lOMoARcPSD|20985858
All-CE-Formulas - Board Exam CE Formulas
BS in Civil Engineering (University of Science and Technology of Southern Philippines)
Studocu is not sponsored or endorsed by any college or university
Downloaded by Ace Juntilla (ace.juntilla@gmail.com)
lOMoARcPSD|20985858
Quadratic Equation
Progression
Trigonometric Identities
Spherical Trigonometry
Form:
2
AM ∙ HM = (GM)2
Squared Identities:
2
2
Sine Law:
Ax + Bx + C = 0
Arithmetic Progression:
Roots:
s 2 2 4AC
2B ± √B
x=
2A
Sum of Roots:
B
x1 + x2 = 2
A
x1 ∙ x2 = +
d = a 2 2 a1 = a 3 2 a 2
a n = a1 + (n 2 1)d
a n = a x + (n 2 x)d
n
Sn = (a1 + a n )
2
C
A
Form:
(x + y)n
rth term:
r th = nCm x n−m y m
where: m=r-1
Worded Problems Tips
 Motion Problems
→a
→s
=0
= vt
Case 1: Unequal rate
rate =
work
time
Case 2: Equal rate
→ usually in project management
→ express given to man-days or man-hours
 Clock Problems
θ=
11M 2 60H
2
Ex-circleIn-circle
+ if M is ahead of H
- if M is behind of H
1 1 1 1
= + +
�㕟 �㕟1 �㕟2 �㕟3
Centers of Triangle
INCENTER
- the center of the inscribed circle (incircle)
of the triangle & the point of intersection of
the angle bisectors of the triangle.
Common Quadrilateral
1
A = bh
2
1
A = ab sin C
2
Square:
CIRCUMCENTER
- the center of the circumscribing circle
(circumcircle) & the point of intersection of
the perpendicular bisectors of the triangle.
Rhombus:
A = ah
A = a2 sin θ
1
A = d1 d2
2
Trapezoid
1
A = (a + b)h
2
Ellipse
A1 n
ma2 + nb 2
=
;w = :
A2 m
m+n
Circumscribing Circle:
Inscribed Circle:
abc
AT =
4R
AT = rs
opposite side
sine of angle
b
c
a
=
=
sin A sin B sin C
Pappus Theorem
Escribed Circle:
AT = R a (s 2 a)
AT = R b (s 2 b)
AT = R c (s 2 c)
NOTE: It is also used to locate centroid of an area.
V = AB Have
LA = PB Have
Frustum of Cone or Pyramid:
CENTROID
- the point of intersection of the medians of
the triangle.
V=
H
(A + A2 + √A1 A2 )
3 1
Prismatoid:
EULER LINE
- the line that would pass through the
orthocenter, circumcenter, and centroid of
the triangle.
H
V = (A1 + 4AM + A2 )
6
Area = n ∙ ATRIANGLE
Deflection Angle, δ:
´ = 180° 2 ³
1
Area = n ∙ R2 sinβ
2
1
Area = n ∙ ah
2
Central Angle, β:
β=
360°
n
Polygon Names
16 - hexadecagon
17 - septadecagon
18 - octadecagon
19 - nonadecagon
20 - icosagon
21 - unicosagon
22 - do-icosagon
30 - tricontagon
31 - untricontagon
40 - tetradecagon
50 - quincontagon
60 - hexacontagon
100 - hectogon
1,000 - chilliagon
10,000 - myriagon
1,000,000 - megagon
∞ - aperio (circle)
3 - triangle
4 - quad/tetragon
5 - pentagon
6 - hexagon/sexagon
7 - septagon/heptagon
8 - octagon
9 - nonagon
10 - decagon
11 - undecagon/
monodecagon
12 - dodecagon/
bidecagon
13 - tridecagon
14 - quadridecagon
15 - quindecagon/
pentadecagon
A = √(s 2 a)(s 2 b)(s 2 c)(s 2 d)
Ptolemy’s Theorem is applicable:
ac + bd = d1 d2
s=
Non-cyclic Quadrilateral:
AB/PB → Perimeter or Area of base
H → Height & L → slant height
AX/PX → Perimeter or Area of crosssection perpendicular to slant height
LA = πrL
LA = PB L
Alune = 2θR2
Spheroid:
Spherical Zone:
4
V = πabc
3
a2 + b2 + c 2
]
LA = 4π [
3
4
1
3
2
V = πR2 h
3
Spherical Segment:
For one base:
4
1
V = πh2 (3R 2 h)
3
For two bases:
about minor axis
Downloaded by Ace Juntilla (ace.juntilla@gmail.com)
1
V = πh(3a2 + 3b2 + h2 )
6
2
Vwedge = θR3
V = Azone R
about major axis
2
3
Vwedge 3 πR
=
θrad
2π
Spherical Sector:
4
1
Spherical Wedge:
Azone = 2πRh
Prolate Spheroid:
V = πaab
3
a2 + a2 + b2
]
LA = 4π [
3
Reg. Pyramid
Alune 4πR2
=
θrad
2π
4
V = πR3
3
LA = 4πR2
1
V = AB H
3
µ
2
Right Circ. Cone
Spherical Lune:
Sphere:
Oblate Spheroid:
Pointed Solid
V = AB H = AX L
LA = PB H = Px L
Spherical Solids
V = πabb
3
a2 + b2 + b2
]
LA = 4π [
3
a+b+c+d
2
A = :(s 2 a)(s 2 b)(s 2 c)(s 2 d) 2 abcd cos 2
v
Pappus Theorem 2:
Truncated Prism or Cylinder:
(n 2 2)180°
n
Prism or Cylinder
SA = L ∙ 2πR
Special Solids
³=
Cyclic Quadrilateral: (sum of opposite angles=180°)
Pappus Theorem 1:
V = A ∙ 2πR
ORTHOCENTER
- the point of intersection of the altitudes of
the triangle.
1 knot =
1 nautical mile
per hour
General Quadrilateral
Triangle-Circle Relationship
diameter =
a2 + b2
2
A = πab C = 2π:
1 statute mile =
5280 feet
Interior Angle, ɤ:
Parallelogram:
a+b+c
2
1 nautical mile =
6080 feet
n-sided Polygon
Rectangle:
A = bh
A = ab sin θ
1
A = d1 d2 sin θ
2
1 minute of arc =
1 nautical mile
1
πR3 E
V = AB H =
3
540°
A = bh
A = s2
P = 2a + 2b
P = 4s
d = √2s d = √b 2 + h2
A = √s(s 2 a)(s 2 b)(s 2 c)
d=
180°
Spherical Pyramid:
2
Triangle
s=
Spherical Polygon:
πR2 E E = spherical excess
AB =
E = (A+B+C+D…) – (n-2)180°
sin 2A = 2 sin A cos A
cos 2A = cos 2 A 2 sin2 A
cos 2A = 2 cos 2 A 2 1
cos 2A = 1 2 2 sin2 A
2 tan A
# of diagonals:
tan 2A =
n
1 2 tan2 A
d = (n 2 3)
1 sin B sin C
A = a2
2
sin A
 Work Problems
Cosine Law for angles:
cos ý = 2 cos þ cos ÿ + sin þ sin ÿ cos ÿ
Double Angle Identities:
r = a 2 /a1 = a 3 /a2
a n = a1 r n−1
a n = a x r n−x
1 2 rn
Sn = a1
12r
a1
S∞ =
12r
 Age Problems
→ underline specific time conditions
cos ÿ = cos Ā cos ā + sin Ā sin ā cos ý
sin (A ± B) = sin A cos B ± cos A sin B
cos (A ± B) = cos A cos B ∓ sin A sin B
tan A ± tan B
tan (A ± B) =
1 ∓ tan A tan B
Geometric Progression:
Binomial Theorem
Cosine Law for sides:
Sum & Diff of Angles Identities:
Harmonic Progression:
- reciprocal of arithmetic
progression
Product of Roots:
sin ÿ sin Ā sin ÿ
=
=
sin ý sin þ sin ý
sin A + cos A = 1
1 + tan2 A = sec 2 A
1 + cot 2 A = csc 2 A
3
lOMoARcPSD|20985858
Archimedean Solids
Analytic Geometry
- the only 13 polyhedra that are
convex, have identical vertices, and
their faces are regular polygons.
E=
Nn
2
V=
s
Nn
v
Slope-intercept form:
y = mx + b
Point-slope form:
where:
E → # of edges
V → # of vertices
N → # of faces
n → # of sides of each face
v → # of faces meeting at a vertex
Conic Sections
Distance from a point to a line:
y 2 y1
m=
x 2 x1
Two-point form:
y2 2 y1 y 2 y2
=
x 2 2 x1 x 2 x 2
Point-slope form:
General Equation:
Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0
Based on discriminant:
B 2 2 4AC = 0 ∴ parabola
B 2 2 4AC < 0 ∴ ellipse
B 2 2 4AC > 0 ∴ hyperbola
Distance from a point to another point:
d = √(y2 2 y1 )2 + (x2 2 x1 )2
x y
+ =1
a b
d=
|Ax + By + C|
√A2 + B 2
Parabola
Angle between two lines:
m2 2 m1
1 + m1 m2
- the locus of point that moves such
that its distance from a fixed point
called the center is constant.
General Equation:
x 2 + y 2 + Dx + Ey + F = 0
Standard Equation:
2
(x 2 h) + (y 2 k)2 = r 2
Line Tangent to Conic Section
To find the equation of a line
tangent to a conic section at a
given point P(x1, y1):
k=
[1 +
y"
3
(y′)2 ]2
1 revolution
= 2π rad
= 360˚
= 400 grads
= 6400 mills
Tetrahedron
2
H = a√
3
√2
V=a
12
3
þ"
(+) minima
(-) maxima
Point of inflection:
=1
Location of foci, c:
c 2 = a2 2 b2
Length of LR:
2b2
LR =
a
(y 2 k)2 (x 2 h)2
2
=1
a2
b2
covers A = 1 2 sin A
Integral Calculus-The Cardioid
r = a(1 2 cos θ)
r = a(1 + cos θ)
1 2 cos A
2
exsec A = sec A 2 1
 Inflation:
�㕖f = �㕖 + f + �㕖f
 Break-even analysis:
cost = revenue
 Rate of return:
annual net profit
RR =
capital
Annual net profit
= savings – expenses
– depreciation (sinking fund)
RP =
1
RR
n2m+1
dm = (FC 2 SV) [
]
∑ years
∑nn2m+1 x
Dm = (FC 2 SV) [
]
∑n1 x
 Declining Balance (Matheson):
BVm = FC(1 2 k)m
SV = FC(1 2 k)n k → obtained
Dm = FC 2 BVm
 Service Output Method:
F = P(1 + �㕖)
r mt
F = P (1 + )
m
r m
I
ER = = (1 2 ) 2 1
m
P
 Continuous Compounding Interest:
rt
F = Pe
ER = er 2 1
 Annuity:
2
(1 + �㕖)n 2 1
]
�㕖
where:
F → future worth
P → principal or present worth
A → periodic payment
i → interest rate per payment
n → no. of interest periods
n’ → no. of payments
2
(1 + �㕖)n 2 1
]
P = A[
�㕖(1 + �㕖)n
 Perpetuity:
where:
FC → first cost
SV → salvage cost
d → depreciation
per year
n → economic life
m → any year before n
BVm → book value
after m years
Dm → total depreciation
CALTECH:
Mode 3 3
x
y
(time)
(BV)
0
FC
n
SV
n+1
SV
P=
A
= F(1 + �㕖)2n
�㕖
C = FC +
AC = C ∙ �㕖
RC 2 SV
OM
+
(1 + �㕖)n 2 1
�㕖
AC = FC ∙ �㕖 + OM +
where:
C → capitalized cost
FC → first cost
OM → annual operation
or maintenance cost
RC → replacement cost
SV → salvage cost
AC → annual cost
(RC 2 SV)�㕖
(1 + i)n 2 1
 Single-payment-compound-amount factor:
n
(F/P, �㕖, n) = (1 + �㕖)
 Single-payment-present-worth factor:
2n
(P/F, �㕖, n) = (1 + �㕖)
 Equal-payment-series-compound-amount factor:
CALTECH:
Mode 3 6
x
y
(time)
(BV)
0
FC
n
SV
 Double Declining Balance:
BVm = FC(1 2 k)m
k = 2/n k → obtained
Dm = FC 2 BVm
where:
F → future worth
P → principal or present worth
i → interest rate per interest period
r → nominal interest rate
n → no. of interest periods
m → no. of interest period per year
t → no. of years
ER → effective rate
 Capitalized Cost:
(1 + i)n 2 1 21
d = (FC 2 SV) [
]
�㕖
m
(1 + i) 2 1
]
Dm = d [
�㕖
FC 2 SV
Qn
D = dQ m
where:
m is (+) for upward asymptote;
m is (-) for downward
m = b/a if the transverse axis is horizontal;
m = a/b if the transverse axis is vertical
c
e=
a
F = A[
CALTECH:
Mode 3 2
x
y
(time)
(BV)
0
FC
n
SV
 Sinking Fund:
d=
Same as ellipse:
Length of LR,
Loc. of directrix, d
Eccentricity, e
y 2 k = ±m(x 2 h)
Eccentricity, e:
 Compound Interest:
n
Half versed sine:
FC 2 SV
d=
n
Dm = d(m)
Eq’n of asymptote:
I = P�㕖n
F = P(1 + �㕖n)
Versed cosine:
Exsecant:
c 2 = a2 + b2
a
d=
e
 Simple Interest:
vers A = 1 2 cos A
hav A =
Location of foci, c:
Loc. of directrix, d:
Engineering Economy
 Sum-of-the-Years-Digit (SYD):
ý2 þ
= y" = 0
ýý 2
A = 1.5Ãa2
P = 8a
r = a(1 2 sin θ)
r = a(1 + sin θ)
Standard Equation:
(x 2 h)2 (y 2 k)2
2
=1
b2
a2
Elements:
Versed sine:
 Straight-Line:
Ä=
dd
LR = 4a
Unit Circle
BVm = FC 2 Dm
Radius of curvature:
3
[1 + (y′)2 ]2
df
Length of latus
rectum, LR:
Depreciation
Maxima & Minima (Critical Points):
ýþ
= y2 = 0
ýý
e=
(x 2 h) = ±4a(y 2 k)
(y 2 k)2 = ±4a(x 2 h)
Differential Calculus
Curvature:
Eccentricity, e:
y + Dx + Ey + F = 0
x 2 + Dx + Ey + F = 0
SA = a √3
ý 2 → ýý1
þ 2 → þþ1
ý + ý1
ý→
2
þ + þ1
þ→
2
ýþ1 + þý1
ýþ →
2
Ax 2 2 Cy 2 + Dx + Ey + F = 0
Elements:
Elements:
General Equation:
2
2
In the equation of the conic
equation, replace:
General Equation:
Ax 2 + Cy 2 + Dx + Ey + F = 0
- the locus of point that moves such that it is always equidistant from a
fixed point (focus) and a fixed line (directrix).
Standard Equation:
2
Circle
General Equation:
(x 2 h)2 (y 2 k)2
+
=1
b2
a2
√A2 + B 2
Based on eccentricity, e=f/d:
þ = 0 ∴ circle
þ = 1 ∴ parabola
þ < 1 ∴ ellipse
þ > 1 ∴ hyperbola
- the locus of point that moves such
that the difference of its distances
from two fixed points called the foci
is constant.
Standard Equation:
|C1 2 C2 |
tan θ =
Hyperbola
- the locus of point that moves such
that the sum of its distances from
two fixed points called the foci is
constant.
(x 2 h)2 (y 2 k)2
+
=1
b2
a2
Distance of two parallel lines:
d=
Ellipse
2
(1 + �㕖)n 2 1
(F/A, �㕖, n) = [
]
�㕖
 Equal-payment-sinking-fund factor:
2
21
(1 + �㕖)n 2 1
(A/F, �㕖, n) = [
]
�㕖
 Equal-payment-series-present-worth factor:
where:
FC → first cost
SV → salvage cost
d → depreciation per year
Qn → qty produced during
economic life
Qm → qty produced during
up to m year
Dm → total depreciation
Downloaded by Ace Juntilla (ace.juntilla@gmail.com)
2
(1 + �㕖)n 2 1
]
(P/A, �㕖, n) = [
�㕖(1 + �㕖)n
 Equal-payment-series-capital-recovery factor:
2
(1 + �㕖)n 2 1
]
(A/P, �㕖, n) = [
�㕖(1 + �㕖)n
21
lOMoARcPSD|20985858
Statistics
Fractiles
Transportation Engineering
Traffic Accident Analysis
Measure of Natural Tendency
 Range
Design of Horizontal Curve
 Accident rate for 100 million
vehicles per miles of travel in a
segment of a highway:
 Mean, x̅, μ → average
→ Mode Stat 1-var
 Coefficient of Range
→ Shift Mode ▼s Stat Frequency? on
→ Input
→ AC Shift 1 var x̅
when n is even
1
n+1
2
1 n
n
= [( ) + ( + 1)]
2 2
2
Q1 = n
Me th =
Me
ý�㕎Ā�㕔þāĂ ý�㕎Ăăþ 2 āþ�㕎ýýþāĂ ý�㕎Ăăþ
=
ý�㕎Ā�㕔þāĂ ý�㕎Ăăþ + āþ�㕎ýýþāĂ ý�㕎Ăăþ
 Quartiles
 Median, Me → middle no.
th
= ý�㕎Ā�㕔þāĂ ý�㕎Ăăþ 2 āþ�㕎ýýþāĂ ý�㕎Ăăþ
4
when n is odd
Q1 =
 Mode, Mo → most frequent
2
3
Q2 = n
Q3 = n
4
4
1
1
1
(n + 1) ; Q1 = (n + 1) ; Q1 = (n + 1)
4
4
4
 Interquartile Range, IQR
Standard Deviation
 Population standard deviation
→ Mode Stat 1-var
→ Shift Mode ▼ Stat Frequency? on
→ Input
→ AC Shift 1 var σx
 Sample standard deviation
→ Mode Stat 1-var
→ Shift Mode ▼ Stat Frequency? on
→ Input
→ AC Shift 1 var sx
NOTE:
If not specified whether population/sample
in a given problem, look for POPULATION.
Coefficient of Linear Correlation
or Pearson’s r
= ý�㕎Ā�㕔þāĂ ÿă�㕎ĀĂ�㕖ýþ 2 āþ�㕎ýýþāĂ ÿă�㕎ĀĂ�㕖ýþ
= Q3 2 Q1
 Coefficient of IQR
ý�㕎Ā�㕔þāĂ ÿă�㕎ĀĂ�㕖ýþ 2 āþ�㕎ýýþāĂ ÿă�㕎ĀĂ�㕖ýþ
=
ý�㕎Ā�㕔þāĂ ÿă�㕎ĀĂ�㕖ýþ + āþ�㕎ýýþāĂ ÿă�㕎ĀĂ�㕖ýþ
Q3 2 Q1
=
Q3 + Q1
 Quartile Deviation (semi-IQR) = IQR/2
 Outlier
→ extremely high or low data higher than
or lower than the following limits:
Q1 2 1.5IQR > x
Q 3 + 1.5IQR < x
Variance
 Z-score or
standard score
or variate
 standard deviation = σ
 variance = σ2
Mean/Average Deviation
1
mv =
∫ f(x)dx
b2a a
R → minimum radius of curvature
v → design speed in m/s
g → 9.82 m/s2
Power to move a vehicle
P = vR
P → power needed to move vehicle in watts
v → velocity of vehicle in m/s
R → sum of diff. resistances in N
Design of Pavement
 Rigid pavement without dowels
Discrete Probability Distributions
t=√
3W
4f
(at the center)
W
2r
t=√
�㔋f1
Walli’s Formula
an =
Period, Amplitude & Frequency
Period (T) → interval over which the graph of
function repeats
Amplitude (A) → greatest distance of any point
on the graph from a horizontal line which passes
halfway between the maximum & minimum
values of the function
Frequency (ω) → no. of repetitions/cycles per unit
of time or 1/T
1
√5
SR =
f∙i
f∙i∙p
f → fatal
i → injury
p → property damage
Us =
∑d
n
=
∑t ∑ 1
( )
U1
 Time mean speed, Ut:
d
∑ U1
Ut = t =
n
n
∑
Ʃd → sum of distance traveled by all vehicles
Ʃt → sum of time traveled by all vehicles
Ʃu1 → sum of all spot speed
1/Ʃu1 → reciprocal of sum of all spot speed
n → no. of vehicles
 Rate of flow:
q = kUs
 Spacing of vehicles (km)
= 1/k
expansion pressure
pavement density
 Stiffness factor of pavement
Es
SF = √
Ep
3
[(
s
[(m 2 1)(m 2 3)(m 2 5) & (1 or 2)][(n 2 1)(n 2 3)(n 2 5) & (1 or 2)]
∙α
(m + n)(m + n 2 2)(m + n 2 4) & (1 or 2)
Tip to remember:
n
n
1 + √5
1 2 √5
) 2(
) ]
2
2
x = r cos θ
y = r sin θ
r = x2 + y2
y
θ = tan21
x
 Peak hour factor (PHF)
= q/qmax
ES → modulus of elasticity of subgrade
EP→ modulus of elasticity of pavement
Fibonacci Numbers
 Poisson Probability Distribution
x −¼
 Severity ratio, SR:
 Spacing mean speed, US:
3W
t=√
f
α = π/2 for m and n are both even
α =1 otherwise
)
A → no. of accidents during period of analysis
ADT → average daily traffic entering all legs
N → time period in years
t=
NOTE:
 Geometric Probability Distribution
x−1
A (1,000,000)
ADT ∙ N ∙ 365
 Minimum time headway (hrs)
= 1/q
0
where:
p → success
q → failure
R=
 Thickness of pavement in terms
of expansion pressure
∫ cosm θ sinn θ dθ =
P(x) = C(n, x) p q
 Accident rate per million entering
vehicles in an intersection:
bet. z & axis → Q(
→ Input
right of z → R(
π
2
 Binomial Probability Distribution
x n−x
A → no. of accidents during period of analysis
ADT → average daily traffic
N → time period in years
L → length of segment in miles
q → rate of flow in vehicles/hour
k → density in vehicles/km
uS → space mean speed in kph
P(x g a) = e−»a
P(x f a) = 1 2 e−»a
P(a f x f b) = e−»a 2 e−»b
b
1
∫ f(x)2 dx
RMS = √
b2a a
A (100,000,000)
ADT ∙ N ∙ 365 ∙ L
f1 → allow bearing pressure of subgrade
r → radius of circular area of contact
between wheel load & pavement
Exponential Distribution
 Mean value
Period
2π/B
2π/B
π/B
v
gR
R=
→ AC Shift 1 Distr
left of z → P(
x → no. of observations
μ → mean value, x
̅
σ → standard deviation
 Mean/average value
b
Function
y = A sin (Bx + C)
y = A cos (Bx + C)
y = A tan (Bx + C)
→ Mode Stat
x2μ
z=
σ
 relative variability = σ/x
μ e
x!
Impact factor =
 Flexible pavement
 Population standard deviation
P(x) =
 Centrifugal ratio or impact factor
2
(at the edge)
-1 ≤ r ≤ +1; otherwise erroneous
P(x) = p(q
R → minimum radius of curvature
e → superelevation
f → coeff. of side friction or
skid resistance
v → design speed in m/s
g → 9.82 m/s2
t → thickness of pavement
W → wheel load
f → allow tensile stress of concrete
Normal Distribution
NOTE:
v
g(e + f)
3W
t=√
2f
m
(n)
im =
10 or 100
→ Input
→ AC Shift 1 Reg r
R=
 Rigid pavement with dowels
 Decile or Percentile
→ Mode Stat A+Bx
 Minimum radius of curvature
2
�㕥 2 2 �㕥 2 1 = 0
Mode Eqn 5
�㕥 =
1 ± √5
2
Amplitude
A
A
A
Downloaded by Ace Juntilla (ace.juntilla@gmail.com)
lOMoARcPSD|20985858
Measurement
Corrections
measure
too long add
too short subtract
Due to temperature:
(add/subtract); measured length
C = αL(T2 2 T1 )
(P2 2 P1 )L
C=
EA
too short
CD = MD (1 2
E = 0.6745√
∑(x 2 x̅)
n21
w 2 L3
24P 2
∑(x 2 x̅)
= 0.6745√
Em =
n(n 2 1)
√n
E
(subtract only); measured length
�㕤 ∝
Normal Tension:
0.204W√AE
1
ý2
�㕤 ∝
1
�㕑
�㕤 ∝ �㕛
Area of Closed Traverse
√PN 2 P
A=
Symmetrical:
L
H = (g1 + g 2 )
8
L 2
x 2 ( 2)
=
L
y
H 1
=
Error of Closure
Perimeter
1 acre =
4047 m2
Horizontal:
Elevþ = Elevý + �㔵�㕆 2 þ�㕆
Reduction to
Sea Level
CD
MD
=
R
R+h
Inclined Upward:
error/setup = 2eBS + eFS
Subtense Bar
Inclined Downward:
error/setup = +eBS 2 eFS
D = cot
Total Error:
eT = error/setup ∙ no. of setups
θ
2
Double Meridian Distance Method DMD
DMDÿ�㕖ÿĀā = Depÿ�㕖ÿĀā
DMDÿ = DMDÿ−1 + Depÿ−1 + Depÿ
DMD�㕙�㕎Āā = 2Dep�㕙�㕎Āā
2A = Σ(DMD ∙ Lat)
d
[h + hn + 2Σh]
2 1
Double Parallel Distance Method DPD
d
A = [h1 + hn + 2ΣhĀýý + 4Σhþ�㕣þÿ ]
3
Relative Error/Precision:
D2
(h 2 h2 ) 2 0.067D1 D2
D1 + D2 1
Leveling
Simpson’s 1/3 Rule:
= √ΣL2 + ΣD2
from South
Stadia Measurement
Trapezoidal Rule:
Error of Closure:
Parabolic Curves
h = h2 +
Area of Irregular Boundaries
Lat = L cos α
Dep = L sin α
Azimuth
hcr = 0.067K 2
D = Ks cos θ + C
H = D cos θ
V = D sin θ
E=error; d=distance; n=no. of trials
C 2 = S 2 2 h2
e
)
TL
Effect of Curvature & Refraction
Inclined:
Proportionalities of weight, w:
Due to slope:
e
)
TL
D = d + (f + c)
�㕓
D = ( )s +C
�㕖
D = Ks + C
Probable Error (mean):
(subtract only); unsupported length
PN =
CD = MD (1 +
Probable Error (single):
Due to sag:
C=
too long
Probable Errors
Due to pull:
(add/subtract); measured length
lay-out
subtract
add
Note: n must be odd
Simple, Compound & Reverse Curves
DPDÿ�㕖ÿĀā = Latÿ�㕖ÿĀā
DPDÿ = DPDÿ−1 + Lat ÿ−1 + Lat ÿ
DPD�㕙�㕎Āā = 2Lat �㕙�㕎Āā
2A = Σ(DMD ∙ Dep)
Spiral Curve
Unsymmetrical:
H=
L1 L2
(g + g 2 )
2(L1 +L2 ) 1
g 3 (L1 +L2 ) = g1 L1 + g 2 L2
Note: Consider signs.
Earthworks
A=
�㕑�㔿 0 �㕑�㕅
±�㕓�㔿 ±�㕓 ±�㕓�㕅
f
w
(d + dR ) + (fL + fR )
2 L
4
T = R tan
L
Ve = (A1 + A2 )
2
L = 2R sin
L
VP = (A1 + 4Am + A2 )
6
L
(c 2 c2 )(d1 2 d2 )
12 1
Volume (Truncated):
VT =
2
I
Σh
= A( )
n
I
2
2
v2
S = vt +
2g(f ± G)
a = g(f ± G) (deceleration)
v
(breaking time)
tb =
g(f ± G)
f
Eff =
(100)
fave
v → speed in m/s
t → perception-reaction time
f → coefficient of friction
G → grade/slope of road
L=
L<S
A(S)2
200(√h1 + √h2 )
L → length of summit curve
S → sight distance
h1 → height of driver’s eye
h1 = 1.143 m or 3.75 ft
h2 → height of object
h2 = 0.15 m or 0.50 ft
L3
6RLs
Y=L2
Ls =
L5
40R2 Ls
2
2
0.036k 3
R
0.0079k 2
R
L
D
=
DC Ls
e=
2
200(√h1 + √h2 )
L = 2(S) 2
A
θ
Ls 2
; p=
3
24R
Ls
I
+ (R + p) tan
2
2
I
Es = (R + p) sec 2 R
2
Parabolic Summit Curve
L>S
L2 180°
∙
2RLs π
Ts =
A
(Σh1 + 2Σh2 + 3Σh3 + 4Σh4 )
n
Stopping Sight Distance
x=
π
Lc = RI ∙
180°
20 2πR
=
360°
D
1145.916
R=
D
Prismoidal Correction:
VT = ABase ∙ Have
i=
I
m = R [1 2 cos ]
Volume (Prismoidal):
VP = Ve 2 Cp
2
E = R [sec 2 1]
Volume (End Area):
CP =
I
θ=
LT → long tangent
ST → short tangent
R → radius of simple curve
L → length of spiral from TS to any point
along the spiral
Ls → length of spiral
I → angle of intersection
I c → angle of intersection of the simple
curve
p → length of throw or the distance from
tangent that the circular curve has been
offset
x → offset distance (right angle
distance) from tangent to any point on
the spiral
xc → offset distance (right angle
distance) from tangent to SC
Ec → external distance of the simple
curve
θ → spiral angle from tangent to any
point on the spiral
θS → spiral angle from tangent to SC
i → deflection angle from TS to any point
on the spiral
is → deflection angle from TS to SC
y → distance from TS along the tangent
to any point on the spiral
Parabolic Sag Curve
Underpass Sight Distance
Horizontal Curve
L>S
L>S
L>S
A(S)2
L=
122 + 3.5S
L<S
122 + 3.5S
L = 2(S) 2
A
A → algebraic difference
of grades, in percent
L → length of sag curve
S → sight distance
A(S)2
L=
800H
L<S
H= C2
800H
L = 2(S) 2
A
h1 + h2
2
A → algebraic difference of
grades, in percent
L → length of sag curve
L=
A(K)2
395
Downloaded by Ace Juntilla (ace.juntilla@gmail.com)
For passengers comfort,
where K is speed in KPH
R=
L<S
R=
S2
8M
L(2S 2 L)
8M
L → length of horizontal
curve
S → sight distance
R → radius of the curve
M → clearance from the
centerline of the road
lOMoARcPSD|20985858
Properties of Fluids
s Mg
W=
M
; Ä=
V
pg
ɤ = Äg =
RT
V 1
s. v. = =
M Ä
ɤ
Ä
s. g. =
=
ɤĀ ÄĀ
W
ɤ=
V
F1 = ɤAh1 = ɤh1 2
hĀ = s. g.1 h1
h2
RM = W1 (X1 ) + W2 (X2 )+. . . +WĀ (XĀ ) + F2 ( )
3
h
1
2
OM = F1 ( ) + U1 ( B) + U2 ( B)
3
2
3
Ig
Aӯ
e=
e=
F = ɤh̅A
2
F = : F/ + Fÿ
Rotation:
a
tan θ =
g
ω2 x
tan θ =
g
a
p = ɤh (1 ± )
g
V=
RM
OM
2 2
ω x
2g
1 2
Ãr h
2
2
2
r
x
=
h
y
;
z1 +
H. L.T = H. L.1 = H. L.2 = H. L.n
volume flow rate → m3/s
weight flow rate → N/s
mass flow rate → kg/s
Q → discharge
→ flow rate
→ weight flux
Constant Head Orifice
Falling Head Orifice
Without headloss:
Time to remove water from h1 to h2 with constant cross-section:
v = Cv √2gh
Q = CA o √2gh
C = Cc C v
a
Cc =
A
v
Cv =
vt
H. L. =
v2 1
[
2 1]
2g Cv 2
H. L. = &H[1 2 Cv 2 ]
y=
x
2
4Cv 2 h
2As
CAo √2g
(√h1 2 √h2 )
t=∫
h1
h2
As dh
CAo √2gh
(As1 )(As2 )
(√h1 2 √h2 )
CAo √2g (As1 + As2 )
2
H. L. =
Hydrodynamics
Force on Curve Vane/Blade:
Force on the Jet
(at right angle):
∑ Fx = ÄQ(v2x 2 v1x )
F = ÄQv
∑ Fy = ÄQ(v2y 2 v1y )
L v2
D 2g
2
Manning’s Formula:
H. L. =
10.29 n2 L Q2
D16/3
Hazen William’s Formula:
10.64 L Q1.85
H. L. = 1.85 4.87
C
D
1 atm
= 101.325 KPa
= 2166 psf
= 14.7 psi
= 760 mmHg
= 29.9 inHg
Open Channel
Trapezoidal:
x = y1 + y2
(rigid pipes)
EB
c=:
Äw
(non-rigid pipes)
0.0826 f L Q
D5
b = 2d
d
R=
2
EB
c=:
E D
Äw (1 + B )
Et
Water Hammer
&Pmax = Äcv
tc =
b = 2d
A = d2
θ = 90°
rapid/instantaneous
&P = &Pmax
Slow Closure
tc
)
&P = &Pmax (
t actual
B. TYPE of closure:
Partial Closure (vf ≠ 0)
&P = Äc(vi 2 vf )
Total Closure (vf = 0)
&P = Äcvi
Specific Energy:
2
Manning Formula:
v = C:RS
Bazin Formula:
Theoretically:
C=:
8g
f
Kutter Formula:
TRAPEZOIDAL:
For minimum seepage:
b = 4d tan
Force on Pipe’s Bend & Reducer:
(same as on Curve Vane/Blade)
Downloaded by Ace Juntilla (ace.juntilla@gmail.com)
θ
2
2L
c
A. TIME of closure:
v
+d
E=
2g
d
R=
2
Q max if d = 0.94D
Vmax if d = 0.81D
Celerity (velocity of sound)
Darcy Weisbach Eq’n:
H. L. = f
sg m
A
sg l tot
sg m
=
V
sg l tot
Abel =
BF = ɤĀ Vý Vbel
Major Losses in Pipes
Rectangular:
Circular:
vs
I
=
VD sin θ VD
BF = W
St = tensile stress
p = unit pressure
D = inside diameter
t = thickness of wall
s = spacing of hoops
T = tensile force
Most Efficient Sections
d = r (full)
r
R=
2
tan2 θ
B2
[1 +
]
2
12D
Buoyancy
Semi-circular:
Time in which water surfaces of two tanks will reach same elevation:
t=
MB�㕂 =
Triangular:
Time to remove water from h1 to h2 with varying cross-section:
MB�㕂 =
pD
2t
2T
s=
pD
with turbine:
Parallel Connection:
RM or OM = Wx
= W(MG sin θ)
St =
Pump → Output & Turbine → Input
Q T = Q1 = Q 2 = Q n
Use (-) if G is above BO and (+) if G is below BO.
Note that M is always above BO.
Stresses/Hoops
P1 v1 2
P2 v2 2
z1 + +
+ HA = z2 + +
+ H. L.
2g
2g
ɤ
ɤ
Q = Av
MG = MB�㕂 ± GB�㕂
Rý
with pump:
Fluid Flow
MG = metacentric height
μRþ
output
QɤE
efficiency =
; HP =
input
746
H. L.T = H. L.1 + H. L.2 +. . . +H. L.n
With headloss:
2
P1 v1 2
P2 v2 2
+
= z2 + +
+ H. L.
2g
2g
ɤ
ɤ
Series Connection:
t=
U2 = (h1 2 h2 )ɤB
P2 v2 2
P1 v1 2
2 HE = z2 + +
+ H. L.
z1 + +
2g
2g
ɤ
ɤ
Series-Parallel Pipes
v = √2gh
2
1
Bernoulli’s Energy Theorem
Ã
rad/sec
1 rpm =
30
Q T = Q1 + Q 2 +. . . +Q n
& FS�㕆 =
;
F2 = ɤAh2 = ɤh2 2
z = elevation head; P/ɤ = pressure head; v2/2g = velocity head
Horizontal Motion:
Vertical Motion:
2
h̅ = ӯ (for vertical only)
y=
;
B
þ = | 2 x̅|
2
R �㕦
6þ
B
[1 ± ]
þ< ; q=2
B
B
6
2R �㕦
B
þ> ; q=
6
3x̅
R �㕦
B
þ= ; q=2
6
B
2R �㕦
þ = 0; q =
B
NOTE:
ħ = vertical distance from cg of
submerged surface to liquid surface
Relative Equilibrium of Fluids
ah
tan θ =
g ± av
FS�㕂 =
F/ = ɤh̅A
Fÿ = ɤV
2
Rx̅ = RM 2 OM
ɤIg sin θ
F
On curved surfaces:
pd
σ=
4
4σcosθ
h=
ɤd
Inclined Motion:
U1 = ɤh2 B
On plane surfaces:
μ L2
=
Ä T
1
pÿĀý = pýÿýþ + pÿþÿ
Hydrostatic Forces
1
&P
; β=
&V
EB
V
ý�㕦 FT
μ=Ç
=
ý�㕉 L2
Stability of Floating Bodies
1
p = ɤh
s. g.1
h2 =
h
s. g.2 1
EB = 2
È=
Dams
Pressure
C=
C=
1 1/6
R
n
87
m
1+
:R
0.000155
1
+ 23 +
S
n
C=
n
0.000155
1+
(23 +
)
S
:R
If C is not given, use Manning’s in V:
v=
1 2/3 1/2
R S
n
lOMoARcPSD|20985858
Unit Weight:
ɤs
Gs =
ɤw
(Gs + Gs ω)ɤw
1+e
(Gs + Se)ɤw
ɤ=
1+e
ɤ=
Weight
g = Gs (1 2 n)
Vv
e=
Vs
Ww
ω=
Ws
Vv
n=
V
Vw
S=
Vv
e=
n
12n
(Gs 2 1)ɤw
1+e
Gs ɤw
=
1 + Gs ω
ɤsub =
ɤzav
PI = LL 2 PL
ɤd ÿ�㕎�㕥
SI = PL 2 SL
LL 2 ω
CI =
LL 2 PI
1
1
2
ɤd ÿ�㕖Ā ɤd
Dr =
1
1
2
ɤd ÿ�㕖Ā ɤd ÿ�㕎�㕥
Stratified Soil
Dr (%)
0 – 20
20 – 40
40 – 70
70 – 85
85 – 100
v
∆h
; v�㕠 =
v = ki ; i =
n
L
Q = vA = kiA
Ac =
Description
Very Loose
Loose
Medium Dense
Dense
Very Dense
Constant Head Test:
QL
k=
Aht
Pumping Test:
k = c ∙ D10 2
k = 1.4e2 k 0.85
Kozeny-Carman: Samarasinhe:
2
e
1+e
Stresses in Soil
k = C3 ∙
k eq =
n
e
1+e
k eq =
NOTE:
Quick
condition:
Effective Stress/
Intergranular Stress:
pE = 0
pE = pT 2 pw
Capillary Rise:
Pore Water Pressure/
Neutral Stress:
C
hcr =
eD10
pw = ɤw hw
Total Stress:
pT = ɤ1 h1 + ɤ2 h2 +. . . +ɤn hn
Lateral Earth Pressure
ACTIVE PRESSURE:
h1 k1 + h2 k 2 +. . . +hn k n
H
Flow Net / Seepage
k o = 1 2 sin Ø
For Inclined:
cos ´ 2 √cos 2 ´ 2 cos 2 Ø
cos ´ + √cos 2 ´ 2 cos 2 Ø
For Horizontal:
If there is angle of friction ³ bet. wall and soil:
2
ka =
cos ³ [1 + :
PASSIVE PRESSURE:
pP =
cos Ø
sin(Ø + ³) sin Ø
]
cos ³
1
k ɤH 2 + 2cH√k P
2 P
2
k P = cos ´
cos ´ +
2
cos ´ 2 √cos 2 ´ 2 cos 2 Ø
For Horizontal:
kP =
1 + sin Ø
1 2 sin Ø
kP =
θ = 45° +
Ø
2
TRI-AXIAL TEST:
σ1 → maximum principal stress
→ axial stress
△σ → additional pressure
→ deviator stress
→ plunger pressure
σ3 → minimum principal stress
→ confining pressure
→ lateral pressure
→ radial stress
→ cell pressure
→ chamber pressure
r
x + σ3 + r
c
tan Ø =
x
sin Ø =
 Unconsolidatedundrained test:
If there is angle of friction ³ bet. wall and soil:
2
cos Ø
cos ³ [1 2 :
sin(Ø 2 ³) sin Ø
]
cos ³
2
9
Ө → angle of failure in shear
Ø → angle of internal friction/shearing resistance
C → cohesion of soil
r
sin Ø =
σ3 + r
Ø
4 56 7 8
Shear Strength of Soil
 Cohesive soil:
cos 2
3
Nf → no. of flow channels [e.g. 4]
Nd → no. of potential drops [e.g. 10]
 Normally consolidated:
For Inclined:
√cos 2 ´
4
2
Description
0
1-5
5-10
10-20
20-40
>40
Non-plastic
Slightly plastic
Low plasticity
Medium plasticity
High plasticity
Very High plastic
Sieve Analysis
D60
D10
Cc =
D60 ∙ D10
Sorting
Coefficient:
So = :
D75
D25
Swell Index, CS:
Cs =
1
C
5 c
10
e 2 e′
H (for one layer only)
1+e
Cc H
∆P + Po
�㕙Ā�㕔
S=
1+e
Po
S=
With Pre-consolidation pressure, Pc:
when (△P+Po) < Pc:
S=
∆P + Po
Cs H
�㕙Ā�㕔
Po
1 + eo
when (△P+Po) > pc:
S=
Cs H
Pc
Cc H
∆P + Po
�㕙Ā�㕔 +
�㕙Ā�㕔
1+e
Po 1 + e
Pc
Over Consolidation Ratio (OCR):
OCR =
pc
;
po
OCR = 1 (for normally consolidated soil)
Coefficient of Compressibility:
av =
∆e
∆P
△e → change in void ratio
△P → change in pressure
Coefficient of Volume Compressibility:
mv =
∆e
∆P
1 + eave
Coefficient of Consolidation:
Hdr → height of drainage path
Hdr 2 Tv
→ thickness of layer if drained 1 side
Cv =
→ half of thickness if drained both sides
t
Tv → factor from table
Coefficient of Permeability: t → time consolidation
k = mv Cv ɤw
DIRECT SHEAR TEST:
σn → normal stress  Normally consolidated soil:
σs → shear stress
σS
c=r
 Unconfined
compression test:
σ3 = 0
PI
For normally consolidated clay:
2
Equipotential line ----
1
Class
AC < 0.7
Inactive
0.7 < AC < 1.2 Normal
AC > 1.2
Active
Cc = 0.009(LL 2 10%)
e 2 e′
Cc =
∆P + Po
�㕙Ā�㕔
Po
3
Nf
q = √k x k z H
Nd
AT REST:
1
Flow line ----
Non-Isotropic soil:
1
pa = k a ɤH 2 2 2cH√k a
2
1 2 sin Ø
ka =
1 + sin Ø
Nf
Nd
Ac
Compression Index, CC:
Isotropic soil:
q = kH
State
LI < 0
Semisolid
0 < LI < 1 Plastic
LI > 1
Liquid
Compressibility of Soil
r
Q �㕙ÿ 1
r2
k=
2πt(h1 2 h2 )
h1 h2
h
+ +. . . + n
k1 k 2
kn
LI
1
1
3
+
+
Sn = 1.7:
2
2
(D20 )
(D10 )2
(D50 )
Confined:
H
1
2 SL
SR
Coeff. of Gradation
or Curvature:
(D30 )2
Uniformity
Coefficient:
Cu =
1
GI = (F 2 35)[0.2 + 0.005(LL 2 40)]
+0.01(F 2 15)(PI 2 10)
Suitability Number:
r1
r2
k=
π(h1 2 2 h2 2 )
for Perpendicular flow:
Casagrande:
Hazen Formula
k a = cos ´
μ = % passing 0.002mm
Q �㕙ÿ
for Parallel flow:
aL h1
�㕙ÿ
At h2
k = C1 ∙
qu
PI
; St = und
μ
q u rem
Unconfined:
Falling/Variable Head Test:
k=
ω 2 PL
LL 2 PL
LI =
Relative Density/
Density Index:
eÿ�㕎�㕥 2 e
Dr =
eÿ�㕎�㕥 2 eÿ�㕖Ā
ɤsub = ɤsat 2 ɤw
ɤ
ɤd =
1+ω
e
n=
1+e
(Gs + e)ɤw
=
1+e
ɤsat
W
V
WS
ɤd =
V
0<n<1
Permeability
When S=100%:
ɤ=
0<e<∞
R=
G�㕠 =
Atterberg Limits
Relative Compaction:
ɤd
Gs ɤw
ɤd =
1+e
Se = Gs ω
SL =
Bulk Specific Gravity:
When S=0:
Volume
m1 2 m2 V1 2 V2
2
ɤw
m2
m2
e
m2
SL =
; SR =
Gs
V2 ɤw
Specific Gravity of Solid:
Downloaded by Ace Juntilla (ace.juntilla@gmail.com)
tan Ø =
σN
 Cohesive soil:
tan Ø =
σS
c
=
x + σN x
σS = c + σN tan ∅
lOMoARcPSD|20985858
Terzaghi‘s Bearing Capacity (Shallow Foundations)
 General Shear Failure
(dense sand & stiff clay)
Square Footing:
qult = 1.3cNaSSSSSSSSSSSSSSSSSSSS
c + qNq + 0.4ɤBNɤ
Circular Footing:
qult = 1.3cNc + qNq + 0.3ɤBNɤ
 Local Shear Failure
(loose sand & soft clay)
′
qult = 1.3c′Nc + qNq + 0.4ɤBNɤ
Circular Footing:
Factor of safety against sliding (without seepage)
Nɤ = (Nq 2 1) tan 1.4Ø
Factor of safety against sliding (with seepage)
Nc = (Nq 2 1) cot Ø
qult → ultimate bearing capacity
qu → unconfined compressive strength
c → cohesion of soil
qult = cNc + qNq + 0.5ɤBNɤ
′
 Analysis of Infinite Slope
Ø
Nq = tan2 (45° + ) eπ tan Ø
2
 Parameters
Strip Footing:
Square Footing:
Soil Stability
 Bearing Capacity Factor
qu
c=
2
′
(for no water table)
qult Pallow
qallow =
=
FS
A
qult 2 q
qnet =
FS
Strip Footing:
qult = c′Nc ′ + qNq ′ + 0.5ɤBNɤ ′
EFFECT OF WATER TABLE:
tan ∅
C
+
ɤ H sin �㗽 cos �㗽 tan �㗽
β
where:
C → cohesion
β → angle of backfill from horizontal
Ø → angle of internal friction
H → thickness of soil layer
ɤ′ tan ∅
C
+
FS =
ɤýÿþ H sin �㗽 cos �㗽 ɤýÿþ tan �㗽
 Analysis of Finite Slope
Factor of safety against sliding
q = ɤDf
qult = 1.3c′Nc ′ + qNq ′ + 0.3ɤBNɤ ′
FS =
FS =
Ff + Fc
W sin �㔃
β
Maximum height for critical equilibrium
(FS=1.0)
Hcr =
4�㔶
sin �㗽 cos ∅
[
]
ɤ 1 2 cos(�㗽 2 ∅)
Stability No.:
Stability Factor:
C
m=
ɤH
SF =
θ
where:
Ff → frictional force; Ff = μN
Fc → cohesive force
Fc = C x Area along trial failure plane
W → weight of soil above trial failure plane
1
m
Capacity of Driven Piles (Deep Foundations)
H
H
2
= BC
tan �㔃 tan �㗽
 Pile in Sand Layer
Case 1
Case 3
Case 2
q = ɤDf
3rd term ɤ = ɤ′
q = ɤ(Df 2 d)+ɤ′d
3rd term ɤ = ɤ′
NOTE:
ɤ′= ɤýÿĀ = ɤ 2 ɤ�㕤
q = ɤDf
3rd term ɤ = ɤave
for d ≤ B
ɤave ∙ B = ɤd + ɤ′(B 2 d)
for d > B
ɤave = ɤ
Group of Piles
 Alternate Equation for Group
Efficiency (sand only)
 Group Efficiency (sand or clay)
Eff =
Q des−group
Eff =
Q des−indiv
2(m + n 2 2)s + 4d
mnÃD
where:
m → no. of columns
n→ no. of rows
s → spacing of piles
D → diameter of pile
Q f = PAkμ
Q tip = pe Nq Atip
Neglecting velocity of approach:
2
Q = C√2g L H 3/2
3
Considering velocity of approach:
va 3/2
va 3/2
Q = m L [(H +
)
2g
2(
2g
Neglecting velocity of approach:
3/2
Q=mLH
)
]
Q des =
Q = 1.84 L′ [(H +
)
2g
2( )
2g
Neglecting velocity of approach:
3/2
Q = 1.84 L′ H
NOTE:
for suppressed
for singly contracted
for doubly contracted
Time required to discharge:
t=
2As 1
1
[
2
]
mL √H2 √H1
where:
W → channel width
L → weir length
Z → weir height
H → weir head
PARAMETERS:
C → coefficient of discharge
va → velocity of approach m/s
m → weir factor
]
θ
8
C√2g tan H 5/2
2
15
Q = m H 5/2
Q=
When θ=90°
Q = 1.4H 5/2
 Cipolletti (symmetrical, slope 4V&1H)
θ = 75°57’50”
3/2
Q = 1.859 L H
 with Dam:
Neglecting velocity of approach:
3/2
Q = 1.71 L H
where:
c → cohesion
Nc → soil bearing factor
Atip → Area of tip
QTIP
Critical depth, dc:
Q T = Q f + Q tip
 Triangular (symmetrical only)
Francis Formula (when C and m is not given)
Considering velocity of approach:
va 3/2
va 3/2
Q tip = cNc Atip
(AKA Qbearing)
where:
pe → effective pressure at bottom
Nq → soil bearing factor
Atip → Area of tip
 Rectangular
va 3/2
2( ) ]
2g
dc
Qf
Q T = Q f + Q tip
Loose 10 (size of pile)
Dense 20 (size of pile)
QT
F. S.
Q des =
Q2 ∙ B c
NF = : 3
Ac ∙ g
For all sections:
NOTE:
E is minimum for critical depth.
For rectangular sections ONLY:
NF = 1
NF < 1
NF > 1
Reynold’s Number
NR =
Dv DvÄ
=
È
μ
Laminar Flow (NR ≤ 2000)
64
hf =
NR
Turbulent Flow (NR > 2000)
2
L v
hf = f
D 2g
hf =
2
0.0826 f L Q
D5
Boundary Shear Stress
Ç = ɤRS
Boundary Shear Stress
(for circular pipes only)
Ço =
f
Äv
8
Downloaded by Ace Juntilla (ace.juntilla@gmail.com)
where:
Q → flow rate m3/s
g → 9.81 m/s2
AC → critical area
BC → critical width
Q2 Ac 3
=
Bc
g
Take note that it is only derived from
the critical depth equation.
Critical Flow
Subcritical Flow
Supercritical Flow
QT
F. S.
Critical Depth
where:
v → mean velocity (Q/A)
g → 9.81 m/s2
dm → hydraulic depth (A/B)
B → width of liquid surface
Q = C√2g L [(H + )
2g
3
where:
C → cohesion
L → length of pile
α → frictional factor
P → perimeter of pile
(AKA Qbearing)
Froude Number
v
NF =
√gdm
Considering velocity of approach:
va 3/2
2
Q f = CLαP
Q
where:
P → perimeter of pile
A → area of pressure diagram
k → coefficient of lateral pressure
μ → coefficient of friction
Weirs
L’ = L
L’ = L – 0.1H
L’ = L – 0.2H
 Pile in Clay Layer
3 q2
2
dc = : = Ec
3
g
Q
B
v2
+ dā
Eā =
2g
q=
vc = √gdc
where:
q → flow rate or discharge
per meter width
EC → specific energy at
critical condition
vC → critical velocity
Hydraulic Jump
Power Lost:
Height of the jump:
∆d = d2 2 d1
Length of the jump:
L = 220 d1 tanh
Solving for Q:
NF1 −1
P = QɤE
22
For all sections:
P2 2 P1 =
P = ɤh̅A
ɤQ
(v 2 v2 )
g 1
For rectangular sections ONLY:
q2 1
= (d1 ∙ d2 )(d1 + d2 )
g
2
lOMoARcPSD|20985858
Load Combinations
Strength Reduction Factors, Ø
→ choose largest U in design
Basic Loads:
Ā = 1.4ÿ + 1.7�㔿
With Wind Load:
Ā = 0.75(1.4ÿ + 1.7�㔿 + 1.7�㕊)
Ā = 0.9ÿ + 1.3�㕊
Ā = 1.4ÿ + 1.7�㔿
With Earthquake Load:
Ā = 1.32ÿ + 1.1�㕓1 �㔿 + 1.1Ā
Ā = 0.99ÿ + 1.1Ā
With Earth Pressure Load:
Ā = 1.4ÿ + 1.7�㔿 + 1.7�㔻
Ā = 0.9ÿ
Ā = 1.4ÿ + 1.7�㔿
With Structural Effects:
Ā = 0.75(1.4ÿ + 1.7�㔿 + 1.4ÿ)
Ā = 1.4(ÿ + ÿ)
Internal Couple Method:
Factor k:
k=
n
n+
Factor j:
1
j= 12 k
3
fs
fc
Moment Resistance Coefficient, R:
1
R = fc kj
2
Moment Capacity:
1
Mc = C ∙ jd = fc kdb
2
∙ jd = Rbd2
Ms = T ∙ jd = As fs ∙ jd
Provisions for Uncracked Section:
 Solve for inertia of gross section, Ig.
 Solve for cracking moment, Mcr.
 Solve for actual moment, Ma:
2
wL
Ma =
8
(a) Flexure w/o axial load ……………………… 0.90
(b) Axial tension & axial tension w/ flexure .… 0.90
(c) Axial comp. & axial comp. w/ flexure:
(1) Spiral ……………………………….………. 0.75
(2) Tie …………………….…………….………. 0.70
(d) Shear & torsion ……………………….………. 0.85
(e) Bearing on concrete ……………….…,……. 0.70
Design Conditions
Values
Over-reinforced:
→ concrete fails first
→ fs < fy
(USD)
→ Ms > Mc (WSD)
Choose Smaller Value/
Round-down
→ Moment Capacity
→
→
Under-reinforced:
→ steel fails first
→ fs > fy
(USD)
→ Ms < Mc (WSD)
Balance Condition:
→ concrete & steel
simultaneously fail
→ fs = fy
(USD)
→ Ms = Mc (WSD)
Choose Larger Value/
Round-up
→
→
 Solve for effective moment of inertia, Ie:
Mcr 3
Mcr 3
) ∙ Ig + [1 2 (
) ] ∙ Icr
Ie = (
Ma
Ma
409.6.2.4. For simply supported, Ie = Ie (mid)
For cantilever, Ie = Ie (support)
Ie mid + Ie support
Ie =
2
409.6.2.5. Factor for shrinkage & creep due
to sustained loads:
time-dep factor, ξ:
�㔆 =
�㔉
1 + 50�㔌′
5 yrs +
12 mos
6 mos
3 mos
2.0
1.4
1.0
1.0
 Solve for instantaneous deflection:
4
δi =
5wL
384Ec Ie
 Horizontal members
(i.e. beam, slab, footing, etc.)
fc = 0.45 f’c
fs = 0.50 fy
fc = 0.25 f’c
fs = 0.40 fy
where:
f’c → compressive strength of concrete at 28 days
fy → axial strength of steel
Structural Grade
ASTM Gr.33 / PS Gr.230
Intermediate Grade ASTM Gr.40 / PS Gr.275
High Carbon Grade ASTM Gr.60 / PS Gr.415
fy = 230 MPa
fy = 275 MPa
fy = 415 MPa
424.3.2 for fy = 275 MPa; fs f 140 MPa
for fy = 415 MPa; fs f 170 MPa
Modular Ratio, n (if not given):
n=
Estronger
Esteel
200,000
=
=
Econcrete 4700√fc′
Eweaker
Solutions for Cracked Section (Doubly):
 Location of neutral axis, NA:
 Location of neutral axis, NA:
Ay̅above NA = Ay̅below NA
x
bx ( ) = nAs (d 2 x)
2
x → obtained
Ay̅above NA = Ay̅below NA
x
bx ( ) + (2n 2 1)A′s (x 2 d′ ) = nAs (d 2 x)
2
x → obtained
 Solve transferred moment of inertia at NA:
bx 3
+ nAs (d 2 x)2
INA =
INA
3
→ obtained
 Solve for Stresses or Resisted Moment:
For concrete:
Mc ∙ x
INA
fc =
For tension steel:
fs Ms ∙ (d 2 x)
=
INA
n
Solutions for Gross Section (Singly):
 Solve transferred moment of inertia at NA:
bx 3
+ (2n 2 1)A′s (x 2 d′ )2 + nAs (d
INA =
INA
3
→ obtained
2 x)2
 Solve for Stresses or Resisted Moment:
For concrete:
fc =
Mc ∙ x
INA
For tension steel:
fs Ms ∙ (d 2 x)
=
INA
n
For comp. steel:
Ms′ ∙ (x 2
fs′
2n
Solutions for Uncracked Section (By Sir Erick):
=
INA
 Location of neutral axis, NA:
 Location of neutral axis, NA:
yt =
h
; y → obtained
2 t
 Solve moment of inertia of gross section at NA:
3
bx
12
Ig → obtained
Ig =
 Solve for cracking moment:
Mcr ∙ yt
Ig
→ obtained
Ay̅above NA = Ay̅below NA
x
d2x
bx ( ) = b(d 2 x) (
) + (n 2 1)As (d 2 x)
2
2
x → obtained
 Solve transferred moment of inertia at NA:
3
3
bx
b(d 2 x)
+
+ (n 2 1)As (d 2 x)2
3
3
→ obtained
INA =
INA
 Solve for Stresses or Resisted Moment:
For tension steel:
For concrete:
fc =
Mc ∙ x
INA
fr = 0.7√fc′ =
Mcr
(for uniformly distributed load)
 Solve for additional deflection:
δadd = δsus ∙ �㔆
δadd = (% of sustained load)δi ∙ �㔆
Say, 70% of load is sustained after n yrs.
δadd = 0.7δi ∙ �㔆
 Solve for final deflection:
δfinal = δi + δadd
 Vertical members
(i.e. column, wall, etc.)
Solutions for Cracked Section (Singly):
409.6.2.3. if Ma < Mcr, no crack; Ig = Ie
if Ma > Mcr, w/ crack; solve for Ie
3
Allowable Stresses (if not given):
424.6.4 n must be taken as the nearest whole number & n g 6
424.6.5 for doubly, use n for tension & use 2n for compression
(for simply supported beam)
 Solve for inertia of cracked section:
bx 3
Icr =
+ nAs (d 2 x)2
Working Strength Design (WSD)
or Alternate Strength Design (ASD)
Downloaded by Ace Juntilla (ace.juntilla@gmail.com)
fs Ms ∙ (d 2 x)
=
INA
n
d′)
lOMoARcPSD|20985858
Ultimate Strength Design
Steel Ratio
 Based in Strain Diagram:
 Ultimate Moment Capacity:
εs
0.003
=
c
d2c
d2c
εs = 0.003 (
)
c
d2c
)
fs = 600 (
c
Mu = ∅Mn
Mu = ∅R n bd2
10
Mu = ∅fc′ bd2 ω(1 2 ω)
17
fy
ω=Ä ′
fc
a
 Coefficient of resistance, Rn:
= β1 c
a → depth of compression block
c → distance bet. NA &
extreme compression fiber
Provisions for β1:
0.65 f β1 f 0.85
* 1992 NSCP
β1 = 0.85 2 0.008(fc′ 2 30)
* 2001 NSCP
β1 = 0.85 2
* 2010 NSCP
β1 = 0.85 2
0.05
7
0.05
7
(fc′ 2 30)
Rn =
Mu
Rn =
∅bd2
2
10
17
ω)
Äb =
0.85fc′ β1 600
fy (600 + fy )
Maximum & Minimum steel ratio:
Ämax = 0.75Äb
As max = 0.75As b
Ämin
1.4
=
fy
Ämin
Minimum Concrete Covers:
20 mm → slab
√fc′
=
4fy
(choose larger between the 2)
Singly or Doubly ?
 Steel reinforcement ratio, Ä:
As
Ä=
bd
 Combined Ä & Rn:
Ä=
(fc′ 2 28)
fc′ ω(1
Steel ratio for balance condition:
2R n
0.85fc′
[1 2 √1 2
]
0.85fc′
fy
Singly Reinforced Beam
INVESTIGATION
Singly Reinforced Beam
DESIGN
Computing MU with given As:
Computing As with given WD & WL:
40 mm → beam
→ column
75 mm → column footing
→ wall footing
→ retaining wall
Singly Reinforced Beam (SRB)
Balance Condition for Doubly
Ä < Ämax (rectangular only)
As < As max (any section)
Äb �㕑 = Äb �㕠 +
Doubly Reinforced Beam (DRB)
Ä > Ämax (rectangular only)
As > As max (any section)
Doubly Reinforced Beam
Investigation
if SRB or DRB:
As′
bd
Ä�㕚ÿ�㕥 �㕑 = 0.75Äb �㕠 +
As �㕚ÿ�㕥 = Ä�㕚ÿ�㕥 �㕑 bd
�㕑
As ′
bd
Doubly Reinforced Beam
INVESTIGATION
Computing MU with given As:
(1st) Compute for ab:
(1st) Compute for a:
(1st) Compute ultimate moment, Mu:
C=T
0.85fc′ ab = As fs
WU = 1.4WD + 1.7WL
WU L2
(for simply supported)
MU =
8
(assume tension steel yields fs=fy)
0.85fc′ ab = As fy
(2nd) Check if assumption is correct:
a = β1 c
c → obtained
Rn =
MU
∅bd2
Ä=
(2nd) Solve for Asmax:
0.85fc′
2R n
[1 2 √1 2
]
fy
0.85fc′
C=T
0.85fc′ a Ā b = As Ā fy
As Ā → obtained
Check:
If fs > fy, tension steel yields; correct a.
If fs < fy, tension steel does not yield;
compute for new a.
(2nd-b) Recomputation:
C=T
0.85fc′ ab = As fs
0.85fc′ β1 cb = As ∙ 600 [
c → obtained
c
]
a = β1 c
a → obtained
(3rd) Solve for Moment Capacity:
a
Mu = ∅(C or T) [d 2 ]
∅(0.85fc′ ab) [d
a
Mu = ∅(As fs ) [d 2 ]
2
2
a
2 ]
2
Ämin f Ä f Ämax
As max = 0.75As Ā
If Ämin < Ä < Ämax, use Ä.
If Ämin > Ä, use Ämin.
If Ä > Ämax, design doubly.
(2nd) Solve for given As
& compare:
(4th) Solve for area of steel
reinforcement, As and required no. of
bars, N:
d−c
or
As = Äbd
As
Äbd
N=
=
2
Ã
Ab
d
4 b
If As > As max
Computing As with given Mu:
(1st) Solve for nominal M1:
(4th) Solve for # of tension bars:
As As1 + As2
=
N=
à 2
Ab
d
4 b
(5th) Solve for fs’:
Äb =
fs ′ = 600 [
a
M1 = (As1 fy ) [d 2 ]
2
(2nd) Solve for nominal M2:
MU
2 M1
M2 =
∅
If As < As max
Solve the given beam
using SRB Investigation
procedure.
Solve the given beam
using DRB Investigation
procedure.
Doubly Reinforced Beam
DESIGN
0.85fc′ β1 600
fy (600 + fy )
Ämax = 0.75Äb
As1 = 0.75Äb ∙ bd
600d
600 + fy
cĀ → obtained
cĀ =
a Ā = β1 cĀ
a Ā → obtained
(3rd) Solve for steel ratio, Ä:
d2c
]
fs = 600 [
c
fs → obtained
Mu =
Thus,
(2nd) Solve for coeff. of resistance, Rn:
a → obtained
d 2 cĀ
]
fs = fy = 600 [
cĀ
(3rd) Solve for As2:
M2 = (As2 fy )[d 2 d′]
As2 → obtained
c 2 d′
]
c
If fs’ > fy, compression steel yields;
As’ = As2.
If fs’ < fy, compression steel does not
yield; Use fs’ to solve for As’.
(6th) Solve for As’:
As ′fs ′ = As2 fy
(7th) Solve for # of compression bars:
N=
As
As′
=
Ab à d 2
4 b
Downloaded by Ace Juntilla (ace.juntilla@gmail.com)
(1st) Compute for a:
Cc + Cs = T
0.85fc′ ab + As ′fs ′ = As fs
(assume tension steel yields fs=fs’=fy)
0.85fc′ ab + As ′fy = As fy
a → obtained
(2nd) Check if assumption is correct:
a = β1 c
c → obtained
d2c
]
c
fs → obtained
fs = 600 [
If fs > fy, tension steel yields; correct a.
If fs < fy, tension steel does not yield;
compute for new a.
c 2 d′
]
c
fs ′ → obtained
fs ′ = 600 [
If fs’ > fy, compression steel yields;
correct a.
If fs’ < fy, compression steel does not
yield; compute for new a.
(2nd-b) Recomputation:
C=T
0.85fc′ ab + As ′fs ′ = As fs
NOTE: Use fs & fs’ as
fs = 600 [
d−c
]
c
c−d′
fs ′ = 600 [
c
c → obtained
a = β1 c
a → obtained
]
(3rd) Solve for Moment Capacity:
a
Mu = ∅Cc [d 2 ] + ∅Cc [d 2 d′]
2
a
Mu = ∅(0.85fc′ ab) [d 2 ]
2
+ ∅(As ′fs ′)[d 2 d′] or
a
Mu = ∅T [d 2 ]
2
a
Mu = ∅(As fs ) [d 2 ]
2
lOMoARcPSD|20985858
Design of Beam Stirrups
T-Beam
Thickness of One-way Slab & Beam
NSCP Provisions for effective flange width:
NSCP Provisions for minimum thickness:
i. Interior Beam:
(1st) Solve for Vu:
NSCP Provisions for
max. stirrups spacing:
ΣFv = 0
Vu = R 2 wu d
wu L
Vu =
2 wu d
2
2Vc =
smax =
Vu = ∅(Vc + Vs )
Vs → obtained
n
smax =
b
Both
Ends
L/10
L/20
L/24
L/28
L/8
L/16
L/18.5
L/21
Factor: [0.4 +
fy
700
] [1.65 2 0.0003�㔌�㕐 ]
(for lightweight concrete only)
Minimum Steel Ratio
For one-way bending:
k → steel ratio
i. fy = 275 MPa,
k = 0.0020
ii. fy = 415 MPa,
k = 0.0018
iii. fy > 415 MPa,
n
k = 0.0018 [
400
fy
]
For two-way bending:
ρ → steel ratio
fyn → steel strength for shear reinforcement
Av → area of shear reinforcement
n → no. of shear legs
Av =
One
End
Beams
d
or 300mm
4
3Av fy
Simple
Support
Slab
iii. & not greater than to:
(4th) Theoretical Spacing:
Vs
NOTE:
d
or 600mm
2
Cantilever
L
bf = bw +
12
s1
bf = bw +
2
bf = bw + 6t f
ii. when Vs > 2Vc,
(3rd) Solve for Vs:
s=
1
√f ′b d
3 c w
smax =
1
Vc = √fc ′bw d
6
dAv fy
s1 s2
bf = bw + +
2
2
bf = bw + 8t f
i. when Vs < 2Vc,
(2nd) Solve for Vc:
ii. exterior Beam:
L
bf =
4
Ämin =
à 2
d ∙n
4
1.4
√fc′
Ämin =
4fy
fy
(choose larger between the 2)
Design of One-way Slab
LONGITUDINAL OR MAIN BARS
(1st) Compute ultimate moment, Mu:
(6th) Compute steel ratio, ρ:
WU = 1.4WD + 1.7WL
WU L2
MU =
8
Ä=
Ämin =
(3rd) Solve for effective depth, d:
db
2
As
bd
(11th) Solve for As:
As = kb⫠ h
NSCP Provision for k:
i. fy = 275 MPa, k = 0.0020
ii. fy = 415 MPa, k = 0.0018
iii. fy > 415 MPa, k = 0.0018 (400/fy)
(7th) Check for minimum steel ratio:
(2nd) Solve for slab thickness, h:
See NSCP Provisions for minimum thickness.
d = h 2 cc 2
TEMPERATURE BARS/
SHRINKAGE BARS
√fc′
1.4
& Ämin =
4fy
fy
(12th) Determine # of req’d temp. bars:
If ρmin < ρ, use ρ.
If ρmin > ρ, use ρmin & recompute As.
N=
(8th) Determine # of req’d main bars:
(4th) Solve for a:
a
N=
Mu = ∅(C) [d 2 ]
2
a
Mu = ∅(0.85fc′ ab) [d 2 ]
2
a → obtained
(13th) Determine spacing of temp. bars:
s=
(9th) Determine spacing of main bars:
s=
(5th) Solve for As:
C=T
0.85fc′ ab = As fy
As → obtained
As
As
=
2
Ã
Ab
d
4 b
As
As
=
Ab à d 2
4 b
b
N
b
N
(14th) Check for max. spacing of temp. bars:
(10th) Check for max. spacing of main bars:
smax = 3h or 450mm
smax = 5h or 450mm
Design of Column
TIED COLUMN
SPIRAL COLUMN
P = PC + PS
P = 0.85fc′ (Ag 2 Ast ) + Ast fy
PN = 0.8P
PU = ∅0.8P ; ∅ = 0.7
PU = (0.7)(0.8)[0.85fc′ (Ag 2 Ast ) + Ast fy ]
PN = 0.85P
PU = ∅0.85P ; ∅ = 0.75
PU = (0.75)(0.85)[0.85fc′ (Ag 2 Ast ) + Ast fy ]
Ä=
Thus,
Ast
Ag
P
Ag =
′
0.85fc (1 2 Ä) + Äfy
0.01Ag < Ast < 0.08Ag
Design of Footing
qA = qS + qC + qsur + qE
qE =
where:
P
A ftg
;
qU =
PU
Aftg
qA → allowable bearing pressure
qS → soil pressure
qC → concrete pressure
qsur → surcharge
qE → effective pressure
qU → ultimate bearing pressure
Ø = 0.85
No. of main bars:
Spacing of bars:
N is based on Pu.
s = 16db
s = 48dt
s = least dimension
Ast
N=
Ab
Äs = 0.45
NOTE: If spacing of main bars < 150mm, use 1 tie per set.
fc′ Ag
volume of spiral
[ 2 1] =
fy Ac
volume of core
Ã
(dsp )2 ∙ Ã(Dc 2dsp ) 4Asp
=
s=4 Ã
Dc Äs
(D )2 ∙ Äs
4 c
WIDE BEAM SHEAR
PUNCHING/DIAGONAL TENSION SHEAR
VU1 = qU (B)(x)
VU2 = PU 2 qU (a + d)(b + d)
VU1 ≤ ∅Vwb = ∅
τwb =
VU1
∅Bd
τwb(allw) =
√fc′
6
√fc′
Bd
6
VU2 ≤ ∅Vpc = ∅
τpc =
VU2
∅bo d
τpc(allw) =
√fc′
b d
3 o
√fc′
3
Downloaded by Ace Juntilla (ace.juntilla@gmail.com)
BENDING MOMENT
x
MU = qU (B)(x) ( )
2
** design of main bars and
temperature bars –
Same as slab.