Load reduction factor_ǿ

Load Reduction Factor Ǿ
ACI Code Section 9.3 specifies the following values to be used:
Tension controlled section Ǿ = 0.90
Compression controlled section
With Spiral Reinforcement Ǿ = 0.75
Other Reinforcement members Ǿ = 0.65
Plain Concrete Ǿ = 0.60
Shear and Torsion Ǿ = 0.75
Bearing on Concrete Ǿ = 0.65
Strut and tie models Ǿ = 0.75
WSD
=
=
= 2 + ( ) −
= 1 −
= 0.45
= 0.4
Resisting moment of concrete
=
Resisting moment of steel
=
=
= 0.85 × = 0.85 × 0.85 × ×
.
. .
= 0.013
= 0.75
<

So the beam will fail by yielding
= .
= ∅ −
Design steps of USD Beam
= 0.85 × = 0.003, = 0.005
= =
∅ .
Minimum =
=
∅
∅ = 0.90
= .
General solution of
+ + = 0
=
±√
( ) = 0.85∅ 0.85 − +
Axial capacity of tied column
( ) = 0.85∅ 0.85 − +
∅ = 0.75
Axial capacity of spiral column
( ) = 0.80∅ 0.85 − +
∅ = 0.65
Design a footing of column by USD method considering that the length of the footing
is 1.5 times of width of the footing.
Given
= 200
= 160

= 5 5 ℎ
= 3 and = 60
Column size = 16 ℎ
Solution
= (200 × 1.2 + 160 × 1.6) = 496
Assume self weight 3%
= (200 + 160) × 1.03 = 370.8
=
.
= 74.16
Now
L = 1.5B
1.5B = 74.16
B = 7.03 ft L = 10.55 ft
q =
.
= 6.69 k/ft
Assume t = 23” d = 23 – 3 =20”
Punching Shear
V = (DL × 1.2 + LL × 1.6)– (a + d) × q = 496 −
( )
×
6.69 = 442.3 k
Resisting Shear
= 4∅ = 4 × 0.75√3000 ×
× ×
= 473.23 > 442.3
Wide beam shear
=
.
× 1 × 6.69 = 19.68
= 2∅ = 2 × 0.75√3000 ×
×
= 19.72 > 19.68
Moment calculation
= =
. ×
.
= 71.04 − /
= =
. ×
.
= 27.14 − /
= 0.85 × = 0.85 × 0.85 × ×
.
. .
= 0.0135

= =
∅ .
=
. ×
. × . × ×
. × . × = 10.77" <
20"
Minimum = =
× ×
= 0.8
=
∅
=
. ×
. × ×
= 0.81
=
.
=
. ×
. × ×
= 1.59 ℎ
Provide ∅ 16 @4.5" / .
=
∅
=
. ×
. × ×
. = 0.306
=
.
=
. ×
. × ×
= 0.6 ℎ
Provide ∅ 16 @4.5" / .
Pre-stressed Concrete:
Concrete in which there have been introduced internal stresses of such magnitude and
distribution that the stresses resulting from given external loadings are counteracted to a
desired degree. In reinforced concrete members the pre-stress is commonly introduced by
tensioning the steel reinforcement.
Losses of pre-stressing
Losses due to
 Elastic shortening
 Creep of concrete
 Shrinkage of concrete
 Steel relaxation
 Anchorage slip
 Frictional loss
 Bending of member

1. Classify soil Based on grain size.
Classification System or
Name of the organization
Particle size (mm)
Gravel Sand Silt Clay
Unified 75 – 4.75 4.75 – 0.075 Fines (silts and clays) < 0.075
AASHTO 75 – 2 2 – 0.05 0.05 – 0.002 < 0.002
MIT > 2 2 – 0.06 0.06 – 0.002 < 0.002
ASTM > 4.75 4.75 – 0.075 0.075 – 0.002 < 0.002
Permeability
=
=
= ( − 35)[0.2 + 0.005( − 40) + 0.01( − 15)( − 10)]
Uniformity Coefficient
=
Coefficient of Curvature
C =
.

0
10%
20%
40%
30%
50%
60%
%FinerbyMass
70%
80%
90%
100%
Grain Size, D (mm)
10 1
10D
60D
30D
0.1
The moisture contents of a soil at the points where it passes from one stage to the next are
called consistency limits or Atterberg limits
PI = LL – PL.
=
=
∆
Where,
=
=
=
= ℎ
=
∆ = ℎ
Compaction Consolidation
 It is a dynamic Process  It is a static Process
 Volume reduction by removing of air
voids from soil grains
 Volume reduction by removing of
water from soil grains
 It is almost instantaneous
phenomenon
 It is time dependent phenomenon
 Soil is Unsaturated  Soil is always saturated
 Specified Compaction techniques are
used in this process.
 Consolidation occurs on account of a
load placed on the soil

PercentPassing
40
30
20
10
0
100
90
80
70
60
50
0.061 0.6 0.2 0.10 0.02 0.01 0.005 0.002
Well Graded
Uniform Graded
Gap Graded
Particle Size in mm (log Scale)
Open Graded
Dense Graded
Void ratio:
Void ratio (e) is defined as the ratio of the volume of voids to the volume of solids.
Mathematically
=
Porosity:
Porosity (n) is defined as the ratio of the volume of voids to the total volume.
Mathematically
=
The relationship between void ratio and porosity
e = = = =
n =
Degree of saturation
Degree of saturation (S) is defined as the ratio of the volume of water to the volume
of voids.
=
The degree of saturation is commonly expressed as a percentage.

Moisture Content
Moisture content (w) is also referred to as water content and is defined as the ratio of
the weight of water to the weight of solids in a given volume of soil.
Mathematically
= × 100
Unit weight
Unit weight (γ) is the weight of soil per unit volume.
=
The unit weight can also be expressed in terms of weight of soil solids, moisture
content, and total volume.
= =
+
=
1 +
=
(1 + )
Density Index or Relative Density
The term relative density is commonly used to indicate the in situ denseness or
looseness of granular soil. The ratio between the minimum density to the maximum
density of granular soil is defined as relative density.
=
−
−
Where
= ,
= ℎ
= ℎ ℎ
= ℎ ℎ
= = + +
= = 1.3 + + 0.4
= = 1.3 + + 0.3
Where
q = Ultimate bearing capacity
N ,N , N = Bearing capacity factor depends on angle of friction ∅
= ℎ ℎ
=
= ℎ
= ℎ

N ,N , N is called Terzaghi bearing capacity factor.
N = Cohesion factor
N = Surcharge factor
N = Unit weight factor
Laboratory Tests of Soil
Properties Test
Grain size distribution Sieve analysis and hydrometer test
Consistency Liquid limit
Plastic limit
Plasticity index
Compressibility Consolidation
Compaction Characteristics Standard proctor, Modified proctor
Unit Weight Specific Gravity
Shear Strength
1. Cohesive Soils
2. Non-cohesive soils
3. General
Corresponding Tests:
1. Unconfined Compression test
2. Direct Shear test
3. Tri-axial test
Field Tests of Soil
Properties Test
Compaction control  Moisture – Density relation
 In place density
Shear Strength – (Soft Clay) Vane shear test
Relative Density – (Granular Soil) Penetration test
Field density  Core Cutting
 Sand replacement
Permeability Pumping test
Soil Sampling and resistance of the soil
to penetration of the sampler
Standard Penetration test
Split Barrel Sampling
Bearing Capacity
 Pavement
 Footing
Corresponding Tests
 CBR, Plate Beating test
 Plate Bearing test
Piles
 Vertical Piles
 Batter Piles
Corresponding Tests
 Load Test
 Lateral Load Test
2. Example
Determine the net ultimate bearing capacity of a mat foundation measuring 15 m ×
10 on saturated clay with = 95 / , ∅ = 0, = 2 .

Solution:
( ) = 5.14 1 +
0.195
1 + 0.4
( ) = 5.14 × 95 × 1 +
0.195 × 10
15
1 + 0.4
2
10
= 595.9 /
The mat has dimension of 30 × 40 . The live load and dead load on the mat are
20MN. The mat is placed over a layer of sot clay. The unit weight of het clay is
.
. Find the for a fully compensated foundation.
Solution:
= =
200 × 10
(30 × 40)(18.75)
= 8.89
Chemical oxygen demand (COD)
Chemical oxygen demand (COD) is a measure of the quantities of such materials present in
the water. COD, however, as measured in a COD test, also includes the demand of
biologically degradable materials because more compounds can be oxidized chemically than
biologically. Hence, the COD is larger than the BOD.
The amount of oxygen required by micro-organisms to oxidize organic wastes aerobically is
called biochemical Oxygen demand (BOD).
Why COD is greater than BOD?
Because BOD contains only biodegradable but whereas COD includes both biodegradable
and non biodegradable that is the reason why cod is larger than BOD.
1. Example:
At 25℃, hydrogen ion concentration of a solution is 0.001M. Determine the of the
solution.
Answer:
Given, [ ] = 0.001 = 10
We know,
= − log[ ]
= − log10
= 3.00

2. Factors influencing water use:
• Size of city
• Climate and location
• Industrial development
• Habits and living standards
• Parks and gardens
• Water quality
• Water pressure
• Cost of water
3. Essential elements of water supply
 Source of supply
 Collection system
 Treatment plant
 Distribution system
4. The most common water treatment methods are
 Plain sedimentation
 Sedimentation with coagulation
 Filtration
 Disinfection
Sewer
A sewer is a conduit through which wastewater, storm water, or other wastes flow.
Sewerage is a system of sewers. The system may comprise sanitary sewers, storm
sewers, or a combination of both. Usually, it includes all the sewers between the ends
of building-drainage systems and sewage treatment plants or other points of disposal.
 Sanitary or separate sewer
o Sanitary sewage
o Industrial sewage
 Storm sewer
 Combined sewer
5. Name deferent types of test for environmental engineering
 Determination of Iron Concentration of Water
 Determination of Sulfur from a Soluble Sulfate Solution
 Determination of of water
 Determination of Total Dissolved Solid (TDS)
 Determination of Alkalinity of Water
 Determination of Ammonia in an Ammonium Salt
 Determination of Chlorine Concentration of Water
 Determination of Arsenic
 Determination of Hardness of Water
 Determination of Dissolved Oxygen
 Determination of Biochemical oxygen demand (BOD)

 Determination of Chemical oxygen Demand (COD)
 Determination of Turbidity of Water
 Correction for pull
=
( − )
Where,
=
= ℎ ℎ
= − ℎ
= ℎ
 Correction for sag
=
24
Where
= ℎ ℎ
= ℎ ℎ
=
 Correction for slope or vertical alignment
=
ℎ
2
If slopes are given in terms of vertical angels
= 2
2
Where,
ℎ = ℎ ℎ ℎ ℎ
= ℎ ℎ
= ℎ
Es or G =

=
( )
Cement Compound
Weight
Percentage
Abbreviation
Chemical Formula
Tri calcium silicate 50 % C3S Ca3SiO5 or 3CaO.
SiO2
Di calcium silicate 25 % C2S Ca2SiO4 or 2CaO.
SiO2
Tri calcium aluminate 10 % C3A Ca3Al2O6 or 3CaO .
Al2O3
Tetra calcium
aluminoferrite
10 %
C4AF Ca4Al2Fe2O10 or
4CaO.
Al2O3
.
Fe2O3
Gypsum or Calcium
Sulphate
5 % CaSO4
.
2H2O

1. Write the standard of strength testing of cement according to ASTM C 109.
American Society for Testing Materials Standard (ASTM C-109)
3 -days 1740 psi (12.0 MPa)
7 -days 2760 psi (19.0 MPa)
28 -days 4060 psi (28.0 MPa)
2. Write allowable slumps for various constructions
Type of Construction
Slumps
mm Inch
RCC Foundation walls & Footings 25 – 75 1 – 3
Plain Footings, caissons & substructure walls 25 – 75 1 – 3
Slabs, beams & reinforced walls 25 – 100 1 – 4
Building columns 25 – 100 1 – 4
Pavements 25 – 75 1 – 3
Heavy mass constructions 25 - 50 1 – 2
Sand is commonly divided into five sub-categories based on size:
a) Very fine sand (1/16 - 1/8 mm)
b) Fine sand (1/8 mm - 1/4 mm)
c) Medium sand (1/4 mm - 1/2 mm)
d) Coarse sand (1/2 mm - I mm), and
e) Very coarse sand (I mm. - 2 mm).
3. Example
The fineness modulus of two different types of sand is 2.84, and 2.24 respectively. The
fineness modulus of their mixture is 2.54. Find the mixing ratio.
Assume
= 2.84, = 2.24
= =
. .
. .
= 1
R: 1 = 1:1
Cul-de-sec
Local Street
Collector Street
Major Arterial
Expressway
Freeway
No through
traffic
No through
traffic
Unrestricted
access
Increasing proportion of though
traffic; increasing speed
Increasinguseofstreet
foraccesspurposes;
parking,loading,etc.
Decreasingdegreeof
accesscontrol
Compleate access control

RoadWay
10 m
Slope (2:1)
3 m
Berm
10 m
Borrow pit
10 m
Slope (2:1)
3m
Berm
10 m
Borrow pit
10 m
Road MarginRoad Margin
Right of Way
2:1 2:1
1 m
1 m 1 m
Section of National Highway
1. What are the lab testing of Aggregates of roadway.
Ans
o Los Angeles Abrasion test
o Aggregate Impact value
o Aggregate Crushing Value
o Soundness Test
o Gradation test
o Unit weight and Void test
o Flakiness Index
o Elongation Index
o Angularity Number
2. What are the laboratory test for bituminous materials
Ans
o Specific Gravity of Semi-Solid Bituminous Materials
o Loss on Heating test
o Penetration test
o Softening Point test
o Solubility test
o Ductility test
o Flash And Fire Points test
o Spot test
o Specific Gravity test
o Distillation test

Load reduction factor_ǿ

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