The document summarizes load reduction factors from the ACI code for different structural members and conditions. It also provides equations and steps for designing reinforced concrete beams, columns, footings, and prestressed concrete members.

1. 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 =
𝐴 𝑆
𝑏𝑑
𝑛 =
𝐸 𝑆
𝐸𝑐
𝑘 = √2𝜌𝑛 + ( 𝜌𝑛)2 − 𝜌𝑛
𝑗 = 1 −
𝑘
3
𝑓𝑐 = 0.45𝑓′
𝑐
𝑓𝑠 = 0.4𝑓𝑦
Resisting moment of concrete
𝑀𝑐 =
1
2
𝑓𝑐 𝑗𝑘𝑏𝑑2
Resisting moment of steel
𝑀𝑠 = 𝐴 𝑠 𝑓𝑠 𝑗𝑑
𝜌 =
𝐴 𝑆
𝑏𝑑
𝜌 𝑏 = 0.85𝛽1
𝑓′ 𝑐
𝑓𝑦
×
𝜖 𝑢
𝜖 𝑢+𝜖 𝑦
= 0.85 × 0.85 ×
3
60
×
0.003
0.003+0.005
= 0.013
𝜌 𝑚𝑎𝑥 = 0.75𝜌 𝑏
𝜌 < 𝜌 𝑚𝑎𝑥

2. So the beam will fail by yielding
𝑎 =
𝐴 𝑠 𝑓𝑦
0.85𝑓′
𝑐 𝑏
𝑀 𝑑 𝑜𝑟 𝑀 𝑢 = ∅𝐴 𝑠 𝑓𝑦 (𝑑 −
𝑎
2
)
Designsteps of USD Beam
𝜌 = 0.85𝛽′ 𝑓𝑐
′
𝑓𝑦
×
𝜖 𝑢
𝜖 𝑢+𝜖 𝑦
𝜖 𝑢 = 0.003, 𝜖 𝑦 = 0.005
𝑑 = √
𝑀 𝑢
𝑅𝑏
= √
𝑀 𝑢
∅𝜌𝑓𝑦𝑏(1−0.59𝜌
𝑓𝑦
𝑓 𝑐
′ )
Minimum 𝐴 𝑠 =
200𝑏𝑑
𝑓𝑦
𝐴 𝑠 =
𝑀 𝑢
∅𝑓𝑦( 𝑑−
𝑎
2
)
∅ = 0.90
𝑎 =
𝐴 𝑠 𝑓𝑦
0.85𝑓𝑐
′
𝑏
General solution of
𝑎𝑥2
+ 𝑏𝑥 + 𝑐 = 0
𝑥 =
−𝑏±√4𝑎𝑐
2𝑎
𝑃 𝑛(𝑚𝑎𝑥) = 0.85∅[0.85𝑓𝑐
′
(𝐴 𝑔 − 𝐴 𝑠𝑡 ) + 𝐴 𝑠𝑡 𝑓𝑦]
Axial capacity of tied column
𝑃 𝑛(𝑚𝑎𝑥) = 0.85∅[0.85𝑓𝑐
′
(𝐴 𝑔 − 𝐴 𝑠𝑡 ) + 𝐴 𝑠𝑡 𝑓𝑦] ∅ =
0.75 𝑓𝑜𝑟 𝑡𝑖𝑒𝑑 𝑐𝑜𝑙𝑢𝑚𝑛
Axial capacity of spiral column
𝑃 𝑛(𝑚𝑎𝑥) = 0.80∅[0.85𝑓𝑐
′
(𝐴 𝑔 − 𝐴 𝑠𝑡 ) + 𝐴 𝑠𝑡 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

3. 𝑞 𝑎 = 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
′ 𝑏0 𝑑 = 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
′ 𝑏0 𝑑 = 2 × 0.75√3000×
12×20
1000
= 19.72 𝑘𝑖𝑝 > 19.68 𝑘𝑖𝑝
Moment calculation
𝑀𝑓 𝐿
=
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
𝜌 𝑚𝑎𝑥 = 0.85𝛽′ 𝑓𝑐
′
𝑓𝑦
×
𝜖 𝑢
𝜖 𝑢+𝜖 𝑦
= 0.85 × 0.85 ×
3
60
×
0.003
0.003 +0.005
= 0.0135

4. 𝑑 = √
𝑀 𝑢
𝑅𝑏
= √
𝑀 𝑢
∅𝜌𝑓𝑦𝑏(1−0.59𝜌
𝑓𝑦
𝑓 𝑐
′ )
= √
71.04×12
0.9×0.0135 ×60×(1−
0.59×0.0135×60
3
)
= 10.77" <
20"
Minimum 𝐴 𝑠 =
200𝑏𝑑
𝑓𝑦
=
200 ×12×20
60000
= 0.8 𝑖𝑛2
𝐴 𝑠 𝐿
=
𝑀 𝑓 𝐿
∅𝑓𝑦( 𝑑−
𝑎
2
)
=
71.04×12
0.9×60 ×( 𝑑−
1
2
)
= 0.81 𝑖𝑛2
𝑎 =
𝐴 𝑠 𝑓𝑦
0.85𝑓𝑐
′
𝑏
=
0.81×60
0.85×3×12
= 1.59 𝑖𝑛𝑐ℎ
Provide ∅ 16 @4.5" 𝑐/𝑐.
𝐴 𝑠 𝑠
=
𝑀 𝑓 𝐿
∅𝑓𝑦( 𝑑−
𝑎
2
)
=
27.14×12
0.9×60×( 𝑑−
.6
2
)
= 0.306 𝑖𝑛2
𝑎 =
𝐴 𝑠 𝑓𝑦
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

5. 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 = 𝐶𝐷10
2
𝐺𝐼 = ( 𝐹 − 35)[0.2 + 0.005( 𝐿𝐿 − 40) + 0.01( 𝐹 − 15)( 𝑃𝐼 − 10)]
Uniformity Coefficient
𝐶 𝑢 =
𝐷60
𝐷10
Coefficient of Curvature
Cc =
D30
2
D60 .D10

6. 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.
𝐿𝐼 =
𝑊𝑐−𝑃𝐼
𝐿𝐿−𝑃𝐼
𝑆 = 𝐶𝑐
𝐻
1+𝑒0
𝑙𝑜𝑔
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

7. 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
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
=
(
Vv
V
)
1−(
Vv
V
)
=
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.

8. 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.
𝐷 𝑟 =
𝑒 𝑚𝑎𝑥 − 𝑒
𝑒 𝑚𝑎𝑥 − 𝑒 𝑚𝑖𝑛
Where
𝐷 𝑟 = 𝑅𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑑𝑒𝑛𝑠𝑖𝑡𝑦, 𝑢𝑠𝑢𝑎𝑙𝑙𝑦 𝑔𝑖𝑣𝑒𝑛 𝑎𝑠 𝑎 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒
𝑒 = 𝐼𝑛𝑠𝑖𝑡𝑢 𝑣𝑜𝑖𝑑 𝑟𝑎𝑡𝑖𝑜 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑜𝑖𝑙
𝑒 𝑚𝑎𝑥 = 𝑉𝑜𝑖𝑑 𝑟𝑎𝑡𝑖𝑜 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑜𝑖𝑙 𝑖𝑛 𝑡ℎ𝑒 𝑙𝑜𝑜𝑠𝑒𝑠𝑡 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛
𝑒 𝑚𝑖𝑛 = 𝑉𝑜𝑖𝑑 𝑟𝑎𝑡𝑖𝑜 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑜𝑖𝑙 𝑖𝑛 𝑡ℎ𝑒 𝑑𝑒𝑛𝑠𝑒𝑠𝑡 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛
𝐿𝑜𝑛𝑔 𝑜𝑟 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑜𝑢𝑠 𝐹𝑜𝑜𝑡𝑖𝑛𝑔 = 𝑞 𝑢 = 𝑐𝑁𝑐 + 𝛾𝐷𝑁 𝑞 +
1
2
𝛾𝐵𝑁 𝛾
𝑆𝑞𝑢𝑎𝑟𝑒 𝐹𝑜𝑜𝑡𝑖𝑛𝑔 = 𝑞 𝑢 = 1.3𝑐𝑁𝑐 + 𝛾𝐷𝑁 𝑞 + 0.4𝛾𝐵𝑁 𝛾
𝐶𝑖𝑟𝑐𝑢𝑙𝑎𝑟 𝐹𝑜𝑜𝑡𝑖𝑛𝑔 = 𝑞 𝑢 = 1.3𝑐𝑁𝑐 + 𝛾𝐷𝑁 𝑞 + 0.3𝛾𝐵𝑁 𝛾
Where
qu = Ultimate bearing capacity
Nc,Nq,Nγ = Bearing capacity factor depends on angle of friction ∅
𝑐 = 𝐶𝑜ℎ𝑒𝑠ℎ𝑖𝑜𝑛 𝑜𝑓 𝑠𝑜𝑖𝑙
𝑞 = 𝛾𝐷 𝑓
𝐷𝑓 = 𝐷𝑒𝑝𝑡ℎ 𝑜𝑓 𝑓𝑜𝑢𝑛𝑑𝑎𝑡𝑖𝑜𝑛
𝛾 = 𝑈𝑛𝑖𝑡 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑜𝑖𝑙

9. 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 𝑐 𝑢 = 95 𝑘𝑁/𝑚2
, ∅ = 0, 𝑎𝑛𝑑 𝐷𝑓 = 2 𝑚.

10. Solution:
𝑞 𝑛𝑒𝑡 (𝑢) = 5.14𝑐 𝑢 [1 + (
0.195𝐵
𝐿
)] [1 + 0.4
𝐷𝑓
𝐵
]
𝑞 𝑛𝑒𝑡 (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 𝑃 𝐻
of the
solution.
Answer:
Given, [ 𝐻+] = 0.001 𝑀 = 10−3
𝑀
We know,
𝑃 𝐻
= −log[𝐻+
]
= − log10−3
= 3.00

11. 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 𝑃 𝐻
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)

12.  Determination of Chemical oxygen Demand (COD)
 Determination of Turbidity of Water
 Correction for pull
𝐶 𝑃 =
( 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 =
𝑤2
𝑙2
24𝑃2
Where
𝑤 = 𝑊𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑎𝑝𝑒
𝐿 = 𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑎𝑝𝑒
𝑃 = 𝑃𝑢𝑙𝑙 𝑎𝑝𝑝𝑙𝑖𝑒𝑑
 Correction for slope or vertical alignment
𝐶 𝑉 =
ℎ2
2𝑙
If slopes are given in terms of vertical angels
𝐶 𝑉 = 2𝑙 𝑠𝑖𝑛2
𝜃
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

13. 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
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

14. Road Way
10 m
Slope (2:1)
3 m
Berm
10 m
Borrow pit
10 m
Slope (2:1)
3 m
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