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Casagrande:
k = 1. 4 e
2
k
- 85
Dr (%) Description
0 – 20 Very Loose
20 – 40 Loose
40 – 70 Medium Dense
70 – 85 Dense
85 – 100 Very Dense
LI State
LI < 0 Semisolid
0 < LI < 1 Plastic
LI > 1 Liquid
Volume
e =
V
v
V
s
n =
V
v
V
S =
V
w
V
v
Weight
ω =
W
w
W
s
0 < e < ∞
e =
n
1 − n
0 < n < 1
n =
e
1 + e
Se = G
s
ω
ɤ =
W
V
ɤ
d
W
S
V
ɤ
d
=
ɤ
1 + ω
Unit Weight:
ɤ =
(G
s
s
ω)ɤ
w
1 + e
ɤ =
(G
s
w
1 + e
When S=0:
ɤ
d
=
G s
ɤ w
1 + e
When S=100%:
ɤ sat
=
(G
s
w
1 + e
ɤ
sub
= ɤ
sat
− ɤ
w
ɤ
sub
=
(G
s
− 1 )ɤ
w
1 + e
ɤ
zav
=
G
s
ɤ
w
1 + G
s
ω
Specific Gravity of Solid:
G
s
ɤ
s
ɤ
w
Bulk Specific Gravity:
g = G
s
( 1 − n)
Relative Compaction:
R =
ɤ
d
ɤ
d
𝑚𝑎𝑥
Relative Density/
Density Index:
D
r
e
𝑚𝑎𝑥
− e
e
𝑚𝑎𝑥
− e
𝑚𝑖𝑛
D r
=
1
ɤ
d 𝑚𝑖𝑛
−
1
ɤ
d
1
ɤ
d
𝑚𝑖𝑛
−
1
ɤ
d
𝑚𝑎𝑥
Atterberg Limits
PI = LL − PL
LI =
ω − PL
LL − PL
SI = PL − SL
CI =
LL − ω
LL − PI
SL =
m
1
− m
2
m
2
V
1
− V
2
m
2
ɤ
w
SL =
e
G
s
; SR =
m
2
V
2
ɤ
w
G
𝑠
SR
− SL
GI = (F − 35 )[ 0. 2 + 0. 005 (LL − 40 )]
A
c
=
PI
μ
; St =
q u und
q u rem
μ = % passing 0.002mm
PI Description
0 Non-plastic
1 - 5 Slightly plastic
5 - 10 Low plasticity
10 - 20 Medium plasticity
20 - 40 High plasticity
40 Very High plastic
Ac Class
AC < 0.7 Inactive
0.7 < AC < 1.2 Normal
AC > 1.2 Active
Sieve Analysis
Uniformity
Coefficient:
Cu =
D
60
D
10
Coeff. of Gradation
or Curvature:
Cc =
(D
30
2
D
60
∙ D
10
Sorting
Coefficient:
So = √
D
75
D
25
Suitability Number:
Sn = 1. 7 √
(D
50
2
(D
20
2
(D
10
2
Permeability
v = ki ; i =
∆h
L
; v
𝑠
=
v
n
Q = vA = kiA
Constant Head Test:
k =
QL
Aht
Falling/Variable Head Test:
k =
aL
At
h
1
h
2
Stratified Soil
for Parallel flow:
k
eq
h
1
k
1
2
k
2
+... +h
n
k
n
H
for Perpendicular flow:
k
eq
H
h
1
k
1
h
2
k
2
h
n
k
n
Pumping Test:
Unconfined:
k =
Q 𝑙𝑛
r
1
r
2
π(h
1
2
− h
2
2
Confined:
k =
Q 𝑙𝑛
r
1
r
2
2πt(h
1
− h
2
Hazen Formula
k = c ∙ D 10
2
Samarasinhe:
k = C
3
∙
e
n
1 + e
Kozeny-Carman:
k = C 1
∙
e
2
1 + e
Stresses in Soil
Effective Stress/
Intergranular Stress:
p
E
= p
T
− p
w
Pore Water Pressure/
Neutral Stress:
p
w
= ɤ
w
h
w
Total Stress:
p
T
= ɤ
1
h
1
2
h
2
+... +ɤ
n
h
n
Flow Net / Seepage
Isotropic soil:
q = kH
N
f
N
d
Non-Isotropic soil:
q = √
k
x
k
z
H
N
f
N
d
Compressibility of Soil
Compression Index, CC:
C
c
= 0. 009 (LL − 10%)
C
c
e − e′
∆P + P
o
P
o
For normally consolidated clay:
S =
e − e′
1 + e
H
S =
C
c
H
1 + e
∆P + P
o
P
o
With Pre-consolidation pressure, Pc:
when (△P+Po) < Pc:
S =
C
s
H
1 + e
o
∆P + P
o
P
o
when (△P+Po) > pc:
S =
C
s
H
1 + e
P
c
P
o
C
c
H
1 + e
∆P + P
o
P
c
Over Consolidation Ratio (OCR):
OCR =
p
c
p
o
; OCR = 1
Coefficient of Compressibility:
a
v
∆e
∆P
Coefficient of Volume Compressibility:
m
v
∆e
∆P
1 + e
ave
Coefficient of Consolidation:
C
v
H
dr
2
T
v
t
Coefficient of Permeability:
k = m
v
C
v
ɤ
w
Equipotential line ----
Flow line ----
Lateral Earth Pressure
ACTIVE PRESSURE:
p
a
k
a
ɤH
2
− 2cH √
k
a
For Inclined:
k
a
= cos β
cos β − √cos
2
β − cos
2
Ø
cos β + √cos
2
β − cos
2
Ø
For Horizontal:
k
a
1 − sin Ø
1 + sin Ø
If there is angle of friction α bet. wall and soil:
k
a
cos
2
Ø
cos α [ 1 +
sin(Ø + α) sin Ø
cos α
]
2
PASSIVE PRESSURE:
p
P
k
P
ɤH
2
P
For Inclined:
k
P
= cos β
cos β + √cos
2
β − cos
2
Ø
cos β − √
cos
2
β − cos
2
Ø
For Horizontal:
k
P
1 + sin Ø
1 − sin Ø
If there is angle of friction α bet. wall and soil:
k
P
cos
2
Ø
cos α [ 1 −
sin(Ø − α) sin Ø
cos α
]
2
AT REST:
k
o
= 1 − sin Ø
(for one layer only)
Swell Index, C S
:
C
s
C
c
(for normally consolidated soil)
△e → change in void ratio
△P → change in pressure
Hdr → height of drainage path
→ thickness of layer if drained 1 side
→ half of thickness if drained both sides
Tv → factor from table
t → time consolidation
Shear Strength of Soil
Ө → angle of failure in shear
Ø → angle of internal friction/shearing resistance
C → cohesion of soil
θ = 45° +
Ø
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
Normally consolidated:
sin Ø =
r
σ
3
Cohesive soil:
sin Ø =
r
x + σ
3
tan Ø =
c
x
Unconsolidated-
undrained test:
c = r
Unconfined
compression test:
σ
3
DIRECT SHEAR TEST:
σn → normal stress
σs → shear stress
Normally consolidated soil:
tan Ø =
σ
S
σ
N
Cohesive soil:
tan Ø =
σ
S
x + σ
N
c
x
σ
S
= c + σ
N
tan ∅
Nf → no. of flow channels [e.g. 4]
Nd → no. of potential drops [e.g. 10]
1
2
3
4
1 2 3 4 5 6 7 8 9 10
NOTE:
Quick
condition:
p
E
Capillary Rise:
h cr
=
C
eD
10
aSSSSSSSSSSSSSSSSSSSS
Terzaghi‘s Bearing Capacity (Shallow Foundations)
General Shear Failure
(dense sand & stiff clay)
Square Footing:
q
ult
= 1 .3cN
c
q
ɤ
Circular Footing:
q
ult
= 1 .3cN
c
q
ɤ
Strip Footing:
q
ult
= cN
c
q
ɤ
Local Shear Failure
(loose sand & soft clay)
Square Footing:
q
ult
= 1 .3c′N
c
′
q
′
ɤ
′
Circular Footing:
q
ult
= 1 .3c′N
c
′
q
′
ɤ
′
Strip Footing:
q
ult
= c′N
c
′
q
′
ɤ
′
EFFECT OF WATER TABLE:
Bearing Capacity Factor
N q
= tan
2
(45° +
Ø
2
) e
π tan Ø
N c
= (N q
− 1 ) cot Ø
N
ɤ
= (N
q
− 1 ) tan 1 .4Ø
Parameters
q ult
→ ultimate bearing capacity
q u
→ unconfined compressive strength
c → cohesion of soil
c =
q
u
q = ɤD
f
(for no water table)
q
allow
q
ult
FS
P
allow
A
q
net
q
ult
− q
FS
Weirs
Rectangular
Considering velocity of approach:
Q =
C√2g L [(H +
v
a
2g
3 / 2
v
a
2g
3 / 2
]
Neglecting velocity of approach:
Q =
C√2g L H
3 / 2
Considering velocity of approach:
Q = m L [(H +
v
a
2g
3 / 2
v
a
2g
3 / 2
]
Neglecting velocity of approach:
Q = m L H
3 / 2
Francis Formula (when C and m is not given)
Considering velocity of approach:
Q = 1. 84 L′ [(H +
v
a
2g
3 / 2
v
a
2g
3 / 2
]
Neglecting velocity of approach:
Q = 1. 84 L′ H
3 / 2
NOTE:
L’ = L for suppressed
L’ = L – 0.1H for singly contracted
L’ = L – 0.2H for doubly contracted
Time required to discharge:
t =
2 A
s
mL
[
√H
2
√H
1
]
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
Triangular (symmetrical only)
Q =
C
2g tan
θ
H
5 / 2
Q = m H
5 / 2
When θ=90°
Q = 1. 4 H
5 / 2
Cipolletti (symmetrical, slope 4V&1H)
θ = 75°57’50”
Q = 1. 859 L H
3 / 2
with Dam:
Neglecting velocity of approach:
Q = 1. 71 L H
3 / 2
Capacity of Driven Piles (Deep Foundations)
Pile in Sand Layer
Q
f
= PAkμ
where:
P → perimeter of pile
A → area of pressure diagram
k → coefficient of lateral pressure
μ → coefficient of friction
Q
tip
= p
e
N
q
A
tip
(AKA Qbearing)
where:
p e
→ effective pressure at bottom
Nq → soil bearing factor
A tip
→ Area of tip
Q
T
= Q
f
+ Q
tip
Q
des
Q
T
F. S.
Pile in Clay Layer
Q
f
= CLαP
where:
C → cohesion
L → length of pile
α → frictional factor
P → perimeter of pile
Q
tip
= cN
c
A
tip
(AKA Qbearing)
where:
c → cohesion
N c
→ soil bearing factor
Atip → Area of tip
Q
T
= Q
f
+ Q
tip
Q
des
Q
T
F. S.
Group of Piles
Group Efficiency (sand or clay)
Eff =
Q
des−group
Q
des−indiv
Alternate Equation for Group
Efficiency (sand only)
Eff =
m + n − 2
s + 4d
m n π D
where:
m → no. of columns
n→ no. of rows
s → spacing of piles
D → diameter of pile
Soil Stability
Analysis of Infinite Slope
Factor of safety against sliding (without seepage)
FS =
C
ɤ H sin 𝛽 cos 𝛽
tan ∅
tan 𝛽
Factor of safety against sliding (with seepage)
FS =
C
ɤ
𝑠𝑎𝑡
H sin 𝛽 cos 𝛽
ɤ′
ɤ
𝑠𝑎𝑡
tan ∅
tan 𝛽
Analysis of Finite Slope
Factor of safety against sliding
FS =
F
f
+ F
c
W sin 𝜃
Maximum height for critical equilibrium
(FS=1.0)
H
cr
ɤ
[
sin 𝛽 cos ∅
1 − cos(𝛽 − ∅)
]
where:
C → cohesion
β → angle of backfill from horizontal
Ø → angle of internal friction
H → thickness of soil layer
Stability No.:
m =
C
ɤH
Stability Factor:
SF =
m
Case 1
q = ɤ(D f
− d)+ɤ′d
3
rd
term ɤ = ɤ′
Case 2
q = ɤD
f
3
rd
term ɤ = ɤ′
Case 3
q = ɤD
f
3
rd
term ɤ = ɤ ave
for d ≤ B
ɤ
ave
∙ B = ɤd + ɤ′(B − d)
for d > B
ɤ
ave
= ɤ
NOTE:
ɤ′= ɤ
𝑠𝑢𝑏
= ɤ − ɤ
𝑤
β
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
β θ
H
tan 𝜃
H
tan 𝛽
= BC
Q
QTIP
Q f
d c
Critical depth, dc:
Loose 10 (size of pile)
Dense 20 (size of pile)
Reynold’s Number
N
R
Dv
υ
Dvρ
μ
Laminar Flow (N R
≤ 2000)
hf =
N
R
Turbulent Flow (N R
hf = f
L
D
v
2
2g
hf =
- 0826 f L Q
2
D
5
Boundary Shear Stress
τ = ɤRS
Boundary Shear Stress
(for circular pipes only)
τ
o
f
ρv
Froude Number
N
F
v
√gd
m
where:
v → mean velocity (Q/A)
g → 9.81 m/s
2
d m
→ hydraulic depth (A/B)
B → width of liquid surface
N
F
= √
Q
2
∙ B c
A c
3
∙ g
Take note that i t is only derived from
the critical depth equation.
Critical Flow NF = 1
Subcritical Flow N F
< 1
Supercritical Flow NF > 1
Critical Depth
For all sections:
Q
2
g
A
c
3
B
c
NOTE:
E is minimum for critical depth.
For rectangular sections ONLY :
d
c
q
2
g
3
2
3
E
c
q =
Q
B
E
𝑐
v
2
2g
𝑐
v
c
= √gd
c
where:
Q → flow rate m
3 /s
g → 9.81 m/s
2
A C
→ critical area
BC → critical width
where:
q → flow rate or discharge
per meter width
E C
→ specific energy at
critical condition
v C
→ critical velocity
Hydraulic Jump
Height of the jump:
∆d = d
2
− d
1
Length of the jump:
L = 220 d
1
tanh
N F
− 1
22
Solving for Q:
For all sections:
P
2
− P
1
ɤQ
g
(v
1
− v
2
P = ɤh
A
For rectangular sections ONLY :
q
2
g
(d
1
∙ d
2
)(d
1
2
Power Lost:
P = QɤE
s
Quadratic Equation
Form:
Ax
2
Roots:
x =
−B ± √B
2
− 4AC
2A
Sum of Roots:
x 1
2
B
A
Product of Roots:
x 1
∙ x
2
C
A
Progression
AM ∙ HM = (GM)
2
Arithmetic Progression:
d = a
2
− a
1
= a
3
− a
2
a
n
= a
1
a
n
= a
x
S
n
n
(a
1
n
Harmonic Progression:
progression
Geometric Progression:
r = a
2
/a
1
= a
3
/a
2
a
n
= a
1
r
n− 1
a
n
= a
x
r
n−x
S
n
= a
1
1 − r
n
1 − r
S
∞
a
1
1 − r
Trigonometric Identities
Squared Identities:
sin
2
A + cos
2
A = 1
1 + tan
2
A = sec
2
A
1 + cot
2
A = csc
2
A
Sum & Diff of Angles Identities:
sin
A ± B
= sin A cos B ± cos A sin B
cos
A ± B
= cos A cos B ∓ sin A sin B
tan (A ± B) =
tan A ± tan B
1 ∓ tan A tan B
Double Angle Identities:
sin 2A = 2 sin A cos A
cos 2A = cos
2
A − sin
2
A
cos 2A = 2 cos
2
A − 1
cos 2A = 1 − 2 sin
2
A
tan 2A =
2 tan A
1 − tan
2
A
Spherical Trigonometry
Sine Law:
sin 𝑎
sin 𝐴
sin 𝑏
sin 𝐵
sin 𝑎
sin 𝐴
Cosine Law for sides:
cos 𝑎 = cos 𝑏 cos 𝑐 + sin 𝑏 sin 𝑐 cos 𝐴
Cosine Law for angles:
Spherical Polygon:
A
B
πR
2
E
Spherical Pyramid:
V =
A
B
H =
πR
3
E
Binomial Theorem
Form:
(x + y)
n
r
th
term:
r
th
= C
m
x
n−m
y
m
where: m=r- 1
n
Worded Problems Tips
Age Problems
→ underline specific time conditions
Motion Problems
→ a = 0
→ s = vt
Work Problems
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 − 60H
cos 𝐴 = − cos 𝐵 cos 𝐶 + sin 𝐵 sin 𝐶 cos 𝑎
n-sided Polygon
Interior Angle, ɤ:
γ =
(n − 2 )180°
n
Deflection Angle, δ:
δ = 180° − γ
Central Angle, β:
β =
n
E = spherical excess
E = (A+B+C+D…) – (n-2)180°
Area = n ∙ A TRIANGLE
Area = n ∙
1
2
R
2
sinβ
Area = n ∙
1
2
ah
Polygon Names
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
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)
Triangle
A =
1
2
bh
A =
1
2
ab sin C
A =
1
2
a
2
sin B sin C
sin A
A = √s(s − a)(s − b)(s − c)
s =
a + b + c
2
Common Quadrilateral
Square:
A = s
2
P = 4s
d = √ 2 s
Parallelogram:
A = bh
A = ab sin θ
A =
1
2
d
1
d
2
sin θ
Rhombus:
A = ah
A = a
2
sin θ
A =
1
2
d
1
d
2
Trapezoid
A =
1
2
(a + b)h
A
1
A
2
=
n
m
; w =
√
ma
2
2
m + n
General Quadrilateral
Cyclic Quadrilateral: (sum of opposite angles=180°)
A = √
s − a
s − b
s − c
s − d
Ptolemy’s Theorem is applicable:
ac + bd = d
1
d
2
Non-cyclic Quadrilateral:
A = √(s − a)(s − b)(s − c)(s − d) − abcd cos
2
ε
Triangle-Circle Relationship
Circumscribing Circle:
A
T
=
abc
4R
diameter =
opposite side
sine of angle
d =
a
sin A
=
b
sin B
=
c
sin C
Inscribed Circle:
A
T
= rs
Escribed Circle:
A
T
= R
a
(s − a)
A
T
= R
b
(s − b)
A
T
= R
c
(s − c)
Ex-circle-
In-circle
1
𝑟
=
1
𝑟
1
1
𝑟
2
1
𝑟
3
s =
a + b + c + d
2
1 minute of arc =
1 nautical mile
1 nautical mile =
6080 feet
1 statute mile =
5280 feet
1 knot =
1 nautical mile
per hour
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.
CIRCUMCENTER
- the center of the circumscribing circle
( circumcircle ) & the point of intersection of
the perpendicular bisectors of the triangle.
ORTHOCENTER
- the point of intersection of the altitudes of
the triangle.
CENTROID
- the point of intersection of the medians of
the triangle.
EULER LINE
- the line that would pass through the
orthocenter, circumcenter, and centroid of
the triangle.
Pappus Theorem
Pappus Theorem 1:
SA = L ∙ 2πR
Pappus Theorem 2:
V = A ∙ 2πR
NOTE: It is also used to locate centroid of an area.
Prism or Cylinder
V = A
B
H = A
X
L
v
LA = P
B
H = P
x
L
Pointed Solid
V =
1
3
A
B
H
AB/PB → Perimeter or Area of base
H → Height & L → slant height
A X
/P X
→ Perimeter or Area of cross-
section perpendicular to slant height
Right Circ. Cone
LA = πrL
Reg. Pyramid
LA =
1
2
P
B
L
Special Solids
Truncated Prism or Cylinder:
V = A
B
H
ave
LA = P
B
H
ave
Frustum of Cone or Pyramid:
V =
H
3
(A
1
2
1
A
2
)
Prismatoid:
V =
H
6
( A
1
M
2
)
Spherical Solids
Sphere:
V =
4
3
πR
3
LA = 4πR
2
Spheroid:
V =
4
3
πabc
LA = 4π [
a
2
2
2
3
]
Prolate Spheroid:
V =
4
3
πabb
LA = 4π [
a
2
2
2
3
]
Oblate Spheroid:
V =
4
3
πaab
LA = 4π
[
a
2
2
2
3
]
about minor axis
about major axis
Spherical Lune:
A lune
θ rad
=
4πR
2
2π
A lune
= 2θR
2
Spherical Wedge:
V
wedge
θ
rad
=
4
3
πR
3
2π
V
wedge
=
2
3
θR
3
Spherical Zone:
A
zone
= 2πRh
Spherical Sector:
V =
1
3
A
zone
R
V =
2
3
πR
2
h
Spherical Segment:
For one base:
V =
1
3
πh
2
(3R − h)
For two bases:
V =
1
6
πh( 3 a
2
2
2
)
Rectangle:
A = bh
P = 2a + 2b
d =
√ b
2
2
Ellipse
A = πab C = 2π
√
a
2
2
2
of diagonals:
d =
n
2
(n − 3 )
s
Tetrahedron
H = a
SA = a
2
V = a
3
Circle
- the locus of point that moves such
that its distance from a fixed point
called the center is constant.
General Equation:
x
2
2
Standard Equation:
(x − h)
2
2
= r
2
Analytic Geometry
Slope-intercept form:
y = mx + b
Point-slope form:
m =
y − y
1
x − x
1
Two-point form:
y
2
− y
1
x
2
− x
1
y − y
2
x − x
2
Point-slope form:
x
a
y
b
Archimedean Solids
- the only 13 polyhedra that are
convex, have identical vertices, and
their faces are regular polygons.
where:
E → # of edges
V → # of vertices
N → # of faces
n → # of sides of each face
v → # of faces meeting at a vertex
E =
Nn
2
V =
Nn
v
Distance from a point to another point:
d = √(y
2
− y
1
2
2
− x
1
2
Distance from a point to a line:
d =
Ax + By + C
√A
2
+ B
2
Distance of two parallel lines:
d =
|C
1
− C
2
√A
2
+ B
2
Angle between two lines:
tan θ =
m
2
− m
1
1 + m
1
m
2
Conic Sections
General Equation:
Ax
2
2
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
Parabola
- the locus of point that moves such that it is always equidistant from a
fixed point (focus) and a fixed line (directrix).
General Equation:
y
2
x
2
Standard Equation:
(x − h)
2
= ±4a(y − k)
(y − k)
2
= ±4a(x − h)
Elements:
Eccentricity, e:
e =
d
f
d
d
Length of latus
rectum, LR:
LR = 4a
Ellipse
- the locus of point that moves such
that the sum of its distances from
two fixed points called the foci is
constant.
General Equation:
Ax
2
2
Standard Equation:
(x − h)
2
a
2
(y − k)
2
b
2
= 1
(x − h)
2
b
2
(y − k)
2
a
2
= 1
Elements:
Location of foci, c:
c
2
= a
2
− b
2
Length of LR:
LR =
2 b
2
a
Loc. of directrix, d:
d =
a
e
Eccentricity, e:
e =
c
a
Hyperbola
- the locus of point that moves such
that the difference of its distances
from two fixed points called the foci
is constant.
General Equation:
Ax
2
− Cy
2
Standard Equation:
(x − h)
2
a
2
−
(y − k)
2
b
2
= 1
(y − k)
2
a
2
−
(x − h)
2
b
2
= 1
Elements:
Location of foci, c:
c
2
= a
2
2
Eq’n of asymptote:
y − k = ±m(x − h)
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
Same as ellipse:
Length of LR,
Loc. of directrix, d
Eccentricity, e
Line Tangent to Conic Section
To find the equation of a line
tangent to a conic section at a
given point P(x 1 , y 1 ):
In the equation of the conic
equation, replace:
2
1
2
1
1
1
1
1
Engineering Economy
Simple Interest:
I = P𝑖n
F = P( 1 + 𝑖n)
Compound Interest:
F = P( 1 + 𝑖)
n
F = P ( 1 +
r
m
mt
ER =
I
P
r
m
m
Continuous Compounding Interest:
F = Pe
rt
ER = e
r
Annuity:
F = A [
n
′
]
P = A [
n
′
n
]
Perpetuity:
P =
A
= F( 1 + 𝑖)
−n
Capitalized Cost:
C = FC +
OM
RC − SV
n
AC = C ∙ 𝑖
AC = FC ∙ 𝑖 + OM +
RC − SV
( 1 + i)
n
Single-payment-compound-amount factor:
(F/P, 𝑖, n) = ( 1 + 𝑖)
n
Single-payment-present-worth factor:
(P/F, 𝑖, n) = ( 1 + 𝑖)
−n
Equal-payment-series-compound-amount factor:
(F/A, 𝑖, n) = [
( 1 + 𝑖)
n
′
− 1
𝑖
]
Equal-payment-sinking-fund factor:
(A/F, 𝑖, n) = [
( 1 + 𝑖)
n
′
− 1
𝑖
]
− 1
Equal-payment-series-present-worth factor:
(P/A, 𝑖, n) = [
( 1 + 𝑖)
n
′
− 1
𝑖( 1 + 𝑖)
n
]
Equal-payment-series-capital-recovery factor:
(A/P, 𝑖, n) = [
( 1 + 𝑖)
n
′
− 1
𝑖( 1 + 𝑖)
n
]
− 1
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
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
where:
C → capitalized cost
FC → first cost
OM → annual operation
or maintenance cost
RC → replacement cost
SV → salvage cost
AC → annual cost
Depreciation
BV
m
= FC − D
m
Straight-Line:
d =
FC − SV
n
D
m
= d(m)
Sinking Fund:
d =
( FC − SV
) [
( 1 + i)
n
− 1
𝑖
]
− 1
D
m
= d [
( 1 + i)
m
− 1
𝑖
]
Sum-of-the-Years-Digit (SYD):
d
m
= (FC − SV) [
n − m + 1
years
]
D
m
= (FC − SV) [
x
n
n−m+ 1
∑ x
n
1
]
Declining Balance (Matheson):
BV
m
= FC( 1 − k)
m
SV = FC( 1 − k)
n
k → obtained
D
m
= FC − BV
m
Double Declining Balance:
BV
m
= FC( 1 − k)
m
k = 2 /n k → obtained
D
m
= FC − BV
m
Service Output Method:
d =
FC − SV
Q
n
D = dQ
m
where:
FC → first cost
SV → salvage cost
d → depreciation
per year
n → economic life
m → any year before n
BV m
→ book value
after m years
Dm → total depreciation
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
D m
→ total depreciation
CALTECH:
Mode 3 3
x
(time)
y
(BV)
0 FC
n SV
n+1 SV
CALTECH:
Mode 3 6
x
(time)
y
(BV)
0 FC
n SV
CALTECH:
Mode 3 2
x
(time)
y
(BV)
0 FC
n SV
Inflation:
f
= 𝑖 + f + 𝑖f
Break-even analysis:
cost = revenue
Rate of return:
RR =
annual net profit
capital
Annual net profit
= savings – expenses
- depreciation (sinking fund)
RP =
1
RR
Differential Calculus
Curvature:
k =
y"
[ 1 + (y′)
2
]
3
2
Maxima & Minima (Critical Points):
= y
′
Point of inflection:
2
2
= y
"
Versed sine:
vers A = 1 − cos A
Versed cosine:
covers A = 1 − sin A
Half versed sine:
hav A =
1 − cos A
Exsecant:
exsec A = sec A − 1
Unit Circle
Radius of curvature:
ρ =
[ 1 + (y′)
2
]
3
2
(+) minima
(-) maxima
Integral Calculus-The Cardioid
A = 1 .5πa
2
P = 8a
r = a( 1 − sin θ) r = a( 1 − cos θ)
r = a( 1 + sin θ) r = a( 1 + cos θ)
1 revolution
= 2π rad
= 360 ˚
= 400 grads
= 6400 mills
s
Statistics
Measure of Natural Tendency
Mean, x̅, μ → average
→ Mode Stat 1 - var
→ Shift Mode ▼ Stat Frequency? on
→ Input
→ AC Shift 1 var x̅
Median, Me → middle no.
M e
th
=
n + 1
2
M e
th
=
1
2
[(
n
2
) + (
n
2
Mode, M o
→ most frequent
Standard Deviation
Population standard deviation
→ Mode Stat 1 - var
→ Shift Mode ▼ Stat Frequency? on
→ Input
→ AC Shift 1 var σ x
Sample standard deviation
→ Mode Stat 1 - var
→ Shift Mode ▼ Stat Frequency? on
→ Input
→ AC Shift 1 var s x
NOTE:
If not specified whether population/sample
in a given problem, look for POPULATION.
Coefficient of Linear Correlation
or Pearson’s r
→ Mode Stat A+Bx
→ Input
→ AC Shift 1 Reg r
NOTE:
- 1 ≤ r ≤ +1; otherwise erroneous
Population standard deviation
Variance
standard deviation = σ
variance = σ
2
relative variability = σ/x
Mean/Average Deviation
Mean/average value
mv =
b − a
∫ f(x)dx
b
a
Mean value
RMS = √
b − a
∫ f(x)
2
dx
b
a
Walli’s Formula
∫ cos
m
θ sin
n
θ dθ =
[(m − 1 )(m − 3 )(m − 5 ) … ( 1 or 2 )][(n − 1 )(n − 3 )(n − 5 ) … ( 1 or 2 )]
( m + n
)( m + n − 2
)( m + n − 4
) … ( 1 or 2 )
π
2
0
∙ α
NOTE:
α = π/2 for m and n are both even
α =1 otherwise
x = r cos θ
y = r sin θ
r = x
2
2
θ = tan
− 1
y
x
Discrete Probability Distributions
Binomial Probability Distribution
P(x) = C(n, x) p
x
q
n−x
where:
p → success
q → failure
Geometric Probability Distribution
P
x
= p(q
x− 1
Poisson Probability Distribution
P(x) =
μ
x
e
−μ
x!
Transportation Engineering
Design of Horizontal Curve
Minimum radius of curvature
R =
v
2
g(e + f)
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/s
2
Centrifugal ratio or impact factor
Impact factor =
v
2
gR
R → minimum radius of curvature
v → design speed in m/s
g → 9.82 m/s
2
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
t =
3W
f
Rigid pavement with dowels
t =
3W
2f
t =
3W
4f
(at the edge) (at the center)
t → thickness of pavement
W → wheel load
f → allow tensile stress of concrete
Flexible pavement
t = √
W
𝜋f
1
− r
f 1
→ allow bearing pressure of subgrade
r → radius of circular area of contact
between wheel load & pavement
Thickness of pavement in terms
of expansion pressure
t =
expansion pressure
pavement density
Stiffness factor of pavement
SF = √
E
s
E
p
3
E S
→ modulus of elasticity of subgrade
EP→ modulus of elasticity of pavement
Traffic Accident Analysis
Accident rate for 100 million
vehicles per miles of travel in a
segment of a highway:
R =
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
Accident rate per million entering
vehicles in an intersection:
R =
A ( 1 , 000 , 000 )
ADT ∙ N ∙ 365
A → no. of accidents during period of analysis
ADT → average daily traffic entering all legs
N → time period in years
Severity ratio, SR:
SR =
f ∙ i
f ∙ i ∙ p
f → fatal
i → injury
p → property damage
Spacing mean speed, U S
:
U
s
d
∑ t
n
U
1
Time mean speed, Ut:
U
t
d
t
n
∑ U
1
n
Ʃd → sum of distance traveled by all vehicles
Ʃt → sum of time traveled by all vehicles
Ʃu 1
→ sum of all spot speed
1/Ʃu 1
→ reciprocal of sum of all spot speed
n → no. of vehicles
Rate of flow:
q = kU
s
q → rate of flow in vehicles/hour
k → density in vehicles/km
u S
→ space mean speed in kph
Minimum time headway (hrs)
= 1/q
Spacing of vehicles (km)
= 1/k
Peak hour factor (PHF)
= q/q max
s
Fractiles
Range
= 𝑙𝑎𝑟𝑔𝑒𝑠𝑡 𝑑𝑎𝑡𝑢𝑚 − 𝑠𝑚𝑎𝑙𝑙𝑒𝑠𝑡 𝑑𝑎𝑡𝑢𝑚
Coefficient of Range
=
𝑙𝑎𝑟𝑔𝑒𝑠𝑡 𝑑𝑎𝑡𝑢𝑚 − 𝑠𝑚𝑎𝑙𝑙𝑒𝑠𝑡 𝑑𝑎𝑡𝑢𝑚
𝑙𝑎𝑟𝑔𝑒𝑠𝑡 𝑑𝑎𝑡𝑢𝑚 + 𝑠𝑚𝑎𝑙𝑙𝑒𝑠𝑡 𝑑𝑎𝑡𝑢𝑚
Quartiles
when n is even
Q
1
1
4
n Q
2
2
4
n Q
3
3
4
n
when n is odd
Interquartile Range, IQR
= 𝑙𝑎𝑟𝑔𝑒𝑠𝑡 𝑞𝑢𝑎𝑟𝑡𝑖𝑙𝑒 − 𝑠𝑚𝑎𝑙𝑙𝑒𝑠𝑡 𝑞𝑢𝑎𝑟𝑡𝑖𝑙𝑒
= Q
3
− Q
1
Coefficient of IQR
=
𝑙𝑎𝑟𝑔𝑒𝑠𝑡 𝑞𝑢𝑎𝑟𝑡𝑖𝑙𝑒 − 𝑠𝑚𝑎𝑙𝑙𝑒𝑠𝑡 𝑞𝑢𝑎𝑟𝑡𝑖𝑙𝑒
𝑙𝑎𝑟𝑔𝑒𝑠𝑡 𝑞𝑢𝑎𝑟𝑡𝑖𝑙𝑒 + 𝑠𝑚𝑎𝑙𝑙𝑒𝑠𝑡 𝑞𝑢𝑎𝑟𝑡𝑖𝑙𝑒
=
Q
3
− Q
1
Q
3
+ Q
1
Quartile Deviation (semi-IQR) = IQR/
Outlier
→ extremely high or low data higher than
or lower than the following limits:
Q
1
− 1 .5IQR > x
Q
3
Decile or Percentile
i
m
m
10 or 100
(n)
Q
1
=
1
4
(n + 1 ) ; Q
1
=
1
4
(n + 1 ) ; Q
1
=
1
4
(n + 1 )
Normal Distribution
Z-score or
standard score
or variate
z =
x − μ
σ
x → no. of observations
μ → mean value, x̅
σ → standard deviation
→ Mode Stat
→ AC Shift 1 Distr
left of z → P(
right of z → R(
bet. z & axis → Q(
→ Input
Fibonacci Numbers
a
n
[(
n
n
]
Tip to remember:
𝑥
2
− 𝑥 − 1 = 0
Mode Eqn 5
𝑥 =
1 ± √ 5
2
Exponential Distribution
P
( x ≥ a
) = e
−λa
P(x ≤ a) = 1 − e
−λa
P
( a ≤ x ≤ b
) = e
−λa
− e
−λb
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
Function Period Amplitude
y = A sin (Bx + C) 2 π/B A
y = A cos (Bx + C) 2 π/B A
y = A tan (Bx + C) π/B A
Design of Beam Stirrups
(1st) Solve for Vu:
ΣF
v
V
u
= R − w
u
d
V
u
w
u
L
− w
u
d
(2nd) Solve for Vc:
V
c
√f
c
′b
w
d
(3rd) Solve for Vs:
V
u
= ∅(V
c
+ V
s
V
s
→ obtained
(4th) Theoretical Spacing:
s =
dA
v
f
y
n
V
s
NOTE:
fy n
→ steel strength for shear reinforcement
Av → area of shear reinforcement
n → no. of shear legs
A
v
π
d
2
∙ n
NSCP Provisions for
max. stirrups spacing:
2V
c
√f
c
′b
w
d
i. when Vs < 2Vc,
s
max
d
or 600mm
ii. when Vs > 2Vc,
s
max
d
or 300mm
iii. & not greater than to:
s
max
3 A
v
f
y
n
b
T-Beam
NSCP Provisions for effective flange width:
i. Interior Beam:
b
f
L
b
f
= b
w
s
1
s
2
b
f
= b
w
f
ii. exterior Beam:
b
f
= b
w
L
b
f
= b
w
s
1
b
f
= b
w
f
Thickness of One-way Slab & Beam
NSCP Provisions for minimum thickness:
C anti- S imple O ne B oth
lever Support End Ends
Slab L/10 L/20 L/24 L/
Beams L/8 L/16 L/18.5 L/
Factor: [ 0. 4 +
f
y
700
]
[
𝑐
]
(for lightweight concrete only)
Design of One-way Slab
(1st) Compute ultimate moment, Mu:
W
U
= 1 .4W
D
+ 1 .7W
L
M
U
W
U
L
2
(2nd) Solve for slab thickness, h:
See NSCP Provisions for minimum thickness.
(3rd) Solve for effective depth, d:
d = h − cc −
d
b
(4th) Solve for a:
M
u
= ∅(C) [d −
a
2
]
M
u
= ∅( 0. 85 f
c
′
ab) [d −
a
]
a → obtained
(5th) Solve for As:
C = T
- 85 f c
′
ab = A
s
f
y
A
s
→ obtained
LONGITUDINAL OR MAIN BARS
(6th) Compute steel ratio, ρ:
ρ =
A
s
bd
(7th) Check for minimum steel ratio:
ρ
min
f
y
& ρ
min
√f
c
′
4f
y
If ρ min
< ρ, use ρ.
If ρmin > ρ, use ρmin & recompute As.
(8th) Determine # of req’d main bars:
N =
A
s
A
b
A
s
π
d
b
2
(9th) Determine spacing of main bars:
s =
b
N
(10th) Check for max. spacing of main bars:
s
max
= 3h or 450mm
TEMPERATURE BARS/
SHRINKAGE BARS
(11th) Solve for As:
A
s
= kb
⫠
h
NSCP Provision for k:
i. fy = 275 MPa, k = 0.
ii. fy = 415 MPa, k = 0.
iii. fy > 415 MPa, k = 0.0018 (400/fy)
( 12 th) Determine # of req’d temp. bars:
N =
A
s
A
b
A
s
π
d
b
2
(13th) Determine spacing of temp. bars:
s =
b
N
(14th) Check for max. spacing of temp. bars:
s
max
= 5h or 450mm
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,
k = 0. 0018 [
400
f
y
]
For two-way bending:
ρ → steel ratio
ρ
min
f
y
ρ
min
√f
c
′
4f
y
(choose larger between the 2)
Design of Column
P = P
C
+ P
S
P = 0. 85 f c
′
(A
g
− A
st
) + A
st
f
y
ρ =
A
st
A
g
Thus,
A
g
P
- 85 f
c
′
( 1 − ρ
y
0 .01A
g
< A
st
< 0 .08A
g
TIED COLUMN
P
N
= 0 .8P
P
U
= ∅0.8P ; ∅ = 0. 7
P
U
= ( 0. 7 )( 0. 8 )[ 0. 85 f
c
′
(A
g
− A
st
) + A
st
f
y
]
No. of main bars:
N =
A
st
A
b
N is based on Pu.
NOTE: If spacing of main bars < 150mm, use 1 tie per set.
SPIRAL COLUMN
P
N
= 0 .85P
P
U
= ∅0.85P ; ∅ = 0. 75
P
U
= ( 0. 75 )( 0. 85 )[ 0. 85 f
c
′
(A
g
− A
st
) + A
st
f
y
]
ρ
s
f
c
′
f
y
[
A
g
A
c
− 1 ] =
volume of spiral
volume of core
s =
π
(d
sp
2
∙ π(D
c
−d
sp
π
(D
c
2
∙ ρ
s
4 A
sp
D
c
ρ
s
Spacing of bars:
s = 16d
b
s = 48d
t
s = least dimension
Design of Footing
q A
= q
S
C
sur
E
q E
P
A
ftg
; q
U
P
U
A
ftg
where:
qA → allowable bearing pressure
q S
→ soil pressure
qC → concrete pressure
q sur
→ surcharge
qE → effective pressure
qU → ultimate bearing pressure
Ø = 0.
WIDE BEAM SHEAR
V
U
= q
U
(B)(x)
V
U
≤ ∅V
wb
f
c
′
Bd
τ
wb
V
U
∅Bd
τ
wb(allw)
f
c
′
PUNCHING/DIAGONAL TENSION SHEAR
V
U
= P
U
− q
U
(a + d)(b + d)
V
U
≤ ∅V
pc
f
c
′
b
o
d
τ
pc
V
U
∅b
o
d
τ
pc(allw)
f
c
′
BENDING MOMENT
M
U
= q
U
(B)(x) (
x
** design of main bars and
temperature bars –
Same as slab.
