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Civil Engineering Formula, Study notes of Civil Engineering

Civil ENgineering Formula Civil ENgineering Formula

Typology: Study notes

2019/2020

Uploaded on 12/02/2020

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Casagrande:
k=1.4e2k0.85
Dr (%) Description
0 20 Very Loos e
20 40 Loose
40 70 Medium Dense
70 85 Dense
85 100 Very Dense
LI State
LI < 0 Semisolid
0 < LI < 1 Pla stic
LI > 1 Liquid
Volume
e=Vv
Vs
n=Vv
V
S=Vw
Vv
Weight
ω=Ww
Ws
0 < e < ∞
e= n
1n
0 < n < 1
n= e
1+e
Se=Gsω
ɤ=W
V
ɤd=WS
V
ɤd=ɤ
1+ω
Unit Weight:
ɤ=(Gs+Gsω)ɤw
1+e
ɤ=(Gs+Sew
1+e
When S=0:
ɤd=Gsɤw
1+e
When S=100%:
ɤsat =(Gs+ew
1+e
ɤsub =ɤsat ɤw
ɤsub =(Gs1w
1+e
ɤzav =Gsɤw
1+Gsω
Atterberg Limits
PI=LLPL
LI=ωPL
LLPL
SI=PLSL
CI=LLω
LLPI
SL=m1m2
m2V1V2
m2ɤw
SL=e
Gs ; SR=m2
V2ɤw
G𝑠=1
1
SRSL
GI=(F35)[0.2+0.005(LL40)]
+0.01(F15)(PI10)
Ac=PI
μ; St=quund
qurem
μ = % passing 0.002mm
PI Description
0 Non-plastic
1-5 Sl ightly 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=D60
D10
Coeff. of Gradation
or Curvature:
Cc=(D30)2
D60D10
Sorting
Coefficient:
So=D75
D25
Suitability Number:
Sn=1.7√ 3
(D50)2+1
(D20)2+1
(D10)2
Permeability
v=ki ; i =∆h
L ; v𝑠=v
n
Q=vA=kiA
Constant Head Test:
k= QL
Aht
Falling/Variable Head Test:
k=aL
At𝑙𝑛h1
h2
Stratified Soil
for Parallel flow:
keq =h1k1+h2k2+...+hnkn
H
for Perpendicular flow:
keq =H
h1
k1+h2
k2+...+hn
kn
Pumping Test:
Unconfined:
k= Q 𝑙𝑛r1
r2
π(h12h22)
Confined:
k= Q 𝑙𝑛r1
r2
t(h1h2)
Hazen Formula
k=cD102
Samarasinhe:
k= C3en
1+e
Kozeny-Carman:
k= C1e2
1+e
Stresses in Soil
Effective Stress/
Intergranular Stress:
pE=pTpw
Pore Water Pressure/
Neutral Stress:
pw=ɤwhw
Total Stress:
pT=ɤ1h1+ɤ2h2+...nhn
Flow Net / Seepage
Isotropic soil:
q=kHNf
Nd
Non-Isotropic soil:
q=kxkzHNf
Nd
Compressibility of Soil
Compression Index, CC:
Cc=0.009(LL10%)
Cc=ee′
𝑙𝑜𝑔P+Po
Po
For normally consolidated clay:
S=ee′
1+eH
S= CcH
1+e𝑙𝑜𝑔∆P+Po
Po
With Pre-consolidation pressure, Pc:
when (P+Po) < Pc:
S= CsH
1+eo𝑙𝑜𝑔∆P+Po
Po
when (P+Po) > pc:
S= CsH
1+e𝑙𝑜𝑔Pc
Po+CcH
1+e𝑙𝑜𝑔∆P+Po
Pc
Over Consolidation Ratio (OCR):
OCR=pc
po; OCR = 1
Coefficient of Compressibility:
av=∆e
∆P
Coefficient of Volume Compressibility:
mv=∆e
∆P
1+eave
Coefficient of Consolidation:
Cv=Hdr2Tv
t
Coefficient of Permeability:
k=mvCvɤw
Equipotential line ----
Flow line ----
Lateral Earth Pressure
ACTIVE PRESSURE:
pa=1
2kaɤH22cHka
For Inclined:
ka=cosβcosβcos2βcos2Ø
cosβ+cos2βcos2Ø
For Horizontal:
ka=1sinØ
1+sinØ
If there is angle of friction α bet. wall and soil:
ka=cos2Ø
cos α [1 + sin + α) sin Ø
cosα ]2
PASSIVE PRESSURE:
pP=1
2kPɤH2+2cHkP
For Inclined:
kP=cosβcosβ+cos2βcos2Ø
cosβcos2βcos2Ø
For Horizontal:
kP=1+sinØ
1sinØ
If there is angle of friction α bet. wall and soil:
kP=cos2Ø
cos α [1 sin α) sin Ø
cosα ]2
AT REST:
ko=1sinØ
(for one layer only)
Swell Index, CS:
Cs=1
5Cc
(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
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
Normally consolidated:
sinØ= r
σ3+r
Cohesive soil:
sinØ= r
x+σ3+r
tanØ=c
x
Unconsolidated-
undrained test:
c=r
Unconfined
compression test:
σ3=0
DIRECT SHEAR TEST:
σn → normal stress
σs shear stress
Normally consolidated soil:
tanØ=σS
σN
Cohesive soil:
tanØ= σS
x+σN=c
x
σS=c+σNtan
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:
pE=0
Capillary Rise:
hcr =C
eD10
pf3
pf4
pf5
pf8
pf9

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Casagrande:

k = 1. 4 e

2

k

  1. 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

  • G

s

ω)ɤ

w

1 + e

ɤ =

(G

s

  • Se)ɤ

w

1 + e

When S=0:

ɤ

d

=

G s

ɤ w

1 + e

When S=100%:

ɤ sat

=

(G

s

  • e)ɤ

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 )]

    1. 01 (F − 15 )(PI − 10 )

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

  • h

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

  • ɤ

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

  • 2cH√k

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

  • r

 Cohesive soil:

sin Ø =

r

x + σ

3

  • r

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

  • qN

q

    1. 4 ɤBN

ɤ

Circular Footing:

q

ult

= 1 .3cN

c

  • qN

q

    1. 3 ɤBN

ɤ

Strip Footing:

q

ult

= cN

c

  • qN

q

    1. 5 ɤBN

ɤ

 Local Shear Failure

(loose sand & soft clay)

Square Footing:

q

ult

= 1 .3c′N

c

  • qN

q

    1. 4 ɤBN

ɤ

Circular Footing:

q

ult

= 1 .3c′N

c

  • qN

q

    1. 3 ɤBN

ɤ

Strip Footing:

q

ult

= c′N

c

  • qN

q

    1. 5 ɤBN

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 =

  1. 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

  • d

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

  • d

2

Power Lost:

P = QɤE

s

Quadratic Equation

Form:

Ax

2

  • Bx + C = 0

Roots:

x =

−B ± √B

2

− 4AC

2A

Sum of Roots:

x 1

  • x

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

  • (n − 1 )d

a

n

= a

x

  • (n − x)d

S

n

n

(a

1

  • a

n

Harmonic Progression:

  • reciprocal of arithmetic

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

  • if M is ahead of H
  • if M is behind of H

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

  • nb

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

  • A

2

  • √A

1

A

2

)

Prismatoid:

V =

H

6

( A

1

  • 4 A

M

  • A

2

)

Spherical Solids

Sphere:

V =

4

3

πR

3

LA = 4πR

2

Spheroid:

V =

4

3

πabc

LA = 4π [

a

2

  • b

2

  • c

2

3

]

Prolate Spheroid:

V =

4

3

πabb

LA = 4π [

a

2

  • b

2

  • b

2

3

]

Oblate Spheroid:

V =

4

3

πaab

LA = 4π

[

a

2

  • a

2

  • b

2

3

]

about minor axis

about major axis

Spherical Lune:

A lune

θ rad

=

4πR

2

A lune

= 2θR

2

Spherical Wedge:

V

wedge

θ

rad

=

4

3

πR

3

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

  • 3 b

2

  • h

2

)

Rectangle:

A = bh

P = 2a + 2b

d =

√ b

2

  • h

2

Ellipse

A = πab C = 2π

a

2

  • b

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

  • y

2

  • Dx + Ey + F = 0

Standard Equation:

(x − h)

2

  • (y − k)

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

  • (x

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

  • 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

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

  • Dx + Ey + F = 0

x

2

  • Dx + Ey + F = 0

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

  • Cy

2

  • Dx + Ey + F = 0

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

  • Dx + Ey + F = 0

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

  • b

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

  • 1 )]

 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

  • y

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

  • 1 .5IQR < x

 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

  • 8 t

f

ii. exterior Beam:

b

f

= b

w

L

b

f

= b

w

s

1

b

f

= b

w

  • 6 t

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

  1. 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

  1. 85 f

c

( 1 − ρ

  • ρf

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

  • q

C

  • q

sur

  • q

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.