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
1 minute of arc =
1 nautical mile
n-sided Polygon
2
 Age Problems
→ underline specific time conditions
 Work Problems
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 = vt
Spherical Polygon:
πR2 E E = spherical excess
AB =
E = (A+B+C+D…) – (n-2)180°
Spherical Pyramid:
Triangle
→a=0
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
√A2 + B 2
Standard Equation:
Standard Equation:
Distance of two parallel lines:
(x − h)2 (y − k)2
+
=1
a2
b2
(x − h)2 (y − k)2
−
=1
a2
b2
(x − h)2 (y − k)2
+
=1
b2
a2
(y − k)2 (x − h)2
−
=1
a2
b2
d=
Point-slope form:
|C1 − C2 |
√A2 + B 2
Angle between two lines:
x y
+ =1
a b
tan θ =
m2 − m1
1 + m1 m2
Parabola
- the locus of point that moves such that it is always equidistant from a
fixed point (focus) and a fixed line (directrix).
Elements:
General Equation:
2
1 revolution
= 2π rad
= 360˚
= 400 grads
= 6400 mills
Tetrahedron
Line Tangent to Conic Section
To find the equation of a line
tangent to a conic section at a
given point P(x1, y1):
In the equation of the conic
equation, replace:
Eccentricity, e:
e=
2
H = a√
3
y"
3
[1 + (y′)2 ]2
V=a
3 √2
12
LR = 4a
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
r = a(1 − cos θ)
r = a(1 + cos θ)
(1 + 𝑖)n − 1
]
𝑖
CALTECH:
Mode 3 2
x
y
(time)
(BV)
0
FC
n
SV
 Sinking Fund:
 Perpetuity:
P=
where:
FC → first cost
(1 + i)n − 1 −1 SV → salvage cost
depreciation
d = (FC − SV) [
] dper→year
𝑖
n
→
economic
life
(1 + i)m − 1
m → any year before n
Dm = d [
]
BV → book value
𝑖
after m years
A
= F(1 + 𝑖)−n
𝑖
 Capitalized Cost:
C = FC +
OM
RC − SV
+
𝑖
(1 + 𝑖)n − 1
AC = C ∙ 𝑖
m
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
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:
FC − SV
Qn
D = dQ m
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
d=
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
′
 Sum-of-the-Years-Digit (SYD):
Integral Calculus-The Cardioid
 Annuity:
(1 + 𝑖)n − 1
P = A[
]
𝑖(1 + 𝑖)n
Dm → total depreciation
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)
=1
Length of latus
rectum, LR:
cost = revenue
Maxima & Minima (Critical Points):
𝑑𝑦
= y′ = 0
𝑑𝑥
dd
Location of foci, c:
SA = a √3
Differential Calculus
Curvature:
df
Elements:
Elements:
2
𝑥 2 → 𝑥𝑥1
𝑦 2 → 𝑦𝑦1
𝑥 + 𝑥1
𝑥→
2
𝑦 + 𝑦1
𝑦→
2
𝑥𝑦1 + 𝑦𝑥1
𝑥𝑦 →
2
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
n+1
2
1 n
n
th
Me = [( ) + ( + 1)]
2 2
2
Q1 = n
1
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
Due to sag:
(subtract only); unsupported length
C=
w 2 L3
24P 2
Due to slope:
(subtract only); measured length
E = 0.6745√
CD = MD (1 +
CD = MD (1 −
∑(x − x̅)
n−1
Probable Error (mean):
∑(x − x̅)
Em =
= 0.6745√
n(n − 1)
√n
E
Proportionalities of weight, w:
𝑤∝
Normal Tension:
0.204W√AE
1
𝐸2
𝑤∝
1
𝑑
𝑤∝𝑛
Area of Closed Traverse
√PN − P
Symmetrical:
L
H = (g1 + g 2 )
8
L 2
x 2 ( 2)
=
L
y
H 1
Error of Closure:
Error of Closure
Perimeter
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:
1 acre =
4047 m2
h = h2 +
Simpson’s 1/3 Rule:
= √ΣL2 + ΣD2
Azimuth
hcr = 0.067K 2
Trapezoidal Rule:
Lat = L cos α
Dep = L sin α
=
e
)
TL
Effect of Curvature & Refraction
Area of Irregular Boundaries
A=
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 =
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=
θ
Ls 2
; p=
3
24R
x=
L3
6RLs
I
I
E = R [sec − 1]
L
Ve = (A1 + A2 )
2
I
2
m = R [1 − cos ]
2
Volume (Prismoidal):
L = 2R sin
L
VP = (A1 + 4Am + A2 )
6
L
(c − c2 )(d1 − d2 )
12 1
VP = Ve − Cp
Y=L−
2
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
VT = ABase ∙ Have = A ( )
n
A
VT = (Σh1 + 2Σh2 + 3Σh3 + 4Σh4 )
n
e=
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
L=
L5
I
π
Lc = RI ∙
180°
20 2πR
=
D
360°
1145.916
R=
D
Prismoidal Correction:
v → speed in m/s
t → perception-reaction time
f → coefficient of friction
G → grade/slope of road
L2 180°
∙
2RLs π
2
Volume (End Area):
CP =
θ=
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:
Q = CA o √2gh
C = Cc C v
a
Cc =
A
v
Cv =
vt
2As
h1
As dh
h2
CAo √2gh
t=∫
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
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
Vbel =
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
When S=0:
Relative Compaction:
ɤd
W
V
WS
ɤd =
V
n
1−n
g = Gs (1 − n)
0<n<1
e
n=
1+e
R=
(Gs − 1)ɤw
1+e
Gs ɤw
ɤzav =
1 + Gs ω
ɤsub =
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
Capillary Rise:
Pore Water Pressure/
Neutral Stress:
C
hcr =
eD10
pw = ɤw hw
Total Stress:
h1 h2
h
+ +. . . + n
k1 k 2
kn
Flow Net / Seepage
Isotropic soil:
q = kH
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
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:
cos β + √cos 2 β − cos 2 Ø
cos β − √cos 2 β − cos 2 Ø
For Horizontal:
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
kP =
4 56 7 8
Nf → no. of flow channels [e.g. 4]
Nd → no. of potential drops [e.g. 10]
 Normally consolidated:
PASSIVE PRESSURE:
k P = cos β
3
Shear Strength of Soil
θ = 45° +
cos α [1 + √
3
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
Compression Index, CC:
10
With Pre-consolidation pressure, Pc:
when (△P+Po) < Pc:
For Inclined:
ka =
1
D75
D25
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
D60 ∙ D10
So = √
Compressibility of Soil
Confined:
for Perpendicular flow:
Hazen Formula
Cc =
3
1
1
Sn = 1.7√
+
+
2
2
(D50 )
(D20 )
(D10 )2
r
Q 𝑙𝑛 1
r2
k=
π(h1 2 − h2 2 )
h1 k1 + h2 k 2 +. . . +hn k n
H
D60
D10
Sorting
Coefficient:
Suitability Number:
Unconfined:
for Parallel flow:
Class
AC < 0.7
Inactive
0.7 < AC < 1.2 Normal
AC > 1.2
Active
Description
0
Non-plastic
1-5
Slightly plastic
5-10 Low plasticity
10-20 Medium plasticity
20-40 High plasticity
>40 Very High plastic
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
SL =
Bulk Specific Gravity:
(Gs + e)ɤw
ɤsat =
1+e
ɤ=
0<e<∞
e=
Gs =
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
′
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
q = ɤDf (for no water table)
qult Pallow
qallow =
=
FS
A
qult − q
qnet =
FS
Circular Footing:
FS =
EFFECT OF WATER TABLE:
FS =
Ff + Fc
W sin 𝜃
Hcr =
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
4𝐶
sin 𝛽 cos ∅
[
]
ɤ 1 − cos(𝛽 − ∅)
Stability No.:
Stability Factor:
C
m=
ɤH
SF =
θ
β
Maximum height for critical equilibrium
(FS=1.0)
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:
3 q2
Take note that it is only derived from
the critical depth equation.
dc = √
Critical Flow
Subcritical Flow
Supercritical Flow
q=
NF = 1
NF < 1
NF > 1
Reynold’s Number
Dv Dvρ
=
υ
μ
g
2
= Ec
3
Q
B
v2
E𝑐 =
+ d𝑐
2g
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
NR =
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
1
j= 1− k
3
f
n+ s
fc
Moment Resistance Coefficient, R:
1
R = fc kj
2
Moment Capacity:
1
Mc = C ∙ jd = fc kdb ∙ jd = Rbd2
2
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
(for uniformly distributed load)
384Ec Ie
 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 − x)2
 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 =
3
INA → obtained
Mc ∙ x
INA
Solutions for Gross Section (Singly):
3
INA → obtained
Mc ∙ x
INA
For tension steel:
fs Ms ∙ (d − x)
=
n
INA
For comp. steel:
fs′
Ms′ ∙ (x − d′)
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
Mcr → obtained
fr = 0.7√fc′ =
 Solve transferred moment of inertia at NA:
3
3
bx
b(d − x)
+
+ (n − 1)As (d − x)2
3
3
INA → obtained
INA =
 Solve for Stresses or Resisted Moment:
For tension steel:
For concrete:
fc =
Mc ∙ x
INA
fs Ms ∙ (d − x)
=
n
INA
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
 Coefficient of resistance, Rn:
 a = 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
(fc′ − 30)
(fc′ − 28)
 Combined ρ & Rn:
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:
Maximum & Minimum steel ratio:
ρmax = 0.75ρb
As max = 0.75As b
1.4
ρmin =
fy
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=
C=T
0.85fc′ ab = As fs
d−c
0.85fc′ β1 cb = As ∙ 600 [ ]
c
c → obtained
a = β1 c
a → obtained
(3rd) Solve for Moment Capacity:
a
Mu = ∅(C or T) [d − ]
2
a
Mu = ∅(0.85fc′ ab) [d − ] or
2
a
Mu = ∅(As fs ) [d − ]
2
(4th) Solve for area of steel
reinforcement, As and required no. of
bars, N:
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
20 mm → slab
Balance Condition for Doubly
C=T
0.85fc′ ab = As fs
0.85fc′ ab = As fy
√fc′
ρmin =
4fy
Minimum Concrete Covers:
Singly Reinforced Beam (SRB)
(1st) Compute for a:
(assume tension steel yields fs=fy)
0.85fc′ β1 600
fy (600 + fy )
Singly or Doubly ?
As
ρ=
bd
ρ=
ρb =
(choose larger between the 2)
Mu
Rn =
∅bd2
 Steel reinforcement ratio, ρ:
0.65 ≤ β1 ≤ 0.85
β1 = 0.85 −
10
R n = fc′ ω(1 − ω)
17
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.
BY: NTDEGUMA
REINFORCED CONCRETE DESIGN FORMULAS AND PRINCIPLES
HISTORICAL BACKGROUND
TYPES OF PROBLEMS
WORKING STRESS DESIGN (WSD)
CONCRETE
a mixture of sand, gravel, crushed rock or other
aggregates held together in a rock-like mass with a
paste of cement and water.
1. Design - given the load, determine the size
DESIGN OF BEAMS FOR FLEXURE
ADMIXTURES
materials added to concrete to change certain
characteristics such as workability, durability and time
of hardening.
MODES OF FAILURE IN BENDING
JOSEPH ASPDEN
2. Yielding of Steel - when the actual tensile stress of
an English bricklayer who obtained a patent for
Portland cement
JOSEPH MONIER
a Frenchman who invented reinforced concrete a
received a patent for the const. of concrete basins and
tubs and reservoirs reinforced w/ wire mesh and iron
wire in 1867.
DESIGN METHODS:
1. WSD - Working Stress Design, Alternate
Stress Design,or Straight-Line Design
2. USD -
Ultimate Stress Design or Strength Design
2. Investigation - given the size, determine the load
b
- 0.45 fc' ( beams/slabs/footings)
- 0.25 fc' ( columns)
fc' - specified compressive strength of conc.
at 28 days curing (MPa)
gconc.- unit weight of concrete
- 23.54 KN/m 3
Ec - modulus of elasticity of concrete
- 4700
fc' (MPa)
STEEL :
fs - allowable tensile stress of steel (MPa)
fs - 0.50 fy ( beams/slabs/footings)
fs - 0.40 fy ( columns)
fy - yield stress of steel (MPa)
gsteel - unit weight of steel
- 77 KN/m3
Ec - modulus of elasticity of concrete
fc = 0.45 fc'
C
h
1. Crushing of Concrete - when the strain concrete
d-kd
reaches the ultimate strain of 0.003 mm/mm.
3. Simultaneous crushing of concrete and
Yielding of Steel
As
z
Beam Section
nAs
Transformed
Section
fc/n
T
TYPES OF DESIGN
1. Overreinforced - when failure is due to crushing of
concrete.
2. Underreinforced - when failure is initiated by yielding
of steel.
3. Balanced Design - when failure is caused by
simultaneous crushing of concrete and
yielding of steel
Mc (kd)
IN.A.
where:
Mc - resisting moment
of concrete
Stress of Steel
fs = Ms (d - kd)
n
IN.A.
where:
Ms - resisting moment
of steel
Stress Diagram
where:
Compressive force of Concrete
C = 1/2 fc kd b
h = overall depth of the beam (mm)
z = steel covering (measure from the centroid of bar)
d = effective depth of the beam (mm)
d = h -z
As = area of the reinforcement ( square millimeters)
fc' = compressive strength of concrete (MPa)
fs = tensile strength of steel (MPa)
b = base of the beam (mm)
n = modular ratio(always a whole number)
n = Es /Ec
Location of the neutral axis (kd)
Tensile force of Steel
T = As fs
Moment Arm ( jd )
d = jd + kd/3
j = 1 - k/3
S MN.A. = 0
FACTORED LOAD COMBINATION
fc - allowable compressive stress of conc.
fc =
N.A.
PROPERTIES OF MATERIALS:
CONCRETE:
b
Stress of Concrete
kd
steel "fs" reaches the yield stress "fy"
BY: NTDEGUMA
REINFORCED CONCRETE DESIGN FORMULAS AND PRINCIPLES
b(kd)(kd/2) - nAs (d - kd) = 0
kd = --------
(NSCP C101-01)
Constant ( k )
k =
Moment of Inertia of the Transformed Section
1. U = 1.4DL + 1.7LL
2. U = 0.75(1.4DL + 1.7LL + 1.7 W)
U = 0.90DL + 1.3W
> (# 1)
I N.A.= (1/3)(b)(kd)³ + nAs (d - kd)²
n
n + fs/fc
( For Design Only )
k = 2rn + (rn)² - rn (For Investigation)
Only )
3. U = 1.1DL + 1.3LL + 1.1E)
U = 0.90DL + 1.1E
> (# 1)
4. U = 1.4DL + 1.7LL + 1.7H
U = 0.90DL
> (# 1)
Resisting Moment of Concrete:
Steel Ratio
Mc = C(jd)
Mc = fc/2 (b)(kd)(jd)
Mc = (1/2)(fc)(kj)(bd²)
DL - Dead Load
E - Earthquake Load
LL - Live Load
H - Earth Pressure
r =
As
bd
Resisting Moment of Steel:
W - Wind Load
Ms = T jd
Ms = As fs jd
- 200,000 MPa
CECC-3
DESIGN AND CONSTRUCTION
1/20
CECC-3
DESIGN AND CONSTRUCTION
2/20
BY: NTDEGUMA
REINFORCED CONCRETE DESIGN FORMULAS AND PRINCIPLES
ULTIMATE STRENGTH DESIGN (USD)
A. BEAMS (FLEXURAL STRESS)
2. Doubly Reinforced Rectangular Beam
(Reinforced in Tension/Compression)
1. Singly Reinforced Rectangular Beam
(Reinforced in Tension Only)
b
0.85 fc'
C = 0.85 fc'ab
a
c
b
0.85 fc'
As'
C'
C
a
d
d
Mn
N.A.
As
d - a/2
b1 = 0.85 - 0.008 (fc' - 30)
C. SHEAR STRESS AND DIAGONAL
TENSION
d'
As' fs'
C'
fy ( 600 + fy)
r fy
If r < rb steel yields , proceed to step III
If r > rb steel does not yield , proceed to
step IV
III. r < rb
Mu2
Vn = Vc + Vs
r fy
fc'
2
Mu = f fc' bd w ( 1 - 0.59 w )
IV. r > rb (fy = fs)
b
where f = 0.85
d - d'
As2
c
Vc =1/6 fc' bd
(Resisting Moment)
fc' w ( 1 - 0.59 w)
2Ru
10.85 fc'
fy
r bd
Balanced Steel Ratio
( rb )
rb = 0.85 fc'b1 600
fy ( 600 + fy)
Mn
d - a/2
d-c
1st STAGE
2nd STAGE
Forces:
Forces:
S=
C 1 = 0.85 fc' ab
T = As 1 fs
C' = As' fs'
T1 = As 2 fs (d-d')
Resisting Moment:
Resisting Moment:
Es = 200,000
Solve for fs from the strain diagram:
Av fy d
Vs
fs/Es
=
d-c
NSCP/ACI Code Specs:
Mu1 = f 0.85fc'ab (d-a/2)
Mu2 = f As'fs' (d-d')
Mu1 = f As1 fs (d-a/2)
Mu2 = f As2 fs(d-d')
If Vs < 1/3 fc' bd ,Smax = d/2 or 600mm
If Vs > 1/3 fc' bd , Smax = d/4 or 300mm
Av min = bS/3fy
TOTAL :
T = T1 + T2
Maximum and Minimum Steel Ratio
rmax = 0.75 r
Vu =
factored or ultimate shear
Vc =
shear force provided by conc.
Vn =
nominal shear
Avmin = area of steel to resist shear
b
0.003
c
; fs = 600
d-c
c
Solve for c by summing up forces along hor.
T = C ; a = b1 c
2
600 As (d-c) = 0.85 b1 f'c b c
Use quadratic formula to solve for "c"
Then, solve for fs and "a" with known "c"
fs = 600
A S = As 1 + As2
e = fs/Es
T = As fy
Spacing of Stirrups:
MU= MU1 + MU2
d-c
;
c
a = b1c
Finally, solve for Mu:
M u = f 0.85fc'ab (d-a/2) or
M u = f As fs (d-a/2)
= 2 Asteel
rmin = 1.4 / fy
DESIGN AND CONSTRUCTION
As
0.003
c
d
N.A.
2nd STAGE
2
0.85 fc'
C = 0.85 fc'ab
a
T2 = As 2 fs'
fc'
r = 0.85 fc' 1 -
CECC-3
rb = 0.85 fc' b1600
Vu = f Vn
As'
As fy
0.85 fc' b
bd
II. Check if steel yields by computing rb
w =
T1 = As 1 fs
1st STAGE
but should not be less than 0.65
As =
f = strength reduction factor
r = As
I. Solve for
r = ratio of tension reinforcement = As/bd
rb= balance steel ratio
Mu1
As1
For fc' < 30 MPa , use b1 = 0.85
For fc' > 30 MPa ,
Ru =
steel area (As)
Mu = factored moment at section, (N-mm)
a/2
d - a/2
Mu = f Ru bd
A. Computing Mu with given tension
Mn = nominal moment, (N-mm)
a = b1 c
w =
FOR SINGLY REINFORCED BEAM
fc' = specified compressive stress of concrete (MPa)
0.85 fc'
b
C
a =
a = depth of equivalent stress block
As = area of tension reinforcement, square millimeters
b = width of the compression face of member
c = distance from extreme compression fiber to N.A. (mm)
d = distance from extreme compression fiber to
centroid of tension reinforcement (mm)
fy = specified yield strength of steel (MPa)
T = As fy
Elevation
Beam Section
T
As
where:
d' = thickness of concrete cover measured from extreme
tension fiber to center of the bar or wire, (mm)
N.A.
BY: NTDEGUMA
REINFORCED CONCRETE DESIGN FORMULAS AND PRINCIPLES
3/20
CECC-3
DESIGN AND CONSTRUCTION
4/20
BY: NTDEGUMA
REINFORCED CONCRETE DESIGN FORMULAS AND PRINCIPLES
B. Computing the required tension steel
area (As) of beam with given Mu
IV. Verify of compression will yield.
ec
b
I. Solve for rmax and M umax
c
rmax = 0.75 0.85 fc' b1600
=
fy ( 600 + fy)
As'
fs' = 600 c - d'
c
where:
es' = fs'/Es
c-d
If fs' < fy, proceed to step V.
es
MU= MU1 + MU2
MU = f As1 fy (d-a/2) + f As' fy(d-d')
If fs' > fy, proceed to step V.
2
M umax = f fc' bd w ( 1 - 0.59 w )
V. Since fs' < fy, assumption is wrong
If fs' < fy, proceed to step VI.
If Mu < Mumax design as Single Reinforced
then, proceed to step II.
d' C2 = As'fs'
(compression steel yields)
II. Solve for r
2
1-
fy
2Ru
0.85 fc'
C2 = T2
fs' = 600
As' fs' = As 2 fy
= ____
[ C1 + C2 = T ]
r bd = _________
0.85 fc' ab + As' fs' = As fy
a = b1 c
b
Reinforced Beam with given Mu.
C
a
d
As'
C
d
C'
a/2
Mu1
d'
C'
d-d'
d-a/2
Mu = f 0.85fc'ab (d-a/2) + f As' fs' (d-d')
T1 = As 1 fy
T2 = As fy
I. Assume Compression steel yield
T2 = As 2 fs'
I. Solve for As 1 = r maxbd
II. Solve for "a" and "c": C 1 = T1
0.85 f'c ab = As1 fy ; a = ____
a = b1 c ; c = ______
III. Solve for MU1, MU2 and As2
If Vu < f Vc , but Vu > 1/2 f Vc
proceed to to Step V
III. Calculate the shear strength Vs to be
provided by the stirrup.
2. Vs = Vn - Vc = Vu /f - V c
If VS < 2/3 fc' bW d , proceed to IV.
If VS > 2/3 fc' bW d , adjust the size
of the beam
IV. Spacing of stirrups:
Spacing, S = Av fy d
Vs
If S < 25mm, increase the value of Av.
either by bigger bar or shear area.
Maximum spacing, s:
If Vs < 1/3 fc' bd ,Smax = d/2 or 600mm
d-d'
Mu2
T1 = As 1 fs
a/2
d'
As
d-a/2
Solve for c by quadratic formula
Solve for fs' and "a"
Solve for Mu :
As' fs'
0.85 fc'
As'
As' fs'
0.85 fc'
c - d'
0.85 fc' b1 c b + As' 600 c = As fy
Beam with given As and As'
A. Computing As and As' of a Doubly
c - d'
c
From stress diagram.
As' = ________
FOR DOUBLY REINFORCED BEAM
Vu > f Vc , stirrups is necessary,
proceed to to Step III.
1. Vn = Vu /f
T = As fy
B. Computing Mu of a Doubly Reinforced
As
c
d - a/2
(compression steel will not yield)
(Solve for Ru)
r = 0.85 fc' 1 -
b
C1
NA
As' = As2
VI. fs' < fy, then use fs'
Mu = f Ru bd
As =
a
Vc =1/6 fc' bd
If
If Vu < 1/2 fV c , stirrups are not needed
0.85 fc'
V. fs' > fy, then use fs' = fy
If Mu > Mumax design as Doubly Reinforced
II. Calculate the shear strength provided
by concrete, Vc
IV. Since fs' > fy, compression steel yields
Es = 200,000
c - d'
fs' = 600
c
fc'
VERTICAL STIRRUP DESIGN
I. Compute the factored shear force, Vu
If fs' > fy, proceed to step IV.
As
Mumax = with considered factored load
r fy
w =
= ________
III. Verify if Compression steel will yield
c-d'
N.A.
r
d'
BY: NTDEGUMA
REINFORCED CONCRETE DESIGN FORMULAS AND PRINCIPLES
If Vs > 1/3 fc' bd , Smax = d/4 or 300mm
(fs' = fy)
V. If Vu < f Vc , but Vu > 1/2 f Vc
As 2 = As' = _______
As 1 = As - As' = _______
Av min = bw S /3fy
where S = d/2 or 600mm (whichever
is smaller
II. Solve for a and c:
[ C1 = T1 ]
0.85 fc' ab = As1 fy ; a = ____
a = b1 c ; c = ______
Mu1 = f As1 fy ( d-a/2 )
Mu2 = MU1 - MU
Mu2 = f As2 fy(d-d') ; As2 = ____
CECC-3
DESIGN AND CONSTRUCTION
5/20
CECC-3
DESIGN AND CONSTRUCTION
6/20
BY: NTDEGUMA
REINFORCED CONCRETE DESIGN FORMULAS AND PRINCIPLES
TYPICAL RESISTANCE FACTORS ARE AS FOLLOWS:
SHEARING STRESS OF RC BEAMS
For members subjected to significant axial tension:
SITUATION
Flexure, without axial load
0.90
Axial tension and axial tension w/ flexure
0.90
Shear and torsion
0.85
Compression members, spirally reinforced
0.75
Other Compression members
0.70
Bearing on concrete
0.70
Plain Concrete: flexure, compression, shear
and bearing
0.65
Vn = Vc + Vs
rmax = 0.75 rb
To avoid sudden tensile failure : rmin = 0.25 fc'
To control deflection:
>
fy
r < 0.18 fc'
fy
BALANCED STEEL RATIOS
1.4
fy
1. BEAM REINFORCED FOR TENSION
fy ( 600 + fy)
where:
Checking Ductility
if
where:
r'= As'
r < r , tension steel yields fs = fy
bd
rw = As
bw d
1. In T-beam construction, the flange and web
shall be built integrally or otherwise effectively
bonded together.
2. The width of slab effective as a T-beam shall
not exceed 1/4 of the span of the beam, and
effective overhanging flange on each side of the
web shall not exceed:
bw d
rlim = 0.85 b fc' d' 600 + r'
fy d (600-fy)
r < rlim , compression steel yields fs = fy
choose the
smallest
4. Isolated beams in which T-shape are used to
provide a flange for additional compression
area shall a flange thickness not less than 1/2
the width of the web and an effective flange
width not more than four times the width of
the web.
b
t
b1
a) 8 times the slab thickness and
b) 1/2 the clear distance to the next web
Mu = factored moment ocurring
simultaneously w/ Vu
3. For beams with slab on one side only, the
effective overhanging flange shall not exceed:
b2
bw
a) 1/12 the span length of the beam,
b) 6 times the slab thickness
c) 1/2 the clear distance to the next web
For members subjected to axial compression:
t > bw /2
Mm = Mu - N u
For compression steel
if
Code Requirements for T-beams
Vc = 0.30 fc' bw d
Vu d < 1.0
Mu
2. BEAM REINFORCED FOR COMPRESSION
r = rb + r '
fc' b w d 1 + Nu
14Ag
fc' + 120 rw Vu d
Mu
1) b = L/4
2) b = 16t + b w
3) b = center-center
spacing of beams
T - BEAMS
but shall not be greater than
rb = 0.85 fc'b1 600
choose the
smallest
For Symmetrical Interior Beam
For members subjected to shear and flexure:
1
7
1) b' = L/12 + b'w
2) b' = 6t + b'w
3) b' = S3 /2 + b'w
Distance of Stirrups from support:
fc' bw d
where:
Ag = gross area of section in sq.mm
Nu = factored axial load occurring with Vu
(- ) for compression, (+) for tension
Nu/Ag = expressed in MPa
Vc =
choose the
smallest
For End Beam
a. 0.50 S from face of column support
b. 0.25 S from face of beam support
For members subjected to axial compression:
Vc = 1
6
1) b = L/4
2) b = 16t + bw
3) b = S1 /2 + S2 /2 + b w
0.30N u
Ag
Nu/Ag = expressed in MPa
Nu is negative for tension
For shear reinforcement, fy < 414 MPa.
For members subjected to shear and flexure only:
Vc = 1
6
fc' bw d 1 +
For Interior Beam
where:
where:
CODE PROVISIONS: FOR DESIGN OF
SINGLY-REINFORCED BEAMS
To ensure yield failure:
Vc = 1
6
Nominal Shear Strength Provided by Concete:
Vn = nominal shear strength of RC section
Vc = nominal shear strength provided by concrete
Vs = nominal shear strength of the shear reinforcement
BY: NTDEGUMA
REINFORCED CONCRETE DESIGN FORMULAS AND PRINCIPLES
4h - d
8
b
but shall not be greater than
Vc = 0.30 fc' bw d
b1
1+
0.30N u
Ag
b'
b2
b < 4b w
b3
t
S1
Substitute Mm for Mu and Vud/Mu not limited to 1.0
bw
S2
Interior Beam
S3
bw'
End Beam
where, h = overall thickness of member
CECC-3
DESIGN AND CONSTRUCTION
7/20
CECC-3
DESIGN AND CONSTRUCTION
8/20
BY: NTDEGUMA
REINFORCED CONCRETE DESIGN FORMULAS AND PRINCIPLES
A. Steps in determining the Tension Steel
Area As of a T-Beam with given Mu
I. Assume that the entire flange is in compression
and solve for Mu1:
Compression force in concrete:
B. Steps in Determining Mu of a T-Beam
with given As.
III. a > t
b
1
t
d
z
a
0.85 fc'
C1 t/2
d-t/2
d'
As
T1 = As 1 fy
bw
Mu1
T2 = As 2 fy
Mu2
Mu1 = _____________
If Mu1 > Mu, then a < t, proceed to Step II
Mu2 = Mu - Mu1
If Mu1 < Mu, then a > t, proceed to Step III
Mu2 = f C2 (d'-a/2)
C=T
0.85 fc' Ac = As fy
Ac = _____
t
b
z = _______
a
d
If fs > fy, steel yields (correct assumption)
If fs < fy, steel does not yield (seldom happen)
Mu1 = f C1 (d - t / 2)
Mu1 = f 0.85 fc' A f (d - t / 2)
Mu2 = f C2 (d' - z / 2)
Mu2 = f 0.85 fc' bw z (d' - z / 2)
II. a < t
t
0.85 fc'
(see Steps I for values of Ac and Af)
Verify if steel yields:
c = a / b1 = ______ fs = 600(d-c)/c = _____
Area of compression flange, Af = bf t
If Ac < Af, a < t, proceed to Step II
If Ac > Af, a > t, proceed to Step III
Mu2 = f 0.85 fc' bw z (d'-z/2)
b
Ac = A f + bw z
a = t + z = _____
d'-z/2
Mu = Mu1 + Mu2
Mu1 = the same value in Step 1
II. a < t
Solve for z:
C2
2
C = 0.85 fc' b f t
Mu1 = f C(d - t/2)
Mu1 = f 0.85 fc' bf t(d - t/2)
I. Assume steel yields (fs = fy) and compute the
area of compression concrete, Ac
0.85 fc'
BY: NTDEGUMA
REINFORCED CONCRETE DESIGN FORMULAS AND PRINCIPLES
0.85 fc'
a
C =0.85 fc' ab
d
Mu = Mu1 + Mu2
d-a/2
C =0.85 fc' ab
T=C
d-a/2
T = As fy
Solve fo r min = 1.4 / fy and compare with As
bw d
If As
bw d
T=C
As fy = 0.85 fc' ab
As = _______
If As <
bw d
c = a / b1
r min , use r = r min(seldom)
Solve fo r min = 1.4 / fy and compare with As
bw d
If As
> r min , design is OK!
bw d
If As
< r min , use r = r min (seldom)
bw d
1. Two or more spans
2. Loads are uniformly distributed
3. Beams or slabs are prismatic
4. L - S < 20%S
5. 1.7 wll < 3.0
1.4 wdl
fs = 600 (d-c) / c
If fs > fy, steel yields (correct assumption)
If fs < fy, steel does not yield (seldom happen)
Solve for Asmax .
If As
> r max , beam needs compression
bf d
steel (seldom happen)
Requirements:
Verify if steel yields(this may not be necessary)
As = r min b w d
As
bf d
If As
< r max , design is OK!
bf d
As = r min b w d
> r min , design is OK!
Beams and Slabs
Ac = bf x a
a = ____
Mu = f As fy (d-a/2)
As = _______
Mu = f 0.85 fc' ab (d-a/2)
a = _______
ACI/NSCP Coefficients for Continuous
Solve for a:
As fy = 0.85 fc' b t + 0.85 fc' bw z
Solve for a:
Mu = f C (d-a/2)
Solve for r max and compare with
T = As fy
As fy = C1 + C2
w (kN/m)
III. a > t
600 d
a = b1
600 + fy
As max = 0.75 A sb
0.85 fc' ( b f t + (a-t) bw
As max = 0.75
fy
b
1
t
d
z
0.85 fc'
a
C1 t/2
d-t/2
As
bw
If As < As max , value is OK
wL1 /2
wL/2
wL/2
d'-z/2
Mu1
If As > As max , the beam needs
compression steel (seldom happens)
1.15wL/2
V-D
-wL/2
-1.15wL/2
T1 = As 1 fy
L4
0.85 fc'
C2
2
d'
L3
L2
L1
T2= As 2 fy
Mu2
2
wL1 /14
2
-wL /10
2
-wL1 /16
2
wL2 /16
2
-wL/2
2
wL3 /16
-wL4 /2
2
wL4 /11
2
2
-wL /11
-wL /11
-wL4 /24
2
2
2
-wL /11
-wL /11
-wL /10
Note:
L = the average span between adjacent
spans in shear and negative moment
CECC-3
DESIGN AND CONSTRUCTION
9/20
CECC-3
DESIGN AND CONSTRUCTION
10/20
BY: NTDEGUMA
REINFORCED CONCRETE DESIGN FORMULAS AND PRINCIPLES
COLUMNS
ECCENTRICALLY LOADED COLUMN
2. SPIRAL COLUMNS
Classification of column as to:
Minimum spiral steel ratio
A. REINFORCEMENT
Dc
P
: rs
ex
D
ey
rs = 0.45 [ (D/Dc)2 - 1 ] fc'/fy
1. TIED COLUMNS
Pu
s
s
s
y
rs = 4 A sp (Dc - db )
longitudinal bars
2
s Dc
P P
P
P
P
where:
Resisting Axial Load:
Pu = f 0.80 Ag [0.85fc'(1- rg )+ rg fy)]
f = 0.70 for tied column
ey
Mx = Pey
P
P
rg = 0.01 - 0.08
My = Pex
rg = gross steel area
= As/Ag
As = total steel area
db = bar diameter
2. Minimum side cover = 40 mm
3. Minimum vertical bars
4 - 16mm dia. - for rec. section
6 - 16mm dia. - for round section
4. Minimum lateral tie bar dia.
10mm dia.- for < 32 db main bar
12mm dia.- for > 32 db main bar
5. Spacing of lateral ties (use the smallest)
a. 16 vert. bar diameter
b. 48 lateral tie bar diameter
c. least column dimension
6. Minimum side dimension of
column = 200 mm
7. Clear distance between longitudinal bars
a) 1.5 times bar diameter
b) 1.5 times max. size of coarse aggregate
8. Minimum covering of ties
a) 40 mm for interior columns
b) 50 mm for exterior columns
c) 1.5 times max. size of coarse aggregate
9. When there are more than four vertical bars,
additional ties shall be provided so that every
longitudinal bar will be held firmly in position. No
bar can be located at a greater distance than 150 mm
clear in either side from a laterally supported bar.
CECC-3
Applied Axial Load:
Pu = 1.4 DL + 1.7 LL
ACI Code specs:
1.
ex
Asp = area of the spiral reinforcement
rs = spiral steel ratio
Dc = core diameter (mm)
Note:
To be safe, Pu act. < Pu res.
DESIGN AND CONSTRUCTION
x
S = spacing of spiral ties
Applied Axial Load:
Pu = 1.4 DL + 1.7 LL
BY: NTDEGUMA
REINFORCED CONCRETE DESIGN FORMULAS AND PRINCIPLES
3. COMPOSITE COLUMN
Resisting Axial Load:
Pu = f 0.85 Ag [0.85fc'(1- rg )+ rg fy)]
A. Compression plus Uniaxial
Bending
Steel
Section
f = 0.75 for spiral column
My = Pex
emin = 0.10 h for rectangular section
emin = 0.05 D for circular section
To be safe, Pu act. < Pu res.
where:
ACI Code specs:
1.
rg= 0.01 - 0.06
B. SLENDERNESS
1. Short Column
Klu/r < 34 - 12 M1/M2
2. minimum diameter = 250 mm
3. min. vertical bars = 6-16 mm
4. minimum spiral = 10 mm
5. clear distance between vertical bars
a) 1.5 times bar diameter
b) 1.5 times max. size of coarse aggregate
6. spacing of spirals
a) not more that 75 mm
b)not less than 25 mm
c) not less than 1.5 times coarse aggregate
d) not more than one-sixth
2. Slender Column
Klu/r > 34 - 12 M1/M2
1. Square/rectangular
2. Round/Circular
D. LOAD
4A sp
rs Dc
1. Axially Loaded
Pu
1. e = 0
c.g.
C. SECTION
7. Spacing of spiral tie:
s =
h = column dimension parallel to
eccentricity (mm)
D = column diameter (mm)
Axially Load:
Pu = f 0.80 Ag [0.85fc'(1- rg )+ rg fy)]
2. Eccentrically loaded
a. Uniaxial bending
b. Biaxial bending
11/20
CECC-3
DESIGN AND CONSTRUCTION
Pu = f 0.85 Ag [0.85fc'(1- rg )+ rg fy)]
12/20
BY: NTDEGUMA
REINFORCED CONCRETE DESIGN FORMULAS AND PRINCIPLES
2. e = e min
e
c.g.
Compression plus Uniaxial Bending:
Pu
e
As
pc
emin
Pn
Pu
ACI Moment Magnifier Method
Factored Design Moment:
As'
Mc = dbM 2b + ds M2s
where:
pc
Axially loaded (Neglect the effect of moment)
Pn
e
Pu = f 0.80 Ag [0.85fc'(1- rg )+ rg fy)]
BY: NTDEGUMA
REINFORCED CONCRETE DESIGN FORMULAS AND PRINCIPLES
b
As
b = bending
s = sidesway
d = moment magnification factor
As'
Moment Magnifiers
Pu = f 0.85 Ag [0.85fc'(1- rg )+ rg fy)]
0.85fc'
d
T1 T2
C1'
0.85fc'
3. e min < e < eb
T
Eccentrically loaded
Consider effect of moment
c.g.
emin
a
c
c.g.
Eccentrically loaded
Consider effect of moment
fs' = fy
fs = fy
Pu
e
c.g.
Eccentrically loaded
Consider effect of moment
eb
Failure initiated by yielding
of tension steel
fs = fy
6. e b <<< e
CECC-3
eb
DESIGN AND CONSTRUCTION
ds =
Cm
> 1.0
1 - S Pu
fS Pc
Pny - nominal load cap.of column at
e & e =0
A. Columns braced against sidesway
1. When Klu/r < 34 -12 M1 /M 2 , column is short.
2. When Klu/r > 34 -12 M 1 /M 2 , column is slender.
Ag = bh
B. Unbraced Columns
1. When Klu/r < 22, column is short.
Mn = Pn (e)
Mu = f Mn
2. When Klu/r > 22, column is slender.
SHORT ECCENTRICALLY LOADED
ROUND COLUMNS
Column Interaction Eqtn: (Homogenous Mat'l.)
Effective length factor, k
Condition
Value of k
pinned at both ends
1.0
fixed at both ends
0.5
fixed at one end, pinned at the other
0.7
fixed at one end, free at the other
2.0
k 1.0 for braced frames, no sidesway
k > 1.0 for unbraced frames, with sidesway
k = 1.0 for compression members in frames braced against
sidesway unless a theoretical analysis shows that a
lesser value can be used.
For slender columns (to consider PD - effect
or secondary moment)
Pu Very large moment and
negligible axial load
Column behaves like a
beam
Cm
1 - Pu
f Pc
> 1.0
Cm = 0.60 + 0.40 M1/M2 > 0.40
c
fa + fbx + fby < 1.0
Fbx
Fby
Fa
e
c.g.
d-c
Mn = nominal moment
Mu = ultimate moment
5. e b < e
db =
SLENDER COLUMNS
rg = (As + As') / Ag
eb
c
ec
Pn - nominal load cap.of column at
ex and ey
Pno - nominal load cap.of column at
e=0
Pnx - nominal load cap.of
ey and ex
d'
Gross Steel Ratio:
Pu
es'2
where:
es'
fs' = fy
fs < fy
4. e = e b
d-c
es
Failure by crushing of concrete
e
es2
es'1
Pu
e
es1
C'
C
C
C2'
Bresler's Eqtn: (Reinf. Conc.-Composite Mat'l.)
Pn
Pn
Pn
+
+
< 1.0
Pno
Pnx Pny
1. When Mu(A) < Pu(15 + 0.03h),
use Mu = Pu (15 + 0.03 h)
2. When Mu(A) > Pu(15 + 0.03h),
use Mu = Mu(A)
13/20
CECC-3
DESIGN AND CONSTRUCTION
(for braced without transversed loads)
Cm = 1.0
(for all other cases)
M1 /M 2 = smaller end moment
bigger end moment
where: = + for single curvature
= - for double curvature
Pc =
p 2 EI
(Klu) 2
EI =
Ec Ig / 2.5
1 +bd
where:
Ec = 4700 fc' (MPa)
Ig = bh3 /12
factored axial dead load
bd =
factored axial total load
Klu/r = slenderness ratio
r = 0.30h for rectangular
= 0.25D for round column
Pu = Pdl + Pll
14/20
BY: NTDEGUMA
REINFORCED CONCRETE DESIGN FORMULAS AND PRINCIPLES
A. BEARING ON SOIL
FOOTINGS
q = P / Af
Types of Footing:
D. TWO -WAY OR PUNCHING SHEAR
For top bars:
Applied Punching Shear Force:
Ld is multiplied by a factor 1.4
2
1. Spread Footing (Isolated Footing)
2. Wall Footing
3. Combined Footing
4. Mat and Raft Foundation
5. Footing on Piles
q = bearing stress on soil (MPa)
Ld =
Vp = vpc (Ap)
q all = allow. bearing stress on soil (MPa)
Resisting Shear stress of Concrete in Punching:
vpc = f [ 1 + 2 / bc ] 1/6 fc' < f1/3 fc'
B. BENDING OR FLEXURE
1. Bearing of soil
2. Bending or Flexure
3. One-way Shear or Beam Shear
4. Two-way Shear of Punching Shear
For 45 mmØ bars
25 A b fy
fc'
where:
2
Mu = qu ( Lx ) / 2
L = side dimension of footing (m)
c = column dimension (mm)
qu = net upward soil bearing stress or pressure (MPa)
Mu = f As fs (d-a/2)
1.4 PDL + 1.7 PLL
qu =
Af
Resisting Moment of Concrete:
2
For 55 mmØ bars
Ld =
40 fy
fc'
Minimum Ld = 300 mm
CHECK DEVELOPMENT LENGTH
To be safe, Mu act < Mu resist.
Minimum Ld = 300 mm
Ap = 4 (c + d) d
Mu = f r fy bd [ 1 - 0.59 r fy / fc']
C. ONE -WAY OR BEAM SHEAR
t
Minimum Ld = 300 mm
Ld =
Applied Moment:
SPREAD FOOTING (ISOLATED FOOTING)
d
0.02 Ab fy
fc'
To be safe, Vp < vpc
Resisting Moment of steel:
P
For 35 mmØ and smaller bars
Resisting Shear force of Concrete:
where:
Af = area of soil in contact with bearing
stress of soil (mm 2)
Modes of failure:
2
Vp = qu [ L - (c + d) ]
To be safe, q < q all
P = column load
SPREAD FOOTING
BY: NTDEGUMA
REINFORCED CONCRETE DESIGN FORMULAS AND PRINCIPLES
0.02 Ab fy
fc'
Ldreqd =
Applied Ultimate Shear:
q
Vu act = qu ( H z )
B. STEEL IN COMPRESSION
Ld =
0.02 Ab fy
fc'
Minimum Ld = 0.04 dbfy or 300 mm
c+d
c.s.for bending
Vu act - critical shear force 'd" from the
face of support
c.s.for punching shear
Resisting Ultimate Shear Force of concrete
DEVELOPMENT LENGTHS
d/2
H c+d
c
d/2
c.s.for beam shear
dz
x
L
A. STEEL IN TENSION
f Vc = f 1/6 fc' bd
Ld
where:
=
0.02 Ab fy
fc'
f = capacity or strength reduction
factor
Minimum Ld = 0.06 dbfy or 300 mm
= 0.85 for shear and torsion
Vc = nominal shear force capacity
of concrete
b,d = beam dimensions (mm)
CECC-3
DESIGN AND CONSTRUCTION
To be safe, Vu act <
Vc
15/20
CECC-3
DESIGN AND CONSTRUCTION
16/20
BY: NTDEGUMA
STEEL DESIGN FORMULAS AND PRINCIPLES
EULER'S CRITICAL LOAD AND STRESS
For Hinged-Ended Columns:
p EI
Pc =
p E
(L / r) 2
5. Allowable Shear Stress
fv1
bf
Fp = 0.35 fc'
2
2. Average Shearing Stress in the Web
On full area of a concrete support
Stress
Load
Maximum Allowable compressive stress of conc.
BY: NTDEGUMA
STEEL DESIGN FORMULAS AND PRINCIPLES
a. When h/tw < 998/ Fy
fv2
Allowable shear stress
tf
2
Fa =
L2
On less than the full area of a concrete support
d
h
tw
b. When h/tw > 998/ Fy
d
Fp = 0.35 fc' A 2 /A 1 < 0.70 fc'
For Fixed - Ended Columns:
Fv = 0.40 Fy
tw
Allowable shear stress
B. BENDING OF BASE PLATE
Fv = Fy Cv/2.89
Stress
Load
4p EI
L2
4p2 E
2
Pc =
Fa =
If n > m,
If m > n,
(L / r) 2
3 fp m
Fb
t =
where:
2
t =
fvave
3 fp n 2
Fb
fv1 =
kL/r = max. effective slenderness ratio
where:
k = effective length factor
k = 1 for columns hinged at both sides
fp = P/ A B
Fb = 0.75 Fy
k = 0.50 fixed-fixed
VQ1
Ib
where:
VQ2
Ib
fvave =
fv2 + 23 (fv1 - fv2 )
SHEARING STRESS OF BEAMS
SPACING OF RIVETS OR BOLTS
Q1 = Q f + Q w
Q2 = Q f
fv2 =
k = 0.70 hinged-fixed
COLUMN BASE PLATE:
S =
RI
VQ
where:
R = shear capacity of each bolts
V = maximum shear of beam
Q = statical moment area
MOMENT REDUCTION DUE TO THE PRESENCE
OF HOLE IN BOTH FLANGE
1. Maximum Web Shear Stress
3. Maximum Vert./Hor. Shear Stress
m
bf
tf
VQ
- holes in beam generally will reduce its capacity. When the
holes are located in the beam web, it reduces its shear
capacity while holes in the beam flanges reduce its moment
capacity.
fvh = I b
0.95D
bf
Afn = net flange area
Afg = gross flange area
Ah = Ag - area of holes
Area of hole = (Dh)(tf)
Dh = db + 3 mm
db = diameter of the bolt
tw
d
Dh
Dh
where:
tw
D
V = maximum shear of beam
Q = statical moment area
I = moment of inertia (mm^4)
b = base sheared
m
n 0.80B
n
A. BEARING ON CONCRETE
fv =
Actual/Applied Bearing stress:
P
4. Shear flow
d tw
V = max. shear force
d = depth of the beam
tw = web thickness
VQ
I
tf
When 0.50 Fu Afn > 0.60 Fy Afg
2. Reduction of holes must be considered
When 0.50 Fu Afn < 0.60 Fy Afg
where:
q = shear flow (N/m)
P = column load (kN)
Effective tension flange section:
Afe =
Ap = contact surface between the
base plate and conc. pedestal
DESIGN AND CONSTRUCTION
Dh
1. Reduction of hole is neglected
q =
where:
CECC-3
Dh
NGCP SPECS:
Vmax
where:
fp = Ap
< 0.40 Fy
5 Fu Afn
6 Fy
Afn = Afg - area of holes
5/16
CECC-3
DESIGN AND CONSTRUCTION
6/16
BY: NTDEGUMA
STEEL DESIGN FORMULAS AND PRINCIPLES
C. BENDING/FLEXURAL MEMBERS
fb =
Mc
I
=
2. When Lb > Lc and Lb > Lu
4. When Lb < Lc
Fb = 0.66 Fy
Actual/Applied bending stress:
M
S
BY: NTDEGUMA
STEEL DESIGN FORMULAS AND PRINCIPLES
703000Cb
Fy
200 bf
Lc =
Fy
< Lb
>
rt
2. SHEARING STRESS
fV =
3520000 Cb
Fy
Fb =
1170 x 10 Cb
ALLOWABLE STRESSES:
(Lb/rt)
2
3. USING INTERACTION EXPRESSION
use bigger value of Fb
but should be
a.
3
5. When Lb > Lu
1. Compact Sections
Fb =
83 x 10 Cb
Lb(d/Af)
Lu =
6. When Lb > Lc and Lb < Lu
Flange width - thickness ratio
<
170
2tf
Fy
Fb = Fy ( 0.79 - 0.00076
tw
d
bf
2tf
Cb < 2.3
Fy
d
<
M2 = bigger end moment
tf
1680
B. LATERALLY UNSUPPORTED BEAMS:
Fy
tw
Note: Consideration should be given to the question of
lateral support for the compression flange which
will indicate wether compact or non-compact
sections.
M1 = smaller end moment
Web depth - thickness ratio
< 1.0
For doubly symmetrical I and H shape members with
compact flanges continuously connected to the web and
bent about their weak axis, the allowable bending stress is
0.75 Fy.
Cb = 1.75 + 1.05 (M1 / M2 ) + 0.30 (M1 /M 2 )
bf
< 1.0
f bx
f by
+
0.66Fy
0.60 F y
2
bf
f by
Fby
+
b. For compact laterally supported shapes:
where:
rt = radius of gyration of a section
comprising the compression flange plus
1/3 of the compression web about the
vertical axis.
137,900 Af
Fy d
f bx
Fbx
< 0.60 Fy
Fb = 0.60 Fy
Fb = 0.66 Fy
Vy Qy
Iy b
3
Lb = unbraced length of compression flange
A. LATERALLY SUPPORTED BEAMS:
Vx Q x
±
Ix b
M1 /M2 = negative (-) for single curvature
TENSION WITH BENDING
M1 /M2 = positive (+) for double curvature
1. When Lb > Lc and Lb > Lu
2. Non-compact Sections
Fb = 0.6 Fy
when
BENDING IN BOTH AXIS
bf
>
2tf
170
Fy
Fb =
3. Partially compact Sections
Fb = Fy ( 0.79 - 0.00076
703000Cb
Fy
bf
2tf
Fy
< Lb <
rt
3520000 C b
Fy
Beams Bending in Both Axis
(Unsymmetrical Bending)
2
1. BENDING STRESS
Fy ( Lb/rt )
2 6
3
10.55 x 10 Cb
Fy
83 x 10 Cb
Lb(d/Af)
fb =
M x Cx
±
Ix
fb =
Mx
Sx
±
M y Cy
Iy
My
Sy
Mx
>
2tf
Web depth - thickness ratio
bf
2tf
CECC-3
>
1. BENDING IN ONE AXIS ONLY
Use biggest value of Fb but should be
P
fb =
< 0.60 Fy
Mx
Sx
My
±
Sy /2
Mx
250
note:
Fy
DESIGN AND CONSTRUCTION
T
MC
±
A
I
b. If lateral loads applied at the top flange and does not
passes thru the centroid of the beam section
170
Fy
f =
Members subject to both axial tension and bending shall be
proportioned at all points along their length to satisfy the
following equation:
P
Flange width - thickness ratio
bf
T
a. If lateral loads pass thru the centroid of the beam section
3
Fb =
T
Only one half of the section modulus about
the y-axis is considered
7/16
CECC-3
DESIGN AND CONSTRUCTION
fa
Ft
+
f bx
Fbx
< 1.0
where:
fa = computed axial stress
fa = T/A
fb = computed bending stress
Ft = allow. tensile stress
= 0.60 Fy
Fbx = allow. bending
stress
8/16
BY: NTDEGUMA
STEEL DESIGN FORMULAS AND PRINCIPLES
BY: NTDEGUMA
STEEL DESIGN FORMULAS AND PRINCIPLES
2. BENDING IN BOTH AXIS
WEB CRIPPING
LOCAL WEB YIELDING
fa
Ft
+
f bx
Fbx
+
f by
Fby
< 1.0
where:
Ft = allowable tensile stress
= 0.60 Fy
Fbx = 0.66 Fy (for compact section)
Fbx = 0.60 Fy (for non-compact section)
Fby = 0.75 Fy
- occurs when heavy concentrated loads produces stress at
the junction of the flange and web of the beam where
the load is being transferred from the relatively wide
flange to the narrow web.
SIDESWAY WEB BUCKLING
A. When the concentrated load is applied at a distance
not less than d/2 from the end of the member.
R
- the web will be subjected to compression if a compressive
force will be applied to braced the compression flanges as
a result the tension flange will buckle
bf
tf
R
web toes
of fillets
web toes
of fillets
K
R
2.5K
N
dc
2.5K
h
d
d
tw
k
2.5K
N
2.5K
tw
d
tw
K
critical
section
A. If the loaded flange is restrained against rotation and
1.5
N + 5K
R = 177.2 tw² 1 + 3
R
N
N
d
dc/tw
L/bf
Fyw tf
tw
tw
tf
is less than 2.30
2.5K
3
where:
Fyw = specified minimum yield stress of beam
web in MPa
a. Stress at the end of the member
R
tw ( N + 2.5 K)
<
0.66 Fy
R = 46880 tw² 1 + 0.4
h
B. When the concentrated load is applied at a distance
less than d/2 from the end of the member.
B. If the loaded flange is not restrained against rotation
and
b. Stress at the concentrated load
R
tw ( N + 5 K)
dc/tw
L/bf
dc/tw
L/bf
is less than 1.70
d
<
tw
0.66 Fy
K
3
R = 46880 tw² 0.4
h
Bearing stiffeners shall be provided if the compressive stress
at the web toe of the fillets resulting from concentrated loads
exceeds 0.66 Fy.
R
N
where:
R = concentrated load or reaction in Newtons
tw = thickness of wed in mm
N = length of bearing (not less than K for end
reactions)
K = distance from outer face of flange to web
toe of fillet in mm
dc/tw
L/bf
2.5K
1.5
R = 89.3 tw² 1 + 3
N
d
tw
tf
Fyw tf
tw
NSCP Specs: If stiffeners are provided and extend at
least one half the web depth, equations A
and B need not to ckeck.
CECC-3
DESIGN AND CONSTRUCTION
9/16
CECC-3
DESIGN AND CONSTRUCTION
10/16
BY: NTDEGUMA
STEEL DESIGN FORMULAS AND PRINCIPLES
BEARING PLATES
Magnification Factor
AXIAL LOAD WITH BENDING
- beams maybe supported by connections to other
structural members or they may rest on concrete or
masonry supports such as walls. When the support is
weaker than steel, it is usually necessary to spread the
load over a larger area so as not to exceed the allowable
bearing stress of the weaker material.
MF =
A. DESIGN FOR AXIAL COMPRESSION AND
BENDING
f =
P
A
±
f =
P
A
M x Cx
M y Cy
±
±
Ix
Iy
MC
I
BY: NTDEGUMA
STEEL DESIGN FORMULAS AND PRINCIPLES
ECCENTRICALLY LOADED COLUMNS USING
SECANT FORMULA
Cm
1 - fa / Fe'
> 1.0
Critical Column Stress
( Bending in one axis only)
smax = AP
Reduction Coefficient (Modification factor)
ALLOWABLE BEARING STRESS OF CONCRETE WALL:
( Bending in both axis)
On full area of a concrete support
B. NSCP SPECS FOR AXIAL COMPRESSION
AND BENDING
Fp = 0.35 fc'
On less than the full area of a concrete support
Fp = 0.35 fc' A 2 /A 1 < 0.70 fc'
L
Cm = 0.60 - 0.40 (M 1 /M 2 )
q = 2r
Cm = 0.85 - for members whose ends are
ec
unrestrained against rotation in the
plane of bending
r=
fa
Fa
N =
R
0.66 Fy tw
P
- 2.5K
B. LARGE AXIAL COMPRESSION ( fa/Fa > 0.15 )
N
d
tw
tf
Fyw tf
tw
M2
fp
Fy
fy
Elastic Distribution
of Stress
Plastic Distribution
of Stress
M2
M2
M1
P
AF =
Fe' =
1
1 - fa / Fe'
M1
P
P
M1 = 12 M2
Cm = 0.40
M1/M2 is positive
M1 = 0
M1 = M2
Cm = 0.60
Cm = 1.0
M1/M2 is negative
Reversed Curvature
Single Curvature
where:
fa = computed axial stress
fb = computed bending stress
Fa = allowable axial stress
Fb = allowable bending stress if bending
moment alone existed
K = effective length factor
Lb = actual unbraced length in the plane of
bending
rb = corresponding radius of gyration
12p 2 E
23 (KLb/rb) 2
- the plastic neutral axis of a section is the line that
divide the section into two equal areas.
Yield Moment
- moment that will just produce the yield stress in the
outermost fiber of the section
My = S Fy
DESIGN AND CONSTRUCTION
11/16
CECC-3
DESIGN AND CONSTRUCTION
where:
S = section modulus
Plastic Moment
- moment that will produce full plasticity in a member
cross section and create plastic hinge.
Mp = Z Fy
where:
Z = plastic section modulus
Shape Factor
Shape factor =
CECC-3
fy
fy
Plastic Neutral Axis
Amplification Factor
B. THICKNESS OF BEARING PLATE:
fy
P
Section
fbx
fby
fa
+
+
< 1.0
Fbx Fby
0.60 Fy
where:
Fyw = Fy if not specified
t = 2n
P
Strength interaction criterion:
1.5
(radius of gyration)
PLASTIC ANALYSIS AND DESIGN
Cm fb
Cm fb
fa
+
+
< 1.0
(1 - fa/Fe') Fb
(1 - fa/Fe') Fb
Fa
x
y
2. Due to web yielding
R = 89.30 tw² 1 + 3
M1 = smaller moment
M2 = bigger moment
fbx
fby
+
+
< 1.0
Fbx
Fby
I
A
L = unsupported length of column
where:
A. MINIMUM WIDTH OF BEARING PLATE : (N)
1. Due to web yielding
P
EA
P = total axial load
Cm = 1.0 - for members whose ends are
A. SMALL AXIAL COMPRESSION ( fa/Fa < 0.15 )
ec
sec q
r²
= eccentricity ratio
r²
restrained against rotation in the plane
of bending
1+
Z
S
12/16
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
Due to sag:
(subtract only); unsupported length
C=
w 2 L3
24P 2
Due to slope:
(subtract only); measured length
E = 0.6745√
CD = MD (1 +
CD = MD (1 −
∑(x − x̅)
n−1
Probable Error (mean):
∑(x − x̅)
Em =
= 0.6745√
n(n − 1)
√n
E
Proportionalities of weight, w:
𝑤∝
Normal Tension:
0.204W√AE
1
𝐸2
𝑤∝
1
𝑑
𝑤∝𝑛
Area of Closed Traverse
√PN − P
Symmetrical:
L
H = (g1 + g 2 )
8
L 2
x 2 ( 2)
=
L
y
H 1
Error of Closure:
Error of Closure
Perimeter
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:
1 acre =
4047 m2
h = h2 +
Simpson’s 1/3 Rule:
= √ΣL2 + ΣD2
Azimuth
hcr = 0.067K 2
Trapezoidal Rule:
Lat = L cos α
Dep = L sin α
=
e
)
TL
Effect of Curvature & Refraction
Area of Irregular Boundaries
A=
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 =
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=
θ
Ls 2
; p=
3
24R
x=
L3
6RLs
I
I
E = R [sec − 1]
L
Ve = (A1 + A2 )
2
I
2
m = R [1 − cos ]
2
Volume (Prismoidal):
L = 2R sin
L
VP = (A1 + 4Am + A2 )
6
L
(c − c2 )(d1 − d2 )
12 1
VP = Ve − Cp
Y=L−
2
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
VT = ABase ∙ Have = A ( )
n
A
VT = (Σh1 + 2Σh2 + 3Σh3 + 4Σh4 )
n
e=
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
L=
L5
I
π
Lc = RI ∙
180°
20 2πR
=
D
360°
1145.916
R=
D
Prismoidal Correction:
v → speed in m/s
t → perception-reaction time
f → coefficient of friction
G → grade/slope of road
L2 180°
∙
2RLs π
2
Volume (End Area):
CP =
θ=
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
BY: NTDEGUMA
THEORY OF STRUCTURES (MODULE 2)
THEORY OF STRUCTURES
BY: NTDEGUMA
P
Values of 6Aa and 6Ab of Common loadings:
L
L
Beam Loading
6A1a1 /L
6A1b1 /L
P
A
2
2
2
Pa ( L2 - a2 )
L
b
L
L
MB =
P
L
MB =
wL
4
L
wL
4
MB =
w (N/m)
w (N/m)
B
b
4
B
=
wL
6EI
=
ML
EI
L
R
B
A
FULLY RESTRAINED BEAM FORMULAS
B
L
2
2
Mmax = - M ;
w L / 20
ML
2EI
=
max
;
B
P
M
A
B
b
L
A
MA = Mb (3a/L - 1)/L
2
MB = Ma (3b/L - 1)/L
5w L / 96
Mmax = -
wL
6
max =
wL
30EI
3
;
B
wL
24EI
=
L/2
A
3
A
7wL
60
w (N/m)
a
2
Mmax =
wL
;
8
4
max
=
5wL
384 EI
max
=
wL
24EI
A
w (N/m)
L
R =
Pa ( 3L - a)
2L 3
A
PL
4
3
;
max
=
L/2
L
PL
48EI
max
=
R =
5P
16
MA = -
wL
8
L
PL
16EI
A
L
R =
3wL
16
B
R
P
M
a
b
L
-M ( 3a2 - L2 )
L
+ M ( 3b2 - L2 )
L
A
Mmax = - PL ;
CECC-3
DESIGN AND CONSTRUCTION
max
=
PL3
3EI
2
;
B
=
max
PL
2EI
L
MA = -
wL
15
R =
wL
10
R
5/12
CECC-3
5wL
;
96
DESIGN AND CONSTRUCTION
mid
=
wL
768EI
A
B
L
7 wL
max =
3840EI
MA = -
2
wL
30
2
2
; MB = -
wL
20
4
A
B
L
mid
=
wL
768EI
2
B
4
2
MA = -
w (N/m)
B
L
wL
12
; MB = -
4
w (N/m)
A
=
wL
384EI
w (N/m)
w (N/m)
w (N/m)
2
;
MA = B
2
wL
12
MA = MB 3PL
16
2
=
2
R
B
L
3
Mmax =
B
P
L/2
5wL
32
B
L
MA
4
A
3
;
P
3
PL
PL
; MB = 8
8
3
PL
192EI
=
=
2
P( b a + a b /2 )
L2
b
A
B
L
2
MA = -
R
L
5wL
32
B
PROPPED BEAM FORMULAS
P
3
MA
L/2
L
max
w (N/m)
8wL
60
Pba
L2
=
Pb (3L - 4b)
mid =
48EI
P
4
;
Pab
; MB
L2
2
B
B
L
2
=
b
L
B
2
2
2
MA
a
A
SIMPLE AND CANTILEVER BEAM FORMULAS
w (N/m)
2
wa
2
3
wb (4L - b)
R = 8L 3
MA = RL B
a
3
;
w (N/m)
MA = -5w L / 96
3
R =
R
2
2
w L / 12
L
Pa
2EI
=
B
A
wL
8 EI
=
w (kN/m)
A
3
max
MA = -w L / 30
a
w (N/m)
wL
;
2
7w L
120
11wL
40
2
Pa (3L -a)
;
6EI
L
MA = B
L
M
L
A
B
2
3 PL 2
8
Mmax = -
w (kN/m)
2
3 PL 2
8
A
M B = PL / 8
w (kN/m)
A
B
M A = -PL / 8
MA = -w L / 12
L/2
=
max
2
Pb ( L2 - b2 )
L
A
L/2
Mmax = - P(a) ;
L/2
L
A
B
2
B
L
2
L/2
MB = Pba / L
P
b
A
P
b
L
MA = -Pab / L
a
a
FIXED END MOMENTS OF COMMON LOADINGS
a
w (N/m)
2
5wL
11wL
; MB = 192
192
2
L/2
ma
A
L/2
=
B
L
6/12