is 58.8 X 1,400 = 82,320 in. -lb. Ans.
{b) From formula 74,
i^=§i- = 82,320. 5, = 12,500;/= 8.
_ , I 82,320X8 _.y.
Therefore, - = — ' ^ ^,^,^ — = 52. G8.
c 12,500
But, - = -^^y- = -7:-, and d= 2^ b, or /; = -— .
Hence, L^^L^1. = ^1 = 52,68. rt^' = 52.68 X 15 = 790.2.
c lo
^ = -^'79(h2 = 9.245". b = ^^- =3.7", nearly. Ans.
(598) Referring to the table of Moments of Inertia,
{bd'- b^d;y -4rbdb^d^{d- d^y _
U{bd-b^d^)
[8 X 10' - 6 X {S^yy - 4 X 8 X 10 X G X 8i (10 - 8^)' ^
12 (8 X 10 - 6 X 8i) â–
280.466.
d b,d/ d-d, \
" "^ 2 \bd-bjj~
2
10 6 X 8^
2 "^ 2
(.x^l^f^^) - -
{a) From the table of Bending Moments, the maximum
. Wl
bending moment is — 7—.
4
5, = 120,000; /= 7; /= 35 X 12 = 420 in.
Using formula 74, M = —r- = —~-^
or
„. ^SJ 4 X 120,000 X 280.466 „ ^ _ ,, .
W— , ' = ,^r. m jTrrrr: = 7.246 lb. Ans.
Ifc 420 X 7 X G.319
226
STRENGTH OF MATERIALS.
(d) In this case /= 5, and the maximum bending
Hence, from formula T4,
moment is
M:
wr
ST %ST
^, , ,„ , %ST 8X120,000X280.466
Therefore, W = w I = , ). — ,^^ , iti^tt.
' Ifc 420 X 5 X 6,319
20,290 lb. Ans.
(599) {a) According to formula 72,
/= A r\ or r = 1^ = j/|^ == 4/3 =
1.732. Ans.
Scale of forces 1=960 lb.
Scale of distance 1=8'
PiG.se.
(d) From the table of Moments of Inertia, 7=—^ d^'
72 ; A=bd=^\. Dividing, ^^
36 ; ^ = 6' and b = 4'. Ans.
bd
= 24'°'" 12''
= 3. </• =
STRENGTH OF MATERIALS. 227
Tzd'
{c) As above, r=>|/^=/iL = /|=^. Ans.
i)d
(600) Using formula 69, pd—^^tS, we have t = ^.
Using a factor of safety of 0,
^, 4/5 ^ epd 6X100X8 _, .
^^=^'^^^^ ir= 4X20,000 =-^^- ^"^-
(601) The graphic solution is shown in Fig. 52. The
uniform load is divided into 14 equal parts, and lines drawn
through the center of gravity of each part. These loads
are laid off on the line through the left reaction, the pole P
chosen, and the rays drawn. The polygon b c d e f a is then
drawn in the usual manner. The shear diagram is drawn
as shown. The maximum shear is either t T ox r v ^=- 540 lb.
The maximum moment is shown by the polygon to be at
/"r vertically above the point ?/, where the shear line crosses
the shear axis. The pole distance /*7 is 1,440 1b. to the
scale of forces, and the intercept/"^ is 14 inches to the scale
of distances. Hence, the bending moment is 20,160 in. -lb.
(602) From formula 74,
^=-p*- = 20,160. 5. = 9,000;/= 8.
_,, / 20,160X8 _ ^-
^^^"'r= 9,000 =^^-^^'
But, — = ^^. . z=i —bd"^ for a rectangle.
c ^d 6
Hence, i^^' = 17.92, ot bd' = 107.52.
Any number of beams will fulfil this condition.
Assuming d=iQ', b— — -^ — = 3', nearly.
Assuming d=h\ b = — ^ — = 4.3'.
(603) Using the factor of safety of 10, in formula 71,
y~ 10 Id 108x2.5
228
STRENGTH OF MATERIALS.
(604) Using formula 87,
* 12,000 ^ 12,000
(605) The radius r of the gear-wheel is 24'
Using
formula 80, ^=c {TPr = . 297 f 350 x 24 = 2. 84'. Ans.
(606) Area of cylinder = .7854 x 12' =113.1 sq. in.
Total pressure on the head = 113.1 X 90 = 10,179 lb.
Pressures on each bolt = — ^^ = 1,017.9 lb.
Using formula 65,
/* 1 017 9
F= A S, or A=^= ' :!^ = .5089 sq. in., area of bolt.
.5089
Diameter of bolt = |/ ' = .8", nearly. Ans.
(607) (a) The graphic solution is clearly shown in Fig.
53. On the vertical through JF., the equal loads F^ and F^
1 Ton. . 1 Ton,
CB ■IS'— ^
Scale of forces 1=2000 lb.
Scale of distance 1^6^
Fio. 53.
are laid off to scale, 0-1 representing F^ and 1-2 representing
STRENGTH OF MATERIALS. 229
F^. Choose the pole P, and draw the rays P 0, P 1, P 2.
Draw a b between the left support and F^ parallel X.o P 0\
b c between F^ and F^ parallel to P i, and c d parallel to
P 2, between F^ and the right support. Through P draw a
line parallel to the closing line a d. 0-1 = 1-2; hence, the
reactions of the supports are equal, and are each equal to 1
ton. The shear between the left reaction and F^ is nega-
tive, and equal to F^-=\ ton. Between the left and the
right support it is 0, and between the latter and F^ it is posi-
tive and equal to 1 ton. The bending moment is constant
and a maximum between the supports. To the scale of
forces /* i = 2 tons = 4,000 lb., and to the scale of distances
a f= 30 in. Hence, the maximum bending is 4,000 X 30
= 120,000 in. -lb. Ans.
{b) Using formula 74,
^ /
J/=-^ - = 120,000. 5, = 38,000; /= 6.
^. / 120,000X6 3G0 ,^ ,
^^^^' 7 = 38,000 = 19- = 1^' """""^y-
nd*
-, / 64 :zd'
But. - = —J- = -—-
c d 32
Hence, — -— = 19, or rf* =
32 ' 3.1416*
d^V%
'Im! = 'â– ''''â– ^-
(608) Since the deflections are directly as the cubes of
the lengths, and inversely as the breadths and the cubes
of the depths, their ratio in this case is
18' 12» 27 9 ,^
or - : :r = 12.
2 X 6' • 3 X 8" 2 8
That is, the first beam deflects 12 times as much as the
second. Hence, the required deflection of the second beam
is .3 ^12 = .025'. Ans.
230 STRENGTH OP MATERIALS.
(609) The key has a shearing stress exerted on two
sections; hence, each section must withstand a stress of
20,000 ,^^^^ ,
— ^^r = 10,000 pounds.
2
Using formula 65, with a factor of safety of 10,
r, AS^ . 10 P 10X10,000
Let d = width of key ;
/ = thickness.
Then, ^ / = ^ = 2 sq. in. But, from the conditions of the
problem,
Hence, di = ^d' = '-Z; d' = 8; b = 2.828'
^ Ans.
(610) From formula 75, the deflection S = a-^ry,
and, from the table of Bending Moments, the coefficient a for
the beam in question is - .
48
JV= 30 tons = GO, 000 lb. ; /= 54 inches; £ = 30,000,000;
/ =
64
Tj c IX 60,000 X 54» ^_, .
Hence, 5 = o .^.q ^ .o^ =-0064 m.
48 X 30,000,000 X ,, Ans.
d4
(611) {a) The circumference of a 7-strand rope is 3
times the diameter; hence, C = 1^ x 3 = 3|'.
Using formula 86, /"= 1,000 T" = 1,000 X (3i)' =
14,062.5 1b. Ans.
(d) Using formula 84,
P=100C%orC=/^- = i/^-^^j^=5.9r. Ans.
STRENGTH OF MATERIALS.
231
(612) Using formula 70,
/ =
5, / 120,000
'^' | + «
= 12,000 lb. Ans.
(613) The construction of the diagram of bending
mZ
Scale of forces 1~1600 lb.
Scale of distance 1^32^
Fig. 54.
232 STRENGTH OF MATERIALS.
moments and shear diagram is clearly shown in Fig. 54. It
is so nearly like that of Fig. 40 that a detailed description
is unnecessary. It will be noticed that between k and k'
the shear is zero, and that since the reactions are equal the
shear at either support = ^ of the load = 2,400 lb. The
greatest intercept is ^ c' = f/^' = 30 ft. The pole distance
H — 2,400 lb. Hence, the bending moment = 2,400 X 30 =
72,000 ft. -lb. = 72,000 X 12 = 864,000 in. -lb.
SURVEYING.
(QUESTIONS 614-705.)
(614) Let ;tr = number of degrees in angle C \ then,
2x = angle A and '6x = angle B. The sum of A, B, and C
[sx + 2x-^dx=6x = 180°, or x-'dO° = C; 2-r = 60° = A,
and dx = 90° = B. Ans.
(615) Let -f = number of degrees in one of the equal
angles; then, 2x = their sum, and 2x X 2 = 4iX = the greater
angle. 'Zx -\- Ax = 6x = sum of the three angles = 180°;
hence, x — 30°, and the
greater angle = 30^^
120°. Ans.
X 4 =
(616) ^^ in Fig. 55
is the given diagonal 3.5 in.
3. 5' = 1 2.25 -f- 2 = 6.125 in.
/6.125 = 2.475 in. = side of
the required square. From
A and B as centers with
radii equal to 2.475 in., de-
scribe arcs intersecting at
C and D. Connect the ex-
tremities A and B with the
points C and D by straight
lines. The figure A C B D is the required square.
(617) Let A B, Fig. 56, be the given shorter side of the
rectangle, 1.5 in. in length. At A erect an indefinite per-
pendicular A C to the line A B. Then, from B as a. center
with a radius of 3 in. describe an arc intersecting the per-
pendicular A C in the point D. This will give us two
234
SURVEYING.
Fig. 56.
adjacent sides of the required rectangle. At B erect an
indefinite perpendicular B E to A B, and at D erect an in-
definite perpendicular DF to AD. These perpendiculars
will intersect at G^ and the
resulting figure A BCD will
be the required rectangle. Its
area is the product of the
leng th A D by the_width^ B.
TD" = BTf - ~AB\ 'BT>^ =
9 in. ; A B =2.25 in. ; hence,
'AD^ = 9 in. - 2.25 in. = 6.75
in. 4/6.75 = 2.598 in. = side
A D. 2.598 in. X 1.5 = 3.897 sq. in., the area of the required
rectangle.
(618) (See Fig. 57.) With
the two given points as cen-
ters, and a radius equal to
3.5'^ 2 = 1.75'= If', describe
short arcs intersecting each
other. With the same radius
and with the point of intersec-
tion as a center, describe a
circle; it will pass through
the two given points.
Fig. 57
(619) A B \n Fig.
C
58
Fig. 68.
is the given line, A the given
point, A C and C B = A B.
The angles A, B, and C are
each equal to 60°. From B
and C as centers with equal
radii, describe arcs intersect-
ing at D. The line A D bi-
sects the angle A ; hence,
angle B A D = 30°.
(620) Let A B in Fig. 59 be one of the given lines,
whose length is 2 in., and let A C, the other line, meet A B
j^t A^ forming an angle of 30°. From A and B as centers,
SURVEYING.
235
with radii equal to A B, describe arcs intersecting at D.
Join A D and B D. The triangle A B D is equilateral;
hence, each of its angles, as A, contains 60°. From Z> as a
Fig. 59.
center, with a radius A D, describe the arc A B.
A C xs tangent to this arc at the point A.
The line
(621) A B in Fig. 60 is the given line, C the given
point, C D = C E.
From D and E as cen- \Q
ters with the same .F^
radii, describe arcs in-
tersecting 2X F.
Through F draw C G.
Lay oK C G and C B
equal to each other,
and from B and G as
centers with equal
radii describe arcs intersecting at H.
angle B CH=^b°.
Draw C H. The
236
SURVEYING.
(622) 1st. A B in Fig. 61 is the given line 3 in. long.
At A and B erect perpendiculars A C and B D each 3 in.
long. Join A D and B C.
'K
il>
\
;<
/
X
9€t
0>^v
These lines will intersect at
some point E. The angles
EA ^and E B A are each 45°,
and the sides A E and E B
must be equal, and the angle
A E B= 90°.
2d. A B \n Fig. 62 is the
\ i g^ven line and c its middle
_ ^A^ point. On A B describe the
Fig. 61. semicircle A E B. At r
erect a perpendicular to A B, cutting the arc A E B in-E.
Join A E and E B. The angle
A EB is 90°.
f
(623) See
and Fig. 237.
Art. 1181
(624) See Art. 1187.
(625) (See Art. 1195
and Fig. 245.) Draw the line
A B, Fig. 63, 5 in. long, the length of the given side. At
A draw the indefinite line A C, making the angle B A C
equal to the given angle of 30°. On ^ Clay off ^ i? 1.5
in. long, the given difference between the other two sides of
SURVEYING.
237
the triangle. Join the points B and Z? by a straight line,
and at its middle point E erect a perpendicular, cutting the
line A C in the point F. Join B F. The triangle A B F is
the required triangle.
(626
(627
(628
(629
(630
(631
(632
(633
(634
(635
(636
(637
(638
See Art. 1198.
See Art. 1 200.
SeeArt. 1201.
See Art. 1204.
See Art. 1205.
See Arts. 1205 and 1206.
See Art. 1 206.
See Art. 1 204.
SeeArt. 1207.
See Art. 1 207.
See Arts. 1209 and 1213.
SeeArt. 1211.
{a) In this example the declination is east, and
Magnetic
Bearing.
True
Bearing.
N 15° 20' E
N 18° 35'
E
N 88° 50' E
S 87° 55'
E
N 20° 40' W
N 17° 25'
W
N 50° 20' E
N 53° 35'
E
for a course whose magnetic bearing is N E or S W, the
238 SURVEYING.
true bearing IS the sum oi the magnetic bearing and the
declination. For a course whose magnetic bearing is N W
or S E, the true bearing is the difference between the mag-
netic bearing and the declination.
As the first magnetic bearing is N 15° 20' E, the true
bearing is the sum of the magnetic bearing and the declina-
tion. We accordingly make the addition as follows :
N 15° 20' E
3° 15'
N 18° 35' E,
and we have N 18° 35' E as the true bearing of the first
course.
The second magnetic bearing is N 88° 50' E, and we add
the declination of 3° 15' to that bearing, giving N 93° 05' E.
This takes us past the east point to an amount equal to the
difference between 90° and 92° 05', which is 2° 05'. This
angle we subtract from 90°, the total number of degrees be-
tween the south and east points, giving us S 87° 55' E for
the true bearing of our line. A simpler method of deter-
mining the true bearing, when the sum of the magnetic
bearing and the declination exceeds 90°, is to subtract that
sum from 180° ; the difference is the true bearing. Apply-
ing this method to the above example, we have 180° —
92° 05' = S 87° 55' E.
The third magnetic bearing is N 20° 30' W, and the true
bearing is the difference between that bearing and the
declination. We accordingly deduct from the magnetic
bearing N 20° 40' W, the declination 3° 15', which gives
N 17° 25' W for the true bearing.
{b) Here the declination is west, and for a course whose
magnetic bearing is N W or S E the true bearing is the su7n
of the magnetic bearing and the declination. For a course
whose magnetic bearing is N E or S W, the true bearing is
the difference between the magnetic bearing and the
declination.
SURVEYING.
239
The first magnetic bearing is N 7° 20' W, and as the
declination is west, it will be added. We, therefore, have
for the true bearing N 7° 20' W + 5° 10' = N 12° 30' W.
Magnetic Bear-
ing.
True
Bearing.
N 7° 20' W
N 12° 30' W
N 45° 00' E
N 39° 50' E
S 15° 20' E
S 20° 30' E
S 2° 30' W
S 2° 40' E
The next two bearings the student can readily determine
for himself.
The fourth magnetic bearing is S 2° 30' W, and to obtain
the true bearing we must subtract the declination, i. e., we
must change the direction eastwards. A change of 2° 30' will
bring us due south; hence, the bearing will be east of south
to an amount equal to the difference between 2° 30' and the
total declination 5° 10', which is 2° 40'. The true bearing
is, therefore, S 2° 40' E.
(639) See Art. 1216.
(640) See Art. 1217.
(641) See Art. 1219.
(642) A plat of the accompanying notes is given in
Fig. 64 to a scale of 600 ft. to the inch. The order of work
is as follows :
First draw a meridian N S (see Fig. 64), and then assume
the starting point A, which call Sta. 0. Through A draw
a meridian A B parallel to A^ S. Then, placing the
center of the protractor at A, with its zero point in the
line A By lay off the bearing angle 10° 10' to the right of
240
SURVEYING.
Co.
A B, as the bearing is
N E. Mark the point of
angle measurement care-
fully, and draw a line join-
ing it and the point A.
This line will give the di-
rection of the first course,
the end of which is at
Sta. 5 + 20, giving 520
ft. for the length of that course. On this
line lay off to a scale of 600 ft. to the inch the
distance 520 ft., locating the point C, which is
Sta. 5 + 20. Through Cdraw a meridian C D, and
with the protractor lay off the bearing angle