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

𝜌 = 𝐴𝑆

𝑏𝑑

𝑛 = 𝐸𝑆

𝐸𝑐

𝑘 = √2𝜌𝑛 + (𝜌𝑛)2 − 𝜌𝑛

𝑗 = 1 −𝑘

3

𝑓𝑐 = 0.45𝑓 ′𝑐

𝑓𝑠 = 0.4𝑓𝑦

Resisting moment of concrete

𝑀𝑐 = 1

2𝑓𝑐𝑗𝑘𝑏𝑑2

Resisting moment of steel

𝑀𝑠 = 𝐴𝑠 𝑓𝑠𝑗𝑑

𝜌 = 𝐴𝑆

𝑏𝑑

𝜌𝑏 = 0.85𝛽1 𝑓′𝑐

𝑓𝑦×

𝜖𝑢

𝜖𝑢+𝜖𝑦= 0.85 × 0.85 ×

3

60×

0.003

0.003+0.005= 0.013

𝜌𝑚𝑎𝑥 = 0.75𝜌𝑏

𝜌 < 𝜌𝑚𝑎𝑥

So the beam will fail by yielding

𝑎 = 𝐴𝑠𝑓𝑦

0.85𝑓′𝑐 𝑏

𝑀𝑑 𝑜𝑟 𝑀𝑢 = ∅𝐴𝑠 𝑓𝑦 (𝑑 −𝑎

2)

Design steps of USD Beam

𝜌 = 0.85𝛽′ 𝑓𝑐′

𝑓𝑦×

𝜖𝑢

𝜖𝑢+𝜖𝑦 𝜖𝑢 = 0.003, 𝜖𝑦 = 0.005

𝑑 = √𝑀𝑢

𝑅𝑏= √

𝑀𝑢

∅𝜌𝑓𝑦𝑏 (1−0.59𝜌𝑓𝑦

𝑓𝑐′ )

Minimum 𝐴𝑠 =200𝑏𝑑

𝑓𝑦

𝐴𝑠 =𝑀𝑢

∅𝑓𝑦(𝑑−𝑎

2) ∅ = 0.90

𝑎 = 𝐴𝑠𝑓𝑦

0.85𝑓𝑐′𝑏

General solution of

𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0

𝑥 =−𝑏±√4𝑎𝑐

2𝑎

𝑃𝑛(𝑚𝑎𝑥) = 0.85∅[0.85𝑓𝑐′(𝐴𝑔 − 𝐴𝑠𝑡 ) + 𝐴𝑠𝑡 𝑓𝑦]

Axial capacity of tied column

𝑃𝑛(𝑚𝑎𝑥) = 0.85∅[0.85𝑓𝑐′(𝐴𝑔 − 𝐴𝑠𝑡 ) + 𝐴𝑠𝑡 𝑓𝑦] ∅ =

0.75 𝑓𝑜𝑟 𝑡𝑖𝑒𝑑 𝑐𝑜𝑙𝑢𝑚𝑛

Axial capacity of spiral column

𝑃𝑛(𝑚𝑎𝑥) = 0.80∅[0.85𝑓𝑐′(𝐴𝑔 − 𝐴𝑠𝑡 ) + 𝐴𝑠𝑡 𝑓𝑦] ∅ =

0.65 𝑓𝑜𝑟 𝑠𝑝𝑖𝑟𝑎𝑙 𝑐𝑜𝑙𝑢𝑚𝑛

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

𝐷𝐿 = 200 𝑘

𝐿𝐿 = 160 𝑘

𝑞𝑎 = 5 𝑘𝑠𝑓 𝑎𝑡 5′𝑑𝑒𝑝𝑡ℎ

𝑓𝑐′ = 3 𝑘𝑠𝑖 and 𝑓𝑦 = 60 𝑘𝑠𝑖

Column size = 16 𝑖𝑛𝑐ℎ 𝑠𝑞𝑢𝑎𝑟𝑒

Solution

𝑃𝑢 = (200 × 1.2 + 160 × 1.6) = 496 𝑘𝑖𝑝

Assume self weight 3%

𝑃𝑢 = (200 + 160) × 1.03 = 370.8 𝑘𝑖𝑝

𝐴 = 370.8

5= 74.16 𝑓𝑡2

Now

L = 1.5B

1.5B2 = 74.16

B = 7.03 ft L = 10.55 ft

qunet=

496

74.16= 6.69 k/ft2

Assume t = 23” d = 23 – 3 =20”

Punching Shear

V = (DL × 1.2 + LL × 1.6)– (a + d)2 × qunet = 496 −

(16+20)2

144×

6.69 = 442.3 k

Resisting Shear

𝑉𝑎 = 4∅√𝑓𝑐′𝑏0𝑑 = 4 × 0.75√3000 ×

36×4×20

1000= 473.23 𝑘𝑖𝑝 > 442.3 𝑘𝑖𝑝

Wide beam shear

𝑉 =35.3

12× 1 × 6.69 = 19.68 𝑘𝑖𝑝

𝑉𝑎 = 2∅√𝑓𝑐′𝑏0𝑑 = 2 × 0.75√3000 ×

12×20

1000= 19.72 𝑘𝑖𝑝 > 19.68 𝑘𝑖𝑝

Moment calculation

𝑀𝑓𝐿=

𝑤𝐿2

2=

6.69×(55.3

12)

2

2= 71.04 𝑘𝑖𝑝 − 𝑓𝑡/𝑓𝑡

𝑀𝑓𝑠=

𝑤𝐿2

2=

6.69×(34.18

12)

2

2= 27.14 𝑘𝑖𝑝 − 𝑓𝑡/𝑓𝑡

𝜌𝑚𝑎𝑥 = 0.85𝛽′ 𝑓𝑐′

𝑓𝑦×

𝜖𝑢

𝜖𝑢+𝜖𝑦= 0.85 × 0.85 ×

3

60×

0.003

0.003 +0.005= 0.0135

𝑑 = √𝑀𝑢

𝑅𝑏= √

𝑀𝑢

∅𝜌𝑓𝑦𝑏 (1−0.59𝜌𝑓𝑦

𝑓𝑐′ )

= √71.04×12

0.9×0.0135 ×60×(1−0.59×0.0135×60

3)

= 10.77" <

20"

Minimum 𝐴𝑠 =200𝑏𝑑

𝑓𝑦=

200 ×12×20

60000= 0.8 𝑖𝑛2

𝐴𝑠𝐿=

𝑀𝑓𝐿

∅𝑓𝑦(𝑑−𝑎

2)

= 71 .04×12

0.9×60 ×(𝑑−1

2)

= 0.81 𝑖𝑛2

𝑎 = 𝐴𝑠𝑓𝑦

0.85𝑓𝑐′𝑏

= 0.81×60

0.85×3×12= 1.59 𝑖𝑛𝑐ℎ

Provide ∅ 16 @4.5" 𝑐/𝑐.

𝐴𝑠𝑠=

𝑀𝑓𝐿

∅𝑓𝑦(𝑑−𝑎

2)

= 27.14×12

0.9×60×(𝑑−.6

2)

= 0.306 𝑖𝑛2

𝑎 = 𝐴𝑠𝑓𝑦

0.85𝑓𝑐′𝑏

= 0.306×60

0.85×3×12= 0.6 𝑖𝑛𝑐ℎ

Provide ∅ 16 @4.5" 𝑐/𝑐.

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

𝑄 = 𝑘𝑖𝐴

𝑘 = 𝐶𝐷102

𝐺𝐼 = (𝐹 − 35)[0.2 + 0.005(𝐿𝐿 − 40) + 0.01(𝐹 − 15)(𝑃𝐼 − 10)]

Uniformity Coefficient

𝐶𝑢 = 𝐷60

𝐷10

Coefficient of Curvature

Cc = D30

2

D60 .D10

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.

𝐿𝐼 = 𝑊𝑐 −𝑃𝐼

𝐿𝐿−𝑃𝐼

𝑆 = 𝐶𝑐𝐻

1+𝑒0𝑙𝑜𝑔

𝜎0′ +∆𝜎′

𝜎0′

Where,

𝑆 = 𝐶𝑜𝑛𝑠𝑜𝑙𝑖𝑡𝑎𝑡𝑖𝑜𝑛 𝑆𝑒𝑡𝑡𝑙𝑒𝑚𝑒𝑛𝑡

𝐶𝑐 = 𝐶𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛 𝐼𝑛𝑑𝑒𝑥

𝑒0 = 𝑁𝑎𝑡𝑢𝑟𝑎𝑙 𝑉𝑜𝑖𝑑 𝑅𝑎𝑡𝑖𝑜 𝑜𝑓 𝑆𝑜𝑖𝑙 𝑆𝑎𝑚𝑝𝑙𝑒

𝐻 = 𝑇ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 𝑜𝑓 𝑆𝑜𝑖𝑙 𝐿𝑎𝑦𝑒𝑟

𝜎0′ = 𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝐸𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑆𝑡𝑟𝑒𝑠𝑠

∆𝜎 ′ = 𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝐸𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑆𝑡𝑟𝑒𝑠𝑠

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

Per

cen

t 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 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

𝑒 =𝑉𝑣

𝑉𝑠

Porosity:

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

𝑛 =𝑉𝑣

𝑉

The relationship between void ratio and porosity

e =Vv

Vs=

Vv

V−Vv=

(VvV

)

1−(VvV

)=

n

1−n

n = e

1+e

Degree of saturation Degree of saturation (S) is defined as the ratio of the volume of water to the volume

of voids.

𝑆 =𝑉𝑤

𝑉𝑣

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

𝑤 =𝑊𝑤

𝑊𝑠

× 100

Unit weight

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

𝛾 = 𝑊

𝑉

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

𝛾 = 𝑊

𝑉=

𝑊𝑠 + 𝑊𝑤

𝑉=

𝑊𝑠 [1 +𝑊𝑤

𝑊𝑠]

𝑉=

𝑊𝑠(1 + 𝑤)

𝑉

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.

𝐷𝑟 = 𝑒𝑚𝑎𝑥 − 𝑒

𝑒𝑚𝑎𝑥 − 𝑒𝑚𝑖𝑛

Where 𝐷𝑟 = 𝑅𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑑𝑒𝑛𝑠𝑖𝑡𝑦, 𝑢𝑠𝑢𝑎𝑙𝑙𝑦 𝑔𝑖𝑣𝑒𝑛 𝑎𝑠 𝑎 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒

𝑒 = 𝐼𝑛𝑠𝑖𝑡𝑢 𝑣𝑜𝑖𝑑 𝑟𝑎𝑡𝑖𝑜 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑜𝑖𝑙 𝑒𝑚𝑎𝑥 = 𝑉𝑜𝑖𝑑 𝑟𝑎𝑡𝑖𝑜 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑜𝑖𝑙 𝑖𝑛 𝑡ℎ𝑒 𝑙𝑜𝑜𝑠𝑒𝑠𝑡 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛

𝑒𝑚𝑖𝑛 = 𝑉𝑜𝑖𝑑 𝑟𝑎𝑡𝑖𝑜 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑜𝑖𝑙 𝑖𝑛 𝑡ℎ𝑒 𝑑𝑒𝑛𝑠𝑒𝑠𝑡 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛

𝐿𝑜𝑛𝑔 𝑜𝑟 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑢𝑠 𝐹𝑜𝑜𝑡𝑖𝑛𝑔 = 𝑞𝑢 = 𝑐𝑁𝑐 + 𝛾𝐷𝑁𝑞 + 1

2𝛾𝐵𝑁𝛾

𝑆𝑞𝑢𝑎𝑟𝑒 𝐹𝑜𝑜𝑡𝑖𝑛𝑔 = 𝑞𝑢 = 1.3𝑐𝑁𝑐 + 𝛾𝐷𝑁𝑞 + 0.4𝛾𝐵𝑁𝛾

𝐶𝑖𝑟𝑐𝑢𝑙𝑎𝑟 𝐹𝑜𝑜𝑡𝑖𝑛𝑔 = 𝑞𝑢 = 1.3𝑐𝑁𝑐 + 𝛾𝐷𝑁𝑞 + 0.3𝛾𝐵𝑁𝛾

Where

qu = Ultimate bearing capacity

Nc,Nq ,Nγ = Bearing capacity factor depends on angle of friction ∅

𝑐 = 𝐶𝑜ℎ𝑒𝑠ℎ𝑖𝑜𝑛 𝑜𝑓 𝑠𝑜𝑖𝑙

𝑞 = 𝛾𝐷𝑓

𝐷𝑓 = 𝐷𝑒𝑝𝑡ℎ 𝑜𝑓 𝑓𝑜𝑢𝑛𝑑𝑎𝑡𝑖𝑜𝑛

𝛾 = 𝑈𝑛𝑖𝑡 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑜𝑖𝑙

Nc,Nq ,Nγ is called Terzaghi bearing capacity factor.

Nc = Cohesion factor

Nq = Surcharge factor

Nγ = 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 𝑚 on saturated clay with 𝑐𝑢 = 95 𝑘𝑁/𝑚2, ∅ = 0,𝑎𝑛𝑑 𝐷𝑓 = 2 𝑚.

Solution:

𝑞𝑛𝑒𝑡 (𝑢) = 5.14𝑐𝑢 [1 + (0.195𝐵

𝐿)] [1 + 0.4

𝐷𝑓

𝐵]

𝑞𝑛𝑒𝑡 (𝑢) = 5.14 × 95 × [1 + (0.195 × 10

15)] [1 + 0.4

2

10] = 595.9 𝑘𝑁/𝑚2

The mat has dimension of 30 𝑚 × 40 𝑚 . 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

18.75𝑘𝑁

𝑚3 . Find the 𝐷𝑓 for a fully compensated foundation.

Solution:

𝐷𝑓 =𝑄

𝐴𝛾=

200 × 103

(30 × 40)(18.75)= 8.89 𝑚

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 𝑃𝐻 of the

solution.

Answer:

Given, [𝐻+] = 0.001 𝑀 = 10−3 𝑀

We know,

𝑃𝐻 = − log[𝐻+]

= − log 10−3

= 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 𝑃𝐻 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

𝐶𝑃 = (𝑃 − 𝑃0)𝐿

𝐴𝐸

Where,

𝑃 = 𝑃𝑢𝑙𝑙 𝑎𝑝𝑝𝑙𝑖𝑒𝑑 𝑑𝑢𝑟𝑖𝑛𝑔 𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑚𝑒𝑛𝑡

𝑃0 = 𝑃𝑢𝑙𝑙 𝑎𝑡 𝑤ℎ𝑖𝑐ℎ 𝑡𝑎𝑝𝑒 𝑖𝑠 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑𝑖𝑠𝑒𝑑

𝐴 = 𝐶𝑟𝑜𝑠𝑠 − 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑎𝑝𝑒

𝐸 = 𝑀𝑜𝑑𝑢𝑙𝑢𝑠 𝑜𝑓 𝑒𝑙𝑎𝑠𝑡𝑖𝑐𝑖𝑡𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑎𝑝𝑒 𝑚𝑎𝑡𝑒𝑟𝑖𝑎𝑙𝑠

Correction for sag

𝐶𝑠 = 𝑤2𝑙2

24𝑃2

Where

𝑤 = 𝑊𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑎𝑝𝑒

𝐿 = 𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑎𝑝𝑒

𝑃 = 𝑃𝑢𝑙𝑙 𝑎𝑝𝑝𝑙𝑖𝑒𝑑

Correction for slope or vertical alignment

𝐶𝑉 = ℎ2

2𝑙

If slopes are given in terms of vertical angels

𝐶𝑉 = 2𝑙 𝑠𝑖𝑛2𝜃

2

Where,

ℎ = 𝐷𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑖𝑛 ℎ𝑒𝑖𝑔ℎ𝑡 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑡ℎ𝑒 𝑒𝑛𝑑𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑙𝑜𝑝𝑒

𝑙 = 𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 ℎ𝑡𝑒 𝑠𝑙𝑜𝑝𝑒

𝜃 = 𝐴𝑛𝑔𝑙𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑙𝑜𝑝𝑒

Es or G = 𝑆ℎ𝑒𝑎𝑟𝑖𝑛𝑔 𝑆𝑡𝑟𝑒𝑠𝑠

𝑆ℎ𝑒𝑎𝑟𝑖𝑛𝑔 𝑆𝑡𝑟𝑎𝑖𝑛 =

𝐸

2(1+𝜇)

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

𝐹1 = 2.84, 𝐹2 = 2.24

𝑅 = 𝐹1 −𝐹𝑐𝑜𝑚

𝐹𝑐𝑜𝑚−𝐹2=

2.84 −2.54

2.54 −2.24= 1

R: 1 = 1:1

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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

2:1 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