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

휌 =

푛 =

푘 = 2휌푛 + (휌푛) − 휌푛

푗 = 1 −

푓 = 0.45푓

푓 = 0.4푓

Resisting moment of concrete

푀 = 푓 푗푘푏푑

Resisting moment of steel

푀 = 퐴 푓 푗푑

휌 =

휌 = 0.85훽 × = 0.85 × 0.85 × × .. .

= 0.013

휌 = 0.75휌

휌 < 휌

So the beam will fail by yielding

푎 = .

푀 표푟푀 = ∅퐴 푓 푑 −

Design steps of USD Beam

휌 = 0.85훽 × 휖 = 0.003,휖 = 0.005

푑 = =∅ .

Minimum 퐴 =

퐴 =∅

∅ = 0.90

푎 = .

General solution of

푎푥 + 푏푥 + 푐 = 0

푥 = ±√

푃 ( ) = 0.85∅ 0.85푓 퐴 − 퐴 + 퐴 푓

Axial capacity of tied column

푃 ( ) = 0.85∅ 0.85푓 퐴 − 퐴 + 퐴 푓 ∅ = 0.75푓표푟푡푖푒푑푐표푙푢푚푛

Axial capacity of spiral column

푃 ( ) = 0.80∅ 0.85푓 퐴 − 퐴 + 퐴 푓 ∅ = 0.65푓표푟푠푝푖푟푎푙푐표푙푢푚푛

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

퐷퐿 = 200푘

퐿퐿 = 160푘

푞 = 5푘푠푓푎푡5 푑푒푝푡ℎ

푓 = 3푘푠푖 and 푓 = 60푘푠푖

Column size = 16푖푛푐ℎ푠푞푢푎푟푒

Solution

푃 = (200 × 1.2 + 160 × 1.6) = 496푘푖푝

Assume self weight 3%

푃 = (200 + 160) × 1.03 = 370.8푘푖푝

퐴 = . = 74.16푓푡

Now

L = 1.5B

1.5B = 74.16

B = 7.03 ft L = 10.55 ft

q =.

= 6.69k/ft

Assume t = 23” d = 23 – 3 =20”

Punching Shear

V = (DL× 1.2+ LL× 1.6)–(a+ d) × q = 496 − ( ) ×6.69 = 442.3k

Resisting Shear

푉 = 4∅ 푓 푏 푑 = 4 × 0.75√3000 × × × = 473.23푘푖푝 > 442.3푘푖푝

Wide beam shear

푉 = . × 1 × 6.69 = 19.68푘푖푝

푉 = 2∅ 푓 푏 푑 = 2 × 0.75√3000 × × = 19.72푘푖푝 > 19.68푘푖푝

Moment calculation

푀 = = . × .

= 71.04푘푖푝 − 푓푡/푓푡

푀 = = . × .

= 27.14푘푖푝 − 푓푡/푓푡

휌 = 0.85훽 × = 0.85 × 0.85 × × .. .

= 0.0135

푑 = =∅ .

= . ×

. × . × × . × . × = 10.77" <

20"

Minimum 퐴 = = × × = 0.8푖푛

퐴 =∅

= . ×. × ×

= 0.81푖푛

푎 = .

= . ×. × ×

= 1.59푖푛푐ℎ

Provide ∅16@4.5"푐/푐.

퐴 =∅

= . ×

. × × . = 0.306푖푛

푎 = .

= . ×. × ×

= 0.6푖푛푐ℎ

Provide ∅16@4.5"푐/푐.

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

푄 = 푘푖퐴 푘 = 퐶퐷

퐺퐼 = (퐹 − 35)[0.2 + 0.005(퐿퐿 − 40) + 0.01(퐹 − 15)(푃퐼 − 10)]

Uniformity Coefficient

퐶 =

Coefficient of Curvature

C = .

0

10%

20%

40%

30%

50%

60%

% F

iner

by

Mas

s

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.

퐿퐼 =

푆 = 퐶 푙표푔 ∆

Where,

푆 = 퐶표푛푠표푙푖푡푎푡푖표푛푆푒푡푡푙푒푚푒푛푡

퐶 = 퐶표푚푝푟푒푠푠푖표푛퐼푛푑푒푥

푒 = 푁푎푡푢푟푎푙푉표푖푑푅푎푡푖표표푓푆표푖푙푆푎푚푝푙푒

퐻 = 푇ℎ푖푐푘푛푒푠푠표푓푆표푖푙퐿푎푦푒푟

휎 = 퐼푛푖푡푖푎푙퐸푓푓푒푐푡푖푣푒푆푡푟푒푠푠

∆휎 = 퐶ℎ푎푛푔푒푖푛퐸푓푓푒푐푡푖푣푒푆푡푟푒푠푠

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

Perc

ent P

assi

ng

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 GradedUniform 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

푒 =푉푉

Porosity: Porosity (n) is defined as the ratio of the volume of voids to the total volume. Mathematically

푛 =푉푉

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.

푆 =푉푉

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

푤 =푊푊 × 100

Unit weight Unit weight (γ) is the weight of soil per unit volume.

훾 = 푊푉

The unit weight can also be expressed in terms of weight of soil solids, moisture content, and total volume.

훾 = 푊푉 =

푊 + 푊푉 =

푊 1 + 푊푊

푉 = 푊 (1 + 푤)

푉

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.

퐷 = 푒 − 푒

푒 − 푒

Where 퐷 = 푅푒푙푎푡푖푣푒푑푒푛푠푖푡푦, 푢푠푢푎푙푙푦푔푖푣푒푛푎푠푎푝푒푟푐푒푛푡푎푔푒 푒 = 퐼푛푠푖푡푢푣표푖푑푟푎푡푖표표푓푡ℎ푒푠표푖푙 푒 = 푉표푖푑푟푎푡푖표표푓푡ℎ푒푠표푖푙푖푛푡ℎ푒푙표표푠푒푠푡푐표푛푑푖푡푖표푛 푒 = 푉표푖푑푟푎푡푖표표푓푡ℎ푒푠표푖푙푖푛푡ℎ푒푑푒푛푠푒푠푡푐표푛푑푖푡푖표푛

퐿표푛푔표푟푐표푛푡푖푛푢표푢푠퐹표표푡푖푛푔 = 푞 = 푐푁 + 훾퐷푁 + 퐵푁

푆푞푢푎푟푒퐹표표푡푖푛푔 = 푞 = 1.3푐푁 + 훾퐷푁 + 0.4훾퐵푁

퐶푖푟푐푢푙푎푟퐹표표푡푖푛푔 = 푞 = 1.3푐푁 + 훾퐷푁 + 0.3훾퐵푁

Where

q = Ultimatebearingcapacity

N ,N , N = Bearingcapacityfactordependsonangleoffriction∅

푐 = 퐶표ℎ푒푠ℎ푖표푛표푓푠표푖푙

푞 = 훾퐷

퐷 = 퐷푒푝푡ℎ표푓푓표푢푛푑푎푡푖표푛

훾 = 푈푛푖푡푤푒푖푔ℎ푡표푓푠표푖푙

N ,N , N iscalledTerzaghibearingcapacityfactor.

N = Cohesionfactor

N = Surchargefactor

N = Unitweightfactor

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 15m×

10푚on saturated clay with 푐 = 95푘푁/푚 ,∅ = 0,푎푛푑퐷 = 2푚.

Solution:

푞 ( ) = 5.14푐 1 +0.195퐵

퐿 1 + 0.4퐷퐵

푞 ( ) = 5.14 × 95 × 1 +0.195 × 10

15 1 + 0.42

10 = 595.9푘푁/푚

The mat has dimension of 30푚 × 40푚. 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 퐷 for a fully compensated foundation.

Solution:

퐷 =푄퐴훾 =

200 × 10(30 × 40)(18.75) = 8.89푚

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 푃 of the solution.

Answer:

Given, [퐻 ] = 0.001푀 = 10 푀

We know,

푃 = − log[퐻 ]

= − log 10

= 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 푃 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

퐶 = (푃 − 푃 )퐿

퐴퐸 Where,

푃 = 푃푢푙푙푎푝푝푙푖푒푑푑푢푟푖푛푔푚푒푎푠푢푟푒푚푒푛푡

푃 = 푃푢푙푙푎푡푤ℎ푖푐ℎ푡푎푝푒푖푠푠푡푎푛푑푎푟푑푖푠푒푑

퐴 = 퐶푟표푠푠 − 푠푒푐푡푖표푛푎푙푎푟푒푎표푓푡ℎ푒푡푎푝푒

퐸 = 푀표푑푢푙푢푠표푓푒푙푎푠푡푖푐푖푡푦표푓푡ℎ푒푡푎푝푒푚푎푡푒푟푖푎푙푠

Correction for sag

퐶 = 푤 푙24푃

Where 푤 = 푊푒푖푔ℎ푡표푓푡ℎ푒푡푎푝푒

퐿 = 퐿푒푛푔푡ℎ표푓푡ℎ푒푡푎푝푒

푃 = 푃푢푙푙푎푝푝푙푖푒푑

Correction for slope or vertical alignment

퐶 = ℎ2푙

If slopes are given in terms of vertical angels

퐶 = 2푙푠푖푛휃2

Where, ℎ = 퐷푖푓푓푒푟푒푛푐푒푖푛ℎ푒푖푔ℎ푡푏푒푡푤푒푒푛푡ℎ푒푒푛푑푠표푓푡ℎ푒푠푙표푝푒

푙 = 퐿푒푛푔푡ℎ표푓ℎ푡푒푠푙표푝푒

휃 = 퐴푛푔푙푒표푓푡ℎ푒푠푙표푝푒

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 퐹 = 2.84,퐹 = 2.24 푅 = = . .

. .= 1

R: 1 = 1:1

Cul-de-secLocal Street

Collector Street

Major Arterial

Expressway

Freeway

No throughtraffic

No throughtraffic

Unr

estri

cted

acce

ss

Increasing proportion of thoughtraffic; increasing speed

Incr

easi

ng u

se o

f stre

etfo

r acc

ess p

urpo

ses;

park

ing,

load

ing,

etc

.

Dec

reas

ing

degr

ee o

fac

cess

con

trol

Compleate access control

Road Way10 m

Slope (2:1)3 m

Berm10 m

Borrow pit10 m

Slope (2:1)3 m

Berm10 m

Borrow pit10 m

Road MarginRoad Margin

Right of Way

2:12:1

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