Cirril
Engineerirf.g
R-efer.e11ce
For Licensure Examinations
Volume 4
DIEGO INOCENCIO
' T. GITLESANIA
Civil Engineer
BSCE, EVSU
(LIT) - Magna Cum Laude
Sth Place, PICE National Students'Quiz, 1989
Awardee, Most Outstanding Student, 1989
3rd Place, CE Board November 1989
Review Director & Reviewer [all Subjects)
Gillesania Engineering Review Center
Author of Various Engineering Books
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Answer Key
1A
ZB
3C
4B
5A
68
7D
BB
9A
108
IIA
TzD
138
L4A
15C
21A 31 C
22D 328
23C 33 B
24A 34C
258 358
16D 26C
278
17 A
1BA zAC
19D 29C
20D 308
36D
37A
38A
39 B
40D
7ID
72C
73D
74C
75D
81A
BzD
9l
83 B
93
54B
55D
61D
628
63C
64C
65B
B4C
94
85D
95
46C
56A
664
478
s7B 67D
58A 68A
59D 69A
70c
60 c
76D
77D
78A
79A
BOA
86A
87 C
BBD
B9A
90
96
97
98
99
100
4tB
42 A
43D
44D
458
514
48 B
49F
50A
S2B
53 B
92
Solutions to May ZO\S Examination
$
$
E
ii
;.
::.,r
6
Rate oI Mr. Curry = | /g
Rate of Mr. Thompson = 1/12
Situation 1 (1 to 3)
Let "t" be the required time to finish the job
m1
(L/6)xr+)Q!/!l)xr=!
H
r=4days
it
tr
i,ru-r,=15s0
k=1
| ).1;7
Let A and B be the times the tanks A and B be filled, respectively.
i1
nth term, a" = (nz _ 2)/3
6th term, a6 = (gz _ Z)/3 = 34/3
i
,.i
Given: (2x- 7/x)10
LL]]B
Let h, t, and u be the hundred's digit, ten's digit, and unit's digit, respectively.
rth term
of(a + [;n.
ar=IlCr-t xan-r+1br-1
I
Nurnber=100h+10t+u
:t
4th
E
term = lZ0 (L28 x7) (_l /xz)
4ttr term = -15360 xa
10Ce (2 1)ro_s
4th
iE
$
,',8
teim =
1/A= 2 (1/B) or B = 2A
{1,/A)6+(1/8)6=t
6/A+6/(2A)=1
A=9hoursandB=lBhours
Ll3
if
B:
Rate of A = twice that of
Wz
[_1/x]:
The digits are in A.P.
u-t=t-h
h-2t+u=.0
)Eq.(11
74h-16t-25u-0
)Eq.(21
Sum of digits = h + t + u
m4
Given: f[x)
= 3yz
(31=0
-
hx
+x-
_100h+10t+u=26
7h
h+t+u
3(S;z-3h+3-7h=0
h=3
(100h + 10t + u) + 198 = (100u + 10t + h)
99h-99u=-198
ms
h-u=-2
+2x+L A B
x, -x-2
x_L x+2
3x2 +Zx+L A
B
(x+2J(x-1)- -t ---2
)Eq.[3]
3x2
3x2 +
2x+
1=
Solving for h, t,.and u:
h=2;t=3;u=4
r119
A(x + 2) + B[x - 1J
)
Given:
identity
Setx = 1;
3(7)2+2(1)+1=A[1 +2J+B[1-1J
A=2
Set x = -2:
3(-2)2 + 2(-2) +
B=-3
n
I
=
A(-2 + Z) + B(-2
_
Average of90 real numbers = 70
Sum of90 real number =90 x 70 = 6300
1)
When two numbers namely 2B and 68 are removed:
6300,
Newaverage
=
90
-28-68 =ro;,
-2
LL-J
tri13
10
Let
H=b+d
x = speed ofairplane in still air in kph
y = speed of the wind in kph
40=atan18"+atan40o
a=
34.3637 m
500
Against with the wind:
" l+45/60
x+V=
"
With the wind:
x = 342.86 kph
y = 57.74 kph
h=d-c
h=atan40"-atan29"
h = 10.563 m
500
1+15/60
t'.114
-
Given: a=300m
a=26"
LU 11
Time for the inlet pipe to fill the empty tank = 6 hours
Time for the outlet pipe to empty the full tank = 12 hours
[1/6)t- (1,/1,2)t=
t = 12 hours
a=AC-BC
a=hcotcr-hcotp
300=h (cot26" -cot56')
1
h = 218.056 m
A
rl,
w\12
15
Triangles EAC and EBD are similar:
- 18;
18 BD
AC
In triangle
In triangle
AB=24m
BD=324
h = 24 tan20"
h = 8.735 m
AC
BAC:
tan 0 =
ABD:
tan
cr =
AC
AC = 8.735 cot 30o
AC = 15.L3 m
36
BD
36
Triangle ABC is a right triangle:
BCZ=242-1_5.132
BC = 18.63 m
But cr = 20
tan c{. = tan 2e
tano=
2tan0
t]:{ 16
Given:
1- tan'e
_ 2{AC/36)
36 1(Ac/3q'z
324/AC _ z(AC/36)
36 7-(AC /3q'Z
a= 25"
Bp
r$
it
ril
,$
tE
t
,;l
AC = 1,2 m
&BD = 324/12 = 27 m
AB = 300 m
P=50"
SSm
AC=Hcotct=
2.1,445H
BC=HcotB=0.8391H
trflz=trfz1fifz
3ggz = (2.1445U12
H = 13O.27 m
+,O.Urrtnr
M17
'
ilil 21
r=6in
Given: ICAD=0=39'
P=2r+C=19in
C=7in
1C.OD=2e =78".
Area=a/zCr=L/z{7)(6)
Area =
,
2lin2
In triangle CDO:
i
mlB
I*$
t
2a+2Q=180"
or
,trza
= U.BA. - 78")/2
o=51o
Given:
a=5cm,b=7cm
I t22
c=10cm
Given: r=13cm
iii
ititii
a+f2=ft
;t
11
:i
)
-r2 =5
Eq. (1)
12=a2+[a+bJ2
b+13=11
11-13=7
H
'nl:
12
lli
ii
b=7
)Eq.(2)
cm
)
Eq. [3.)
A2
Solving using calculator:
rr=11cm; rz=6cm; rs=4
z
1/2b2
Az = 1/z(7)2 = 24.5 cm2
N
Given: r = 25
= 1/z(13)2 144.7 6" x [n/180"J - sin 44.7 6"]
At= 6.51259
CD=48
c=39
\c
h=l/zCD=24cm
r= '[7 +' =z
a=c-b=32cm
,,/;1ll
.'.
t\
By integration:
Equation of circle: x2 + Yz -
i,ll
1
bl
""{-'- ---'-'c
"' l--;"
./i
nr"r=
J,,,,(r
a)dx =
Area= 3L.Ol26
,L
lJ 23
Given: r=15m
Given: r=4ft
12=a2+ta/2)2
0=30"
lgz=azaazf!
42=Ar".to.-Ar
Az =
rz 0, r(r /2) sin 0
A2 = t/2 yz (0. - % sin 0)
Az = 1/z(4)2 [30" x [n/180"J Az = 2.7189 ft2 = 315.19 inz
7/z
a
= 13.4164 m
1/z
Apothem of octagon:
1/z
sin 30ol
cm2
Area = Ar + Az = 31.O126 cmz
\r ln\
l--'.8 .{ i
{a
= 40 cr4
Diameter = 2R = B0 cm
m20
\
At = yz r2 (0. - sin 0)
m19
=
b
arcsin (a/r) = 22.6199"
0 = 90' - 2a = 44.7 603"
cr =
+r3 =c
Y2 +r3 -7
R
-\, n
n -;-
732=52+(5+bJz
|{
ii
^ffi :,\
a=5cm
h= a= 13.4764m
732
or y =
'lT6r=
I;[v[6r=-sld-
Number of sides, n =
I
B
t\27
Length of arc = r x central angle
The area of regular polygon in terms of its apothem:
the length of arc is directly proportional
to the central angle
area= lxh2tan[180'/nJ
4
a.",
=
* - 15o;a=Zoo
4x 3x
I2' x (L3.4164)ztan (180"/B)
Area= 149.12m2
Angle PQO:
ffi24
g=(180"-o")/2=BO"
Given: a=Bcm, b=15cm
c=12cm, d= lBcm
NIU28
Area=m
,li
Given d=25cm
IACD=q=30o
5=[a+b+c+d)/2
il
ICDB=F=20"
iri
s = (B + 15 + 72 +
1B)/2 = 26.5 cm
Note: the angle subtended by an arc of
$
a circld from any Point on .the
circumference of the circle are the
same. This angle is alwaYs half the
angle ofthe arc.
Area = 161.93 cmz
m2s
Given: a=2in,b=4in
c=6in, d=8.224in
Thus, ZBAC = 9 = 20" and IABD = cr = 30o.
In right triangle
ACB:
ar62=@
BC = d sin B = 25
BC = 8.55 cm
s-(a+b+c+d)/2
s = (2 + 4 + 6 +
lrea
8.224)/2 = 10.t12 in
I
llr 29
Given: IBDC=0=15"
AD=7cm
=
Diameter = 25 cm
Area = 19.619 inz
In right triangle ADB:
m26
r----:-----=
Arc BAD = 340 o
Arc BC = 360o - 340o = 2O"
Arcs AB, BC, and CD are equal, i.e. 20o
Thus, arc AD = 60o
Angle DEA = r/z Arc AD = %[60'J = 30o
A
dr = ^J25' -7' = 24 cm
cr = arcsin (7 /25) = 16.26"
In triangle BCD:
F=90" +a=\06.26o
0=180'-0-F=58.7+"
sin 30'
By sine
law: cD sin g
d1
sin
r33
Given: a=12/2=6cm
B
24
'co
= sin 106.26'
CD
=21.37 cm
b=2012 = 10 cm
sin 58.74"
h=3.6cm
zu30
Edge
ofregular tetrahedron, a = 4 ft
.
Volume of regular tetrahedron;
v=
V=
i:
6,lz
a3
-6,12-
=7.54ft3
i--_
i-.*--
\
Il[:a'
V
=
,,
V
n(3.6)
_
= .:;::1
r
6L
3b' + h1
1
L3(6)z
|
J
+3(10): + t3.6),'l
V=793.5cm3
till 34
Volume of spherical pyramid,' V
E=
R=
0=30'
E
540"
v_n(9.2)3(28')
540u
2R= 75 cm; R = 37.5 cm
rcR0
"_ 180'_
r[37.5)[30.)
180'
C=6.25ncm
L=
Zrtr
V=
A]
tt
Given:
Algebraic form of coniplex number, a + bi = 5
C
ExPonential form ofa +
6.25x = 2nr
r = 3.1.25 cm = 31.25 mm
As=Abase*Asides
49.48=fDz+nD[1.5D)
D=3ft
36in
[i - r sxi
J/*b' = "@*n' =n
.=
x = arctan (b/a)
x = arctan (12/5) = 67.38" = 1,.1,76 rad
ExPonential felm = lJ sr.rzei
H=1.5D
As= zf2arrPg
126.85 cm3
ul 35
ru32
D=
nRi
spherical excess = 2Eo
radius of sphere = 9.2 cm
v= nR3E
m31
=
,36
Given: [x+yi)(1 -2i)=7 - 4t
x+Yt= 7-4i =3+2i
r
Thus,x=
zi
3
andy= 2. Thenx +y= 5
+
t2i
(x, * y, + 2x - 4Y - 20)
m37
Value
of r/i
- (xz + Yz - 10x+5y+25)=0
12x-9y-45=0
:
4x-3y-15=0
i = 1290"
|t.42
Ji = (719Q")t/z = 1245"
t.22
Vt = -+ --F
tl2 '12
L+i
s
,,=
,fy
I
I
I
-'l
E
.l
i
l
I
tE 38
(1+ilo=[(1+iJ:]z=-$i
m39
Given
line: x+3y=0
Point: (3,2)
)
Slope,
(4 * y), = (x- 6)z+ (B -
m= -1/3
16 + By + y2 = x2
x2
-lZx-,24y
yJ2
- !2x + 36 + 64 -
+84=O
16Y +
Yz
aParabola
The slope of a line perpendicular to this line is m = 3
Equation of required line:
y-yl=m(x*xrl
u;nl43
y-2=3(x-3)
3x-Y
m40
'
Given
line:
Parabola:
-Bx+
6Y + 17 = 0
2x-y-!3=0
=7
5x + 4y + 3 = 0
Points of intersection:
y= (y + 73)/2
Y'z-B[[Y+13)/2]+6Y+17 =o
Y=5 &-7
Equation of line with origin translated to [1, 2J:
5(x'-1J+4(y'-2)+3=0
5x'-5+4y'-B+3=0
5x'+ 4Y'= fQ
y2
Line:
A=
I (x*
-
O
xr)dy
Jv,
m41
xo= [y +
Given circles:
xr=
x2+y2+2x-4y-20=0
x2+y2-10x+5y+25=0.
The common tangent of
these
circle is the radical axis of these
circle. The equation ofthis axis is
obtained by eliminating the
second-degree term of the given
equations.
o=
l3)/2
(Vz + 6y +
v, -ov,
,1', '-;
r.Iv+13
1
17)/B
LTfay
=36squareunits
(lvil
47
Equation of curve in polar form: r2 - csc 20 = 0
Equation of ellipse:
x' v'
-+1=1
b' a'
Note: x=rcos0
y2ay2=y2
Y=rsine
a=12
b=Tz{18)=9
1
rz = csc 20
f).-
_
-
sin20
x' v'
-+1==1
9' 1,2'.
2
12= ---]2(y / r)(x
sine cos0
/r)
=
"
2xy
ZxY=1
Aty = -4
x' * [-4]'_,
92
4B
. Given:
122
x = 8.485 in
!22 + 65 = 0
Reduce to standard form:
(xz + 2x + 1z) + (yz - 16y + 82) + (22
(x + 1lz + (Y - B)z + (z - 6.)2 = 36
2x= 16.97 in
:45
Given hYperbola: x2
- 4y2 -2x - 63 = 0
x2-2x+1-4yz=63+1
(x-t)z-4yz=64
(x-11' v'
82
xz + Zx +y2 - 1,6y + zz -
- 122 + 62) = - 65 a lz a $z a $z
Center: (-1,8,6)
Radius:
OR Center: x,= -(2/2) = -1
yc= -(-16/2) = B
42
a=Bandb=4
6
bl
y,= -(-12/2) = 6
_T
Center at (1, 0)
49
Given: Planel: x+4y-z+3=0
Plane2: x-12Y+22-7=O
0 = arctan [b/a)
0 = 26.565'
The vectors contained in the given planes are:
E0 46
Given:
Note:
r=2sin0+2cos0
sin.0 = y/r and cos 0 = x/r
Angle "0" between vectors:
12=2y+2x
+y2=2x+2y
(xz
-2x+
1) + (yz
-2y
(x-t)z+(y-1)z=l
+
cos0=
l)=0+z
Thus the curve r = 2 sin 0 + 2 cos 0 is a circle of radius
1.)
Area= nrz = 2n
Yt'Yz
l;;lj;,1
cos0=
center at [1,
)VctA
)VctB
y2-y2ay2
r=2(y/r)+2(x/r)
x2
vr=i+4j-k
yz=i-1,2)+2k
J7 with
VctA
o
VctB
Abs(VctAJ AbsIVctB)
cos 0 = -0.9461,6
0 = 16r.17
$fi:f=,,T=ffiffi;i
B:il: ., r ::
"tr l
Work= Force x distance
,Io
limln' '
z+0
ii'
9
=0
work
=
ru51
[' org, * y;
Jv,
work =
a=3.8m
',a
1l l,
Find dx/dt when
55
s=6m
_=-n!
ncr=
[n - r)l r!
I
dx
dr
dx
-dt
=
= 1.051
(n
-
101(n
(n-10J(n-rL)=6
9*9
n=B&13
'16'3.8'
=L.L63m/s
Jlro**xx
r
'il.
- 11)9=-nX
vl7-i7
^
,1
9z$l
Number of waYs 14 PICE member can
choose 1 President, 1 Vice President
and 1 secretarY, i.b. n = 14,r =3
N = nPr = 14P3 =2lB4ways
,i
"s5s*s?63?i.:
I rlr
m53
_11
t"-1oIto! " (n-12)ll2l
Jt' -3€'
y's@s/dt)
_
-
zu52
r'^
/
nC10 = 22 nC72
x=.6'-l
L:."
!" t-"O.++-v'/)(o.B+v)dY
Work = 44.37gkN-m = 44.379k|
ds/dt = 0.9 m/s
il
112
57
Number of waYs to hire 3 out of 10
aPPlicants, i.e. n = 10 & r = 3
Bxcos2x dN = 1.14
N=nCr
N = 10C3
N=120
ms4
.
Given:
r = 7.2m
a=0.80m
|w = 9.81 kN/m3
Equation of circle:
X2+y2=f2=\,22
xz _ 7.44 _yz
dV = n x2 dy = n(1.44 - y2) dy
dF = y. dV = y* n(7.44 - yz) dy
irtSB
Probability of selecting a packet with
less than 20 candies' p = 40/1000 = 0'04
ProbabilitY of selecting a Packet
with 20 candies,
P=
960/1000 = 0'96
The probability of selecting a packet with less than 20 candies (in 3 packetsJ
P=pxqxq+qxpxq+qxqxp
P
OR
=(0.96 x 0.96 x 0.04)(3) = 0'1106 = ll'O60/o
P = nCrprqn-rlvhsrsn = 3
P = 3C1 [0.0+1t [0.96]3 1 =
andr=
1
0'1106 = 1l'o60/o
,
Probability of failing once = 0.422
Probability of failing twice = 0.141
Probability of failing thrice = 0.0.016
Probability of failing once
OR
27,37, 40,28,23,30,35, 24,30,32'31,28
Arranging in ascending order:
Since there are even number of data, the
median of the first half of the data.
Given 71 students;
1
0 are ch ines e,
(12 datal
23. 24. 27, 28. 28. 30, 30. 31.. 32. 35' 37, 40
twice = 0.422 + 0.016 = 0.438
mi 60
'
data:
Given
24 are f apanese, and 37 are Filipinos.
The median of the first
Probability that three are Chinese is:
First quartile,
P=19, 9,, B =o.oo21
71, 70 69
Xo.zs =
lower quartile [first quartile) is the
half:23,24,27.28'28'30"
27.5
This means that 25% of .the data is lower than 27'5o/o
';67
The second quartile, X050 = (30 + 30)/2 =30
P=
D
Student [agree or neutral) x Teacher (agree or neutral)
_ 132+54 ..
x--.- S+1,2 186x_22
L32+78+54 S+1,4+LZ=_264 36
P=
The third quartile, Xo.zs = (32 + 35)/2 = 33.5
lrl65
Given
0.4306
Mean value, px = 84
4.5, 12.7, 28.5, 25.6, 52.6, 45.4,
1'B'5,
Standard variate, z =
- _B0-84
_
27'5, 125
52.6
(9 datal
Since there are odd number of data, the median (second quartile) is the sth
data (i.e. 27.5). The first quartlle is the median of the data below the fifth data.
Standard deviation, o.= 4
x-trr,
Yn.r5
o
'
BB_84
22= _
=l
_1
4
-Lt,,
3
Arranging from smallest to highest:
t1.s. tzl . $.s. zs.o. 27.5, 28.5. 34.5. 45'4,
w62
z1
data:
= {12.7 + 18.5)
/2
= 15.6
This means that 250k of the data is less than 75'6
The third quartile, 1o15 = (34.5 + 45.4)/2 = 39'95
4
tl66
P=
I e'"/zdz
-: J"
Jz"
Given: r = 5o/o
=O.682g
Effective rate, ER = er - 1 = eo os - 1
Effective rate, ER = 0.0513 = 5.13o/o
m63
Mean value, px= 17
Standard deviation, o = 3
Standard variate, z =
71 -
1,4
-
t)\ 67
x
*
The future'of "n" equal payments after
l-r,
n'periods is:
17
--'l
22=
20
3
p
=
:
f'
Jzn J,, "
- A[(t*!_itr
o
-L7
3
,'/, dz
=o.6BZe
-
tn=
=
r
i),
7
1,500,000
-i
A[[1+ 0.08)3 - 1]
'
0.0
A = P 396,133.6
f
1+ 0.0Bls
*m*'
o1'23qk]
"
Fs =
f.iiM
j
AAA
= 1,500,000
I
L I _L,-*F,
Itr. < ar t\
M68
Given:
million
SV= P600,000
FC = P1.20
D3 =
n=5
i=1.0.1.o/o
* --9Y,
)
+ -FC-SV
[1+i)'-1 [lriJ'1
100'000
Capitalized cost, K = 1,200,000 *
[1 r 0.101]'
1
*
2(1s)
= P720,000
BV: = 1,200,000 - 720,000
BV: = P4BO,0O0
)71
Given: FC=P1,800,000 n=5
SV=P300,000
- 600,000
(1+ 0.101)s - 1
1,200,000
m=3
Using the constant percentage method:
Capitalized cost, K = P2,469,954
u69
- 300,0001' "\'^^1-'.=l-' 't
BV:=FC-D:
OM: = P100,000 (m = 3)
Capitalized cost, K= p6
(1,200,000
BV.=FC[1-k]. k=1- t6viFc
D-=FC-BV.
Given:
FC =
SV=
P1,500,000
P600,000
BVz =P870,000
n=?
Using SOYD Method:
Depreciation charge, D. = FC
5rn-, =
p,
k=
t<1: = 1,800,000(1
BV: = P614,309.35
BV: = FC (1 -
- BV.
111*n;
'
-
0.30117)3
D: = FC - D: = 1,800,000 - 614,309.35
D: = F1,185'690'65
2'
= (FC - SV)
1 i500!00/1€00o00 =0.30717
t[2':I:-l]
2Sum
l)t; 72
Given:
Dz = FC= BVz =
a=28m
b=38m
630,000
pr=IFC-tYr##
Lcos = 158.82 m
Lcoe = R 0.
2'
630,000 = (1,500,000 - 600,0001
n = 4 years
m70
2(2n-2+7)
,R
158.82
\
H'r
;-t.
\ ..:Pl
n[1 + nJ
b=R-RcosP
^'r - - :O
]-Yf
I r ss.ez )
Given: FC=P1,200,000 n=5
SV =
P300,000
BV: =?
Using SOYD Method:
38=R-Rcos _
[R /
|
R=325.35m&B=27.969"
28 = 325.25(7
a = 23.948o
0=F-0
0=27.969" -23.948'
BV*=FC-D.
D. = f FC - rr1
m(2n
2Sum
sum=Itl+nl=15
2'
-m
t 1)
-
a=R-Rcoscr
cos
crJ
0 = 4.021.'
Chord AB = 2R sin (0/2J
Chord AB = 2(325.35J sin (4.02L/2)
Chord AB = 22.83 m
|,"76
m73
I
Given:
Given:
l
Central angle ofthe curve, I = 30o
a=25m
t,
l
CosH=
0
External
S
i
R
= 19.989'
415
s = 736.78
Sr=L
0.08 0.08 + 0.02
s=R0
f.i
h = 0.08[L/2) = 6.4 m
v 0.08
A
L = _;y=0.055
aS,
36.78=792.18e
iIh:
ii
I
m
e = 0.1914 rad = 10.965'
. x=R-Rcos0
A, =
PC
At=2.7
x=792.78(7-cos10.965')
x = 3.51m
6l
Given: R=200m
TD=MD+E
^_ nRO _ n(200)(r2)
1800
1800
il
c = 41.89 m
Sta.q=1+441,89
TD=MD+eN
TD = 51.2.2r + [+0.0.4) [5 12 .27 / 50)
TD = 512.6198 m
Q:
tt
{
Number of tapelengths, N = MD/LI
Length of curve from
to
- 2.7 = 18.57 m
Measured length, MD = 512.21, m
Length oftape, Lt = 50 m
Error per tapelength, e = +0.04 m
Sta.PC=7+400
PC
r
7A
cr=5o
hI
i,t
riil
m
Elt='14.87+h-Ar
0=2a=12o
ilI
-'
2 rll41
Ele= 1.4.87 + 6.4
:t
r{
il
o'o8*Y
m75
iil
h
a=Sr-L/4=BBm
0+736.78
- 700 = 36.78
1l
L=160m
Sr=128m
Sta.A=
-
I
&.
PC
Sta.PC=0+700
,I
- l)
i)\i77
I
Given: R=192.18m
$
i{
(1/2)
i
I
IS
EA74
t"
E = R[sec
R=425.2L6rrr
s= l44.7Bm
fl
distance:
15 = R[sec (30" /Z)
rRO nf415)119.989''l
_______L
1BO'
1BO"
;I
;l
External distance, E = L5 m
,R=415m
I
,
,79
Given:
Stadia intercept, S = 0.60 m
Distance from transit to stadia, D = 67.20 m
Instrument constant, f + c = 0.30 m
Required: Stadia constant
:t
'iu
I
I
D = kS + (f +
cJ
61..20 = k(0.60J + 0.30
k = 101.5 m
EEI
r,, tl:l
80
Given data:
. Route
A
Differ'ence of Elevation
Frequency
Line
35 L.1
2
5
7
AB
+2.O0
N
54.32'
E
BC
L2.1.0
S
38'45'
CD
18.50
s
352.3
351.3
351.7
B
C
D
6
DMD
18.059
37.979
37.91.9
684.796
E
-9.437
7.574
83.4L2
-787.1.25
34'56', W
t5.t67
10.594
a0392
-27.040
42'.758
-37t.21.4
-7.779
7.938
119.988
Weighted mean =
DE
28.40
s72't2','W
-8.682
Weighted mean = 351.8
EF
t7.00
N27" 14', W
75.t1.6
IFrequency onJ
Area
{il
Shift-Stat-Var- x
84
Impact factor, I. =
{
gR
v = 100 kph + 3.6 = 27.778 m/s
R=550m
m81
fr
Weighted mean
Ii
=
I,_D x
W.
527,62(2) + 527,51,(5) + 521,,1,3(1) + 521,,93(8)
IW
Impactfactor
2+5+L+8
,-=
Weighted mean = S2L.7lm
i$
.:l
This problem is
similar to Problem
87.
Brakingdistance, BD
Compass rule:
Correction in
latitude
Total error in latitude
I
Length of line
Perimeter of traverse
C, 48'1'2 ; Cr = 0.2747 m
L.67= 292.52
0.143
m=
';ll 85
mr82
I
t2t9.277
1572.83t
Using MODE-3-2
ir
2A
Dep
Lat
Bearing
Length
=
v'
Zgfu+C)
v = 100 kph + 3.6 = 27.778 m/s
5 = +0.02
u = 0.60
BD= v+ G]
BD=
2g[p
latitude exceed the positive latitude, the
error is added to the positive latitude and subtracted'from the
Since the negative
negative latitude.
Corrected latitude = +21.09 + 0.2747
=21.36m
BD =
27.7782
2[e.81)(0.60 + 0.02J
63.45 m
,786.416
EQ
86
il
W=200kN
r=165mm
Fp
lact area
lilr
89
EoH = Economic overhaul
Cost
Stress on effective area:
.W
J,
=Fn
zoo.oto
15;
n[r+tJ'
t=
W
Formula: t
=
,F,
he,lllo{i
Dooor,o
t= l--165
\/n(0.15)
-r
l\
rl6$.?rail *trk.ajl
The frog is a device by means of which the rail at the turnout curve crosses
the rail of the main track.
Frog number, FN = 7z cot (F/2),whereF = frog angle.
7/z
cot
(F
/2);
F=
m88
5.725"
y:
Given:
F = 43' 20'
0 = 47'20'
Grade ofdrift
=2/100
o=0-F=4"
tan
y = arctan
=
Ct = cost of
Cr,
x 20
(1 station = 20 mJ
borrow Per station
Cost of overhaul, Cl = 0.25
/mz
FEH = Free Haul Distance = 50 m
486.47 mm
t=486.47 mm
a = 100
f
ofborrow
LEH = 450 m
-=0
70 =
Cost'of overhaul
Effective area
= 0.15 MPa
fD=
.A _
costofborrow
-
cr =
6.9927
(2/a) = 15.96o
dip angle
LEH = FHD +
EOH
450 = 50
.'#
Cr = P5.00
per meter station
,,i.,r_,::.
:
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,
Solutions to May ?OLS Examination
I
l
atm = 101.325 kPa = 760 mmHg.
.)Z
Surface tension
3
Atmospheric pressure, put- = 101.325 kPa
P"t*
h. =
6. =
101'325
9.81
= 10.329 m
t4
2.5_pr- _ 2.5(101.325J
=zB.4m
h. =
v
9.81
r5
Given:
h*r
= 800 mmHg
*]&,r
h^z = 700 mmHg
y"i. = 12 N/m3
s* = 13.6
,,
ll
..
-
Y^
u
-3rr
[h*t - h.r)
-
y"r
eB10[13.6][0.8-0.7)
= 1111.8 m
1,2
t6
-
Since the
P
=
unit weight of mud varies with depth, then
rh
J"
rlh)dh
Y(h) = 10+ o'5n
n=
r5
lf (to+o.sh)dh
= 56.25 kPa
Given: h=0.60m
Fo=pcgxArea
F. = 1.695 kN
p=yh[1-alg)
7
=
9.Bl kN/m3
p = 9.81[0.6)(7
P=
In water:
p.g = [9.81x0.8)(0.6]
+ [9.81x0.3)
P'c=
^(
a=2m/sz
['o = (9.81x0.80]0.30x(1.2x0.60)
n = 0.025
Given: Q=0.003m3/s
'652kPa
L=1km=1000m
D=100mm
Fw=pcsxArea
F.=7.652 x [1.20x0.60J
F. = 5.509 kN
ur _
10.2s (o.o2r2
HL=@
10.29n2 LQ2
J)
Lo/
r
t,\ t2
MB
Given: D=0.15m
Given:
n = 0.015
Vr = 500 m3
st = 0.92
sr = 1.02
HL=
V*Vs=
-
lo.2gn2LOZ
b=
500 =
tt
9Ev
=5100
VelocitYhead
rl
=*=
L=500m
Wr=20N/s
Displacement (weightJ of ship, W' = 24,000,000 kg
Draft is seawater, dr* - 10.4 m
Specific gravity of seawater, Ssw = 1,03
Cross-sectional area of ship, A, = 3,000
Wr=ye
mi
3'252
2[e.81]
=
= 0.80
f = 0.025
so
26 = [9810 * 0.80)
Q=
volume displaced in seawater, ' vr,
Volume displaced in fresh water,
=
W = !i'11,9'999 = 23300,97 m3
= p.,
VDr
1000(1.03)
=
w. _ 24,ooo,ooo
_.
JU\'!#Y
1000
= 24,ooo
= 0.233 m
Draft in fresh water = d'* - Ah = 10.4 + 0.233 = 10.633 m
mi
o.538 m
lll 14
Given: D=75mm
Given:
*
-ll5r^
0.01023 m3ls
lh 13
me
EE9
change in draft, on =
10.2e(o.o1sF [I ooo)Q'
5V
st
V-
L=1km=1000m
Q=
st
000x0.003y
HL=6m
----
Totalvolume,Y=ry's+Vn
5v
(1
HL= 12.47 m
Ratio = F*/Fo = 5.509 /1.695 = 3.25
vo=
-2/s'81)
4.686 kPa
,r=
ugff:ry
0'002548 m3ls
HL=
HL=2.827 m
Q
Given: y^=
The most efficient triangular
section is the half- s qu are
ti
G+GMC
Ym= --Yw
'
Ross-sectional area, A = d2 = 9 m2
Perimeter, P =2dA = 8.485 m
Hydraulicradius,R=
li
'$
f
:il
+
Ysat
'
Q = $.$
d=1.2m
G+e
=
1+e
-
Q=A
i!i
1i
1
n
Rz/3
G=
6:/5
385
G+(0.4882G-1) (e.81)
1+ [0.4882G
-
1)
2.74
0'338
= 1+ 0.338
x 100% =25'25o/o
22
0.015
S=
n
1+e
-
=o.6g6m
I
8.5 = 3.84
t!
e
n=
3.2+2(1.2)
st/2
1o.sr1
e = 0.338
A = bd = 3.2(1.20) = 3.84 m2
lli
G*flo.o8l
22..57 =
Yw
n = 0.015
R=A= A P b+2d
=
e=0.4882G-1
=1.061 m
Given: b=3.2m
;!
21.7
1+e
16
a
lil
Bolo
,]
2,12
rLUr
MC =
kN/m3
Y,^t= 22.57 kN/m:
on its apex.
$
21,.7
Given:
(0.686)2/3
-
st/2
0.00182
G. =
2'40
x
Largest unit weight possible = ]w x Gs = 9'$t 2'40
kN/ma
23'544
possible
=
weight
Lar[est unit
irl
iLll 17
Nozzle diameter, Dn = 50 mm
Flowrate, Q = 30 L/s = Q.Ql[ 6s/5
Given:
Weigh!W=250N
Vn-
a
A,
-
Saturated unit weight, y.u,
;[0.0s)'z
v"=
Saturated unit weight, y,^r=
15.279 m/s
Cl rr
-'* v=W
o
b
0.03(9810)
9.81
Given:
v = 250
v = 8.33 m/s
yz=ynz-lgl1
8.332 = L5.27 92
h=8.36m
18
19
m 20
zu
Degree ofsaturation
LLI
Porosity
Moisture content
2'9019
" ,* =
10 . o'sr
'"
1+0.40
1+e
2l'O2l kN/m:
G'
*
-
Dynamic force:
Fn=
S = 20o/o
'-
o.o3
..
Vn-
G. = 2.60
e = 4oo/o
G. = 2.60
S = 20o/o
e = 400/o
Dryunirweight,]sat=
- 2(9.81)h
Dry unit weight,
1},
=
ffi"0'at
kN/ms
Ysat = 7A.219
pr= po + Ap = 180 kPa
Given:
o: = 14 kPa
oa = 40 kPa
C,/\ P'l
toel
AH=H
L+e. _\p"/
^H
=
3ooo
o'30 t"n f lry l
1.30 " \125 )
1+
AH = 61.97 mm
[, I Situation 1
Given:
Weight of stone in air, W. = 350 N
Weight of stone in water, W. = 240 N
w)26
Given:
Confining pressure;
oa=BOkPa
tlJ
O
R
-
ys =
sln0-
40
Ws
sp.sr.= Y' = -#
T
Given: H=10mm
Ae = eo
1.8
AE
1+e,
er = a/z eo =
0.90
AH =
1o
*
9.81
=3.tBZ
lllSituation2
o'90
.1+ 1.8
AH = 3.214 mm
Newthibkness = H - AH = 6.786 mm
28
Given: H=3m
eo
Formula
I
- er= 0.90
AH=H
Lt_l
€
y,
350 = (31,274)Vs
V. = 0.01121 m3
=|sVs
-
E)27
=
t
31.214 kN/m3
100
0 = 23.57a"
en
l\
_ y.
240 y" -9.81
Center,C=o:+R=100kpa
stnQ=
Vs
350
60 + B0 = 140 kpa
= 40 kpa
7/z cta
Radius, R =
-|*J
w.= y.v. tr---- ,, I
y-)v. '
w,
-t,*
Plunger pressure;
o'1 = 03 + oo =
W* = (y.
Yi/. = ys Vs
or = 60 kPa
= 1.30
C. = 0.30
p" = 125 kPa
Ap; 55 kPa
n
;f
I
t_l
l.2m
Part
1
Part
3
lt
t=
,i:l
Part 1:
l'6=ywhA
,I
Fr, =
Fr' =
E
Ll
Situation 4
9.81[1.5] (3 x 1.2)
B=4.20m
b=0.60m
52.974 kN
E
*
-tj
Psin0x3=Fr,(3-1J
Psin45'x3=52.974x2
iT
P=
XMo=0
rii
t
3t
,*
$
1;1:
Part2:
Fh=y*[A
$
I
$:
84.758 kN
L.2,3)3 /1,2
AY
(1.2x3)Q.a)
= 0.3 13 m
,l
IMo=0
6
P
rA
f
=72.42kN
XMo=0
n=
Fr, =
F
Fu =
176.58
1
RM=Wry1 +Wzxz
lz(4.2- 0.6)[7)] x [(2 ft)@.2 - 0.6)l
9.81(2.1 / 2)(2.1 x 7.2)
25.957 kN
FS" =
RM/OM = 1118.88/353.16 =
- 0.6/2\
3.168
)
Part2
)
Part 3
Psin0x3=Fn[3-0.7J
P=
llISituation 5
Given: R=6m
L=10m
28.14 kN
0=60o
M Situation 3
Gatewidth,b=2m
T-
h=Rsin0
l
h = 5.196 m
pr=y*[A=9.81[3)[6x2J
Fr = 353.16 kN
P6=y* hA
y* [h/2J(h LJ
9.8L (5.L96/2)(5'196
Fn = 1324.3 kN
Fh =
Fr, =
P, = y* hA = 9.81[1.5J(3 x 2)
88.29
kN
XMs=0
P(3)=r,111-Pr121
3P = 353.16(1) - BB.2e(2)
P=
Part
+ [24 x (0.6 x 7J] x la.2
RM = 111B.BB kN-m
Psin45'x3=25.957 x2.3
Fz =
kN )
p14 =
[24 x
Part 3:
pn=y.hA
e.B1(*J[6 * 1]
OM = 353.16kN-m
Psin0x3=Fl(1.5+e)
P sin 45o x 3 = 84.758 x [1.5 + 0.313)
:li
[A
Fu =
OM=Fux(h/3J
oM=176.58x16/3)
)ti
!li
h=6m
y. = 24 kN/m3
y* = 9.81 kN/m:
I's=y*
Fn=9.81(2.4)(3xr.2)
c
iil
:ll1
.l;
H=7 m
,
49.94 kN
Fr, =
ar:!
b
Given:
58.86 kN
6m
2m
Fu
- y*
Venc
V,qsc=AescxL
n
x 101
AABC
R+BC
- 'i----:- xh
-
2
4or. =
6l36s.roo.;
2'
nR2o
360'
BC=R-Rcoso=3m
n[6)'z[60'J
= 4.533 m2
360"
Vasc =
Fu =
4,533 x 10 = 45.33 m3
Part 3:
For cylindrical rotating vessel
with vortex of paraboloid
9.81 x 45.33 = 444.7 l<N
Part 3:
since the total hydrostatic force on cyri.dricaI sections passes
through the center of the circle, the moment due to Fr, and Fu about the
center ofthe circle are equal.
F, x = Fr,
(h13)
444.7 x = 1324.3(5.196 /3)
x = 5.158 rn
below the vessel, the
relationship between
H, h, and y is:
H=,W
where:
h = depth of liquid
[+] Situation 5
t
before rotation
H = depth of
D=1.7m
I
Fart
1:
or,= 90
rpm = 9.425
x=r=0.85m
rad,/s
' 9:t
2s
Ileightorparaboroid,y=
Part 2l
V"i.r = ir[0.85)2{2.1
v^,,r=
=
l
? 2!toss) =3.271 m
2(9.81)
Volume of liquid unchanged.
- 1.4) = 1.5889
ff (yz-yr2)
m3
In this problem,y =3.271. m and h =
u^nr=
ffil3.zztz-t.L7121
Vai,z = 3.2365 m3
Vspilled = Vair2
-
Vair
liquid
during rotation
Y = height of Paraboloid
rl
1
Vspiirea
= 3.2365
Vspiref =
-
rhus,
H=
JN
= lrllA)(3.rU
H = 3.026 m
1.5889
1.648 mz
1
4m
M Situation
7
0 = arctan
(\/2)
= 26.565
Top widl.h, T = 5 + I x 11.2' 2J = 9.8 m
g = 5+9'B 1.2r = 8.88 m,
2-
P=5+2
x 1.2 csc0 = 10.367 m
R = A/P = 8.88/L0.367 = 0.857 m
v = Q/A = 30/B.BB = 3.378 m/s
Specific energy:
ll
Given: ht=
hz= 3.7 m
i
C"=
l
h:=1.2m
6.7 m
!l
H=
!
)a
+d
1
Slope of channel bed:
^
1
v-
Head,H=hr-hz=3m
R2/3S1/2
n
t:
ii
vo =
C,
J2gH
Vo =
2(e.81)
3.378= L
:l
1g.g571r,rs',,,
0.013'
S=
2.l29.u)(3) =7.672m/s
0.00237
Boundary shearing stress:
'ii
:]
t onoZ
f{= Jrro +1.2=l.7B2m
0=45'
.:
,:i
I,,2
Part 1:
to=Y*RS
v'
i
r-
'?o
=
9810[0.857)(0.00237)
t"= 19.925Pa
zo.Br)
Y=1.50m
Situation 9
El 100m
Part 2:
U=
t"
(7.672sin45")'z
-(hz + h3) = -zl.Q rn
gx2
-:-"y=xtan0- zvn'cos'O
9.81x2
-4.9=xtan45o2(7
.674'z cos' 45o
x=9.197 m
Part 3r
X=Voxt
9.L97 = 7.672 cos 45" x t
t = 1.695 sec
Qa
hr=KQ2
__
m Situation
B
0.0826fL
D5
Given:
Q=
500 m,250 mm
f = 0.0178
= 0.06 m3ls
lQ
6:/s
Kar =
n = 0.013
b=5m
Side slope, ss = %
Depth,d=1.2m
Knn =
0.0826(0.022s)(1s00)
0.45s
0.0826(0.01781 (s00J
0.25s
= 151.1
= 753
Koc =
0.0826[0.0168) [7s0]
hfz = Kos
0,3s
hlr=-0.0826 f L,O,
'''
= 428.3
'z
D,,
0.0826[0.02J [1s000) (0.e4seJ'Z
hfz -- 753[0.06)2 = 2.711. m
Qe2
=200-91.56
D.,t
Dr = 0.73 m
0.0826 f 1..o,.'z
Qs=Qa-Qe=Qe-0.06
= 428.3(QA - 0.06)2
hfr = Keo Qe2 = 15 1.1 Qaz
hf: = Krc
Q3'z
Ela*hfr-hf:=Elc
100 - 151.1 QA'z Qa =
htt=
-
0.06)2 = BZ
-
Dr = 0.602 m
0.06 = 0.159 m3ls
Els=Ela-hfr*hfz
Els = 100 - 151.1(0.279),
Els = 90.04 m
-
Z.Z
tt
hf:=0.0826f r. o-'/
Dr'
=91.56 -15.21
0.08r6(0"021t6000)(0.5208):
D.t
tUJ
Dr =
Situation 10
Lr = 15000 m
Lz = 9000 m
L: = 6000 m
= 35.56
Dru
0.219 m3ls
Qc = 0.219
91.56 - 56
D"'
0.0826[0.02J Ie000] [0.434J'z
428.3(QA
= 1,08"44
= 76.35
0.512 m
+200
Situation 11
Given:
G,=2.67
S=
40%
e = 0.45
f = 0.02
Part 1:
C +Se
2.67 ra.4{t{0.45)
/m_--rw
7r, =
1+e
Y-Demand per capita = 150
lit/day
Population ofC = 250,000
Population ofC = 300,000
Part 2:
+15.21
y,l
=
'
I +e
-_
0.15 x 250,000 = 37,500 m3 /day
Qz = 37500 + (24 x 36001 = 0.434 m3 /s
Q: = 0.15 x 300,000 = 45,000 m3/day
45,000 + (24 x 3600) = 0.5208 m3ls
Qs =
Q1 = Qz + Q3 =
0.9549 m3/s
Part 3:
G+e
I+e
f)s.rt =
14).2,82
[e"81]
kN/m:
?77 sg.611'
,,,=
' i+0.45'
7,,
7a
Q2 =
1+ 0.45
]N
= 18.064 kl{/rm3
Ys.,r
=
+0.45 .
_2.67_-_
((r.[J1l
1+ 0.45
2'1..L!
kN/rn:
'1sat=
,u
'til
m Situation
12
lri
Sieve
No.
Diameter
[mmJ
4
4.760
2.380
6
10
20
40
60
100
200
2.000
0.840
0.420
0.250
0.749
0.o74
Soil
A:
Percentage finer than No. 200 = 5o/o < 50% (Coarse GrainedJ
Percentage finer than No.4 = 90% > 50% (SandJ
Dao = 2.2 mm; D:o = 0.66
Dro = 0.16
Cu=Dao/Dro =3.66<4
C.= D:o/[Dro x DooJ = 1.9 (between 1&3)
This soil is SW
Soil
B:
Percentage finer than No. 200 = 33o/o < 50% (Coarse GrainedJ
Percentage finerthan No.4 = 1o0o/o> 50% [Sand)
Percent Fines [No. 200) = 33o1o r rr
The soil is either SM or SC
SOIL SAMPLE
A
B
C
Percent Pass ng
90
100
100
64
90
100
54
77
9B
34
59
92
22
B4
L7
9
51
42
35
5
JJ
63
42
29
47
PL=29
24
PI=LL-PL
79
70
PLASTICITY CHART
LL= 42
LL
PL
Y:
ai
PI=13
50
C]H
&
x 'r0
a
430
,tNt
20,)
CI,
izo
Ei
?ro
J
A.
MH
OH
1.3- -
"l:,7 \,lLi OL
"0 10 20 30 40 50 60 70 80 90
0
LIQUTD
UI{IT (LIi,
%
The Atterberg Limit is above A-line with PI > 7.
This soil is
Soil
C:
SC
Percentage finer than No. 200
=
630/o>
50% (Fine Grained)
LL=47<50 [ML,CL,orOL]
PI=LL-PL=47-24=23
This soil is CL
PI,ASTICITY CHART
::
50
*
I+o
C}I
,ll\
a
I
:,1fi,1
=Bo
F
M}I OH
.CL
c20
2o)
FI
t10
,f
ai
g
0lU
MI
ol,i
10 20 30 40 50 60 70
I,IQUID LIMIT (LL).96
80
100
l^00
m Situation
13
,b
Recha
0=
Lll.
?r"r
u
\
\
\\
\}\1
El..{i5*
tr = 49.3 kPa
cir=C-Rsin$,
oi = 300
'
\_-#
\
\
!
Given:
Ita
:
Confining pressure;
o: = l-10 kPa
Plunger pressure;
oa= l70kPa
m
k = 35 m/day
n = 25o/o
01 = 03 + oa = 110 + 170 = 280 kPa
Radius, R = 7/z oa = 85 kPa
width,w=4km=4000m
Partl: Flow,q=P14
3:
t= 4000 \
stnq=
4 = 16,000 m2
Q=
3s[0.01](16,000J
Q=
5600 ms/day
R
SlnQ=
'
e = 45" +
-
tl/2
0=
45'+25.842"/2
e = 57.92\"
Vs=
3
5r0.011
-
D Part
)
Part 2
Effective stress at point of maximum shear,
Situation 16
1'-sindr
*"_
4000
=
1+sin<f
Lt,U0'
C
= 195 kPa
=0.4903
1+sin20"
G = I.65
* * 0.50
\:.4
LLll
Part 1:
Situation 14
Y'"=
Given:
= 250 kPa
oa = 350 - 250 = 100 kPa
R= 7/z oa = 50 kPa
C=o:+oa=300kPa
lv =
G+GMC
j1;i-7*
or = 350 kPa
o:
1
Part 3:
0.25
vs= 1.4 m/day
Time for water to travel 4 km downstream.
Dis tan ce
B5
195
b =25.a42"
Seepagevelocity:
vki
nn
Part
center,c=o:+R=1g5kPa
I = 75-65 =0.0t
L 1000
Cross-sectional area,A = w x
50 sin 9.59'
Situation 15
,'.,.
10OO
-
-
or=291.67 kPa
t
Hydraulicgradient,i=
9.5s'
tr=Rcosd
ls
Part2:
sin $ = 56736s
= R/C
Tn, =
l0m
2.65+2.65(0.70) x
9.81
1+ 0.50
y^= 1e.064l*r-"
'(,
Fat=TzKuY,.H2
P
= l7lo.49
031
^,
F"r = 467'35 kN
(19'064) [1 0)'z
1i.)0,/r
$=2oo
t'.
Part2:
2'65
va=
'
G
Yd= lYw
'
1+e
I +0.50
* 9.8'L
F
F
oz
^z
= a/z(O.49031 (17.33
= 424.9 kN
1J (1
'
0),
Ns
ip, =
Part 3:
yn= 2'65-r ,9.8-l
'
lYw-1
1b=
'
G-
1+e
,
1+ 0.50
-
3t1o)s2n(10)'
5000
;;:
=
=1om
0.4775
x 0.477 5 = 23.87 5 kPa
Situation 18
Given:
10.791kN/m:
ya =
R='l?+* = r6+1d
m=:
- 2nR'
ya= 1,7.33L kN/m:
Faz=YzK^yaH2
o
lNa
AP,=
Footingdiameter, D = 4m
Load, Q = 1.6 MN = 1600 kN
F
7/z Ka
at =
r/z
]u H2 + y* H2
F
^s
=
1/z
(0.49 o3
) ( 1 0. 79
1 J
[ 1 0)2
+ a/z(9.8 I) (1 0)2
Partl: t= rt
F,r = 755.04 kN
q=
Q=
Increase in force:
AF,=755.04-467.35
AFa=Fas-Fa1
AFa =
m Situation
Part2:
287.69 kN
Pressure
"(N
l27.32kPa
atz= 4m belowthe center offooting:
Q,=Qxle
17
Is=1-1lN
Given:
Q=
5000 kN
N=
Hr=Bm
Hz='J-2m
N=
B=4m
t=Hz-
1600
Hr
Ap=5000
4,
AP = 312.5 kPa
fi
Part2:
t1
7=111+t/2
z=B+2=10m
Part 3: At
a
.t
t,)"
-iJ/2
-r) :
r
R=D/2=2m
P/?
l(2/4)"+r.]
=1.3e75
q,= 727.32 x 0.2844 = 36.2lkPa
il
9
Q
ap,=
' (B+ z).
L(R
Ie=1-1/1.3975=0.2844
=4m
Part 1:
ao=
r82
r.
ap=
5ooo-
(4 + t0)'
Part3:
H,
Depth"2" whereq,= %q
Q,=exr,=0,[r
rrl
%q=q "
[, U*y*U,,.1
r1l
1/6 -= | 1-"" L
z= 5.56 m
=25.5rkpa
point below the center of footing, r = 0
.o#_,yl
l1z1z1"
11'''
1
M
.i
:ti
l:l ii,
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r. trl:i1d i.il.t: liir;;l'i
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:;
tr,
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1'::',
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t
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.lolutions to May ZOLS Examination
[, , Situation 1
-
0.60 In
m1
An isotropic mqterial has a stress-strain relationship that is independent of
the orientation ofthe coordinate system at a point. An isotropic material is the
one having the same elastic properties (E, p) in all directions at any one point
ofthe body.
A material is said to be homogeneo.rs
are the same at all points in the body.
m2
il the material elastic properties
(E, p)
ebsticigt
LE3
work hardening, also known as strain hardening or cord working, is the
strengthening of a metal by plastic deformation. This strengthening occurs
because of dislocation movements and dislocation generaiion within the
crystal structure ofthe material.
BLICKLING - 1. Failure by lateral or torsional instability of a structural
member, occurring with stresses below the yield of ultimate values; 2. pulling
away of a panel edge from its support structure.
tu4
Y
XMe=0
Re(1.5J = P(0.60) = 1s00(0.60J
Rs=600N
IFv=0
Pa+Rr=P=1500
Re=900N
|t;
tit
,in
;lr
t7?
It:
i
aii
ri*
,t+
.fli
i:l
trl
XMe=0
r:lti
$ll!
:x:!.i
,}*
r;i
REF=
;iid
iL.j
i lr':
l,iI
,,9
o
RsF
XF"=0
P=REF+W*
P = 2.446 + 20 sin 0
P=
i.l
:i$
2o
rr.zt
= 9.81
'
REF = 2.446 kN
;i4
::,j
ti:4
[a
7.623 kN
Ro[1.5)=1500[15-06]
Ro=900N
m Situation 2
I
Situation 3
L
L=20 m
W=50kN
d=1.5m
240 kN
240 kN
Part 1:
For equal tensions at AC and
BC,xr=xz=L/2=10m
Parts2&3:xr=5m
xz=20-5=15m
'
IF
I
l;,
Figure A
0r = arctan (7.5/5) = 16.699"
0z =
or=90o-0r=73.301"
az=90" -02=84.289o
cr: = 1B0o - ct1
-
arctan (1.5/15) = 5.711."
c\2 = 22.4.1,"
By sine law:
F*-W
sin cr, sin
I:.
I ii'
ril
Fr.
cx,3
= sin(22.41")
--50
Fec =
i:{
tr
x sin 73.301"
125.62 kN
1;
,'i
ria
Length of cable:
.
L=Lac+Lec
L=xrsec0r+xzsec0z
ilii
L = 5 sec
;ii.
L=20.295m
::a
,i.,
{16.699") + 1S sec {5.711.)
In Figure A:
IMe=
0
B'(6s) = 90(s) + 2a0Qs) +240(4s) + 90(5s)
B" = 330 kN
ttill
tli*
lj-
tr*
{
By symmetry, Au = Bu = 330 kN
In Figure B:
:t
IMc=0
240[s) +90[2s) + BH(HJ -B,[3s) = 6
190 kN
$
it.+
r$
]$
ii
,:1
i'{
i;
r.i
XFv=0
Cv+Bv-240-90=0
Cv=0
By symmetry, Au = Bu = 190
kN
m Situation
4
Notes:
P,,
l
- Cables carry tensile force only.
- For two diagonal cables, both cannot be in tension at the
same time.
t,
v
x
rt
,H
+r
I
1.5
kN
.aq
r)
ll , *.-..*.{r>
il ,..
dj
+i-
P,
AD
Es = 1'5
I
t*
"ttr
kN
L-J'
v
Ev = 2'3
LMr=0
Bv(7.5J
kN
- 1[10.s) - 3(4.s] -3(2'2s) - 1's(31= 0
Bv = 4.7 kN
Wind pressure,p = L.44kPa
1.5
Pr =
L.44[0.8][6 x 6) =
41.472kN
kN
kN
Pzu = 1.44(0.1)(3 x 6) = 2.592
Pzv = 1.44{0.\ I (4.5 x 6) = 3.BBB
t 6) =7.776kN
1.44(0.2J[3 x 6) = 5.184 kN
= L.44(0.4){6 x 6) = 20.736 kN
Psv
= L.44(0.2)@.5
i=
Pzs =
Ps
i=
XMa=0
[Pr + P+)(3) + [P:H
Bv = 16.092 kN
)l-v=0
-
Pzu)[7.5) -Pzv(2.25)
Bv+Pzv+P:v-Av=0
27.756kN
Av =
Fn=Pr+P+-PzH+P:u
Fs = 64.8
-
Psv[6.75)
- Bv[9) = g
kN
4.7
Section x-x
Part 3:
kN
kN
lr
Spacing of frames, s = 6 m
q
'ni
=i
t
3
kN
2.3
kN
Section Y-Y
Section x-x:
The required cable force for equilibrium is 3.7 kN downward, cable BG
cannot support this load because it wiil introduce compression on that cable,
hence Fsc; = 0.
At joint B, Fsr = Bv = 4.7 kN
Section y-y:
tan
cx,
=
3
/2.25; o = 53.13.
Fcu
The required netvertical cable force for equilibrium
is 0.7 kN upward.
Thus FcH = 0.
Foc sin
Ar
joinr
c = 0.7
Fnc = O.B7S
sin o. = 3
Fou=3-0.875sino,
FoH =
6
kN
D:
FoH + Fnc
[U Situation
(3) + 1(2.25) = 1.5(3)
0.75 kN
Fos =
2.3 kN
.
Part 1: Maximum positive moment occurs when the
loads are within span BC.
Smaller load, P. = Wr = 19.6 kN
Total load, p = Wr + Wz = 98.2 kN
Span length, L = 22 m
d=4.3m
-P,
jvlmax= tPL
dy
4PL
lsa.z1zz1-ts.6(4a]'z
f,i
l'lmax-- -4(e8.2)(22)
M*u" = 498'78 kN-m
.5
Bv=2kN
1,5
kN
At joint A:
Fap
sin 45" = 1
= 1.414 kN,
Fcu = 0
Foc sin cr = 1
Foc = 1.25 kN
The maximum negative moment occurs at B when the heaviest load (Wz)
!,
is at point A as shown.
M*u, = -Wz (3)
Section Y-Y:
The required net vertical cable
force is I kN upward.
Thus
moment:
Part2: Maximum negative
Et= 1kN
0=45'; g=53.13o
Fae
kN
M.*
= -78.6[3)
Mm,r
='235.8 kN'm
Part 3: Maximum shear:
w,
0
Section Y-Y
1kN
when the heavier load is at that point.
x
Rc 22 = W t(17.7) + W z(22)
The maximum shear is at
IMe=0
C
R*
Rc = 94.369 kN
Vmax
= Rc =
R = wL/10
3w U8
IH
94.369 kN
Llll Situation B
'll:i
= 36 kNlm
z
_lt
Fixerd
fl,
,
Ro= 1wl,
o
o
= o [36')[Bl
Ra= 10BkN
RA
Ms=ReL-wL2/2
Me = 108(8)
Me = -2BB
-
54 kNhn
,._J
il,I.,
) Part3
36(B)'z
kN-m
/2
Situation 10
)
Part 1
Part 3:
Location of maximum moment, C (zero shear):
Ra-wx=0
108-36[x)=a
x=3m
t_
r
v
Ull Situation 9
Part
1;
Mr= 1/z(3)(3), 3 t3l + 1/z(3)(54)
><
+ (3)
Parts 2 & 3:
R, =
1*rL *
810
a-,
?1
RB=:(3)(3)+ ^ (sl)[3]
L
Rs = 18.675 kN
3(3)[1.5)
Mmax = Ma = 18.67 5(3)
Mmax = Me = -33.975 kN-m
-
Vnrax
V-,*
F-
f
p,#
Me = 90 kN-m
- Yz{3)(51)1
= VA = 3(3) + 1/z(3)151) - 1'8'67 5
= V,q = 66.825 kN
x
-l-
--
F,',.
(2)
P=20kN
Compression parallel to grain, p = 13.2 MPa
Compression perpendicular to grain, Q = 4'3 MPa
Shear parallel to grain, F" = 2.32 MPa
A
ti
z3
Fixed
Part
1; Axial stress of ntember AC:
At joinr C:
By symmetry, Fnc = Fsc
XFu=0 2Facsin0=p
Part 1: Minimum thickness of column:
2 Fec sin 30" = 20
= 20 kN
F,qc
f^.=*
.IAC=
20,000
_=--
fac =
1.33 MPa
100(1s0)
Part2t
Minimum value of x:
In Figure (21:
pu
1200x103=60xA.
P=F.xA.
A. = 20,000 mmz
,
Ac=
20,000=t(2802-d2)
;(D2-d2)
^
Thickness, t = Yz(D - d) = 1/zl290
Thickness, t=24.96mm
Part2: Minimum diameter o[ base
= Fac"
Fv Av
= Fac cos 0
Minimum value of y:
In Figure (3J:
D = 280 mm; t = 20 mm; K = 1.0 (hinged at both endsJ
d = D - 2t = 280 - 2(20) = 240 mm
Rc=F.xArea
=
13.2(4.3)
13.2sin'?30.
r
4.3cos2 30o
F. =B;699 MPa
=
F,lr.r
Fc x Area = Fec cos 0
8.699 x [100 x y) = 20,000 cos 30o
Y=
[i] Situation
plate
Part 3: Effective slenderness ratio ofthe column
3:
Rc
19'91 mm
I= *[D4-d4J
I=
A= +(Dr-dr)
A = + (2802
*
[(280)4
r
92.195
L=3.5m
D=2B0mm
Situation 12
R=
Colun"rn:
F. = 60 MPa
v55'*1*
=39.825
omax=65+R=104.825MPa
Concrete:
omin=65-R=25.175MPa
= 9.6 MPa
rmax=R=39.825MPa
-
Base plate
13B.B5B x 106 mma
- 2402) = 16,336.28
1_6,336.28
KL _ 1[3,s00)
=37.e6
P = 1200 kN
- (2+01t1=
mm2
138.858x106 _
92.195mm
,= ,lt/ a
Effective slenderness ratio:
11
Fp
230.07 7)
Dp
x=74.66mm
Pq
E"- psin'zo+qcos'zo
-
1200x1Oz=Q.$x iDp2
= 398.94 mm
P=Fp^Ap
2.32 x IQA{y) = 20,000 cos 30o
Part
d=230.077 mm
Situation 14
r
Given:
Ag= J$QQ
30
t*
rnn'tz
=12 mm
xr=50mm
T
R
yr=60mm
Allowable stresses of plate:
19
*ru
250 MPa
400 MPa
Fv =
F,, =
Bolt:
Gross area, Ft = 0,60 Fy = 150 MPa
Net area areai Ft = 0.50 F, = 200 MPa
Shear, Fu = 0.30 F" = 120 MPa
Bearing, Fp = 7.20 F" = 480 MPa
19
dr=25mm
hole = 27 mm
F" = 90 MPa
x
Part X:
P
based on bolt shear [single shear):
P=FuAu
P=e0x +t25)r[B)
P=
100
Part2:
p=
m Situation
P
based on bearing:
FpAp
P=480 x [25 x 12J[B]
P=
13
Part 3:
353.4 kN
P based on
1152 kN
block shear:
6
t-
.-
WZr'"
\a
7nr
t..
o*
&,
'w,
30
x]
-tr
Xr
Xr
Block Shear
t"y is in counterclockwise sense, thus "x" is represented by point
Thus, o, = 90 MPa, oy = 30 MPa
R
=
fr0140'
Shear area
6, = (5xr - 2.5 dr,l x tw x 2
6, = [5(50) - 2.5(27)l\z) x 2 = 4,380 mmz
Tension area
At=lZyr-2dr,+
B.
2
x
"2
-lL]
t*
'lY
= 5o MPa
At = [2[60)
t
=
- 2(27) * z, ].1'^- l(12)
4[60) "
1042 mm2
P = Ft
At +
Fu
Au
lli
lii
l,i
ll
2,
lf
Q
lt
i,:
Iri
I --rzs'
,..] lzso'+[xb'Fyb')l
. 2 ,..]
'"'l ^- 1112
'l=2x1251ltz +[x,'+y,')l+250
i
t=2* t25ll?!^--l tz *ts r.zs'+tzs'1l*
'l ruo['Jo'*t3r.2s' r0')l'l
Part 1: P based on gross area yielding:
P=FtxAe
P=150x774
P=
J
116.10 kN
= 6.O2Z v lQc
Part 3: Allowable value of
Parts 2 & 3: Values ofLtandLz:
!iil
Total weld length,
P
At corner 1:
#ti
P=Fu*xA"
Direct
Y = 71 + Lz
P=Fu-x0.707tL
i,
176,100 = 1.24 x 0.2 07 (6)L
L = 220.72 mm
itrl
= 9.5 mm.
P
L = 500
P
fTx-
h{23)=Lz(75-23)
Lzx.z
Lt = 2.2609
lil
i,l
Rttr=
'l
75 rnm
Ltxt -
t.
T = P x s = P x [250
,
;;i
Roy =
if
Torque: xr = 125 - 31.25
t,
i
load:
L-Lt+Lz
-
=o.o02P
= 93.75 mm
31.25) = 218.75P
-n"Y -
21.8.75P[93.75]
6..o22r1ru
Rrrv = 0.00341
Lz
220.72=2.2609L2+Lz
Lz= 67.7 mm
Lr = 153.03 mm
'li
It2
pp+/6m
llr I
li
= 31.25 mm
I =>L(L2/12+ (xz + yzJl
Fv
iir:
itl
/2)-2s0(0)
500
yl;1.1,z
tl,
ili
lli
1125)(1.2s
x"=125/2-x=31'25mm
Gross area, Ag =
= 250 MPa
F,* = 124 MPa
t = 6mm.
Ft = 0.6 Fy = 150 MPa
ii
,_
m Situation 15
xr=23mm
rl
0.5[400)[1,042) + 0.3(400)[4,380)
P=734kN
I
1l
Part 2; Polar moment of inertia
P = 0.5F, At + 0.3Fu Au
P=
Rrr" =
TY,
I
Rrr, =
P
21.8.75P(125\
---#
6.022, t0"
Rrr* = 0.00454
P
,i,
il
ir
)
lUl Situation 16
Rr =
Fr- x 0.707
t-
[1mm)
Rr = 973.89 N
250 mm
Fu* = 145 MPa
R,,
P=160kN
[Rrr*J2 +(Rrr, +RrrJ2
r{1r
&''
Part 1: Direct load
L=250+125x2
L=500mm
P=
F*
x 0.707
t*L
160,000 ='1.45 x 0.707 t*(500)
= 3.12 mm
t*
)
^' *,.,
R, 'tI\
\
R",t \n
t\
Rr=145x0.707(9.5J[1]
s73.Bs=@r;'
P
= 137.9 kN
E0 Situation 17
Column length, L = 12 m
Braced length:
F.,
lv
300 mm x 16 mm
L"= 72m
0.746')
z+a
2 /l- l.Bes
F" = 94.45 MPa
P=94.45x19,200
Patt=FaxArea
P=
,
1,813.4 kN
16) = 19,200 mmz
1
t- =
F.= l-l
I
Ly=L/3=4m
k*=ftr=1
A = a[300
I
[
I a')
F,-11--l'
l. 2)FS
x 332: _
*(300
1.2'
268,
*-x
i
3003J
,I
Situation 18
E
o
I* = 311.859 x 106 mma
1-*----*l
e
I,'
=
o
o
(332 x 3oo: - 3oo x 2683J
.1.
12'
lv = 265.779 x lQ6 6ry1+
r..
f - r-E rsse
- rr
10,
a ! io,zoo = 127.447 mm
/r,,- r_
lzos.izy n"_
r:
a ! ro,zoo =117.655mm
V
r..-
V
rl
L
Parts 1 & 2: Slenderness ratios:
--,,-,----,,----,!
254
SR*
SRy
=
k. L-
5p.
rx
KI,
t
= --lL-
=
5p*=
v
1[12,ooo)
L27.447
1[4,ooo]
117.655
= 94.157
Solve for y & INa using MODE-3-2:
Part 3:
Maximum effective slenderness ratio,KL/r = 94.157
E?n
uc- /_
VF,
ZTE'z(2oo,ooo)
248
Cl=
'
r
/18
2
6002
/18
3
15gz
/t6
c. 125.17
53cr'
F)= - +
38B
= L26.17
= 0.746
0.7
FS=
!
*
JO
FS
= 1.895
1rc.2+ol
-n(150)2
/4
162,328.54mm2
y =3lB.4B mm = yt
INa = Ix + n Yon2 =5,319'71' x 100 mma
Modulus of rapture:
f. = smaller of (0.7
94.157
a/2'350){600)
1/z(250)(600)
A=n=
KLfr<Cc
KL /
1
Uggz
Freq
v
400
200
300
x
Area
=33.998
to.74q3
tF\
f,
)
and [0 7 ^
=3.7o4MPa
f. = 3.528 MPa
1B
f't )
0.7 '1'.8 f.t = 3.528 MPa
Part 1: Cracking moment:
M61 =
fI 'n'
-t
_ 3.828(5,319.7 1.106)
r,,r
rvr at -
y,
318.48
ti
M* = 58.93 kN-m
';i
Part2:
,lc _-
M*Y.
f _ 58.929x106[600-318.48]
_
l*o
5,319.7Lx106
f. = 3.119 MPa
!l
,!
fal
I
,.i
Part 3:
Positive moment due to weight of beam:
wb = yc Area
wa = 24 x (162,328.54 + 100021
wr, = 3.896 kN/m
M5=
wL'L
3.896[6)'z
^,. _
o
o
Mr, =
17.531kN-m
Additional moment the beam can support:
Madd = Mcr
M,ao =
-
Mr, = 58.93
41.399 kN-m
-
17.531
.
Assuming tension steel yields and compression steel will not:
i.e. f, = fy and f. < fy
Additional weighr:
Madd =
wooo
L'
4\'399=
w"aa
(6)'
f.
c-d'
= 600
B
.wadd
LL-ll
= 9.2
f. = 415 MPa
C
kN/m
d'= 60 mm
Br = 0.85
T=A'fy
Situation 19
f,=28MPa; fy=415Mpa
Column dimension = 400 mm x 400 mm
h=350mm
Parts 1 &2: At support, the top is in tension and the bottom is in
Cc
]
x
fc (Brcl
C'. = A'. f's = A's
"
b
600
.c"
c-d
c'.
c
b=320
(zolz= 2513 mmz
Compression A'. = 4
aata
tata
= 0.85 f'. a b
C. = 0.85
compression.
TensionAs = B x
320 mm
+ (28), = 2463 mmz
d=h-65=350_65=2B5mm
IFH=0
1-[.+C's
A, fv = 0'85 f'. [Br c) b + A', x 600
t
d
C
2513(415)= 0.85(28)[O.BSc)[320) + 2463' 600
c = BB.L7 mm
t-60
Check if assumptions are correct:
c-d 6oo BB'17-60
c =
BB.l7
d-c
t = 600
600285-88'17
c =
BB..J.7
f', = 6o0
Ifl
If,i
fi
2 = Br c =
Mn = Cc x (d
x (d - a/2) + C'. (d - d')
M, = 0.85 (28) [0.85 xBB.17] [320)(435
Mn = Cc
= 19L.72 MPa < fy IOKJ
= 1339 MPa >
fy
(OKl
L
- a/2)+
C,. (d
M. = 0.85 [2BJ [0.85 x BB.
M, = 247.535 kN-m
il[i
fii
ffi
*:l
llr
F
Iltl
1
-
7J I
O
ZiOl Zes _ 7 4.s s / 2) +
24 63 (1.e 1,.7 2)
(285 _ 60)
(f
'o
,]
-c
M, = 0.90(247.535) = 2ZZ.7BI kN_m
h=250mm
tow,LnzfL0.
PART
Ln = L
- column width
=6
Mus0Mn
I\4,
=
Ln'
10
w,
-
0.4 = 5.6 m
222.781-
w,
P=
Part3: h=500mm
d=500-65=435mm.
Assuming t'. 600 t d'
=
,a:=
600 kN; Loss =
15%o
P"n =
600
- 15%(600J = 510 kN
Part
1:
f. = -
P"n
A, fy = 0.85
"
Part
2:
f. = -
,nd f, = 415 Mpa
P"u
A
6P"ne
bh'
f.
[pr
cJ
b + A,. x 600
Part3:
5d'
6[s10,000)[100J
2so{4sq'Z
510,000
,
= 450/6 = 75 mm
Situation 21
Load:
p
= 147 kN
w, = 19.6 kN/m
Pu
Check if assumptions are correct:
= 1,97.72 MPa <
= 2360 MPa >
0.85[88.17] = 74.95 mm
_
"--zso(+soJ
To produce zero stress atthe bottom,e
e
mm
f. = 6o0 t-d' = 600 BB'17-60
c
88.17
d-t
f, = 600
c = 600435-BB.l7
88.1.7
c
f. = -10.578 MPa
251.3[4ts)= 0.B5t2B)(6.bsc][320) + 2463" 600 c-60
a = Br c =
250[450J
f. = -4.533 MPa
C
c = 88.17
-510.000
i-
A
T=C.+C,,
,
PART 3
PART 2
1
(5.6J'z
'10
w, = 71"04 kN/m
lli
ir,
Situation 20
d,)
The maximum factored negative moment in beam
EFGH is
at exterior face offirst interior support, and is equal
lji
ti
4.95 / 2)
il
Design st-rength,
,TI
II
7
0.85[88.17) = 74.95 mm
lir,
14i
-
2)[435 - 60)
M. = 403.99 kN-m
+ 2463(1e1.7
fy
fy
(OKJ
tOKl
Beam:
L=7.5m
b=260mm
h = 600 mms
d=540mm
dr.
= 10 mm
ry*
(
\
M*
=h/6
f', =
2l
fyn =
275 MPa; fy = 415 MPa
MPa
s=
A, = 2 An = 2 x + (70)2 = 1.57.7 mmz
L)
Re= TzI2(147J + 19.6(7.5J1
415 MPa
fyn=275 MPa
fy =
V,=209.92/0.75
F". = 0.BB MPa
Vuv = 480 kN
V" = 279.89 kN
0.17r
fl
b.
d
v.=0.17(1)
Ji
Qeq$+o)
V. = 109,377 kN
V.=Vn-V.
Auf'nd
^
,f i
Ua
170,51,0
s=
136.8 mm
at third point:
Yor = 220.5 - 79.6{7
Vn,r = 171.5 kN
V.=Fu.b-d
V. = 0.BB[600J[384) = 202'7 52 kN
Vn =,Vuv/0
V, = 480/0.85 = 564.706 kN
y.=!n_Vc
Y"=564.706-202.752
V. = 361.954
.5
/3)
s=
A" f"h d
"' -
V.
Vnuz=Vorr-Pu
1"
A,=4x +(12)'=452.4mmz
_ 157.1(27s)(s4))
V.
Part2: Spacing of stirrups
Vn,r=Rr_y7,x[L/3)
d'= 66 mm
d=3B4mm
V'=279.89 - 109.377
s*a*= d/2 = 270 mm
=
d'=40+ 12+1/2(2A)
d=b-d'=450-66
V. = 170.51 kN < 0.33
5
195.5 mm
h=600mm=b*
f,=27.5MPa
V, = 209.92 kN
=
s=
{,lSituation22
b=450mm
Part 1: Spacing of stirrups near the support:
Vr=Re-wud
Y"=220.5-19.6(0.54)
v.
---Tts6o--
27O mm
Part 3: Spacing of stirrups atmidspan [zero shear] = d/2 =
Re = 220.5 kN
V.=V,/0
1s7.7(27s)(s4o)
s=
V"
-
tr= I (normal weightconcrete); 0=0.75
Rd= 7/z(2P" + w,
Au frh d
Yo"z=177.5-L47
kN
)
Part 1
)
Part 2
452.4(275)1384)
361,954
s=132mm
Vouz = 24.5 kN
Part 3:
Use Vu = 171.5 kN
V.
= V,/0
s = 100
A" frh d
Y"= 177.5/0.75
V"
Y"=228.67 kN
V'=Vn-V.
kN < 0.33
' -100=4s2.4(27$(384)
v.
Y,= 477.734kN
Y,=228.67 -109377
v,='tt9.29
mm
Vn-V'+Vc
,l{
ua
Vu-$Vn
y,= 477.734 + 202.752
V'
= 680.486 kN
V" = 0.85[680.486)
V, = 578.4 kN
M Situation
23
b=450mm
h=600mm
a,n = o.oe
sh
f'
33e.3
f
sz
f,=27.5MPa
fv =
415 MPa
s; (3sBJ[27'5)
275
= 105.3 mm
Payt 3: Maximum spacing of lateral reinforcement:
fyt = 275MPa
dr, =
=o.oe
Maximum spacing is the smallest of:
1. b/4 = 450/4 = 112.5 mm
2. 6 ds= 6(28) = l"68 mm
28 mm
dt=12mm
d,= 40 + dt+ dr/Z
d'= 66 mm
i
Ae=bh=450[600J
Ag= IIQ,QQS
N
* 350-h'
3.
1oo
smax
= 112.5 mm
= 1oo
J
+ 350-199
3
= 150.3
Situation 24
111m2
4,,= ](12)2= 113mm2
br=b- 2x40=370mm
h.*=hr-12=50Bmm
h.r=5r-12=35Bmm
hr=h-2x40=520mm
ht - 1,92,400 mm2
-2d')/2 + du + dt = 199 mm e h,
- 2d') +du + dt= 196 mm
Acr = br x
h,r = [b
h,z = [h
Part 1: Shear parallel to short direction:
h. = h.* = 508 mm
Asir=4x Art=452.4mm2
A,r,=0.3lh.i-
[L-r']
f"h ( A,,
452.4
=0.,
)
sr = 73.6
Ash =
sh
s, (5oB)[27.5J ( zzoooo .,1
_
f'
0.09 ---!-q
fyn
452.4
sz =
mm
=n.n,
27s
€
(
t92400 )
Ans
s' (5oBJ(27'5J
275
dr = cover + tie +
98.9 mm
Shear parallel to long direction:
h. = h.y = 358 mm
Asl=3xAst=339.3mm2
shf'(n
-;...
.--.-1 I
rrr i\ 4,,
)
339.3 =
main
bar
mm
Au
=
;QS)'=
490.9 mmz
dr = 40 + l0
Arz= + (28)'= 615.8 mmz
d? = 250 - 62.5 = 187.5 mm
d: = 250 + 350 - [40 + 10 + Tz(28)) = 536 mm
Part2:
Ash = 0.3
1/z
+ 1/z(25) = 62.5
,.,
sr = 78.3
s,
[358)(27.5) (zzoooo _11
mm
275
€
Ans
\ 192400
)
fv=
41.5
MPa,.f. = 28 MPa
Part 1: Geometric centroid using MODE-3-2:
Area
x
v
1
0
t25
2
0
425
lilSituation26
Freq = Area
500t'2501
35012001
Given:
xr=2m
xz=0.60m
d=0.60m
Geometric centroid = y = 232,69 mm
Part2:
Area
x
V
Freo = Force
0
dt = 62.5
0
dz = 187.5
d: = 536
1,25
4 Aat f,
2 Anfv
4 Aazf
s00[250] 0.85 f',
425
350t2001 0.85 f'.
(4-25mm\
2 t2-25 mm)
3 {4-2Bmm)
4 [concrete')
5 (concreteJ
0
0
0
i--
pr=430kN
PL=3BOKN
pr=220kN
Plastic centroid using MODE-3-2:
I
f_-
x,l
Plastic centroid = y = 254.9 mm
Part 3:
P, = 3500 kN
Eccentricity, e = 460 - 270 = 190 mm
Moment=Pxe=3500x0.19
Moment = 665 kN-m
Me
= 170
I
I
I
;<l
kN_m
I
I
I
Load
combinatic-^'
)ns:
t
I
*'l
load:
.71,
Dead & Live
lJ = 'l .4D +
t-
1
Dead, Live, & Earthquake:
U ='1.32D + 1.1L + 1.1E
Dead + Live:
P,r= 1.4(430J +
P,r = 1,248 kN
1.7(380) oL
,f
Mur=0
Utl Situation 25
Parts 1 & 2:
The geometric and plastic centroids are along the x-axis. Thus its location
from x-axis is zero.
Part 3:
Pu =
4600 kN
Dead + Live + Earthquake:
P"z= r.32(430J + 1.1(380) + 1.r(220)
P"z = 1,227.6kN
M'z = l.L(l7O) = lB7 kN-m
e=26mm
Pile reaction due to dead and live loads only (no moment):
Mu=P,,xe
M, 4600 x 0.026
-=
M, = 119.6 kN-m
Rp=P*r=5=249.6kN
Pile reaction due to dead, live, and earthqualie loads:
ly=laz
Iy=4" (xt/2)2
ly=4,(2/7)'z=4
P M.x
u"
= rl +
'5t'54
Ror
R"t
- l'227'6 + 187(l)
v
Rpt = 292.27 kN
IilSituation27
R,r=&
,5
B=3m
B=3m
=245.52kN
d
Part 1: Wide beam shear at critical section:
v,
D+L:
Ei
= 2 Rpr = 2 x 292_27 = 584.54 kN
-;;;l
,q
D+L+E:
v,
AI
il
= 2 Rpz = 2 x 245.52= 491.04 kN
,.1
Vu= $Vn
I
I
I
I
t.
rE
600 mm
600+d
lg
q
l3
6I
t-3
lt
:!
I€
_____l
Fl
l',
lci
l,i
I
I
I
I
I
I
I
I
vn=
bd
v.
Part2
=
3200[600]
0.358 MPa
-
Punching shear
D+L:
V, = 4
D+L+E
I
I
Nominal wide-beam shear stress: (b* = xr + Zxz = 326 m)
V
687,694
Rp = aQag.6) = 998.4 kN
- Rpz = 1227.6 245.52 =
Vu = Puz
-
982.08
lll
vu-ovn
lll
998.4 = 0,85 V"
V. = 11,74.588 kN
Nominal punching shear stress: I b" = 4 x [400 + 600J = 4966
V
1,t74,588
vn=
bd
vn =
4000(600)
xm = 0.5
xr
-
l----J-l
--
--]
rAtrl;
PARTS1&2
0.489 MPa
- section:
Part 3: Required moirnent at critical
Mu=2RprxXm
0.5(0.4J = 0.8 m
M" = 2{292.271 (0.81 = 467.632 kN,m
d=650mm
P, = 3600 kN; Mu = 660 kN-m
Parts 1 & 2: Wide beam and punching shears due to axial load:
P, = 3600 kN
q. = P,/[BL)
/ (3 x 2.5)
480 kPa
qu = 3600
eu =
Wide beam shear:
- 0.6) Vu-quxArea
xr = 0.5(3
ro
lg
E\E%'i*
li
*i_
_
584.54 = 0.85 Vn
V. = 687,694 kN
d ix,
t- x
0.65 = 0'55 m
Vu=4B0xLx1
V, = 480[2.5J[0.55] = 660 kN
V.=V,/0
V, = 660
vn=Vn/b*d
v" = 77 6,47 0 / {2500 x 65A)
v,, = 0.478 MPa
/
0.85 = 77 6.47 kN
Punching shear:
[u=qrxArea
V =Qr xltsl.-(0.6+d)zl
V, = 480 x [:-t[2.5],- (0.6 + 0.65J21
V,, = 2,850 I<N
V"=V,/0
V, = 2,850/0.85 = 3352.94 kN
vn=Vn/bod
vt=Ynl[(4[600+dJxd]
v" = 3,352,940
/
v" = 1.032 MPa
[4(600
+ 650J x 6gQ]
Part 3: Wide beam shear stress due to axial load ancl moment:
e,='
( t 6Mt'
LB LB'
o', =-16oo- -+ 6(660)
2.s[3)
2.5G1
qu=-480t176
q,r = -656 kpa &
xz
quz = _304 kpa
= t/z(O 6) + 0.65 = 0.95 m
Qu3=Qu2.
q,:
= 304
S+9*t B/2.x,)
+
B
656-304
J
(1.5
+ 0.951
= 59r.467 kpa
V" = 7/z(qa+ q":][1.5 - xz) x I
V" = 1/z(556 + 591.467)(1.5
V" = 857.63 kN
vn=Vn/b*d
/
vn =857,630 [2500 x 650]
v,, = 0.528 MPa
- 0.95) x 2.5
i:"!:':.l;;1
€ret
rt:+itt
,::,iiliii
::tii::!:i
!ir:liiiii
:i
li:liill
l ii.l;i.i
:;iiiilt
p.s,ti
:l:r:!i:ii
i::i:!ii:
irr:ir!ii:
il,ilili
:,iiii:iir
:i:,,!ifl
Solutions to November ZOIS Examination
r-
\lallzJa" =z
m1
-,_1
Given:
xa
_y2 +y -yz
The factor ofxa
$
$
Then;
$
I
-yz is fxz + y) (xz -y)
(xz + y) (x2 _y) + y _ yz
[x2 + yJ 82 -y) _ txz _ y)
Given: E=w+20G
(xz_y)(xz+y_t)
{
$
a=9m12 &w=10mph
m2
Given:
{
{
f[xJ = ;s
(31
y=(2)z=$
rr=w3
E = 10+
-
Zxz + 5x - B
= [3)3 _ 2(312+ 5[3] _ B = 16
20J9 =r/6
6
f
$
*
B
Given:
[83
Given:
256
Factor;
"16(16
-
-
t2)
Speed V varies inversely
= t6(4+ t)(4 - t)
V=
k
T
or
VT=constant
VrTr=VzTz
M4
urven:
2
;(x-A)
2x-72 > 5x3x=-7
i9
Given:
5
f[xJ = [x - 4J(x + 3J + 9
When f(xJ is divided by (x _ kJ, the remainder
is k
(k)
= (k - 4)(k +
k2-k-LZ+9=k
k2-2k-3=0
(k-3)(k+1J=0
k=3&-1
First term, at = 54
Comiron difference, d = 58
on = &1
rus
Remainder =
Qgxg=\/2xQ
Vz=72kph
> x_L
x<-7/3 or -cr,-7f3
Given:
with time T.
V=48kphwhenT=6sec
l6t2
3.) +
9=k
* [n -1)d
-
54
=4
an=54+(nan=50+4n
L)4
t10
Given:
Rate ofA = 6 km every hour
Rate of B = 4 km for. the first hour and gaining 1 km every
B left 2 hours after A
hour
At any time "n" alter B left:
Distance travelled byA, Sn =
6(n+2)
Distance travelled by B:
Se=4+ 5 + 6+....
sumofAPwithar =4andd=
5u=
n;2r, (n-I)dI
?,
1
su
=
![z*nl
|tzt+)*tn-1J[1)] = 2-
Given: P=Pn
eors'
P" = 5,000
6(n+2)= lgz+n)
Se=Se
In B months, P= 5000 e0.rsrBl
2'
P=
n=B&-3
m11
In an A.P. of 13 terms, the
always twice the middle
7th
Let c, u, and e be the percentage of cellular subscriber to the World population
of China, USA, and Europe, respectively.
term is the middle term. The sum any A.p. is
term.
,
Thus;
S,,
ol AP,
S,, of
ap,
=
2A,
2a,
=
16,600
Together, they contiibute 34.\a/o of the total subscribers:
A,
c+u+e=0.348
a7
Thus, the ratio of their corresponding middle terms
equals Lhe ratio of their sum.
)Eq.[1J
1%o less than 4 times the percentage of the world
subscribers in Europe:
China is 3.
)
c=4e-0.031
rm 12
Given:
Population at the beginning ofday 1 (t = 01, po = 10
The population doubles every day
USA is 4.3% more than the percent of the
)
u=e+0.043
Population growth, P =
world subscribers in Europe:
Eq. (3J
Po srr
"
The population doubles every day
p = po grt
Eq. (2)
Solving: c = 19.3o/oi u = 9.9o/oi e =
5.60/o
2Pe = po gr(r1
er=2
Given
P=Pn2r
The population atthe beginning of day 7
{t=
Let t = time required to do the job working together
6)
P=10 \26=640
Using MODE-STAT-e^x:
Mary can do the job in 4 hours
|ohn can do the same job in 7 hours
]-,.1t= l; t =2-6/tthours
47
6y
iN
16
Given:
Ana was four times as old as Billy B years ago:
A-B=4[B-B)
Ana will be twice as old as Billy eight years hence:
A+B=2[B+BJ
Solving:
A=40&B=16
2x+2Y=l0O
m17
y=50-x
Let vr and va be the rates of Albert and Bernard, respectively.
Cost = 2500x + 500(2y) = 95,000
2500x + 1000(50 - x) = 95,000
Albert can walk 4 km with the same time as Bernard can walk 5 km
45 --i
---=
VA
vs=7.25vn
'
VB
x= 30 m &Y =20m
.21
Albert takes 3 minutes more than it takes Bernard to walk a kilometer
1.1.3
vA vB 60
1.
13
vA L.25v o
Revenue = 4500x
'
122
Given:
60
Desired interest on investmenl =
Tax on income = 45%o
[U 18
P+
Given:
lll23
Let x = width ol the garden
, Given:
Price increase'= L00/o
Sale discount = 12%
Length=x+9
A=xfx +9)=736
x=23 m
Let P be the original Price:
Final Price = P x (1 + 10%) x lL
Final Price = 0.968P
width=23+9=32m
x (Length,L) _ 7
Width=2L-7
A=wxL
15=(2L_7)(L)
24
L
Given: Area,A=15m2
I
Given:
20
ts,1
cn
Perimeter = 100 m
Cost of fencing per meter;
Along side "x" = P2500
Along side "Y" = P500
Total fencing cost = P95,000
f=
P12,500
TotalCost,C=P25,500
Area = 15 m2
il
Let N = required number of kilometers
B
C
t,
Given:
Fixed charge,
- 1T)
Variable charge, r = P10 Per km
tr-
. L=5mandw=3m
EEl
r x P x [1 - 0.45J = P x[1 + 0'07J
r=0.!273=12.73o/o
+9
Area,A =736m2
Width, w =
7o/o
Let P be the amount of investment and r be the rate of return
ve=4kph&ve=5kph
M19
)Eq.(1)
=
f+
,i125
Given:
rN
25,500 = 12,500 + 10N
N = 1,300 km
+5 s]
@ T t = 7:04:00 pm, the watch is seconds ahead (er =
ahead
is
9,5
seconds
watch
= +9'5
pm,
the
[ez
71:20:00
Tz
@
=
-error, e. = €.-er
-,,-T,
Rate of
(0:0:9.5)-[0:0:51
[11:20:00)-[7i04:00)
Rate of error, e. = 0.00094401
sJ
Error at Ts = 9:1"3:26
e.==L
T, -T'
o.ooog44o1- "er-(-o:o:51
tan 30 = h/a
tan 30 = h/10
p:13:26)-{7,o4,ool
es = 0:0:7.275
t
h/ I ot =
3(h/sol-(h /sil3
r-3[h/sO]'?
h=1B.B9Bm&0=20.705"
Corrected time = T: + ez = 9:13:].B.72S
30 = 62.72o
E0 26
Given: F=5i+6j-3k
f
tt
Magnitude, lFl =
E
.p+6, +f-31,
Given:
=
8.367
h=200m
s
*
!!
cr
Mzz
f
Given: A=4i+2j+0
i!
B=ai+bj+0
;:
i!
AxB=o
w
$
:l!1
l
AoB=30
fr
AxB=0
A.B=30
iB
In triangle CDB:
0l.-!: -?
BC=h/sin0
:o=rl^=o
BC =
!:
412.967 m
h
Y=180'-cr-0
aQb)+2b=30
tl
!i
2B' 5B'
P=50"
4a+2b=30
lri
= 50o
0=
Y
b=3anda=6
= 101.033'
0=180"-V-F=28.967"
It!x
Thus;B=6i+3i
i
ln triangle ABC:
AC_BC
EE
sinB
28
6g
sinQ
sin 5o'
AC= 653.21 m
Given:
a=10m
b=50m
In right triangle CEA:
tan 3o = tan [o + 2e) = tane+tan20
' 1.,tan0tan20
tan 30 =
tan0-tan30+2tan0
1
tan 3e
412'goz
= sin29.967'
-
tan.2 0
-2tan2
3tane-tan3e
=
-
H=ACsino
2tan0
tanH+
1
1-tano
H= 653.27 sin 50o
H = 5O0.39
-tan2 0
2tano
-1-tanz0
Given:
m
. C=102.8' c=3.2cm
A = 46.5'
0
B=180'-A-C=30.7"
1-3tan'z0
Sine law:
tan 0 = h/b
tan 0 = h/50
b= c
A
sinC
b=
_
sinB
3.2
sin 102.8"
sin 30.7o = 1.6754 cm
m31
In triangle ADB:
Given:
'
F=80"
0=42"
a+2$ +Y= 1B0o
a + 2(75" - 2cr) + 96" = 180.
a=22"&P=31"
cr=180'-p-e=58"
Angle ACB =
If P is the midpoint of arc AB, then
angle PAB = 0/2
Thus, angle PAC = 0/2 + a
0 = 180' - 2d - 2p = 74o
)34
Given: AngleC=70'
=
79o
Angle A = 45"
Sidec=150cm
M32
fl = lB0" -70"
-
45"
= 65"
From the figure shown:
'
Solve lor side a:
The measure of angle ApB
is one-half the arc AB.
a=
1/z(2a)
b=
0=o
a
c
sinA
sin
c
C
- 150
= sin70"
_:
sin 45o = 172.873;
a/2 = 56.436
a= 90" - 0/2
In triangle ABE:
rhus,l?lqoi oA
= c2 + (a/2)2 - 2(c)(a/2) cos p
= 1502 + (56.436/2)2 - 2(t5O)(56.436/Z) cos 65"
m"= 136.L24 cm
tr1a2
ma2
0=52o
d=90" -52"f2=64o
Using complex mode ftriangle ABEJ
ma= 156.436 _ IS0265"l
m33
Given:
Line AD is an angle bisector of angle A
Line BE'is an angle bisector ofangle B
AngleBEC
=t=75"
R=6cm
[R+r)z=(R-rJz+[R-r)z
Rz +
0=180'-e=105"
i{ty
t_\
i--"'
y=180"-6=96o
In triangle ABE:
2Rr + 12 = 2Rz _ 4Rr + 2rz
12_6Rr+Rz=0
r--------:----
r=
[6RJ-V36R'-4R',
2
r=3R-2'l2R
2u+B+0=180'
-2a )
Given
/gr
AngleADC=5=84.
F=75"
illt s5
Eq.
[1]
Fi:-rJa{
€
Formura
r=(3-zJ2Xq=r.o2ecm
,l
trE' {-*S'{SficES],,
' ,, ' 136,Xl4m4gtl
m36
The given triangle is not possible because the sum ofthe sides (5 + 6) is
less
than the third side.
ED
Given:
One side oftriangle = 20 cm
Area of triangle, At = 96 cm2
DC
37
Since one side ofthe triangle equals
the diameter of the circle, the tiiangle
is a RIGHT TRIANGLE as shown.
From the figure:
a=90" -75"
d,
radius, r = 10 cm
= 15o
D=20cm
8t
Ar= r/z ab = 96
a=
Dz=az+bz
202 = (l$) /l)z .+ lz
b=1,2cm&16cm
L92/b
60"
A
m38
When b = 12 cm, a = 16
Given:
a=38cm
.
c=32cm
. .4t
Given:
Formula:
A.
Volume
a
+b+c
2
=
Coneradius, r= 6 cm
Height,h=Bcm
s*b
A,
(and vice versaJ
Thus, the perimeter is 12 + 1,6 + 20 = 48 cm
b =22 cm
rb=
cm
46_22
Volume
=
jn to)z tal
Volume = 96n
=46cm
$[-alJs-5;65-g)
351.636
=! nrzh
= 351.636 cml
..42
Given:
AB=30cm
= 14.65 cm
m3e
Given:
BC=40cm
AC=50cm
ha=18cm
hc= 27 cm
Volume = 74,400 cms
ho
A
a=OA=30
c=BC=32
Formula:
OA2=OBxOC
302=bx[b+cJ
The base is a right triangle,
Au*" = %[30)(40) = gQQ 662
Volume=Ab"se x h,v"
.
B
14,400= 600 xh,""
hur" = 24 cm
b=18
1',,u"=(he+he+hc)/3
2a=gB+he+27)/3
hs=27 cm
El 43
Y=mx+b
Given: a=18cm
m
x
Volume, V =
i;
i:
a3
Total surface area, A,= 6a2
t:
i
T
ti
Ratio= . = _
I
A
rit
il
li
=a/6
ha'
it
*
Using the STAT MODE (MODE 3-2J
:E 44
I
Normal
equation:
Givenline: x+y=Bo. I * I
P
A=150,900 B=1,4,220
xcos B +ysin p - p = Q
XV
r
=1
p / cosB -p/sinB
t,
cos B
SHIFT-STAT.REG
(Cube)
1
Ii
Thus, y = L4,22Ox+ 150,900
Regular hexahedron
Ratio= 18/6=3
ir
1s0,000 =14,220(0)+b
b = 150,000
Y=A+Bx
,.
=
P
sinB
p=BcosF
47
The semi-transverse and
semi-conjugate axis of equilateral
1
=a
y= 150,900 +L4,22Ox
=B
p=Bsin0
hyperbola xy = k is a = b = J2k
For the equilateral
hYPerbolaxy=9,k=9
e
gl9lp
tanB=1;P=45'
P SsinB'
=
p=BsinF=a"
b= Jzk
b= vEi, =4.243
1
l=+JZ
r12
conjugate axis = 2b
conjugate axis = 8.485
G4s
Given:
E0
f(x'1 =
275.5* + 2860.2
This function is a straight line with a slope of 275.5.
This curve is an ellipse.
Given:
An ellipse is the locus of a point
that moves such that its ratio
from a fixed point (the Focus)
and a straight line [the directrix)
is constant and is less than 1.
+o
In 2001 the price is $ 150,000 point 1:
[0, 150900J
tn 2006theprice is $222,000 point2: (S,ZZZOOOi
.r_.
y"
Xz
-y,
-X,
222,000-150,900
5_0
= 14,220
In this problem, dz =
Yz
dt
"Eccentricity, e = dz/ dr =
L
/Z
P(x, y)
ilIJ 51
Given:
Equation of circle:
x2+y2-4x-5=0
(xz-4x+4)+yz=5+4
, .B
Center @ [2, 0J
radius =
3
h,
Distance from (2, 0J to [4, 6) "AB":
AB=
AB =
Je-4)\04)'
v7
\80
N
t\
Length of tangent:
T
= JAB'-
At minimum cost,
(4, 6),1-
(x-Z1z+Y2=9
/'
i\
,\
o i
,dx
\
= 0.
. =16x-L92=0;x=12
dx
T
rlrsi
Given:
(2, o)
d+.
{
dC
i'\
Ai:--r
l
Cost,C = Bx2-792x+ 1800
d
.
dq
/
C(ql = Se + L64q- O.0O02qz
C(qJ
-0 + 1.64-0.0004q=a
q = 4'100
IA
C(4,
r= {(J+o) -3'z =5.568
|ll
100) = 98 + 1.64(41.001
-
0.0002
(
4700)2 = 3,460
s3
m50
Civen:
Given: r -
Volume,V=864in:
?
1-sin0
v=x2y
Covert this polar equation to rectangular lorm [to identify the curve)
r-rsin0=2
wherer =
x'+y' -y=2or
Squaring both
Ei
sides:
JZ.t'
Y=Y/xz
Ar=x2+4xy
andrsin0=y
As=Xz+
dA,
4V
-x
dx =z*- 4
x"
=z*v
=o
x3=2V=2{B6a)
x2 + y2 = 4 +
4y + yz
orx=12inandy=6in
4Y=Yz-4 )aParabola
NOTE: For an open-top rectangular box (with square basel ofgiven volume,
the surface area is minimum if its base is twice its height or x = 2y.
For a parabola Y2 = lc< or xz = ky, k is the length of the latus rectum
Y=r/+(x2-1)
Length of latus rectum, LR=Y+
s4
Given
x-component of motion, x= 3t2 +
y-component of motion, y = 1a
2
x-component of velocity:
v, = dx/dt = 6t
when t =4, v* = 24
Given:
r-^ s=,/x.rh-
Radius, R = 10 ft
Q = 314 1:/min
ds
dr=
ds
dt
/x(dx / dr)
4216.e44)
V67
,142'+B'
=6.822m/s= 24.558 kph
Q=Asv
314 = n(10)z v
v = 1ft/min
dy/dt = v
I I.U
58
Given: r(*) = i [3x: - 7xJ dx
f(l=a
Required: f[0) = 2
;:s6
r
3xn 7x'
f(x)=J(3x:-7x)dx= 4 - Z *'
Given: L=3m
dx/dt
= 2 cm/s
f(2)=Y
Findd0/dtwhenx=2m
ry
+c=4;
c=6
Then, f[OJ = 6
Whenx =2m,0=48'!9"
coso=
u59
I
'
Given:
L
de
dt
-qi60-=-- -'
1
dx
300 dt
-sin46.re. 99 =
Number of different CD players = 9
Number of different sPeakers = 10
Number of different aryPlifiers = 4
x
Possible numbers ofsound system = 9' L0 4 =36O
lqzl '
dt 300'
99
dt
=
r60
-o.oog9 4rad.f sec
Given:
Probability to quit smoking, P = 0'25
Probability not to quit smoking' a = 0'75
quit:
Probability that exactly r = B out of n = L5 smokers will
ms7
Given:
h=Bm
= 25 kPh = 6'944
v = dx/dt
Find ds/dtwhenx= 42m
P=
m/s
c(n, r) p'
qn-'
P=
c[15, B) [0'25)8 [0'751ts
P=
0.0131
lr lt 61
Given: Probability of hitting
the target, p = 0'30
Probability of missing the target,
P=1-Q
0'75=1-q'
0.75=l-0'7"
n=3.9say4
q
= 0'70
-
a
U]I68
Giveni
parts within a power tolerance
5 parts are not within power tolerance
1B
.
Given:
=
c[l
B'
3)
c[23, 3)
=
2,100,000
BVz=FC-Dz
Dz =
Dz =
2,L00,000
660,000
m[2n-m+11
D-=IFC-SV] '
5gy=11r*r,1
2STJM
Given:
Cost of insurance, C = P7,000
Minor accident claims, Cr = P10,000
Major accident claims, Cz = P50,000
@m =
2:
Probability of having minor accident, Pr = 0.20
Probability of having major accident, Pz = 0.05
,
Expectation = C - Cr Pt
Expectation =P2,5OO
-
CzPz= 7000
660,000 = 1,26o,ooo
2(2n-2+1)
zlU+n)
n=6
- 10000[0,20) - 50000(0.05]
.
il.t 69
Given (for machine A)
First cost, FC = 5,000
'
Given:
- 1,440,000
0.4607 6
m63
U64
1,440,000
FC* SV= 1,260,000
BVz = P
840,000
BV-=FC-D-
If three parts are selected at random, the probability that all
three are within the power tolerance is:
p
FC = P
SV= P
Salvage value, SV = 600
Annual maintenance, OM = 500
Usefullife,n=5
Interestrate,i=0.08
Number of defective components per packet
l,=0.0020 x200=0.4
EUAC=FCi+oM+
Probabiliry that a packet contains 2 defective units (x = 2)
o' (0.4)'
e-r ),' _ e
P
=
xl
=
2!
0.0536
EUAC
(FC-SV)i
(1+ iJ"
-1
f5.000 600tr0.081
- 5,000(0.08J + 500 + -ffi -
EUAC =
1650
m65
Frequency, x
Probabilitv. Pl'xl
Given
Expectati on = 2(1
lM
66
m67
'
/3)
+ 3(1
/2)'+
2
3
11
r/3
L/2
1/76
77(7 / 6) =
4
ILJ
70
Given:
Volume of sales, N = 500,000
Selling price per unit, p = P5.00
Fixed expense per year, f = P800,000
Let a = cost per unit
Answer: D
To break-even, Cost = Revenue
Given:
Nominal rate = 9o/o compounded continuously
From
800,000 + a[500,000J = 5(500,000)
a=P3.40
F=Pert
Compound amount factor =
f+aN=pN
F
-P
= et =
eooelsl =
1.568
Given:
,,.74
Revenue,R=B0N
Given: A= 1.25 m2
Cost,C=60N+10,000
Period = 0:5
To break-even, Cost = Revenue
Formula;
BON=60N+10,000
N = 500 units
where
EE72
Given:
i = L60/o
FC = P6M
SV = P600,000
n=5
s
Period =
g = acceleration due to gravity
p = density of liquid in kg/ms
A = cross-sectional area in mz
M = mass in kg
Maintenance flump sumJ, MC = P400,000
a*
0.5 =
Annual cost.= Annual interest on FC + Annual maintenance
+ Annual depreciation
JroooJo.zslqr .zs) /M
M=77kg
Annual interest on investment, FC x i = 6,000,000 x 0.16 = P960,000
Annual interest of lump sum [after 5 yearsJ
(MC)i 400'000(q'-161
maintenance=p58.163.75
(1 r i)'-1
(1 r 0.16)'- 1
Annual depreciation
=
Annual depreciation =
!l! ..:ull -
Given:
Errors, 10.015 m, t0J22 m, 10.029
g,= 1Jon15' +on
(6'000'000 -600'000)(0'161
[1+iJ"-1
P7
:,u75
Et =
(1+0.16J5-1
+0.039 m
85,210.66
Annual cost = 960,000 + 58,1.63.75 + 785,21,0.66
Annual cost = P1,803,374.4L
ull 76
Given:
Total distance, S = 1,000 ft
Total error, E = 0.1 ft
Length of tape used = I 00 ft
LQ 73
Number oftape lengths, N = 1,000/100 = 10
Let e = error per tape length
Given:
Diameter,D=0.2m
Tangential speed, v = 24 m/s
E=
r=0.5D=0.1
v=rxcD
r !6-Ej
m
0.1=t\FEj
Et
24=0.1
xa
ot = 24O
rad,/s
=
foE,'
= +0.0316
t77
Given: xt= 826.52;xz=826.68;
Mean,
i
xa = 826.1L
= (xr + xz + xz)/3 = 826.437
t
+o.wg'
Lzu81
Probable error ofthe mean:
PE. = *0.6745
Given:
ftJ
.l n[n-1)
1"
!
Pull,P=60N
14 m
lensrh. u,
Error per suppa.rrted
--- .-..o---,
-- = i
$, S?45rcXc*+{(*,:,,,
-,r. r,,
r,
&;
i
..:
II4*ffi5f{S
Error per tape length,
i
AH = 1.91
Numberoftapelengths,
-
*'1..' - - t0'498)'z(1-00)r
24(60).
24p,
= -0.04488 m
4 x e1 = -0.17953 m
=
T
=
#
= 24.326
Correct distance, TD = MD + E = MD + €t x N
Correct distance, TD = 2,432.65 - 0.17953(24.326) = 2,428.28 m
- 1.83 = 0.08 m
[!
SPaces, s = 5
AH O.OB
Sensitivity,0
' = =sD= =-=
5[100]
Q
er =
l
Given: D=100m
Sensitivity,
(@ mid and quarter points)
Weight of tape, w = Mt g/L- = (5.08 x 9.81J/100 = 0.498 N/m
USING THE STAT MODE: MODE 3-1
m78
m
Measured distance, MD = 2,432.65
3(3 l)
!
= +0.1
Supported length, L" = L/4 = 25
,
pE* = 10.6745 l(826.437 -826.52)'z +@26.437 -826.6q'z +(826.437 -826.1.D'z
PE.
Length oftape used, L- = 100 m
Mass of tape, Mt = 5.08 kg
= 0.00016 vlfl x 1BO"fn
= 0.009167" x 3600 = 33"
82
Civen:
Radius of curve, R = B0 m
Length ofcurve, L. = 90 m
' Required sight distance, S = 60
m
I*
When S < L.
LCI
79
D_
Given: D=100m
s2
Bm
AH = 1.91- 1.83 = 0.08 m
i*liue-of-s-iglrJ:- .1-- " ---
SPaces, s = 6
When
Sensitivity,e=
+sD= **
6[100)
S > Lc
L.[2s-1.)
Bm
=0.0001333 raiJxtB}"/x
sensitivity, 0 = 0.007639" x 3600 = 27.5"
*4,..
Since S < L.,
m8o
i
Given:
D
Length oftape, Lt = 50 m
Recorded distance, MD = 685.24 m
Error per tape length, q = +0.014 m
TD=MD+Et
TD=MD+etN
TD = MD + Ct (MD/LI)
TD = 685.24 + 0.01a(685.2a / 50)
TD = 6a5.432m
s2
Bm
Bo
=
6o'
Bm
m = 5.625 m
--
--- ---
--\
Offset from TS to any point on the spiral:
Given:
f
AB=Ti+Tz=520m
753
l--
6R. L"
Dt=3o
6[300x100]
h=50"
Iz =
D
_
x=2,34m
35.
l,t85
Given:
3600
fiD
Rl
=
3600
_
,r[3J
Length of spiral, L,=2,200m
Radius ofcentral curv€, Rc = 320 m
Required: Degreeofspiralatthird-quarterpoint
=381,.972m
i.e.@ L =3/+x2,200 = 1,650 m
The degree of spiral at any distance ,,L,, from TS is given by
T=Rtan(l/2J
Tt = 38'1.972 tan (50' /2)
Tt = 178.1'J.6 m
Degree of cenrral curve, D. =
Tz=AB-Tr
Tz= 520 - Tr = 341.884 m
Tz = Rz tan(lz/2)
34l.BB4 = Rz tan (35" /Z)
Rz = 1084.316 m
i--I,
it) -\' ,'
D
:/
3.581
l,/
t,
l/'
d
1650
_.
D=
2200'
D=L
D. L.
-99 - 3600 = 3.581
trR. n(320)
2.6A6
1,186
Given:
Length of curve 2,Lcz = RzIz = L084.316(35" r/tBO.)
"
Length of curve 2,L,2= 662.37 m
Length ofspiral, L. = 160 m
Radius of central curv€, Rc = 320 m
Required:
Offset distance from TS to the third-quarter point.
L= %(160) = 72Om
offsetdistance,x= Lt -
m84
6R.1,
1203
6(320)(160)
Offset distance, x = 5.625 m
tilBT
Given:
Degree of cenl.ral curve, D. = 6o
Maximum design velocity, v = B0 kph
Desirable length of spiral
[based on rate of change of centripetal acceleration)
Ls=
0.036v3
where v in in kph and R in meters
R
Given:
Length ofspiral, L, = 100 m
.Radius ofcentral curve, Rc = 300 m
L = 3/s L, = 75 m (third-quarter pointJ
D_ 3600
_ 3600
(6)
fiD.
,Ls- _ 0.036[80]3
-- 190.986
=
190.99 m
96.51m
m88
I
I]tr
89
Given:
D=B+3C
Ar= 36m2
Az=72m2
L=75m
B=10m
'
B
A=BxC+r/z(C)l!.5C)"2
A=BC+1.5C2
Atsection
hr=1.4m
Ar= L0Cr + 7.5Ci=36;
[1):
h=2.592m
Dr=B+3h=77.776m
hz=2.8m
L=50m
B=14m
Atsection
(2):
7.5C22=72; Cz=4.355rh
Az=10C2+
.
Dz=B +3az=23.065m
a=52m
Ground slope,
Gs
End-area volumel
= 0.05
Yp,
h'=h'
x1
x2
1.4
60-x,
I
h=
+ AzJL = 1/z(36
Prismoidal correction
x2
vr, =
?9 g2)=2.2a^
Area of cut = B x h + 2 x 1/z(2h)h = Bh + 2hz
Area of cut = M(2.24 + 2(2.24)2 = 41.395 mz
:
#1z.ssz
- 4.3ss)ltz
-
-
23.06s)
58.27 2
Given:
.
ir
i1
= 399L.728 ms
Number of accidents in 6 years = 5892
Average daily traffic ADT = 47 6
Total vehicles entering in 6
Szears =
58g2
Accidentrate- :--
x106
476 x 365 x 6 = 7,042,440
1,042,440
it
= sl.27z m3
ra 90
f;
il
.72 6
Corrected volume = Ve - Vpc = 4050
il
ii
+ 72)(75) = 4050 m3
vr.= *(cr-cz)(Dr-Dz)
l2
b=a-x1=52-20-32m
b",
lz(At
2.8
xz'= 40 m and xr = 20 m
!=&
=
Accident rate = 5652 per million vehicles
fi'
ll,r
Y?'
-
'::l;' r-rr! :i.; ii:.:il ':.
l_i:.:. ri, .,-,il -. l
,i l: i :l /.:.ri:'.,:.1
lr.
.:r,,.I
I'
7,
V!
V
,i
fi
j
'rt
.,,1t-ld
i
;i.':ii];g
I r".:.r,
;,,
.r
r"r,i
!
-1
:'
.--:r,,.,':
1l:i
iJ
t,'.r,; lf
.i
i r,
:.'.ir',il.l)i'1+;ri:'i'iiil::,iiil:ji'1il:; rli'i;,'rll:l:r:' ':li;lri'
-' i- {-r',' iir :l:i ;lir:;:i ;:l:;i ii r'' -'li:-l
irl'j':l*,:ii;tll,,l1
Iii:,t .tir::-iti't 1.'a-rli:11ll,l;,1:,:l!:li.l l-.i.'llr"
:\
:r
rli l-ii::iir' rl
1...,\.. it:.
:.-t ;:,
:ll. i...]!
:.,1:
':,1
_i'
.'l
Lr .'''l *i
r
l.'r:.i,ti-)
.,.';..,:.1'1y
r 'i
'-,
Lt
f:l.j
'i'1,: rt;,rilri
,irr
i.1.. ,li
i'r.
:.'t.:i::
l-r
'r t ]ii
::l
il tliil.
i.;
i
l
r
l-]'.!:,: l I
I't!i:.ii
t
;rir
i-i a:l:::iri-'il;r;
:.1.lt'a:ilir i, r li
i)ti,
!'t
l:i
irij:r.)iil:1r:: liliri
r::
:,
l
i!.
;i,.
itirr':r.
l.,jlr,
i;"lii
,,1.1:i
t1fl1
't'iI;
ll-;.:
a::i
1;;1li'l:ll'i
Answer Key
L4B
158
2tA
22A
23C
24D
25A
318 4tB
32C 42D
33 D 43D
34A 44C
35C 45A
51A
52D
53C
54D
55C
6D L6C
78 174
BD 18A
9D LgC
10 B 208
26C
27C
2BA
29C
30A
36
B 464
374 474
3BC 4BB
39B 49C
40D 50B
56
1D
2C
38
4C
58
1lD
lzD
13C
61A
62 A
63D
64D
65C
D 66C
578 678
58 D 68A
59D 694
60B 708
7IB
72A
73D
74D
75C
81
A2
83
A4
85
9L
768
778
78B
B6
87
96
97
BB
9B
79D
B9
BOA
90
99
100
92
93
94
95
ffi
m1
Solutions to November ?OLS Examination
At standard temperature and pressure,[0 'C and 100 kPaJ, dry air has
IL]U
5
Given:
a
Height of mountain, H = Ah- x y^ f !"i,
density af. At20 'C and 101.325 kPa, dry air has a densi'ry of 1,.2041t<g/m1.
Heightof mountain, H
m2
atmosphere = 101.325 kPa
Equivalent height of water,
m3
Given:
Air pressure,
paus
h.
p _ 2.55x'I0L.325
y*
9.81
=
Given:
=26.34m
Fr = 400 N
Fz=10kN
Dr=25mm
dePth,h=1.5m
Unit w'eight, y* = 9.81 kN/m: [water)
Atmospheric pressure, p"t. = 101.325 kPa
,
i
-
LOt.325
F.E
A, A2
400
10,000
;(2s)'
XD,'
.---L = ---L
Dl = p2
= 40 kPaa
The gage pressure is, p = 54.715
-
(0'536-0'a6-l!9810x13'6J
tll6
The absolute pressure at 1.5 m b'elow the surface is:
pr s = 4o o.or6l.s1 = 54.71,5kPaa
Dz=125mm
-
lt.l 7
Given:
='46.6lkPa
b = 7.2m
. h=2.4m
m4
Sum-up pressure from
C
y(+ [9.81 x 10)t0.4s] - 9.81[0.6)
pe = 38.259
kPa
tr = Y =ttls
A= r/zbh
le=bh3/36
to A:
=
pr
l
_t
150 mm
450 mm
[h.o _h_,)y."..,,
Tatr
L'Z7\Akglm: x g = 12'51 N/m:
1
Reading at mountain top, h.t = 460 mm = 0.46 m
Reading at base, h-r, = 536 mm = 0.536 m
Unitweightof air,yur.= 12 N/m3
a=
bh3
AY
"-
e
=
/36
(+bhlth/31
=h/6
e=2.4/6=9.4*=400mm
=
845 m
m Situation 1
F=y-or h A
P
Given:
b=3m
L= 7.2m
I
e
0 = 60deg
hr= 2 m
blt
=
'L2
e=
= ---=
.AY
= 0.432 ma
o-032206
0.636(2.85)
-
e = O.432
yr=D/2ie=0.4678m
= hr + (L/ZJ sin 0
i
= 2 + (1..2/2) sin 60" =
IMo=0
PxD=Fxyr
Px0.90=24.9x0.4678
P=
V = [/sinQ=2.909m
l=y- h A
F=
F=
e.B1[2.51e6)[3.6)
88.98 kN
o'q32
I"
e= --=
AY
yz=L/Z-e
"= (3.6)(2.e0e)
yz =
1.2/2 - 0.04725 = 0,559 m
yt=L/2+e=0.641,25m
PxL=Fxyr
P(1.2) = e6.eB[0.64125J
P=
47.55 kN
Part 1
-:
Given: D=1.5m
e=45"
' =10m
h
)
Part 2
A=
ht =2.4m
sg = 1'40
h =hr+D/2=2.85m
A= i(0.9)2=0.636m2
lc= x(0,9)a/64
lc= 0.032206 m+
V= h=2.85m
LADL
=
le= *D4
9n/16
=
m2
8Ln/7024
Y = [/rino=14.r42
p=y-
[
A
F = e.B1(101[ en/76)
F = 173.357 kN
l"
2
Given:
D= 0.90 m
12.94 kN
Situation 3
= 0.04125 m
Part 3:
XMo=0
)
e=
-
)
Part 1
)
Part 2
=0.0178m
Part 3:
h
m situation
kN
Location of force from the bottom = D /2
A=bL=3.6m3
L
= [e.81 x 1.aJ(2.85)[0.636)
F = 24.9
etn / toz4
w16)(14.144
0.009944 = 9.944 mm
yr=D/2-e=L.5/2 - 0.00994 =O.74rn )Part2
Part 3:
yz=D/2+e=0.76m
pxD=Fxy2
XMo=0
P = 773.357(0.76)/L.5
P = 87.83 t<N
m
m18
Given:
h=5m
a=0.60m
y,= 23.5 kN/ms
y.- = 9.81 x 1.03 = 10.1043 kN/ms
.
Since there is no seepage at the bottom,
there will be no buoyant force. The
force F may be taken as the sum of
the weight of concrete and of water
above the block.
F=W.+W*
.
F-y.V.+|swVsw
F=23.5(0.6)3+10.1043[0.6),(5)
F = 23.26 kN
x area= (yc x a +ysw x hJ x
+
10.1043(5)l(0.61'z
= 23.26 kN
[23.5(0.6]
OR F=pressureatblockbottom
F=
ffi
a2
Situation 4
Given:
Weight of stone (in air), W" = 400 N
Weight of stone in water, Wr. = 240 N
Formula (derivation):
Let Vo = volume of the object
so =
sL =
specific gravity ofthe object
specific gravity ofthe liquid
Weight of object: Wo = y* x So x Vo
Weight of object submerged in liquid:
Wr. = [yo
-
yr-) x Vo = yw(so.- sr.) x
W"
Y-s,%
q - y-[s.-s,)V"
ttton"
=
|stone =
W"r.," S-,,",
w*-. -w*
|w
X Sstone =
ry'o
)
_
400(1)
400 -240
_ 2.s
9.87 x 2.5 = 24.525
kN/ms
Formula
)
Part
3
) PafiZ
Elll
Part 3:
Situation 6
Given:
xt=20cm=0.2m
Horizontalacceleration, a=Yz g= 4.905 m/s2
0
=
0)'
y-y1 =H
tan 0 = a/g
tan 0 = 4.905/9.8L
Ah
26.57"
x'
)a
(')'
(0.s2
2(e.Br)
Angle with the
ro
vertical = 90o - 0 = 63.43"
co
, -H
0)z x-2
2o
-
0.22) = 2.s
= 15.28 rad/s x 30 /n
= 145.9 rpm
Part 3:
L=4m
Ull Situation B
Ah = L tan 0 = 4 tan 26.57o = 2 m
lPartsl&2
Ap = Y,- Ah = 9.81{2) = 19.62 kPa
I
tm situation 7
Given
D=1m
H=2.5m
h = (3/slH = 1.5 m
r=0.5D10.5m
a=H-h=1m
x=r=0.5m
oJ2
'
x2
?o
Part 1 u-a--.t
1l
"I
IL
I
(rl
'fi
Diameter = 2 m
h= 1/2H= 2 m
Y=2a=2m
oJ2
"
xz
Part 2
)o
^
ot
"-I*1
Part
3
*-I-i
I
co'10.51'
<f-=
2(e.Bl)
oil
I
i
x,lx,
Part2:
Y=H=2'5m
o)2 x2
)o
,l
at_
L..t -
a=H-h=2m
Parts 1 & 2:
Y=2a=4m
a2 x2
'
?o
,
+=
i
= 12.528 rad/s
119.64rpm
"
x=1m
H=4m
Part 1:
rtr'r0.51'
2(e.Bt)
a = 14.007 radls = 133.74 rpm
-
co
(n'(l)'
)o
rad/s x 3O/n
84.58 rpm
= 8.857
or =
Height of water at the sides when the depth of water
at center is zero = 4 m
Part 3:
= 90
'co
'
t!
rpm = 9.42.5 rad/s
(D'x'
'v=
?o
9.4252(tI'z
"
3s
Given:
Discharge, Q = 18Z
=4.529m
2(9.81)
yt=y-H=0.529m
-
Volume of paraboloid = Yz
Volume of paraboloid =
xx2y
where vz = )gryf
at2
HL
3V,
(0'
#r,- #u,
Volume of air
ff
=
(y,
- vrr)
_ 0.084f
fr
l+.Szo,
-
0.0826[0.021)
nL=T
LQ'z
E
D=200mm
= Q.Q/$ rna/s
f = 0.015
0.5292) = 7.0193 m3
h=500sin0
h= 86.824 m
0.0826(0.0 1s) (s00J [0.07s)'z
rrl=T
,,, _ 0.0826fLQ'Z
rru-
Another solution:
--lT-
The relationship
between H, y, and hr
in the open cylindrical
vessel shown is:
HL = 10.9 m
Energy equation between B and A:
initial liquid
level
Ee-HL=Ea
liquid level
r
rotation
.Po +ze
9 .w +zs-HL=g
t2g
4g
t
P,
Pe
4 = Jzh@.szs)
_ Po
^t
.=Ze-Zs+HL=h-0+HL
- Pe = 86.824
+L0.9
9.8I
h=1..7664m=hr
ps
-
p6 =
958.67 kPa
Initial depth, hi'= 2 m
= 7Ix2Ah
= n(1)2(2
Vspirred = 0.734 =
m37
Vspilled
-
1,.7
B2)'?
M36
FORMULA
Volume spilled = Final volume of air - Initial volume of air
Volume spilled = 7.0193 - = xx2 a= 7.0793 - n(1)z(2)
Volume spilled = 0.736 ms = 736 Iiters
Vspiued
(B4s) (0.1
For these reservoirs, the difference in elevation is the head loss = 20 m
Q=75L/s
volume of air =
0'1'BZm3/s
HL = 19.99 m
Volume of air = Volume of big paraboloid - Volume of small paraboloid
vorumeof air=
L/s=
PiPe diameter, D = 300 mm = 0'30 m
PiPe length, L = 845 m
PiPe friction, f = 0.021
664)
734 liters
Given:
VelocitY of flow, v = 4'2 m/s
PiPe length, L = 400 m
PiPe diameter, D = 150 mm = 0'l-5 m
Hazen William's coefficient, Cr = 110
Q= Q.Q'/{ll
HL =
Power input, Pi = Q y-
Q= L4(0.15)2[4.2)
Q=Av
10.671.O18s
----r==--i:C,"" D.''
HL=
10.
H
0.429[9.81)[30]
Pi= 726.2kW
;1;1z/g
67[400J [0. 07 422)1
Pr =
Power output, Po = Pi x efficiency
Power outpu! P" = 726.2 x 0.70 = BB.3 kW
8s
(110)1Bs (0.15)48?
HL = 59.78 m
ll
1,ll1
Situation 10
.
m38
Given:
fL
HL=
Flow velocity , v = 2.5 m/s
Pipe length, L = 360 m
Pipe diameter, D = 100 mm = 0.10 m
Friction factor, f = 0.026
,,,
_
lllr
-
v2
nE
0.026(360J
2.52
0.10
2(9.81)
Given:
HL=9m
C= 120
510 m, 250 mm
Q1 = Qn =
hf
=
0.05 m3ls
10.57 L018s
hfn =
10.67 (3oo)[0.05)13s
[120)'Bs (0.2)n"
-
LU 39
HL
=
hfe
Flow velocity, v = l.B m/s
Pipe length, L = 480 m
Pipe diameter, D = 150 mm = 0.15 m
Roughness coefficient, n = 0.012
6.3si]"'
[0.
1
=
hfc =
,,, _ 6.35(0.012)'z(48o)(1.8)'z
5J*/'
Line
C
c1€s D4:r?
Hl.=29.82m
Given:
Iine B
m3/s
Q = 0.05
hfn =
t0.67(266)Q^18s
.----
r:;,-j_::,-
= 4.53 m
= 3037.36
[120]',- [0.16]*',
10.67(t90)Q.'"
=
(L2oJ'BsD.o"
10.67[51oXo.o5)1.8s
(L2O)1Bs (0.25)48?
o.2BB7
D.aez
QBr.Bs
o.,.r,
-"
=2.6m
Hl.= 77.8'4m
HL=hfe+hfe+hfo
m situation
9= 4.53 +hfa+2.6
hfs = 1.87 m
9
Given:
D=150mm
Head,H=30m
-
Surge tank
=
Theoretical discharge,
Q1
=
i
Q.15)2(24.26)
Qt= O.429 mz/s
Qc=0.05-0.0185
Qc = 0.0315
lEtrsqt3ol
vt=24.26m/s
= A vt
3037.36 QB1.B5 t.B7 = 3037.36 Qst.as
Qr = 0.0185 6:/s = 1B.5 L/s
Qn+Qc=0.05
Theoretical velocity,
.,r,=.EH
hfB =
Penstock
hfc = hfe
0.2887
D.'"'
Dc =
ec1.Bs = 1.87
0.1831m = 183.1 mm
Cohesion
Given: Canalwidth,b=6m
Canal depth, d = 1.5 m
Slope ofchannel bed, So = 0.002 =
Rbughness coefficient, n = 0.013
-s4
A=6(1.5)=9m2
Given:
P=b+2d
P=6+2(1.5)=9m
R=9/9=1m
Voidratio,"=
f,[.$$ p:/5
Q=
= Bm
SPecific energy, H = 2.2 m
Given:
/3 =
2(2.2)
/3
square opening
particles of rockthatwill pass a72-in. (30O-mm] square
opening and be retained on a 3-in. (75-mm)
particles of rock that will pass a 3-in. (75-mm) sieve and be
retained on a No. 4 (4.75'mm) U.S. standard sieve with the
Cobble
Gravel
t,
following subdivisions:
= ?l'q
V
Coarse
e.81
q = 5.56 m:/s per meter
:
Fine
passes 3-in. [75-mm) sieve and retained on3/4-in.
(19-mmJ sieve, and
passes 3/4-in. (19-mml sieve and retained on No. 4
(4.75-mm) sieve
5.56 = v, t7.47)
v,=3.7BZm/s
Nlll
Part 3: Critical slope:
A=bd=BU.a7)=1_1.76m2
R = A/P = 17.76/(8 +2 x 1.47) = 1.075 m
, = Lgr1r
partilles of rock that will not pass a 12-in. (300-mm)
Boulders
Part 2t Critical velocity:
Q=vd
y- = 10 (1 + 0.15J
y- = 11.5 kN/ms
[]J 57
= r.47 m
'147
=0.3333=33.33o/o
Dry unit weight, ya = 10 kN/m3
Moisture content, MC = 15o/o
ym=ydx[1 +MCJ
total energy
Part 1: Critical depth
For reitangular canal,
the critical depth is (2/3)H,
d, = 2H
n - 0'25
n 1.-0.25
uu ss
Situation 11
Given: Channelwidth,b
Porosity, n=0.25
1-
1
O=9
-- 0.013 (t)2/3(o.ooz)t/z
e=Av=41pr7r5r7,
gr1,
3.782
S=
is
S
A=bd
R=A/P
.ffi
is the component of shear strength of a rock, or soil that
independent of interparticle friction.
= 0.013
-J- -11.g751rr= grt,
0.0022
Situation 12
Given:
H:=5m
Hr=3m
Hz=Bm
yt = 25 kN/m:
yz = 20 kN/m:
Total stress at A:
pt = |w Hr + y7H2 + yz (0,5HaJ
pt= 9.81(3J + 25[B) + 20[0.5 x 5]
pt=279.43kPa
[I,
0.5
Part
Pore water pressure at A:
p- = y- [Hr + Hz + 0'5Hs)
pw = 9.81[3 + B + 0.5 x 5] = 132.435 kPa
1:
Part2:
Effective stress at A:
pe = pt
* p* = 279.43
-
Flow, Q = 1i 4
_
pe= 147 kPa
15,094 mz/day
vki
nn
40(0.003774)
0.25
vs =
Part
m62
Given:
a0(0.003774)(100,000)
Q=
Seepagevelocity:
vs
132.435
Q=
3:
Time for water to travel 4 km downstream.
Distance
Diameter of sample, D = 150 mm = 15 cm
Length of sample, L = 300 mm
Constanthead difference, h = 500 m
0.604 in/day
,=
L-
V,
ffi
=6,62sdays
Time,t=5min=300sec
Volume collected, V- = 360 cm3
.
Flow, Q
=Vol/t= 360/300
t.l
,L! 66
Given:
grnz
Q=kiA
7.2 = k(s
SamPle diameter, D = 50 mm
Load,F=120N
f sec
Hydraulic gradient, i=h/L= 500/300 = 5/3
Cross-sectional area, A = ; (15), = 77 6.77 cm2
=
Strain at failure, s = 150/o = 0.15
Original cross-section al area, Ao = iD2 =
/3)(17 6.71)
Cross-sectional area at failure, e,
k= 0.004074 cm/s = 4.O7 x l0;t cm/s
.
m Situation 13
t" = 1'!t,u= 1-e
1-0.15
F
=
t20
n ffi
Cohesion, cu =
1/z
gu
= o'o5L948MPa=51'e5kPa
= VzlSl.95) = 25.975 kPa
Situation 14
Given:
+T
oi = 50 kPa
rr= 30 kPa
Part 1:
R=
k= 40
4U m/day
m/Oay
1325 m
tan$=
n=
n = 25o/o
width,w=4km=4000m
0=
a
30
of
50
30.96+'
lo,
C
Hvdraulicsradient.i=
'
h 65-60
L = 1,325 =0.003774
Cross_sectional area,A = w x
t= 4000
x 25 = 100,000 m2
= 2309.99 mm2
Unconfined compression strength:
9'=
l[!
i $O)' = 1963'5 mm2
---'-.-,
PartZz
Angle of failure plane, 0 = 45' + $/2 = 6O.5'
l
Part 3:
R=ti/cos0
Part2t Water table is at base
of footing:
= 16.7 kN/m:
Y" = fsat - ]w = 2O - 9.81= 10.19 kN/ms
/
R = 30 cos 30.964"
R = 34.986 kPa
C=or+trtanQ
C=
T = ym
qu = 1.3 6 l\q + yDrNq + 0.4 ye B N7
(1 0J [1 2.e) + 1 6.7 (1.2)
50 + 30 tan 30.964.
q, = 1.3
C=68kPa
U.\
q"=276.26kpa
or=C+R
or=68+34.9.86
Part2t Water table is at ground surface:
y = ye = |sat - ]w = 20 - 9.87 = 10.79 kNTm:
or = 102.986 kPa
q, = 1.3
m70
o: = 18
q, =
kPa
6
5.+yDrNq + 0.4y" B Ny
1.3 [1 0J [ 1 2.e)
+ 10.7e (1.2) (4.4) +
q, = 241.88 kPa
oa = 34 kPa
R=oa/2 = 17kPa
Given:
sin S = P76
sinQ=17735
0.ag\.te) [2) (2.s)
..)74
C=03+R=35kpa
q=
+ 0.4 (I0.te) (2) (2.s)
o
6"
L
29.06"
odl
ol
->O
Pile cross section, a x a = 400 mm x 400 mm
Pile length, L = 20 m
Unconfined compression strength, q, = 110 kpa
Adhesion factor, o = 0.75
ar=crcupL
cu = 0.5qu =
55 kPa
p=4xa=a(0.0=1.6m
@ Situation 15
Given:
B=2m
Qf =
0.7s[ss](1..6)(20) = 1320 kN
Dr= 1..2m
c=10kPa
[n 75
y^ = 1,6.7 kN/m:
f
sat
= 20
kN/m:
0=15"
N. = 12.9
Nq =
NY
Given:
"L
4'4
Pile cross section, a x a = 300 mm x 300 mm
Pile length, L = 10 m
Unconfined compression strength, q. = 110 kpa
Adhesion factor, cr. = 1
= 2'5
qu = 1.3 s
I{.
+
yDrNq + 0.4ye B Ni
Part 1: Water table is way below the bottom of footing:
T = T" =
yn = 16.7 kNTm:
qu = 1.3 s
q, = 1.3
I{.
+
[1 0) (1
yDrNq + 0.4y" B Ny
2.e) + 16.7 (1.2)@.\ + 0.4(16.7 ) (2)(2.s)
q"=289.276kpa
::tr,:.v!2.!g-Part I
Qr=oc"pL
cu
= 0.5q, = 55 kPa
p=4xa=a[0.3) =1..2m
Qr= 1[55J[1.2)(10) = 660 kN
m76
.
2O1O NSCP:
An increase of 2oo/o is allowed for eAch additional 300 mm of width and/or
depth to a maximum value of three times the designated value.
302.4 Fills
Part 1:
302.4.1 General
Uniess otherwise recommended in the approved geotechnical engineering
report, fills shall conform to the provisions ofthis section. In the absence of
Footing width, B = 1,200 mm
Footing depth, d = 300'mm
[300+3x300); r=3
an.approved geotechnical engineering report, these provisions may be
waived for minor fills not intended to support structures.
eal=Qo(1+0.20xr1
q,l=75(1 +0.2x3)
Fills to be used to support the foundations of any building or structure shall
be placed in accordance with accepted engineering practice. A geotechnical
investigation report and a report of satisfactory placement of fill, both
acceptable to the building official, shall be submitted When required by the
building official.
No
fill or oiher surcharge loads shall be placed adjacent to any building or
structure unless such building or structure is capable ofwithstanding the
additional vertical and horizontal loads caused by the fill or surcharge.
Fill slopes shall not be constructed on natural slopes steeper than 1 unit
vertical in 2 units horizontal [50% slope), provided further that ben'bhes
shall be made to key in the subsequent fill material.
EA77
2O1O NSCP:
302.2 Cuts
302.2.2 Slope
The slope of cut surfaces shall be no steeper than is safe for the intended
use and shall be no steeper than 1 unit vertical in 2 units horizontal [5 0%
slopel unless a geotechnical engineer, or both, stating that the site has
been investigated, and giving an opinion that.a cut at a steeper slope will
be stable and not create a hazard to public or private property,
is
submitted and approved. Such cuts shall be protected against erosion or
degradation
by sufficient cover, drainage, engineering
and,for
biotechnical means.
-
Situation 16
Designated allowable foundation pressure, q. = 75 kpa for footing having a
minimum footing width of B. = 300 mm and minimum depth of d = 300 mm
into the natural grade.
q"rr
Part 2:
Footing width, B = 1,200 mm
Footing depth, d = 900 mm
qau =
qo[1 + 0.20 x [rr + rz)]
Part 3:
Q=qaux82
= 120 kPa
[300 + 3x300J; rr = 3
[300 + 2x300); rz=2
qar=75[1+Q.lx[3+2J]
q"l = 150 kPa
Q= t$Q x t.2z
Q=216kN
l,
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.Endspans
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'
niscqntinx{rusendunrestl?ined,,,:....,,,,,^:.;.:,...".,,,,,,........vr,,L,?1.11
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...,...,,",,,,.....r,y,,L,i:
'
r
,
'
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y' .l_{}
Faee of ,'
fir,rlixlter:ior$rri:i:crt..,..".....,-.....
Shear in'end menrbr:rs at
'
Wher"e l,n - c'lear span 1..,r pcsitive
negafrve rn$l11cfit.
D. 215
:r i:j
tsirrt3,$2.triNl
,
,
74:
.............,,.",....,,..."..l15v.2.,[.,,,/,2:.
i,:
75il
/I
m*meitt *r shear and avcrage r..ri'acijaceut rlear spanr i'or
situation 34 -Ttr* sectiirn of ii:T"bearlr is shown iilFigure cog6-4s0?,rThe:b.eam.is
reirifcr"ceci rvith six 28:rnm;diarneter.teniion bars ar"ld
4.1s:kilt
$ [i.{s
f*ur ?B-rum.diarnetirr
carnpression hars with fy =r'415 MPa. The stir-rups pr*videcl are 1,,0 mm in
diaineter witir fyr. *'?75 MPa", clear concrete {ol/eris 4o mnl. f. * z1 Mpa. The
no,irrinai sirear,stressaf,concrele section is 0.BE MFt.
.st!fi
lE,ihii,ir'#:,iii:,r
1
I
l(ii)
tur-r:
{,
.10* *r
i*$0 mxr
FiSure CO96-4502
v1..
::::::::
Whar is the'minimum value of "a" agcording tolNSCF?
,
A. '53 mnr
C. 11 mrn
B" 56 rnm
'il, Sq mm
it
:rr:
.(, .3*rIiid*i:i
s,i t{1S.,kt{..i
$.,:$s$,kN:1
:
r
L,,
it:,.;,ri,,r,,:...,t,,.;j...r1r..11..1
kN
73. If the slirrups are spaced at 100 mm on cenlers,.calculate the
'
Shear at faee o.tall'{rthersirFp$trts.;.;-.,.:.;]..;...i-..:.:r......"...:.r..,,;.............;,,,.tt,
.,;,,:.,r.,i,:,,rCir:.il:tr,.k$:,..ii:ii:t.i
/ lll
Whelc:Srrppnrt is a sparldrel bcam.,,.;;.:....,,,.,,,..;,;....;..,.,,.........".r,v,, L,,t i 24
Wherr ,sirpport .is,a e(}[uintr....,,;::.:..i^;..-.,..........,,i.,:,. ..."....,,,,.,,.,.,,.."w,11.,1i2. I i 4
:
kN
strength of the beam.
N*gative mi)ffeni aI intsrj$r lace of B{teiior ' i .:
slippor! for n:embers built integraily with.sr.rppcrts
,
A. rsa
*isccnilnirouse{diutegralwjthsupport,..,.;.;....,.....;..;w,[,,.::y'X4.,
lnt*riorspar1s ,..:......".-...,..".,...,...", ...,.,.:..;..,......-....r-yrL,": / i{:. '
i\egative t]lomen!,at exl€rjorf,farrrof fifst iBtsl'j,or.slrppor.t , , ,
Twcspans,;.,,.'':..,-...-..,-.....'.',',,'
,iliore [h6n hryo spaus,.,-.--,........,,,;,,:..:.r].:,.;.,,,.......r;,.._-..*.,.]",.;",.,,,., w, ],r,:
'
.
calculate the norninil shear strength (n,tPa) provided by concrete ilthe efiecbive
depth d = 520 mm.
:
clesign shear
--
i
.l
i
,..
l.-
R
Solutions to November }OLS Examination
^[re
n*O
.11
From the influence
line shown, the reaction
at B is maximum when the
moving uniform load
is acts on the entire
span of 10 meters.
I
1t
??l
r.ir
Influence Line for Rn
fl-lll
Situation 1
Parts 1 & 2: Neglecting the weight of the boom, boom
a
BC may be considered
two-force member. The forces can be solved by method of joints.
tanB=914'
F=66.038'
Considering the forces at joint
tancr=9/1,2;a=36.87"
C:
0=0-d=29.168"
y=90'+u=126.87"
0=180'-0-y=23'962"
Part 1: W= 24 kN
Fu.-W
siny sin0
Fsc
24
= sin 29.168"
sin 126.87o
Fsc = 39.4 kN
Ra
= Fsc
Rs =
39.4 kN
\,
A
\"
&"---'----
t.**r,_*t
*-*-*t;
Part}:T = 48 kN
w
T
sinO= sin$
B,;
- --5&-
w=
48
sin23.962"
W = 57.6 kN
sinz9.\69"
4m
Part 3;
lU Situation
T=50kN, Wu=BkN
3
b*-:
s->
3
I
IMe=0
W[4J+Wr[z)=Tsino(BJ
H,=lkN
I
I
I
4W = 50 sin 36.87o (Bl _ B(Z)
W= 56kN
:
l
9m
l
Fn,
:i'"!'
^I
I
I
0
itt"B
tr
3m
(1
I
o
.-. .-t -
E =1kN
2.45 m
I
I
i
m Situation 2
P,
=
1.5
L-
kN
Pr=13kN
Pz
>a
B" = 2.5 kN
= 6.5 kN
0=45'
"
=
r."r"
IMs
=
(312.25) = 53.13'
0
Bv(7.5) = 1.5(10.s) + 1[3)
Bv = 2.5 kN
xFv=0
IFu=0
1.5 p
At joint
E:
D
Ev=Bv-1.5=1kN
EH=1kN
Fau sin cr = Ev
Fr;lt= 1 /sin 53.13o =
(4a)z = 32 + 62 ; a ='1.677 m
Section a-a:
Part 1: Wn = 0
XMa =
0
'
Part}:Wo
LFv=0
72Bv= pr(a) + p{2a) + pr(3al + pz[4aJ
12 Bv = 13[1..677) (L + 2 + 3] + 6.5[4 x L677)
Bv = 14.534 kN
= 27.5 kN
Bv=74.534+1/z(57.5)
Bv =
28.28 kN
Part 3: Wo = 27.5 kN
At joint D, Fco = Wo = 27.5 kN
Fcnsino=1.
Fon =
1.25 kN
1'25 kN T
&.=1kN
Section b-b:
Cross-sectional area of AL:
Aec = 3B[76J - 26(64) = L,224 mm2
IMc=0
Lx=L/
Frc[3)+2.5[3)=1.5(6)
Fec=0.5kN=50ONT
cose= L732m
Part 1:
Stress in AC = Pac
/
24=Pec/7,224
A,qc
Pec =
29.376 kN
W=Prcsin0
At joint A:
W = 29.379 x sin 30o = 14.69 kN
Part2:
1.5
Strain = 6,rc/Lac = 0.0002
kN
i:
6o.
P.^ L._
OAC=
Ao.
Situation 4
Lo.
E
Po.
Ao.
E
P.^
-
Refer to the previous Situation (See Figure above):
r,224(200,000)
Fsc=0
Pac =
At joint A:
Far sin 0 =
1.5;
Fns = FeF cos
Fec
]
Situation
Part 3:
0;
= Fae = 1.5kN
Fes = 1.5
kN
C
W=B0kN
W=Pacsin0
B0 = pec x sin 30"
C
Pec
At joint
5
IlTfr--{i:l
!;ti
[":jo)
l!l'h
li I
I
t::t
tlti
l;!l
t:.1
l!!t
l!i.i
lai
!!r
i:n
38 mm
I
I
BH
W = 48.96 x sin 30o = 24.48 kN
W=Plcsin0
Fer = 2.121 kN
48.96 kN
By
l^
--"t{-----__]<.-
C:
CH = Pec cos 0
= 160 kN
= 138.564 kN
t[E Situation 6
Lr=1.8m
Lz=7.2m
Lcn=3m
dco = 36 mm
E = 200 GPa
L=Lr+Lz=3m
Aco=
f
(36)2= 1,018mm2
AB
6"
L=1.5m
0=30"
;
Part 1: W = B0 kN
TxLr=WxL
T = B0[3) / 1.8 = 133.333 kN
XMa=0
6.
=
TL.,
A.,
^ 133.333r30001
dc'
1018(200,000)
E
situation
B
Given:
D=300mm
t=6mm
L=3m
H=3kN
T=25kN-m
G=7BGPa
=1.965mm
d=D-ztr3o0-2(6)
dt-o.
'LL1
aB
Part2z Allowable stress,
Fco
T=FcoxAco
-
1'965[3)
1.8
= 3.27s mm
t=
= 124Mpa
T = 124 x 1,018 = 126.232 kN
TxLr=WxL
XMa=0
W = 726.232 x l.B
Part 3: Strain,
CCD
6-^
L.,
= -s
eco
d=2BBmm
/
3=
75.74 kN
$ro^
d,
=
+(3004
-
2BB+)
] = 119.8 x 106 mma
Part 1:
e=
= 0.002
TL
e=
-IG
0=
T
A.,
-
E
T = 0.002 x 1,018 x 200,000
T = 407.2 kN
fx[=!!xl,
w = 407.2 x'L.B /
XM,q = 0
25x106[3000)
119.8 x 10" (78,000J
0.4599"
Part2:
"
16TD
n(D4
_
'
-d41
_
-
1.6(25x 106J(300)
nt3oo- -2BBo1
t = 31.302 MPa
3 = 244.3 kN
Part 3:
EE
Situation 7
Q=>Ay
Given:
a=0.4m
b=2.4m
,
Q
pressure = 1.6 kPa
F = 3.84 kN
c = 0.4 +
2.4/2 = t.6 m
Moment, M = F x L = 3.Ba(B) = 30,72 kN-m
Torque, T = F x c = 3.84[1.6) = 6,144 kN-m
Shear,V=F=3.84kN
+3n-1/zxr2x !3n
:22 _
' = 3'[R3 13) = :3'(1503_ 1443)
d=1m
L=Bm
F=p"Area=t.6"e.{.]!)
e=yztcRr'
Q=259,344mm2
t= 2(6) = 1"2mm
V=H=3kN
I= tr f300n-2BB4l
64'
vQ
"" Ir
""
f'
=59.9 x 106mmq
3,000(2s9,344)
59.9x 106 (12J
= 1,082 MPa
1800
m Situation 9
b=6m;c=3m
EJ Situation 10
Pr = 290(3)
Pr = 870 kN
Part 1:
.
Pr = 1200 kN; Pz = 800 kN
P=Pr+Pz=2000kN
For uniform base
p=p1+pz
pressure "q", P must
be atL/2.
P=
P
Location ofP:
P(x)=Pr161+Pz[bJ
x
x
2000(x)=0+800(6)
x=2.4m
2610 kN
= Pr[1.5J
+ Pz[10.5)
=7.5m
L=2i
L/2=c+(b-x)
Pr=
Pz= 580(31
= 1740 kN
Pz
=2(7.5)
L=15m
L/2=3+[6-2.4)
L/2.= 6.6m
a+x=L/2
Part2:
a+2.4=6.6
Ve=0
a=4.2m
Part 3:
w(LJ=P
Parts 2 & 3:
Pr = Pz = 1200 kN
a=3
w =174 kN/m
tr4r=syzf2 - Pr[xx = 8.162 m
q = 200 kN/m
Vsr = 200(3) = 600 kN
Vez=500-1200=-600
V-u* = 600 kN
Location of zero
x=6m
1.5J
0=174fl/)-870(x-L.5)
q= 2400/12
ll
w(1s)=2610
)
L=3+6 +3=12m
il
w(5)-870=0
w = 174 kN/m
moment. I
I
m situation 11
R
m Situation
= 160'+ 40,
12
R= u65,"1#
= 50 Mpa
R = 39.825 MPa
omu=60+R=110Mpa
dmin=60-R=10Mpa
omax=65+R
o.", = 104,825 MPa
tT
--> ('
omln=65-R
o.rn = 25.175 MPa
-lmax -D
- t\
t-,*
m Situation
= 39.825 MPa
13
P=400kN
EI = 75,000 kN-mz
a=3.75m
L=5m
Tangential stress due to internal pressure, o.1 6max
=
= 1j.0 Mpa
Longitudinal stress due to internal pressure, 6z a/z
=
6min=G2-03 10=55-o-:
o.s = 45 Mpa
or= 55 Mpa
-
[compressive stress.due to pJ
Cross-sectional area, A = nDt = ,T(400J[2.5) 1000n
mm2
=
03=;
A
Part 1:
OB=
Tank diameter, D = 400 mm
Wall thickness, t = 2.5 mm
P
' L=5m
.
Pa3
3EI
3(75,0001
-
6s=0.09375m=93.75mm
Part 2:
6-"* = 6c =
r"
6EI
r31-r1
I
P=o:A=45(1000nJ
-\---"'-l
P=
141.4
kN
6**=
$St3(s)-3.zsl
6[7s,000) ' ' '
6max=0.1406m=140.6mm
,
Part 3:
^
oc=
R-
R"L3
3EI
Rc =
o^2
Formula: Rc=
15'13
0.1406 = -----!-:--i3[75,000)
. -
L
400(3.7513
*(3L_a)
2U-
253.1 kN
m Situation 14
Given: t=4m
0lJ
Situation 15
Given:
w = 25 kN/m
6e=I6mm
L=Bm
w (kN/m)
w = 55 kN/m
Part 1:
R, =
RB
_
Part 1:
3*L
B'
3[2s_.)(4]
Me=yzytyxL/3
= 37.5
Me =
Ma =
kN
Yz(ss)(B)(B/3)
586'67 kN-m
z
Part 2:
WL
RR-10_
8EI
0.016
=
Part
Parts 2 & 3:
wla
^ .-dB-
w(4)n
RB=
55(BJ
10
Par-ts 2
1
= ++ r.N
l
BEI
Me = 586.67
EI = 50,000 kN-m2
Yr
=
r/zwl -
-
44{B) = 234.67 kN-m
Rs =
Yr(55)(B) - 44 = 17 6 kN
6z=10mm
.
02=
P.L3
"
o.o1
3EI
=
P'
EEI
[4)'
Situation 16
L=Bm; w=36kN/m
3
[50,000)
Pa = 23.438 kN
Part 3:
By three-moment equation:
Ma=?
MB= _w(L/2)2
w kN/m)
/2
Me= -25[2)2/2 = -50 kN-m
Lr=L/.2=2m
MoLo+2Ma(L"+Lrl+Met-r* 64"%
Lo
wLt
*
6116;
wU
124
124
-O
L1
0+2Mn[0 +2)+(-50)(2)+o+ Z!(2J'
=s
4'
Ma = 12.5 kN-m
Ma =
Rr[L/2] -wL2/2
't2.5=2Rn-25[4)z/z
Rs =
106.25 kN
-wL,2
/12
'wLz
ll2
'wl'j
112
&
3
P=0.5F,xAt+0.3FuxAv
Part 1: Degree of indeterminacy under the given loading:
Numberofreactions, R= 2 + 7+2=5
fnohorizontal reactions)
(excluding horizontal)
Number of equations, E = 2
P=
0.5(a001 x L,272 + 0.3(400J[3,060]
P=621.6kN
x-
Path2:
Degree of indeterminacy = R - E = 3o
r,4-f-+irl_+]
6u=lZxz+xr-2.5dr,]t
6) + 33
= 1,530 mm2
Av = l2(7
Part2:
Y^u"= wL/2
Part 3:
Mna= w
Au
Y^
"=36(8)/2
V*,* = 144 kN
+ 50
At= t,734 mmz
M^^"= 3618)2/12
M-,, = 192 kN-m
P=0.5F,x&+0.3FsxA"
P=
m situation 17
Given:
.
t
- 2.5[23J](12J
6t= lZxz + xa - 2.5dr,]
a= l2(76)
L2/12
- 2.5[23)](12)
P=
.xr = 33 mm, x2 = 76 mm, x3 = 50 mm
Plate thickness , t = 12 mm
Bolt diameter, dr, = 20 mm
Hole diameter. dr, = 23 mm
m situation
0.5[400J x r,734 + 0.3(400)(1,530J
530.4 kN
18
Gross area' Ac = 930 mmz
Allowable tensile stress on gross area = 0'6Fv = 148'8 MPa
Allowable tensile stress on net area = 0'5F' = 200 MPa
Allowable shear stress on net area = 0'3Fu = 120 MPa
Part 1:
Gross width, W e = 2 xz + 2 xz = 2(50) + 2(7 6) = 252 rurm
Gross area, Ae = Wc x t= 252(12) = 3,024 mmz
Part 1r
' P=0.6FyxAe P=0.6(248) x3,024
based on gross area:
P
P = Fte'x Ag
P=450kN
P=
P=
148.8[930)
138.38 kN
Part2:
Net width, W. = We - Xholes = 252 - 3[23] = 1B3 mm
Net area, A, = Wn x t = 183 [12) = 2,L96 mmz
P=0.5FuxAn
Part 2:
P=FtnxAu
P=0.5(400) x2,196
P=
based on netarea [U = 0'85)
P
439.2 kN
x" x"
|.--"*t.---*|
Path 1:
x,
Part 3:
P
based on blockshear.
Shear area:
.N = 2 x lZxz + xt- 2.5dr,]t
N = 2 x l2(7 6) + 33 - 2.s{23)l(72)
tru=[Lr+Lz)t
,.,1
"l
shear
6"=[60+115)6
Au =
1050 mm2
-,1
gr= l2xz - 2dr,]
6,=
t
12(76) -2(23)l(72)
At= 1,272mm2
I
x (U x An)
P=200x[0.85x930)
P=
Part 3:
Au = 3,060 mmz
P = Ftn
Tension area:
At= 76(6) = 455 mmz
158.1kN
where An = Ae
x,
x,
P=FuA"+.FtAt
Part 3:
P=120[1050]+200(4561
P
P*=H
=217.2kN
F*"ra 0.707 tw L* = H
124 x 10.707(B) Lwl = 300,000
L*
m Situation
Fv
19
= 248 MPa
Fuuort
Fvweld
= 140 MPa
= 1,24MPa
.
*
=
+
H
428 mm
Situation 20
BeamsPan,L=9m
Beam depth, d = 482 mm
Moment of inertia,
P = 1200 kN
H=300kN
L = 911.5 x lQ6
B=2B0mm
W= 450 mm
d=381 mm
p6a
Web thickness, tw = 17 mm
Allowable bending stress,
' Fr = 164 MPa
Allowable web shear stress,
F" = 98 MPa
Part 1: Plate thickness
Plate thicknesr,, =
^' P
BW
Part 1:
1200.000
482
280[4s0)
Fu = 0.75Fy
= 186 MPa
x = 0.5(W
dJ = 0.s[asO
17.25
-
-Mc <l'b
fb=
- I*.
Fol* 164[911.5x 106)
14=
=
"8"'
IFb
nt=x/2=
- 381] = 3a.5
M=
620.274kN-m
M=
3P
/2
620.274 =3P
P=
206,76 kN-m
nz=B/4 = 70 mm
n = max [nr, nz) = 70 mm
Plate thicknes
r,r=,@'^)Uof
Y 186
Part 2:
-V
tv=
= 27.44mm
dt-
<
Fv
-
V=Fvxdt* =98x(4B2xL7)
V = 803.012 kN
Part 2:
V=H
FvboltxAlott=H
140 x + (2212
"10
N = 5.6 say 6
H
= 300,000
bolts
,,twt
V
V=P
P=803.012kN
x
wll
Part 3:
,
Given: P = 278 kN, tp = 16 mm
M = 3P = 834 kN-m
c = d/2 + tp= 482/2+ 16=257 mm
Fh=
,rna -
Mc
I
-
(1668'6s+76'64l$)'Z
B.
o
flrr=
jM _ (7508.7s
164
l"^ = 1306.94
I 106 mm4
*rL'
Mv=
=
o
ffa,r-or;
1306.g4x 10o-911.5 x 10b+
,S= ---Y -
fh,
dt=d+2to=514n*
I $143 t2'
4823)
=
344.BBJ N-m
.t.,t.3s +
S.S7Z
Ut7 '2s+]9'2)(6)2 =
eB77.4s+ 86.4) N-m
o'
M
Ina=lx*
+344.88Jx103
6.19x104
S-
B34x n6 (257)
= ffsll.zs+
o
(1877.4s+86.4)x 103
1.382lOa
v
=
t36.04s + 6.26
*
136'04s+6'26
Interaction equation:
12L'3s+5'572
1o * j!L- =,
x = 199.2 mm
148
Fo, Fo,
*,
1,48
s=0.529m=529mm
ie Situation 21
Given:
Lengthofpurlin,L=6m
Dead load, pa= 720 Pa
Live load, pr. = 1000 Pa
Wind load, p- = 1.44 kPa
Wind pressure coefficients:
Windward, c*w = 0.60
leeward, cr. = 0.20
Channel section properties:
S, = 6.19 x l[a 663
Sv=1'3BxfQa66:
w. = 79 N/m
Part2:
Spacing of purlins due to dead, live, and wind load on leeward side:
w1=(720 + 1000Js +79=1720s+79
wz= t440(0.2)s = 2BBs
Normal load, wn = w1 cos 0 + wz = (7720s + 79) cos 14.036" + 2BBs
Normal load, wn = (1956.6s + 7 6.64) N /m
Tangential load, wt = wr cos O = (1720s+ 79) sin 14.036'
Tangential load, wi = (477 .2s + 19.2) N/m
y* =
Allowable stresses:
,
Fb' = FIY = 146 P1O'
tan0=1/a;0=14.036"
Part 1: Spacing of purlins due to dead and live load only:
wt= (720 + 1000Js
+ 79 = 1720s + 79
,,=
yJ- - 9956.6s+76'64)(6)'z = [8804.7s + 344.BBJ N-m
BB
[8804.7s+344.88Jx10'
ro-= & =r42.24s+s.s72
6.19xL04
S-
# =W#M
fn,=
M,
Sy
=(t877.4s+86.4JN-m
(1877.4s+86.4)x703
1'38 x 100
=
136.04s + 6.26
wz=0
.
Normal load, wn = w1 cos e = G72Os+ 791 cos 14.036.
Normal load, wn = [1668.6s + 7 6.64) N /m
Tangential load, wt = w1 cos e = (1720s + 79) sin 74.036"
Tangential load, wt = {477.2s t 19.2) N/m
Interaction equation: (Note: the allowable stresses are increased by 1/3)
142.24s+5.572 . 736.04s+6.26 _,
fu, fo,
+Fo. +Fo,
^
+(148)
s=0.667m=667mm
+(148)
Part 3: Spacing of purlins due to dead, live, and wind load on windward side:
wr = (720 + 1000)s + 79 = 1720s + 79
'
w2 = 1440[0.6)s = B64s
Normal.toad, wn = w1 cos 0 + tt12 = f17[0s+ 791 cos 1.4.036. + B64s
Normal load, wn = (2532.6s + 26.64)N/m
.Tangential load, wr = wr cos O = (l7ZOs+ 79) sin t4.036"
Tangential load, w1 = (417 .Zs + I9.2) N /m
,.=
=
S='3m
Column = 0.4 m
po = 4.8 kPa
pr = 2.9 kPa
A
L,=B-0.4
D
E
P
6
il{
1
L"= 7.6m
- Q532.6s;16.64)(6-)'z =(tt3e6.Js+344.88) N-m
Y
fhx
El Situation 22
Lr=Lz=Bm
M
----!
S,
'B Y'f
Mu=
=
c_y _M
tbv_
's
_
(11396.7s + 344.88Jx r03
6.1 9 x 10a
( 417.2s+19.2)(6\2
:________:______
B'
-
1.38 x 10a
s.s72
= (1877 .4s + 86.4J N_rn
(1877 'a?!86.a_)x1'03
v
=.1,84.12s +
= r36.04s + 6.26
Part 1:
pu=1.4pn+7.7pr
pu = 11.65 kPa
Wu-puxS
w, = 11.65[3J
w, = 34.95 kPa
Interaction equation: [Note: the allowable stressesare increased by t/3)
fo,
fo*
1,84.72s+5.572 736.04s+6.26
*
*Fo-
=l
1Fo,
iu49)
i
*,,L,'lta
w,,i,,,'/L4
,dffiffim)-*,L"1$
'w,,f,3fg
+t148l
s=0.579m=579mm
Part 2:
'
w 1,2
Mc = - ----g---L-
.16
tr,
_
34.95(7.q'z
t6
Mc= -126.L69 kN-m
Part 3:
Mr=-w,Ln'
9
lvlc=-
34.95(7.q'z
9
Mc= -224.301 kN-m
-
##trffift*'-w,l-"'lts
=
A.-pbd
Situation 23
A. = 0.00718[1000][75)
A, = 538,5 mmz
L=7.5m
S=3m
pa = 4.8 kPa
'
1000Ah 1000x [(10)'z
Spaclng,= A = 53Bs
ns
B
lA
pr = 2.9 kPa
bw=300mm
Spacing,s=145.8mm
q
Ln=S-b*=2.7m
0
p,=1.4p0+1.7pr
Part 3:
According to NSCP [section 407.7.5) in wa]ls and slabs other than
concrete joist construction, primary flexural reinforcement shall not be
spaced farther apart than three times the wall or slab thickness, nor farther
'than 450 mm.
D
p, = 11.65 kPa
wu=puxb.=11.65(1J
w, = 11.65 kN/m
F
Thus, s.," = 3(1001 = 300 mm
I
b"=1m
Part 1:
w
'14= J,
Moo.
1,2
lm situation24
Part 1:
Section 407.7.2 '
17.65(2.T'z
tr,
,rrror _
14_
Mpos =
6.066 kN-m
w"L"it<
M,,
r*r1\\r,/';\,\
Moment
Part2:
Diagram
w 1,2
= u'
9
M"-
L1"65(2'7)'z
9
.w,t,.'/z+
4'Z* **A
parallel
Where
reinforcement is placed
w"L,u/14
-w*t,,19
w,,L,,"/za
= 9.437 kN-m
in two or more
layers,
bars in the upper layers
shall be placed directlY
above bars in the
bottom layer with clear
distance between
layers not less than 25
6-28mm0
mm.
Effective depth, d = slab thickness - cover * y2 db
Effective depth, d = 100 - 20 - y2(L0) = 75 mm
b*
-
$50 mm
a=25+2x(28/2)=53mm
b=1m=1000mm
Parts 2 & 3:
Mq=$R.bdz
F". = 0.BB MPa
9,437 x 106 = 0^90 R. [1000J[75J,
R. = 1.864 MPa
0.85f
P- --
-
t
v
'
['-
r_-4-.l
0.85 r'.
_
.J
p = 0.0071 B
pmi"= 1.4/fy = La/275 = 0.0051
0.Bs(20.7)
275
(r- "
I
y.=p.*b*d
,,(13641 )
0.Bs[20.7)
V. = 0.BB(350)[520)
V.=
160.16kN
)Part2
J
Av=2 x An
A,
Use P = 0.00718
s = l-00 mm
d=520mm
frh
S
d
{'=2x
Vs=
[(70)2='],57mmz
ts7(27s)(s20)
100
V. = 224.51 kN
Vn = Vc +
Vs
Vn = 160.16 + 224.51
Part 3:
b* = 600 mm
V. = 384.67 kN
d=400-d'
Design shear strength = $Vn = 0.85(384.67J = 327 kN
d=334mm
Av= 4 x
m Situation 25
Given:
b* = 600 mm
Dimension, B x h = 400mm x 600mm
f c =28 MPa; fy = 415 MPa; fyn=275MPa
Main bar = L0 - 28 mm O
Hoop diameter = l0 mm
+(t2),
A"= 452.4mm2
Spacing ofhoops
s=100mm
F,. = 0.BB MPa
d
Part 1:
Ag =
Afd
400 x 600 = 240,000 mm2
x + (28)2 = 6L57 mm2
vs
A.t = 10
Pn = 0.8 [0.85 f,. (Ac - A",.) + l, A,rl
P" = 0.8 [0.85 (28) (2 40,000 - 6,Ls7
P" = 6496.7 kN
-
) + a15 @,757)]
6110
Part2:
= Fu'
4s2.4(27s)(334)
100
V. = 415.53 kN
b-'
Vn = V. +
mm
Vs=
s
Vs
X:: ?#!urof#uuu,
Vn
= 176.35+ 415.53
V" = 591.88 kN
d'=40+12*14{uTI
d'= 66 mm
b*=400mm
E
ts
v,..
d=600-66 I
d=534mm ll *
=
Situation 26
Given:
;a
Part
A,=3x +(tZ),
Au = 339.3 mmz
L
Afd
"Yn
ps =
0.025;
Ast = 0.025 Ae
pu=Qpn
d'
pu
= Q 0.85 [0.85fc (Ae - As!) + fy A't]
2,900,000 = 0.75(0.85)[0.85(21)[As
Spacing of hoops, s = 100 mm
\/ .S
Factored axial load, Pu = 2,9O0 kN
Concrete strength, f. = 21 MPa
Steel yield strength, fy = 415 MPa
,, _ 339.3(27 5)[534)
100
V' = 498.25 kN
V.=Fu.b.d
V. = 0.BB(a00)[534J
V. = 187.97 kN
Vn=V.+V.
V"= L87.97 + 498.25
V. = 686.22 kN
As= t$J,'/$Q rnrnz
Ag-
iDz
763,759= iDz
D=457mm
-
0.025As) + a15[0.025Ag)]
PartZz
p. = 0.02; Ast = 0.02 Ag; Number of bars, N = 6
Bottom
fiber:
'P e ca / I
.rhor =- 1,100,000
2oopoo'-
frot = -P/A
P,-OP"
0.85 [0.85f. [As - A,, + fy A*]
2,900,000 = 0.7s [0.85] [0.85 (2 1) [As
Pu = 0
Ag=
l'/$,f$$
-
0.02As) + 41 5[0.0zAs)]
fuot =
1,100,000('t92)1267)
rsgrld
-35.495 MPa
rnrnz
Part 3: Additional load to "zero" the stress at the bottom:
A* = 0.02(1.7 6,366) = 3,527 mmz
A,t=NxAu
3,527-6x
dr =
Part3:
D = 500
27
.
Net fiber stress due to prestress at service loads (loss = 20Vo):
Top fiber = +4.386 x (l - 20o/o) = +3.509 MPa
Bottom fiber = -35,495
id,oz
. - Mrco
As= + [500), = 196,350 mm2
A,t = 6 x + (28), = 3,695 mmz
th -
I
OPn
Mw=
wl:
-
3,169 kN
28.396-
M*(267)
1.BBv
10'
M. = 199.942 kN-m
-
0P" = O 0.85 [0.85f. (Ag * A.t) + fy A,t]
$P" = 0.75(0.85) t0.85(21)t196,350 - 3,69s) + a15(3,695)l
QP, =
= -28.396 MPa
To "zero" the stress at the bottom, the load must cause
a tensile stress of 28.396 MPa at the botLom.
mm,db= 28mm, N = 6
Design strength,
r [1 - 20o/o)
.4 mm say 28 mm
B
199.942
wff.5)z
.8
w = 28.436 kN/m
Total floor load (pressure), pt:
Ml Situation 27
Given:
mmz
mma
A = 200,000
I = 1.BB x 10e
ct = BB
mm
cr=267mm
e=267 -75 =
w'Ptx2.4
192 mm
P = 1100 kN
loss = 20%o
Additional load = pt - (DL + LLJ
Additional load = 11.848 - (?.3 + 6.2) = 3'349
L=7.5m
Dead load, DL = 2.3 kPa
Live load, LL = 6.2kPa
D
P/A
2.40 m
*Pecr/I
Water-cement ratio, rwc = 0.41
Mass of water per cubic meter of concrete, Mw = 180 kg
Coarse aggregate, Wca = 10.1 kN per cubic meter of concrete
Entrapped air, Vair = 1%o of volume of concrete
Specific gravities:
'
P/A
1,100,000
{
i
,!,
!:
ftop =
200,000
+4.386 MPa
-
f
Cement, sgcem = 3.15
Fine aggregates,sgra = 2.64
Coarse aggregates, sgca= 2.68
'Peeu/1
Calculate the required component quantities for one (1) m: of concrete:
ftop=-P74+Pect/l
llon--
Lat
Situation 28
Given:
Parts 1 & 2: Stresses due to initial prestress
Top fiber:
28.436=ptx2_4
pt = 1,1.848 kPa
1,100,000(1e2) [BB)
1.BB
x 10e
Vc=l-m3
Water: V- = M-/p- = 180/1000
= 0.18 m3
Cement: rwc = W*/Wce., W... =W*
W."*
'
-.
[180 " 9.81J
- -y.".
f
rwc
Mass of cement:
Iwc = Mw/Mcem
0.48=LBO/M*-
0.4L = 4306.8 N
M."- = 375 kg
w-
.
Air:
/
4306.8
9810[3.15)
vorume of
cement:
v."- = b
p."*-
375
1000(3.15)
= 0.119 m3
Va,r= Io/o (1J = 0.01 m:
Volume of air
Vair=1o/ox1m3-0.01
m3
Coarse aggregates:
V, =W,ufy,u= 1,0.1/(g.Bl ,2.68)
= 0.384 m3
Volume of walter, cement, water, and air:
V.*u = 0.119 + 0.18 + 0.01 = 0.309 m:
Fine aggregates:
Vfa = Vc
Vr, =
1-
- V* - V."- -
Vca
-
-
0.18
0.139
- Vri,
0.384 - 0.01 = 0.287 mz
Volume of aggregates,
Vugg
Volume of fine aggregates,
Part
1:
Volume of cement, water, and course aggregate
per cubic meter of concrete = 0.139 + 0.18 + 0.384 = 0.703 m:
ParLZt Weight
of cement for 10 m: of concrete = 4306.8 x 10 = 43,068 N
Weight of cement for 10 m: of concrete = 43,068 kN
Part
3:
m Situation 29
Given:
Vru =
Volume of coarse aggregates,
rr x Vacc = 0.33 x 0.691 = 0.228 ms
Vcu =
0.691
- 0.228 = 0,463
Part 1: Mass of cement and water:
M.* = 375 + 180 = 555 kg
Mr" = pr, x fsx = (1000 x 2.68)(0.228) =
611 kg
Part 3: Mass ofcoarse aggregates:
Mca = pca x Vgz = [1000 x 2.64)(0.463) = 7222
Water-cement ratio, r*. = 0.48
.Mass of water per cubic meter of concrete, M* = 180 kg
Entrapped air, Vair = 1%o of volume of concrete
Fine aggregates, rr = 33%o (oftotal aggregatesl
Specific gravities:
Cement, sgcem = 3.L5
' Fine aggregates, sgr" = 2.68
Coarse aggregates, sg,"= 2.64
1 m3
Volume of
il
- 0.309 = 0.69L m:
Part?':, Mass of fine aggregates:
Volume of fine aggregates for 0.55 ms of concrete:
Volume = 0.287 x 0.55 = 0.158 ms
Calculate the required component quantities for one [1) m3 ofconcrete:
V.=
=1
water:
V. = M*/p* = 180/1000
=
0.lB
m3
l<g
m3
Notes
f a::aili:.i*a:.!l :1:ii::it:iiitilr
ii::ili:r5}+i;ii1i,:r
44. A
il{}ntain$ $4 rsd .marirles, 14 yeil*w marllles, 3? trlue ri.rarbl*s, al} the
are ol'the same size., Three :iarbles aredrfiwg at,randr:m from r-he hox,
: 'firsl or:+, then a secorid, and tlien a t}:ird: *cl+rmine the prr*bahiiity *f ge$ing
, firsf. red marbl*; sec*nd yelir.rw marille and thir,d
}:rlue ryt;il"ble with replacernen{.
a. 0.0?42
c.0.0-128
r.r-r:x
' m;lrbies
E.
,0.0342
50, :G.Eill{ C$ri:ora[iuu:s stock wl]iei]i] currenttry:se]is firr F508,per sha]:e' has been
pe! shara an$ inereasinS in value,at au al/erage
'
,: ,paying a P3S arrntlai,diviriend
S ytra::s. ,ltiq expeited:tlia! thi].doypaly's stolk
per
*v-er,rhglast
yea,fi
*}
S%
,, ,qati
(sm:Jlany s
. will rtaintain this per:frirn:aueelovr*r the'next 5:year$, ltrhal,ii tite
cost of the capital raised through the selling ot this stock?
n,
O;et4sz
*rhen, s}re talles
45. "{n induo{rial engineer }+as fr,:lund that
* samp}e cF sjze S ot a
p.roduct, 900S p{.ihe'tima thele.are nc,clefectives ill ii:le samptre, 3Yn olthe time
ther*,is ,l,detectiva; 2!6 of the time there ar*,? delectivcs, ?% of the tlme ihcre
ai'e,3 llet*ctives, ?*,4,of the tinle lhere are ,4 deteclives, and tYr: eii'the Lii:re tlrere
5 defectives^ Wltat is the probabiliqr of taking a setmple that has ]east 3
d*i'eclives,in,tlre
5
A:
a
l'}
A.
4{i.
1494
C,. SYO
H,. 7q6
*. 904
fine ,tho*.sanrl copp*r r*ds:hat e the fol}*wing proFe,-ties:
:t 1,.
Diameler
To* thin
Lqxg'tt
'Foo
sh*it
?q:ro
thiclr
40
Tclri lt:ns
a
Z*
the rrld rnEers the lelrgth specllic:rriorrs,,find tlre:prr:babiliry that ihe rotl meets
iamel.er speri fi celi*ns
0.7n
c, $.95
Il. 0 43
$. $.87
A cataputrl is pl*e*d 1{i0 fuet f{or* thq rastie wall,:w}:i*h is 35 feet high, The
stlldier wariis the hurning balc of hay Lo cleal the iop ol'the wall and l:rrC L;fr leet
insj*le * casllo wail. lf the initjal velncity cf tho,bale is ?$ feeilsec*nd, at whar
angiq sllouJd the iialg +f,hay be,iaunchqd s+that it lrav*is "[S$,f,**e and pass ovef
tlre castle wal1" trIse g, 3Z {tfxd.
.
A. 54"5"
C. 58,?"
the
d
A.
47
LQ
$3,052,4?8,614
:'
:
: :
{,
$3,?52,428,6}4
F15"5SS
D, 1,586
c, F256,S00:
4", li.1,34"4{}0
D
}J; P3{}?,4"$S
A,aii.*tanCe wa$ trteasuileil
P254,700
' ' :.
on an,8{% slope:and f'on,nd tn be 25t&"75]nrr Wllat is
llt,. horizontal cli.slance measured itr ntetcrs'l
C., '?463,587
A. ,?589^365
D; t625.365
I*, 2520.59ir
A liqf .of levcis, 6.krn ft.rng ,is rr,in l etweex A. lo B, wlfh aveli:tgs tlackslghr and
fefesighf !.lisrances r.)f, .15g m. :The average ijqcksigh reading is 3,$ m anr1' each
timerit.is,taken, the rcdis inctineri sideward fiol* thever,lcal 5o. Wlrat is the
coi:ieet eievatil:a of B,is'its recorded eldvaticn is 4.J5.56 meters? ' : ',
C 422'89 rn
A 424.8(rm
.
$3,15?,42S,614
s. $3,552,428,61_4
An euglneer has just tro,rrowed.P800,0{}S fi:om a l*caI hank. at,the rate {}f tVo per
rnonth un Lhc unpaid balance. llis contra|t stares lhrt hc lnust r.epay tho Ioan rn
3{i equal rnonth}y instalments. Llow much money must he rcpay eail.r mnnth?
A, P77,587,49
C. I126;.57L45
E. P30"556.12
D, P25,365"78
B.
c^
serviee; itwas&se :f,or: j.Z;0{i$hoiirs.lfarrhe,erdof t}lese{ondyeaf,'itwaq r.rsed
ft:r: "t 5,*t]&'honrs find ille delireeiation'at the stld,bf the'Seconri'yeat' '
1.67-6
4,,
2I.r bir thdays?
s", 1.444
I): 39,2"
Peter l,4inuit cilnvinced -qhe Wappingerlnriia*s ro reli hii:t Manhaftan
island: tur $24. {f the Natirre:Alnerirana had pu,i t}ie $24 int* a ltafilr account
paying:50/+ irltei:est, how muuih wnuld the inve'irnrent worfh ill the yeai Z${lO i;
ln
cq]lepie
eduqation. ,Morte)l eair:l.le,deptrsiLed in a bqnh a*r:cllrtt::ihal pays,p% pqr year.
con:1rcuni!*d aulr*al,ly,,l'\nfhatr*quatr,depal{ts,shsu1d he made'hy,therfatl:er, on
his saxls tlihthroirgl: t"?iL}, liirthiiavs, irr oider ic pla",rid* F80,fl*O orr ]ris son's 1 8s,
:l^SVo o{ its cost at t}re
54" An *quiprnent ccst P4i){j,fit]0 and,has & salYage va:,li*e rlf
enr,llof lis lifuof 3S,*0$ cp*rating,h*trrs in a periud:of 5 yrs' lnthe first:year of
E. llfi,E.
48.
urlly.
3J" A:sr*all e$ffep1'*fieux invs51cd, a capi{ai erf F90,S$0 ior.a iiuy anci sell husiness.
He esl.irrraterl iii have a gross incoi.:le of PZE.0O0 annnallyqnd an *peratlng cosl
af Ps,fiCIo annually. it ifassu.rnecl tlre busldess tq ha\re a life sf l'0 year$.; If the
r'.rtc oi interisr is l. ?0/o curxpulc the henefit cost ratio.
C.', 1:338
A. .1"136
.
'
10.6SYo
C; i:4,L25;0$O
A. F3,810,0{iS
D,'rX4,33&{J**
E, F3,54,0,00fi
a fathir w;inis:'t* sgt asider'mon61, foi"his ,S-;rearr o!id';saq's.lilfql:e
A.r P12547
s112
4
rul
l$rrr, lSLir. and
10
OK
1f
f)l{
C.
L2.3b*1t
D. 10.4&9rii
B,',14,75%
51, lleter:nin*,tfue *pprO4imate.siz{i of:an annuai pay-rile,nt,fieeded,fo reqir*Pp0r0O0f
Uho lh ironds tsiri*d 6yh cifirt*,fuut]ri a ttam: ?heb'imrir.mult-,q-g,repair'l *ve1.1
$A-year pariod,,*rrd they,ear*,i:lteSest at: all.a::nual rate of 6Yi c*r*pounded
,
5t.
m: r
..
i,,
'
D:. 425.12,n: ',
.,
ts., 4I4.lT
A srr"ldent was askecl rc,m*ke a,365.24-m 1*ng llne using a 25-n] rape that is
0,t:024* toc lang, Whal is'tile reqrlir€d rrteasr*letrtent,t ,, , ,'
C;' ,365,.1.5* m
A. 365;2O$ rn
. ,
E: '365"458
n't
I),
3d5;275 nr
ffi
n
ILJ
Solutions to May 2OL6 Examination
1
Given:
B
out of l0homes are heated by etectricity.
For 36,000 homes, the number of homes heated by electricity is:
36,000 x (B/25) = 17,52O homes
l,t
2
Given: logzx-logz5=3
I =3
log,-5
a, -
X
5
x=5xB=40
Llr3
Given series: L, 1,1/2,1./6,1/24, ...a^
The nth term is, an
a7=
1,
(7 -1)t
=
1
- 1)l
:
[n
=l/72O
-
zu4
Given:
Rate of decrease,r = -5.75%o
Initial forest size = Po = P
Final forest size = P = 0.25Po
P=Pnen
0.25 P" = Po s-00s7st
t= 24.lLyears
tus
Given: Y=-x2+2x+27
Whenx=7,y=28m
m6
Given:
log [1 - k) = -0.3/H
Half-life,H=Bdays
log [1 - k) = -0.3/B; k= o.oa27
m12
Given:
Given:
SPeed,
d=5sin (2n
'13 t)*g
Time=
a-
High tide, t = 13 /4:
Lowtide, t=39/4:
rl=5sin (!
,13 Y)+9=L4m
4'
39
d=5sin(-''13 4' )+!=4m
21
m13
Given:
Given: W=0.12 ]nP+0.84
P=192 W =1,.47 m/s
NewYork, P =7343 W= 1.908 m/s
fackson,
15,000 = 6,000 + Sales x 0,15
P60,000
Final alloy
y lbs
x lbs
1001bs
,&r
ffiffi
W
ofthe plane in still air, x = 150 kph
ofwind, Y = 30 kPh
Travel time (round triP) = 4 5tt
Distance traveled due south = (.L50 + 301(1.61 = 2BB
Copper:
Tin:
km
10s
mi/s
Time of travel, t = 5 x 102 s
Distance =v x t= 1'86 x 10s (5 x 102J
Distance = 93 x 100 miles
tin
0.20(1001 + 1 x = 0,30[100 + x + Y)
, 0.7x- 0.3y = 16
)
Eq.
0.05[100) + 1y = 0.1[100
0.1x
Solving
m11
100%
100% copper
6%tin
Distance traveled with the wind = distance traveled against the wind
[150 + 30)t = [1s0 - 30][4 - t]
t = 1.6 hrs
-
0.9Y =
1,00+x+y
ffi=
+ ffi+
"*
20% copper
Speed
SPeed
,v = 1.86 x
Fixed monthly incorne, F = P6,000
Commission tate,r = 15o/o
Desired monthly income, I = P15,000
Given alloy
v=x/2-5/2
Speed of light
=201'.75325hrxld,ay/24hr
m14
Y- 5
y=y/2_S/Z
Given:
3107
Sales =
Inverse function: 2x=
.
-
t
[=F+Sa]esxr
Givenfunction: Y=2x+5
M10
S
75.4
Time = 8'40639 day = g days + (0.40639 x24) = B days, 9'7533hr
Time = B days, t hrs, (0,7533 x 60J = B days, thrs, 45'2 min
1:)
EOB
me
v = 15.4 mi/hr
Distance,S=3107mi
-$ )
Eq.
[1]
+
x + YJ
[2)
x=.L7.Slbs and Y = 7.5 lbs
30% copper
10% tin
m18
x & y be the number of hours each person can do the
x is
work ALONE
Given:
for the faster and y is for the slower person
Working
together:
Working
alone:
5-
*
5-
X = u!
=r
,
)
)
Area of tile border, Aa = 102 mz
X
c{
go= (12 + 2x)(B + 2x)
- 1.2(B)
792={L2. + 2x)(B +2x)-96
Eq. (1)
Eq. (2)
+
00
x = 2.106 m
Substitute x in Eq. (21 to Eq. (1):
12+ 2x
55
_+__l
y-2
m1e
v
y = 11.099 hours; y = 9.099 hours
1
1.1 hours for the slower person and 9.1
Food A
liours for the faster person
Calcium
30
Food B
Food C
=16
10
10
10
20
20
Vitamin A
10
30
20
Let A, B, an C be required the quantities of each food:
Required man-days to finish the project = 75 x 20 = 300 man-days
10 men started the work for 6 days, then 10 men are added until finish:
10(6J+(10+10Jx=300
x= 12 days
Calcium:
30A + 108 + 20C = 310
Iron:
Vitamin
10A+10B+20C=190
A:
10A + 30B + 2OC = 250
Solving the three unknowns:
A = 6 ounces, B = 3 ounces, C = 5 ounces
Total number ofdays to finish the project is 6 + 12= 18 days.
Delayed by 3 days
IEE
m17
IInits ner Ounce
lron
20
Demand = -0.0Lx2 - 0.2x+ 9
Given:
SuPPIY=0.01x2-0'1x+3
Perimeter,P=1,000m
Cost of fenci ng:
Alongx: c,
=
cy =
At equilibrium, Demand = Supply
P1500/m
P500/m
-0.01x2 - O.2x + 9 = 0.01x2
Along y:
Total fencing cost, C = P650,000
P=2x+2y
1000 = 2x+2y
x = 500 -y
) Eq. [1]
C=C*xX+Cy>.2y
650,000 = 1500(x) + 500(2y)
650,000 = 1500[500 -y) + 100Qy
y=200m; x=300m
Dimension: 300 m x 200 m
-
0.1x + 3
x--15
m21
)
The fundamental frequency of a signal is the greatest common divisor IGCDJ
of all the frequency components contained in a signal, and, equivalently, the
fundamental period is the least common multiple (LCM) of all individual
periods of the components.
(tl
= S cos 20nt + 2 cos 40nt + cos BOd
at = 2Orc; t:sz = 40n; o:: = B0rc
Frequencies,
f=
i :
fr =
10; fz=20;
fz= 40
Given:
Price ofproperty = P9,500 per square meter
Lot dimension [triangular)
a=7om,b=72m,c=49m
Fundamental frequency, f" = GCD of fu, fz, and fz
Fundamental frequency, f""= GCD (L0,20, aOl = 16
Lot area:
s=Yz(a+b+cJ=95.5m
4192 = GG:aXs-bxs-c) = 163t.294 m2
M22
h = 85.41 {q
0 = 47" 52'
cr = 39' 36'
Price = 1,631.294 x 9,500 =P15,497,291.OO
0=0-o=8.267'
F=90'+a=729.6"
lLUt
25
AP
civen:
y=90o-0=42.133"
In triangle ABT:
ABh
siny sinQ
AB
1:
=
PB1
AO?
oc5
PB=API3
AB=AP+PB=4AP/3
85'41
= sinB.267o
AC=AO+OC=3.5AO
sin 42.733"
A*o=lzAOxAPsin0
Alr,c=
AB = 398.492 m
In right triangle ACB:
H = AB sin o = 398.492 sin
(39'
7/z
AB x AC sin 0
A.-^
Aor.
36'J
H=254m
-l
AOxAPxsin0
jABxACxsin0
Ratio=3/1.4=3t14
m23
Given sides, a = 235 m, b = 280 m, c = 429 m
Another solution:
Semi-perimeter,s = Tz [a + b + c) = 472 m
A
Ratio -
Area =
Area =
47 2(47 2
-
23 s) (47 2
-
28 0X47 2
A
-
429) = 30,390 mz
Ratio
=
Ratio
=
Ratio=
'rAPo
Aor.
lAO,APxsin0
z
jABxACxsin0
AoxAP
ACXAB
2(3)
7(4)
=stt+
AOxAP
"/FZAo
Areaoca = Area of arc AEC
EA26
From the figure shown:
Areaoca
'" ='n
hb
h.
as=
a.=13
Areaocn
OEC
nR'0 -yzaxRxsino
360'
r(17J'Z(35.4'l
.
AreaocA
360'
32.5 ._
(20J
50'
- Area of triangle
- Vr(7)U7) sin 35.4'
= 54.87 cmz
El 30
UA27
C=3x-50"
Radius,R=2500ft
TrainsPeed,v=Bmph
D=2x-20"
Time,t=1min
Given angles:
A=x+10o
2x+ 20"
B=
The sum of the interior angles of a quadrilateral is 3 60'.
Note:
A+B+C+D=360'
[x + 10") + (2x+ 20'J + [3x
- 50']
+
l
mile = 5280 ft
s=vt=Bx[1/60)
(2x- 20") = 360'
s=
x=50o
2/15 mi x 5280
s=704ft
Angle A = 60o
Angle B = 120"
Angle C = 100"
Angle D = B0o
s=rx0
704=25OOx0
'
0 = 0.2816 rad
LU 28
Given:
R=L/z(28) = 14inches
O=3O. =n/6
s= R0= AQt/6)
s=
7.33 in
0=
16.13'
x
180"
-
1l
LA 31
Given:
Sum ofinterior angles ofpolygon = 3600o
Formula:
w12s
For intersecting chords ofa circle:
OAxOB=OCxOD
- 2J
3600'=180'[n-2J
n=22
Sum = 180'[n
10xOB=!2x20
OB=24cm
H-U
Diameter, AB = 10 + 24 = 34 cm
Radius,R=17cm
In triangle OEC:
a=R-10=7
oc2=R2+az-2Racos0
122 = 772 + 7z
0 = 3-5.4o
- 2(17)t7) cos 0
32
Given: x=4t+3
y=16t2-9
4t=x-3; t=x/4-3/+
y = l6t2
-9
y =16{x/4 -3/q)2
-9
-" - -9
v=16" -6x+9
xz
'16
Y=x2-6x
,-i33
Given:
Equation ofparabola: y2 = Bx
Equation ofline: x -y= { (slope, m = 4J
PointA (1,5,
Plane:4x
Slopeofchords,m=4
-31
+y+Bz+33=0
Distance from point
,
Parabola:
(xr,yt,zlto
plane Ax + By + Cz + D = 0 is:
lA*, +By,+Cz,+Dl
JA'+B' + c'
Y2=Bx
Differentiate:
,2v v' = B;
wherey'=m=4
Equation of diameter:
IE 36
2Y(4) = B
RadiusofsPh€r€,r=3
Y=l
Ifthe sphere is to be tangent to the three coordinate planes and with its center
in the first octant, the center ofthe sphere is at [3, 3, 3)
ul34
Given:
Center ofcircle: (3, -27)
Equation of line: 3x + 4y
-
Equation of sphere:
[x - trlz + [y - kJz + (z -l)z = 12
[x - Slz + (y -3)2 + lz-3)z =32
26 =0
- 6x + 9 + yz - 6y + 9 + it2.- 62 + 9 = 9
xz + yz + zz - 6x - 6y - 6z+ 1B = 0
x2
The radius ofthe circle
is the distance from the
center to the line.
m37
r=
l-+
Given point in cylindrical
coordinate, P(8, 30", 5J
ls1:1++J 27)-261
32
r=B
e=30'
z=5
+42
r=25
Let P(xr, yr) be the point of
tangency, then yr = (26 - 3xt) / 4
-
x.coordinate: x=rcos 0=6.928
y-coordinate: Y= r sin 0 = 4
{r,0,
,''
i
''.
The distance from [3, -27) to P(xr, yrJ is 25.
-ylz
| ._ 26
- 3x, \,
l-2,r4)
-
252=(3 -xrJ2a (-27
025=13-xrJz*
xr
=18&yr=-7
Point of tangency: (1B, -7)
|
J38
Given:
Equation of curve: y.= x2 + 6x - 4
The slope of the curve at any point is dy/dx = 2x + 6
At [0, -4],
dY
/dx
= 210) + 6 = 6
z)
x=t2+2t dx=2t+2
y=2t3_6t dy=612_6
dy _
dx
Given:
Probability that neither is defective:
-6 _ 3t2 -3
Zt+Z t+1
6t2
.
Whent=0,dyldx=_3
Whent=2,dy/dx=3
en t = 5, dy/dx= t2
,
Bt2
195 194
/
200 199 -o.gsos
Draw three marbles at random with replacement:
First draw red, R = 54 /7OO
Second draw Yellow, Y = 74/1OO
Third draw blue,
velocity of the particle, v = dx/dt =
4t3
- l6t
Times when the particle is at rest [v = 0):
v=4t3_16t=0
=
Total number of marbles in the box = 100
54 Red, 14 Yellow, 32 Blue
.
-
p
ml44
[0 40
Given:
Position, x(t) = t4
Box with 5 defectives and l-95 non-defective batteries'
wo (2) batteries are selected at random without replacement'
4t(t2-4)=0
4t[t+2J[t-2)=o
,
u45
Probability=
Civen:
B = 32/1,00
Z
..
# #'
=o.orn
ln a sample of 5, the probability that:
no defective, Po = 90o/o
- l defective, Pr=3o/o
'- 2 defectives, Pz 2o/o
=
Between t = 0 and t = 4, the particle is at restwhen t Z
=
,il:[:it;::'i:-iT;
t=2
Probability that has at least 3 defectives = P: + P+ + Ps
Probability that has at least 3 defectives = 2o/o + lo/1t a !o/n =
t=0
Distance traveled between t = 0 and t = 2:
xr=x[2)-x(0)=-16
rEn
$o/o
46
Diameter
Distance traveled between t= 2 andt= 4:
xz = x(4) - x(2) = 128 - (-t6) = I44
Lensth
Too short
Total distance traveled = 76 + I44 = 16O
Too lons
,i,r0l*
Too thin
OK
10
5
li.4S,
'rg6}!
4
20
Too thick
5
l
i!r?iir
7
Since the rod meets tt,e length specifications, then the sample taken are:
too thin = 40, OK =902,and too thick = 7
m41
fe dx
I --:
J,
J1*J,
= 5.333
Thus, 40 + 7 = 47 does not meet the diameter specifications
and902 meets the diameter specifications
P-..t-
902
-"- =0.95O47
902+47
WJ
47
Given:
Given:
xr = 100 ft
xz=50ft
h=35ft
vo = 70 ft/s
Principal amount, P = 800,000
Monthly interest rate, i = 10%
Number of payments (monthsl, n = 36
Formura: p=
4H$#
I
X1
Horizontal range ofburning bale, x
y = x tan
g ---gL-,
z v.- cos- 0
= x1 + xz
800,000
Xg
= 1S0 ft
A=P26,571.45
wherey = Q
mso
Given:
x= o 'tan0cosze
$o = ry-9ltan
Current price ofstock, P" = P500
Annualdividend,D=P30
Annual increase in price ofstoch r
o cos2 o
= 39.2" and 50.8.
The expected price ofthe
5 years from now is:
Using 0 = 39.2o and x = xr = 100 ft
y=xtan
O ;**"
=27.19tt<h
zvo-cos'o
-*=40.9tr>h
z vo- cos' B
= 5% = 0.05
stock
0
F=p"(1+r)s
F=
F=
can,tclearthetopoJ.thewall
- -"- --r
1
DDDDD
500(1 + 0.05Js
P638.14
a\
*t"r*
P,,
cost ofcapital, I tD-)
P,1
Let i be the required
Using 0 = 50.8. and x = xr = 100 tt
y=xran0
1]
(1 + 0.0I)'" 0.01
ou2
2v2
e
_ A[[1+ 0.01J36 -
Po=Pl+Pr
. _ D[(l*i)"-1]
[1+ iJ'i
Okay
Thus, the required angle is 50.8o
F
(1+iJ'
30[[1+i)s-u 638.14
(I -i)'
[1+i)'i
JUU=---
m48
Given:
i=0.1048=l0.4$o/o
p = $24
Nominal interest rate, r = 5yo compounded monthly
t = 2000 _ 1,626 = 37 4 years
tf,
51
Given:
i = 0.05/12
Face value ofthe bond, P = 60,000,000
Term ofbond, t = 50 years
Annual interest rate, i = 60lo
n=12xt=4488
F=P[1+i]"
F=24(t+0.05/12)4488
F=
$3,052,428,613.85
Redemption value ofthe bond after n = 50:
F = P(1 + i)n = 66,969,000[1 + 0.061s0 = 7,105,209,256.50
F
Required annual payment (sinking fundJ to raise this amount:
Fi
a_A[(1+i)'.1]
i
Present worth of all the benefits:
. _ A[[1+iJ' -1] _ 23,000[[1+0.12)10
e=
'-
[1+i)" -1'
1,105,209,256.50(0.06J
^
(1
.
A = P80,000
t^,*,_fr
i = Bo/o
:s4
Itlrl
lllll
Given:
I
tr, _ D.
ro-rA
D.
'"
iJ4
-
Total depreciation, D=d
1l
<
m
-
Given:
-1] _
-1]
-1] [1+0.08J4[[1+0.08J1, -1]
80,000[[1+0.08J4
Slope = B%
Slope distance,S = 2528.75
Angle, 0 = arctan (SlopeJ = 4.5739'
Horizontal distance, H =
S
r
cos 0 =
ms3
Capital, C = P90,000
Annual operating cost, OM = P5,000
Ben€fit: Annual income, Ai = 28,000
Annual interest, i = 1,2o/o
Life of investm ent, n = 12o/o
Net annual
income:
A = Ai - OM = 28,000
A = P23,000
- 5,000
11.2 <27,OOO
rlL 55
d =P13,962.6O
'
36,000
d = PLL.2/hr
1]
g+iyi
[1+i)4[[1+iJ1'
Given: Costs:
480,000 - 76,800
After two years:
m = 12,000 + 15,000 = 27,000 hrs
(ll)ti
_
d=
"
12
d[(1+iJL ll _ Al(1+i)4=
d
FC_SV
Depreciationrate,d=
lltt
A[[1+
_
-
A[[1+iJ4
First cost, FC = P480,000
Salvage value, SV = L6o/oFC = P76,800
Useful operating hours, n = 36,000 hours
dddd
n=
d[[1+i)l'Z-1-I
-rd=-4..i
-1]
-0.r4*io.rz1
Benefit-Cost Ratio = P /C = L29,955.13190,000
Benefit-Cost Ratio = 1.444
IAAAA
lltrr
1r
P = L29,955.13
ru52
ddddd
+D\ -
+0.06)so-1
A= 3,aO6,657.20
Given:
tr
u56
Length ofline, L = 6000 m
distance = 150 + 150 = 300 m
Average backsight, BSav" = 3.8 m
Angle of sideward inclination, cr = 5o
Elevation of point B = 425.56 m
BS + FS
Correct
backsight:
BS. = BS cos cr
BS. = 3.8 cos 5'
BS. = 3.78554 m
2520.697 m
=P302,400
Error in backsight reading,
e = BS
-
BS. =
+0.01446m
Given:
Number ofbacksight readings, N = 6000/300 = 20
Total error in backsight = +0.0t446
Distance traveled during PIEV time, S = 72.2 m
r 20 = 0.2892 m
Note:
Note: A positive error in backsight yields to a higher elevation
Correct elevation of B = 425.56
-
O.2BTZ
E=Rsec(e/2)-R
TD=MD+E
MD
36s.24= MD
L, "
+
MD
m
R=
ro.oorot
sec[0/2J
25'
msB
3600
Stadia intercept = 1.25 m
Stadia interval factor, f /i=k700.42
Stadia constant, f+ c = 0.3
1
7'54
R'
MD = 365.2O5 m
Given:
72.2=0.47704xt
t= 2.4A5 seconds
Central angle,0 = 26'
External distance, E = 7.54 m
Length oftape, Lt = 25 m
Error per tape length, e = +0.0024 m
*
mph = 1.60934 kph = 0.44704 m/s
m61
True distance,TD = 365.24 m
'l'D = MD
l-
S=vt
= 425.2708
m57
Given:
Speed, v = 65 mph
nD
=286.642m
= 286 647
cr----.-
D=40
Line ofsight is horizontal
D = kS +
[f + c)
D=100.42(1.25) + 0.3
D = 125.825 m
m62
Given:
Degreeofcurve,D=12"
Design speed, v = BB kph
Coefficient of friction, f = 0.40
us9
Mode 3-2:
x
Y
1
t/6.3
2
1.
428.635
426.325
429.524
3
/5.7
L/3.6
-
Radius of curve (arc basis), o
Formula:
e+f=
= 'u9o = ry$ =95.49
rD r(12)
Y
where e is the superelevation, v is the
1,27R
Shift-Stat-Sum:
AH=
speed in kph, and R is the radius in meters
I vrr
-"r
Xx'
=428.376
e+0.4=
BB2
t27les.4e)
Elev. of B = 48.254 + 428.37 63
Elev. of B = 476.63O3 m
e
= 0.2386 =23.860/o
Given:
Given:
h=7.5m
L=75 m
R=200m
E
t.
t-.
ra\
Lr=L/2
1
L. = 85 m
Desirable length of spiral (based on rate of change of centripetal acceleration)
'r
I_
0.036v3
l
H
R]
,'"*'ti "s.b!r
tu=
1rt -
-66
o.s(L
/
2)-Lh
4h
4(7.s)
Source: Road Safety Design Manual (DPWH)
The warrant for the use of safety barriers can be established considering:
. Fore slope or back slope steepness and height;
. Unforgiving hazards within the dear zone; and
. Water hazards within the clear zone.
The two triangles shown are similar,
R
o'o36vr
: v=77.B7koh
rnn
-
The warrant for barrier systems can be determined by a risk assessment
taking into account the various issues'
Equal-radius reversed
curve with parallel tangent
R -L
0.5 L1 h
L'
p= = 75' =187.5
where v in in kph and R in meters
R
C
012
m
LU 67
Source: Road Signs and Pavement Markings Manual (DPWH)
m64
Length ofchord
from
PC
to PT
L=140m
E
?,
3'.
r\
h=12m
n=
R=
R=
#
R
4h
Messages when painted on pavement should be
limited to three words or
less.
They shall only be used to supplement other traffic control devices. The
distance between words is variable depending on the message and location at
which it is based fusually twice the length of the word if achievable)'
The first word of the message is to be nearest the motorist on rural roads. In
urban low speed areas, the order is optional.
1402
Messages are white in color. Letters or numerals used on roads in urban
areas shall be at least 2.5 m. On high speed highways, they must be at least
4(1?)
5m.
408.33
Equal-radius reversed
curve with parailel tangent
m70
Source: Road Signs and Pavement Markings Manual (DPWH)
Flow (q) is the rate at which vehicles pass a fixed point (vehicles per hourl
Pavement and curb markings
Density [ConcentrationJ ftJ is number of vehicles [N) over a stretch bf
.
roadway (LJ (in units of vehicles per kilometer]'
Longitudinal lines which are those laid in the direction of travel. These
include: Center Line; Lane Line; Double Yellow Line;'NoPassing' Zone
Markings; Pavement Edge Line; Continuity Lines; and, Transition Line.
Transverse Lines which are laid across the direction oftravel, These indude
Stop Line; Give Way Lines; Pedestrian Crossing Markings; and,
.
Volume is the actual number ofvehicles observed to pass a given point on the
highway in a given time.
Capacity is the maximum hourly rate at which vehicles can be reasonably
expected to traverse a point or a uniform segment ofa lane or rohdway during
a given time period under prevailing roadway, traffic and control conditions.
Roundabout Holding Lines;
.
Other lines; which include: Turn Lines; Parking Bays; Painted Median
Islands; and, Bus & PUj lane Lines; and,
. Other markings which
include: approach markings
to
islands and
obstructions; Chevron marking; diagonal markings; Markings on Exit and
Entrance Ramps; Curb markings for Parking restrictions; Approach to
Railroad crossing; Messages and Symbols; and, Pavement Arrows.
EE
LU 71
includes any stage, stair, landing place, landing stage, jetty, floating barge or
pontoon, and any bridge or other works connected therewith.
69
Climbing lanes are
a
Port - a place where ships may anchor or tie up for the purpose of shelter,
repair, loading or discharge of cargo, or for other such activities connected
with waterborne commerce and including all the land and water areas and the
structures, equipment, and facilities related to these functions'
roadway lane design
typically used on Interstate highways.
They allow slower travel for
Pier- any structure built into the sea but not parallel to the coast line and
large
vehicles, such as large trucks or Semitrailer trucks, ascending a steep grade.
Since climbing uphill is difficult for these
vehicles, they can travel in the climbing
lane without slowing traffic.
Lighthouse - a tower, building, or other type of structure designed to emit light
from a system of lamps and lenses, and to serve as a navigational aid for
maritime pilots at sea or oh inland waterways.
continuous structure built parallel to or along the margin of the sea
or alongside riverbanks, canals, or waterways where vessels may lie alongside
to receive or discharge cargo, embark or disembark passengers, or lie at rest.
Wharf
Overtaking lane or passing lane is the part
-a
of a main road that is used for passing
other vehicles and is nearest the center of
Lhe road.
VA72
Source: Road Safeqt Design Manual [DPWH]
These lanes consist of three stages. The
initial diverge tape, the auxiliary lane, and
the end of merge taper.
The vertical alignment of a road consists of a series of straight grades joined
by vertical curves. A vertical curve is expressed as a K value, which is the
length ofvertical curve in meters for !o/o change in grade. In the final design,
the vertical alignment should fit into the natural terrain considering
Merging and Diverging for Auxiliary Lanes (Road SafeQt Design Manual)
The design ofovertaking lanbs and climbing lanes requires the consideration
of the:
'
Initial diverge taper;
o Auxiliary lane length; and
o
.
End or merge taper.
earthworks balances, appearance and the maximum and minimum vertical
curvature allowed.
Large K value curves should be used where they are reasonably economical.
Minimum K value vertical curves should be selected on the basis of three
controlling factors:
. Sight distance is a requirement in all situations for driver safety;
. Appearance is generally required in low fill and flat topography
situal.ions; and
. Riding comfort is a
with specific need
general requirement
on
approaches to a floodway where the length of depression needs to be
minimized.
ru73
Time period
- 6:45
- 7:00
- 7:1.5
7:1.5 - 7:30
7:30 - 7:45
7:75 - B:00
15-min
volume
6:30
75
6:45
7:00
100
-K12=J-
pro
Ru
=6
490
65
555
6
*1
tth
cycle
t"h
!r4sl
*1
mln
= 10 .cYtl"'
min
6
L75
300
410
BO
125
min
cycle
s
Meter
10
72
12
10
110
Service rates:
cycles
Cumulative
volume
cycle
, 1 u"h
cycle
u75
Draft = 7.5 m
Design low tide = -0.35
6=_[7.5+0.35)
h = -7.85 m
,I
x 15 min = 75 vehicles
x 15 min = 90 vehicles
x 15 min = 150 vehicles
Cycles
per min
10
6
5
5
6
10
fi!ii
:!iiii:iil;:;:!:i:L#i.li:14ffi
i:iiri:iiili:r;i':ir:ts1,:ir$;
I
(
H-
Solutions to May 2OL6 Examination
*u
lfi
1
Given:
DePth of Point, h = 0.25 m
Consta n ts:
Unit weight of mercury, Y- = Y* x sg = 9.81 x 13.6 = 1.33.41'6 kN/m:
Standard atmospheric pressure, p,t- = 101.325 kPa
pabs
=
pgage
*
patm
p"r,s
= (133.416 x 0'25) +
pnr. = 134.68 kPa
tu2
Given:
sp. gr. of
oil = 0.84
Hr=3m
Hz= 4m
.l'
p1=p2
yod=f*[Hr-Hz]
,
e.B1(4
- 3l
I
.l_
e.B1[0.84)
d = 1.786 m
lll Situation
w-s-
1
Given:
b=1.5m
d=3m
hr=2m
y* = 9.8L kN/ms
y = h =hr+0.5d=3.5m
A=1.5x3=4.5m2
Part 1:
p
=y. h A
F=e.Bl(3.s)(a.s)
F=
154.508 kN
.-",
water
l0t325
Part2l.
^s
Diameter = 5.8 m [radius, r = 2.9 m)
Unit weight of gas inside the balloon,
llQ:,
e= --724.s(3.s)
I
AY
e=
O.2l4m
Volumeof
balloon,
Part 3:
YP=
ILU
i
+e
= 5 N/m3
v=
!nrr=
33
!n12.g1=
V = L02.16m3
Yp=3.5 +0.21,4
Yp = 3.7L4 m
weight:
Total
W = Wen. + Woaa
W=yeV+Wtoad
W=5xL02.L6+Wroad
W=510.8+Woaa
Situation 2
Given:
r-,x
= 3.5
= 13.6
Vo = 0.02 m3
Object
so
-sg=35
sr
Buoyant force:
I'
',:1-----*;--/ f - \'
-- 13.6
*d**L*
Part 1: Weight of object:
r4y'
Ye
Unitweight of surroundingair,yu = 12 N/m3
= yo Vo
tue fhBE*d?dlr
PafiZ
Part
\02.16
510,8+Woad=1225.92
715'12 kN
W=BF'
Wloaa =
L'.-_*-*J
Liqu.icl. sg
1Y= [e.B1x 3.5)[0.02)
W = 0.6867 kN
BF - yoi. xV = 12 x
BF = 1225.92 kN
3
[f,i 10
L
Given:
Part2t Fraction
B,=6m
of volume exposed:
Fractionofvolumeclisplaced=
'
to
35
r, = 13.6
L=15m
= 0.2574
Fraction of volume exposed = 1- - 0.2574 = 0.7426
Fraction of volume exposed = 74,260/o
H=3m
Wu = 350 kN
sg'* = 1.03
wb+w"
Load:
3000 bags
Part 3: Additional force to completely submerge the object:
l'=
[y1
- y"] V"
F=
F=
@
lkg/bag
[9.81 x 13.6 - 9.81 x 3.5][0.02J
1.982 kN
Weightof cement: Wc = M x g = [40 x 3000) (9.81] = L,t77,200
W,= 1,177.2kN
gp=\^/.+Wrr
|swVo=Wt+W.
(9.81 x 1.03)[15 x 6 x d] = 350 + 1,177.7
d.= 7.679 m
N
LUI
LU 11
Situation
3
rIL
W = 25,000 metric tons = 25,000,000 kg
Draft in sea water, h = 8,4 m
Area,A=3000m2
1{
I
Hil
fi
11
waterline section
initiai condition
Area, A
vn, =
l{
p..
=
214994!9
1000[1.03)
Height Hz to avoid spillage
Given:
waterline section
Volume displaced in sea water:
vn,=
I pm
1000
+
h= (2/3)(2.1) = r.4m
H=2.Lm
ro =
90 rpm x n/30 = 9.4248 rad/s
a=H-h=0.7m
Parts 1 & 2: Height of paraboloid: [x = r = 0.65
=25,ooom3
o)'x'
'
Difference in volume displaced, AVo = 25,000
Difference in draft, on =
r = L.3/2 = 0.65 m
=z4,z71.B4smz
Volume displaced in fresh water:
25'ooo'ooo
l'l
=
ffi#
-
)o
..
Y
mJ
_ (9.424q'zQ.6r'z
-
ze.B:-)
Y = 1.913 m
24,271''845 = 728'155 mz
Drop of water, nh = y/2 - a = 1.913/2
= 0.243 m
-
0.7 = 0.2565
Volume spilled = nr2 x Ah = r(0.65)2 (0.2565) = 0.3405 m3
Volume spilled = 340.5 liters
Since the ship floats deeper in fresh
water, then draft = 8.4 + 0.243 = 8.643 m
Part 3: Minimum height of tank to avoid spillage:
Hmin =
h + y /2 = 1.4 + 1'.973 /2 = 2.357 m
[f=I17
Velocity head, vz/29, varies directly to the square of velocity. Thus, if the
velocity is increased three times, the velocity head will be increased 32 = 9
times.9x5=45m.
LL] 19
0.082
x=30m
Y =2m
6 f 1Q'z
60.e7
D5
0=30o
=
o.og26(0.02)(e,150){0.o$q'
Drt
Dz=O.2157 m
Reservoir D [Pipe 3]:
Q: = 30,000 x 0.15 = 4500 mz/day = Q.Qglt
hn = 9L.46
,-
ou2
Y=xtan0-
vo
Ltr.l
=
75.49 = 75.97 m
0.0826f
e'91(39)'
- 2v'cos'30"
2 = 3o tan 30"
,%'."r'o
-
6:/5
LQ'z
.7< o.7
_
0.0826(0.021(6' 100)(0.0521),
D,
D: = 0.2048 m
L9.597 m/s
Reservoir A IPipe
Situation 4
l):
Qr = Qz + Q: = 0.0434 + 0.0521 = 0.0955
hrr = L50
150 m
-
91.46 = 58.54 m
.hrr=0.0826tt,o,
r tr/ _ 0.0826(0.02)(15,200)(0.0955),
'
5r...D'
Dru
Dr = 0.33 m
91.46 m
-l'-1
B
€
o$l
N
\\
.9.
h,,
\\o,
30.49 m
\\\a
Q,
15.49 m
Flowrate,
C
Population = 25,000
Water ciemand = 150 liters per capita per day
Friction factoq f= 0.02 for all pipes
Reservoir
C
fPipe 2):
Qz
hn = 9 1.46
-
30.4g = 60.97 m
Q=A x
v = 0.5(2.5J =
u24
A=
B'B*4
2'
rr.zr
A =.7.68 m2
P=4+2(2.683)
Qz=Populationxdemand
= 25,000 x 0.15 = 3750 m3/day =
Mean velocity of flow, v = 2.5 m/s
Cross-sectional area, A = 1 x 0.5 = 0.5 m2
Q.QQ\ Q
sz /5
P=
9.366 m
R=
A/P = 7;68/9.366 = 0.82 m
v = Q/A = 35 /7.68 = 4.557 m/s
l.lg
lrs/s= 7g
6:/min
u=
4.5s7= 1 '1g.g21r1rs,1,
0.015
1y1r1rg',.1,
S=
M Situation
'
U
Situation 6
0.00609
t..t 6m
5
Given:
ll
b=2d
S=S"=0.001
n = 0,025
*."::i-:
Parts
*"&- *:r: -**1-*::*..:
Given:
A'7 L)
A= "'''ro.st
2'
Part
2
3 (M.E.S.)
Slope ofchannel bed, S = S" = 0.001
Roughness coefficient, n = 0.013
Area,A =6(L)=6m2
Wetted perimeter, P = 6 + 2 x 1 = B m
Hydraulic radius, R = A/P = 0.75 m
A = 3.015 m2
Perimeter, P = 2 + 2 x 1..6225 = 5.245 m
Parts 1 & 2: Mean velocity of flow and discharge
Hydraulic radius, R = A/p = 3.0L5 / 5.2a5
Hydraulic radius, = O.5748 m
v
&
---
Area:
Part2:
1
.,,r= 1pr7r5r7,
, = -l-
lo.7 5121s (o.oo r.; iz
r
0.013 '
v = 2.008 m/s
Mean velocity:
- R2/3 St/z
- 1_
n
u
=
I
=j-,
0.025
Q=Av
(o.O)l)ttz
@.S74BlzB
'
Q=
v = 0.8745 m/s
b=2d
A=bd=
Dutyofwater,q=z L/s
hectare
Q=A
3.015[0.8745) = 2.6366 mt/s = 2636.61/s
ServiceA.""=
9q3
12.048 m3ls
Part 3: Using the most efficient section:
Part 3:
Q = Av =
Q=6x2.008
R= d/2
2d2
y -ll R2/3 St/2
n
12.048 = (2dz)- x
d.=
l.67lm
__: @/2)2/3
'
0.013'
2636'6
Service Area = 878.9 hectares
[l]
31
Given:
Water content,MC = 15o/o
Moist unit weight y," = 18 kN/m3
Specific gravity of solids, G = 2.61,
y. =
qlqyg
l+e
y*
tB
- 2'61+2'6u0'1s)
1-+e
e=
0.661
xs.Bt
(0.00tJ
rrz
M Situation 7
Given:
Given:
Dry unit weight, y6 = 17 kN/m:
Unit weight of water, y* = 9.81 kN/m:
'
17= G" x9.81
1+ 0.6
G, = 2.773
)
G"
1+e
-\w
= 1.5; H = 15 m; er =
aH=H Ae
Voidratio,e=0.60
\d=
eo
1+
eo
7/z eo
= 0.75
AH=1515-075
1+ 1.5
AH=4.5m
Part
1
m38
Given:
Part2:
Diameter of sample, D = 5 cm
Saturated unit weight:
G+e
]sat = ;L
l+e
|w
Length of sample, L = 20 cm
w,=
Diameter of standpipe, d = 1 cm
Initial and final heads: hr = 90 cm, hz = 50 cm
Time of observation, t = 1 min = 60 sec
Volume of water collected, V = 1.5 liters = 1500 cm:
*?!Jl!!a,o.at
yvt= 2O.679 kN/mr
Part 3: Critical hydraulic gradient:
2.773-1
. G"-1
rtr=
r(r=
1+e
1+ 0.6
i- = 1.108
-
coefficienr of permeabilirv. k =
Average Undrained Shear Strength of Clay:
Soft Clay c = 72 to 24 KPa
Medium Clay c = 24to 48KPa
Stiff Clay c = 48 to 96 KPa
Very Stiff Clay c = 196 to 792 KPa
Hard Clayc =92to 383 KPa
m36
Given: Initial
void ratio,
eo = 2
Thickness,H=12.5m
Final void ratio, e = r/z en = 7
Reduction of
thickness:
-e 12.52-1
1+eo=
L+2
LH = 4.167 n
AH = H
.
€o
I-
I
D'r,, IL
I h,
,1
'
coefricient of permeabiliry, k
-
M35
{
rI
Situation
B
=
1'ge]
t,
= 0.00784 cmls
( so I/
s'[60) f::
m43
Coefficient of permeability,k= 4 m/d.ay
Number of flow lines, Nr = 4
Number ofpressure droPs, Na = 10
Given:
Diameter of sample, D = 0.05 m
Axial load at failure, P = 225 N
Head,H=20m
unconfined compressive srrengrh, q, =
Part 1: Seepage per meter length oldam
Cohesion, cr= Yz eu= Yz(114.592) = 57.3 kPa
4
o=4.20\
'10
-
=kH\
'Nd
o
q= 32
m3
/day per meter
"
1000
24(60)
q= 22.22lit/min Per meter
PartZt Uplift pressure at heel.(point
"a", 1 potential drop)
Pressure head drop per equipotential line, Ah = H/Na = 20/10 = 2 m
Pressure head
at"a":
Uplift pressure,
ha = H
Pa = Yw
- Ah x
l
= 20
* = ##
- Z x 1' = 1'B m
ru Situation 9
Given:
Load,q=9kN/m
Footing width, B = 0.6 m
L
rz12
N=zl1+lll
L \z/ l
I
Ap
x hu = 9.81 x 18 = 176.58 kPa
=0.$7IN-
0'637q
-
,lt*U /4')'
Part 1: Bearing pressure at base of footing:
Part 3: Uplift pressure at the toe [point "b", 9 potential drops]
Pressure head at "b": hu = H
-
Ah x 9 = 20
-
2 x 9 ='2 m
'
o
B
9kN
/m
0.6m
Uplift pressure, pb = |w x ht = 9.81 x 2 = 19.62 kPa
Part 2: Stress at depth of z = 2xB = 7.2 m and r
o'63'9=
or=
' zlt+1r 1 z1')'
[a 42
Given:
o: = 26 kPa
R= Yzoa = 20 kPa
C=o:+R=46kPa
R20
sln0=
- =
'c46
b = 25.77'
.\n=j'6!Z(21
l.z[r + o]'
AP.=
oa = 40 kPa
-
= 0
4.7BkPa
Part 3: Stress at depth of z = 2 m and r = 3 m
0.637o
AP=...................-.''....-_
zll+(r /z)' )
Ap =
0.637(9)
------:--r ;
-zlt+12121')'
Ap = 0.271kPa
= 114.6kpa
lLlJ
Situation 10
Given:
Footing width,B = 4 m
Footing dePth, Dr = 1.2 m
Unit weight of soil, Y- = 20 kN/m3
Cohesion,c=10kPa
Angle ofinternal friction,0 = 20"
Bearing capacity factors for $ = 36" [from the given table]
' N. = 17.69; Nq= 7.44; Ny = 3.64
Terzaghi's equation for square footing:
q, = 1.3 c N. + y. DrNq + 0.4ym B NI
Part 1: Contribution ofcohesion strength:
1.3 c Nc = 1..3(1.0){1.7.69) =
-
229.97 kN
Part 2: Contribution ofsoil overburden:
Y- Dr Nq = 20{1'2)U 'aal = 178'56 kw
Part 3: Contribution of footing dimension:
0.4
y-
B Nr =
0.a[20][a)G.64) = 116.48 kN
LE 50
Given: Pilecross-section= ax a = 0.3 m x 0.3 m
Penetration length, L = 15 m
Unconfined compression strength, q, = 110 kPa
Adhesionfactor,o=1.0
Pile Perimeter, P = 4a = 7'2 m
Unconfined shear strength, cu = Yz q, = 55 kPa
pL
= 1.0(55)(1.2)[15)
Friction caPacitY, Qr = = 990 kN
Friction capacity,
Friction capacity,
Qr = cr cu
Qr
$iffi
ii""ii;'#ii;
liiii4:r"
+tlf
iis{t
1i
',,i':.' :.:'i I L".'::i ';'..
"
i' ; ,u. ii i
j',;
i,';,;.,,
:
' ;. i ;i
l',
,
:
,;
I
,;
i
,
;i,
::i
.,:::',,
:
.;
:i ;:i, i,,: ,i ; ;r
:
,:'i
,,
I
:,:;
.l;';
;'
,',;
i::;
;.,:
;
;' ii';' ;":i' ' ';';.'"
'.
,;'; .',;;;
,.'l:1.",: l;:: I;. ;::.,
,,,' ; ,:'.
I .:.:;
,.;
;;,:': ::,:'::
!:;,:;,:i;''i'
;
,
';
.!!i,
: ,'"
i., ,,i:''
:,'.'..,";,,';".'',,,;;, f' ;;..;,"i'',
;i:';.;
.: . ;;.' ;*,aruuq$.ig
*.A
i;' :,,
.":.
-
".
i.. '
':!
:
i::
..
raiiFB,tln
i:i'
':''.'i;;,i
:.::;;
:.;,
:;;
i',;,u,;;.i,:,;i";,:;,:i;,$i:"*:r41
:' .:, :; ;:;.';;", : ;i;.:.' ...,,; ;,,i,
.: ,
':;.
l:::' , ;; i : :,;;,";":
. ,, ,i,''
;;:
,:: .i.iir;;
,, :. l:; ,,i'**,.,i-m*;
. .i. i.,i;.i: :;,'i!ir;:,, t2.l
,;', .'i.'.u.', ;""'",:,'.;s:"'?n";
:ilfidi&
r'",':
i'.:.
"'i:i
r,
.
,, 3S:
viia
;;i',':::r;i.'!.;',;*ri*tJrit',
:.'' ''::,: iii"ii't:'
:: '': .::."' : , li,il
';
.: ,
'
:
'il::,zii;
: ;;
i'r+r
.' .;t:;. ;91,;31
," ::':;.';,
l:
;.i ii i:',
.;
: : ..,
'
*g:
.:i..ii sau:,cAf,ira,ie
i:, ';' ,,''r'irti6 viialt.i
;;,.1
.,.' .i:
.
.:8..':
: ":,$imluo" ili.l'i
'
""
, .;'::' "; ;ieitti'rid,
'"
,i.
i"
a
- zugr"iF'I
f.i.'::i.i:,:'i
'.,,
, ....;', :..,;"
,:
.,:
i.'
i ifrFzu,qerl
'
;; ::. ::". .. 37- Gakulbte
i;' ,,' ;,,
i'"irsqP
;i"
,,. i'
;
' i' ,il ;':
i" .'ig"qi:"'.".4."0:ll
'c$!cu{;fd
,",'',.i,
:i "'";.'':
i;,
-'::,,,:;
,,,'.;' :,
';"';,'
:;,;;;,;.;
:'. ." .:,r;
-:'
",'f,,i*;
.'
:;
:'
,;:
,,h,:'ji*
;:;;1iii;;
;,,1:;
''';':
':tr'"" "'
:,t ,: ,
;:'.,i
;;,
ffi
*
Solutions to May z}l6Examination
Situation 3
Given:
a=L.2m, b=0.3m, c=2.4m
W=2kN
xMg=0
m
Ar,(c)=w(a/z)
Au(2'4) = 2(1'2/2)
Ar, = 0.5
IFr,=0
a
iA.
kN
Br,=Ar,=0,5kN
Bu=A,=%W=1kN
R,=
vE;,+B-
nr= Jo.s,
Rs
tan 0g =
B
___-1L
#
= 1.12 kN
tan0s=
Bh
0s =
1
0.5
63.435'
M Situation 4
L=60m
w = 2.8 kN/m
T."" = 1100 kN
Wt=wL/Z
wr = 2.8(60)/2
Wr=84kN
i
i
Part 1:
IMa=0
Given:sag,y=2m
B"=2W/3
tan0= Y - 2
Ll4 60/4
stn
U
W-
= ---l
T
0=
XF,=0
7.595'
-rT
tan0= I
L/4
A,=W/3
In Figure (BJ, the required vertical internal force to balance the external force
^84
is upward. Cable DE cannot support this force because it will induce
sin7.595'
T = 635.55 kN
compressive
force. Thus, only cable
CF
will
act.
Parts 1 & 2:
Parts 2 & 3:
Given: T=Tmax=1100kN
srnH=
Bv[3s]=W[zsJ
srnU=
Given:
Fco = 8.9 kN
In Figure
(A):
=
0
Fco
[h) = (2W /3)(s)
B.e(41 =
ffO,;0=4.38"
(2w/3)(4)
W = 13.35 kN
lFu = 0
y = (60/4) tan 4.38o
Y=
IMr
B4
Fcr sin 0 + (2W /3) =W
Fcr sin 45o =W /3 = 13.35/3
Fcp =
1.145 m
6.29 kN
Part 3:
Civen: W=20kN
H
= Jl,Loo, -84,
H = 1,096.788
kN
A, =W /3 = 6.667 kN
At joint A:
XFu=0
@ Situation 5
Facsin0=W3
Fec
sin 45" = 6.667
Fec =
Figure [A)
Given: s=4m,h=4m,
tan0=h/s =4/4;0=45.
Figure IB)
9.428 kN
fl
A','
Y r F^"
l*o
m Situation
7
Design
-.- Situation 6
Support reactions and angles (solution not shown anymore):
Bv=3'3kN
Ev = 2.7 kN
Er+
0=45o
cr
tributary width, s = 6 m
Wind pressure,P = 1.44kPa
= 53.13o
F,,
Wind pressure
= 1.5 kN
coefficients:
P,
,Cr = 0.8
Cz = -0.1
= 1.5 kN
Frr€
H^
C: = -0.5
C+ = -0.4 H,
f.,
D
.''
F,
C,IY
,]
C,
+
'Bn
A,
2.25
Eu
I
b
P,=3kN
P.=3kN
= 1.5 kN
E, = 2.7 kN
At joint B:
IFv=0
Ft=pxCtxHrxs
Fzr=pxCzxHzxs
Fzr= p x
XMs=
Fsr=3.3kN
Cz
x Lr x
0
(Ft+Ft)(Ht/2) + [F:rr- Fzr,)[Hr +Hz/Z)
Lt/2) + Fz'(Lz/2) - A,[Lr + Lz') = 0
(27.648 + r3.Bza)(2) + (8.64 - 1'728)(4 + 1)
+ 5.184[6 + 3) + 25.92(3] - A"(6 + 6) = 0
memberJ.
N=2O.16 kN downward
Fcrsin0=3.3
Fcr = 4.667
kN
Considering section b-b:
The required force to be supported by the cables is a downward force of
(3.3 - 3 = 0.3 kN), hence diagonal CH cannot support this load.
IFv=0
XFu=0
Au+Br=Fzv+F:v
20.1,6 + B, = 5.184 + 25.92
1.0.944 kN downward
B'=
Fco sin cr = 3.3 - 3
Fco = 0.375 kN
Refer to Figure next page:
IMo.igLr = 0
B,(Lz) + Br,(Hr+ Hz)-Fz,(Lz/Z)-F:.;,(Hz/2)- Fa(Hz +Ht/2)=O
10.e 44(6) + Eir(4 + 2) - 25.92(3) - 8.64[1J - 1.3.824 (4) = 0
Fcosincr+FrH=3
XFr, = 0
Br= 12.672kN
At joint D:
XFv=0
Fr = 1.44[0.8)t4X6) = 27.648 kN
Fzn= 1.44(0.71(2)t6l = 1.728 kN
Fz"= 1.44(0.1)(6)t6) = 5.184 kN
F:r, = 1.44(0.5)(2)(6) = 8.64 kN
F:"= 1.44[0.5)(6]t6) = 25.92 kN
F+ = 1.44(0.4)t4)t6l = 13.824 kN
+ Fz,(Lz +
Considering section a-a:
To resist the vertical force Bv = 3.3 kN, a downward force is needed. Thus,
diagonal BG cannot support the Ioad (this will induce compression to this
XFv=0
s
F:n=pxC:xHzxs
F:u=pxC:xLzxs
F+=pxCexHrxs
Fns = 2'7 kN
95+Al-P1 +F++F:r,-Fzn
L2.672 + h,= 27.648 + 13.824 + 8.64
Ar, =
35.712 kN
F,
- 1.728
Part 3:
D
F",
D,
r
I
IC,
!r-+-
A'
lo
&"+
DI
D-
+-o
I
rlv -
IF,
C,
"'*F.,,
I
'BD
-A
a - L
, ..
nv^
Y
D.
C
F
"
76,368
lv = _
628.31,9
C,
ftzol,
-"rrm\a
\N
tr\,,
= 628.32mmz
= 121.5MPa
F,
+-
Bl
m Situation
9
Given:
Vessel diameter, D = 500 mm
m Situation
Parts 1 & 2:
B
Considering post AC:
IMe=0
Fgn
Internal pressure, p = 4000 kPa = 4 MPa
Allowable tensile stress of plate, Ft = 138 Mpa
C;r--"*
il, t'= 18 kN
sin 45" (11 = 1B[3]
Part 1: Cylindrical vessel
Fen = 76.368 kN
Since member BD is a
Ro = Feo =
two-force member,
76.368 kN
ft= PD <Ft
,t2t
L=
oD
2F,
.
_
4(s001
2[138)
t = 7.25 mm
Part 2: Spherical vessel
,oD
It= L
Part2z
fsp
4t
F,
= Auo
44
4r5001
4{738)
Aso= 1es2_632)
t=3.62mm
4
Aen =
nD
SF,
1300.619 mmZ
fuo= 76'368 =5B.7ZMPa
1300.619
Part 3:
t=
72 mm; Ft = 120 MPa
-oD <Ft
tt-'
2t
2Ft
n=
,D '
^ _ 2(120)(12)
Psoo
P=
5.76 MPa
m Situation
m situation
10
11
0,
,C
@
Axial tension, F
jtn:
10.5
12
Load diagram (kN)
oc
-*@"w-:
oc
Stress due Stress due
to internal
tensile
pressure force
105
0.5
Os
Resultant
stress as
presented in
the Mohr Circle
Part 1: Stress due to
p:
oD
oiB= L
2t
1o
P=
-
p(3oo)
2{3)
0.2 MPa
Part2l. Stress due to
oc=
Moment diagram &N'm)
F:
F
A,,
110*
@ Situation 12
F
'r(300J(31
F=311kN
o-;. = 110
Part 3: Maximum principal stress = 110 MPa
Part
l"
Part
2
wr = 34 kN/m
L=Bm
wz = 136 kN/m
EI = 400,000 kN-m2
Part 3
Part 1:
6[].3.286.106)
.to -
Me = wr(L) (L/2) + 1/z(wz - wr)(Ll(L/31
M,q = 3a[B)(B/2) * y,(t36 - 34)(B](B/31
Ma = 2176 kN-m
Tro*y
fu= 14.17
MPa
)
Part
)
Part 2
1
Part2:
Vbu," = Fu = 16.608 kN
p=1w,1+11w2-w,JL
810
p=
1 1:+11e;.
fr
trru - 34)tB)
= 183.6 kN
_3V
fv= _
c
2bt
_
3[16,608J
2(1000)(7s)
f, = 0.332 MPa
Ma
= 2176 - 183.6[8) = 707.2 kN-m
LU Situation 14
Given:
Part 3:
6t = 6 due to w1 + 5 due
(wr-
w, Ln
to (wz - wrJ -
wr)f
6 due
pL'
" _ BEI , 3OEI 3EI
(34)t8r + [136-34J[B)4
0.035
=B[400,000J 30[400,000)
P=
to p
= 248 MPa
F, = 400 MPa
Fv
PIBf
-
3(400,000)
101.57 kN
m Situation
13
Given:
H=2.4m
y" = 77.3
K^=
t/3
Plate and angle thickness, t = B mm
bolt shear, Fuu = 68 Mpa
bearing stress, Fp = 1.2Fu = 480 Mpa
tensile stress on nbt area of angle, Ft = 0.5F, = 200 Mpa
shearing stress on net area angle, Fu = 0.3Fu = 120 Mpa
Fr = 10.4 MPa
0.8 MPa
t=75mm
Part 1:
=
a/2
l(uy'
!12
Mbase=FaxH/3
F
= Yz(l / 3) (77 .s) (2.4)2
^
F" = 16.608 kN
Mu."" = 16.608(2.4/3)
Mu".e = 13.286
P based
on bolt shear (double shearJ
P=F,uxA,
Consider 1 meter width of wall (b = 1 mJ
2
Block shear ofangle
Allowable
Allowable
Allowable
Allowable
Fu =
F
I
Bolt diameter, dr = 16 mm
Hole diameter, dr, = 18 mm
kN/m:
kN-m
P=68x[!1e1rx3lx2
P=
Part2:
P based on
P=FpxAp
82.03 kN
bearing stress (bearing on gusset]:
P=Fpx(drt)x3
P=480x[16[8)]x3
P=
184.32 kN
Part 3:
based on block shear:
P
Resultant load, R = 750 N/mm
P=FtxAL+FvxA,
Weld
capacity:
Tension: At = [s4 - 0.5dr,J t x 2
a, = [40 - 0.5[18]lB x 2 = 496 mmz
Shear: 4,= (sr +s2 +s3- 2.5dhl tx 2
4.= [50 + 35 + 50 - 2.5[1BJ] B x 2 =
R = Fu* x 0.707
750 = 93 x 0.707
t
t = 11.41 mm
1440 mm2
Ja Situation 16
'Given:Lr=9m
P=200x496+120xL440
P =272kN
t
L,=3m
'L,=9m
w = 12 kN/m
Lz=3m
wo = 7 kN/m
wr = 5 kN/m
m situation 15
Part 1:
Given:
a=200mm
b=500mm
Fu =
93
MPa
P=360kN
W=WD+WL
w = i.2 kN/m
r-1-
L
Part 1: Directload (due to
-Lt+Lz
L= 72m
PJ
D
R,=
-L,I
Lr=2b= 1000mm
- _- 360,000
^'
looo
R, = geO
N/mm
IMa=0
w
PrY,
Resultant load of weld:
M
nx_
M.r*c
fr, =
M=Pa=360[0.2)=72kN-m
- ^*
Rt
72x106
93,33333
r----;-:
= v/R,'+Ry"
soo'z
J
fr, =
=83,333.33mm2
= 864
L
I
I
N/mm
p,
kN-m
^tb-__
.I
-S-
s-= 4 -
slz/2
Rs=96kN
Mmu, = 96
Part2:
=
Rn(e) = 72(12)2/2
= Jlso+]'+(360)'z
Rt =
936 N/mm
96x706[350
1.6x 108
105 MPa
given: d = 350 mm, I, = 1.6 x 108 mm4
/2)
LU
12
kN/m
Situation 17
Given:
br= 193 mm
tr=19mm
d=465mm
I" = 445 x 10e mma
248 MPa
Fv =
rt=50mm
ma=9T kg/m
pr = 3.6
kpa
pa. = 5 kPa
t_
2m
2m
2m
Part 1: Maximum bending stress in beam BF:
Dead
load:
wD = superimposed dead load + weight of beam
wo = 5(2)
Live
Part 2:
Load pattern, dead load on all span & live load on alternate span.
XMa=0
Rn[eJ =
t2(e)(4.s) + 7[3J[e + 1.5)
Rs = 78.5 kN
From the moment diagram, M-a" = 106.25 kN-m
^
M,*.
I..
*
_
,'o _- 106.25rrc6350/2)
16 t, 1oe
,o
=
116.27
load:
w.
= 3.6(2) = 7.2
-V
dt*
-
.lv =
60,000
3s0(B)
f"=2l.43MPa
-
kN/m
Maximum bending moment [assuming simply supported)
*L
18.151(10),
M-r" =
o
M
",_
o
o
M."" = 226.89 kN-m
.Mc
tb =
th
- I-
_
226.89x,106{465
445xL06
fu =
Maximum shear, V-,, = 60 kN (from Part 1)
= 10 es1 kN/m
Total uniform load, w = 10.951 + 7.2 = 18.L5lkN/m
Part 3:
lv =
. '|t::A
1000
/2)
.
118.54 MPa
Part 2: The maximum allowable bending stress is 0.66Fy. This can be utilized
if the braced length Lr S Lc, where L. is the smaller value of:
-| zOOb,
, 2OO(1s3) =2,4Slmmr'
J{,
a4B
1.37.900
^ 137,900
=
(d / At)Fy
ffiqQ+q
Thus, L. = 2,45L mm = 2.45
= 4,385 mm
m
[choose 2 mJ
Part 3:
Lr= 1/zL= 5 m > L.
j
Cr =
To speed-up the solution, we will first compute the bending stress caused by
a unit normal and unit tangential loads. Tdke note that all loads pass through
the centroid ofthe section.
1, rt= 50
li:
LRT =
Lrlrt
=
5000/50 = 100
Due to
LRT1 =
li
Folztoq
!F,\248
Fonzool
= ^t-
unit normal load, wn =
= JJ.aJ
;
ll
3,516,330(1).
LRT2 =
'tl
248
1
kN/m'
*nL
M,=
99 =l(6)'=4kN-m
4^106.
r*= & =81.967Mpa
S, 4.BB t 10"
= 179.075
rl
Due to
unit tangential load, wt =
LRT2>LRT>LRT1
fi
]i
ij
)li
Slope: 0 = arctan (1/3) =18.435'
Stress due to beam weight:
wa = 7'1. N/m = 0,071 kN/m
i1
lr
ii
Fr,
= larger value of Fur & Fr: = 130,5 MPa
Normal:
m Situation
1,8
= 0.067357 kN/m
fa"= 81.967(0.067357)
fr,' = 5.521 MPa
the moment diagrams for
simple beam with and
Tangential:
Fig. A: Moment diagram without,sag rod
In this problem, the
swLzlstz
I
I
i
moment is wL2/9 as
shown in Fig. A, and for
bending about the y-axis
the maximum moment
Stress due to dead and Iive loads:
wd*r = [1.2 + 0.576)s = 1,.776 s
Normal:
atL/3.
u2
u2
Fig. B: Moment diagram with sag rod atIJZ
2wLzt225
wi,'lloo
s is
the purlin spacing
wn = wd+l cos 0 = 1.685s kN/m
f*= 81.967(L 685) = 138.103 s MPa
Tangential: wt = wd+r
fuv =
sin 0 = 0.562
:3.9,t,, '562) = 19'038
s MPa
Part 1: SpacingduetoD+Lonly:
fo,
L/3
[kN/m) where
2wL2lzzi
,r*1_z/90.
U3
wr = wb sin9 = 0.022452
fr,y= 33.898(0.022452) = 0.761MPa
L
purlin is provided with
two lines of sag rods
For bending on the
x-axis, the maximum
wn = wb coS 0
w,
wL'/8
The figure to the right shows
without sag rods.
kN/m
=o.4kN-m
90 =ry
90
M
fi411o6
fby=i=ffi=33.BeBMPa
Mr=
lt FfLRTl'z'1
Fnr=l:- Y' i l=107.04MPa
L3 10.55xi0"C0.]
B2,74OCb _ 82,74.0(l)
D.
r'
^_
rb3=
Lbd - 5000(465) = 130.5 Mpa
193(1,
b, t,
ir
lli
1
*='-t'
U3
Fig. C: Moment diagram with sag rods at L/3
F,
f,
+ oY -1
Fo,
5.521
- 138.103s
207
s= 1.277'm
0.761 + 19.038s
,
207
(Choose 1.2 m)
Part 2: Spacing due to D * L * Wwindwardi
ww= p x cw X S = 1.44 x 0.2x s = 0.288
fw=8\.967(0.2BBs) = 23.306 s
f,o"
1
+
Fo,
f.
"t =1
Live load, pr = 4 kPa
Unit weight of concrete, y, = 23.5 kN/ms
+F.,
5.521+ 138.103s + 23.306s
i{207)
s=1.49m
.
Superimposed dead load, pa' = 2.6 kPa
Lr=Lz=L=6m
s=3m
b=0.3m
h=.42m
t=0.1 m
s
(Choose 1.4 m)
w = p x crw X S = 1.4a x (0.6) x 0.75
ft' = 81.967[-0.648) = -53.115 Mpa
fu =
Wd=Wc+pdsxS
Uniform dead load:
Wteeward
y7.=y.[st+b(h-tJl
w. = 23.5[3(0.1) + 0.3[0.a2 - 0.1)]
w. = 9.306 kNlm
+(207)
Part 3:. Total flexural stress due to D + L *
Total stress,
Weight of concrete:
Wc=|cVc
0.761+ 19.038s
+=-=-'--1
wa=9.306+2.6x3
with s = 0.75 m
wa = 17.106
fr, due to beam wbight, dead + live, and wind
+ frv due to beam weight and dead + live
1
Part 3:
= [ 5.521 + [138.103 x 0.75J - 53.1 15 ]
+ [0.76L + [19.038 x 0.75)l
wu = 1.4 wa + 1.7wr
w" = 1.4(17 -106) + 1.7(72)
w,
fa=7l.O23MPa
Concentrated load at E:
.I
Part
w-prxs
w=4x3=12kN/m )PartZ
Uniform live Ioad:
Factored uniform load:
fb
kN/m )
= _0.648
= 44.348
kN/m
Rr-wuxL
Rn=44.348x6=266'09kN
Situation 19
q-;rFl
illr,
L:_-J]
lLUl
Situation 20
w (kN/m)
Two spans loaded
)
3wU8
b
5wL/4
Moment diagram
wL2l8
3wU8
Part 3: Maximum positive moment in
DE
For maximum
One span loaded
)
wl,i16
wr, (kN/m)
dead load on both spans
and. live load on one span.
wau = 24
kN/m
Rr
wr, = 12.2 kN/m
Moment diagram
wL2l16
Rr=
lwa,L* '
816
pa' = 3.2 kPa
pr = 3.6 kPa
p, =
:? [za)[7.s)
Given:
Lt=Lz=L=7.5m
S=2.5m
*
Beam,bxh=400mmx600mm
Slab thickness,
w,, (kN/m)
positive moment in DE,
the load arrangement is
t = 100 mm
3wo.,L/8
+ 7w,,L,/16
w,,L
7
|let gz.z11t.s1
Rr = 107.531 kN
Dead loads:
SuperimPosed =3.2x2.5 = BkN/m
weigtrt of concrete = 2412.5(0-7) + 0.4[0.6
-
0.1J] = 10'B kN/m
x=
Total dead load, wo = 1B.B kN/m
wdu
=3.6 x2.5
=9
,
-
kN/m
107.531
+wru
Mpo.=
,2 Rr
Live load:
wL
R
x-
24+L2.2
=2.9/m
(wu, t-w,,)x'
Mpo' = 159.71
=
rc7.53112.97)
-
(24 + 12.2)(2.97)2
kN-m
Factored load, w, = 1.4wo + L.7wL= 1'4(18.8) + 7J[9) = 41'62 kN/m
Part 1: Maximum moment at
E:
For maximum negative moment [at EJ, the load arrangement is deadload
and live load on both sPans.
wt, ,2
Mr-r*= a
m Situation 21
Given;
4t.52fi.5)2
lvlu max -
B
Mumax =
o
Beam dimension, b x h = 300 mm x 450 mm
Effective depth, d = 380 mm
f. = 30 MPa; fy = 415 MPa
Unit weight of concrete, y, = 24 kN /mz
Part 1:
292.64kN-m
Simplespan,L=5m
Superimposed dead load, wa. = 16 kN/m
at interior support "E"'
For maximum reaction at interior support, the load arrangement is dead
load and live load on both sPans.
Part 2: Maximum reaction
Rr-r*=
l5wL
^
t(E m:x =
B
RE
Live load, wr = 14 kN/m
Weight of beam, w5 = yc b h = 24[0.3)[0,45J = 3.24
Total dead load, wa = \6 + 3.24 = 19.24kN/m
s(41.62)(7.s)
Factored
B
ma = 390.19
kN
load:
p*r-
w, = 1.4wa +'l,.7wr= LaO,g.2a) + L7$a)
w. = 50.736 kN/m
Maximum factored moment:
lvlu =
w,,
L2
50.736[5)'z
BB
t
ul
M" = 158.55
kN-m
E
lrr
o.i.=4=0.00337
'f
IJ
Pl
Part
2:
r
/
v
6,=pbd
Design (fadtoredJ moment, M, = 200 kN-m
Assuming "singly-reinforced"
M,=0Rnbdz
A' = 0.00399[300][380)
A' = 455 mm2
Number of 16-mm bars:
:
200 x 100 = 0.90
Rn
p= A =
(300)(380),
A,"
R" = 5.13 MPa
=Li
!-??,,
X
(
l6)'
=z.zo
(choose3bars)
Situation 22
Given:
s=2.6m
L=5.3m
b=250mm
h=400mm
slab,t=100mm
ob
-
pmax
o'B5f '' Br 60o
= o.o3oB7
f,(600 t f,)
pmax,
d'= 75 mm
Superimposed dead load, pa. = 2.6 kPa
Live load, pr = 3.6 kPa
Parts 1 & 2:
= 0.75 pr = 0.023
Since p <
,
y,= 23.6 kN/m:
Ln=L-b=5.3-0.25
[he beam may be singly-reinforced.
Ln=5.05m
B
WD=Wslab+Wds
4.=pbd
A' = 0.01394[300][380)
A. = 1589 mmz
(choose 8 bars)
=7.9
Part 3:
1.2wo +
l.6wr
w"= 1.2(12.896)
w,
=
M"
-
+ L.619.36)
30.451kN/m
Maximum span moment:
., Factored concentrated load at midspan, P, = 50 kN
rvruR,
wo = 23.6(2.6J [0.1) + 2.6(2.6) = 12.8e6 kN/m
wL= pr x s = 3.6{2.6)= 9.36 kN/m
Number of 16-mm bars:
--N= As = 1589
A'u iG6)'
WD=YcSt+pdsxS
=
P,
L
sots)
- 44
Mu.
obd'
-
1\4u
w,
=
L"''
BB
M, = 97.07 kN-m
=62.5kN-m
62 5x 106
0.e[300)(380)'
-
= 1.603 MPa
30.451(5.05f
Maximum shear at face of support:
,,vu _- w, Ln
22
-
.,vu _- 30.451(5.05)
V, = 76.889 kN
-
b*=b=250mm
d=h-d'=400-75=325mm
M*=T{d-
23
A,r ft (d - a/2)
= 2552(41s] (40s
M.r = 349.23 kN-m
a/2)
MnL =
1{n=[{n1 +Mn2
b=b-=400mm
h=500mm
4BB.B9
- 1,50.s /2)
=349.23 +M"z
M"z = 139.66 kN-m
fyn= 275 MPa
415 MPa
fv =
M"z=Tz[d-d')
=20.7 MPa
Mn2 = As2 fy (d
- d')
739.66 x 106 = A.z[415)(405
A'z = 990 mt42
Part 1:
0.021
M" = 440 kN-m
pu =
d=500-95=405mm
0 = 0'eo
p.,*
= 0.0158
Assuming the beam to be singly-reinforced:
M,=0R"bd2
= 150.5 mm
M*
Given:
,
0.85f. ab=A'rfy
a
V. = 62.563 kN
f
= 0.0158[400)[405J = 2552 mm2
C.=Tr
v,= 0.77(250)(325)
V.=Frcb-d
m situation
.""
Asr = As
Part 3:
- 65)
A'= 2552 + 990
A. = 3542 mmz
As=Asr+Asz
Part2t
440 x lle = 0.90 R, (400)(4051'z
Rn = 7.451 MPa
V, = 280 kN
0=0.85
F". = 0.76 MPa
b* = 400 mm
Av=2x i12)'z=226.2mm2
V" = V,/0 = 280/0.85 = 329.412 kN
=
0.0258 > 0.75pu
Vc = Fu.
b* d = 0.76{4001(405) = 123.12 kN
V. = Vn
-
A;f*d
s=
Thus, the beam shall be doubly-reinforced.
V, = 329.41.2
- 1.23.12 = 206.292kN
226.2(27s)(405)
206,292
v.
s=
l22rnm
b=400
40
Part 3:
T, = 180 kN-m
e, =
t
i
i
i
Mo
Mor
A-za
&
00
e
<t
II
>.
0=45o
y = 308[a0B) = 125,664
A" = 0.85 Aor, = 106,814 mmz
Aoh = x
pn =
M. = M"/0 = 4a0/0.90 = 4BB.B9 kN-m
[!']."t,
s [t/
p,
46
2(x + Y) = 1432 mm
b=400
Tn=! -
o
Tn
=
1Bo
o.Bs
E2AAf,
ss
cot
0
21,1.77 x lQo
=
2(106'8L4)A'(275)
Maximum eccentricity about the strong axis is e = 10%h 50 mm
=
Maximum design moment, Mu = Pu x € = 3,272 x 0.05 = 163.6 kN-m
cot 45o
Nominal moment, Mn = Mu/d = 163.6/0.65 = 251.7 kN-m
&
o' =
Part 3: Column dimension, b x h = 400 mm x 500 mm
=211.77kN-m
t
*s []1."u
(t"'
= 3.605 mm
Ar = 3.605[1,4
o,
34#
J
coo 45o
Situation 25
Tp prevent uplift, the soil pressure
isasshown. i.e.esB/6.
ot=3,4zommz
Part 1:
M = 126 kN-m
m Situation 24
Given:
P=2B0kN
kN
PL = 845 kN
,=
27.5 MPa
Po = 1600
f
0 = 0.65
U=1.2D+1.6L
fv =
e
415 MPa
=M/P
= 0.45
m
B=6e=2.7m
T
tl
L
P, = 1.2[1600] + 1.6[845] = 3,272kN
Part
1:
ratio, pc= 3o/o = 0.03; Ast = 0.03Ac
Column width, b = 350 mm
Steel
P, = $ 0.80 [0.85 f. (Ag - A.t] + fy A.tl
3,27 2 x 103 = 0.65 (0.801 [0.85 [27.5] (As
As= tf),!Ql rnmz
Ae=b"h
0.03AsJ + 415 (0.03AgJl
l7g,I47=350<h
h = 511.8
Part?t
-
mm
choose 500 mm
Part2:
RM = 440 kN-m
OM = 260 kN-m
R=P=265kN-m
Number ofvertical bars, N = 16
RM_OM
Ae=bxh=400mmx500mm
440
RM
R
.
Ag =
200,000 mmz
x=
A.r = 0.03Ac = 0.03[200,000]
A,t = 6000 mm2
A*=NxAr,
21,85
265
0.679 m
B=3x=2.04m
6,000 = 16x
dr, =
-260
Id*
mm
choose 25 mm
):^,
el8/6
Notes
Part 3:
B=3m
RM = 500 kN-m
OM = 265 kN-m
R=P=335kN
8/6
= 0.5 m
RM_OM
R
500
- 265
--/
JJJ
x = 0.701 m
e
=B/2
-x
e = 0.799
2R
0=;
m> 8/6
'
2[33s]
3[0.701]
q = 318.4 kPa
Notes