· 2019-09-24 · Minimum Shear Reinforcement •Clause 26.5.1.5 of IS 456 stipulates that the maximum spacing of shear reinforcement measured along the axis of the member shall

Cement Concrete

Civil Engineering by Sandeep Jyani

𝑀5 → 1: 5: 10

𝑀10 → 1: 3: 6

𝑀7.5 → 1: 4: 8

𝑀15 → 1: 2: 4

𝑀20 → 1: 1.5: 3

𝑀25 → 1: 1: 2

Young’s modulus of Elasticity of Concrete

As per IS 456: 2000, young’s modulus of elasticity of concrete is …

𝑬𝒄 = 𝟓𝟎𝟎𝟎 𝒇𝒄𝒌

For example, for M25,

So, 𝑬𝒄 = 𝟓𝟎𝟎𝟎 𝒇𝒄𝒌

𝑬𝒄 = 𝟓𝟎𝟎𝟎 𝟐𝟓

𝑬𝒄 = 𝟓𝟎𝟎𝟎 × 𝟓 = 𝟐𝟓𝟎𝟎𝟎 𝑵/𝒎𝒎𝟐

Mix Characteristic compressive strength (N/mm2)


As per IS specifications, the nominal cover for moderate exposure should not be less than

a) 20 mm

b) 30 mm

c) 45 mm

d) 50 mm


As per IS specifications, the nominal cover for moderate exposure should not be less than

a) 20 mm

b) 30 mm

c) 45 mm

d) 50 mm


Nominal Value of Cover in “mild exposure”

1. Slab 20mm

2. Beam 25mm

3. Column 40mm

4. Foundation 50mm


Control of Deflection


30. If ratio of span to overall depth does not exceed 10, the stiffness of the beam will ordinarily be satisfactory in case of a

a) Simply supported beam

b) Continuous beam

c) Cantilever beam

d) None of these


30. If ratio of span to overall depth does not exceed 10, the stiffness of the beam will ordinarily be satisfactory in case of a

a) Simply supported beam

b) Continuous beam

c) Cantilever beam

d) None of these


Span to effective depth ratio for spans upto 10m,

Cantilever 7Simply supported 20Constinuous 26

The maximum depth of neutral axis for a beam with Fe 415 bars in limit state method of design

a) 0.46 d

b) 0.48 d

c) 0.50 d

d) 0.53 d


The maximum depth of neutral axis for a beam with Fe 415 bars in limit state method of design

a) 0.46 d

b) 0.48 d

c) 0.50 d

d) 0.53 d


fy N/mm2 𝒙𝒖 𝒍𝒊𝒎

Fe 250 0.53 d

Fe 415 0.48 d

Fe 500 0.46 d

Moment of Resistance

• 𝑀𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑅𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑓𝑟𝑜𝑚 𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛 𝑠𝑖𝑑𝑒• 𝑴𝑶𝑹 = 𝟎. 𝟑𝟔 𝒇𝒄𝒌 𝑩 𝒙𝒖(𝒅 − 𝟎. 𝟒𝟐𝒙𝒖) Valid for Balanced and over

reinforced section

• 𝑀𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑅𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑓𝑟𝑜𝑚 𝑇𝑒𝑛𝑠𝑖𝑜𝑛 𝑠𝑖𝑑𝑒• 𝑴𝑶𝑹 = 𝟎. 𝟖𝟕𝒇𝒚 𝑨𝒔𝒕(𝒅 − 𝟎. 𝟒𝟐𝒙𝒖) Valid for Balanced and under

reinforced section


Effective Span of Beam 1) Simply Supported beam or Slab: The effective span of a simply

supported member is taken lesser of the following:I. 𝒍 = 𝑳𝒄 + 𝒅II. 𝒍 (centre to centre distance between supports)Where 𝒍 = centre to centre distance between supports

𝑳𝒄 = clear span𝒅 = 𝒆𝒇𝒇𝒆𝒄𝒕𝒊𝒗𝒆 𝒅𝒆𝒑𝒕𝒉 𝒐𝒇 𝒃𝒆𝒂𝒎 𝒐𝒓 𝒔𝒍𝒂𝒃

𝐿𝑐

𝒍


Effective Span of Beam 2) Continuous Beam or Slab: The effective span of a continuous member

is taken as:i. If width of support 𝒕𝒔 ≤

𝑳𝒄

𝟏𝟐, then effective span is taken as lesser of …

a) 𝒍 = 𝑳𝒄 + 𝒅b) 𝒍 (centre to centre distance between supports)

Where 𝒍 = centre to centre distance between supports𝑳𝒄 = clear span

𝒅 = 𝒆𝒇𝒇𝒆𝒄𝒕𝒊𝒗𝒆 𝒅𝒆𝒑𝒕𝒉 𝒐𝒇 𝒃𝒆𝒂𝒎 𝒐𝒓 𝒔𝒍𝒂𝒃

ii. If width of support 𝒕𝒔 ≥𝑳𝒄

𝟏𝟐, then effective span is taken as lesser of …

a) 𝒍 = 𝑳𝒄 + 𝟎. 𝟓 𝒅b) 𝒍 = 𝑳𝒄 + 𝟎. 𝟓 𝒕𝒔

𝐿𝑐

𝒍 𝒍

𝐿𝑐𝒕𝒔 Civil Engineering by Sandeep Jyani

For M20 grade of concrete, the maximum

shear stress shall not exceed

a) 1.6 𝑁/𝑚𝑚2

b) 1.9 𝑁/𝑚𝑚2

c) 2.8 𝑁/𝑚𝑚2

d) 2.2 𝑁/𝑚𝑚2


For M20 grade of concrete, the maximum

shear stress shall not exceed

a) 1.6 𝑁/𝑚𝑚2

b) 1.9 𝑁/𝑚𝑚2

c) 2.8 𝑵/𝒎𝒎𝟐

d) 2.2 𝑁/𝑚𝑚2


𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠 𝑠ℎ𝑜𝑢𝑙𝑑 𝑛𝑜𝑡 𝑏𝑒

𝑔𝑟𝑒𝑎𝑡𝑒𝑟 𝑡ℎ𝑎𝑛 2.8𝑁

𝑚𝑚2

Minimum Shear Reinforcement

• Clause 26.5.1.5 of IS 456 stipulates that the maximum spacing of shear reinforcement measured along the axis of the member shall not be more than 0.75 d for vertical stirrups and d for inclined stirrups at 45° , where d is the effective depth of the section.

1. Sv < 0.75 d Vertical shear Reinforcement

2. Sv < d Inclined shear Reinforcement

3. However, the spacing shall not exceed 300 mm in any case.


Steps for Design

1. Find maximum shear force

• SSB = 𝑾𝑳

𝟐𝒐𝒓

𝑾𝒖𝑳

𝟐

2. Nominal Shear Stress

• 𝝉𝒗 =𝑽

𝒃𝒅𝒐𝒓

𝑽𝒖

𝒃𝒅and compare with 𝝉𝒄 𝒎𝒂𝒙

3. Shear Strength of Concrete (𝝉𝒄)

4. Net shear force resisted by Shear reinforcement• 𝑽𝒔 = 𝝉𝒗 − 𝝉𝒄 𝑩𝒅

5. Design of Shear reinforcement


Steps for Design

6. Minimum shear reinforcement

7. Maximum spacing in shear reinforcement1. Sv < 0.75 d Vertical shear Reinforcement

2. Sv < d Inclined shear Reinforcement

3. However, the spacing shall not exceed 300 mm in any case.

(whichever is minimum out of these)


Side Face Reinforcement

1. When depth of the beam is greater than 750mm (and beam is not subjected to torsion)

2. When beam is subjected to torsion, and depth of the beam is greater than 450mm, then side face reinforcement is also provided

• In above both cases, side face reinforcement at 0.1 % of the web area (B x d) is provided on both the faces respectively

Vertical shear reinforcement

Side face reinforcement


Design Bond Stresses for Mild steel bars as per IS 456:2000 and LSM

• For deformed bars conforming to IS 1786 these values shall be increased by 60 percent

• For bars in compression, the values of bond stress for bars in tension shall be increased by 25 percent.

Deformed bars are rods of steels provided with lugs, ribs or deformation on the

surface of bar, these bars minimize slippage in concrete and increases the bond

between the two materials.

Deformed bars have more tensile stresses than that of mild steel plain bars.


Anchoring Reinforcing Bar (26.2.2) a) Deformed bars may be used without end

anchorages provided development length requirement is satisfied. Hooks should normally he provided for plain bars in tension.

b) Bends and hooks - Bends and hooks shall conform to IS 2502i. Bends-The anchorage value of bend shall be taken

as 4 times the diameter of the bar for each 45°bend subject to a maximum of 16 times the diameter of the bar.• 4∅

• 𝐌𝐚𝐱𝐢𝐦𝐮𝐦 𝟏𝟔 ∅

ii. Hooks-The anchorage value of a standard U-type hook shall be equal to 16 times the diameter of the bar

The minimum value of n for mild steel is 2 and for other steels is 4


Effective Length of column


Effective Length of column


For columns, As per IS 456:2000, CL 26.5.3

1. Minimum percentage of steel= 0.8%

2. Maximum percentage of steela) 4% if bar are lapped

b) 6% if bars are not lapped

3. Minimum diameter of longitudinal bars = 12mm

4. Minimum number of bars a) For circular=6

b) For Rectangular=4

5. Maximum spacing of longitudinal bars = 300mm


As per IS code6. Load Carrying capacity of column is increased by 5%

percent when helical reinforcement is provided as a transverse reinforcement

7. Maximum compressive strain in concrete in axial compression is taken to be 0.002

8. Pitch and diameter of lateral tiesA. The pitch of transverse reinforcement shall not more than

the least of the following distances:a) The least lateral dimension of the compression members;b) Sixteen times the smallest diameter of the longitudinal

reinforcement bar to be tied, andc) 300 mm

B. Diameter-A. The diameter of the polygonal links or lateral ties shall not be less

than one fourth of the diameter of the largest longitudinal bar and in no case less than 16mm i.e.

• Greater of ൝𝑑𝑖𝑎 𝑜𝑓 𝑚𝑎𝑖𝑛 𝑏𝑎𝑟

4

6 𝑚𝑚


IS code Recommendations1. Minimum area of reinforcement

• 𝑭𝒐𝒓 𝑭𝒆 𝟒𝟏𝟓 = 𝟎. 𝟏𝟐% 𝒐𝒇 𝒕𝒐𝒕𝒂𝒍 𝒂𝒓𝒆𝒂 = 𝟎. 𝟏𝟐% × 𝑩 × 𝑫

• 𝑭𝒐𝒓 𝑭𝒆 𝟐𝟓𝟎 = 𝟎. 𝟏𝟓% 𝒐𝒇 𝒕𝒐𝒕𝒂𝒍 𝒂𝒓𝒆𝒂 = 𝟎. 𝟏𝟓% × 𝑩 × 𝑫

2. Maximum diameter of steel bar in slab

• ∅𝒎𝒂𝒙 =𝑻𝒉𝒊𝒄𝒌𝒏𝒆𝒔𝒔 𝒐𝒇 𝒔𝒍𝒂𝒃

𝟖

3. Maximum spacing of the reinforcement

a) Main bar = 𝐥𝐞𝐬𝐬𝐞𝐫 𝒐𝒇 ቊ𝟑𝒅

𝟑𝟎𝟎 𝒎𝒎

b) Distribution bar = 𝐥𝐞𝐬𝐬𝐞𝐫 𝒐𝒇 ቊ𝟓𝒅

𝟒𝟓𝟎 𝒎𝒎‘d’ is the effective depth of slab.


Carbon Content

28

𝑃𝑖𝑔 𝑖𝑟𝑜𝑛 (4−5%)> 𝐶𝑎𝑠𝑡 𝐼𝑟𝑜𝑛(2−4.5%)> 𝐶𝑎𝑠𝑡 𝑆𝑡𝑒𝑒𝑙 (>2%)> 𝐶𝑎𝑟𝑏𝑜𝑛 𝑠𝑡𝑒𝑒𝑙 (𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 2%)

> 𝐻𝑖𝑔ℎ 𝑐𝑎𝑟𝑏𝑜𝑛 𝑠𝑡𝑒𝑒𝑙 (0.6−1.4%)>𝑀𝑒𝑑𝑖𝑢𝑚 𝑐𝑎𝑟𝑏𝑜𝑛(0.25−0.6%)>𝑙𝑜𝑤 𝑐𝑎𝑟𝑏𝑜𝑛 𝑠𝑡𝑒𝑒𝑙 (𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 0.25%)> 𝑊𝑟𝑜𝑢𝑔ℎ𝑡 𝐼𝑟𝑜𝑛 (𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 0.1%)>𝑃𝑢𝑟𝑒 𝑖𝑟𝑜𝑛 (0%)


Some Important Codes

• IS 456: 2000 RCC

• IS 800: 2007 Steel (2007-LSM, 1984-WSM)

• IS 1343 Pre Stress Concrete

• IS 10262 Design Mix

• IS 383 Fine and Coarse Aggregate

• IS 875 Design Load for buildings and structures

29Civil Engineering by Sandeep Jyani

IMPORTANT TERMS

1. PITCH – It is the distance between two consecutive/continuous rivetsmeasured parallel to the direction of force. It is denoted by ‘𝒑’.

2. END DISTANCE – It is the distance between centre of rivet and edge/end of the plate element, measured parallel to the direction of force.

3. GAUGE DISTANCE - It is the distance between two continuous rivets measured perpendicular to the force of direction.

4. EDGE DISTANCE - It is the distance between centre of rivet and edge/end of the plate element, measured perpendicular to the force of direction.

30

P

P𝒑𝒊𝒕𝒄𝒉

𝒆𝒏𝒅𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆

𝒈𝒂𝒖𝒈𝒆

𝒆𝒅𝒈𝒆𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆

N/mm2 Axial Tension

Shearing Bearing

PDS 100 100 300

PDF 90 90 270

HDF 80 80 250

31

CONNECTIONSi. Power shop rivetsii. Power driven field rivetsiii. Hand driven field rivets

CONNECTIONS1. RIVETED CONNECTIONS:

• The nominal dia of rivet is said to be shank dia under cold condition, and gross dia of rivet is taken as dia of hole.

• The strength of a rivet is based on its gross diameter under the assumption that rivet fills the hole completely.

• For ease in connection dia of hole is taken larger than nominal dia of rivet thus as per IS: code:• For nominal dia ≤ 25 mm

• Gross dia = nominal dia + 1.5 mm,

• For nominal dia > 25 mm

• Gross dia = nominal dia + 2 mm,

32

shank

head

∅

𝒅𝒊𝒂 𝒐𝒇 𝒉𝒐𝒍𝒆 = ∅ + 𝟏. 𝟓

𝒅𝒊𝒂 𝒐𝒇 𝒉𝒐𝒍𝒆 = ∅ + 𝟐

∅ = 𝒏𝒐𝒎𝒊𝒏𝒂𝒍 𝒅𝒊𝒂

𝟏. 𝟔 ∅

𝟎. 𝟕∅

SPECIFICATIONS AS PER IS 800 - 1984

• MINIMUM END AND EDGE DISTANCE• This recommendation is provided to prevent three types of

failure in plates:i. Splitting failure of plate

ii. Shearing failure of plate

iii. Bearing failure of plate

• Edge distance and end distance(minimum) • = 1.5 gross dia of rivet (machine cut element)

• The above provision is valid for the end distance and edge distance is done by machine cut element.

33


• MINIMUM END AND EDGE DISTANCE• Edge distance and end distance(minimum)

• = 1.7 gross dia of rivet (hand driven elements)

• The above provision is valid for the end distance and edge distance is done by hand driven elements.

• But for analysis and design purpose, we adopt edge distance and end distance(minimum) • = 2.0 gross dia of rivet.

34

SPECIFICATIONS AS PER IS 800 - 1984• PITCH

• Minimum pitch of rivet is 2.5 nominal dia of rivet.

• Maximum pitch of rivet or weld• IN COMPRESSION

• The maximum pitch provision is provided to ensure the prevention of buckling between the connections

• Maximum pitch = min(12t or 200mm) where t is thickness for thinner plate

• IN TENSION• The maximum pitch provision is provided to ensure the

prevention of separation of plates between the connections

• Maximum pitch = min(16t or 200mm) where t is thickness for thinner plate

35


• NOTE : • If the rivets are staggered (not in the same line)and of the

gauge distance smaller than 75mm, then above recommended values in compression and tension zone for maximum pitch are increased by 50%, i.e.,

• For compression –• Maximum pitch = 18t or 300mm (minimum of the two)

• For tension –• Maximum pitch = 24t or 300mm (minimum of the two)

36

Unwin’s formula

• It is used when diameter of rivet is not known∅ = 𝟔. 𝟎𝟒 𝒕

Where t is thickness of thinner plate

37

Effective Length• In SOM, we use Theoretical

Value, and in Design we use IS recommended values

• For Laced Columns, above values are increased by 5%

• For battened column above values are increased by 10%

• Effective length in IS Code is slightly larger than the theoretical value to account for the lack of 100% fixity at support

38

𝟎. 𝟓𝟎 𝒍 𝟎. 𝟔𝟓 𝒍

𝟎. 𝟕𝟎 𝒍 𝟎. 𝟖𝟎 𝒍

𝟏. 𝟎𝟎 𝒍 𝟏. 𝟎𝟎 𝒍

𝟏. 𝟎𝟎 𝒍 𝟏. 𝟐𝟎 𝒍

− 𝟏. 𝟓𝟎 𝒍

𝟐. 𝟎𝟎 𝒍 𝟐. 𝟎𝟎 𝒍

𝟐. 𝟎𝟎 𝒍 𝟐. 𝟎𝟎 𝒍

Slenderness Ratio

Slenderness ratio 𝝀 =𝑬𝒇𝒇𝒆𝒄𝒕𝒊𝒗𝒆 𝒍𝒆𝒏𝒈𝒕𝒉

𝒓𝒂𝒅𝒊𝒖𝒔 𝒐𝒇 𝑮𝒚𝒓𝒂𝒕𝒊𝒐𝒏

Civil Engineering by Sandeep Jyani 39

Radius of Gyration• It is distance such that its square

multiplied by area gives Moment of inertia about the given axis

𝜸𝒎𝒊𝒏𝟐 × 𝑨 = 𝑰

𝜸𝒎𝒊𝒏 =𝑰

𝑨

CG

Given axis

KArea A

𝝀 =𝑳𝑬𝒇𝒇𝒆𝒄𝒕𝒊𝒗𝒆

𝜸𝒎𝒊𝒏

40

𝝀 =𝑳𝑬𝒇𝒇𝒆𝒄𝒕𝒊𝒗𝒆

𝜸𝒎𝒊𝒏

LACINGS

• Lacing member are idealised as truss element, i.e., they re subjected either to tension or compression.

• B.M. in lacing member is zero, to ensure that bending moment is zero, provide only one rivet at each end as far as possible.

• Maximum slenderness ratio 𝝀 for lacing member is limited to 145.

• The angle of lacing w.r.t. vertical is 40°to 70° (welding 60° to 90°)

LACINGS

• If θ decreases, length of lacing will increase.

• Effective length• For single lacing 𝑳𝒆𝒇𝒇 = 𝑳𝟏

• For double lacing 𝑳𝒆𝒇𝒇 = 𝟎. 𝟕𝑳𝟏

• For welded lacing 𝑳𝒆𝒇𝒇 = 𝟎. 𝟕𝑳𝟏

• Minimum thickness of lacing member

• 𝒕𝒎𝒊𝒏 = 𝒍𝟏

𝟒𝟎(for single lacing)

• 𝒕𝒎𝒊𝒏 = 𝒍𝟏

𝟔𝟎(for double lacing)

BATTENS

• It behave like very small beam member and subjected to bendingmoment.

• The effective length of battened column should be increased by 10%.

• Minimum number of battens provided = 4

• Provide batten on opposite faces such that one should be the mirror image of other.

43

d

C S

de

s

𝑬𝒏𝒅 𝒃𝒂𝒕𝒕𝒆𝒏

𝑰𝒏𝒕𝒆𝒓𝒎𝒆𝒅𝒊𝒂𝒕𝒆𝒃𝒂𝒕𝒕𝒆𝒏

𝜶

SHAPE FACTORSHAPE SHAPE FACTOR

Triangle 2.343

Triangle 2.0

Rhombus 2.0

Rectangle and Square 1.50

Circle 1.698

Hollow circle 1.273

44

Some Important Terms

1. Plinth Area:• The built-up covered area of a building

measured at floor level of any storey is called Plinth area

✓Wall area, porch, internal shaft(<2m2), ducts, lift, Staircase room, Machine room etc are included

❖Cantilever porch, external staircase, Terrace at first floor, Additional floor for seating in assembly buildings, theatres, auditoriums


𝑷𝒍𝒐𝒕 𝑨𝒓𝒆𝒂

𝑷𝒍𝒊𝒏𝒕𝒉 𝑨𝒓𝒆𝒂

2. Carpet area the covered area of the usable spaces of rooms at any floor. It is measured between walls to walls within the building• It is the sum of the actual areas of the rooms where

you can carpet.• Following areas are not included in Carpet area:

1. Wall area2. Veranda3. Corridor and passage4. Entrance hall and porch5. Staircase and stair cover6. Lift shaft and machine room for lift7. Bathroom and lavatory8. Kitchen and pantry9. Store10. Canteen11. Air conditioning duct and plant room12. Shaft for sanitary piping13. Stilted floor and garage



𝑪𝒂𝒓𝒑𝒆𝒕 𝑨𝒓𝒆𝒂

3. Floor Area• 𝑃𝑙𝑖𝑛𝑡ℎ 𝑎𝑟𝑒𝑎 − 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑤𝑎𝑙𝑙𝑠

4. Set Back Area• Setback can be explained as the minimum

open space required around any building or structure.

• Municipal regulations make it mandatory for a specific distance to be maintained between a building and the boundary of the plot on which the building is being constructed on, in order to keep it away from roads, water bodies or other buildings

5. Circulation Area• Area that is helpful in movement of people

through, around and between buildings and other parts of the built environment

• Lobbies, corridors, stairs, lift, landings, etc.



𝑭𝒍𝒐𝒐𝒓 𝑨𝒓𝒆𝒂

Que 1. What Is the Floor Area Ratio (FAR)?

Answer:

The floor area ratio (FAR) is the relationship between the total amount of usable floor area that a building has, or has been permitted to have and the total area of the lot on which the building stands. The ratio is determined by dividing the total or gross floor area of the building by the gross area of the lot.

𝐹𝑙𝑜𝑜𝑟 𝐴𝑟𝑒𝑎 𝑅𝑎𝑡𝑖𝑜 =𝑇𝑜𝑡𝑎𝑙 𝐵𝑢𝑖𝑙𝑑𝑖𝑛𝑔 𝐹𝑙𝑜𝑜𝑟 𝐴𝑟𝑒𝑎

𝐺𝑟𝑜𝑠𝑠 𝑃𝑙𝑜𝑡 𝐴𝑟𝑒𝑎


𝟔𝟎 𝒄𝒎

The above figure represents plan and section of an excavation

layout. The volume of earthwork in excavation of foundation trench

is:

a) 6.528 Cu. m

b) 8.064 Cu. m

c) 8.832 Cu. m

d) 9.600 Cu. m

𝟖𝟎𝒄𝒎section

𝟔𝒎

𝟒𝒎

𝟔𝟎 𝒄𝒎

The above figure represents plan and section of an excavation

layout. The volume of earthwork in excavation of foundation trench

is:

a) 6.528 Cu. m

b) 8.064 Cu. m

c) 8.832 Cu. m

d) 9.600 Cu. m

𝟖𝟎𝒄𝒎section

𝟔𝒎

𝟒𝒎

𝑳𝒆𝒏𝒈𝒕𝒉 𝒐𝒇 𝒄𝒆𝒏𝒕𝒓𝒆 𝒍𝒊𝒏𝒆 = 𝟑. 𝟐 + 𝟓. 𝟐 × 𝟐 = 𝟏𝟔. 𝟖𝒎

𝑪𝒓𝒐𝒔𝒔 𝒔𝒆𝒄𝒕𝒊𝒐𝒏 𝒂𝒓𝒆𝒂 = 𝟎. 𝟖 × 𝟎. 𝟔 = 𝟎. 𝟒𝟖𝒎𝟐

𝑽𝒐𝒍𝒖𝒎𝒆 𝒐𝒇 𝒆𝒙𝒄𝒂𝒗𝒂𝒕𝒊𝒐𝒏 = 𝟎. 𝟖 × 𝟎. 𝟔 × 𝟏𝟔. 𝟖 = 𝟖. 𝟎𝟔𝟒𝒎𝟑

4. Compressibility (𝜷) and Bulk Modulus (K)• Compressibility is the reciprocal of Bulk Modulus

• 𝜷 =𝟏

𝑲


Properties of Fluid

𝑩𝒖𝒍𝒌 𝑴𝒐𝒅𝒖𝒍𝒖𝒔 𝑲 = −∆𝑷

∆𝑽𝑽

⇒ 𝑲 = −𝑷𝟐 − 𝑷𝟏

𝑽𝟐 − 𝑽𝟏

𝑽𝟏

𝑷+∆𝑷

𝑷

𝑷

𝑷

𝑷

𝑷

+∆𝑷

+∆𝑷

+∆𝑷

+∆𝑷

+∆𝑷


5. Viscosity:Properties of Fluid

⇒ 𝝁 =𝝉

𝒅𝜽𝒅𝒕

⇒ 𝝁 =𝝉

𝒅𝒖𝒅𝒚

𝝉 = 𝝁𝒅𝒖

𝒅𝒚𝝉 = 𝝁

𝒅𝜽

𝒅𝒕

𝑼𝒏𝒊𝒕𝒔 𝒐𝒇 𝑽𝒊𝒔𝒄𝒐𝒔𝒊𝒕𝒚:

𝝁 =𝑵/𝒎𝟐

𝒎/𝒔𝒆𝒄𝒎

= 𝑵𝒔/𝒎𝟐

𝝁 = 𝑷𝒂. 𝒔

𝝉 =𝑭

𝑨= 𝝁

𝒖

𝒚

𝑭 =𝝁𝑨𝒖

𝒚


• 𝑼𝒏𝒊𝒕𝒔 𝒐𝒇 𝑽𝒊𝒔𝒄𝒐𝒔𝒊𝒕𝒚:

• 𝝁 =𝑵/𝒎𝟐

𝒎/𝒔𝒆𝒄

𝒎

= 𝑵𝒔/𝒎𝟐

𝝁 = 𝑷𝒂. 𝒔

=𝑵

𝒎𝟐𝒔𝒆𝒄

=𝒌𝒈 𝒎

𝒔𝒆𝒄𝟐 𝒎𝟐𝒔𝒆𝒄

=𝟏𝟎𝟎𝟎𝒈

𝒔𝒆𝒄 × 𝟏𝟎𝟎 𝒄𝒎

=𝟏𝟎𝒈

𝒄𝒎 − 𝒔𝒆𝒄

𝟏 𝑷𝒂. 𝒔 = 𝟏𝟎𝑷𝒐𝒊𝒔𝒆

6. Kinematic Viscosity

𝝂 =𝝁

𝝆

• SI Unit = 𝒎𝟐/𝒔𝒆𝒄

• In CGS, 𝒎𝟐

𝒔𝒆𝒄=

𝟏𝟎𝟎×𝟏𝟎𝟎𝒄𝒎𝟐

𝒔𝒆𝒄= 𝟏𝟎𝟒 𝒄𝒎𝟐

𝒔𝒆𝒄= 𝟏𝟎𝟒 stokes



𝟐𝝈

𝒅

𝟖𝝈

𝒅

𝟒𝝈

𝒅

𝑩𝒖𝒃𝒃𝒍𝒆: 𝑫𝒓𝒐𝒑𝒍𝒆𝒕: 𝒍𝒊𝒒𝒖𝒊𝒅 𝒋𝒆𝒕

𝒂𝒊𝒓

𝒂𝒊𝒓

𝟒 ∶ 𝟐 ∶ 𝟏

Expression for Capillary Rise and Capillary FallFor capillary fall,

⇒ 𝐹𝑠 𝑐𝑜𝑠 𝜃 = 𝑃 ×𝜋

4× 𝑑2

⇒ 𝜎 × 𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 × 𝑐𝑜𝑠 𝜃 = 𝜌𝑔ℎ × 𝐴

⇒ 𝜎 × 𝜋𝑑 × 𝑐𝑜𝑠 𝜃 = 𝜌 ×𝜋

4𝑑2 × ℎ × 𝑔

⇒ ℎ =4𝜎 𝑐𝑜𝑠 𝜃

𝜌𝑔𝑑


𝑭𝒔𝑭𝒔

𝜽 𝜽 𝒉

𝑷

CAPILLARY RISE FOR PARALLEL PLATEFor equilibrium,

𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑓𝑙𝑢𝑖𝑑 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 = 𝐹𝑠 cos 𝜃𝐹𝑠 = 𝜎 × 2𝑏 cos 𝜃

⇒ 𝑊 = 𝜎 × 2𝑏 cos 𝜃⇒ 𝜌 × 𝐴 × ℎ × 𝑔 = 𝜎 × 2𝑏cos 𝜃

⇒ 𝜌 × 𝑏 × 𝑡 × ℎ × 𝑔 = 𝜎 × 2𝑏cos 𝜃

⇒ ℎ =2𝜎 cos 𝜃

𝑊𝑡

⇒ 𝒉 =𝟐𝝈 𝐜𝐨𝐬 𝜽

𝝆𝒈𝒕


𝒕

𝑭𝒔

𝒉

𝑭𝒔

𝒃

TYPES OF PRESSURE1. Atmospheric pressure

• Atmospheric air exerts a normal pressure upon all surface with which it is in contact and it is called atmospheric pressure.

• Measured by barometer.

2. Absolute pressure• Pressure which is measured about with reference

to complete vacuum pressure.• It is also called actual pressure at a given point.

3. Gauge and Vacuum pressure• Vacuum pressure is the pressure below

atmospheric pressure.• Gauge pressure is the pressure measured above

atmospheric pressure.• Gauge pressure is measured with the help of

pressure measuring instrument in which atmospheric pressure is taken as datum i.e. atmosphere pressure on the scale is marked as 0.

Atmospheric pressureGauge pressure

Absolute pressure

Vacuum pressure

Pressure

Absolute zero Pressure

A

B

𝑷𝒂𝒃𝒔 = 𝑷𝒂𝒕𝒎 + 𝑷𝒈𝒂𝒖𝒈𝒆


METACENTRE• It is the point of intersection of normal axis with

the new line of action of force of buoyancy, when a small displacement is given.

• It is a point about which a body starts to oscillate when it is tilted by small angle.

𝑮. 𝑴. = 𝑩. 𝑴 − 𝑩. 𝑮

𝑮𝑴 =𝑰

𝑽− 𝑩𝑮

• Where 𝑰𝒙𝒙 =𝒃𝒅𝟑

𝟏𝟐; 𝑰𝒚𝒚 =

𝒅𝒃𝟑

𝟏𝟐;

𝑰𝒎𝒊𝒏

𝑽= metacentric radius

• For stability G.M. should be positive, i.e., B.M>B.G


• From designing point of view, • the moment of inertia 𝑰 is taken about the top view at the

surface of liquid.• Oscillation about longitudinal axis is called rolling.• Oscillation about transverse axis is called pitching.

• TIME PERIOD FOR OSCILLATION:

𝑻 = 𝟐𝝅𝑲𝟐

𝑮𝑴. 𝒈

• Where K is radius of gyration• GM is metacentric height• T is time of one complete oscillation


• STABILITY CONDITION FOR FULLY SUBMERGED BODY:

• A completely submerged body will be in Stable Equilibrium when centre of buoyancy is above centre of gravity.

• If centre of buoyancy is below centre of gravity, then completely submerged body will be in Unstable Equilibrium.

• If centre of buoyancy and centre of gravity coincides, then fully submerged body will be in Neutral Equilibrium.

Stable Equilibrium

Unstable Equilibrium

𝑩 𝑮

Neutral EquilibriumCivil Engineering by Sandeep Jyani 61

• STABILITY CONDITION FOR PARTIALLY SUBMERGED BODY:• A partially submerged body will be in

Stable Equilibrium when metacentre is above centre of gravity.

• If metacentre is below centre of gravity, then partially submerged body will be in Unstable Equilibrium.

• If metacentre and centre of gravity coincides, then partially submerged body will be in neutral equilibrium.

• The distance between C.G. and metacenter is called metacentric height.

M

G

B

M

G

B

Neutralequilibrium

G,B

Unstableequilibrium

Stableequilibrium

Stable Equilibrium

Unstable Equilibrium.


CENTRE OF PRESSURE

BODY CENTRE OF PRESSURE (h)

d

d

d

d

d

Centre of pressure h = ഥ𝒙 +𝑰𝑮

𝑨ഥ𝒙

⇒ 𝐡 =𝒅

𝟐+

𝒃𝒅𝟑

𝟏𝟐

(𝒅×𝒃)(𝒅

𝟐)

b

⇒ 𝐡 =𝒅

𝟐+

𝒅

𝟔

⇒ 𝐡 =𝟐𝒅

𝟑Civil Engineering by Sandeep Jyani 63

CENTRE OF PRESSURE

BODY CENTRE OF PRESSURE (h)

𝟐

𝟑𝒅

𝟏

𝟐𝒅

𝟑

𝟒𝒅

𝟓

𝟖𝒅

d

d

d

d


SUMMARY


𝒅𝒙

𝒖=

𝒅𝒚

𝒗Equation of STREAM LINE

Velocity potential Function

𝒖 = −𝝏∅

𝝏𝒙

𝒗 = −𝝏∅

𝝏𝒚

𝒘 = −𝝏∅

𝝏𝒛

𝝎𝒛 =𝟏

𝟐

𝝏𝒗

𝝏𝒙−

𝝏𝒖

𝝏𝒚𝒓𝒐𝒕𝒂𝒕𝒊𝒐𝒏 𝒊𝒏 𝒙 − 𝒚 plane

𝑽𝒐𝒓𝒕𝒊𝒄𝒊𝒕𝒚 = 𝟐𝝎

Stream Function

𝒗 =𝝏𝜳

𝝏𝒙𝒖 = −

𝝏𝜳

𝝏𝒚

⇒𝝏𝜳

𝝏𝒚=

𝝏∅

𝝏𝒙 Cauchy Reiman Equation𝝏𝜳

𝝏𝒙= −

𝝏∅

𝝏𝒚

𝝏𝒖

𝝏𝒙+

𝝏𝒗

𝝏𝒚+

𝝏𝒘

𝝏𝒛= 𝟎 CONTINUITY EQUATION

𝐁𝐞𝐫𝐧𝐨𝐮𝐥𝐥𝐢′𝐬 𝐞𝐪𝐮𝐚𝐭𝐢𝐨𝐧

⇒𝑷

𝝆𝒈+ 𝒛 +

𝒗𝟐

𝟐𝒈= 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕

•𝑷

𝝆𝒈= Pressure energy per unit weight of fluid or 𝑷𝒓𝒆𝒔𝒔𝒖𝒓𝒆 𝒉𝒆𝒂𝒅

• 𝒛 = potential energy per unit weight or 𝑷𝒐𝒕𝒆𝒏𝒕𝒊𝒂𝒍 𝑯𝒆𝒂𝒅

•𝒗𝟐

𝟐𝒈= 𝒌𝒊𝒏𝒆𝒕𝒊𝒄 𝒆𝒏𝒆𝒓𝒈𝒚 𝒑𝒆𝒓 𝒖𝒏𝒊𝒕 𝒘𝒆𝒊𝒈𝒉𝒕 𝒐𝒓 𝑲𝒊𝒏𝒆𝒕𝒊𝒄 𝑯𝒆𝒂𝒅

• 𝐁𝐞𝐫𝐧𝐨𝐮𝐥𝐥𝐢′𝐬 𝐓𝐡𝐞𝐨𝐫𝐞𝐦:• It states that in a STEADY, IDEAL flow of an incompressible fluid, the total energy at

any point of the fluid is constant (pressure, kinetic, and potential energy)


𝑷𝒓𝒆𝒔𝒔𝒖𝒓𝒆 𝒉𝒆𝒂𝒅 𝑲𝒊𝒏𝒆𝒕𝒊𝒄 𝑯𝒆𝒂𝒅

𝑷𝒐𝒕𝒆𝒏𝒕𝒊𝒂𝒍 𝑯𝒆𝒂𝒅

𝐁𝐞𝐫𝐧𝐨𝐮𝐥𝐥𝐢′𝐬 𝐞𝐪𝐮𝐚𝐭𝐢𝐨𝐧

⇒𝑷

𝝆𝒈+ 𝒛 +

𝒗𝟐

𝟐𝒈= 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕

• 𝐁𝐞𝐫𝐧𝐨𝐮𝐥𝐥𝐢′𝐬 𝐓𝐡𝐞𝐨𝐫𝐞𝐦:• It states that in a STEADY, IDEAL flow of an incompressible fluid, the total

energy at any point of the fluid is constant (pressure, kinetic, and potential energy)

• Assumptions of Bernoulli’s Equation1. Fluid is ideal (viscosity is zero)2. The flow is incompressible3. Flow is steady4. The flow is irrotational


𝑷𝒓𝒆𝒔𝒔𝒖𝒓𝒆 𝒉𝒆𝒂𝒅 𝑲𝒊𝒏𝒆𝒕𝒊𝒄 𝑯𝒆𝒂𝒅

𝑷𝒐𝒕𝒆𝒏𝒕𝒊𝒂𝒍 𝑯𝒆𝒂𝒅

PRACTICAL APPLICATIONS OF BERNOULLI’s EQUATION

1. Venturimeter


⇒ 𝑸 = 𝒂𝟐𝒗𝟐 = 𝟐𝒈𝒉𝒂𝟏

𝟐

𝒂𝟏𝟐 − 𝒂𝟐

𝟐

⇒ 𝑸 = 𝒂𝟐𝒗𝟐 = 𝒂𝟏𝒂𝟐

𝟐𝒈𝒉

𝒂𝟏𝟐 − 𝒂𝟐

𝟐

⇒ 𝑸𝒕𝒉 =𝒂𝟏𝒂𝟐

𝒂𝟏𝟐 − 𝒂𝟐

𝟐

𝟐𝒈𝒉 𝑻𝒉𝒊𝒔 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝒈𝒊𝒗𝒆𝒔 𝒗𝒂𝒍𝒖𝒆 𝒐𝒇 𝒅𝒊𝒔𝒄𝒉𝒂𝒓𝒈𝒆 𝒖𝒏𝒅𝒆𝒓𝒊𝒅𝒆𝒂𝒍 𝒄𝒐𝒏𝒅𝒊𝒕𝒊𝒐𝒏𝒔, 𝒔𝒐 𝒊𝒕 𝒊𝒔 𝒄𝒂𝒍𝒍𝒆𝒅 𝒂𝒔 𝑻𝒉𝒆𝒐𝒓𝒆𝒕𝒊𝒄𝒂𝒍 𝑫𝒊𝒔𝒄𝒉𝒂𝒓𝒈𝒆.𝑨𝒄𝒕𝒖𝒂𝒍 𝒅𝒊𝒔𝒄𝒉𝒂𝒓𝒈𝒆 𝒘𝒊𝒍𝒍 𝒃𝒆 𝒍𝒆𝒔𝒔 𝒕𝒉𝒂𝒏 𝑻𝒉𝒆𝒐𝒓𝒆𝒕𝒊𝒄𝒂𝒍 𝒅𝒊𝒔𝒄𝒉𝒂𝒓𝒈𝒆

⇒ 𝑸𝒂𝒄𝒕𝒖𝒂𝒍 = 𝑪𝒅 ×𝒂𝟏𝒂𝟐

𝒂𝟏𝟐 − 𝒂𝟐

𝟐

𝟐𝒈𝒉

𝑨𝒄𝒕𝒖𝒂𝒍 𝒅𝒊𝒔𝒄𝒉𝒂𝒓𝒈𝒆

𝑻𝒉𝒆𝒐𝒓𝒆𝒕𝒊𝒄𝒂𝒍 𝒅𝒊𝒔𝒄𝒉𝒂𝒓𝒈𝒆𝑪𝒅 = 𝐜𝐨𝐞𝐟𝐟𝐢𝐜𝐢𝐞𝐧𝐭 𝐨𝐟 𝐝𝐢𝐬𝐜𝐡𝐚𝐫𝐠𝐞 𝐚𝐧𝐝

𝐢𝐭𝐬 𝐯𝐚𝐥𝐮𝐞 𝐢𝐬 𝐥𝐞𝐬𝐬 𝐭𝐡𝐚𝐧 𝟏 (𝟎. 𝟗𝟒 − 𝟎. 𝟗𝟖)

𝑪𝒅 =𝑨𝒄𝒕𝒖𝒂𝒍 𝑫𝒊𝒔𝒄𝒉𝒂𝒓𝒈𝒆

𝑻𝒉𝒆𝒐𝒓𝒆𝒕𝒊𝒄𝒂𝒍 𝑫𝒊𝒔𝒄𝒉𝒂𝒓𝒈𝒆



2. Orificemeter• Coefficient of Contraction

𝐶𝑐 =𝑎2

𝑎0• Coefficient of discharge for

Orificemeter is much smallerthan that for a Venturimeter as the area reduction is sudden in Orificemeter, the losses are more, hence the 𝐶𝑑 of Orificemeter is less than 𝐶𝑑 of Venturimeter

• 𝐶𝑑 = 0.68 − 0.74

• 𝑸𝒂𝒄𝒕𝒖𝒂𝒍 =𝑪𝒅𝒂𝟏𝒂𝟎

𝒂𝟏𝟐−𝒂𝟎

𝟐𝟐𝒈𝒉

𝒂𝟎

𝒂𝟐

𝒂𝟐 = 𝒗𝒆𝒏𝒂 𝒄𝒐𝒏𝒕𝒓𝒂𝒄𝒕𝒂 𝒂𝒓𝒆𝒂

𝒂𝟎 = 𝒐𝒓𝒊𝒇𝒊𝒄𝒆 𝒂𝒓𝒆𝒂

𝒂𝟏

𝟏

𝟐

𝟏

𝟐

𝒂𝟎

𝒂𝟏 = 𝒂𝒓𝒆𝒂 𝒐𝒇 𝒑𝒊𝒑𝒆



⇒ 𝒗𝟏 = 𝟐𝒈𝒙𝒔𝒎𝒂𝒏𝒐𝒎𝒆𝒕𝒓𝒊𝒄 𝒇𝒍𝒖𝒊𝒅

𝒔𝒇𝒍𝒖𝒊𝒅 𝒊𝒏 𝒑𝒊𝒑𝒆− 𝟏 𝟏

𝟐

𝒎𝒂𝒏𝒐𝒎𝒆𝒕𝒓𝒊𝒄 𝒇𝒍𝒖𝒊𝒅

𝒙

WEIRS AND NOTCHES• Rectangular notch or weir:

𝑸𝒕𝒉𝒆𝒐𝒓𝒆𝒕𝒊𝒄𝒂𝒍 =𝟐

𝟑𝟐𝒈 × 𝑳𝑯𝟑/𝟐

𝑸𝒂𝒄𝒕𝒖𝒂𝒍 =𝟐

𝟑𝑪𝒅 𝟐𝒈 × 𝑳𝑯𝟑/𝟐

Where -𝑸 is the total discharge 𝑪𝒅 is coefficient of discharge𝑳 is length of notch or weir𝑯 is head of water over the crest


WEIRS AND NOTCHES

• Triangular notch or weir:

𝑸𝒕𝒉𝒆𝒐𝒓𝒆𝒕𝒊𝒄𝒂𝒍 =𝟖

𝟏𝟓𝟐𝒈 tan

𝜽

2𝑯𝟓/𝟐

𝑸𝒂𝒄𝒕𝒖𝒂𝒍 =𝟖

𝟏𝟓𝑪𝒅 𝟐𝒈 tan

𝜽

2𝑯𝟓/𝟐

Where -𝑸 is the total discharge 𝑪𝒅 is coefficient of discharge𝜽 is the angle of notch𝑯 is head of water above the V-notch

𝜽 𝑯


CLASSIFICATION3. Laminar flow and turbulent flow –

• If Reynolds no. is less than 500, in an open channel then flow is laminar.

• If Reynolds no. is more than 2000, in an open channel then flow is turbulentin open channel flow.

• If 𝑹𝒆 lies between 500 and 2000, it is in transition state

𝑹𝒆 =𝝆𝑽𝑹

𝝁

𝑹 =𝑪𝒓𝒐𝒔𝒔 𝑺𝒆𝒄𝒕𝒊𝒐𝒏 𝒂𝒓𝒆𝒂 𝒐𝒇 𝒇𝒍𝒐𝒘 𝒏𝒐𝒓𝒎𝒂𝒍 𝒕𝒐 𝒕𝒉𝒆 𝒅𝒊𝒓𝒆𝒄𝒕𝒊𝒐𝒏 𝒐𝒇 𝒇𝒍𝒐𝒘

𝑾𝒆𝒕𝒕𝒆𝒅 𝒑𝒆𝒓𝒊𝒎𝒆𝒕𝒆𝒓

𝝆 = 𝒅𝒆𝒏𝒔𝒊𝒕𝒚

𝑹 = 𝒉𝒚𝒅𝒓𝒂𝒖𝒍𝒊𝒄 𝒓𝒂𝒅𝒊𝒖𝒔 𝒐𝒓 𝒉𝒚𝒅𝒓𝒂𝒖𝒍𝒊𝒄 𝒎𝒆𝒂𝒏 𝒅𝒆𝒑𝒕𝒉


CLASSIFICATION4. Sub critical, critical and super critical flow

𝑭𝒆 =𝑽

𝒈𝑫

𝑫 − 𝒉𝒚𝒅𝒓𝒂𝒖𝒍𝒊𝒄 𝒅𝒆𝒑𝒕𝒉 =𝒂𝒓𝒆𝒂

𝒕𝒐𝒑 𝒘𝒊𝒅𝒕𝒉

𝑭𝒆 < 𝟏; 𝒔𝒖𝒃 𝒄𝒓𝒊𝒕𝒊𝒄𝒂𝒍 𝒇𝒍𝒐𝒘

𝑭𝒆 = 𝟏; 𝒄𝒓𝒊𝒕𝒊𝒄𝒂𝒍 𝒇𝒍𝒐𝒘

𝑭𝒆 > 𝟏; 𝒔𝒖𝒑𝒆𝒓 𝒄𝒓𝒊𝒕𝒊𝒄𝒂𝒍 𝒇𝒍𝒐𝒘


HYDRAULIC JUMP

• HYDRAULIC JUMP is the rise of water which takes place due to transformation of unstable shooting flow (super critical flow) to stable shooting flow (sub critical flow)

• When a hydraulic jump takes place, loss of energy due to eddy formation and turbulence occurs.

𝒅𝒆𝒑𝒕𝒉 𝒐𝒇 𝒉𝒚𝒅𝒓𝒂𝒖𝒍𝒊𝒄 𝒋𝒖𝒎𝒑 𝒊𝒏 𝒕𝒆𝒓𝒎𝒔 𝒐𝒇 𝒇𝒓𝒐𝒖𝒅𝒆 𝒏𝒐. =𝒅𝟏

𝟐𝟏 + 𝟖 𝑭𝒆 𝟏

𝟐 − 𝟏

𝒆𝒏𝒆𝒓𝒈𝒚 𝒇𝒐𝒓 𝒍𝒐𝒔𝒔 𝒅𝒖𝒆 𝒕𝒐 𝒉𝒚𝒅𝒓𝒂𝒖𝒍𝒊𝒄 𝒋𝒖𝒎𝒑 =(𝒅𝟐−𝒅𝟏)𝟑

𝟒𝒅𝟏𝒅𝟐

𝒅𝟏 = depth of flow before hydraulic jump

𝒅𝟐 = depth of flow after hydraulic jump

𝒅𝒆𝒑𝒕𝒉 𝒐𝒇 𝒉𝒚𝒅𝒓𝒂𝒖𝒍𝒊𝒄 𝒋𝒖𝒎𝒑 = 𝒅𝟐 − 𝒅𝟏


HYDRAULIC JUMP• Froude’s number of incoming flow should be

greater than 1 for Hydraulic jump to occur

1. Undular Jump (1.0 < 𝐹1 ≤ 1.7)• Water surface is undulating with small ripples on

the surface

2. Weak Jump(1.7 < 𝐹1 ≤ 2.5)• A series of small rollers forms on the jump surface

but the downstream water surface remains smooth


HYDRAULIC JUMP3. Oscillating Jump (2.5 < 𝐹1 ≤ 4.5)

• Entering jet of water oscillates in a random manner between bed and surface.

• These oscillations are very common in canals and can travel considerable distances and damaging earthen banks

4. Steady Jump(4.5 < 𝐹1 ≤ 9)• The jump is well established, the roller and jump

action is fully developed to cause appreciable energy loss (downstream surface smooth)

• It is always preferred to have steady jump


HYDRAULIC JUMP5. Strong or Choppy Jump (9 < 𝐹1)

• During this jump, water surface is very rough and choppy, which continues downstream for a long distance

• Energy dissipation is very efficient


Turbine Turbine Type Head Specific SpeedRpm

Flow type Discharge Speed Ratio

Pelton Single Jet

Impulse High Head300-2000m

10-35 Tangential Flow

Low 0.43 to 0.48

Pelton Multiple Jet

35-60

Francis Reaction Medium Head60-300mm

60-300 Mixed Flow (radially in and axially out)

Medium 0.6 to 0.9

Kaplan (adjustable blades, mostly used in India)

Reaction Low HeadUpto 60m

300-1000 Axial Flow high 1.4 to 2


The value for lateral coefficient of friction as per IRC is

a) 0.007

b) 0.01

c) 0.15

d) 0.30

The value for lateral coefficient of friction as per IRC is

a) 0.007

b) 0.01

c) 0.15

d) 0.30

Quality of Irrigation water3. Proportion of Sodium Ion concentration

• When percentage of Na ions in the total exchangeable cations exceed to 10%, aggregation of soil grains breakdown and hence soil becomes less permeable and of poor tilth

• Proportion of Na ion concentration is generally measured by SAR(Sodium absorption ratio):

𝑆𝐴𝑅 =𝑁𝑎

+

𝑀𝑔+ 2 + 𝐶𝑎

+ 2

2𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑖𝑠 𝑒𝑥𝑝𝑟𝑒𝑠𝑠𝑒𝑑 𝑖𝑛 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑐𝑒 𝑝𝑒𝑟 𝑚𝑖𝑙𝑙𝑖𝑜𝑛

𝐸𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑐𝑦 𝑝𝑒𝑟 𝑚𝑖𝑙𝑙𝑖𝑜𝑛 =𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛 𝑖𝑛 𝑝𝑝𝑚

𝐸𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑤𝑒𝑖𝑔ℎ𝑡82

SAR values

Sr No. Classification SAR Uses

1 Low sodium water 0-10 For all crops on all soils

2 Medium sodium water

10-18 Can be used in coarse grain soil

3 High sodium water 18-26 Can be used if leaching is done and

requires good draingage system

4 Very high sodium water

>26 Not suitable


Relationship between Duty (D) and Delta ∆• Let ‘D’ hectare area is irrigated by supplying 1 m3/sec of water for the base period of

‘B’ days such that total water stored in the root zone is ∆ m

• Therefore, volume of water supplied = B Days x 1 m3/sec

= B x 24x60x60 sec x 1 m3/sec

= 68400 B m3 … (1)

• Amount of water stored = D (ha) x ∆ (m)

= D x 104 m2 x ∆ m

=D x ∆ 104 m3 …..(2)

Equating (1) and (2),

D x ∆ 104 m3 = 68400 B m3

=> D x ∆ = 8.64 B

84

D =ha/cumec ∆ = m B= days

Important Definition

1. Crop ratio/Kharif-Rabi ratio

• Crop = 𝑨𝒓𝒆𝒂 𝒊𝒓𝒓𝒊𝒈𝒂𝒕𝒆𝒅 𝒊𝒏 𝑲𝒉𝒂𝒓𝒆𝒆𝒇 𝒔𝒆𝒂𝒔𝒐𝒏

𝑨𝒓𝒆𝒂 𝒊𝒓𝒓𝒊𝒈𝒂𝒕𝒆𝒅 𝒊𝒏 𝑹𝒂𝒃𝒊 𝒔𝒆𝒂𝒔𝒐𝒏= 2 (approx)

2. Paleo irrigation• It is defined as watering done prior to the sowing

of crops.

• It is done to make area suitable for sowing as it becomes very dry and to add sufficient moisture to the soil which would be required for initial growth


Important Definition 3. Kor watering, Kor depth, Kor Period:

• Kor Watering is the first watering after sowing of crops when crop is few ‘cm’ high

• This irrigation depth is the maximum of all the waterings and this depth is called as KOR depth

• The portion of Base period in which KOR watering is needed is called as ‘Kor period’

4. Capacity factor:

= 𝐌𝐞𝐚𝐧 𝐬𝐮𝐩𝐩𝐥𝐲 𝐝𝐢𝐬𝐜𝐡𝐚𝐫𝐠𝐞 𝐟𝐨𝐫 𝐚 𝐜𝐚𝐧𝐚𝐥 𝐟𝐨𝐫 𝐚 𝐜𝐞𝐫𝐭𝐚𝐢𝐧 𝐝𝐮𝐫𝐚𝐭𝐢𝐨𝐧

𝐌𝐚𝐱𝐢𝐦𝐮𝐦 𝐝𝐢𝐬𝐜𝐡𝐚𝐠𝐞 𝐜𝐚𝐫𝐫𝐲𝐢𝐧𝐠 𝐜𝐚𝐩𝐚𝐜𝐢𝐭𝐲 𝐨𝐟 𝐜𝐚𝐧𝐚𝐥

= 𝐐𝐫𝐞𝐪𝐮𝐢𝐫𝐞𝐝

𝐐𝐝𝐞𝐬𝐢𝐠𝐧𝐞𝐝


Important Definition 5. Time Factor

Time Factor = No.of days canal has actually run

No.of days of irrigation period

6. Crop Calendar

Crop calendar is a tool that provides information about planting, sowing, and harvesting period of locally adopted crops in an area.

87

Important Definition 7. Gross Command Area (GCA)

• It is the total area with in the limit of an irrigation project.

• It includes cultivable as well as non-cultivable area like road, Building, Pond etc.


Important Definition 8. Cultivable Command Area (CCA);

• It is that portion of GCA which is cultivable and it denotes the area on which actually cultivation is to be done.

• In absence of data we may assume, CCA = 80% of GCA.


Important Definition 9. Intensity Of Irrigation /Annual Intensity Of Irrigation

• Intensity of Irrigation = sum of seasonal intensity of irrigation in a year.

• Seasonal intensity of Irrigation is that percentage of CCA on which actually the irrigation is done in a particular season.

90

ANNUAL INTENSITY = 90%+70% = 160%


Important Definition 10. Irrigation efficiency;• In general,

irrigation efficiency = Water available for use

water applieda) Water conveyance efficiency (nc)

NC = 𝑤𝑓

𝑤𝑟

× 100

where, wr is water released from river reservoir for the field.wf is water given to the field

This efficiency account for water which possess seepage loss & evaporating loss during conveyance from river (or) reservoir to the field.


Important Definition 10. Irrigation efficiency;

b) Water application efficiency (na)

Na = ws

wf×100

where, ws is the water stored in the field i.e, in the root zonewf is the water given to the field

This efficiency accounts for water losses such as surface run off (Rf) & Deep Percolation (Df) which occurs during application of water in the field.

Wf = WS + Rf + DF



c) Water use efficiency (nu)

Nu = Wu

wf×100

where, wf is the water given to the fieldwu water use consumptivelywu = leaching requirement + presowingrequirement + water stored



d) Water storage efficiency (ns)

Ns = ws

wn× 100

where,

ws is the water stored in the field, i.e, in the root zone

wn is the amount of water to be stored in the field such that moisture content is raised to field capacity (FC)

wn is the total water required up to Fc i.e., available moisture content before irrigation.


CROSS DRAINAGE WORKS

• A cross drainage work is a structure constructed for carrying a canal across a natural drain (𝒐𝒓) river intercepting the canal so as to dispose the drainage water without interrupting the continuous canal supplies.

• These are unavoidable in any type of canal system.

• In order to minimize the no. of cross drainage works, the alignment of canals should be generally along the watershed so that we have less no. of natural drains.


TYPES OF CROSS DRAINAGE WORKS

1. Cross drainage works carrying the canal over the natural drain

i. AQUEDUCT

• An Aqueduct is a hydraulic structure which carries a canal (through a trough (or) a duct) across and above the drainage similar to a bridge in which instead of road (𝒐𝒓) a railway, a canal is carried over a natural drain.

• In the case of an aqueduct, HFL (Highest flood level) of the drainage should main lower than the level of the underside of the canal trough.

• The canal water is taken across the drain in a trough supported on piers



1. Cross drainage works carrying the canal over the natural drain

ii. SYPHON AQUEDUCT

• A syphon aqueduct is a cross drainage structure similar to an aqueduct except that the stream bed is depressed locally where it passes under the trough of the canal and the barrels discharges the stream flow under pressure.

• A syphon aqueduct is constructed where the water surface level of the train at high flood is higher than the canal bed.



2. Cross drainage works carrying the natural drain over canal

i. SUPER PASSAGE

• A super passage is also similar to a bridge in which the natural drain is carried over the canal.

• A super passage is reverse of an aqueduct.

ii. SYPHON

• A syphon is similar to a syphon aqueduct with the difference that in the case of a syphon, the canal water is carried through the barrels-under the drain.

• The barrels in this case also act as inverted syphons through which the canal water flows under pressure.



3. Cross drainage works admitting the drain water into the canal

• In this type of cross drainage works, the canal water and the drain water are allowed to intermingle with each other.

• This may be achieved by the following two types of the Cross-drainage works:

i. Level Crossing

ii. Inlet and Outlet


TYPES OF CROSS DRAINAGE WORKS3. Cross drainage works admitting the drain water into the canal

i. Level Crossing –• A level crossing is a cross drainage work in which the

drainage and the canal meet each other at approximately the same level.

• It consists of a regular with quick falling shutters across the drain at its downstream junction with the canal.

• Such an arrangement is adopted when both the canal and the drainage carry considerable discharge, the latter during the high flood season when syphoning either the canal (𝒐𝒓)the stream proves to be extremely costly or else the head loss through the syphon barrels is very high. Arrangement is practically similar to that provided on a canal head work.

• In this arrangement, the perennial discharge is used advantageously in order to increase the canal supplies.


TYPES OF CROSS DRAINAGE WORKS3. Cross drainage works admitting the drain water into the canal

ii. Inlet And Outlet -• An inlet is an open cut (𝒐𝒓) a pipe which is provided in a

canal bank suitably protected by pitching to pass the drain water into the canal.

• This arrangement is provided only where the silt load of the drainage is suitable.

• An inlet may be provided for a small drain coming across a canal it the bed level of drain is slightly higher (𝒐𝒓) lower than the canal F.S.L

• It is not necessary that the no of inlets & outlets should be the same.

• Inlet is a non-regulating structure

• Outlet is another open cut in the canal bank with bed & sides of the cut properly pitched.


RELATION B/W DEGREE OF CURVE AND RADIUS OF CURVE

• Generally, degree of curve is defined for per chain length.

• If we are using 30m chain, then degree and radius can be related by following:

𝑫°

𝟑𝟎=

𝟑𝟔𝟎°

𝟐𝝅𝑹

⇒ 𝑹 =𝟑𝟔𝟎° × 𝟑𝟎

𝟐𝝅𝑫°

⇒ 𝑹 =𝟏𝟕𝟏𝟖

𝑫°≈

𝟏𝟕𝟐𝟎

𝑫°

𝑫

30m

𝑹


RELATION B/W DEGREE OF CURVE AND RADIUS OF CURVE

If we are using 20m chain, then degree and radius can be related by following:

𝑫°

𝟐𝟎=

𝟑𝟔𝟎°

𝟐𝝅𝑹

𝑹 =𝟑𝟔𝟎° × 𝟐𝟎

𝟐𝝅𝑫°

𝑹 =𝟏𝟏𝟓𝟎

𝑫°𝑫

20m

𝑹


The unit weight of RCC in kg/m3 is

a) 1200

b) 1800

c) 2400

d) 3000


SUPERELEVATION/CANT• It is the elevation provided to the outer rail

w.r.t the inner rail at the circular curve to counteract the effect of centrifugal force.

𝑚𝑔 sin 𝜃 =𝑚𝑣2

𝑅cos 𝜃

tan 𝜃 =𝑣2

𝑔𝑅

𝑒

𝐺=

𝑣2

𝑔𝑅

𝑒 =𝐺𝑣2

𝑔𝑅Civil Engineering by Sandeep Jyani

𝑉 𝑖𝑠 𝑠𝑝𝑒𝑒𝑑 𝑖𝑛 𝑘𝑚𝑝ℎ

𝑹 𝒊𝒔 𝒓𝒂𝒅𝒊𝒖𝒔 𝒊𝒏 𝒎

𝑒 𝑖𝑠 𝑐𝑎𝑛𝑡 𝑖𝑛 𝑚

𝑮 𝒊𝒔 𝒈𝒂𝒖𝒈𝒆 𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆 𝒊𝒏 𝒎

𝑒 =𝐺𝑉2

127𝑅

𝒗 𝒊𝒔 𝒔𝒑𝒆𝒆𝒅 𝒊𝒏 𝒎/𝒔𝒆𝒄

𝜽

𝒎𝒗𝟐

𝒓

𝒎𝒈

𝒆𝒎𝒗𝟐

𝒓𝒔𝒊𝒏𝜽

𝒎𝒗𝟐

𝒓𝒄𝒐𝒔𝜽

𝒎𝒈 𝒄𝒐𝒔𝜽

𝒎𝒈 𝒔𝒊𝒏𝜽

The unit weight of RCC in kg/m3 is

a) 1200

b) 1800

c) 2400

d) 3000


Origin of Soil and soil water relationships

• Soil is composed of particles found from the disintegration of rocks.

• Formation of Soil takes place by two methods: 1. Physical Weathering2. Chemical Weathering


Soil Formation

Physical Weathering

Erosion of rock by Wind , water glaciers, alternate freezing and

defreezing

Soil formed retain minerals and composition

Exp- sand and gravel

Chemical Weathering

Chemical actions of acids and alkalies in

water, air, glaciers, etc

Mineral composition is changed

Exp- clay


Soil Deposits

• Residual Soils: Located at location of formation

• Transported Soil: Transported from parent location to a new location

• Alluvial deposits: deposited by river

• Lacustrine deposit: deposited by still water of lakes

• Marine deposit: deposited by sea Water

• Aeolian deposit: deposited by wind , example Loess

• Glacial deposit: deposited by glaciers , example drift, till

• Colluvial deposit: transported by Gravity expect Talus


Some important Soils• Bentonite clay:

• Has high percentage of montmorillonite

• Highly plastic, high swelling and shrinkage

• Formed due to volcanic ash, used as drilling mud

• Black cotton soil: • Contains high percentage of montmorillonite

• Has high swelling and shrinkage potential

• Has very low bearing capacity

• Formed from chemical weathering of basalt

• Loam: mixture of sand silt and clay, known as Garden soil

• Indurated clay: hardening of clay due to heat and pressure

• Organic clays: soil gets mixed with decomposed vegetation and dead and decayed matter

• Muck: inorganic + organic matter

• Peat: fully decomposed organic matter

• Humus: Top soil, it contains partly decomposed organic material


Soil Water Relationships

Phase System

Three phase Two Phase


Soil Water Relationships


Soil Water Relationships• Void ratio (e) is the ratio of the volume of voids

(Vv) to the volume of soil solids (Vs)

• Porosity (n) is the ratio of the volume of voids to the total

volume of soil (V )

• Degree of saturation (S) The volume of water (Vw) in a soil can vary between zero (i.e. a dry soil) and the volume of voids. This can be expressed as the degree of saturation (S) in percentage.


Soil Water Relationships• Air content (ac) is the ratio of the volume of air (Va) to the

volume of voids.

• Percentage air voids (na) is the ratio of the volume of air to the total volume.

• Water content (w): The ratio of the mass of water present to the mass of solid particles is called the water content (w), or sometimes the moisture content. Its value is 0% for dry soil and its magnitude can exceed 100%


Soil Water Relationships• Bulk Unit Weight (ɣt): It is total weight by total volume of soil

• Dry Unit Weight of soil (ɣd): It is weight of solids divided by total volume of soil

• Saturated Unit Weight (ɣsat) of soil:

• Submerged unit Weight of soil (ɣsub) :


Soil Water Relationships• Unit Weight of Solids(ɣs): It is weight of solids per unit

volume of solids

• Unit weight of water(ɣw): It is weight of water per unit volume of water

• True Specific Gravity (Gs):

• Mass Specific Gravity (Gm):


Relation between different terms

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.


Methods of determination of Water content (w)

1. Oven Drying method

2. Pycnometer Method

3. Sand Bath method

4. Calcium Carbide Method

5. Torsion Balance Method

6. Radiation Method

7. Alcohol Method



1. Oven Drying method

• 105-110 ℃ (inorganic)• 60℃ (organic)• 24 hours



2. Pycnometer Method



3. Sand Bath method• Field Method

• Sand bath container maintains temp of 105-110 ℃

4. Calcium Carbide Method•Water in soil reacts with

calcium carbide and forms acetylene•Pressure is co related with w/c



5. Torsion Balance Method• Lab method

• Used for soil which quickly absorbs moisture

• Equipment is costly

6. Radiation Method•Radioactive isotope

Cobalt 60 is used


Methods of determination of Unit Weight (ɣ):

1. Core cutter method

2. Water Displacement Method

3. Sand Replacement Method

4. Water balloon method

5. Radiation method



1. Core cutter method• Wt of empty cutter = W1

• Wt of Cutter+ soil= W2

• Wt of soil= W2-W1

• Vol of soil =1000cc



2. Water Displacement Method

• Wt of sample = W1• Wt of sample + wax = W2• We need to find volume of

sample • Wax coating is done so as to put

it in water and we can find the volume of Wax+ sample

• Then using ‘G’ of wax we find volume of Wax which is then subtracted from Total water displaced



3. Sand Replacement Method


Index Properties of Soil

1. Soil Aggregate Properties• Depends on soil mass as a

whole

2. Soil Grain Properties• Depends upon soil grain size,

shape, etc


Grain Size Distribution

1. Helps to find gradation and uniformity

2. Separates out soil into different fractions based on particle size


Gra

in S

ize

Dis

trib

uti

on

Coarse Grain Sieve Analysis

Fine GrainSedimentation

Analysis



1. Sieve Analysis




Sedimentation Analysis (Pippette Method)

• For particles less than size 75micron (clay and silt)

• Based on Stokes’ law

• Particle size is from 0.0002mm to 0.2mm

•


Consistency

• The consistency of a fine-grained soil refers to its firmness, and it varies with the water content of the soil.

• Used mainly for fine soils

• Water content influences consistency of soil


Consistency Limits

• The three limits are known as the shrinkage limit (WS), plastic limit (WP), and liquid limit (WL) as shown. The values of these limits can be obtained from various methods

• Liquid limit (WL) - change of consistency from plastic to liquid statePlastic limit (WP) - change of consistency from brittle/crumbly to plastic state laboratory tests.

• The difference between the liquid limit and the plastic limit is known as the plasticity index (IP)


Methods to find Liquid Limit

1. Casagrande Method• When Groove of 2mm is

filled with 25 No. of blows, then that water content is called Liquid limit

2. Cone penetrometer Method• 30 sec – 25 mm

• That water content is liquid limit


• Consistency Index:

• Liquidity Index:

• Activity:


• Toughness Index:

• Flow Index:

• Relative Density:

• Thixotrophy : Property by virtue of which loss of shear strength on remolding can be regained if soil is left undisturbed for sometime. The increase in shear strength is due to regain of chemical equilibrium and reorientation of water molecules.

• Sensitivity:


Consistency



Clay Minerals

Clay Mineral atomic structure

Silica Sheet Alumina Sheet


Clay Minerals

• Kaolinite (1:1)• H-bonding between layers• Less swelling and shrinkage• Antidiarrheal medicine

• Montmorrillonite (2:1)• Weak vanderwaal forces• High swelling and shrinkage• Highly plastic


• Illite(2:1)• Isomorphous substitution

• consists of the basic montmorillonite units but are bonded by secondary valence forces and potassium ions

Clay Minerals


• Total StressWhen a load is applied to soil, it is carried by the solid grains and the water in the pores. The total vertical stress acting at a point below the ground surface is due to the weight of everything that lies above, including soil, water, and surface loading. Total stress thus increases with depth and with unit weight.

• Pore Water PressureThe pressure of water in the pores of the soil is called pore water pressure (u). The magnitude of pore water pressure depends on:

• the depth below the water table.

• the conditions of seepage flow.

Effective Stress


• Effective Stress: The principle of effective stress was enunciated by Karl Terzaghi . This principle is valid only for saturated soils, and consists of two parts:

Effective Stress


Effective Stress


Quick Sand Condition

• Effective stress becomes zero

• For Cohesionless soil (c=0), and hence Shear strength becomes zero

• Contact force between grain particles becomes zero.

• Critical Hydraulic gradient:

• Factor of Safety:


• In soils, the interconnected pores provide passage for water. A large number of such flow paths act together, and the average rate of flow is termed the coefficient of permeability, or just permeability. It is a measure of the ease that the soil provides to the flow of water through its pores.

• Darcy stated that discharge in One dimensional flow Q is proportional to hydraulic gradient and area of cross section

• Q= K i A• v = q/A = k.iwhere k = permeability of the soili = Dh/LDh = difference in total headsL = length of the soil mass

Permeability


Permeability


Factors affecting Permeability


Methods to determine Permeability

1. Constant Head Permeability Method

2. Variable/Falling Head Permeability Method


Permeability of Stratified Deposits


Seepage in 2-D

• This is the Laplace equation governing two-dimensional steady state flow

• Flow NetsGraphical form of solutions to Laplace equation for two-dimensional seepage can be presented as flow nets. Two orthogonal sets of curves form a flow net:• Equipotential lines connecting points of equal total head h• Flow lines indicating the direction of seepage down a hydraulic gradient


• The space between two adjacent flow lines is known as a flow channel, and the figure formed on the flownet between any two adjacent flow lines and two adjacent equipotential lines is referred to as a field.

• Total seepage discharge of Flownet:

Seepage in 2-D


Uses of Flownet

• Estimation of seepage losses from reservoirs

• Determination of uplift pressures below dams

• Checking the possibility of piping beneath dams


Compaction

• Compaction is the application of mechanical energy to a soil so as to rearrange its particles and reduce the void ratio

• The objectives of compaction are:• To increase soil shear strength and therefore its bearing

capacity.

• To reduce subsequent settlement under working loads.

• To reduce soil permeability making it more difficult for water to flow through.


Tests of Compaction

• Standard Proctor test:• Wt of hammer = 2.5 kg

• Height of fall= 305mm

• Volume of mould= 944cc

• 3layers->25 no. Of blows

• Modified Proctor test• Wt of hammer = 4.5 kg

• Height of fall= 457.2mm

• Volume of mould= 944cc

• 5layers->25 no. Of blows


Tests of Compaction

• Indian Standard Light Compaction TestSoil is compacted into a 1000 cm3 mould in 3 equal layers, each layer receiving 25 blows of a 2.6 kg rammer dropped from a height of 310 mm above the soil. The compaction is repeated at various moisture contents.

Indian Standard Heavy Compaction TestIt was found that the Light Compaction Test (Standard Test) could not reproduce the densities measured in the field under heavier loading conditions, and this led to the development of the Heavy Compaction Test (Modified Test). The equipment and procedure are essentially the same as that used for the Standard Test except that the soil is compacted in 5 layers, each layer also receiving 25 blows. The same mould is also used. To provide the increased compactive effort, a heavier rammer of 4.9 kg and a greater drop height of 450 mm are used.


Dry Density - Water Content Relationship


Factors affecting Compaction

• The various factors which affect the compacted density are as follows:

(i) Moisture content

(ii) Compactive effort

(iii) Type of soil

(iv) Method of compaction

(v) Addition of admixture.



Field Compaction Equipment






Compression and Consolidation of Soils

• Components of Total SettlementThe total settlement of a loaded soil has three components:

• Elastic settlement/Immediate settlement

• Primary consolidation

• Secondary compression

Compressibility Characteristics


Some Parameters related to Compression

1. Compression Index:

2. Coefficient of compressibility, av.

3. Coefficient of volume compressibility, mv



4. Compression Modulus:

5. Coefficient of Consolidation:

6. Computation of settlement:


7. Overconsolidation ratio (OCR): is defined as the ratio of the preconsolidation stress to the current effective stress.



Time rate of Consolidation

• Terzaghi's one-dimensional consolidation equation

• During the consolidation process, the following are assumed to be constant:1. The total additional stress on the compressible soil layer is assumed to remain constant.2. The coefficient of volume compressibility (mV) of the soil is assumed to be constant.3. The coefficient of permeability (k) for vertical flow is assumed to be constant.


• In solution of the consolidation equation, non-dimensional parameters are provided :1. Drainage path ratio (Z):

H = drainage path which is the longest path taken by the pore water to reach a permeable sub-surface layer above or below.

z= depth at any point from top of clay layer

For single drainage H= D, for double drainage H=D/2



• In solution of the consolidation equation, non-dimensional parameters are provided :2. Time factor



Shear Strength of Soil

• Soils consist of individual particles that can slide and roll relative to one another. Shear strength of a soil is equal to the maximum value of shear stress that can be mobilized within a soil mass without failure taking place.

• The Mohr-Coulomb failure criterion can be written as the equation for the line that represents the failure envelope. The general equation is


Shear Strength Tests1. Direct Shear TestThe test is carried out on a soil sample confined in a metal box of square cross-section which is split horizontally at mid-height

The soil is sheared along a predetermined plane by moving the top half of the box relative to the bottom half. The box is usually square in plan of size 60 mm x 60 mm.

Tests on sands and gravels can be performed quickly


Shear Strength Tests1. Direct Shear Test


Shear Strength Tests2. Triaxial Test

The triaxial compression test consists of two stages:

First stage: In this, a soil sample is set in the triaxial cell and confining pressure is then applied.

Second stage: In this, additional axial stress (also called deviator stress) is applied which induces shear stresses in the sample. The axial stress is continuously increased until the sample fails.


The triaxial test has many advantages over the direct shear test:• The soil samples are subjected to uniform stresses and strains.

• Different combinations of confining and axial stresses can be applied.

• Drained and undrained tests can be carried out.

• Pore water pressures can be measured in undrained tests.

• The complete stress-strain behaviour can be determined


Shear Strength Tests3. Vane Shear TestUsed in plastic cohesive clays where obtaining undisturbed sample is difficult


Vertical Stresses

• As per Boussinesq Equation:

• As per westergaards solution


Vertical Stresses

• Newmarks Influence Chart

• m=no. of concentric circles• n= no of radial lines• q= intensity of load• N= equivalent no. of areas


Constituents of Cement

• 62-65%Lime (CaO)

• 17-25%Silica (SiO2)

• 3-8%Alumina (Al2O3)

• 3-4 %Gypsum (CaSO4)

• 3-4%Iron Oxide (Fe2O3)

• 1-3%Magnesia (MgO)

• 1-3%Sulphur (S)

• 0.2-1Alkalies (K2O, N2O)


Bougues Compounds

1. TriCalcium Aluminate (C3A or 3CaO.Al2O3) • 4-14% by weight

• Formed within 24 hours of addition of water in cement

• Responsible for maximum amount of heat of hydration

2. TetraCalcium Alumino Ferrate (C4AF or 4CaO.Al2O3.Fe2O3)• 10-18% by weight

• It is also formed within 24 hours of addition of water to cement

• Amount of heat of hydration evolved during formation of this compound initially is comparatively more which goes on decreasing with time


Bougues Compounds3. TriCalcium Silicate(C3S or 3CaO.SiO2)

• 45-85% by weight• Formed within a week of addition of water in cement• Responsible for development of early strength of cement in initial stages

4. DiCalcium Silicate(C2S or 2CaO. SiO2)• 15-35% by weight• It is formed very slowly after addition of water in cement and may require a

year of so for its formation• It is responsible for progressive strength of cement in later stages


Note:✓ If early strength is required- increase C3S (emergency road work, prefabricated construction work,

etc.)

✓Strength in required to be increased in later stages- increase C2S

1 . Which of the following shows thecorrect decreasing order of rate ofhydration of Portland cementcompounds?a) 𝐶3𝐴 > 𝐶4𝐴𝐹 > 𝐶3𝑆 > 𝐶2𝑆b) 𝐶3𝐴 > 𝐶4𝐴𝐹 > 𝐶2𝑆 > 𝐶3𝑆c) 𝐶3𝐴 > 𝐶3𝑆 > 𝐶2𝑆 > 𝐶4𝐴𝐹d) 𝐶4𝐴𝐹 > 𝐶3𝐴 > 𝐶3𝑆 > 𝐶2𝑆

1 . Which of the following shows thecorrect decreasing order of rate ofhydration of Portland cementcompounds?a) 𝐶3𝐴 > 𝐶4𝐴𝐹 > 𝐶3𝑆 > 𝐶2𝑆b) 𝐶3𝐴 > 𝐶4𝐴𝐹 > 𝐶2𝑆 > 𝐶3𝑆c) 𝐶3𝐴 > 𝐶3𝑆 > 𝐶2𝑆 > 𝐶4𝐴𝐹d) 𝑪𝟒𝑨𝑭 > 𝑪𝟑𝑨 > 𝑪𝟑𝑺 > 𝑪𝟐𝑺

Important Note

• Rate of Hydration

• Heat Evolution


𝑪𝟒𝑨𝑭 > 𝑪𝟑𝑨 > 𝑪𝟑𝑺 > 𝑪𝟐𝑺

𝑪𝟑𝑨 > 𝑪𝟑𝑺 > 𝑪𝟒𝑨𝑭 > 𝑪𝟐𝑺

Testing of Cement

I. Field Test1. Colour test: Cement should possess uniform grey colour

2. Physical Property Test:• Should feel smooth when rubbed in between fingers

• Cement should sink in water and should not float over the surface

• Sample should be free from presence of any lumps which are formed due to absorption of moisture

Civil Engineering by Sandeep Jyani185

3. Strength Test• Prepare a block of cement to be tested of size

25x25x200 cubic mm

• Immerse in water for 7 days

• Now remove the mould and subject it to point load of 340N by placing it over supports 150mm apart

• Sample should show no sign of failure under the application of this load

Testing of CementII. Lab Test

1. Fineness Test: used to check proper grinding which has direct impact on rate of hydration, rate of gain of strength and evolution of heatA. Sieve Test:

• Take 100gm of cement and place it on IS Sieve no. 9 (90 micron)• Perform sieving for 15mins • Weight of residue should not exceed 10% of original weight for OPC

B. Air Permeability Test: represented in terms of specific surface area (cm2/gm)• Blaines air permeability apparatus is used• Based on relationship between flow of air through cement bed and surface area

of cement particles forming the bed• For OPC, SSA should not be less than 2250cm2/gm


Testing of Cement

2. Standard Consistency Test:• Standard Consistency should be known before we

know about Setting time, soundness

• Standard Consistency-permits penetration of vicatplunger of size 10mm dia and 50 mm height upto a depth of 33-35mm from top into the mould

• We find the water content at which the cement consistency is produced

• Take 500gm of cement and add 23% of water by weight of cement in first trial

• Lower the plunger gently upto surface of mould and release quickly

• The moisture content at which this cement paste of standard Consistency is produced is termed as “P”.


Testing of Cement

3. Setting time Test: carried out to know whether cement is deteriorated due to storage. Two times are associated with setting of cement:

A. Initial Setting Time

B. Final Setting time

A. Initial Setting time: time which is being measured from the instant water is added to cement upto the time cement starts loosing its plasticity.

• Take 500gm cement and gauge it with 0.85P

• Lower the square needle gently and release it quickly

• Note time required by needle to penetrate 33-35mm from top

• This time is called Initial Setting time

✓For OPC, initial setting time is 30 mins


Testing of Cement

B. Final Setting time: time which is being measured from the instant water is added to cement up to the time cement completely looses its plasticity and attains sufficient firmness.

• Take 500gm cement and gauge it with 0.85P

• Lower the annular collar with needle gently and release it quickly

• Final setting time is the time when annular collar fails to make an impression over the mould but needle does show

✓For OPC, final setting time is 10 hours


Testing of Cement4. Soundness Test:It is important that cement after setting does not show any appreciable change in volume as it seriously affects the durability of the structure

• Soundness of Cement is due to LIME, MAGNESIA and SULPHUR

A. Soundness due to Lime• Le Chatlier Apparatus is used• Gauge the cement with 0.78P and fill the paste in

mould• Cover top and bottom with glass plates and immerse

entire assembly in water having temperature 27-32°c for 24hours

• Remove mould and note the displacement of split with help of indicator arms

• Then again we immerse entire assembly, increase temperature in 25-30 mins upto boiling point, and maintain it for next three hours

• Note the displacement of split with the help of indicator arms

✓The difference in reading in both the parts of test should not exceed 10mm for OPC


Testing of Cement4. Soundness Test:

B. Soundness due to Lime and Magnesia

• This test is sensitive to both Lime and Magnesia

• Prepare a mould of lean cement of size 25mm(cube)

• Place in autoclave in which steam pressure is increased upto 21kg/cm2

and is maintained for next three hours

• The mould is removed, percentage expansion of the mould is noted in each direction which should not exceed 0.8% for OPC


• Compaction Equipment1. Tamping or Rodding

• Tamping is the process of compacting the concrete manually with rod in layers of 15 cm for RCC and 30 cm in mass concreting

2. Vibrators1. Needle Vibrator

• Also known as Immersion/Poker vibrator

• It consists of a power unit and long flexible tube at the end of which vibrating head is installed

• Effective for mass concreting

2. Form Vibrator

• It is used for heavily reinforced thin sections where needle vibrators can not reach


Concreting Equipment

Needle Vibrator

Form Vibrator

Compressive strength Test• Cement : annure sand = 1:3

• Wt of sand 550gm and water cement ratio 0.4

• Mortar is filled in size of 70.5mm mould and immersed in water for 7 days curing period

• Compressive strength of cement is tested in Universal Testing Machine (UTM)

✓At 28 days = 33N/mm2

✓At 7 days = 2/3(Strength at 28 days)

✓At 3 days = 50%(Strength at 28 days)


Testing of Cement5. Strength Test:

Lime Cycle


(𝑳𝒊𝒎𝒆𝒔𝒕𝒐𝒏𝒆)𝑪𝒂𝑪𝑶𝟑 𝑪𝒂𝑶(𝑳𝒊𝒎𝒆) + 𝑪𝑶𝟐

Calcination

𝑸𝒖𝒊𝒄𝒌 𝑳𝒊𝒎𝒆 𝑪𝒂𝑶 + 𝑯𝟐𝑶 𝑪𝒂 𝑶𝑯 𝟐 𝑺𝒍𝒂𝒌𝒆𝒅 𝒍𝒊𝒎𝒆/𝒉𝒚𝒅𝒓𝒂𝒕𝒆𝒅 𝒍𝒊𝒎𝒆

𝑺𝒍𝒂𝒌𝒆𝒅 𝒍𝒊𝒎𝒆 𝑪𝒂 𝑶𝑯 𝟐 + 𝑪𝑶𝟐 𝑪𝒂𝑪𝑶𝟑 (𝑳𝒊𝒎𝒆𝒔𝒕𝒐𝒏𝒆)

Stones

• The stones that are suitable for construction of structures such asretaining walls, abutments, dams, roads, etc. are termed as BuildingStones


• Quarrying of Stones: Various methods which are involved in theextraction of stones from rock beds are collectively termed as“Quarrying of Stones”

• Dressing of Stones: Process in which surface of stone is made fit to beused in any construction purpose

Examples of Igneous Rocks

1. Obsidian : Its a volcanic glass without gas bubble and usually black or dark brown in colour.

2. Pumice: it is a volcanic glass filled with gas bubbles.

3. Rhyolite: It is a high silica fine grained rock colour are grey , light brown, pale yellowish, pink and other earth colours.

4. Andesite: It is the midway of Rhyolite & basalt. It is in medium brown colour.

5. Basalt: It is rich in iron and black to brown colours.

6. Granite: it is made up of small crystals & in pink to reddish colour

7. Diorite: It is intermediate between granite and Gabbro, and is described as white granite.

8. Gabbro: It is in dark colour, coarse grained structure.

9. Porphyry: It is made up of mixed grained structure.

10. Pegmatite: It is coarsely crystalline structure.


Examples of Sedimentary Rocks

1. Conglomerate: It is made up of grounded particles cemented together.

2. Breccia: It is made up of angular particles cemented together.

3. Sandstone: It is sand grains cemented together into solid stones.

4. Silt stone: It is made up from silt particles cemented together.

5. Shale: composed of mud that is a mix of flakes of clay minerals and tiny fragments of other minerals, especially quartz and calcite.

6. Halite: known as rock salt, is a type of salt, the mineral form of sodium chloride.

7. Selenite: It is made up of calcium sulphate.

8. Limestone: It is made up of calcium carbonate or calcite.

9. Coal or Lignite: It is compared of organic matter in the form of plant segments.


Examples of Metamorphic Rocks

1. Quartzite: It is a coarse grained metamorphic rock derived from sandstone.

2. Marble: It comes from Limestone

3. Slate: It is fine grained metamorphic rock

4. Phyllite: It is also a fine grained metamorphic rock.

5. Schist: It is a coarse grained metamorphic rock.

6. Gneiss: It is medium to coarse grain metamorphic rock.

7. Laterite: It is rich in iron and formed by the leaching of silica.


Important Points

• Basalt(Igneous rock) → Granulated Basalt (metamorphic rock)

• Limestone(Sedimentary rock) →Marble (metamorphic rock)

• Sandstone (Sedimentary rock)→ Quartzite(meta rock)

• Mud stone(Sedimentary rock)→ Slate (Meta Rock)


Qualities of good Building stones

1. Crushing strength> 100 N/mm2

2. Appearance

3. Durability

4. Hardness • >17 for road work• 14-17 Medium hardness• <14 poor hardness

5. % wear and tear→ less than 3%

6. Resistant to fire

7. Specific gravity > 2.7

8. Water absorption – not more than 5 percent, rejected if more than 10%

9. Toughness index→• >19 high toughness• 13-19 Medium Toughness• <13 low toughness


1. Composition of Brick


•50-60%Silica

•20-30%Alumina

•4-5%Lime

•5-6%Oxide of iron

•1%Magnesia

IV. Quality/Testing of Bricks

1. The bricks should be at least table moulded, well burnt and free from cracks

2. The bricks should be of uniform shape & size✓Conventional size of bricks= 22.4 cm x 11.4 x 7.6 cm✓Standard size of brick= 19cmx 9cm x 9cm✓With mortar = 20cm x 10 cm x 10 cm✓Wt of 1m3 bricks = 1800 kg✓Wt of 1 brick= 3.5- 3.6 kg

3. Sound Test- The bricks should produce clear metallic ringing sound when struck with each other.

4. Structure test- the bricks should have uniform homogenous structure across all the section i.e. it should be free from any voids.


5. Absorption Test: The bricks when immersed in water for 24 hours should not absorb water by more than-• 20% in case of 1st class bricks• 22% in case of 2nd class brick• 25% in case of 3rd class bricks

6. Compressive Strength Test: The bricks should process minimum compressive strength of:

1st class- 10 N/mm2

2nd class- 7.5 N/mm2

3rd class- 3.5 N/mm2

7. Toughness test: The brick should not break into pieces when dropped on levelled ground from the height of 1m.

8. Hardness/ Abrasion: the brick should possess sufficient hardness and does not show any sign of impression when scratched with finger nails.

9. Alkali test: the Bricks when immersed in water for 24 hours should not show any sign of efflorescence and staining.


IV. Quality/Testing of Bricks

• Most common method

• Frustum having top diameter 10cm, bottom dimension around 20 cm, height of 30 cm

• Concrete to be tested in filled in the mould in four layers where each layer is compacted 25 no of times with the help of rod

• Mould is removed immediately by lifting it in upward direction which causes the concrete to subsidize and the subsidence of the concrete is referred as slump which may also defined as difference of the height of the mould and the top level of the subsidize concrete.


1. Slump Test:

• Higher the value of Slump, more is the workability

• This test is not suitable for the concrete which process either very high workability or very low workability.


2. Compaction Factor Test:• This test is used for the concrete

possessing very low workability for which slump test is not suitable.

• Principle of this test is based upon finding the degree of compaction achieved by the standard amount of work done by the concrete when allowed to fall from known height

• This degree of compaction is represented in terms of compaction factor that represents the density ratio : Density of concrete obtained during test to the Density of fully compacted concrete



Slump Test and Compaction Factor Test:

3. Flow Table Test


• This test is done for concrete having very high workability

• Consists of a circular table and mould in form of frustum(top dia 17cm, bottom dia 25cmand height 12cm)

• This mould is then placed on the table and concrete to be tested is filled in it in two layers

• It is raised immediately and table is lifted and dropped by 12.5 mm, 15 times in 15 seconds

• The spread of concrete direction is measured in at least 6 directions and is further used to represents the workability in terms of flow percent.

• This test is used for the concrete that possessed low workability having the value of slump limited to 50 mm.

• In this test mould in the form of frustum in placed inside the cylinder and concrete to be tested is filled in the mould that is lifted immediately after it is filled completely.

• The cylinder is then subjected to vibration and the time required by the concrete to assume the cylinder shape is noted which is further used to represent the workability in terms of Vee –Bee Degrees.


4. Vee Bee Test

1. Base:• A base is a solid substance in a fine state of division and it

forms the bulk of a paint.✓It reduces the shrinkage cracks formed on drying.• Exp : white lead, zinc oxide.

2. Vehicle or carrier: a) Linseed Oil : These Oil dries very quickly and is suitable

for external work. It however requires a thinning agent like turpentine.

b) Tung Oil : This Oil is far superior to linseed Oil and is used for preparing paints of superior quality.


Fundamental components of an Oil Based Paint

3. Driers:• These Substances accelerate the process of drying.• A drier absorbs Oxygen from the air and transfer it to the

linseed oil which in turn gets hardened.• Exp –Litharge, red lead, manganese sulphate.

4. Coloring Pigments: • when it is desired to have a different colour than the base

of a paint, a colouring pigment is to be added.



5. Solvents or thinners: ✓The function of a solvent is to make the paint thin to

that it can be easily applied on the surface✓It also helps the paints in penetrating through the

porous surface.• The most commonly used solvent is spirit of turpentine.• Benzene and Naptha are also used as a substitute



Varnish

• A transparent solution of resinous substance in linseed oil, turpentine or in alcohol is called Varnish.

• Varnish has little or no colour, is transparent, and has no added pigment, as opposed to paints, which contain pigment and generally range from opaque to translucent


Varnish

Resins

Driers

Solvents for Resins

Ingredients of Varnish1. Resins or Resinous Substances:

• The commonly used resins are copal, lac or shellac and rosin.

• The rosin is obtained from pine trees

• Other resins are Amber, mastic, gumdammer etc

2. Drier:• The function of a drier in varnish is to accelerate the process of drying.

exp- litharge, white copper, lead sulphate

3. Solvent for Resins


GLASS

Glass is obtained by fusion of Silica, chalk/lime, and Potash and soda at over 1000°C

• Ingredients of Glass:1. Silica: major constituent of Glass2. Potash: it renders glass infusible and gives fire

resistance properties3. Soda: It accelerates fusion of glass4. Lime: provides durability and toughness to glass5. Cullet: It is broken glass used as raw material to reduce

the costCivil Engineering by Sandeep Jyani 215

Classification of Glass on the basis of Composition

1. Soda Lime/Soda Ash/Soft Glass:• Obtained from fusion of silica, lime, soda and alumina

2. Lead Glass:• Obtained from Silica, Lead and Potash• Shining appearance and takes good polish

• Used for cut glass work, bulbs

3. Boro Silicate Glass• Obtained from Silica, borax, Lime

• It can stand high temperature• Lab equipment and cooking utensils


Special Commercial forms of Glass

1. Sheet Glass: • Made by blowing, available within 2-6mm thickness. • Used for doors, windows panels

2. Plate Glass: • Molten glass is rolled into plate glass with thickness 6 to 25mm. • Glazing work, Table tops, etc

3. Coloured Glass: • Mixing different pigments at fusion state, also known as Stained Glass

4. Wired Glass: • Plate and sheet glass and wire mesh, so also known as Reinforced Glass. • Used in sky lights, high class roofing works, fire proof doors and windows


5. Laminated Glass:• Also known as Safety Glass• Sheets of glass glued together with celluloid layer• Used for wind screen in vehicles

6. Bullet Proof Glass:• Number of glass layers alternating with vinyl resin plastic layers pressed

together

7. Fibre Glass:• Glass Wool• Highly resistance to wear and tear• Thermal and sound insulation

8. Flint Glass:• Glass containing lead- highly shining• Used for table glass, lenses, prism, optical glass etc.


Engineering Properties of Glass

• Transparency - vision of outside world through it.

• Strength- depends on modulus of rupture value of glass

• Workability - can be molded into any shape or it can be blown during melting

• Transmittance - The visible fraction of light that passing through glass is the property of visible transmittance.

• U value- U value represents the amount of heat transferred through glass. If a glass is said to be insulated unit then it should have lower u value.

• Recycle property -Any glass can be 100% recyclable. It can also be used as raw material in construction industry.


II. Strength Of Concrete1. Compressive Strength of

Concrete• Compressive strength of concrete is measured in

UTM using the mould that may be either be cubical, cylindrical, trapezoidal or prismoidal in shape.

• Generally cubical mould of size 150 mm is used if the maximum nominal size of aggregate is greater than 20 mm and of size 100 mm is used if the maximum nominal size of aggregate is less than 20 mm.

• If cylindrical mould is used, height to diameter ratio is kept to be 2:1. Generally height of 30 cm and dia of 15 cm is used.

• The results obtained from cylindrical mould are found to be more precise than the results obtained from cubical mould.

• Strength of the concrete measured using cylindrical mould is approx 0.8 times the strength obtained from the cubical mould as the straining effects of the platens (plates of UTM) is found to be for the entire height of the cubical mould.


Trees1. Exogenous Trees

a) These are the trees that grow in bulk by increasing in outward direction

b) These trees never shed their leaves until new ones are grown- EVERGREEN TREES (Conifer Trees)

c) These trees bear Coned shaped fruits hence this class of trees is known as Conifer Trees


d) These Trees consist of Circular rings along their horizontal section that is added up each year hence is termed as Annular Rings, which is helpful in finding age of tree

Trees1. Exogenous Trees

• The wood used for engineering purposes is mostly derived from these trees only

• These are further divided into two types:

A. Conifer trees (Soft Wood Trees)▪ These trees possess distinct

annular ring and indistinct medullary rays

▪ Light in weight, color hence soft wood

▪ Example- Chir, Deodar, Pine, Spruce



B. Deciduous Trees (Hard Wood Trees)▪ These are also known as

BROAD LEAVED TREES

▪ They shed their leaves in autumn which again grown back in springs

▪ Wood obtained for these trees is Hard, strong, non resinous and dark in color

▪ These possess indistinct annular rings and distinct medullary rays

▪ EXP: Sal, teak, Oak, Babul

1. Exogenous Trees

• These are the trees which grow in Bulk in inward direction and consist of Fibrous marks throughout their longitudinal section

• Wood obtained from these trees have limited Engineering applications

• Example: Bamboo cane, palm tree


2. Endogenous Trees

Trees


Structure of Tree

Pith-> Heart Wood -> Sap Wood->Cambium Layer->Inner bark->

Outer bark

Structure of Tree1. Pith/Medulla:

a) Inner most portionb) Entirely consists of cellulose (which is

used for nourishment of plant in its young age)

c) As Plant grows into Tree, it decays and does not take any active part in growth of the tree

2. Heartwooda) Inner annular ring surrounding the pith

constitutes heartwoodb) It has dark color that represents dead

portion of the treec) Does not participate in growth of tree

but imparts strength and rigidity to itd) Wood used for engineering purposes

majorly


Structure of Tree3. Sap Wood (Alburnum):

a) Outer annual rings between Heartwood and Cambium layer

b) It has light colorc) It plays active role in growth of the

tree

4. Cambium Layera) It is the Sap being converted to

Sapwood, i.e future growth of the tree is dependent upon this Cambium Layer only

b) If due to any reason, bark of the tree is removed exposing the Cambium layer, it can lead to Scissor of Cells of wood in Cambium layer, which can finally lead to death of tree


Structure of Tree5. Medullary Rays:

a) Radial Fibers extending from Pith to Cambium layer holding the annual rings of sapwood and heartwood together is termed as Medullary rays

b) Strength of timber in Transverse direction is dependent on presence of Medullary rays

6. Barka) It is protected covering over the Cambium layer to

safeguard the future growth of the tree


Processing of Timber

1. Falling of Tree2. Seasoning of

Tree3. Conversion of

Timber4. Preservation

of Timber


1. Falling of Trees• Should be done at suitable age (optimum age 50-100yrs)

❖Over Mature: Decayed wood is obtained

❖Under mature: Sap wood is obtained

• Trees should be cut in such a way that maximum wood is obtained from it. Sawing is done at section just above the ground

• Sawing should be done in season when movement of Sap is minimum, hence avoided in Autumn and Spring when Sap movement is maximum through the section of tree

• In hilly areas, cutting should be done in mid summers as in winter- rain fall occurs

• In Plain areas, cutting should be done in winters as excessive loss of moisture may take place in summers



2. Seasoning of TreesNewly fallen trees contain moisture up to 50%, which needs to be removed before using for any Engineering activity, this process of drying the timber to make it fit for Engineering use is called Seasoning

• Water is present in form of Moisture and Sap

• During seasoning, free water is removed from tree and point at which it is completely removed is termed as Fiber Saturation Point

• Seasoning is done upto moisture content of 10-12%

• Practical limit of moisture content achieved is 15%




Seasoning

1. Artificial

Boiling

Electrical

Kiln

Chemical

2. Natural

Air Seasoning

Water Seasoning

2. Seasoning of Timber

I. Artificial SeasoningA. Boiling

• Logs of timber are placed in water, temperature of which is raised upto boiling point and is maintained for 3-4 hours, then dried naturally

• Increase in temperature->Quick Evaporation->Duration of seasoning decreases

• Quickest method of seasoning but costly

B. Electrical Seasoning• Timber is subjected AC(Alternative Currents), as

wood is bad conductor of electricity, produces heat, which is utilized in seasoning of timber



I. Artificial SeasoningC. Kiln seasoning

• Logs of timber are placed in air tight chamber

• Fully saturated air is forced into it at temperature of 35-40°C

• Then temperature is increased gradually and does not result in evaporation, relative humidity is not reduced, hence uniformly drying the timber

D. Chemical seasoning• Logs are immersed in Solution of salts that

increase rate of evaporation and helps bring down the duration of seasoning



II. Natural Seasoning


A. Air Seasoning• In a shed with a platform of height 300 mm

timber is stacked.• Care is taken to see that air can circulate

around each timber balk. Moisture content decreases over a period of time.

B. Water Seasoning• Logs of Timber are immersed in water stream

such that larger portion is placed upstream direction of flow

• The sap is replaced by flowing water, flowing water can be removed in shorter time


3. Conversion of TimberProcess of sawing the timber into sections of suitable shape and size is termed as Conversion of Timber

Sawing

A. Ordinary Sawing

B. Quarter Sawing

C. Tangential Sawing

D. Radial Sawing


3. Conversion of TimberA. Ordinary Sawing

✓Wastage is minimum

✓Most Economical

✓Cuts are made tangential to Annular Rings

• Sections are liable to shrink

B. Quarter Sawing• Cuts are made at right angles to each other

✓Used when annual rings are indistinct

• Sections are liable to bend in transverse direction


3. Conversion of TimberC. Tangential Sawing

• Cuts are made tangential to the annual rings that meet each other at right angle

❖Strength obtained in this section is minimum as medullary rays are also cut that are used for holding the annual rings in position

D. Radial Sawing• Cuts are made parallel to the

medullary rays/ radial directions

• Decorative effect

✓Strongest Section is obtained

❖Wastage is also maximum

4. Preservation of Timber


• Preservation is carried out in order to increase the life of timber structure to make it more durable and increase its resistance against attack of fungi, insects, termites, etc.



A. AsCu Arsenic Copper Treatment:• Timber is coated with solution of

Arsenic pentoxide, Copper sulphate and Potassium Dichromate

✓This treatment increases resistance against attack of white ants.

B. Chemical Salts:• Copper sulphate, Zinc Chloride,

mercury chloride, Sodium Fluoride increase the durability of Structure



C. Preservation with Coal Tar:• Hot tar is applied over the surface✓It increases resistance against fire.

D. Creosote Oil :• Produced by distillation of tar• Application of creosote oil almost doubles

life of timber• Process of application is called Creosotic in

which oil at temperature of 50°C, pressure 0.7-1 N/mm2 is applied uniformly over the timber



E. Oil Paints/Solignum Paints:• Oil and solignum paints increase the

resistance to water

Defects in Timber• Defects can occur due to:


Seasoning

Natural Resources

Conversion of Timber

Fungi

1. Defect due to Seasoning


Defects in Timber

2. Defect due to Natural Resources• Heartshakes and StarShakes: a defect in timber

consisting of shrinkage and separation of tissues across the annual rings usually along the rays


Defects in Timber

2. Defect due to Natural Resources

• Cupshakes and Ring shake: defect in timber consisting of shrinkage and separation of the annual rings


Defects in Timber

3. Defect due to Conversion of Timber

A. Chipmark• Formed due to marks of plaining tools

B. Torn Grain• Depression over timber due to falling of

tools

C. Diagonal Grain• Due to improper cutting

D. Wane• Due to presence of originally rounded

surface on finished timber section


Defects in Timber

4. Defects in Timber due to Fungi

A. Blue stain: fungi carries decomposition of sap and leaving blue color spot over it

B. Brown Rot: caused due to removal of cellulose by fungi resulting in reddish color

C. White Rot: formed due to removal of Lignin(important in the formation of cell walls) leading to development of white spots

D. Dry rot: formed where there is no free circulation of air, fungi converts it to dry powdered form


E. Wet Rot: formed due to alternate wetting and drying of timber structure


4. Defects in Timber due to Fungi

II. Strength Of Concrete2. Tensile Strength of Concrete

• Tensile, strength of the concrete is tested indirectly, by noting its modules of rupture that is determined by preparing a block of size 15 cm × 15 cm ×70 cm if the Maximum nominal size of aggregate is greater than 20 mm; and of size 10 cm × 10 cm × 50 cm if the maximum nominal size of aggregate is less than 20 mm.

• The beam is then placed over the roller supports and is subjected to the load at which its failure takes place that is further used to find its modulus of rupture (stress at which failure takes place)


III. Non Destructive Tests

1. Schmidt Rebound Hammer

• In this test rebound hammer is used to analyse the quality of hardened concrete which consists of spring controlled rebound mass that slides over the plunger in tubular housing

• When the release button is pressed, rebound mass strikes against the surface to be tested and retracks due to spring force carrying along the rider with it in backward direction.

• Displacement of which is noted over the scale and is termed as rebound number that is further used to analyse the quality of structure.

• Higher is the value of rebound number, better is the quality of the structure. The results obtained from the test are dependent upon various factors such as


2. Ultrasonic Pulse velocity test

• This test deals with the measurement of time of travel of electronically generated mechanical pulse through the structure to be tested that is further used to analyse its quality.

• The mechanical pulse is generated with the help of Electro Acoustic Transducers and detected with the help of Transmitter that is further used to analyse the time of travel and the velocity of pulse through the structure so as to represent the quality of structure.


• Compaction Equipment2. Vibrators

3. Surface Vibrator

1. Also known as Screed/Pan Vibrator

2. Used at the time of screeding off the concrete

4. Vibrating Table

• Used for stiff and harsh mix for prefabricated elements

• Also used for laboratory mixes


Concreting Equipment

Surface Vibrator

Vibrating Table

𝝈Stress

Strain 𝜺

Proportional Limit.

Elastic Limit.

C

O

BA

C’

𝝈ult E

Upper Yield Point

Lower Yield Point

Strain Hardening

Yield Plateau

Necking Region

𝝈f

F

D

Stress – Strain Curve for Mild Steel

ElasticRegion

Plastic Region


Application of Hooke’s Law

2) Elongation of Prismatic bar due to Self Weight

𝑾

𝑨

𝑳 𝒅𝒙

𝑾𝒆𝒊𝒈𝒉𝒕 𝑫𝒆𝒏𝒔𝒊𝒕𝒚(𝑾𝒕/𝑽𝒐𝒍𝒖𝒎𝒆) = 𝝀

𝒙

𝚫𝒙 =𝑷𝒙−𝒙 × 𝒅𝒙

𝑨𝑬… (𝟏)

𝒅𝒙

𝑾𝒙−𝒙

𝑾𝒙−𝒙 = 𝑷𝒙−𝒙

𝑬𝒍𝒐𝒏𝒈𝒂𝒕𝒊𝒐𝒏

⇒ 𝑷𝒙−𝒙 = 𝝀 𝑨𝒙 … (𝟐)

𝚫𝒙 =𝝀 𝑨𝒙 × 𝒅𝒙

𝑨𝑬

⇒ 𝚫𝒙 =𝝀 𝒙𝒅𝒙

𝑬

𝑭𝒐𝒓 𝒕𝒐𝒕𝒂𝒍 𝑬𝒍𝒐𝒏𝒈𝒂𝒕𝒊𝒐𝒏 𝒐𝒇 𝒕𝒉𝒆 𝒃𝒂𝒓,

∆𝑳𝒔𝒆𝒍𝒇 𝒘𝒕=𝛌𝑳𝟐

𝟐𝑬

∆𝑳𝒔𝒆𝒍𝒇 𝒘𝒕 = න

𝟎

𝑳

𝚫𝒙 = න

𝟎

𝑳𝝀 𝒙𝒅𝒙

𝑬



3. Elongation of Tapered Bar 𝑳

𝑫𝟐𝑫𝟏

∆𝑳 =𝟒𝑷𝑳

𝝅𝑫𝟏𝑫𝟐𝑬



L

D2

∆𝑳 =4𝑷𝑳

𝝅 D1D2 𝐄

D1

Que. 14 For same elongation, what is the relation between the two (Prismatic Bar and Tapered Bar)?

a) 𝐃 = 𝑫𝟏𝑫𝟐

b) 𝑫 = 𝑫𝟏𝑫𝟐

c) 𝑫𝟏 = 𝑫𝑫𝟐

d) 𝑫𝟐 = 𝑫𝑫𝟏∆𝑳 =

𝑷𝑳

𝑨𝐄

D

D = D1D2



𝑳

𝑫 + 𝒂∆𝑳 =4 PL

𝝅 D1D2 𝐄𝑫 − 𝒂

Que. 15 What is the value of percentage elongation error in Tapered bar?

∆𝑳 =PL

𝑨𝐄D

% 𝒆𝒓𝒓𝒐𝒓 𝒊𝒏 𝒆𝒍𝒐𝒏𝒈𝒂𝒕𝒊𝒐𝒏 = (∆L)Tapered – (∆L)Prismatic

(∆𝑳)Tapered

=4 PL

𝝅 (D+a)(D−a) 𝐄

4 PL

𝝅 (D+a)(D−a) 𝐄

−4PL

𝛑𝐃𝟐𝐄= {

𝟏𝟎𝒂

D𝐦𝐞𝐚𝐧

}2 % of error

X 100

X 100

By using mean dia of Tapered bar, make a prismatic bar of dia equal to mean dia

𝑫𝒎𝒆𝒂𝒏 =𝑫 + 𝒂 + 𝑫 − 𝒂

𝟐= 𝑫

✓𝝁cork = 0

✓𝝁metal = 0.25-0.33

✓𝝁human tissue/foam= negative

✓𝝁𝐫𝐮𝐛𝐛𝐞𝐫 = 0.5

✓𝝁steel = 0.286

✓𝝁concrete = 0.1-0.2

• 𝝁generally 0<= 𝝁<=0.5

• 𝝁ideal -1<= 𝝁<=0.5


Poisson’s Ratio

Volumetric Strain1. If there is uniaxial loading, then

volumetric strain will be

2. If the poisson's ratio of material is 𝝁= 0.5, then volumetric strain will be 0 under any state of loading

3. If the poisson's ratio of any material is less than 0.5 i.e. (𝝁<0.5), the change in volume or volumetric strain will be 0 if sum of all the normal stresses is 0


𝑽𝒐𝒍𝒎𝒆𝒕𝒓𝒊𝒄 𝑺𝒕𝒓𝒂𝒊𝒏 =𝝈𝒙 (𝟏 − 𝟐𝝁)

𝑬

𝒆𝒗 =𝝈𝒙+𝝈𝒚+𝝈𝒛

𝑬(𝟏 − 𝟐𝝁)

Shear Modulus/ Modulus of Rigidity


𝑺𝒉𝒆𝒂𝒓 𝑴𝒐𝒅𝒖𝒍𝒖𝒔 =𝑺𝒉𝒆𝒂𝒓 𝑺𝒕𝒓𝒆𝒔𝒔

𝑺𝒉𝒆𝒂𝒓 𝑺𝒕𝒓𝒂𝒊𝒏

Bulk Modulus

•𝑩𝒖𝒍𝒌 𝑴𝒐𝒅𝒖𝒍𝒖𝒔 =𝒅𝒊𝒓𝒆𝒄𝒕 𝑺𝒕𝒓𝒆𝒔𝒔

𝑽𝒐𝒍𝒖𝒎𝒆𝒕𝒓𝒊𝒄 𝑺𝒕𝒓𝒂𝒊𝒏

𝑮 =𝝉

𝒚

𝑲 =𝝈

∆𝑽𝑽

IMPORTANT RELATIONS

E= 𝟐𝑮 𝟏 + 𝝁

E= 𝟑𝑲 𝟏 − 𝟐𝝁

(E, G, 𝝁, K)


𝑬 =𝟗𝑲𝑮

𝟑𝑲 + 𝑮

For metals

E> 𝑲 > 𝑮

CASE 1: CANTILEVER BEAM

a) Cantilever beam subjected to Point load at the free end


Px

x 𝑥

SFD

BMD

L(𝑺𝑭)𝒙−𝒙= 𝐏

𝑨 𝑩

(𝑺𝑭)𝑨= 𝐏

(𝑺𝑭)𝑩= 𝐏

(𝑴)𝒙−𝒙= −𝐏 × 𝒙

(𝑴)𝑩= −𝐏 × 𝟎 = 𝟎

(𝑴)𝑨= −𝐏 × 𝑳 = −𝑷𝑳

(𝑴)𝑩−(𝑴)𝑨 = 𝒂𝒓𝒆𝒂 𝒐𝒇 𝑺𝑭𝑫 𝒃𝒕𝒘 𝑩 𝒂𝒏𝒅 𝑨

𝐏

L

CASE 1: CANTILEVER BEAMb) Cantilever beam subjected to Uniformly

Distributed Load (UDL)


w N/mx

x x

SFD

BMD

L (𝑺𝑭)𝒙−𝒙= 𝒘𝒙

(𝑺𝑭)𝑨= 𝒘𝑳

(𝑺𝑭)𝑩= 𝟎

(𝑴)𝒙−𝒙= −𝐰 × 𝒙 ×𝒙

𝟐

(𝑴)𝑩= 𝟎 𝒂𝒕 𝒙 = 𝟎

(𝑴)𝑨= −𝒘𝑳𝟐

𝟐𝒂𝒕 𝒙 = 𝑳

𝑾𝒉𝒆𝒏 𝑺𝑭𝑫 𝒊𝒔 𝒓𝒆𝒄𝒕𝒂𝒏𝒈𝒖𝒍𝒂𝒓, 𝑩𝑴𝑫 𝒕𝒓𝒊𝒂𝒏𝒈𝒖𝒍𝒂𝒓

𝑨 𝑩

(𝑴)𝒙−𝒙= −𝒘𝒙𝟐

𝟐

𝑾𝒉𝒆𝒏 𝑺𝑭𝑫 𝒊𝒔 𝒕𝒓𝒊𝒂𝒏𝒈𝒍𝒖𝒍𝒂𝒓, 𝑩𝑴𝑫 𝒊𝒔 𝒑𝒂𝒓𝒂𝒃𝒐𝒍𝒊𝒄

+𝒘𝒙

−

−𝒘𝑳𝟐

𝟐


SFD

BMD

x

x x

M (N-m)CASE 1: CANTILEVER BEAM

c) Cantilever beam subjected to Concentrated Point Moment

L

M (N-m)

SF = 0

When SF = 0 and Bending Moment is constant, this state is said to be Pure Bending

Pure Bending


SFD

BMD

CASE 1: CANTILEVER BEAM

d) Cantilever beam subjected to Uniformly Varying Load

w N/m

L

x

x

x

(𝑺𝑭)𝒙−𝒙=𝟏

𝟐× 𝒙 ×

𝒘

𝑳𝒙

(𝑺𝑭)𝒙−𝒙=𝒘𝒙𝟐

𝟐𝑳𝒑𝒂𝒓𝒂𝒃𝒐𝒍𝒊𝒄

(𝑺𝑭)𝑩= 𝟎

A B

(𝑺𝑭)𝑨=𝒘𝑳𝟐

𝟐𝑳=

𝒘𝑳

𝟐

𝒘𝑳

𝟐 +

(𝑩𝑴)𝒙−𝒙= −𝒘𝒙𝟐

𝟐𝑳×

𝒙

𝟑

(𝑩𝑴)𝒙−𝒙= −𝒘𝒙𝟑

𝟔𝑳𝒄𝒖𝒃𝒊𝒄

(𝑩𝑴)𝑩= 𝟎 (𝑩𝑴)𝑨= −𝒘𝑳𝟐

𝟔−

𝒘𝑳𝟐

𝟔


CASE 2: Simply Supported BEAM

b) Simply Supported beam subjected to Uniformly Distributed Load (UDL)

L

x

x

SFD

BMD

W N/m

A B

Ra Rbx

x

B

Rb = wL/2

x

(M)xx = + Rb× 𝒙 - (w𝒙) ×𝒙

𝟐

(put x=0) (M)B = 0

- wL/2

-

wL/2

+

wx

(put x=L/2) (M)C = wL2

𝟖

(M)xx = wL𝒙

𝟐-

w𝒙𝟐

𝟐

C

(put x=L) (M)A = 𝟎

+Mmax = wL2

𝟖


CASE 2: Simply Supported BEAM

c) Simply Supported beam subjected to Concentrated Moment

SFD

BMD

A BM (N-m)

L/2 L/2

Ra Rb

Ra + Rb = 0 ….(1)

Taking moment about A = 0

(+Rb)xL + M = 0

Rb = -M/L (or we say M/L downward)

Ra= +M/L (from eqn 1)

+ M/LMB = 0

MB – MC = area of SFD between C and B

=> MB – MC = (𝑴

𝑳×

𝑳

𝟐)

MC = -𝑴

𝟐(considering upto length L/2 from B)

Now at Length just beyond C

=> MC’ = -𝑴

𝟐+ M =

𝑴

𝟐

And MA = 0

_

-𝑴

𝟐

+

𝑴

𝟐


CASE 3: OVERHANG BEAM

b) Overhang beam subjected to Uniformly distributed load

L

SFD

BMD

A B

Ra Rb

+

-

w N/ m

aa CD

w×a

−w 𝑳

𝟐

+

w 𝑳

𝟐

-w×a

ME =w𝟖

(L2 – 4a2)

Case i)

If L= 2a

ME =w𝟖

((2a)2 – 4a2)

ME = 0

E

MB = -wa2

𝟐Ma =-

wa2

𝟐

MB = -wa2

𝟐Ma =-

wa2

𝟐

ME = 0




L

SFD

BMD

A B

Ra Rb

+

-

w N/ m

aa CD

w×a

−w 𝑳

𝟐

+

w 𝑳

𝟐

-w×a

ME =w𝟖

(L2 – 4a2)

Case ii)

If L> 2a

ME =𝒑𝒐𝒔𝒊𝒕𝒊𝒗𝒆

We get two points of contraflexure

E

MB = -wa2

𝟐Ma =-

wa2

𝟐

MB = -wa2

𝟐Ma =-

wa2

𝟐

ME > 0




L

SFD

BMD

A B

Ra Rb

+

-

w N/ m

aa CD

w×a

−w 𝑳

𝟐

+

w 𝑳

𝟐

-w×a

ME =w𝟖

(L2 – 4a2)

Case ii)

If L< 2a

ME =𝑵𝒆𝒈𝒂𝒕𝒊𝒗𝒆

E

MB = -wa2

𝟐Ma =-

wa2

𝟐

MB = -wa2

𝟐Ma =-

wa2

𝟐

ME > 0

Que. 63 What percentage of total length should either overhang be so that BM= 0 at the centre ?


A Bw N/ m

aa CD ME =w𝟖

(L2 – 4a2)

ME = 0

ME =w𝟖

(L2 – 4a2) = 0

So L= 2a or

a = L/2

Total length = L+2a = 4a

Therefore, 𝒂

𝟒𝒂× 𝟏𝟎𝟎

= 𝟐𝟓% 𝒐𝒇 (𝑳 + 𝟐𝒂))

Que. 64 If magnitude of bending moment at support is equal to Bending moment at centre, what is the relation between a and L?


A Bw N/ m

aa CD

ME = MB

ME =w𝟖

(L2 – 4a2) = |-wa2

𝟐|

L2/4 - a2 = a2

L = 2 𝟐𝒂

Strain Distribution


N Ay

E GTension

Compression

Tension

Compression

Strain Distribution


N Ay

E GTension

Compression

∈𝑚𝑎𝑥

𝑦𝑚𝑎𝑥

=∈

𝑦∈yymax

∈=∈𝑚𝑎𝑥

𝑦𝑚𝑎𝑥

×y

Section Modulus (z)

• It is the ratio of Moment of Inertia about the neutral axis to ymax

(fiber which is at maximum distance from Neutral Axis) i.e.

𝒛 =𝑰𝑵𝑨

𝒚𝒎𝒂𝒙

• Section Modulus represents Bending strength of the Section

• Greater the value of z, greater the bending strength.

• The value of z depends upon Moment of Inertia and Distribution of Area



Moment of Resistance

N Ay

ymax

𝝈𝒎𝒂𝒙 MR

𝑰𝑵𝑨

= 𝝈𝒎𝒂𝒙

𝑦𝑚𝑎𝑥

MR

𝑰𝑵𝑨

= 𝝈𝑩𝑬𝑵𝑫𝑰𝑵𝑮

𝑦=

𝑬

𝑅


We know that

𝝉

𝑹=

𝑮𝜽

𝑳

𝐝𝐓𝐑 = 𝝉𝒎𝒂𝒙

𝑹x r2 x dA

For total Resisting Moment,

𝐝𝐓𝐑 =𝝉𝒎𝒂𝒙

𝑹x r2 x dA

Second moment of area or Polar moment of Inertia

𝐓𝐑 = 𝝉𝒎𝒂𝒙

𝑹x J

Therefore,𝝉

𝑹=

𝑮𝜽

𝑳=

𝐓𝐑

𝑱

Rr

dF

dF is acting parallel to the plane

dA

Strain Energy


P

L

∆L

Work Done =Net Force x Displacement

Case1 : When Load is applied gradually:


Work Done =(𝑷

𝟐)x ∆L

IRF

Case2 : When Load is applied suddenly:


Work Done =(𝑷)x ∆L

Resilience

U = Work done by IRF

= Work done by force P

U=(𝑷

𝟐)x ∆L


1. Resilience:▪ Total strain Energy stored in body

is commonly known as Resilience.

▪ Whenever straining force is removed from strained body, the body is capable of doing work

▪ Resilience is also defined as capacity of a strained body for doing work on removal of straining force

U=𝟏

𝟐𝑷 x

𝑷𝑳

𝑨𝑬

U= 𝑷𝟐 𝑳

𝟐𝑨𝑬

Resilience

U = Work done by IRF

= Work done by force P

U=(𝑷

𝟐)x ∆L


1. Resilience:

U=𝟏

𝟐𝑷 x

𝑷𝑳

𝑨𝑬

U= 𝑷𝟐 𝑳

𝟐𝑨𝑬

U= 𝟏

𝟐×

𝑷

𝑨×

∆𝑳

𝑳× 𝑨 × 𝑳

U= 𝟏

𝟐× 𝑺𝒕𝒓𝒆𝒔𝒔 × 𝑺𝒕𝒓𝒂𝒊𝒏 × 𝑽𝒐𝒍𝒖𝒎𝒆

U= 𝟏

𝟐× 𝝈 × 𝜺 × 𝑽

U= 𝟏

𝟐× 𝝈 ×

𝝈

𝑬× 𝑽

U= 𝝈𝟐

𝟐𝑬× 𝑽


2. Proof Resilience • Maximum strain energy, stored

in body in called as Proof Resilience

• Strain Energy stored in body will be maximum when body is stressed upto Elastic Limit

• Area of Load vs Deformation curve upto Elastic Limit gives us the value of Proof Resilience

Load

Deformation

P1

P2

PEL or P3

𝜹𝟏 𝜹𝟐 𝜹𝟑

Work done by P1

Resilience upto P1

Work done by P2

Resilience upto P2

Max strain energy in elastic region

3. Modulus of Resilience• It is defined as Proof Resilience of

a material per unit volume


𝑴𝑹 =𝑷𝒓𝒐𝒐𝒇 𝑹𝒆𝒔𝒊𝒍𝒊𝒆𝒏𝒄𝒆

𝑽𝒐𝒍𝒖𝒎𝒆

𝑴𝑹 =𝑴𝒂𝒙𝒊𝒎𝒖𝒎 𝑺𝑬 𝒖𝒑𝒕𝒐 𝒆𝒍𝒂𝒔𝒕𝒊𝒄 𝒍𝒊𝒎𝒊𝒕

𝑽𝒐𝒍𝒖𝒎𝒆

𝑴𝑹 =𝝈𝟐

𝑬𝒍𝒂𝒔𝒕𝒊𝒄 𝑳𝒊𝒎𝒊𝒕𝟐𝑬

𝑴𝑹 =

𝝈𝟐

𝟐𝑬× 𝑽

𝑽

Stress

Strain

𝝈1

𝝈 2

𝝈 EL or 𝝈3

𝜺𝟏 𝜺𝟐 𝜺𝑬𝑳

4. Super Position Theorem

1.

2.


B

W

ymax

𝜽

A 𝒚𝒎𝒂𝒙 =𝑾𝑳𝟑

𝟑𝑬𝑰𝜽𝒎𝒂𝒙 =

𝑾𝑳𝟐

𝟐𝑬𝑰

ByB

𝜽𝑩

A 𝜽𝒎𝒂𝒙 =𝑾𝑳𝟑

𝟔𝑬𝑰𝒚𝒎𝒂𝒙 =

𝑾𝑳𝟒

𝟖𝑬𝑰

Slope Deflection


3.

4.


B

ymax

𝜽

A 𝒚𝒎𝒂𝒙 =𝑴𝑳𝟐

𝟐𝑬𝑰𝜽𝒎𝒂𝒙 =

𝑴𝑳

𝑬𝑰

ByB

𝜽𝑩

A 𝜽𝒎𝒂𝒙 =𝑾𝑳𝟑

𝟐𝟒𝑬𝑰𝒚𝒎𝒂𝒙 =

𝑾𝑳𝟒

𝟑𝟎𝑬𝑰

Slope DeflectionM

W N/m


5.

6.


ymax𝜽𝒚𝑪 =

𝑾𝑳𝟑

𝟒𝟖𝑬𝑰𝜽𝑨 = 𝜽𝑪 =𝑾𝑳𝟐

𝟏𝟔𝑬𝑰

𝜽𝑨 = 𝜽𝑪 =𝑾𝑳𝟑

𝟐𝟒𝑬𝑰𝒚𝒎𝒂𝒙 =

𝟓𝑾𝑳𝟒

𝟑𝟖𝟒𝑬𝑰

Slope DeflectionW

𝑳/2 𝑳/2BA C

ymax𝜽

𝑳/2 𝑳/2

BA C


Effective Length as Per End Conditions

1. Both End Hinged 3. One End Fixed and other is free

2. Both End Fixed 4. One End Fixed and other is Hinged

𝑷𝒄𝒍 =𝝅𝟐𝑬𝑰

𝑳𝟐 𝑷𝒄𝒍 =𝝅𝟐𝑬𝑰

𝟒𝑳𝟐

𝑷𝒄𝒍 =𝟒𝝅𝟐𝑬𝑰

𝑳𝟐𝑷𝒄𝒍 =

𝟐𝝅𝟐𝑬𝑰

𝑳𝟐

L 𝒍𝒆𝒇𝒇 = 𝒍𝒂𝒄𝒕𝒖𝒂𝒍 = 𝑳

𝒍𝒆𝒇𝒇 = 𝑳/𝟐

𝒍𝒆𝒇𝒇 = 𝟐𝑳

𝒍𝒆𝒇𝒇 = 𝑳/√𝟐

Equipments used in Chain Surveying1. Surveying chain:-A. Revenue chain → 33 feet, 16 linkB. Gunter’s chain → 66 feet, 100 linkC. Engineer’s chain → 100 feet, 100 linkD. Metric chain →

a) 20 m, 100 link b) 30 m, 150 link

Note:

• 1 mile = 1.6093 km = 80 Gunter’s chain

• 1 Acre = 4046.856 m² = 10x (Gunter’s chain)² = 40 guntas = 43560 ft²

• 1 yard = 3 feet = 0.9144 m



• Bending Moment in Simply Supported Beam

• Concrete strength data indicate that for cements low in alkali and aluminate content, increasing the SO3 to 3.6%

• Porosity of soil formula

• Glass is prepared by

• Which type of survey involve principles of trigonometry

• Effective length of member with one side free and one side fixed

Limit of chloride in drinking water

Which type of irrigation is called trickling irrigation?• Surface.

• Sprinkler.

• Drip/trickle.

• Subsurface.

Crippling load for Both End Hinged

Aeration process is used to

Which type of estimations is plinth area base

What is valuation

Vehicle in paints

• Extender improvesthe performance of latex and acrylic paints and reduce the appearance of brush marks, enhances elastic properties

Contigency?

Effective cover of footing?

50mm

Floating object ABOVE B below G

Settling of particles in steady state ....?

Slenderness ratio for tension member reversal of stress due to earth quake loads

Strength of class c cement

• This product is intended for use when conditions require high early strength.

• Available in ordinary, O , MSR, and HSR grades

Unit of Dynamic Viscosity?

Chain cannot be shorted by

• Rounding flattened rings

• By heating

Dimensionless quantity

• Viscosity

• Reynolds number

• Specific weight

shear strength for lacing for 1000kn

Slenderness ratio for column fixed at one end and free at another

What is FAR?

Plinth Area estimate definition

Neutral axis depth

BM at end in case of cantilever beam carrying udl

Flow index= ?

Froude no between 4.5-9 what is the type of fall?

Chemical classification of rocks?

G=given; dry density=given; what is the water content

Flexural rigidity direct formula

Porosity is defined as

% of carbon in wrought iron

unit wt of plain cement concrete.

Cement Concrete (Plain) 2400 Kg/ m3

strength of class 3 brick

Early strength in cement in initial stage is due to which component

Delta and duty related numerical

Depth of centre of pressure of a rectangular surface dipped vertically in water

Value of poisson’s ratio for rubber

What's toughness index?

How many types of soil are there ?

Thickness of plastering

Factored load at limit state?

Example of sedimentary rocks?

Water storage efficiency?

shoulder in highway?

Road plan from 1981-2001

When was 4 classification of Indian road decided

Which type of error cannot be adjusted

FREE HOLD PROPERTY?

Elongation due to load where load is 40kn and cross section 20×20 mm.. length 1m

SAR??

Maximum spacing in vertical stirrup is

Autoclave test is done for?

Load carrying capacity of column?

Rate of filtration of slow sand filters?

Bougues compound?

Shift 2

Grade of M20

𝑀5 → 1: 5: 10

𝑀10 → 1: 3: 6

𝑀7.5 → 1: 4: 8

𝑀15 → 1: 2: 4

𝑀20 → 1: 1.5: 3

𝑀25 → 1: 1: 2

Design Bond strength of M20

Development length is inversely proportional to

Effective depth d=400mm, depth of neutral axis for Fe 250 ?

• 0.53d= 0.53 x 400 = 212 mm

Percentage of alumina in Brick?

Shear Stress is which type of stress

• Normal?

• Bending?

•Tangential

Shear Modulus is…?

Total length of bar is hook is provided on both ends?• Generally, hook length is

taken as 9d for one side.

• Where d = Diameter of the bar.

• The total length of stirrups = Total length of the bar + 2 x hook length (for two hooks)

• = L + 2 x 9d

• = L + 18d.

IS code used for Wind loads?

Loads and Forces

IS code for Design mix?

• IS 456: 2000 RCC

• IS 800: 2007 Steel (2007-LSM, 1984-WSM)

• IS 1343 Pre Stress Concrete

• IS 10262 Design Mix

• IS 383 Fine and Coarse Aggregate

• IS 875 Design Load for buildings and structures


IS code for Shape of the brick?

Scrap value of building is generally taken as..?

• 10% of the building cost.

Duty delta related 2 Questions

Compressive strength of grade 53 cement?

• 53 MPa

Unit of Kinematic Viscosity?

Shift of Transition curve

Length of Transition curve depends on…?

Longitudinal friction coefficient…?

Bernoulli’s equation assumption?

Bulk Modulus…?

Deflection due to point load in cantilever beam?

Extra widening provided due to off tracking?

Pointing is measured in…?

One question related to plaster quantity on both sides of wall

Geodetic surveying considers…

• Curved surface

• Inclined surface

• Declined surface

Process of Production of natural stone?

The component of road provided for drainage..?

Formula of Lime ?

• CaO

Plasticity index/Flow index is… ?

Toughness index

Type of cement used in underwater construction?• Portland cement.

• Quick Setting Cement

Coefficient of Venturimeter?

Gunter’s chain in metre?

• 20.1168 m

Edge and End distance for Flame Cut?

Combined correction for curvature and Refraction…?

Shear stress is directly proportion to strain in which fluid?

Soil formed due to glaciers?

• Drift

Maximum discharge for economical circular section

Stale Sewage has colour…?

• In about 6 hours, with all the dissolved for free oxygen present in the sewage being practically exhausted due to aerobic decomposition, the sewage becomes stale and its colour becomes light-to-medium grey

Degree of curve for 30 m chain..?

Deductions for opening..?

Most accurate method for water content in lab?• Oven drying method

Standard deviation is directly proportional to…?• Mean Strength

Longitudinal movement of rail…?

• Creep

Estimation by centre line method?

• Length, Width and height given

Slope of bending moment gives?

In gauge pressure, value of atmospheric pressure is…? 0

Cover for RCC column

Spillway related question

• A dam built with a crest to allow the overflow of water. Also known as overfall dam; spillway dam

Shape of Ridge line

• U or V

Core dia for circular column when eccentricity e is given • D/4

Other Questions

• Water absorptions for roof tiles

• Submerged weight per unit volume is…?

• Offsets are.. (Perpendicular and oblique)

• Kaolinite is an ore of Aluminum

• Liquidity index-plasticity limit?

• Maximum size for spacing of tie bars

• Level Surface Definition

• Book value

• Salvage Value

Documents

· 2019-09-24 · Minimum Shear Reinforcement •Clause 26.5.1.5 of IS 456 stipulates that the maximum spacing of shear reinforcement measured along the axis of the member shall