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 − 4AC
−B ± √B
x=
2A
Sum of Roots:
B
x1 + x2 = −
A
x1 ∙ x2 = +
C
A
(x + y)n
rth term:
th
= nCm x n−m y m
where: m=r-1
cos 𝑎 = cos 𝑏 cos 𝑐 + sin 𝑏 sin 𝑐 cos 𝐴
1
πR3 E
V = AB H =
3
540°
1
A = bh
2
1
A = ab sin C
2
Square:
Case 1: Unequal rate
rate =
work
time
a+b+c
2
s=
 Clock Problems
Trapezoid
θ=
11M − 60H
2
Ex-circleIn-circle
+ if M is ahead of H
- if M is behind of H
1 1 1 1
= + +
𝑟 𝑟1 𝑟2 𝑟3
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.
Ellipse
a2 + b2
2
A = πab C = 2π√
1
Area = n ∙ R2 sinβ
2
1
Area = n ∙ ah
2
β=
360°
n
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
Inscribed Circle:
Cyclic Quadrilateral: (sum of opposite angles=180°)
AT = rs
A = √(s − a)(s − b)(s − c)(s − d)
Escribed Circle:
Ptolemy’s Theorem is applicable:
AT = R a (s − a)
AT = R b (s − b)
AT = R c (s − c)
ac + bd = d1 d2
diameter =
opposite side
sine of angle
a
b
c
=
=
sin A sin B sin C
s=
a+b+c+d
2
Non-cyclic Quadrilateral:
A = √(s − a)(s − b)(s − c)(s − d) − abcd cos 2
Pappus Theorem
Pappus Theorem 1:
Prism or Cylinder
Pointed Solid
SA = L ∙ 2πR
V = AB H = AX L
LA = PB H = Px L
1
V = AB H
3
v
Pappus Theorem 2:
Special Solids
Truncated Prism or Cylinder:
Sphere:
4
V = πR3
3
LA = 4πR2
Frustum of Cone or Pyramid:
Spheroid:
H
(A + A2 + √A1 A2 )
3 1
AB/PB → Perimeter or Area of base
H → Height & L → slant height
AX/PX → Perimeter or Area of crosssection perpendicular to slant height
Spherical Solids
V = AB Have
LA = PB Have
V=
H
V = (A1 + 4AM + A2 )
6
1
2
Spherical Lune:
Spherical Wedge:
Alune 4πR2
=
θrad
2π
3
Vwedge 3 πR
=
θrad
2π
4
2
Vwedge = θR3
Spherical Sector:
1
V = Azone R
3
2
V = πR2 h
3
Spherical Segment:
For one base:
about major axis
4
V = πaab
3
a2 + a2 + b2
]
LA = 4π [
3
LA = PB L
Azone = 2πRh
V = πabb
3
a2 + b2 + b2
]
LA = 4π [
3
Oblate Spheroid:
LA = πrL
Spherical Zone:
4
Prolate Spheroid:
Prismatoid:
Reg. Pyramid
3
V = πabc
3
a2 + b2 + c 2
]
LA = 4π [
3
1
V = πh2 (3R − h)
3
For two bases:
1
about minor axis
V = πh(3a2 + 3b2 + h2 )
6
ε
2
Right Circ. Cone
Alune = 2θR2
4
EULER LINE
- the line that would pass through the
orthocenter, circumcenter, and centroid of
the triangle.
Area = n ∙ ATRIANGLE
δ = 180° − γ
abc
AT =
4R
NOTE: It is also used to locate centroid of an area.
CENTROID
- the point of intersection of the medians of
the triangle.
Deflection Angle, δ:
Circumscribing Circle:
V = A ∙ 2πR
ORTHOCENTER
- the point of intersection of the altitudes of
the triangle.
(n − 2)180°
n
General Quadrilateral
Triangle-Circle Relationship
d=
CIRCUMCENTER
- the center of the circumscribing circle
(circumcircle) & the point of intersection of
the perpendicular bisectors of the triangle.
A = ah
A = a2 sin θ
1
A = d1 d2
2
A1 n
ma2 + nb 2
=
;w = √
A2 m
m+n
1 knot =
1 nautical mile
per hour
Polygon Names
Rhombus:
1
A = (a + b)h
2
1 statute mile =
5280 feet
Central Angle, β:
Parallelogram:
A = √s(s − a)(s − b)(s − c)
Case 2: Equal rate
→ usually in project management
→ express given to man-days or man-hours
γ=
Rectangle:
A = bh
A = ab sin θ
1
A = d1 d2 sin θ
2
1 nautical mile =
6080 feet
Interior Angle, ɤ:
A = s2
A = bh
P = 4s
P = 2a + 2b
d = √2s d = √b 2 + h2
1 sin B sin C
A = a2
2
sin A
 Work Problems
1 minute of arc =
1 nautical mile
n-sided Polygon
2
 Age Problems
→ underline specific time conditions
=0
= vt
180°
sin 2A = 2 sin A cos A
cos 2A = cos 2 A − sin2 A
cos 2A = 2 cos 2 A − 1
cos 2A = 1 − 2 sin2 A
2 tan A
# of diagonals:
tan 2A =
n
1 − tan2 A
d = (n − 3)
Common Quadrilateral
→s
Spherical Polygon:
πR2 E E = spherical excess
AB =
E = (A+B+C+D…) – (n-2)180°
Spherical Pyramid:
Triangle
→a
cos 𝐴 = − cos 𝐵 cos 𝐶 + sin 𝐵 sin 𝐶 cos 𝑎
Double Angle Identities:
Worded Problems Tips
 Motion Problems
Cosine Law for angles:
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
r = a 2 /a1 = a 3 /a2
a n = a1 r n−1
a n = a x r n−x
1 − rn
Sn = a1
1−r
a1
S∞ =
1−r
Form:
Cosine Law for sides:
Sum & Diff of Angles Identities:
Geometric Progression:
Binomial Theorem
r
d = a 2 − a1 = a 3 − a 2
a n = a1 + (n − 1)d
a n = a x + (n − x)d
n
Sn = (a1 + a n )
2
Harmonic Progression:
- reciprocal of arithmetic
progression
Product of Roots:
sin 𝑎 sin 𝑏 sin 𝑎
=
=
sin 𝐴 sin 𝐵 sin 𝐴
sin A + cos A = 1
1 + tan2 A = sec 2 A
1 + cot 2 A = csc 2 A
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
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
General Equation:
Ax 2 + Bxy + Cy 2 + Dx + Ey + F = 0
Based on discriminant:
B 2 − 4AC = 0 ∴ parabola
B 2 − 4AC < 0 ∴ ellipse
B 2 − 4AC > 0 ∴ hyperbola
Based on eccentricity, e=f/d:
𝑒 = 0 ∴ circle
𝑒 = 1 ∴ parabola
𝑒 < 1 ∴ ellipse
𝑒 > 1 ∴ hyperbola
Distance from a point to another point:
d = √(y2 − y1 )2 + (x2 − x1 )2
Point-slope form:
Distance from a point to a line:
General Equation:
General Equation:
Ax 2 + Cy 2 + Dx + Ey + F = 0
Ax 2 − Cy 2 + Dx + Ey + F = 0
y − y1
m=
x − x1
x 2 + y 2 + Dx + Ey + F = 0
Standard Equation:
2
(x − h) + (y − k)2 = r 2
Two-point form:
y2 − y1 y − y2
=
x 2 − x1 x − x 2
|C1 − C2 |
√A2 + B 2
tan θ =
1 revolution
= 2π rad
= 360˚
= 400 grads
= 6400 mills
2
H = a√
3
√2
V=a
12
Elements:
Eccentricity, e:
e=
3
=1
LR = 4a
Location of foci, c:
c 2 = a2 − b2
Length of LR:
2b2
LR =
a
Versed cosine:
covers A = 1 − sin A
exsec A = sec A − 1
 Inflation:
 Rate of return:
annual net profit
RR =
capital
𝑖f = 𝑖 + f + 𝑖f
 Break-even analysis:
Annual net profit
= savings – expenses
– depreciation (sinking fund)
1
RR
𝑦"
(+) minima
(-) maxima
Integral Calculus-The Cardioid
r = a(1 − cos θ)
r = a(1 + cos θ)
CALTECH:
Mode 3 2
x
y
(time)
(BV)
0
FC
n
SV
(1 + i)n − 1 −1
d = (FC − SV) [
]
𝑖
m
(1 + i) − 1
Dm = d [
]
𝑖
n−m+1
dm = (FC − SV) [
]
∑ years
∑nn−m+1 x
Dm = (FC − SV) [
]
∑n1 x
 Declining Balance (Matheson):
BVm = FC(1 − k)m
SV = FC(1 − k)n k → obtained
Dm = FC − BVm
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
 Perpetuity:
P=
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
k = 2/n k → obtained
Dm = FC − BVm
 Service Output Method:
A
= F(1 + 𝑖)−n
𝑖
 Capitalized Cost:
C = FC +
OM
RC − SV
+
𝑖
(1 + 𝑖)n − 1
AC = C ∙ 𝑖
AC = FC ∙ 𝑖 + OM +
where:
C → capitalized cost
FC → first cost
OM → annual operation
or maintenance cost
RC → replacement cost
SV → salvage cost
AC → annual cost
(RC − SV)𝑖
(1 + i)n − 1
 Single-payment-compound-amount factor:
n
(F/P, 𝑖, n) = (1 + 𝑖)
 Single-payment-present-worth factor:
−n
(P/F, 𝑖, n) = (1 + 𝑖)
 Equal-payment-series-compound-amount factor:
CALTECH:
Mode 3 6
x
y
(time)
(BV)
0
FC
n
SV
 Double Declining Balance:
BVm = FC(1 − k)m
FC − SV
Qn
D = dQ m
(1 + 𝑖)n − 1
]
𝑖
′
 Sinking Fund:
d=
 Annuity:
(1 + 𝑖)n − 1
P = A[
]
𝑖(1 + 𝑖)n
 Sum-of-the-Years-Digit (SYD):
𝑑2 𝑦
= y" = 0
𝑑𝑥 2
F = Pe
ER = er − 1
′
BVm = FC − Dm
ρ=
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
 Continuous Compounding Interest:
rt
F = A[
Depreciation
Radius of curvature:
3
[1 + (y′)2 ]2
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
y − k = ±m(x − h)
c
e=
a
F = P(1 + 𝑖)
r mt
F = P (1 + )
m
I
r m
ER = = (1 − ) − 1
P
m
1 − cos A
2
FC − SV
d=
n
Dm = d(m)
Eccentricity, e:
Eq’n of asymptote:
 Compound Interest:
n
Half versed sine:
 Straight-Line:
c 2 = a2 + b2
I = P𝑖n
F = P(1 + 𝑖n)
vers A = 1 − cos A
Exsecant:
Same as ellipse:
Length of LR,
Loc. of directrix, d
Eccentricity, e
a
d=
e
 Simple Interest:
Versed sine:
hav A =
Location of foci, c:
Loc. of directrix, d:
Engineering Economy
Unit Circle
RP =
Point of inflection:
A = 1.5πa2
P = 8a
r = a(1 − sin θ)
r = a(1 + sin θ)
dd
Length of latus
rectum, LR:
cost = revenue
Maxima & Minima (Critical Points):
𝑑𝑦
= y′ = 0
𝑑𝑥
df
Elements:
Elements:
2
Differential Calculus
3
(y′)2 ]2
(y − k)2 (x − h)2
−
=1
a2
b2
m2 − m1
1 + m1 m2
General Equation:
2
𝑥 2 → 𝑥𝑥1
𝑦 2 → 𝑦𝑦1
𝑥 + 𝑥1
𝑥→
2
𝑦 + 𝑦1
𝑦→
2
𝑥𝑦1 + 𝑦𝑥1
𝑥𝑦 →
2
y"
(x − h)2 (y − k)2
+
=1
b2
a2
- the locus of point that moves such that it is always equidistant from a
fixed point (focus) and a fixed line (directrix).
SA = a √3
Curvature:
(x − h)2 (y − k)2
−
=1
a2
b2
Parabola
In the equation of the conic
equation, replace:
[1 +
d=
Standard Equation:
(x − h)2 (y − k)2
+
=1
a2
b2
Angle between two lines:
x y
+ =1
a b
Line Tangent to Conic Section
To find the equation of a line
tangent to a conic section at a
given point P(x1, y1):
Standard Equation:
√A2 + B 2
Distance of two parallel lines:
Point-slope form:
Tetrahedron
k=
d=
|Ax + By + C|
(x − h) = ±4a(y − k)
(y − k)2 = ±4a(x − h)
General Equation:
- the locus of point that moves such
that the difference of its distances
from two fixed points called the foci
is constant.
y = mx + b
Slope-intercept form:
Standard Equation:
2
- the locus of point that moves such
that its distance from a fixed point
called the center is constant.
Hyperbola
- the locus of point that moves such
that the sum of its distances from
two fixed points called the foci is
constant.
y + Dx + Ey + F = 0
x 2 + Dx + Ey + F = 0
Circle
Ellipse
′
(1 + 𝑖)n − 1
(F/A, 𝑖, n) = [
]
𝑖
 Equal-payment-sinking-fund factor:
′
−1
(1 + 𝑖)n − 1
(A/F, 𝑖, n) = [
]
𝑖
 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
(1 + 𝑖)n − 1
(P/A, 𝑖, n) = [
]
𝑖(1 + 𝑖)n
 Equal-payment-series-capital-recovery factor:
′
(1 + 𝑖)n − 1
(A/P, 𝑖, n) = [
]
𝑖(1 + 𝑖)n
−1
Statistics
Fractiles
Transportation Engineering
Traffic Accident Analysis
Measure of Natural Tendency
 Range
= 𝑙𝑎𝑟𝑔𝑒𝑠𝑡 𝑑𝑎𝑡𝑢𝑚 − 𝑠𝑚𝑎𝑙𝑙𝑒𝑠𝑡 𝑑𝑎𝑡𝑢𝑚
Design of Horizontal Curve
 Mean, x̅, μ → average
→ Mode Stat 1-var
 Coefficient of Range
 Accident rate for 100 million
vehicles per miles of travel in a
segment of a highway:
→ Shift Mode ▼s Stat Frequency? on
𝑙𝑎𝑟𝑔𝑒𝑠𝑡 𝑑𝑎𝑡𝑢𝑚 − 𝑠𝑚𝑎𝑙𝑙𝑒𝑠𝑡 𝑑𝑎𝑡𝑢𝑚
=
𝑙𝑎𝑟𝑔𝑒𝑠𝑡 𝑑𝑎𝑡𝑢𝑚 + 𝑠𝑚𝑎𝑙𝑙𝑒𝑠𝑡 𝑑𝑎𝑡𝑢𝑚
→ Input
→ AC Shift 1 var x̅
 Quartiles
 Median, Me → middle no.
when n is even
1
n+1
2
1 n
n
= [( ) + ( + 1)]
2 2
2
Q1 = n
Me th =
Me
th
4
2
3
Q2 = n
Q3 = n
4
4
when n is odd
Q1 =
 Mode, Mo → most frequent
1
1
1
(n + 1) ; Q1 = (n + 1) ; Q1 = (n + 1)
4
4
4
 Interquartile Range, IQR
Standard Deviation
= 𝑙𝑎𝑟𝑔𝑒𝑠𝑡 𝑞𝑢𝑎𝑟𝑡𝑖𝑙𝑒 − 𝑠𝑚𝑎𝑙𝑙𝑒𝑠𝑡 𝑞𝑢𝑎𝑟𝑡𝑖𝑙𝑒
= Q3 − Q1
 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
𝑙𝑎𝑟𝑔𝑒𝑠𝑡 𝑞𝑢𝑎𝑟𝑡𝑖𝑙𝑒 − 𝑠𝑚𝑎𝑙𝑙𝑒𝑠𝑡 𝑞𝑢𝑎𝑟𝑡𝑖𝑙𝑒
=
𝑙𝑎𝑟𝑔𝑒𝑠𝑡 𝑞𝑢𝑎𝑟𝑡𝑖𝑙𝑒 + 𝑠𝑚𝑎𝑙𝑙𝑒𝑠𝑡 𝑞𝑢𝑎𝑟𝑡𝑖𝑙𝑒
Q − Q1
= 3
Q3 + Q1
 Outlier
→ extremely high or low data higher than
or lower than the following limits:
NOTE:
If not specified whether population/sample
in a given problem, look for POPULATION.
Q1 − 1.5IQR > x
Q 3 + 1.5IQR < x
Coefficient of Linear Correlation
or Pearson’s r
 Decile or Percentile
→ AC Shift 1 Reg r
R=
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
 Centrifugal ratio or impact factor
2
Impact factor =
v
gR
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
t=√
3W
4f
(at the center)
 Flexible pavement
 Population standard deviation
 Z-score or
standard score
or variate
 standard deviation = σ
 variance = σ2
→ Mode Stat
→ AC Shift 1 Distr
left of z → P(
x−μ
z=
σ
 relative variability = σ/x
Mean/Average Deviation
right of z → R(
bet. z & axis → Q(
→ Input
x → no. of observations
μ → mean value, x
̅
σ → standard deviation
 Mean/average value
b
1
mv =
∫ f(x)dx
b−a a
b
1
RMS = √
∫ f(x)2 dx
b−a a
Walli’s Formula
 Binomial Probability Distribution
x n−x
∫ cosm θ sinn θ dθ =
π
2
P(x) = C(n, x) p q
0
where:
p → success
q → failure
f → fatal
i → injury
p → property damage
∑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
q → rate of flow in vehicles/hour
k → density in vehicles/km
uS → space mean speed in kph
 Thickness of pavement in terms
of expansion pressure
 Minimum time headway (hrs)
= 1/q
expansion pressure
pavement density
Es
SF = √
Ep
3
q = kUs
 Spacing of vehicles (km)
= 1/k
 Peak hour factor (PHF)
= q/qmax
s
[(m − 1)(m − 3)(m − 5) … (1 or 2)][(n − 1)(n − 3)(n − 5) … (1 or 2)]
∙α
(m + n)(m + n − 2)(m + n − 4) … (1 or 2)
α = π/2 for m and n are both even
α =1 otherwise
 Geometric Probability Distribution
x−1
)
Tip to remember:
Fibonacci Numbers
 Poisson Probability Distribution
x −μ
μ e
x!
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
Period
2π/B
2π/B
π/B
f∙i
f∙i∙p
f1 → allow bearing pressure of subgrade
r → radius of circular area of contact
between wheel load & pavement
NOTE:
Function
y = A sin (Bx + C)
y = A cos (Bx + C)
y = A tan (Bx + C)
SR =
ES → modulus of elasticity of subgrade
EP→ modulus of elasticity of pavement
Discrete Probability Distributions
P(x) = p(q
 Severity ratio, SR:
 Rate of flow:
 Stiffness factor of pavement
P(x ≥ a) = e−λa
P(x ≤ a) = 1 − e−λa
P(a ≤ x ≤ b) = e−λa − e−λb
 Mean value
A → no. of accidents during period of analysis
ADT → average daily traffic entering all legs
N → time period in years
W
t=√
−r
𝜋f1
t=
Exponential Distribution
A (1,000,000)
ADT ∙ N ∙ 365
Us =
(at the edge)
-1 ≤ r ≤ +1; otherwise erroneous
Variance
 Accident rate per million entering
vehicles in an intersection:
 Spacing mean speed, US:
3W
t=√
f
t → thickness of pavement
W → wheel load
f → allow tensile stress of concrete
Normal Distribution
A (100,000,000)
ADT ∙ N ∙ 365 ∙ L
A → no. of accidents during period of analysis
ADT → average daily traffic
N → time period in years
L → length of segment in miles
R=
R → minimum radius of curvature
v → design speed in m/s
g → 9.82 m/s2
3W
t=√
2f
NOTE:
P(x) =
v
g(e + f)
 Rigid pavement with dowels
m
im =
(n)
10 or 100
→ Mode Stat A+Bx
→ Input
R=
Power to move a vehicle
 Coefficient of IQR
 Quartile Deviation (semi-IQR) = IQR/2
→ Shift Mode ▼ Stat Frequency? on
→ Input
→ AC Shift 1 var sx
 Minimum radius of curvature
2
Amplitude
A
A
A
1
√5
n
[(
n
1 + √5
1 − √5
) −(
) ]
2
2
x = r cos θ
y = r sin θ
r = x2 + y2
y
θ = tan−1
x
𝑥2 − 𝑥 − 1 = 0
Mode Eqn 5
𝑥=
1 ± √5
2
measure
too long add
too short subtract
Measurement
Corrections
Due to temperature:
Probable Errors
C = αL(T2 − T1 )
Probable Error (single):
(add/subtract); measured length
(P2 − P1 )L
C=
EA
(subtract only); unsupported length
w 2 L3
24P 2
CD = MD (1 −
∑(x − x̅)
n−1
∑(x − x̅)
Em =
= 0.6745√
n(n − 1)
√n
E
Proportionalities of weight, w:
Due to slope:
(subtract only); measured length
𝑤∝
Normal Tension:
0.204W√AE
1
𝐸2
𝑤∝
1
𝑑
𝑤∝𝑛
Area of Closed Traverse
√PN − P
Error of Closure:
L
H = (g1 + g 2 )
8
L 2
x 2 ( 2)
=
L
y
H 1
Error of Closure
Perimeter
1 acre =
4047 m2
from South
D2
(h − h2 ) − 0.067D1 D2
D1 + D2 1
Stadia Measurement
Leveling
Horizontal:
Elev𝐵 = Elev𝐴 + 𝐵𝑆 − 𝐹𝑆
D = d + (f + c)
𝑓
D = ( )s +C
𝑖
D = Ks + C
Inclined Upward:
Inclined:
Total Error:
Reduction to
Sea Level
CD
MD
=
R
R+h
error/setup = −eBS + eFS
Subtense Bar
Inclined Downward:
error/setup = +eBS − eFS
D = cot
θ
2
eT = error/setup ∙ no. of setups
Double Meridian Distance Method DMD
DMD𝑓𝑖𝑟𝑠𝑡 = Dep𝑓𝑖𝑟𝑠𝑡
DMD𝑛 = DMD𝑛−1 + Dep𝑛−1 + Dep𝑛
DMD𝑙𝑎𝑠𝑡 = −Dep𝑙𝑎𝑠𝑡
2A = Σ(DMD ∙ Lat)
d
[h + hn + 2Σh]
2 1
Double Parallel Distance Method DPD
d
A = [h1 + hn + 2Σh𝑜𝑑𝑑 + 4Σh𝑒𝑣𝑒𝑛 ]
3
Relative Error/Precision:
=
h = h2 +
Simpson’s 1/3 Rule:
= √ΣL2 + ΣD2
Azimuth
hcr = 0.067K 2
Trapezoidal Rule:
A=
Symmetrical:
e
)
TL
Effect of Curvature & Refraction
Area of Irregular Boundaries
Lat = L cos α
Dep = L sin α
Parabolic Curves
e
)
TL
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 − h2
PN =
CD = MD (1 +
Probable Error (mean):
Due to sag:
C=
E = 0.6745√
too long
too short
(add/subtract); measured length
Due to pull:
lay-out
subtract
add
Note: n must be odd
Simple, Compound & Reverse Curves
DPD𝑓𝑖𝑟𝑠𝑡 = Lat𝑓𝑖𝑟𝑠𝑡
DPD𝑛 = DPD𝑛−1 + Lat 𝑛−1 + Lat 𝑛
DPD𝑙𝑎𝑠𝑡 = −Lat 𝑙𝑎𝑠𝑡
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
𝑑𝐿 0 𝑑𝑅
±𝑓𝐿 ±𝑓 ±𝑓𝑅
A=
f
w
(d + dR ) + (fL + fR )
2 L
4
T = R tan
I
m = R [1 − cos ]
L = 2R sin
L3
6RLs
L
(c − c2 )(d1 − d2 )
12 1
VP = Ve − Cp
L5
I
Y=L−
2
π
Lc = RI ∙
180°
20 2πR
=
D
360°
1145.916
R=
D
Prismoidal Correction:
40R2 Ls
2
Ls
I
+ (R + p) tan
2
2
I
Es = (R + p) sec − R
2
Ts =
Ls =
Volume (Truncated):
0.036k 3
R
0.0079k 2
R
D
L
=
DC Ls
Σh
= A( )
n
e=
A
(Σh1 + 2Σh2 + 3Σh3 + 4Σh4 )
n
Stopping Sight Distance
Parabolic Summit Curve
v2
S = vt +
2g(f ± G)
a = g(f ± G) (deceleration)
v
(breaking time)
tb =
g(f ± G)
f
Eff =
(100)
fave
L>S
v → speed in m/s
t → perception-reaction time
f → coefficient of friction
G → grade/slope of road
x=
2
L
VP = (A1 + 4Am + A2 )
6
VT =
θ
Ls 2
; p=
3
24R
2
Volume (Prismoidal):
VT = ABase ∙ Have
i=
I
E = R [sec − 1]
L
Ve = (A1 + A2 )
2
CP =
L2 180°
∙
2RLs π
2
I
Volume (End Area):
θ=
L=
A(S)2
200(√h1 + √h2 )
2
L<S
200(√h1 + √h2 )
L = 2(S) −
A
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
2
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
A(S)2
L=
800H
L<S
L<S
122 + 3.5S
L = 2(S) −
A
800H
L = 2(S) −
A
R=
A → algebraic difference
of grades, in percent
L → length of sag curve
S → sight distance
A → algebraic difference of
grades, in percent
L → length of sag curve
L<S
L=
A(K)2
395
H= C−
h1 + h2
2
For passengers comfort,
where K is speed in KPH
R=
S2
8M
L(2S − L)
8M
L → length of horizontal
curve
S → sight distance
R → radius of the curve
M → clearance from the
centerline of the road
Properties of Fluids
Dams
Pressure
s Mg
W=
p𝑎𝑏𝑠 = p𝑔𝑎𝑔𝑒 + p𝑎𝑡𝑚
W
ɤ=
V
p = ɤh
s. g.1
h2 =
h
s. g.2 1
M
; ρ=
V
pg
ɤ = ρg =
RT
V 1
s. v. = =
M ρ
ɤ
ρ
s. g. =
=
ɤ𝑤 ρ𝑤
∆P
1
; β=
∆V
EB
V
𝑑𝑦 FT
μ=τ
=
𝑑𝑉 L2
Ig
Aӯ
e=
h̅ = ӯ (for vertical only)
U2 = (h1 − h2 )ɤB
2
RM
OM
& FS𝑆 =
ω2 x
tan θ =
g
z1 +
2 2
2
y=
ω x
2g
V=
1 2
πr h
2
2
r
x
=
h
y
;
Stresses/Hoops
pD
2t
2T
s=
pD
St =
H. L. = f
P1 v1 2
P2 v2 2
z1 + +
+ HA = z2 + +
+ H. L.
ɤ
2g
ɤ
2g
H. L.T = H. L.1 + H. L.2 +. . . +H. L.n
Fluid Flow
Most Efficient Sections
Q T = Q1 = Q 2 = Q n
Q = Av
Rectangular:
Q → discharge
→ flow rate
→ weight flux
b = 2d
d
R=
2
Q T = Q1 + Q 2 +. . . +Q n
Constant Head Orifice
Falling Head Orifice
Without headloss:
Time to remove water from h1 to h2 with constant cross-section:
v = √2gh
With headloss:
v = Cv √2gh
t=
H. L. =
v2 1
[
− 1]
2g Cv 2
H. L. = ∆H[1 − Cv 2 ]
y=
x
CAo √2g
(√h1 − √h2 )
Time to remove water from h1 to h2 with varying cross-section:
h1
h2
Q = CA o √2gh
C = Cc C v
a
Cc =
A
v
Cv =
vt
2As
t=∫
As dh
CAo √2gh
Time in which water surfaces of two tanks will reach same elevation:
t=
(As1 )(As2 )
(√h1 − √h2 )
CAo √2g (As1 + As2 )
2
Force on Curve Vane/Blade:
Force on the Jet
(at right angle):
∑ Fx = ρQ(v2x − v1x )
F = ρQv
∑ Fy = ρQ(v2y − v1y )
Force on Pipe’s Bend & Reducer:
(same as on Curve Vane/Blade)
BF = ɤ𝑤 V𝑑 Vbel
Celerity (velocity of sound)
L v2
D 2g
Manning’s Formula:
H. L. =
10.29 n2 L Q2
D16/3
Hazen William’s Formula:
1 atm
= 101.325 KPa
= 2166 psf
= 14.7 psi
= 760 mmHg
= 29.9 inHg
EB
c=√
ρw
EB
c=√
E D
ρw (1 + B )
Et
Water Hammer
∆Pmax = ρcv
tc =
2L
c
A. TIME of closure:
rapid/instantaneous
∆P = ∆Pmax
Slow Closure
tc
∆P = ∆Pmax (
)
t actual
B. TYPE of closure:
Partial Closure (vf ≠ 0)
∆P = ρc(vi − vf )
Total Closure (vf = 0)
∆P = ρcvi
Open Channel
x = y1 + y2
Specific Energy:
2
v
E=
+d
2g
d
R=
2
Triangular:
v = C√RS
b = 2d
A = d2
θ = 90°
Theoretically:
Semi-circular:
Kutter Formula:
C=√
8g
Manning Formula:
C=
1 1/6
R
n
Bazin Formula:
C=
87
m
1+
√R
f
1
0.000155
+ 23 +
n
S
C=
n
0.000155
1+
(23 +
)
S
√R
d = r (full)
r
R=
2
Circular:
(rigid pipes)
(non-rigid pipes)
0.0826 f L Q
D5
Trapezoidal:
Q max if d = 0.94D
Vmax if d = 0.81D
Hydrodynamics
2
4Cv 2 h
H. L. =
sg m
A
sg l tot
sg m
=
V
sg l tot
Abel =
BF = W
10.64 L Q1.85
H. L. = 1.85 4.87
C
D
Pump → Output & Turbine → Input
volume flow rate → m3/s
weight flow rate → N/s
mass flow rate → kg/s
vs
I
=
VD sin θ VD
2
output
QɤE
efficiency =
; HP =
input
746
H. L.T = H. L.1 = H. L.2 = H. L.n
MB𝑂 =
Darcy Weisbach Eq’n:
P1 v1 2
P2 v2 2
z1 + +
− HE = z2 + +
+ H. L.
ɤ
2g
ɤ
2g
Parallel Connection:
B2
tan2 θ
[1 +
]
12D
2
Major Losses in Pipes
with pump:
Series Connection:
MB𝑂 =
Buoyancy
St = tensile stress
p = unit pressure
D = inside diameter
t = thickness of wall
s = spacing of hoops
T = tensile force
P1 v1 2
P2 v2 2
+
= z2 + +
+ H. L.
ɤ
2g
ɤ
2g
Series-Parallel Pipes
Use (-) if G is above BO and (+) if G is below BO.
Note that M is always above BO.
R𝑥
with turbine:
π
1 rpm =
rad/sec
30
MG = MB𝑂 ± GB𝑂
RM or OM = Wx
= W(MG sin θ)
z = elevation head; P/ɤ = pressure head; v2/2g = velocity head
a
tan θ =
g
MG = metacentric height
μR𝑦
Bernoulli’s Energy Theorem
Rotation:
a
p = ɤh (1 ± )
g
2
NOTE:
ħ = vertical distance from cg of
submerged surface to liquid surface
Horizontal Motion:
Vertical Motion:
;
B
𝑒 = | − x̅|
2
R𝑦
B
6𝑒
𝑒< ; q=−
[1 ± ]
6
B
B
2R 𝑦
B
𝑒> ; q=
6
3x̅
R𝑦
B
𝑒= ; q=−
6
B
2R 𝑦
𝑒 = 0; q =
B
F𝑣 = ɤV
Relative Equilibrium of Fluids
ah
tan θ =
g ± av
FS𝑂 =
Fℎ = ɤh̅A
F = √ Fℎ + F𝑣
2
1
Rx̅ = RM − OM
ɤIg sin θ
F
On curved surfaces:
pd
σ=
4
4σcosθ
h=
ɤd
F2 = ɤAh2 = ɤh2 2
h2
RM = W1 (X1 ) + W2 (X2 )+. . . +W𝑛 (X𝑛 ) + F2 ( )
3
h
1
2
OM = F1 ( ) + U1 ( B) + U2 ( B)
3
2
3
F = ɤh̅A
2
1
;
U1 = ɤh2 B
On plane surfaces:
μ L2
=
ρ T
Inclined Motion:
2
Hydrostatic Forces
EB = −
υ=
F1 = ɤAh1 = ɤh1 2
h𝑤 = s. g.1 h1
e=
Stability of Floating Bodies
1
TRAPEZOIDAL:
For minimum seepage:
b = 4d tan
θ
2
If C is not given, use Manning’s in V:
v=
1 2/3 1/2
R S
n
Volume
Vv
e=
Vs
Ww
ω=
Ws
Vv
n=
V
Vw
S=
Vv
Gs =
When S=0:
Relative Compaction:
ɤd
g = Gs (1 − n)
W
V
WS
ɤd =
V
ɤ=
0<e<∞
n
e=
1−n
e
n=
1+e
(Gs − 1)ɤw
1+e
Gs ɤw
=
1 + Gs ω
ɤsub =
ɤzav
Permeability
PI = LL − PL
ɤd 𝑚𝑎𝑥
Dr (%)
0 – 20
20 – 40
40 – 70
70 – 85
85 – 100
∆h
v
v = ki ; i =
; v𝑠 =
L
n
Q = vA = kiA
qu
PI
; St = und
μ
q u rem
μ = % passing 0.002mm
Cu =
k=
aL h1
𝑙𝑛
At h2
k eq =
Casagrande:
k = c ∙ D10 2
k = 1.4e2 k 0.85
Kozeny-Carman: Samarasinhe:
k = C1 ∙
k eq =
n
2
e
1+e
k = C3 ∙
e
1+e
Stresses in Soil
NOTE:
Quick
condition:
Effective Stress/
Intergranular Stress:
pE = 0
pE = pT − pw
C
hcr =
eD10
pw = ɤw hw
h1 h2
h
+ +. . . + n
k1 k 2
kn
Flow Net / Seepage
Isotropic soil:
q = kH
Capillary Rise:
Pore Water Pressure/
Neutral Stress:
Total Stress:
Nf
Nd
ACTIVE PRESSURE:
Nf
q = √k x k z H
Nd
AT REST:
k o = 1 − sin Ø
cos β − √cos 2 β − cos 2 Ø
cos β + √cos 2 β − cos 2 Ø
For Horizontal:
1 − sin Ø
ka =
1 + sin Ø
If there is angle of friction α bet. wall and soil:
2
cos Ø
sin(Ø + α) sin Ø
]
cos α
2
k P = cos β
cos 2
Ø
cos β − √cos 2 β − cos 2 Ø
For Horizontal:
kP =
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
 Cohesive soil:
For Inclined:
−
1 + sin Ø
1 − sin Ø
r
x + σ3 + r
c
tan Ø =
x
sin Ø =
 Unconsolidatedundrained test:
If there is angle of friction α bet. wall and soil:
2
c=r
kP =
 Unconfined
compression test:
cos Ø
cos α [1 − √
9
Ө → angle of failure in shear
Ø → angle of internal friction/shearing resistance
C → cohesion of soil
r
sin Ø =
σ3 + r
1
pP = k P ɤH 2 + 2cH√k P
2
cos β +
4 56 7 8
Nf → no. of flow channels [e.g. 4]
Nd → no. of potential drops [e.g. 10]
 Normally consolidated:
PASSIVE PRESSURE:
√cos 2 β
3
Shear Strength of Soil
θ = 45° +
cos α [1 + √
3
Compression Index, CC:
Swell Index, CS:
Cc = 0.009(LL − 10%)
e − e′
Cc =
∆P + Po
𝑙𝑜𝑔
Po
1
Cs = Cc
5
e − e′
H (for one layer only)
1+e
Cc H
∆P + Po
S=
𝑙𝑜𝑔
1+e
Po
S=
4
2
D75
D25
10
With Pre-consolidation pressure, Pc:
when (△P+Po) < Pc:
For Inclined:
ka =
1
D60 ∙ D10
So = √
For normally consolidated clay:
2
Equipotential line ----
Non-Isotropic soil:
1
pa = k a ɤH 2 − 2cH√k a
2
k a = cos β
1
Flow line ----
pT = ɤ1 h1 + ɤ2 h2 +. . . +ɤn hn
Lateral Earth Pressure
r
Q 𝑙𝑛 1
r2
k=
2πt(h1 − h2 )
H
Cc =
Compressibility of Soil
Confined:
for Perpendicular flow:
Hazen Formula
D60
D10
Sorting
Coefficient:
3
1
1
Sn = 1.7√
+
+
2
2
(D50 )
(D20 )
(D10 )2
r1
r2
k=
π(h1 2 − h2 2 )
Q 𝑙𝑛
h1 k1 + h2 k 2 +. . . +hn k n
H
Non-plastic
Slightly plastic
Low plasticity
Medium plasticity
High plasticity
Very High plastic
Suitability Number:
Unconfined:
for Parallel flow:
Class
AC < 0.7
Inactive
0.7 < AC < 1.2 Normal
AC > 1.2
Active
Description
0
1-5
5-10
10-20
20-40
>40
Coeff. of Gradation
or Curvature:
(D30 )2
Uniformity
Coefficient:
Pumping Test:
Falling/Variable Head Test:
Ac
PI
Sieve Analysis
Constant Head Test:
QL
k=
Aht
State
LI < 0
Semisolid
0 < LI < 1 Plastic
LI > 1
Liquid
LL − ω
CI =
LL − PI
Description
Very Loose
Loose
Medium Dense
Dense
Very Dense
1
− SL
SR
LI
SI = PL − SL
Ac =
1
GI = (F − 35)[0.2 + 0.005(LL − 40)]
+0.01(F − 15)(PI − 10)
ω − PL
LL − PL
LI =
1
1
−
ɤd 𝑚𝑖𝑛 ɤd
Dr =
1
1
−
ɤd 𝑚𝑖𝑛 ɤd 𝑚𝑎𝑥
Stratified Soil
G𝑠 =
Atterberg Limits
Relative Density/
Density Index:
e𝑚𝑎𝑥 − e
Dr =
e𝑚𝑎𝑥 − e𝑚𝑖𝑛
ɤsub = ɤsat − ɤw
ɤ
ɤd =
1+ω
0<n<1
R=
(Gs + e)ɤw
=
1+e
ɤsat
SL =
Bulk Specific Gravity:
When S=100%:
Se = Gs ω
ɤs
ɤw
(Gs + Gs ω)ɤw
ɤ=
1+e
(Gs + Se)ɤw
ɤ=
1+e
Gs ɤw
ɤd =
1+e
Weight
m1 − m2 V1 − V2
−
ɤw
m2
m2
e
m2
SL =
; SR =
Gs
V2 ɤw
Specific Gravity of Solid:
Unit Weight:
sin(Ø − α) sin Ø
]
cos α
2
σ3 = 0
S=
Cs H
∆P + Po
𝑙𝑜𝑔
1 + eo
Po
when (△P+Po) > pc:
S=
Cs H
Pc
Cc H
∆P + Po
𝑙𝑜𝑔 +
𝑙𝑜𝑔
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
tan Ø =
σN
 Cohesive soil:
tan Ø =
σS
c
=
x + σN x
σS = c + σN tan ∅
Terzaghi‘s Bearing Capacity (Shallow Foundations)
 General Shear Failure
(dense sand & stiff clay)
Square Footing:
qult = 1.3cNaSSSSSSSSSSSSSSSSSSSS
c + qNq + 0.4ɤBNɤ
Circular Footing:
Soil Stability
 Bearing Capacity Factor
 Analysis of Infinite Slope
Ø
Nq = tan2 (45° + ) eπ tan Ø
2
Factor of safety against sliding (without seepage)
Nc = (Nq − 1) cot Ø
qult = 1.3cNc + qNq + 0.3ɤBNɤ
Nɤ = (Nq − 1) tan 1.4Ø
Strip Footing:
 Parameters
qult → ultimate bearing capacity
qu → unconfined compressive strength
c → cohesion of soil
qult = cNc + qNq + 0.5ɤBNɤ
 Local Shear Failure
(loose sand & soft clay)
Square Footing:
′
qu
c=
2
′
qult = 1.3c′Nc + qNq + 0.4ɤBNɤ
′
q = ɤDf
qult = 1.3c′Nc ′ + qNq ′ + 0.3ɤBNɤ ′
Strip Footing:
qult = c′Nc ′ + qNq ′ + 0.5ɤBNɤ ′
C
tan ∅
+
ɤ H sin 𝛽 cos 𝛽 tan 𝛽
β
where:
C → cohesion
β → angle of backfill from horizontal
Ø → angle of internal friction
H → thickness of soil layer
Factor of safety against sliding (with seepage)
C
ɤ′ tan ∅
FS =
+
ɤ𝑠𝑎𝑡 H sin 𝛽 cos 𝛽 ɤ𝑠𝑎𝑡 tan 𝛽
 Analysis of Finite Slope
Factor of safety against sliding
FS =
(for no water table)
qult Pallow
qallow =
=
FS
A
qult − q
qnet =
FS
Circular Footing:
FS =
EFFECT OF WATER TABLE:
Ff + Fc
W sin 𝜃
β
Maximum height for critical equilibrium
(FS=1.0)
Hcr =
4𝐶
sin 𝛽 cos ∅
[
]
ɤ 1 − cos(𝛽 − ∅)
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
H
H
−
= BC
tan 𝜃 tan 𝛽
1
m
Capacity of Driven Piles (Deep Foundations)
 Pile in Sand Layer
Case 1
Case 2
Case 3
q = ɤ(Df − d)+ɤ′d
3rd term ɤ = ɤ′
q = ɤDf
3rd term ɤ = ɤ′
q = ɤDf
3rd term ɤ = ɤave
for d ≤ B
ɤave ∙ B = ɤd + ɤ′(B − d)
NOTE:
ɤ′= ɤ𝑠𝑢𝑏 = ɤ − ɤ𝑤
for d > B
ɤave = ɤ
 Alternate Equation for Group
Efficiency (sand only)
Group of Piles
 Group Efficiency (sand or clay)
Eff =
Q des−group
Eff =
Q des−indiv
2(m + n − 2)s + 4d
mnπD
where:
m → no. of columns
n→ no. of rows
s → spacing of piles
D → diameter of pile
 Pile in Clay Layer
Q f = PAkμ
where:
P → perimeter of pile
A → area of pressure diagram
k → coefficient of lateral pressure
μ → coefficient of friction
2
Q = C√2g L H 3/2
3
Considering velocity of approach:
va 3/2
va 3/2
Q = m L [(H +
)
2g
−( )
2g
Q des =
 Triangular (symmetrical only)
Neglecting velocity of approach:
3/2
Francis Formula (when C and m is not given)
Considering velocity of approach:
va 3/2
va 3/2
)
2g
−( )
2g
Neglecting velocity of approach:
3/2
Q = 1.84 L′ H
NOTE:
L’ = L
L’ = L – 0.1H
L’ = L – 0.2H
for suppressed
for singly contracted
for doubly contracted
Time required to discharge:
t=
2As 1
1
[
−
]
mL √H2 √H1
8
θ
C√2g tan H 5/2
15
2
Q = m H 5/2
Q=
Q=mLH
Q = 1.84 L′ [(H +
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
]
]
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:
Loose 10 (size of pile)
Dense 20 (size of pile)
Q T = Q f + Q tip
 Rectangular
Neglecting velocity of approach:
Q tip = cNc Atip
(AKA Qbearing)
where:
pe → effective pressure at bottom
Nq → soil bearing factor
Atip → Area of tip
Q T = Q f + Q tip
QT
F. S.
Q des =
For all sections:
NOTE:
E is minimum for critical depth.
For rectangular sections ONLY:
Take note that it is only derived from
the critical depth equation.
NF = 1
NF < 1
NF > 1
Reynold’s Number
NR =
Dv Dvρ
=
υ
μ
3 q2
2
dc = √ = Ec
g
3
Q
B
v2
E𝑐 =
+ d𝑐
2g
q=
where:
q → flow rate or discharge
per meter width
EC → specific energy at
critical condition
vC → critical velocity
vc = √gdc
Laminar Flow (NR ≤ 2000)
64
hf =
NR
Turbulent Flow (NR > 2000)
2
L v
hf = f
D 2g
Hydraulic Jump
Height of the jump:
Power Lost:
∆d = d2 − d1
P = QɤE
Length of the jump:
L = 220 d1 tanh
NF1 −1
22
Solving for Q:
2
hf =
where:
Q → flow rate m3/s
g → 9.81 m/s2
AC → critical area
BC → critical width
Q2 Ac 3
=
g
Bc
Q2 ∙ B c
NF = √ 3
Ac ∙ g
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
va 3/2
−( ) ]
2g
dc
(AKA Qbearing)
Froude Number
v
NF =
√gdm
Q = C√2g L [(H + )
3
2g
where:
C → cohesion
L → length of pile
α → frictional factor
P → perimeter of pile
Qf
Q tip = pe Nq Atip
Weirs
Considering velocity of approach:
2
va 3/2
Q f = CLαP
Q
0.0826 f L Q
D5
Boundary Shear Stress
For all sections:
P2 − P1 =
ɤQ
(v − v2 )
g 1
τ = ɤRS
P = ɤh̅A
Boundary Shear Stress
(for circular pipes only)
For rectangular sections ONLY:
f
τo = ρv
8
q2 1
= (d1 ∙ d2 )(d1 + d2 )
g
2
Strength Reduction Factors, Ø
Load Combinations
→ choose largest U in design
Basic Loads:
𝑈 = 1.4𝐷 + 1.7𝐿
With Wind Load:
𝑈 = 0.75(1.4𝐷 + 1.7𝐿 + 1.7𝑊)
𝑈 = 0.9𝐷 + 1.3𝑊
𝑈 = 1.4𝐷 + 1.7𝐿
With Earthquake Load:
𝑈 = 1.32𝐷 + 1.1𝑓1 𝐿 + 1.1𝐸
𝑈 = 0.99𝐷 + 1.1𝐸
With Earth Pressure Load:
With Structural Effects:
𝑈 = 0.75(1.4𝐷 + 1.7𝐿 + 1.4𝑇)
𝑈 = 1.4(𝐷 + 𝑇)
Internal Couple Method:
k=
Factor j:
n
n+
1
j= 1− 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:
Values
Over-reinforced:
→ concrete fails first
→ fs < fy
(USD)
→ Ms > Mc (WSD)
Choose Smaller Value/
Round-down
→ Moment Capacity
→
→
Balance Condition:
→ concrete & steel
simultaneously fail
→ fs = fy
(USD)
→ Ms = Mc (WSD)
Choose Larger Value/
Round-up
→
→
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
(for uniformly distributed load)
 Solve for additional deflection:
δadd = δsus ∙ 𝜆
δadd = (% of sustained load)δi ∙ 𝜆
Say, 70% of load is sustained after n yrs.
δadd = 0.7δi ∙ 𝜆
 Solve for final deflection:
δfinal = δi + δadd
fy = 230 MPa
fy = 275 MPa
fy = 415 MPa
424.3.2 for fy = 275 MPa; fs ≤ 140 MPa
for fy = 415 MPa; fs ≤ 170 MPa
Modular Ratio, n (if not given):
n=
Estronger
Esteel
200,000
=
=
Eweaker
Econcrete 4700√fc′
Ay̅above NA = Ay̅below NA
x
bx ( ) + (2n − 1)A′s (x − d′ ) = nAs (d − x)
2
x → obtained
 Solve transferred moment of inertia at NA:
bx 3
INA =
+ nAs (d − x)2
 Solve transferred moment of inertia at NA:
bx 3
INA =
+ (2n − 1)A′s (x − d′ )2 + nAs (d
INA
INA
3
→ obtained
3
→ obtained
 Solve for Stresses or Resisted Moment:
 Solve for Stresses or Resisted Moment:
For concrete:
For tension steel:
For concrete:
fs Ms ∙ (d − x)
=
n
INA
fc =
Mc ∙ x
INA
Solutions for Gross Section (Singly):
Mc ∙ x
INA
For tension steel:
fs Ms ∙ (d − x)
=
n
INA
− x)2
For comp. steel:
fs′
Ms′ ∙ (x −
2n
=
INA
Solutions for Uncracked Section (By Sir Erick):
 Location of neutral axis, NA:
Ay̅above NA = Ay̅below NA
x
d−x
bx ( ) = b(d − x) (
) + (n − 1)As (d − x)
2
2
x → obtained
 Location of neutral axis, NA:
Ig =
𝜉
1 + 50𝜌′
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
Ay̅above NA = Ay̅below NA
x
bx ( ) = nAs (d − x)
2
x → obtained
409.6.2.4. For simply supported, Ie = Ie (mid)
For cantilever, Ie = Ie (support)
𝜆=
where:
f’c → compressive strength of concrete at 28 days
fy → axial strength of steel
 Location of neutral axis, NA:
yt =
409.6.2.5. Factor for shrinkage & creep due
to sustained loads:
time-dep factor, ξ:
fc = 0.25 f’c
fs = 0.40 fy
 Location of neutral axis, NA:
 Solve for effective moment of inertia, Ie:
Mcr 3
Mcr 3
Ie = (
) ∙ Ig + [1 − (
) ] ∙ Icr
Ma
Ma
Ie mid + Ie support
Ie =
2
fc = 0.45 f’c
fs = 0.50 fy
 Vertical members
(i.e. column, wall, etc.)
Solutions for Cracked Section (Doubly):
409.6.2.3. if Ma < Mcr, no crack; Ig = Ie
if Ma > Mcr, w/ crack; solve for Ie
3
 Horizontal members
(i.e. beam, slab, footing, etc.)
424.6.4 n must be taken as the nearest whole number & n ≥ 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 − x)2
Allowable Stresses (if not given):
Solutions for Cracked Section (Singly):
fc =
 Solve for inertia of gross section, Ig.
 Solve for cracking moment, Mcr.
 Solve for actual moment, Ma:
2
wL
Ma =
8
Design Conditions
Under-reinforced:
→ steel fails first
→ fs > fy
(USD)
→ Ms < Mc (WSD)
𝑈 = 1.4𝐷 + 1.7𝐿 + 1.7𝐻
𝑈 = 0.9𝐷
𝑈 = 1.4𝐷 + 1.7𝐿
Factor k:
(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
Working Strength Design (WSD)
or Alternate Strength Design (ASD)
h
; y → obtained
2 t
 Solve moment of inertia of gross section at NA:
3
bx
12
Ig → obtained
 Solve for cracking moment:
Mcr ∙ yt
Ig
→ obtained
fr = 0.7√fc′ =
Mcr
 Solve transferred moment of inertia at NA:
3
3
bx
b(d − x)
+
+ (n − 1)As (d − x)2
3
3
→ obtained
INA =
INA
 Solve for Stresses or Resisted Moment:
For tension steel:
For concrete:
fc =
Mc ∙ x
INA
fs Ms ∙ (d − x)
=
n
INA
d′)
Ultimate Strength Design
Steel Ratio
 Based in Strain Diagram:
 Ultimate Moment Capacity:
εs
0.003
=
d−c
c
d−c
εs = 0.003 (
)
c
d−c
fs = 600 (
)
c
Mu = ∅Mn
Mu = ∅R n bd2
10
Mu = ∅fc′ bd2 ω(1 − ω)
17
fy
ω=ρ ′
fc
a
 Coefficient of resistance, Rn:
= β1 c
a → depth of compression block
c → distance bet. NA &
extreme compression fiber
Provisions for β1:
* 1992 NSCP
β1 = 0.85 − 0.008(fc′ − 30)
* 2001 NSCP
0.05
7
* 2010 NSCP
β1 = 0.85 −
0.05
7
−
10
17
(fc′ − 28)
Maximum & Minimum steel ratio:
0.85fc′
2R n
[1 − √1 −
]
fy
0.85fc′
Singly Reinforced Beam
INVESTIGATION
Singly Reinforced Beam
DESIGN
Computing MU with given As:
Computing As with given WD & WL:
ρmin
Doubly Reinforced Beam (DRB)
ρ > ρmax (rectangular only)
As > As max (any section)
Doubly Reinforced Beam
Investigation
if SRB or DRB:
a = β1 c
c → obtained
(3rd) Solve for steel ratio, ρ:
d−c
fs = 600 [
]
c
fs → obtained
ρ=
(4th) Solve for area of steel
reinforcement, As and required no. of
bars, N:
C=T
0.85fc′ ab = As fs
0.85fc′ β1 cb
= As ∙ 600 [
d−c
c
]
c → obtained
a = β1 c
a → obtained
(3rd) Solve for Moment Capacity:
a
Mu = ∅(C or T) [d − ]
2
Mu =
∅(0.85fc′ ab) [d
a
Mu = ∅(As fs ) [d − ]
2
a
− ]
2
or
As = ρbd
As
ρbd
N=
=
2
π
Ab
d
4 b
If As < As max
Solve the given beam
using SRB Investigation
procedure.
If As > As max
Solve the given beam
using DRB Investigation
procedure.
Doubly Reinforced Beam
DESIGN
Computing As with given Mu:
(1st) Solve for nominal M1:
0.85fc′ β1 600
fy (600 + fy )
ρmax = 0.75ρb
As1 = 0.75ρb ∙ bd
ρb =
M1 = (As1 fy ) [d − ]
2
(2nd) Solve for nominal M2:
MU
M2 =
− M1
∅
(3rd) Solve for As2:
M2 = (As2 fy )[d − d′]
As2 → obtained
Doubly Reinforced Beam
INVESTIGATION
Computing MU with given As:
(1st) Compute for a:
Cc + Cs = T
0.85fc′ ab + As ′fs ′ = As fs
0.85fc′ ab + As ′fy = As fy
a → obtained
a = β1 c
c → obtained
d−c
]
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 − 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
(4th) Solve for # of tension bars:
NOTE: Use fs & fs’ as
As As1 + As2
N=
=
π 2
Ab
d
4 b
fs = 600 [
(5th) Solve for fs’:
c → obtained
fs ′ = 600 [
a
𝑑
(2nd) Check if assumption is correct:
(2nd) Solve for given As
& compare:
(2nd-b) Recomputation:
As 𝑚𝑎𝑥 = ρ𝑚𝑎𝑥 𝑑 bd
a 𝑏 = β1 c𝑏
a 𝑏 → obtained
As max = 0.75As 𝑏
If ρmin < ρ < ρmax, use ρ.
If ρmin > ρ, use ρmin.
If ρ > ρmax, design doubly.
c − 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 ′
bd
(assume tension steel yields fs=fs’=fy)
C=T
0.85fc′ a 𝑏 b = As 𝑏 fy
As 𝑏 → obtained
ρmin ≤ ρ ≤ ρmax
ρ𝑚𝑎𝑥 𝑑 = 0.75ρb 𝑠 +
600d
600 + fy
c𝑏 → obtained
(2nd) Solve for Asmax:
0.85fc′
2R n
[1 − √1 −
]
fy
0.85fc′
Check:
If fs > fy, tension steel yields; correct a.
If fs < fy, tension steel does not yield;
compute for new a.
d − c𝑏
fs = fy = 600 [
]
c𝑏
c𝑏 =
MU
∅bd2
As′
bd
(1st) Compute for ab:
Thus,
(2nd) Solve for coeff. of resistance, Rn:
(2nd) Check if assumption is correct:
75 mm → column footing
→ wall footing
→ retaining wall
ρb 𝑑 = ρb 𝑠 +
WU = 1.4WD + 1.7WL
WU L2
(for simply supported)
MU =
8
Rn =
40 mm → beam
→ column
ρ < ρmax (rectangular only)
As < As max (any section)
(1st) Compute ultimate moment, Mu:
a → obtained
ρmin
20 mm → slab
√fc′
=
4fy
Balance Condition for Doubly
C=T
0.85fc′ ab = As fs
0.85fc′ ab = As fy
1.4
=
fy
Singly Reinforced Beam (SRB)
(1st) Compute for a:
(assume tension steel yields fs=fy)
Minimum Concrete Covers:
ρmax = 0.75ρb
As max = 0.75As b
Singly or Doubly ?
As
ρ=
bd
ρ=
0.85fc′ β1 600
fy (600 + fy )
(choose larger between the 2)
Mu
Rn =
∅bd2
 Combined ρ & Rn:
(fc′ − 30)
ρb =
ω)
 Steel reinforcement ratio, ρ:
0.65 ≤ β1 ≤ 0.85
β1 = 0.85 −
Rn =
fc′ ω(1
Steel ratio for balance condition:
As
As′
=
Ab π d 2
4 b
d−c
]
c
c−d′
fs ′ = 600 [
c
]
a = β1 c
a → obtained
(3rd) Solve for Moment Capacity:
a
Mu = ∅Cc [d − ] + ∅Cc [d − d′]
2
a
Mu = ∅(0.85fc′ ab) [d − ]
2
+ ∅(As ′fs ′)[d − d′] or
a
Mu = ∅T [d − ]
2
a
Mu = ∅(As fs ) [d − ]
2
Design of Beam Stirrups
(1st) Solve for Vu:
NSCP Provisions for
max. stirrups spacing:
ΣFv = 0
Vu = R − wu d
wu L
Vu =
− wu d
2
NSCP Provisions for effective flange width:
NSCP Provisions for minimum thickness:
i. Interior Beam:
ii. exterior Beam:
L
bf =
4
L
bf = bw +
12
s1
bf = bw +
2
bf = bw + 6t f
Cantilever
Simple
Support
One
End
Both
Ends
Slab
L/10
L/20
L/24
L/28
Beams
L/8
L/16
L/18.5
L/21
Factor: [0.4 +
smax =
d
or 600mm
2
] [1.65 − 0.0003𝜌𝑐 ]
(for lightweight concrete only)
Minimum Steel Ratio
For one-way bending:
k → steel ratio
ii. when Vs > 2Vc,
(3rd) Solve for Vs:
smax =
Vu = ∅(Vc + Vs )
Vs → obtained
d
or 300mm
4
i. fy = 275 MPa,
k = 0.0020
ii. fy = 415 MPa,
iii. & not greater than to:
(4th) Theoretical Spacing:
smax =
n
3Av fy
k = 0.0018
iii. fy > 415 MPa,
n
b
k = 0.0018 [
Vs
NOTE:
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 =
fy
700
i. when Vs < 2Vc,
1
Vc = √fc ′bw d
6
s=
Thickness of One-way Slab & Beam
s1 s2
bf = bw + +
2
2
bf = bw + 8t f
1
2Vc = √fc ′bw d
3
(2nd) Solve for Vc:
dA v fy
T-Beam
ρmin =
π 2
d ∙n
4
1.4
√fc′
ρmin =
fy
4fy
(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
ρ=
(11th) Solve for As:
As
bd
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.
ρmin =
(3rd) Solve for effective depth, d:
d = h − cc −
TEMPERATURE BARS/
SHRINKAGE BARS
√fc′
1.4
& ρmin =
fy
4fy
(12th) Determine # of req’d temp. bars:
If ρmin < ρ, use ρ.
If ρmin > ρ, use ρmin & recompute As.
db
2
N=
(8th) Determine # of req’d main bars:
(4th) Solve for a:
a
As
As
=
2
π
Ab
d
4 b
N=
Mu = ∅(C) [d − ]
2
a
Mu = ∅(0.85fc′ ab) [d − ]
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
=
Ab π d 2
4 b
b
N
b
N
(14th) Check for max. spacing of temp. bars:
smax = 5h or 450mm
(10th) Check for max. spacing of main bars:
smax = 3h or 450mm
Design of Column
TIED COLUMN
SPIRAL COLUMN
P = PC + PS
P = 0.85fc′ (Ag − Ast ) + Ast fy
PN = 0.8P
PU = ∅0.8P ; ∅ = 0.7
PU = (0.7)(0.8)[0.85fc′ (Ag − Ast ) + Ast fy ]
PN = 0.85P
PU = ∅0.85P ; ∅ = 0.75
PU = (0.75)(0.85)[0.85fc′ (Ag − Ast ) + Ast fy ]
ρ=
Ast
Ag
No. of main bars:
Thus,
P
Ag =
′
0.85fc (1 − ρ) + ρfy
0.01Ag < Ast < 0.08Ag
Design of Footing
qA = qS + qC + qsur + qE
qE =
P
A ftg
;
qU =
PU
Aftg
where:
qA → allowable bearing pressure
qS → soil pressure
qC → concrete pressure
qsur → surcharge
qE → effective pressure
qU → ultimate bearing pressure
Ø = 0.85
Spacing of bars:
Ast
N=
Ab
ρs = 0.45
s = 16db
s = 48dt
s = least dimension
N is based on Pu.
NOTE: If spacing of main bars < 150mm, use 1 tie per set.
fc′ Ag
volume of spiral
[ − 1] =
fy Ac
volume of core
π
(dsp )2 ∙ π(Dc −dsp ) 4Asp
s=4 π
=
Dc ρs
(D )2 ∙ ρs
4 c
WIDE BEAM SHEAR
PUNCHING/DIAGONAL TENSION SHEAR
BENDING MOMENT
VU1 = qU (B)(x)
VU2 = PU − qU (a + d)(b + d)
x
MU = qU (B)(x) ( )
2
VU1 ≤ ∅Vwb = ∅
τwb =
VU1
∅Bd
τwb(allw) =
√fc′
Bd
6
VU2 ≤ ∅Vpc = ∅
τpc =
√fc′
6
VU2
∅bo d
τpc(allw) =
√fc′
3
√fc′
b d
3 o
** design of main bars and
temperature bars –
Same as slab.