All Formulae Of Civil Engineering 27042022.pdf

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SANDEEP JYANI
❏ CIVIL ENGINEERING
EDUCATOR

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CIVIL ENGINEERING
SANDEEP JYANI

Fineness Modulus Test
Aggregate Type FM
Fine Aggregate (Sand) Fine Sand 2-2.5
Medium Sand 2.5-3.0
Coarse Sand 3.0-3.5
Coarse Aggregate 5.5-8.0
All in aggregate 3.5-7.5
𝑭𝒊𝒏𝒆 𝒂𝒈𝒈𝒓𝒆𝒈𝒂𝒕𝒆𝒔
𝑪𝒐𝒂𝒓𝒔𝒆 𝒂𝒈𝒈𝒓𝒆𝒈𝒂𝒕𝒆𝒔
=
𝑭𝑴𝑪𝑶𝑼𝑹𝑺𝑬 − 𝑭𝑴𝑴𝑰𝑿
𝑭𝑴𝑴𝑰𝑿 − 𝑭𝑴𝑭𝑰𝑵𝑬
𝑊ℎ𝑒𝑛 𝑓𝑖𝑛𝑒 𝑎𝑛𝑑 𝑐𝑜𝑎𝑟𝑠𝑒
𝑎𝑔𝑔𝑟𝑒𝑔𝑎𝑡𝑒𝑠 𝑎𝑟𝑒 𝑚𝑖𝑥𝑒𝑑 𝑡𝑜𝑔𝑒𝑡ℎ𝑒𝑟,

CIVIL ENGINEERING
Civil Engineering by Sandeep Jyani
ALL FORMULAE OF CIVIL ENGINEERING

CIVIL ENGINEERING
𝐹𝑖𝑛𝑒 𝑎𝑔𝑔𝑟𝑒𝑔𝑎𝑡𝑒𝑠
𝐶𝑜𝑎𝑟𝑠𝑒 𝑎𝑔𝑔𝑟𝑒𝑔𝑎𝑡𝑒𝑠
=
𝐹𝑀𝐶𝑂𝑈𝑅𝑆𝐸 − 𝐹𝑀𝑀𝐼𝑋
𝐹𝑀𝑀𝐼𝑋 − 𝐹𝑀𝐹𝐼𝑁𝐸
⇒
𝐹𝑖𝑛𝑒 𝑎𝑔𝑔𝑟𝑒𝑔𝑎𝑡𝑒𝑠
𝐶𝑜𝑎𝑟𝑠𝑒 𝑎𝑔𝑔𝑟𝑒𝑔𝑎𝑡𝑒𝑠
=
6.8 − 5.4
5.4 − 2.6
= 0.5

• This test is not applicable of the
aggregates having size less than 6.3
mm.
• In order to perform this test,
sufficient quantity of aggregates are
taken such that 200 hundred pieces
of each fraction can be tested.
𝐋𝐄𝐍𝐆𝐓𝐇 𝐆𝐀𝐔𝐆𝐄 𝐎𝐅 𝐄𝐋𝐎𝐍𝐆𝐀𝐓𝐈𝐎𝐍 𝐓𝐄𝐒𝐓
𝐓𝐇𝐈𝐂𝐊𝐍𝐄𝐒𝐒 𝐆𝐀𝐔𝐆𝐄 𝐎𝐅 𝐅𝐋𝐀𝐊𝐈𝐍𝐄𝐒𝐒 𝐓𝐄𝐒𝐓
Flakiness Index and Elongation Test
𝑭𝑰 =
𝑾𝒕 𝒐𝒇 𝒂𝒈𝒈𝒓𝒆𝒈𝒂𝒕𝒆 𝐩𝐚𝐬𝐬𝐢𝐧𝐠 𝒕𝒉𝒓𝒐𝒖𝒈𝒉 𝒔𝒍𝒐𝒕 𝒐𝒇 𝐭𝐡𝐢𝐜𝐤𝐧𝐞𝐬𝐬 𝒈𝒂𝒖𝒈𝒆
𝑻𝒐𝒕𝒂𝒍 𝑾𝒆𝒊𝒈𝒉𝒕 𝒐𝒇 𝒔𝒂𝒎𝒑𝒍𝒆
× 𝟏𝟎𝟎
𝑬𝑰 =
𝑾𝒕 𝒐𝒇 𝒂𝒈𝒈𝒓𝒆𝒈𝒂𝒕𝒆 𝐫𝐞𝐭𝐚𝐢𝐧𝐞𝐝 𝐨𝐧 𝐥𝐞𝐧𝐠𝐭𝐡 𝒈𝒂𝒖𝒈𝒆
𝑻𝒐𝒕𝒂𝒍 𝑾𝒆𝒊𝒈𝒉𝒕 𝒐𝒇 𝒔𝒂𝒎𝒑𝒍𝒆
× 𝟏𝟎𝟎

Strength of Concrete
1. Water/Cement Ratio
• Water Cement Ratio means the ratio between the
weight of water to the weight of cement used in
concrete mix.
• Normally water cement ratio falls under 0.4 to 0.6
as per IS Code 10262 (2009) for nominal mix (M10,
M15 …. M25)
• In 1918 Abrams presented his classic law in the
form:
• 𝑆 =
𝐴
𝐵𝑥
• where x =water/cement ratio by volume and for 28
days results the constants A and B are 14,000 lbs/sq.
in. and 7 respectively.

2. Gel/Space Ratio
• Gel space ratio is defined as the ratio of the volume of the
hydrated cement paste to the sum of volumes of the hydrated
cement and of the capillary pores.
• Gel/Space ratio 𝐱 =
𝐕𝐨𝐥𝐮𝐦𝐞 𝐨𝐟 𝐠𝐞𝐥
𝐬𝐩𝐚𝐜𝐞 𝐚𝐯𝐚𝐢𝐥𝐚𝐛𝐥𝐞
• Many researchers argued that Abrams water/cement ratio law
can only be called a rule and not a law because Abrams’
statement does not include many qualifications necessary for its
validity to call it a law.
• Instead of relating the strength to water/cement ratio, the
strength can be more correctly related to the solid products of
hydration of cement to the space available for formation of this
product. Powers and Brownyard have established the relationship
between the strength and gel/space ratio
• 𝐓𝐡𝐞𝐨𝐫𝐞𝐭𝐢𝐜𝐚𝐥 𝐬𝐭𝐫𝐞𝐧𝐠𝐭𝐡 = 𝟐𝟒𝟎 ×
𝐠𝐞𝐥
𝐬𝐩𝐚𝐜𝐞
𝟑
• Where 240 = intrinsic strength of gel in MPa

Que: If gel space ratio of a concrete is
given as 0.756, what is its theoretical
strength?
𝐓𝐡𝐞𝐨𝐫𝐞𝐭𝐢𝐜𝐚𝐥 𝐬𝐭𝐫𝐞𝐧𝐠𝐭𝐡 = 𝟐𝟒𝟎 ×
𝐠𝐞𝐥
𝟑

Que: If gel space ratio of a concrete is
given as 0.756, what is its theoretical
strength?
𝐓𝐡𝐞𝐨𝐫𝐞𝐭𝐢𝐜𝐚𝐥 𝐬𝐭𝐫𝐞𝐧𝐠𝐭𝐡 = 𝟐𝟒𝟎 ×
𝐠𝐞𝐥
𝟑
= 𝟐𝟒𝟎 × 𝟎. 𝟕𝟓𝟔 𝟑
= 𝟏𝟎𝟑. 𝟕𝟐 𝐍/𝐦𝐦𝟐

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II. Strength Of Concrete
2. 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)
𝐌𝐨𝐝𝐮𝐥𝐮𝐬 𝐨𝐟 𝐫𝐮𝐩𝐭𝐮𝐫𝐞 =
𝐩𝐥
𝐛𝐝𝟐
𝒑 = 𝒎𝒂𝒙𝒊𝒎𝒖𝒎 𝒍𝒐𝒂𝒅 𝒂𝒑𝒑𝒍𝒊𝒆𝒅
𝒍 = 𝒍𝒆𝒏𝒈𝒕𝒉 𝒐𝒇 𝒕𝒉𝒆 𝒔𝒖𝒑𝒑𝒐𝒓𝒕𝒆𝒅 𝒔𝒑𝒂𝒏
𝒃 = 𝒘𝒊𝒅𝒕𝒉 𝒐𝒇 𝒔𝒑𝒆𝒄𝒊𝒎𝒆𝒏
𝒅 = 𝒅𝒆𝒑𝒕𝒉 𝒐𝒇 𝒔𝒑𝒆𝒄𝒊𝒎𝒆𝒏
𝑳
𝟕𝟎𝒄𝒎
𝟏𝟓 𝒄𝒎
𝟏𝟓 𝒄𝒎
𝐏
Modulus of Rupture 𝒇𝒄𝒓 = 𝟎. 𝟕 𝒇𝒄𝒌
D𝐢𝐫𝐞𝐜𝐭 𝐓𝐞𝐧𝒔𝒊𝒍𝒆 𝑺𝒕𝒓𝒆𝒏𝒈𝒕𝒉 = 𝟎. 𝟓 𝒕𝒐 𝟎. 𝟔𝟐𝟓 𝒇𝒄𝒓

II. Strength Of Concrete
3. Split Tensile Strength of Concrete
• The length of cylinder varies from one to two
diameters. Normally the test cylinder is 150
mm diameter and 300 mm long
• The test consists of applying compressive line
loads along the diameter until it fails
• 𝑺𝒑𝒍𝒊𝒕 𝑻𝒆𝒏𝒔𝒊𝒍𝒆 𝑺𝒕𝒓𝒆𝒏𝒈𝒕𝒉 =
𝟐𝑷
𝝅𝑫𝑳
𝑷 = 𝒎𝒂𝒙𝒊𝒎𝒖𝒎 𝒍𝒐𝒂𝒅 𝒂𝒑𝒑𝒍𝒊𝒆𝒅
𝑳 = 𝒍𝒆𝒏𝒈𝒕𝒉 𝒐𝒇 𝒔𝒑𝒆𝒄𝒊𝒎𝒆𝒏
𝑫 = 𝒅𝒊𝒂 𝒐𝒇 𝒔𝒑𝒆𝒄𝒊𝒎𝒆𝒏

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Maturity Concept of Concrete
• The strength development of concrete depends on
both time and temperature it can be said that
strength is a function of summation of product of
time and temperature. This summation is called
maturity of concrete
Maturity = Σ (time × temperature)
• Hydration of concrete continues to take place upto
about –11 C. Therefore, –11 C is taken as a datum
line for computing maturity

Maturity Concept of Concrete
• Exp: A sample of concrete cured at 20 C
for 28 days is taken as fully matured
concrete. Its maturity would be equal to…?
= (28 × 24) × [20 – (– 11)]
Maturity = Σ (time × temperature)
= 20832°𝐶 ℎ

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Tensile Strength of Concrete in Flexure
𝒇𝒄𝒕 = 𝟎. 𝟕 𝒇𝒄𝒌
where 𝒇𝒄𝒌 is the characteristic compressive
strength of concrete in N/mm2

Effect of Creep on Young’s Modulus of Elasticity
• Long term Young’s Modulus of Elasticity of concrete
𝑬𝑪𝑳 =
𝟓𝟎𝟎𝟎 𝒇𝒄𝒌
𝟏 + 𝜽
Where 𝜽 is the creep coefficient
Creep Strain is strain which occurs due to continuous loading and temperature
effect for longer duration which may cause permanent deformation
Age of Loading 𝛉
7 days 2.2
28 days 1.6
1 year 1.1

Tensile Strength of Concrete in Flexure
𝒇𝒄𝒕 = 𝟎. 𝟕 𝒇𝒄𝒌

Probabilistic Curve
Frequency
distribution
Or
Probability
distribution
(No. of times test
results obtained)
𝒇𝒄𝒌 𝒇𝒎
𝒇𝒎 − 𝟏. 𝟔𝟒𝝈
5%
𝝈 = 𝒔𝒕𝒂𝒏𝒅𝒂𝒓𝒅 𝒅𝒆𝒗𝒊𝒂𝒕𝒊𝒐𝒏
(𝒅𝒆𝒑𝒆𝒏𝒅𝒔 𝒐𝒏 𝒈𝒓𝒂𝒅𝒆 𝒐𝒇 𝒄𝒐𝒏𝒄𝒓𝒆𝒕𝒆)
𝒇𝒎 = 𝒇𝒄𝒌 + 𝟏. 𝟔𝟒𝝈
=> 𝒇𝒎 = 𝒇𝒄𝒌 + 𝟏. 𝟔𝟒𝝈
𝒇𝒎 = 𝒎𝒆𝒂𝒏 𝒔𝒕𝒓𝒆𝒏𝒈𝒕𝒉
𝒇𝒄𝒌 = 𝒄𝒉𝒂𝒓𝒂𝒄𝒕𝒆𝒓𝒊𝒔𝒕𝒊𝒄 𝒔𝒕𝒓𝒆𝒏𝒈𝒕𝒉

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𝒎 =
𝟐𝟖𝟎
𝟑𝝈𝒄𝒃𝒄
Grade Modular Ratio
M-15 19
M-20 13
M-25 11
M-30 9
M-35 8
M-40 7
Modular Ratio

ANALYSIS OF RCC SECTION
1. 𝑩 × 𝒙𝒂 ×
𝒙𝒂
𝟐
= 𝒎𝑨𝒔𝒕 × 𝒅 − 𝒙𝒂 𝑭𝒊𝒏𝒅 𝒙𝒂
2. 𝑥𝑐 =
m𝝈𝒄𝒃𝒄
m𝝈𝒄𝒃𝒄
+𝝈𝒔𝒕
× 𝑑 (𝑥𝑐 = kd)
3. Different types of Sections
• Balanced Section
• 𝒙𝒂 = 𝒙𝒄
• Ca= C = 𝝈𝒄𝒃𝒄
• ta=t = 𝝈𝒔𝒕
• Under Reinforced Section
• 𝒙𝒂 < 𝒙𝒄
• Ca < 𝝈𝒄𝒃𝒄
• ta= 𝝈𝒔𝒕
• Over Reinforced Section
• 𝒙𝒂 > 𝒙𝒄
• Ca = 𝝈𝒄𝒃𝒄
• ta< 𝝈𝒔𝒕

ANALYSIS OF RCC SECTION
4. Moment of resistance (Balanced Section)
• 𝑀𝑅 𝑐𝑜𝑚𝑝 = 𝑏 × 𝒙𝒂 ×
𝝈𝒄𝒃𝒄
2
× 𝑑 −
𝑥𝑎
3
• 𝑀𝑅 𝑡𝑒𝑛𝑠𝑖𝑜𝑛 = 𝝈𝒔𝒕 × 𝐴𝑠𝑡 × 𝑑 −
𝑥𝑎
3
5. Moment of resistance (Under reinforced Section)
𝑪𝒂
2
× 𝑑 −
𝑥𝑎
3
• 𝑀𝑅 𝑡𝑒𝑛𝑠𝑖𝑜𝑛 = 𝝈𝒔𝒕 × 𝐴𝑠𝑡 × 𝑑 −
𝑥𝑎
3
Always find MR
from tension side because (ta= 𝝈𝒔𝒕)
6. Moment of resistance (Over reinforced Section)
𝝈𝒄𝒃𝒄
2
× 𝑑 −
𝑥𝑎
3
Always find MR
from compression side as (Ca= 𝝈𝒄𝒃𝒄)
• 𝑀𝑅 𝑡𝑒𝑛𝑠𝑖𝑜𝑛 = 𝒕𝒂 × 𝐴𝑠𝑡 × 𝑑 −
𝑥𝑎
3

Design Formula for balanced section
=> 𝑀𝑅 𝑐𝑜𝑚𝑝 =
1
2
𝑐𝑗𝑘 × 𝐵𝑑2
𝑸
𝑸 =
1
2
𝑐𝑗𝑘
𝝈𝒄𝒃𝒄
1 −
𝑘
3
m𝝈𝒄𝒃𝒄
m𝝈𝒄𝒃𝒄 + 𝝈𝒔𝒕

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T - Beam
B
Bf
Flange
Web
Df
Effective Width
Bw

Partial safety factors
2. Design Strength
• Design strength of material is the ratio of characteristic strength of material to
partial factor of safety
𝑫𝒆𝒔𝒊𝒈𝒏 𝒔𝒕𝒓𝒆𝒏𝒈𝒕𝒉 𝒇𝒅 =
𝒄𝒉𝒂𝒓𝒂𝒄𝒕𝒆𝒓𝒊𝒔𝒕𝒊𝒄 𝒔𝒕𝒓𝒆𝒏𝒈𝒕𝒉 𝒐𝒇 𝒕𝒉𝒆 𝒎𝒂𝒕𝒆𝒓𝒊𝒂𝒍 (𝒇)
𝒑𝒂𝒓𝒕𝒊𝒂𝒍 𝒇𝒂𝒄𝒕𝒐𝒓 𝒐𝒇 𝒔𝒂𝒇𝒆𝒕𝒚 (𝜸𝒎)
• Clause 36.4.2 of IS 456 states that 𝛾𝑚 for concrete and steel should be taken as 1.5
and 1.15, respectively when assessing the strength of the structures or structural
members employing limit state of collapse.
• Partial safety factor for steel (1.15) is comparatively lower than that of concrete
(1.5) because the steel for reinforcement is produced in steel plants and
commercially available in specific diameters with expected better quality control
than that of concrete.

Partial safety factors
• 𝑭𝒐𝒓 𝑺𝒕𝒆𝒆𝒍 𝒇𝒅 =
𝒇𝒚
𝟏.𝟏𝟓
= 𝟎. 𝟖𝟕𝒇𝒚
• 𝑭𝒐𝒓 𝑪𝒐𝒏𝒄𝒓𝒆𝒕𝒆 𝒇𝒅 =
𝒇𝒄𝒌
𝟏.𝟓
= 𝟎. 𝟔𝟕𝒇𝒄𝒌
• In case of concrete the characteristic strength is calculated on the basis of test results on
150 mm standard cubes. But the concrete in the structure has different sizes. To take the
size effect into account, it is assumed that the concrete in the structure develops a strength
of 0.67 or (1/1.50) times the characteristic strength of cubes.
• Accordingly, in the calculation of strength employing the limit state of collapse, the
characteristic strength (fck) is first multiplied with 0.67 (size effect) and then divided by 1.5
(𝜸𝒎 for concrete) to have 0.446 fck as the maximum strength of concrete in the stress block.
• 𝑭𝒐𝒓 𝑪𝒐𝒏𝒄𝒓𝒆𝒕𝒆 𝒇𝒅 =
𝒇𝒄𝒌
𝟏.𝟓
× 𝟎. 𝟔𝟕 = 𝟎. 𝟒𝟒𝟔𝒇𝒄𝒌

Assumptions of Limit State Method
5. Tensile strength of concrete is neglected
6. Maximum strain in tension reinforcement in the section at failure
should not be less than
𝜺𝒔 ≥
𝟎. 𝟖𝟕𝒇𝒚
𝑬𝒔
+ 𝟎. 𝟎𝟎𝟐 𝒐𝒓 𝜺𝒔 ≥
𝒇𝒚
𝟏. 𝟏𝟓𝑬𝒔
+ 𝟎. 𝟎𝟎𝟐
D
b
Neutral Axis
Ast d
𝜺𝒔 ≥
𝟎. 𝟖𝟕𝒇𝒚
𝑬𝒔
+ 𝟎. 𝟎𝟎𝟐
0.0035 0.446 fck
C
0.002
𝟎. 𝟖𝟕𝒇𝒚

Limit State Method
Neutral Axis
𝜺𝒔 ≥
𝟎. 𝟖𝟕𝒇𝒚
𝑬𝒔
+ 𝟎. 𝟎𝟎𝟐
0.0035 0.446 fck
0.002
𝟎. 𝟖𝟕𝒇𝒚
𝑥𝑎 ⇒ 𝑥𝑢
𝑥𝑐 ⇒ 𝑥𝑢 𝑙𝑖𝑚
𝑀𝑅 ⇒ 𝑀𝑢
𝜎 ⇒ 𝑓
𝑈𝑠𝑖𝑛𝑔 𝑠𝑖𝑚𝑖𝑙𝑎𝑟 𝑡𝑟𝑖𝑎𝑛𝑔𝑙𝑒 𝑝𝑟𝑜𝑝𝑒𝑟𝑡𝑦,
𝒙𝒖
0.0035
=
𝒙𝟏
0.002
=>𝒙𝟏 =
4
7
𝒙𝒖 , so
=>𝒙𝟐 =
3
7
𝒙𝒖
𝒙𝟐
𝒙𝟏
𝒙𝒖
𝒙𝒖
𝟑
𝟕
𝒙𝒖
𝟒
𝟕
𝒙𝒖
𝑩
𝑫
𝒅
𝑨𝒔𝒕

Analysis of Singly Reinforced Section
𝑵𝒆𝒖𝒕𝒓𝒂𝒍 𝑨𝒙𝒊𝒔
𝜺𝒔 ≥
𝟎. 𝟖𝟕𝒇𝒚
𝑬𝒔
+ 𝟎. 𝟎𝟎𝟐
0.0035 0.446 fck
𝑪𝟏
0.002
𝟎. 𝟖𝟕𝒇𝒚
𝒙𝒖
𝟑
𝟕
𝒙𝒖
𝟒
𝟕
𝒙𝒖 𝑪𝟐
𝑪
𝑩
𝑫 𝒅
𝑨𝒔𝒕
𝑴𝒐𝒎𝒆𝒏𝒕 𝒐𝒇 𝒓𝒆𝒔𝒊𝒔𝒕𝒂𝒏𝒄𝒆 𝒘. 𝒓. 𝒕 𝒄𝒐𝒏𝒄𝒓𝒆𝒕𝒆 𝒄𝒐𝒎𝒑𝒓𝒆𝒔𝒔𝒊𝒐𝒏
𝑴𝒖 𝒄 = 𝒄𝒐𝒎𝒑𝒓𝒆𝒔𝒔𝒊𝒗𝒆 𝒇𝒐𝒓𝒄𝒆 × 𝒍𝒆𝒗𝒆𝒓 𝒂𝒓𝒎
𝑴𝒖 𝒄 = 𝑪 × 𝒍𝒆𝒗𝒆𝒓 𝒂𝒓𝒎
𝒄𝟏 𝒄𝟐
𝒄𝟏 =
𝟑
𝟕
𝒙𝒖 × 𝑩 × 𝟎. 𝟒𝟓𝒇𝒄𝒌 = 𝟎. 𝟏𝟗𝟐𝒇𝒄𝒌𝒙𝒖𝑩
𝒂𝒓𝒆𝒂 𝒂𝒗𝒈 𝒔𝒕𝒓𝒆𝒔𝒔
𝒄𝟐 =
𝟐
𝟑
×
𝟒
𝟕
𝒙𝒖 × 𝑩 × 𝟎. 𝟒𝟓𝒇𝒄𝒌 = 𝟎. 𝟏𝟕𝟏𝒇𝒄𝒌𝒙𝒖𝑩
𝑪 = 𝒄𝟏 + 𝒄𝟐
⇒ 𝑪 = 𝟎. 𝟑𝟔𝒇𝒄𝒌𝒙𝒖𝑩

𝜺𝒔 ≥
𝟎. 𝟖𝟕𝒇𝒚
𝑬𝒔
+ 𝟎. 𝟎𝟎𝟐
0.0035 0.446 fck
𝑪𝟏
0.002
𝟎. 𝟖𝟕𝒇𝒚
𝒙𝒖
𝟑
𝟕
𝒙𝒖
𝟒
𝟕
𝒙𝒖
𝑪𝟐
𝑪
𝒃
𝑫 𝒅
𝑨𝒔𝒕
𝒍𝒆𝒗𝒆𝒓 𝒂𝒓𝒎 = (𝒅 − ഥ
𝒚)
ഥ
𝒚
⇒ ഥ
𝒚 =
𝒄𝟏𝒚𝟏 + 𝒄𝟐𝒚𝟐
𝒄𝟏 + 𝒄𝟐
𝒚𝟏 =
𝟏
𝟐
𝟑
𝟕
𝒙𝒖 =
𝟑
𝟏𝟒
𝒙𝒖
𝒚𝟐 =
𝟑
𝟕
𝒙𝒖 +
𝟑
𝟖
×
𝟒
𝟕
𝒙𝒖 =
𝟗
𝟏𝟒
𝒙𝒖
𝑪𝑮 𝒐𝒇 𝒑𝒂𝒓𝒂𝒃𝒐𝒍𝒂
𝟑
𝟖
𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆 𝒇𝒓𝒐𝒎 𝒕𝒐𝒑
ഥ
𝒚 =
(𝟎. 𝟏𝟗𝟐𝒇𝒄𝒌𝒙𝒖𝑩 )(
𝟑
𝟏𝟒
𝒙𝒖) + (𝟎. 𝟏𝟕𝟏𝒇𝒄𝒌𝒙𝒖𝑩 )(
𝟗
𝟏𝟒
𝒙𝒖)
𝟎. 𝟏𝟗𝟐𝒇𝒄𝒌𝒙𝒖𝑩 + (𝟎. 𝟏𝟕𝟏𝒇𝒄𝒌𝒙𝒖𝑩 )
⇒ ഥ
𝒚 =≈ 𝟎. 𝟒𝟐𝒙𝒖

𝜺𝒔 ≥
𝟎. 𝟖𝟕𝒇𝒚
𝑬𝒔
+ 𝟎. 𝟎𝟎𝟐
0.0035 0.446 fck
𝑪𝟏
0.002
𝟎. 𝟖𝟕𝒇𝒚
𝒙𝒖
𝟑
𝟕
𝒙𝒖
𝟒
𝟕
𝒙𝒖
𝑪𝟐
𝑪
𝑩
𝑫 𝒅
𝑨𝒔𝒕
𝒍𝒆𝒗𝒆𝒓 𝒂𝒓𝒎 = (𝒅 − ഥ
𝒚)
ഥ
𝒚
𝒍𝒆𝒗𝒆𝒓 𝒂𝒓𝒎 = (𝒅 − 𝟎. 𝟒𝟐𝒙𝒖)
𝑴𝒖 𝒄 = 𝟎. 𝟑𝟔𝒇𝒄𝒌𝒙𝒖𝑩 × (𝒅 − 𝟎. 𝟒𝟐𝒙𝒖)
𝑴𝒖 𝒕 = 𝒕𝒆𝒏𝒔𝒊𝒍𝒆 𝒇𝒐𝒓𝒄𝒆 × 𝒍𝒆𝒗𝒆𝒓 𝒂𝒓𝒎
𝑴𝒖 𝒕 = 𝟎. 𝟖𝟕𝒇𝒚𝑨𝒔𝒕 × (𝒅 − 𝟎. 𝟒𝟐𝒙𝒖)

Actual Depth of Neutral Axis
The actual depth of Neutral axis is
calculated by equating the compression
force to tensile force
𝑪 = 𝑻
⇒ 𝟎. 𝟑𝟔𝒇𝒄𝒌 𝒙𝒖 𝑩 = 𝟎. 𝟖𝟕𝒇𝒚 𝑨𝒔𝒕
⇒ 𝒙𝒖 =
𝟎. 𝟖𝟕 𝒇𝒚 𝑨𝒔𝒕
𝟎. 𝟑𝟔 𝒇𝒄𝒌 𝑩

Limiting Depth of Neutral Axis
Since stress diagram is not linear, hence we
use corresponding strain diagram
b
Ast d
𝜺𝒔 ≥
𝟎. 𝟖𝟕𝒇𝒚
𝑬𝒔
+ 𝟎. 𝟎𝟎𝟐
0.0035
0.002
𝒙𝒖 𝒍𝒊𝒎
𝑼𝒔𝒊𝒏𝒈 𝒔𝒊𝒎𝒊𝒍𝒂𝒓 𝒕𝒓𝒊𝒂𝒏𝒈𝒍𝒆 𝒑𝒓𝒐𝒑𝒆𝒓𝒕𝒚,
𝟎. 𝟎𝟎𝟑𝟓
=
𝟎. 𝟖𝟕𝒇𝒚
𝑬𝒔
+ 𝟎. 𝟎𝟎𝟐
𝒅 − 𝒙𝒖 𝒍𝒊𝒎
⇒
𝒅 − 𝒙𝒖 𝒍𝒊𝒎
=
𝟎. 𝟖𝟕𝒇𝒚
𝑬𝒔
+ 𝟎. 𝟎𝟎𝟐
𝟎. 𝟎𝟎𝟑𝟓
⇒
𝒅
=
𝟎. 𝟖𝟕𝒇𝒚
𝑬𝒔
+ 𝟎. 𝟎𝟎𝟐
𝟎. 𝟎𝟎𝟑𝟓
+ 𝟏
⇒ 𝒙𝒖 𝒍𝒊𝒎 =
𝟕𝟎𝟎
𝟏𝟏𝟎𝟎 + 𝟎. 𝟖𝟕𝒇𝒚
× 𝒅
⇒ 𝒙𝒖 𝒍𝒊𝒎 = 𝒌 × 𝒅
d

Maximum Depth of Neutral Axis
fy N/mm2 xulim
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

Design of Beam
Assuming overall depth as 𝐷 = 2𝐵 𝑜𝑟 𝐿/10
Step 1: Load and Bending moment
𝑰𝒇 𝒘 = 𝒘𝒐𝒓𝒌𝒊𝒏𝒈 𝒍𝒐𝒂𝒅
𝑭𝒂𝒄𝒕𝒐𝒓𝒆𝒅 𝒍𝒐𝒂𝒅 ⇒ 𝒘𝒖 = 𝟏. 𝟓𝒘
Note:
(i) Total factored load = 𝑤𝑢
(ii) If total working load = 𝑤, then convert into factored load 𝑤𝑢 = 1.5𝑤
(iii) When only live load is given, then dead load/self weight of the beam is considered to find
working load and convert into factored load i.e.
• 𝒘 = 𝑫𝑳 + 𝑳𝑳
• 𝑭𝒂𝒄𝒕𝒐𝒓𝒆𝒅 𝒍𝒐𝒂𝒅 ⇒ 𝒘𝒖 = 𝟏. 𝟓𝒘
𝑴𝒂𝒙𝒊𝒎𝒖𝒎 𝑩𝒆𝒏𝒅𝒊𝒏𝒈 𝒎𝒐𝒎𝒆𝒏𝒕 =
𝒘𝒖𝒍𝟐
𝟖

Design of Beam
Step 2: Check for depth assumed:
𝑴𝒖 𝒍𝒊𝒎 = 𝑸 𝑩𝒅𝟐
⇒ 𝒅𝒓𝒆𝒒𝒅=
𝑴𝒖 𝒍𝒊𝒎
𝑸𝑩
𝑫𝒓𝒆𝒒𝒅 = 𝒅𝒓𝒆𝒒𝒅 + 𝒆𝒇𝒇𝒆𝒄𝒕𝒊𝒗𝒆 𝒄𝒐𝒗𝒆𝒓
If 𝑫𝒂𝒔𝒔𝒖𝒎𝒆𝒅 > 𝑫𝒓𝒆𝒒𝒅 ⇒ 𝑫𝒆𝒔𝒊𝒈𝒏 𝒊𝒔 𝒐𝒌𝒂𝒚
If 𝑫𝒂𝒔𝒔𝒖𝒎𝒆𝒅 < 𝑫𝒓𝒆𝒒𝒅 ⇒ 𝑹𝒆𝒅𝒆𝒔𝒊𝒈𝒏 𝒐𝒗𝒆𝒓𝒂𝒍𝒍 𝒅𝒆𝒑𝒕𝒉

Design Formula
We always try to design the limiting section
PROOF:
𝑴𝒖 𝒄 = 𝟎. 𝟑𝟔𝒇𝒄𝒌𝑩 𝒙𝒖 𝒍𝒊𝒎(𝒅 − 𝟎. 𝟒𝟐𝒙𝒖 𝒍𝒊𝒎)
= 𝑪 𝑩 𝒌𝒅 (𝒅 − 𝟎. 𝟒𝟐𝒌𝒅)
= 𝑪 𝑩 𝒌𝒅𝟐 (𝟏 − 𝟎. 𝟒𝟐𝒌)
= 𝑪 𝑩 𝒌𝒅𝟐 𝒋
= 𝑪𝒋𝒌 𝑩𝒅𝟐
= 𝑸 𝑩𝒅𝟐
𝒅 =
𝑴𝒖
𝑸𝑩

Note: Value of Q 𝑴𝒖 𝒄 = 𝟎. 𝟑𝟔𝒇𝒄𝒌𝑩 𝒙𝒖 𝒍𝒊𝒎(𝒅 − 𝟎. 𝟒𝟐𝒙𝒖 𝒍𝒊𝒎)
⇒ 𝑴𝒖 𝒄 = 𝟎. 𝟑𝟔𝒇𝒄𝒌𝑩 𝒌𝒅(𝒅 − 𝟎. 𝟒𝟐𝒌𝒅)
⇒ 𝑴𝒖 𝒄 = 𝟎. 𝟑𝟔𝒇𝒄𝒌𝑩 𝒌𝒅𝟐(𝟏 − 𝟎. 𝟒𝟐𝒌)
⇒ 𝑴𝒖 𝒄 = 𝟎. 𝟑𝟔𝒇𝒄𝒌 𝒌 𝟏 − 𝟎. 𝟒𝟐𝒌 𝑩𝒅𝟐
𝑸
𝑭𝒐𝒓 𝑭𝒆 𝟐𝟓𝟎 ⇒ 𝑸 = 𝟎. 𝟑𝟔𝒇𝒄𝒌 (𝟎. 𝟓𝟑) 𝟏 − 𝟎. 𝟒𝟐(𝟎. 𝟓𝟑)
= 𝟎. 𝟏𝟒𝟖𝒇𝒄𝒌
𝑭𝒐𝒓 𝑭𝒆 𝟒𝟏𝟓 ⇒ 𝑸 = 𝟎. 𝟑𝟔𝒇𝒄𝒌 (𝟎. 𝟒𝟖) 𝟏 − 𝟎. 𝟒𝟐(𝟎. 𝟒𝟖)
= 𝟎. 𝟏𝟑𝟖𝒇𝒄𝒌
𝑭𝒐𝒓 𝑭𝒆 𝟓𝟎𝟎 ⇒ 𝑸 = 𝟎. 𝟑𝟔𝒇𝒄𝒌 (𝟎. 𝟒𝟔) 𝟏 − 𝟎. 𝟒𝟐(𝟎. 𝟒𝟔)
= 𝟎. 𝟏𝟑𝟒𝒇𝒄𝒌

Design Formula
Step 3: Area of Steel Ast
𝑴𝒖 𝑻 = 𝟎. 𝟖𝟕 𝒇𝒚 𝑨𝒔𝒕 (𝒅 − 𝟎. 𝟒𝟐𝒙𝒖 𝒍𝒊𝒎)
⇒ 𝑨𝒔𝒕 =
𝑴𝒖 𝑻
𝟎.𝟖𝟕𝒇𝒚(𝒅−𝟎.𝟒𝟐𝒌𝒅)
⇒ 𝑨𝒔𝒕 =
𝑴𝒖 𝑻
𝟎. 𝟖𝟕𝒇𝒚𝒅(𝟏 − 𝟎. 𝟒𝟐𝒌)
⇒ 𝑨𝒔𝒕 =
𝑴𝒖 𝑻
𝟎. 𝟖𝟕𝒇𝒚𝒅 𝒋

Limiting values of Tension steel
• Minimum area of tension reinforcement
should not be less than
𝑨𝟎
𝒃𝒅
=
𝟎. 𝟖𝟓
𝒇𝒚
𝐴0=minimum area of tension r/f
fy = characteristic strength of steel in N/mm2
• Maximum area of tension reinforcement
should not be greater than 4 % of the gross
cross sectional area to avoid difficulty in
placing and compacting concrete properly
in framework

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 + 𝒅
II. 𝒍 (centre to centre distance between supports)
Where 𝒍 = centre to centre distance between supports
𝑳𝒄 = clear span
𝒅 = 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 + 𝒅
b) 𝒍 (centre to centre distance between supports)
Where 𝒍 = centre to centre distance between supports
𝑳𝒄 = clear span
𝒅 = 𝒆𝒇𝒇𝒆𝒄𝒕𝒊𝒗𝒆 𝒅𝒆𝒑𝒕𝒉 𝒐𝒇 𝒃𝒆𝒂𝒎 𝒐𝒓 𝒔𝒍𝒂𝒃
ii. If width of support 𝒕𝒔 >
𝑳𝒄
𝟏𝟐
or 600 mm , then effective span is taken as lesser of
…
a) 𝒍 = 𝑳𝒄 + 𝟎. 𝟓 𝒅
b) 𝒍 = 𝑳𝒄 + 𝟎. 𝟓 𝒕𝒔
𝐿𝑐
𝒍 𝒍
𝐿𝑐
𝒕𝒔

T - Beam
B
Bf
Flange
Web
Df
d
Bw

𝑩𝒇
Flange
Web
𝐷𝑓
Bw
ANALYSIS OF T- BEAM
CASE 1: When actual depth of Neutral
axis lies in the FLANGE PORTION
Step 1: Calculate actual depth of
Neutral Axis
𝑩
𝑨𝒔𝒕
𝐶 = 𝑇
⇒ 0.36 𝑓𝑐𝑘𝑥𝑢𝐵𝑓 = 0.87𝑓𝑦𝐴𝑠𝑡
⇒ 𝑥𝑢=
0.87𝑓𝑦𝐴𝑠𝑡
0.36 𝑓𝑐𝑘𝐵𝑓
(𝑤ℎ𝑖𝑐ℎ 𝑠ℎ𝑜𝑢𝑙𝑑 𝑏𝑒 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 𝐷𝑓)
Step 2: Calculate Limiting depth of Neutral Axis
𝑥𝑢𝑙𝑖𝑚 = ቐ
0.53𝑑 𝑓𝑜𝑟 𝐹𝑒 250
0.48𝑑 𝑓𝑜𝑟 𝐹𝑒 415
0.46𝑑 𝑓𝑜𝑟 𝐹𝑒 500
𝒙𝒖
𝟎. 𝟒𝟓𝒇𝒄𝒌
0.87𝑓𝑦

𝑩𝒇
Flange
Web
𝐷𝑓
Bw
ANALYSIS OF T- BEAM
CASE 1: When actual depth of Neutral
axis lies in the FLANGE PORTION
Step 3: Calculate Moment of Resistance
𝑩
𝑨𝒔𝒕
𝑀𝑢 𝑐 = 𝐵𝑓 × 0.45𝑓𝑐𝑘 ×
3
7
𝑥𝑢 +
2
3
× 0.45𝑓𝑐𝑘 ×
4
7
𝑥𝑢 𝑑 − 0.42𝑥𝑢
𝒙𝒖
𝟎. 𝟒𝟓𝒇𝒄𝒌
0.87𝑓𝑦
⇒ 𝑀𝑢 𝑐= 0.36𝑓𝑐𝑘𝑥𝑢𝐵𝑓 𝑑 − 0.42𝑥𝑢
𝑀𝑢 𝑇 = 0.87𝑓𝑦𝐴𝑠𝑡 𝑑 − 0.42𝑥𝑢
𝟑
𝟕
𝒙𝒖
𝟒
𝟕
𝒙𝒖

𝑩𝒇
Web
𝐷𝑓
Bw
ANALYSIS OF T- BEAM
CASE 2: When actual depth of Neutral axis lies in the
WEB PORTION
Sub Case A: When 𝐱𝐮> 𝐃𝐟 𝒂𝒏𝒅 𝐃𝐟 <
𝟑
𝟕
𝐱𝐮
Step 1: Calculate actual depth of Neutral Axis
𝑩
𝑨𝒔𝒕
𝐶 = 𝑇
⇒ 0.36 𝑓𝑐𝑘 × 𝐵𝑤× 𝑥𝑢 + 0.45𝑓𝑐𝑘(𝐵𝑓 − 𝐵𝑤)𝐷𝑓 = 0.87𝑓𝑦𝐴𝑠𝑡
𝑥𝑢𝑙𝑖𝑚
= ቐ
0.53𝑑 𝑓𝑜𝑟 𝐹𝑒 250
0.48𝑑 𝑓𝑜𝑟 𝐹𝑒 415
0.46𝑑 𝑓𝑜𝑟 𝐹𝑒 500
𝒙𝒖
𝟎. 𝟒𝟓𝒇𝒄𝒌
0.87𝑓𝑦
𝑀𝑢 𝑐 = 0.36𝑓𝑐𝑘𝑥𝑢𝐵𝑤 𝑑 − 0.42𝑥𝑢 + 0.45𝑓𝑐𝑘(𝐵𝑓 − 𝐵𝑤)𝐷𝑓 𝑑 −
𝐷𝑓
2
𝟑
𝟕
𝒙𝒖
𝟒
𝟕
𝒙𝒖

𝑩𝒇
Web
𝐷𝑓
Bw
ANALYSIS OF T- BEAM
CASE 2: When actual depth of Neutral axis lies in the WEB
PORTION
Sub Case B: When 𝐱𝐮> 𝐃𝐟 𝒂𝒏𝒅 𝐃𝐟 >
𝟑
𝟕
𝐱𝐮
According to IS code recommendation, when actual
depth of Neutral Axis lies in web portion and depth of
Flange is greater than
𝟑
𝟕
𝐱𝐮 then in that case equivalent
depth of Flange in stressed diagram
𝒚𝒇 = 𝟎. 𝟏𝟓𝐱𝐮 + 𝟎. 𝟔𝟓𝑫𝒇
𝑩
𝑨𝒔𝒕
𝐶 = 𝑇
⇒ 0.36 𝑓𝑐𝑘 × 𝐵𝑤× 𝑥𝑢 + 0.45𝑓𝑐𝑘 𝐵𝑓 − 𝐵𝑤 𝑦𝑓 = 0.87𝑓𝑦𝐴𝑠𝑡
𝑥𝑢𝑙𝑖𝑚 = ቐ
0.53𝑑 𝑓𝑜𝑟 𝐹𝑒 250
0.48𝑑 𝑓𝑜𝑟 𝐹𝑒 415
0.46𝑑 𝑓𝑜𝑟 𝐹𝑒 500
𝒙𝒖
𝟎. 𝟒𝟓𝒇𝒄𝒌
0.87𝑓𝑦
𝑀𝑢 𝑐 = 0.36𝑓𝑐𝑘𝑥𝑢𝐵𝑤 𝑑 − 0.42𝑥𝑢 + 0.45𝑓𝑐𝑘 𝐵𝑓 − 𝐵𝑤 𝒚𝑓 𝑑 −
𝒚𝒇
2
𝟒
𝟕
𝒙𝒖
Step 1: Calculate actual depth of Neutral Axis
𝒚𝒇
𝟑
𝟕
𝒙𝒖

𝑩𝒇
Web
𝐷𝑓
Bw
Design Formula OF T- BEAM
CASE 2: When actual depth of Neutral axis lies in the
WEB PORTION
Sub Case A: When 𝐱𝐮> 𝐃𝐟 𝒂𝒏𝒅 𝐃𝐟 <
𝟑
𝟕
𝐱𝐮
𝑩
𝑨𝒔𝒕
𝐶𝑤 = 𝑇1
⇒ 0.36 𝑓𝑐𝑘 × 𝐵𝑤× 𝑥𝑢 = 0.87𝑓𝑦𝐴𝑠𝑡𝒘
𝒙𝒖
𝟎. 𝟒𝟓𝒇𝒄𝒌
0.87𝑓𝑦
𝟑
𝟕
𝒙𝒖
𝟒
𝟕
𝒙𝒖
⇒ 𝐴𝑠𝑡𝒘 =
0.36 𝑓𝑐𝑘 × 𝐵𝑤× 𝑥𝑢
0.87𝑓𝑦
𝐶𝑓 = 𝑇2
⇒ 𝐴𝑠𝑡𝒇 =
𝐵𝑓 − 𝐵𝑤 0.45𝑓𝑐𝑘 × 𝐷𝑓
0.87𝑓𝑦

𝑩𝒇
Web
𝐷𝑓
Bw
ANALYSIS OF T- BEAM
CASE 2: When actual depth of Neutral axis lies in the WEB
PORTION
Sub Case A: When 𝐱𝐮> 𝐃𝐟 𝒂𝒏𝒅 𝐃𝐟 >
𝟑
𝟕
𝐱𝐮
According to IS code recommendation, when actual
depth of Neutral Axis lies in web portion and depth of
Flange is greater than
𝟑
𝟕
𝐱𝐮 then in that case equivalent
depth of Flange in stressed diagram
𝒚𝒇 = 𝟎. 𝟏𝟓𝐱𝐮 + 𝟎. 𝟔𝟓𝑫𝒇
𝑩
𝑨𝒔𝒕
𝒙𝒖
𝟎. 𝟒𝟓𝒇𝒄𝒌
0.87𝑓𝑦
𝟒
𝟕
𝒙𝒖
𝒚𝒇
𝟑
𝟕
𝒙𝒖
𝐴𝑠𝑡𝒘 =
0.36 𝑓𝑐𝑘 × 𝐵𝑤× 𝑥𝑢
0.87𝑓𝑦
𝐴𝑠𝑡𝒇 =
0.45𝑓𝑐𝑘 𝐵𝑓 − 𝐵𝑤 𝑦𝑓
0.87𝑓𝑦

Steps for Design of Shear Reinforcement
1. Find maximum shear force
• SSB =
𝑾𝑳
𝟐
𝒐𝒓
𝑾𝒖
𝑳
𝟐
2. Nominal Shear Stress
• 𝝉𝒗 =
𝑽
𝒃𝒅
𝒐𝒓
𝑽𝒖
𝒃𝒅
and 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
reinforcement
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.
(whichever is minimum out of these)

BEAMS:
1. Minimum tension reinforcement:
𝑨𝟎
𝒃𝒅
=
𝟎.𝟖𝟓
𝒇𝒚
2. The maximum reinforcement in tension or compression: should not exceed 0.04bD,
Where, D = overall depth of section (4 % of the gross cross sectional area)
3. Side face reinforcement: If depth of the web in a beam exceeds 750 mm, side-face
reinforcement should be provided along the two faces. The total area of such
reinforcement should not be less than 0.1 % of the web -area. It should be equally
distributed on each of the two faces; The spacing of such reinforcement should not
exceed 300 mm or web thickness whichever is less.
4. 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.
• Sv < 0.75 d Vertical shear Reinforcement
• Sv < d Inclined shear Reinforcement
• However, the spacing shall not exceed 300 mm in any case.
REINFORCEMENT REQUIREMENTS

SLABS:
1. Minimum tension reinforcement: minimum reinforcement in either direction in slabs should not
be less than 0.15% of the total cross-sectional area when using mild steel reinforcement, and
0.12% of the total cross-sectional area when using high strength deformed reinforcement or
welded Wire fabric
• 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. No maximum reinforcement recommendation is given in IS code so we consider the maximum
reinforcement criteria same as in case of beam i.e. 4% of bD
3. Maximum diameter of steel bar in slab
• ∅𝒎𝒂𝒙 =
𝑻𝒉𝒊𝒄𝒌𝒏𝒆𝒔𝒔 𝒐𝒇 𝒔𝒍𝒂𝒃
𝟖
4. Maximum spacing of the reinforcement
a) Main bar = 𝐥𝐞𝐬𝐬𝐞𝐫 𝒐𝒇 ቊ
𝟑𝒅
𝟑𝟎𝟎 𝒎𝒎
b) Distribution bar = 𝐥𝐞𝐬𝐬𝐞𝐫 𝒐𝒇 ቊ
𝟓𝒅
𝟑𝟎𝟎 𝒎𝒎
‘d’ is the effective depth of slab
REINFORCEMENT REQUIREMENTS

Anchorage Bond • Let us consider a uniformly loaded
cantilever beam which has to resist
given BM and SF
• Let us assume that tension
reinforcement consists of single bar of
diameter ∅
• The tensile force at B is equal to
A B C
𝑳𝒅
W
T
↽↽
↼
↽
↼
↼
𝝉𝒃𝒅
𝑻 =
𝝅
𝟒
∅𝟐
𝒇𝒚
𝟏. 𝟏𝟓
• Which must be transmitted to concrete
by bond stress in the embedded length
𝑳𝒅 = 𝑨𝑩
⇒ 𝝉𝒃𝒅 × 𝝅∅ 𝑳𝒅 =
𝝅
𝟒
∅𝟐 𝒇𝒚
𝟏.𝟏𝟓
𝑳𝑺𝑴 𝑳𝒅 =
𝟎.𝟖𝟕 𝒇𝒚 ∅
𝟒𝝉𝒃𝒅
𝑻𝒉𝒆 𝒍𝒆𝒏𝒈𝒕𝒉 𝑳𝒅 𝒊𝒔 𝒄𝒂𝒍𝒍𝒆𝒅 𝒂𝒔
𝑫𝒆𝒗𝒆𝒍𝒐𝒑𝒎𝒆𝒏𝒕 𝒍𝒆𝒏𝒈𝒕𝒉
𝑾𝑺𝑴 𝑳𝒅 =
𝝈𝒔𝒕∅
𝟒𝝉𝒃𝒅
• If 𝝉𝒃𝒅 is the average bond stress acting
over the surface area 𝝅∅𝑳𝒅 then
𝑺𝒕𝒓𝒆𝒏𝒈𝒕𝒉 𝒐𝒇 𝑩𝒐𝒏𝒅 = 𝑺𝒕𝒓𝒆𝒏𝒈𝒕𝒉 𝒐𝒇 𝑺𝒕𝒆𝒆𝒍

Design Bond stress for concrete for PLAIN bars in TENSION
Grade of Concrete WSM LSM
M15 0.6
M20 0.8 1.2
M25 0.9 1.4
M30 1 1.5
M35 1.1 1.7
M40 and above 1.2 1.9
• 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.
• For fusion bonded epoxy coated deformed bars,
design bond stress values shall be taken as 80
percent of the values given in the above table
Fusion Bonded Epoxy is very fast curing, thermosetting Protective Powder Coating. It is based on specially selected Epoxy resins and hardeners. The epoxy
formulated in order to meet the specifications related to protection of steel bars as an anti-corrosion coating

87
CIVIL ENGINEERING

88
CIVIL ENGINEERING
For Fe 415, 𝝉𝒃𝒅 𝒕𝒆𝒏𝒔𝒊𝒐𝒏 = 1.5 × 1.6 = 2.4
(𝐼𝑛𝑐𝑟𝑒𝑎𝑠𝑒 𝑏𝑦 60% 𝑓𝑜𝑟 𝑑𝑒𝑓𝑜𝑟𝑚𝑒𝑑 𝑏𝑎𝑟𝑠 𝐹𝑒 415 𝑖𝑛 𝑐𝑎𝑠𝑒 𝑜𝑓 𝑡𝑒𝑛𝑠𝑖𝑜𝑛)
𝝉𝒃𝒅 𝒄𝒐𝒎𝒑𝒓𝒆𝒔𝒔𝒊𝒐𝒏 = 1.5 × 1.6 × 1.25 = 3 (𝐹𝑢𝑟𝑡ℎ𝑒𝑟 𝐼𝑛𝑐𝑟𝑒𝑎𝑠𝑒 𝑏𝑦 25% 𝑖𝑛 𝑐𝑎𝑠𝑒 𝑜𝑓 𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛)

89
CIVIL ENGINEERING

90
CIVIL ENGINEERING
For Fe 215, 𝝉𝒃𝒅 𝒕𝒆𝒏𝒔𝒊𝒐𝒏 = 1.5
𝝉𝒃𝒅 𝒄𝒐𝒎𝒑𝒓𝒆𝒔𝒔𝒊𝒐𝒏 = 1.5 × 1.25 = 1.87

91
CIVIL ENGINEERING

92
CIVIL ENGINEERING

93
CIVIL ENGINEERING

94
CIVIL ENGINEERING

95
CIVIL ENGINEERING

96
CIVIL ENGINEERING

97
CIVIL ENGINEERING

98
CIVIL ENGINEERING

99
CIVIL ENGINEERING

100
CIVIL ENGINEERING

101
CIVIL ENGINEERING

102
CIVIL ENGINEERING

103
CIVIL ENGINEERING

104
CIVIL ENGINEERING

When Beam is subjected to Bending Moment and Torsional Moment
• Critical Section:
• Sections located less than a distance d from the face of the support may be designed for the same
torsion as computed at a distance d, where d is the effective depth
• Step 2: Equivalent Nominal Shear
Stress 𝝉𝒗𝒆
• 𝝉𝒗𝒆 =
𝑽𝒆
𝑩𝒅
• This value of 𝝉𝒗𝒆 should not exceed
maximum shear stress 𝝉𝒄 𝒎𝒂𝒙
• If 𝝉𝒗𝒆 > 𝝉𝒄 𝒎𝒂𝒙 , then revise the section
• Step 1: Equivalent shear force (41.3.1)
• The equivalent shear force 𝑽𝒆 = 𝑽𝒖 + 𝟏. 𝟔
𝑻𝒖
𝒃
• 𝑉𝑒 = 𝐸𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑁𝑜𝑚𝑖𝑛𝑎𝑙 𝑠ℎ𝑒𝑎𝑟 𝑓𝑜𝑟𝑐𝑒
• 𝑉𝑢 = 𝐹𝑎𝑐𝑡𝑜𝑟𝑒𝑑 𝑠ℎ𝑒𝑎𝑟 𝑓𝑜𝑟𝑐𝑒
• 𝑇𝑢 = 𝑇𝑜𝑟𝑠𝑖𝑜𝑛𝑎𝑙 𝑚𝑜𝑚𝑒𝑛𝑡
• 𝑏 = 𝑏𝑟𝑒𝑎𝑑𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑒𝑎𝑚 (𝑏𝑤 for flanged beam)

Design for Equivalent Shear
• Step 2: Equivalent Nominal Shear Stress 𝝉𝒗𝒆
• 𝝉𝒗𝒆 =
𝑽𝒆
𝑩𝒅
• This value of 𝝉𝒗𝒆 should not exceed maximum shear
stress 𝝉𝒄 𝒎𝒂𝒙 for safety against diagonal compression
failure
• If 𝝉𝒗𝒆 > 𝝉𝒄 𝒎𝒂𝒙 , then revise the section
• If the equivalent nominal shear stress, 𝝉𝒗𝒆 does not
exceed
𝝉𝒄 (
𝑫𝒆𝒔𝒊𝒈𝒏 𝒔𝒉𝒆𝒂𝒓 𝒔𝒕𝒓𝒆𝒏𝒈𝒕𝒉 𝒐𝒇 𝒄𝒐𝒏𝒄𝒓𝒆𝒕𝒆, 𝒕𝒂𝒃𝒍𝒆 𝟏𝟗)
given, minimum shear reinforcement shall he provided
as per 26.5.1.6

Design for Equivalent Bending
• Step 3: Equivalent Moment
• 𝑀𝑒1 = 𝑀𝑢 + 𝑀𝑇
• 𝑀𝑇 =
𝑇𝑢
1.7
(1 +
𝐷
𝐵
)
• 𝑀𝑒 = equivalent bending moment
• 𝑀𝑇 = equivalent moment due to torsion
• 𝑀𝑢 = bending moment at cross section
• 𝑇𝑢 = Torsional moment
• D = overall depth
• b = breadth of beam
• Step 4: Equivalent effective depth
• 𝑑 =
𝑀𝑒
𝜃𝐵

Design of Column
1. Short Column
• Load carrying capacity:
𝑃 = 𝑃𝑠𝑡𝑒𝑒𝑙 + 𝑃𝑐𝑜𝑛𝑐𝑟𝑒𝑡𝑒
= 𝜎𝑠𝐴𝑠 + 𝜎𝑐𝐴𝑐
𝑆𝑖𝑛𝑐𝑒 𝑇𝑜𝑡𝑎𝑙 𝐴𝑟𝑒𝑎 𝐴 = 𝐴𝑠 + 𝐴𝑐
⇒ 𝑷 = 𝝈𝒔(𝑨 − 𝑨𝒄) + 𝝈𝒄𝑨𝒄
𝑨
𝑨𝒄
𝑨𝒔
2. Long Column
• Load carrying capacity:
⇒ 𝑷 = 𝑪𝒓{𝝈𝒔(𝑨 − 𝑨𝒄) + 𝝈𝒄𝑨𝒄}
Where 𝑪𝒓 = 𝟏. 𝟐𝟓 −
𝑳𝒆𝒇𝒇
𝟒𝟖𝑳𝑳𝑫
𝒇𝒐𝒓 𝑹𝒆𝒄𝒕𝒂𝒏𝒈𝒖𝒍𝒂𝒓 𝒐𝒓 𝒔𝒒𝒖𝒂𝒓𝒆 𝒄𝒐𝒍𝒖𝒎𝒏
And 𝑪𝒓 = 𝟏. 𝟐𝟓 −
𝑳𝒆𝒇𝒇
𝒇𝒐𝒓 𝒄𝒊𝒓𝒄𝒖𝒍𝒂𝒓 𝒄𝒐𝒍𝒖𝒎𝒏 (𝜸𝒎𝒊𝒏 = 𝑫/𝟒)

CIVIL ENGINEERING
250𝑚𝑚 × 250𝑚𝑚

CIVIL ENGINEERING
250𝑚𝑚 × 250𝑚𝑚
𝑪𝒓 = 𝟏. 𝟐𝟓 −
𝑳𝒆𝒇𝒇
= 𝟏. 𝟐𝟓 −
𝟒𝟖𝟎𝟎
𝟒𝟖(𝟐𝟓𝟎)
= 𝟎. 𝟖𝟓

Design of Column
3. Circular Column
a) With separate ring used as a stirrup:
𝑷 = 𝑪𝒓{𝝈𝒔𝑨𝒔 + 𝝈𝒄𝑨𝒄}

Design of Column
3. Circular Column
b) With helical reinforcement
𝑷 = 𝟏. 𝟎𝟓 × 𝑪𝒓 × {𝝈𝒔𝑨𝒔 + 𝝈𝒄𝑨𝒄}
For helical reinforcement, following criteria has to be satisfied
𝟎. 𝟑𝟔𝒇𝒄𝒌
𝒇𝒚
𝑨𝒈
𝑨𝒄
− 𝟏 ≤
𝑽𝒉
𝑽𝒄
• 𝑨𝒈 = 𝒈𝒓𝒐𝒔𝒔 𝒂𝒓𝒆𝒂 =
𝝅
𝟒
𝑫𝒈
𝟐
• 𝑨𝒄 = 𝒄𝒐𝒓𝒆 𝒂𝒓𝒆𝒂
• 𝑽𝒉 = 𝒗𝒐𝒍𝒖𝒎𝒆 𝒐𝒇 𝒉𝒆𝒍𝒊𝒄𝒂𝒍 𝒓𝒆𝒊𝒏𝒇𝒐𝒓𝒄𝒆𝒎𝒆𝒏𝒕
= 𝒂𝒓𝒆𝒂 𝒐𝒇 𝒉𝒆𝒍𝒊𝒄𝒂𝒍 𝒓𝒇 × 𝒍𝒆𝒏𝒈𝒕𝒉 𝒊𝒏 𝒐𝒏𝒆 𝒍𝒐𝒐𝒑 × 𝒏𝒐. 𝒐𝒇 𝒕𝒖𝒓𝒏𝒔
=
𝝅
𝟒
∅𝒉
𝟐 × 𝝅𝑫 ×
𝟏𝟎𝟎𝟎
𝒑𝒊𝒕𝒄𝒉
• 𝑽𝒄 = 𝒗𝒐𝒍𝒖𝒎𝒆 𝒐𝒇 𝒄𝒐𝒓𝒆 𝒇𝒐𝒓 𝟏𝒎
= 𝟏𝟎𝟎𝟎𝒎𝒎 × 𝑨𝒄𝒐𝒓𝒆
= 𝟏𝟎𝟎𝟎 ×
𝝅
𝟒
𝑫𝒄
𝟐
• 𝑫𝒉 = 𝒅𝒊𝒂 𝒐𝒇 𝒄𝒆𝒏𝒕𝒓𝒆 𝒕𝒐 𝒄𝒆𝒏𝒕𝒓𝒆 𝒉𝒆𝒍𝒊𝒄𝒂𝒍 𝒔𝒕𝒊𝒓𝒓𝒖𝒑
𝒑 = 𝑷𝒊𝒕𝒄𝒉 𝒐𝒇
𝒉𝒆𝒍𝒊𝒙
𝑫𝒉
𝑫𝒈 𝑫𝒄

Using LSM
• For short axially loaded column with 𝒆𝒎𝒊𝒏 ≤ 𝟎. 𝟎𝟓𝑳𝑳𝑫,
column can be designed as
• 𝑷𝒖 = 𝟎. 𝟒𝒇𝒄𝒌𝑨𝒄 + 𝟎. 𝟔𝟕𝒇𝒚𝑨𝒔𝒄
• For circular column,
• 𝑷𝒖 = 𝟏. 𝟎𝟓(𝟎. 𝟒𝒇𝒄𝒌𝑨𝒄 + 𝟎. 𝟔𝟕𝒇𝒚𝑨𝒔𝒄)

Columns (LSM)
• Assumptions: Clause 39.1, Minimum Eccentricity 39.2,
Short Axially loaded Members in Compression 39.3
1. Minimum Eccentricity (whichever is maximum) (Cl 39.2)
𝑒𝑚𝑖𝑛 ≥
𝑙
500
+
𝐷
30
> 20𝑚𝑚
2. If Minimum eccentricity value as per CL 39.2 is less than
or equal to 0.05 × Least Lateral Dimension, then the
column can be designed as per following equation of Short
Column
𝒆𝒎𝒊𝒏 ≤ 𝟎. 𝟎𝟓 𝑳𝑳𝑫 (𝒂𝒙𝒊𝒂𝒍𝒍𝒚 𝒍𝒐𝒂𝒅𝒆𝒅 𝑺𝒉𝒐𝒓𝒕 𝑪𝒐𝒍𝒖𝒎𝒏)

CIVIL ENGINEERING
350𝑚𝑚 × 650𝑚𝑚

CONTROL OF DEFLECTION
• The deflection of a structure or its part should
not adversely affect the appearance or
efficiency of the structure or finishes or
partitions
• The deflection including the effects of
temperature, shrinkage and creep occurring
after the construction of partitions, and finishes
should not exceed span/350 or 20 mm,
whichever is lesser
• The total deflection due to all loads including
the effects of temperature, shrinkage and creep
should not exceed span/250 when measured
from the as cast level of the supports of floors,
roofs and all other horizontal-members,

Depth of Slab
• The depth of slab depends on bending moment and deflection criterion. the trail
depth can be obtained using: • The effective depth d of two way slabs can also be
assumed using cl. 24.1,IS 456 provided short span
is ≤ 3.5m and loading class is < 3.5KN/m2

CIVIL ENGINEERING
3𝑚 × 3.5𝑚 𝑠𝑖𝑧𝑒

CIVIL ENGINEERING
3𝑚 × 3.5𝑚 𝑠𝑖𝑧𝑒
For slabs spanning in two directions, the
shorter of the two spans should be taken
for calculating the span to effective depth
For 3𝑚 × 3.5𝑚 𝑐𝑜𝑛𝑡𝑖𝑛𝑜𝑢𝑠 slab (using Fe
250),
𝑠𝑝𝑎𝑛
𝑜𝑣𝑒𝑟𝑎𝑙𝑙 𝑑𝑒𝑝𝑡ℎ
= 40
⇒
3𝑚
𝑜𝑣𝑒𝑟𝑎𝑙𝑙 𝑑𝑒𝑝𝑡ℎ
= 40
⇒ 𝑜𝑣𝑒𝑟𝑎𝑙𝑙 𝑑𝑒𝑝𝑡ℎ =
3
40
⇒ 𝑜𝑣𝑒𝑟𝑎𝑙𝑙 𝑑𝑒𝑝𝑡ℎ = 0.075𝑚

Design of One Way Slab
Step 1: Find load over 1m of span
a) 𝐋𝐢𝐯𝐞 𝐥𝐨𝐚𝐝
b) 𝑺𝒆𝒍𝒇 𝒘𝒆𝒊𝒈𝒉𝒕 = 𝑻𝒉𝒊𝒄𝒌𝒏𝒆𝒔𝒔 × 𝟏𝒎 × 𝟐𝟓
c) 𝑭𝒍𝒐𝒐𝒓 𝒇𝒊𝒏𝒊𝒔𝒉𝒊𝒏𝒈 = 𝑻𝒉𝒊𝒄𝒌𝒏𝒆𝒔𝒔 𝒐𝒇 𝒇𝒍𝒐𝒐𝒓 × 𝟏𝒎 × 𝟏𝒎 × 𝟐𝟒
Step 2: Bending Moment
a) 𝑩𝑴𝒎𝒂𝒙 =
𝒘𝑳𝟐
𝟖
( for simply supported beam)
Step 3: Area of Steel
𝑨𝒔𝒕 =
𝟎. 𝟓𝒇𝒄𝒌
𝒇𝒚
𝟏 − 𝟏 −
𝟒. 𝟔 𝑩𝑴
𝒇𝒄𝒌𝑩𝒅𝟐
× 𝑩 × 𝑫
Step 4: Check for Shear
𝝉𝒗 =
𝑽
𝑩𝒅
≯ 𝝉𝒄 𝒎𝒂𝒙
𝑽 =
𝑾𝑳
𝟐

Criteria for Design
1. Depth of footing:
All foundations should be located at a minimum
depth of 0.5m below the ground surface
The depth is primarily governed by availability
of bearing capacity, minimum seasonal variation
like swelling and shrinkage of soil
Using rankine’s formula, minimum depth of
foundation is given by
𝐷𝑓 =
𝑞
𝛾
1 − 𝑠𝑖𝑛∅
1 + 𝑠𝑖𝑛∅
2
𝑞=gross safe bearing capacity
𝛾=unit weight of soil
∅=angle of friction

Design Steps for Footing
1. Size of Foundation:
a) Load from column = 𝑷
b) Weight of foundation =𝑷𝒇 = 𝟏𝟎% 𝒐𝒇 𝑷
c) Total load 𝑷𝒕 = 𝑷 + 𝑷𝒇 = 𝟏. 𝟏 𝑷
d) Area of Foundation
𝑨𝒓𝒆𝒂 𝒐𝒇 𝒇𝒐𝒖𝒏𝒅𝒂𝒕𝒊𝒐𝒏 =
𝑷𝒕
𝒒𝟎
While calculating the area of footing required, self weight of footing is considered
e) Decide size of foundation
Square footing⇒ 𝑨𝒓𝒆𝒂 = 𝑩 × 𝑩
Rectangular Footing⇒ 𝑨𝒓𝒆𝒂 = 𝑩 × 𝑳
(𝒂𝒔𝒔𝒖𝒎𝒆 𝑩 𝒂𝒏𝒅 𝒇𝒊𝒏𝒅 𝑳)
f) Net upward design soil pressure on foundation
𝑾 =
𝑷 (𝒄𝒐𝒍𝒖𝒎𝒏)
𝑨(𝒇𝒐𝒐𝒕𝒊𝒏𝒈)
𝑾𝑺𝑴
𝑾𝒖 = 𝟏. 𝟓𝑾 (𝑳𝑺𝑴)
While calculating the upward soil pressure, the self weight of the footing is not considered
𝑷𝒇
𝒒𝟎
(𝑩𝒆𝒂𝒓𝒊𝒏𝒈 𝑪𝒂𝒑𝒂𝒄𝒊𝒕𝒚
𝒐𝒇 𝒔𝒐𝒊𝒍)
𝒂
𝒃
𝑩
𝑳
𝑺𝒊𝒛𝒆 𝒐𝒇 𝑪𝒐𝒍𝒖𝒎𝒏 = 𝒂 × 𝒃
𝑺𝒊𝒛𝒆 𝒐𝒇 𝑭𝒐𝒖𝒏𝒅𝒂𝒕𝒊𝒐𝒏 = 𝑩 × 𝑳

CIVIL ENGINEERING
Area of Foundation
𝑨𝒓𝒆𝒂 𝒐𝒇 𝒇𝒐𝒖𝒏𝒅𝒂𝒕𝒊𝒐𝒏 =
𝑷𝒕
𝒒𝟎
While calculating the area of footing
required, self weight of footing is
considered
Net design soil pressure on foundation
𝑾 =
𝑷 (𝒄𝒐𝒍𝒖𝒎𝒏)
𝑨(𝒇𝒐𝒐𝒕𝒊𝒏𝒈)
𝑾𝑺𝑴
𝑾𝒖 = 𝟏. 𝟓𝑾 (𝑳𝑺𝑴)
While calculating the upward soil pressure,
the self weight of the footing is not
considered

2. Check for Bending Moment
Critical section for bending moment is at
the face of the column.
➢Bending moment about x1-x1 axis
➢OX1= overhang = (
𝑩−𝒃
𝟐
)
➢Mx= BMx1-x1
=w𝒐 × 𝟏𝒎 ×
𝑶𝒙𝟏
𝟐
𝟐
=w𝒐 × 𝟏𝒎 ×
{𝑩−𝒃
𝟐
}𝟐
𝟐
= 𝒘𝒐 × 𝟏𝒎 ×
𝑩−𝒃 𝟐
𝟖
𝑳
𝒂
𝒃
𝒐𝒗𝒆𝒓𝒉𝒂𝒏𝒈
𝟏𝒎
𝟏𝒎
𝑩
𝒚𝟏
𝒚𝟏
𝒙𝟏
𝒙𝟏

2. Check for Bending Moment
Critical section for bending moment is at the face of
the column.
➢Bending moment about y1-y1 axis
➢My= BMy1-y1
=w𝒐 × 𝟏𝒎 ×
𝑶𝒀𝟏
𝟐
𝟐
=w𝒐 × 𝟏𝒎 ×
{𝑳−𝒂
𝟐
}𝟐
𝟐
= 𝒘𝒐 × 𝟏𝒎 ×
𝑳−𝒂 𝟐
𝟖
➢Maximum BM is maximum value among My and Mx
➢Depth required
d =
𝑩𝑴𝒎𝒂𝒙
𝑸𝑩
where B = 1000mm
𝑳
𝒂
𝒃
𝒐𝒗𝒆𝒓𝒉𝒂𝒏𝒈
𝟏𝒎
𝟏𝒎
𝑩
𝒚𝟏
𝒚𝟏
𝒙𝟏
𝒙𝟏

3. Check for One way Shear
Critical section for one way shear is at
a distance ‘d’ from the face of column.
➢OX2 = (
𝑩−𝒃
𝟐
− 𝐝)
➢OY2 = (
𝑳−𝒂
𝟐
− 𝐝)
Max shear about x2-x2
➢OX2 = (
𝑩−𝒃
𝟐
− 𝐝)
➢V = w𝒐 × 𝟏𝒎 × OX2
➢V = w𝒐 × 𝟏𝒎 × (
𝑩−𝒃
𝟐
− 𝐝) 𝑳
𝒂
𝒃
𝟏𝒎
𝟏𝒎
𝑩
𝒚𝟐
𝒚𝟐
𝒙𝟐 𝒙𝟐
𝒅
𝒅
𝑶𝑿𝟐
𝑶𝒀𝟐

3. Check for One way Shear
Critical section for one way shear is at a distance ‘d’
from the face of column.
• Max shear about y2-y2
➢OY2 = (
𝑳−𝒂
𝟐
− 𝐝)
➢V = w𝒐 × 𝟏𝒎 × OY2
➢V = w𝒐 × 𝟏𝒎 × (
𝑳−𝒂
𝟐
− 𝐝)
• Nominal shear stress
➢𝝉𝒗 =
𝑽
𝑩.𝒅
➢𝝉𝒗 =
𝑽
𝟏𝟎𝟎𝟎×𝒅
➢𝝉𝒗 𝒔hould not be greater than k.𝝉𝒄min
Where 𝝉𝒄min = 0.18N/mm2 (WSM)
𝝉𝒄min = 0.28N/mm2 (LSM)
𝑳
𝒂
𝒃
𝟏𝒎
𝟏𝒎
𝑩
𝒚𝟐
𝒚𝟐
𝒙𝟐 𝒙𝟐
𝒅
𝒅
𝑶𝑿𝟐
𝑶𝒀𝟐
THICKNESS >300 275 250 225 200 175 150
k 1 1.05 1.10 1.15 1.20 1.25 1.30

4. Check for Two way Shear
Punching shear shall be checked around
the column on a perimeter half the
effective depth of footing away from the
face of column
Net punching force
𝑷𝑭 = 𝒘𝟎 𝑳 × 𝑩 − 𝒘𝟎(𝒂 + 𝒅)(𝒃 + 𝒅)
Resisting Area
= 𝟐[ 𝒂 + 𝒅 + (𝒃 + 𝒅] × 𝒅
𝝉𝒗𝒑 (𝒅𝒆𝒗𝒆𝒍𝒐𝒑𝒆𝒅) =
𝑵𝒆𝒕 𝒑𝒖𝒏𝒄𝒉𝒊𝒏𝒈 𝒇𝒐𝒓𝒄𝒆
𝑹𝒆𝒔𝒊𝒔𝒕𝒊𝒏𝒈 𝒂𝒓𝒆𝒂
⇒ 𝝉𝒗𝒑 (𝒅𝒆𝒗𝒆𝒍𝒐𝒑𝒆𝒅)=
𝒘𝟎 𝑳 × 𝑩 − 𝒘𝟎(𝒂 + 𝒅)(𝒃 + 𝒅)
𝟐[ 𝒂 + 𝒅 + (𝒃 + 𝒅] × 𝒅

4. Check for Two way Shear
⇒ 𝝉𝒗𝒑 (𝒅𝒆𝒗𝒆𝒍𝒐𝒑𝒆𝒅)=
𝒘𝟎 𝑳 × 𝑩 − 𝒘𝟎(𝒂 + 𝒅)(𝒃 + 𝒅)
𝟐[ 𝒂 + 𝒅 + (𝒃 + 𝒅] × 𝒅
𝒂𝒏𝒅 𝝉𝒗𝒑 (𝒅𝒆𝒗𝒆𝒍𝒐𝒑𝒆𝒅)𝒔𝒉𝒐𝒖𝒍𝒅 𝒏𝒐𝒕 𝒃𝒆
𝒈𝒓𝒆𝒂𝒕𝒆𝒓 𝒕𝒉𝒂𝒏 𝝉𝒗𝒑 (𝒑𝒆𝒓𝒎𝒊𝒔𝒔𝒊𝒃𝒍𝒆)
• 𝝉𝒗𝒑 (𝒑𝒆𝒓𝒎𝒊𝒔𝒔𝒊𝒃𝒍𝒆) = ቐ
𝑲𝒔 × 𝟎. 𝟏𝟔 𝒇𝒄𝒌 𝑾𝑺𝑴
𝑲𝒔 × 𝟎. 𝟐𝟓 𝒇𝒄𝒌 𝑳𝑺𝑴
Where 𝑲𝒔 = 𝟎. 𝟓 +
𝒃
𝒂
< 𝟏. 𝟎

5. Area of Steel
𝑨𝒔𝒕 =
𝑴𝒙
𝝈𝒔𝒕𝒋𝒅
WSM
𝑨𝒔𝒕 =
𝑴𝒖𝒙
𝟎.𝟖𝟕𝒇𝒚𝒋𝒅
LSM
𝑨𝒔𝒕 =
𝟎. 𝟓𝒇𝒄𝒌
𝒇𝒚
𝟏 − 𝟏 −
𝟒. 𝟔𝑴𝒖
𝒇𝒄𝒌𝑩𝒅𝟐
𝑩𝒅
The above 𝑨𝒔𝒕 is for 𝟏𝒎 width,
𝑻𝒐𝒕𝒂𝒍 𝑨𝒔𝒕 = 𝑳 × 𝑨𝒔𝒕

CIVIL ENGINEERING
𝝉𝒗
𝝉𝒗 < 𝐾𝑠𝝉𝑐 𝝉𝑐
a) 𝟎. 𝟏 𝒇𝒄𝒌
b) 𝟎. 𝟏𝟔 𝒇𝒄𝒌
c) 𝟎. 𝟐𝟓 𝒇𝒄𝒌
d) 𝟎. 𝟒 𝒇𝒄𝒌

CIVIL ENGINEERING
𝝉𝒗
𝝉𝒗 < 𝐾𝑠𝝉𝑐 𝝉𝑐
a) 𝟎. 𝟏 𝒇𝒄𝒌
b) 𝟎. 𝟏𝟔 𝐟𝐜𝐤
c) 𝟎. 𝟐𝟓 𝒇𝒄𝒌
d) 𝟎. 𝟒 𝒇𝒄𝒌

CIVIL ENGINEERING
a) 𝑩𝑴 =
𝒑𝒃(𝒃−𝒂)
𝟒
b) 𝑩𝑴 =
𝒑𝒃(𝒃−𝒂)𝟐
𝟒
c) 𝑩𝑴 =
𝟖
d) 𝑩𝑴 =
𝒑𝒃(𝒃+𝒂)
𝟒

CIVIL ENGINEERING
a) 𝑩𝑴 =
𝒑𝒃(𝒃−𝒂)
𝟒
b) 𝑩𝑴 =
𝟒
c) 𝐁𝐌 =
𝐩𝐛(𝐛−𝐚)𝟐
𝟖
d) 𝑩𝑴 =
𝒑𝒃(𝒃+𝒂)
𝟒

PERMISSIBLE STRESS IN STEEL STRUCTURES
1. As per WSM
i. Maximum permissible AXIAL stress in compression is given
by
𝝈𝒂𝒄 = 𝟎. 𝟔𝟎 𝒇𝒚
• Used in the design of columns and struts.
• Column is a compression member where bending moment
exists while in case of struts, also being a compression
member, bending moment is zero. Because strut is a
component of roof trusses and roof trusses are pin jointed
connection having bending moment equal to zero.
ii. Maximum permissible AXIAL stress in tension is given by
𝝈𝒂𝒕 = 𝟎. 𝟔𝟎 𝒇𝒚
It is used in design of tension members

1. As per WSM
iii. Maximum permissible bending stress in compression is given
• Used in design of flexural (bending) member that is beam, built up beam,
plate girder etc.
𝝈𝒃𝒄 = 𝟎. 𝟔𝟔 𝒇𝒚
iv. Maximum permissible bending stress in tension is given
• Used in the design of beams
𝝈𝒃𝒕 = 𝟎. 𝟔𝟔 𝒇𝒚
v. Maximum permissible average shear stress is given by
𝝉𝒗 𝒂𝒗𝒈 = 𝟎. 𝟒𝟎𝒇𝒚
vi. Maximum permissible Maximum shear stress is given by
𝝉𝒗 𝒎𝒂𝒙 = 𝟎. 𝟒𝟓𝒇𝒚
FOS=2.5 for average shear stress
FOS=2.2 for maximum shear stress

1. As per WSM
vi. Maximum permissible bending stress in column base
is given by
𝝈 = 𝟎. 𝟕𝟓 𝒇𝒚
Increase of permissible stress
• When wind and earthquake load are considered, the
permissible stresses in steel structure are increased by
33.33%.
• When wind and earthquake load are considered, the
permissible stresses in connections (rivet and weld) are
increased by 25%.

PERMISSIBLE DEFLECTION IN STEEL STRUCTURES
• As per WSM, Maximum permissible horizontal and vertical
deflection is given by
𝛅 =
𝒔𝒑𝒂𝒏
𝟑𝟐𝟓
• As per LSM, Maximum permissible horizontal and vertical
deflection is given by
a) If supported elements are not susceptible to cracking
𝛅 =
𝒔𝒑𝒂𝒏
𝟑𝟎𝟎
b) If supported elements are susceptible to cracking
𝛅 =
𝒔𝒑𝒂𝒏
𝟑𝟔𝟎

PERMISSIBLE DEFLECTION IN GANTRY GIRDER
1. For manually operator crane, the
maximum permissible deflection is
𝛅 =
𝒔𝒑𝒂𝒏
𝟓𝟎𝟎
2. For electrically operator crane, the
maximum permissible deflection for a
capacity upto 50T or 500kN
𝛅 =
𝒔𝒑𝒂𝒏
𝟕𝟓𝟎
3. For electrically operator crane, the
maximum permissible deflection for a
capacity more than 50T or 500kN
𝛅 =
𝒔𝒑𝒂𝒏
𝟏𝟎𝟎𝟎
Gantry girders are laterally unsupported beams to carry. heavy loads from
place to place at the construction sites

FACTOR OF SAFETY FOR DIFFERENT STRESSES
Factor of Safety =
𝑦𝑖𝑒𝑙𝑑 𝑠𝑡𝑟𝑒𝑠𝑠
𝑤𝑜𝑟𝑘𝑖𝑛𝑔 𝑠𝑡𝑟𝑒𝑠𝑠
=
𝑓𝑦
𝑓
1. For axial stress, F.O.S. =
𝑓𝑦
0.60𝑓
= 1.67
2. For bending stress, F.O.S. =
𝑓𝑦
0.66𝑓
=
1.50
3. For shear stress, F.O.S. =
𝑓𝑦
0.40𝑓
= 2.50

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CONNECTIONS
1. RIVETED CONNECTIONS:
• Strength of riveted joint
• It is taken as minimum of shear strength, bearing strength and tearing
strength.
• FOR LAP JOINT:
1. FOR ENTIRE PLATE
a) SHEAR STRENGTH OF RIVETS
𝑷𝒔 = 𝒏 ×
𝝅
𝟒
𝒅𝟐𝑭𝒔
Where n → total number of rivets at joint
Fs → permissible shear stress in rivets
Fs = 100MPa (WSM)
Fu = ultimate shear stress in rivet
so in LSM =
𝑭𝒖
𝟑  1.25
d → gross diameter of rivet (hole diameter)
Gross dia = nominal dia + 1.5 mm, for nominal dia ≤ 25 mm
Gross dia = nominal dia + 2 mm, for nominal dia > 25 mm
P B P

CONNECTIONS
• FOR LAP JOINT:
1. FOR ENTIRE LENGTH
b) BEARING STRENGTH OF ALL RIVETS
𝑷𝑩 = 𝒏 × 𝒕𝒅 𝑭𝒃
t → thickness of thinner main plate
Fb → permissible shear stress in rivets (300MPa in WSM)
c) TEARING STRENGTH OF PLATE
𝑷𝒕 = 𝑩 − 𝒏𝟏𝒅 𝒕𝑭𝒕
Where n1 → total number of rivets at critical section 1-1
B → width of plate
Ft → permissible tensile stress (Axial = 0.6fy = 0.6250 = 150MPa)
B
1
1
P
P
1
1
B

CONNECTIONS
• LAP JOINT:
2. FOR GAUGE LENGTH/PITCH LENGTH
𝑷𝒔𝟏 = 𝒏 ×
𝝅
𝟒
Where n → total number of rivets at joint in crossed gauge
length
Fs = 100MPa (WSM)
Fu = ultimate shear stress in rivet so in
LSM =
𝑭𝒖
𝟑  1.25
1
1
P
P g

CONNECTIONS
• LAP JOINT:
b) BEARING STRENGTH OF RIVETS
𝑷𝑩𝟏 = 𝒏𝟏 × 𝒕𝒅 𝑭𝐛
Where n → total number of rivets at joint in crossed gauge length
Fb → permissible bearing stress in rivets (300MPa in WSM)
𝑷𝒕𝟏 = 𝒈 − 𝒅 𝒕𝑭𝒕
Where g → gauge length
Ft → permissible tensile stress in plate(Axial = 0.6fy = 0.6250 =
150MPa)
When pitch distance is given then
𝑷𝒕 = 𝑷 − 𝟑𝒅 𝒕𝑭𝒕
1
1
P
P g
FIRST PLATE
SECOND PLATE
1
B
1
g

CONNECTIONS
• DOUBLE COVER BUTT JOINT:
1. FOR ENTIRE WIDTH OF PLATE
• SHEAR STRENGTH OF RIVETS
𝑷𝒔𝟏 = 𝟐𝒏𝟏 ×
𝝅
𝟒
Fs = 100MPa (WSM)
2 → Double shear
1
1
P
P B
MAIN PLATE
P
P

CONNECTIONS
• BEARING STRENGTH OF RIVETS
𝑷𝑩 = 𝒏 × 𝒕𝒅 𝑭𝒃
t → min of (thickness of thinner main plate,
sum of cover plate thickness)
Fb → permissible bearing stress in rivets
1
1
P
P B
MAIN PLATE
P
P

• TEARING STRENGTH OF PLATES
𝑷𝒕 = 𝑩 − 𝒏𝟏𝒅 𝒕𝑭𝒕
Where n1 → total number of rivets at critical section 1-1
t → min of (thickness of thinner main plate,
sum of cover plate thickness)
B → width of plate
Ft → permissible strength of plate in tearing
1
1
P
P B
MAIN PLATE
P
P
CONNECTIONS

CONNECTIONS
2. FOR GAUGE LENGTH
𝑷𝒔𝟏 = 𝟐 × 𝒏 ×
𝝅
𝟒
Where n → total number of rivets at joint in crossed gauge
length (here 2)
Fs = 100MPa (WSM)
Fu = ultimate shear stress in rivet so in
LSM =
𝑭𝒖
𝟑  1.25
1
1
P
P g B

CONNECTIONS
b) BEARING STRENGTH OF RIVETS
𝑷𝑩𝟏 = 𝒏𝟏 × 𝒕𝒅 𝑭𝐛
Where n → total number of rivets at joint in crossed gauge length
t → min (thickness of thinner main plate, sum of cover plate thickness)
Fb → permissible bearing stress in rivets (300MPa in WSM)
𝑷𝒕𝟏 = 𝒈 − 𝒏𝟏𝒅 𝒕𝑭𝒕
Where g → gauge length
Ft → permissible strength of plate in tearing
n → total number of rivets at in critical section 1-1 in crossed gauge length (here 1)
1
1
P
P g

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• Number of Rivets required at a joint=
𝑻𝒐𝒕𝒂𝒍 𝒇𝒐𝒓𝒄𝒆 𝒂𝒕 𝒂 𝒋𝒐𝒊𝒏𝒕
𝑹𝒊𝒗𝒆𝒕 𝑽𝒂𝒍𝒖𝒆
𝒏 =
𝑭
𝑹𝒗
• Efficiency of joint
𝜼 =
𝒍𝒆𝒂𝒔𝒕 𝒗𝒂𝒍𝒖𝒆 𝒐𝒇𝑷𝒔 , 𝑷𝒃 , 𝑷𝒕
𝑺𝒕𝒓𝒆𝒏𝒈𝒕𝒉 𝒐𝒇 𝑺𝒐𝒍𝒊𝒅 𝑴𝑨𝑰𝑵 𝒑𝒍𝒂𝒕𝒆
× 𝟏𝟎𝟎
𝑷𝒔 = 𝒔𝒉𝒆𝒂𝒓𝒊𝒏𝒈 𝒔𝒕𝒓𝒆𝒏𝒈𝒕𝒉 𝒐𝒇 𝒋𝒐𝒊𝒏𝒕
𝑷𝒃 = 𝒃𝒆𝒂𝒓𝒊𝒏𝒈 𝒔𝒕𝒓𝒆𝒏𝒈𝒕𝒉 𝒐𝒇 𝒋𝒐𝒊𝒏𝒕
𝑷𝒕 = 𝒕𝒆𝒂𝒓𝒊𝒏𝒈 𝒔𝒕𝒓𝒆𝒔𝒏𝒈𝒕𝒉 𝒐𝒇 𝒑𝒍𝒂𝒕𝒆

• Efficiency for entire plate
• We have to ensure that 𝑃𝑡 is less because rivet failure is more dangerous
• For Entire PLATE:
For Gauge Length:
𝜂 =
𝑙𝑒𝑎𝑠𝑡 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓𝑃𝑠 , 𝑃𝑏 , 𝑃𝑡
𝑆𝑡𝑟𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑆𝑜𝑙𝑖𝑑 𝑚𝑎𝑖𝑛 𝑝𝑙𝑎𝑡𝑒
× 100
⇒ 𝜂 =
𝐵 − 𝑛1𝑑 × 𝑡 × 𝐹𝑡
𝐵 × 𝑡 × 𝐹𝑡
× 100
⇒ 𝜂 =
𝐵 − 𝑛1𝑑
𝐵
× 100
⇒ 𝜂 =
𝑔 − 𝑑 × 𝑡 × 𝐹𝑡
𝑔 × 𝑡 × 𝐹𝑡
× 100
⇒ 𝜂 =
𝑔 − 𝑑
𝑔
× 100

Unwin’s formula
• It is used when diameter of rivet is not known
∅ = 𝟔. 𝟎𝟒 𝒕
Where t is thickness of thinner plate

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174
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ASSUMPTIONS IN DESIGN OF RIVETED JOINT
𝒈 − 𝒅 𝒕𝑭𝒕  n𝑹𝒗 (MOST IMPORTANT CONSIDERATION)
1
P
g
P
Where n is the number of rivets in shaded region

Analysis of Eccentric Connection
Step 1: Shear Force (𝐹1) in Rivet due to
Direct load P
𝐹𝑖 =
𝑃
σ 𝐴𝑖
× 𝐴𝑖
If dia of rivets are same, then the cross
section area would also be the same,
there fore direct shear load is
𝐹1 =
𝑃
𝑛 × 𝐴𝑖
× 𝐴𝑖
⇒ 𝐹1=
𝑃
𝑛
𝒆 𝑷
𝑭𝒍𝒂𝒏𝒈𝒆 𝒐𝒇 𝑰 𝒔𝒆𝒄𝒕𝒊𝒐𝒏

Step 2: Shear Force (𝐹2) in Rivet due to
Twisting Moment T
𝐹2 =
𝑇 × 𝑟𝑖
෌ 𝑟𝑖
2 ; 𝑇 = 𝑃𝑒
(𝑣𝑎𝑙𝑖𝑑 𝑤ℎ𝑒𝑛 𝑑𝑖𝑎 𝑜𝑓 𝑟𝑖𝑣𝑒𝑡𝑠 𝑎𝑟𝑒 𝑠𝑎𝑚𝑒)
Where 𝑟𝑖 is the radial distance of each rivet
from centre of Rivet Group
𝒆 𝑷

Step 3: Resultant Shear Force in the
Rivet (𝐹𝑅)
𝐹 = 𝐹1
2
+ 𝐹2
2
+ 2𝐹1𝐹2cos𝜃
𝒆 𝑷

CONNECTIONS
3. WELDED CONNECTIONS:
2. FILLET WELD
• The effective length of fillet weld should not be
less than 𝟒 × 𝒕𝒉𝒊𝒄𝒌𝒏𝒆𝒔𝒔 𝒐𝒇 𝒘𝒆𝒍𝒅(𝒔) i.e.
• 𝒍𝒆𝒇𝒇 = 𝟒 × 𝒕𝒉𝒊𝒄𝒌𝒏𝒆𝒔𝒔 𝒐𝒇 𝒘𝒆𝒍𝒅(𝒔)
• The size of normal fillets shall be taken as the
minimum weld leg size.
• Fillet weld should not be used if the angle
between fusion faces is less than 60° and greater
than and greater than 120° or we can say
• In weld, angle should be between 60° to 120°
• 𝒕𝒉𝒓𝒐𝒂𝒕 𝒕𝒉𝒊𝒄𝒌𝒏𝒆𝒔𝒔 𝒕𝒕 = 𝐊 × 𝒔(𝒔𝒊𝒛𝒆 𝒐𝒇 𝒘𝒆𝒍𝒅)
• 𝒕𝒕 = 𝑲 × 𝒔
𝒕𝒉𝒓𝒐𝒂𝒕 𝒕𝒉𝒊𝒄𝒌𝒏𝒆𝒔𝒔 (𝒕𝒕)
(𝒔) Size or leg of weld
Size or leg of weld
𝑨
𝑩 𝑪
𝑫
𝑫
𝑨
𝑩 𝑪
𝑨𝑩 𝒐𝒓 𝑩𝑪 = 𝒔𝒊𝒛𝒆 𝒐𝒇 𝒘𝒆𝒍𝒅
𝑩𝑫 = 𝒕𝒉𝒓𝒐𝒂𝒕 𝒕𝒉𝒊𝒄𝒌𝒏𝒆𝒔𝒔 (𝒕𝒕)
𝒕𝒐𝒆

IS RECOMMENDATIONS
5. EFFECTIVE CROSS SECTION AREA OF WELD (Throat area)
• Effective cross section area of weld = effective length of weld 
throat thickness
𝑨𝒓𝒆𝒂𝒆𝒇𝒇 = 𝑳𝒆𝒇𝒇 × 𝒕𝒕
6. LOAD CARRYING CAPACITY OF WELD/SHEAR STRENGTH OF
WELD
• P = Permissible shear stress  effective area of weld
• 𝑷 = 𝑭𝑺 × 𝑳𝒆𝒇𝒇 × 𝒕𝒕
• Fs → permissible shear stress
• Fs = 110MPa (WSM)
Fu = ultimate tensile stress in weld metal
so in LSM =
𝑭𝒖
𝟑  1.25
(1.25 for shop weld and 1.5 for field weld)
7. PITCH OF WELD
• For weld in compression zone, max pitch p = 12t or 200mm
• In tension zone, max pitch p = 16t or 200mm

Eccentric Welded Connection
1. IN PLANE ECCENTRIC CONNECTION
a) Direct Shear Stress due to P at 1
b) Torsional Shear Stress at Point 1
c) Calculate Resultant Force
𝒆 𝑷
𝟏
𝟐
𝟑
𝟒
𝑳𝟐
𝑳𝟑
𝑻 = 𝑷𝒆
𝑷
𝑭𝟐
𝑭𝟏
𝑭𝑹
𝐹1 =
𝑃
(𝑙1 + 𝑙2 + 𝑙3)𝑡𝑡
𝑇
𝐼𝑃
=
𝜏
𝑟𝑚𝑎𝑥
𝜏 = 𝑭𝟐 =
𝑇 × 𝑟𝑚𝑎𝑥
𝐼𝑃
𝑳𝟏

Eccentric Welded Connection
1. IN PLANE ECCENTRIC CONNECTION
• The effect of eccentric load at the CG of
weld group will be direct load P and
twisting moment T i.e. 𝑻 = 𝑷𝒆 where 𝒆
is measured from CG of the weld group.
• Due to direct load the direct shear stress
𝑭𝟏 developed at point 𝟏
• Due to twisting moment, the torsional
shear stress 𝑭𝟐 is developed at 𝟏
• Since these are two stresses are shear
stresses, we can find their resultant 𝑭𝑹
𝑭𝑹 = 𝐹1
2
+ 𝐹2
2
+ 2𝐹1𝐹2cos𝜃
𝒆 𝑷
𝟏
𝟐
𝟑
𝟒
𝑳𝟐
𝑳𝟑
𝑻 = 𝑷𝒆
𝑷
𝑭𝟐
𝑭𝟏
𝑭𝑹
𝑭𝑹 should not be greater than 𝜎𝑠 (WSM, 𝜎𝑠 = 110 𝑀𝑃𝑎 )
𝑭𝑹 should not be greater than 𝑓𝑠 (LSM 𝑓𝑠 =
𝑓𝑢
3×1.25
)

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.
• 𝒓𝒎𝒊𝒏 =
𝑰𝒎𝒊𝒏
𝑨
=
𝒃𝒕𝟑
𝟏𝟐×𝒃𝒕
=
𝒕
𝟏𝟐
• 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°)
𝒕
𝒃

Arrangement in A is better than B, because if one rivet fails, spacing of lacing
member does not change in A while in B, spacing will be doubled. Hence there
will be possibility of buckling of connection in B.
c
𝑭 =
𝑽
𝟐
𝒄𝒐𝒔𝒆𝒄𝜽
c
𝑭 = 𝑽𝒄𝒐𝒕𝜽
A B
𝑭 =
𝟐𝑽
𝑵
𝒄𝒐𝒕𝜽

FORCES IN LACING MEMBER
• Lacing system is designed to resist a transverse shear force
of 𝑽 = 𝟐. 𝟓% 𝒐𝒇 𝒄𝒐𝒍𝒖𝒎𝒏 𝒍𝒐𝒂𝒅.
• The transverse shear force 𝑽 is shared by lacing system
both side equally, so the transverse shear force on each
lacing bar is
𝑽
𝟐
• 2 denotes number of parallel planes
• For single lacing system of two parallel force system, the force in
each lacing bar 𝐅 =
𝑽
𝟐 sin 𝜽
• For double lacing system 𝐅 =
𝑽
𝟒 sin 𝜽

Plate Girder
• Compression Flange:
• It consists of flange plate, flange
angle and web equivalent
• Web equivalent is the web area
embedded between two flange angle
• In compression zone flange, web
equivalent is taken as
𝑎𝑟𝑒𝑎 𝑜𝑓 𝑤𝑒𝑏
6
or
𝑎𝑤
6
• Tension Flange:
• It consists of flange plate, angle and
web equivalent
• In tension zone, web equivalent is
taken as
𝑎𝑤
8
𝑾𝒆𝒃 𝑷𝒍𝒂𝒕𝒆
𝑭𝒍𝒂𝒏𝒈𝒆 𝑨𝒏𝒈𝒍𝒆
𝑭𝒍𝒂𝒏𝒈𝒆 𝒄𝒐𝒗𝒆𝒓 𝒑𝒍𝒂𝒕𝒆
𝑭𝒍𝒂𝒏𝒈𝒆 𝑨𝒏𝒈𝒍𝒆
𝒈𝒂𝒑
𝑪𝒍𝒆𝒂𝒓 𝒅𝒆𝒑𝒕𝒉 𝒅𝟏

Case1: LOAD CARRYING
CAPACITY FOR PLATE
• Load Carrying Capacity of tension member:
• Safe load carrying capacity:
•Calculation of 𝑨𝒏𝒆𝒕
• Chain riveting
• 𝑨𝒏𝒆𝒕 = 𝐁 − 𝟑𝐝 𝐭
• Diamond riveting
• 𝑨𝒏𝒆𝒕 = 𝐁 − 𝐝 𝐭

CAPACITY FOR PLATE
• Load Carrying Capacity of tension
member:
• Staggered riveting
• The critical section would be …..
4 − 1 − 3 − 5
4 − 1 − 7
4 − 1 − 2 − 3 − 5
4 − 1 − 2 − 6
𝒈𝒂𝒖𝒈𝒆
4
2
3
7 6
5
1

CAPACITY FOR PLATE
• Load Carrying Capacity of
tension member:
• The critical section would be the
minimum area that would be 4-1-2-3-5
• Along the critical sections, for each
inclined leg, correspondingly
𝒓𝟐
𝒕
𝟒𝒈
term
is added to net area where 𝒓 is the
staggered horizontal distance of the
inclined leg along the critical path or
section and g is the gauge distance
corresponding to the inclined leg
4
2
3
7 6
5
1
4 − 1 − 3 − 5
4 − 1 − 7
4 − 1 − 2 − 3 − 5
4 − 1 − 2 − 6

CAPACITY FOR PLATE
• Load Carrying Capacity of tension member:
• 𝑨𝒏𝒆𝒕 = 𝐁 − 𝐧𝟏𝐝 +
𝒓𝟏
𝟐
𝟒𝒈𝟏
+
𝒓𝟐
𝟐
𝟒𝒈𝟐
𝐭
• 𝒏𝟏 is no. of rivets along critical section
• d is gross dia or hole dia
• 𝒓𝟏 and 𝒓𝟐 are staggered pitch
• 𝒈𝟏 and 𝒈𝟐 are staggered gauge
𝒓𝟏
4
2
3
7 6
5
1
𝒓𝟐

Case 2: LOAD CARRYING CAPACITY FOR ANGLE
A. If single angle tension member is
connected to gusset plate, then …
• Calculation of 𝑨𝒏𝒆𝒕
• For angle
• 𝑨𝒏𝒆𝒕 = 𝑨𝟏 + 𝒌𝑨𝟐
• k =
𝟑𝑨𝟏
𝟑𝑨𝟏+𝑨𝟐
• K= shear lag effect
• Where 𝑨𝟏 is net area of connected leg
• 𝑨𝟏 = (gross area of connected leg - area of
rivet hole)
• 𝑨𝟐 is gross area of unconnected leg/outstand
leg
• 𝑨𝟏 = 𝒂 − 𝒅 −
𝒕
𝟐
𝒕
• 𝑨𝟐 = 𝒃 −
𝒕
𝟐
𝒕
gussete plate
angle
𝒂
𝒃
𝑨𝟏 = (𝒂 − 𝒅 −
𝒕
𝟐
)𝒕
𝑨𝟐 = (𝒃 −
𝒕
𝟐
)𝒕

B. If two angles are placed back
to back and connected to ONE
side of gusset plate
• 𝑨𝒏𝒆𝒕 = 𝑨𝟏 + 𝒌𝑨𝟐 where K= shear
lag effect
k =
𝟓𝑨𝟏
𝟓𝑨𝟏+𝑨𝟐
• If TACK rivets are not provided along
their flange then each angle behaves
independently hence factor k =
𝟑𝑨𝟏
𝟑𝑨𝟏+𝑨𝟐
𝑆𝑡𝑟𝑢𝑐𝑡𝑢𝑟𝑎𝑙 𝑅𝑖𝑣𝑒𝑡
𝑇𝑎𝑐𝑘 𝑅𝑖𝑣𝑒𝑡

C. If two angles are placed back to
back and connected to both sides of
gusset plate
• 𝑨𝒏𝒆𝒕 = 𝑨𝟏 + 𝒌𝑨𝟐 where K= 1
• 𝑨𝒏𝒆𝒕 = 𝑨𝟏 + 𝑨𝟐
• It is the most efficient way of
connecting, then load carrying
capacity is maximum.
• If the two angles do not have rivet,
then each angle behaves
independently hence factor k =
𝟑𝑨𝟏
𝟑𝑨𝟏+𝑨𝟐

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1. Arithmetic Progression Method
• Increase in population from
decade to decade is assumed to be
constant
Population Forecasting Methods
𝑷𝒏 = 𝑷𝟎 + 𝒏ഥ
𝒙
𝑷𝒏 =Projected population after n decades
𝑷𝟎 = initial population/ last census
𝒏 = number of decades between now and future
ഥ
𝒙 =average increase in population per decade

Que. 1 The population of 5 decades are
given. Find out the population after 1 and 6
decades beyond last census by arithmetic
progression method.
Year Population
1930 25000
1940 28000
1950 34000
1960 42000
1970 47000

progression method.
𝑨𝒗𝒆𝒓𝒂𝒈𝒆 𝑰𝒏𝒄𝒓𝒆𝒂𝒔𝒆 =
𝟐𝟐𝟎𝟎𝟎
𝟒
Year Population
1930 25000
1940 28000
1950 34000
1960 42000
1970 47000
𝒙
ഥ
Increase
3000
6000
8000
5000
ഥ
𝒙 = 𝟓𝟓𝟎𝟎
1. 𝑷𝒐𝒑𝒖𝒍𝒂𝒕𝒊𝒐𝒏 𝒂𝒇𝒕𝒆𝒓 𝟏 𝒅𝒆𝒄𝒂𝒅𝒆 𝑷𝟏 = ?
𝑷𝟏 = 𝑷𝟎 + 𝒏ഥ
𝒙
⇒ 𝑷𝟏𝟗𝟖𝟎 = 𝟒𝟕𝟎𝟎𝟎 + (𝟏) × 𝟓𝟓𝟎𝟎
⇒ 𝑷𝟏𝟗𝟖𝟎 = 𝟓𝟐𝟓𝟎𝟎

progression method.
𝑨𝒗𝒆𝒓𝒂𝒈𝒆 𝑰𝒏𝒄𝒓𝒆𝒂𝒔𝒆 =
𝟐𝟐𝟎𝟎
𝟒
Year Population
1930 25000
1940 28000
1950 34000
1960 42000
1970 47000
𝒙
ഥ
Increase
3000
6000
8000
5000
ഥ
𝒙 = 𝟓𝟓𝟎𝟎
2. 𝑷𝒐𝒑𝒖𝒍𝒂𝒕𝒊𝒐𝒏 𝒂𝒇𝒕𝒆𝒓 𝟔 𝒅𝒆𝒄𝒂𝒅𝒆𝒔 𝑷𝟔 = ?
𝑷𝟔 = 𝑷𝟎 + 𝒏ഥ
𝒙
⇒ 𝑷𝟐𝟎𝟑𝟎 = 𝟒𝟕𝟎𝟎𝟎 + (𝟔) × 𝟓𝟓𝟎𝟎
⇒ 𝑷𝟐𝟎𝟑𝟎 = 𝟖𝟎𝟎𝟎𝟎

2. Geometric Progression Method or Geometric Increase Method
• In this method Percentage Increase in
population from decade to decade is
assumed to be constant
𝑷𝒏 = 𝑷𝟎 𝟏 +
𝒓
𝟏𝟎𝟎
𝒏
𝑷𝟎 = population of last known decade
𝒓 =geometric mean rate of increase in population per decade
𝑟 = 𝑚
𝑟1𝑟2𝑟3𝑟4 … 𝑟𝑚
𝑵𝒐𝒕𝒆: 𝑮𝑶𝑰 𝑴𝒂𝒏𝒖𝒂𝒍 𝒓𝒆𝒄𝒐𝒎𝒎𝒆𝒏𝒅𝒔 𝑮𝒆𝒐𝒎𝒆𝒕𝒓𝒊𝒄
𝑷𝒓𝒐𝒈𝒓𝒆𝒔𝒔𝒊𝒗𝒆 𝑴𝒆𝒕𝒉𝒐𝒅 𝒃𝒆𝒄𝒂𝒖𝒔𝒆 𝒊𝒕 𝒈𝒊𝒗𝒆𝒔 𝒕𝒉𝒆
𝒎𝒂𝒙𝒊𝒎𝒖𝒎 𝒗𝒂𝒍𝒖𝒆 𝒐𝒖𝒕 𝒐𝒇 𝒐𝒕𝒉𝒆𝒓 𝒎𝒆𝒕𝒉𝒐𝒅𝒔

Que. 2 Determine the future population of a
town by Geometric Increase method in the
year 2011.
Year Population (in 1000)
1951 93
1961 111
1971 132
1981 161

town by Geometric Increase method in the
year 2011.
Year Population (in 1000)
1951 93
1961 111
1971 132
1981 161
Increase
18
21
29
Percentage Increase
19.35 %
18.91 %
21.96 %
𝟏𝟖
𝟗𝟑
× 𝟏𝟎𝟎 =
𝟐𝟏
𝟏𝟏𝟏
× 𝟏𝟎𝟎 =
𝟐𝟗
𝟏𝟑𝟐
× 𝟏𝟎𝟎 =
𝑷𝒏 = 𝑷𝟎 𝟏 +
𝒓
𝟏𝟎𝟎
𝒏
𝒓 = 𝒎
𝒓𝟏𝒓𝟐𝒓𝟑𝒓𝟒 … 𝒓𝒎
⇒ 𝒓 =
𝟑
𝟏𝟗. 𝟑𝟓 × 𝟏𝟖. 𝟗𝟏 × 𝟐𝟏. 𝟗𝟔
⇒ 𝒓 = 𝟐𝟎. 𝟗 % 𝒑𝒆𝒓 𝒅𝒆𝒄𝒂𝒅𝒆
𝑷𝟐𝟎𝟏𝟏 = 𝑷𝟏𝟗𝟖𝟏 𝟏 +
𝟐𝟎. 𝟗
𝟏𝟎𝟎
𝟑
= 𝟏𝟔𝟏 𝟏 +
𝟐𝟎.𝟗
𝟏𝟎𝟎
𝟑
= 𝟐𝟖𝟒. 𝟓𝟏𝟒

c) Incremental Increase Method
• Combination of Arithmetic and Geometric Increase method
• Actual increase in each decade is found
• Average increment of increases is found
𝒙 +
𝒏(𝒏 + 𝟏)
𝟐
ഥ
𝒚
𝑷𝟎 = population of last known decade
ഥ
𝒙 =average increase of population of known decades
Population after n decades from present is
given by
ഥ
𝒚 =average of incremental increase of known decades

town by Incremental Increase method in the
year 2000.
Year Population
1940 23798624
1950 46978325
1960 54786437
1970 63467823
1980 69077421
𝒙 +
𝒏(𝒏 + 𝟏)
𝟐
ഥ
𝒚

town by Incremental Increase method in the
year 2000.
Year Population
1940 23798624
1950 46978325
1960 54786437
1970 63467823
1980 69077421
Increase
23179701
7808112
8681386
5609598
𝒙 +
𝒏(𝒏 + 𝟏)
𝟐
ഥ
𝒚
Incremental Increase
-15371589
873274
-3071788
ഥ
𝒙 =
𝟒𝟓𝟐𝟕𝟖𝟗𝟕𝟗
𝟒
ഥ
𝒙 =average increase of
population of known decades
ഥ
𝒚 =average of incremental
increase of known decades
ഥ
𝒚 =
−𝟏𝟕𝟓𝟕𝟎𝟏𝟎𝟑
𝟑
ഥ
𝒙 = 𝟏𝟏𝟑𝟏𝟗𝟔𝟗𝟗. 𝟐 ഥ
𝒚 = −𝟓𝟖𝟓𝟔𝟕𝟎𝟏

𝒙 +
𝒏 𝒏 + 𝟏
𝟐
ഥ
𝒚
⇒ 𝑷𝟐𝟎𝟎𝟎 = 𝑷𝟏𝟗𝟖𝟎 + (𝟐)(𝟏𝟏𝟑𝟏𝟗𝟔𝟗𝟗. 𝟐 ) +
𝟐 𝟐 + 𝟏
𝟐
(−𝟓𝟖𝟓𝟔𝟕𝟎𝟏)
⇒ 𝑷𝟐𝟎𝟎𝟎 = 69077421 + (𝟐)(𝟏𝟏𝟑𝟏𝟗𝟔𝟗𝟗. 𝟐 ) +
𝟐 𝟐 + 𝟏
𝟐
(−𝟓𝟖𝟓𝟔𝟕𝟎𝟏)
⇒ 𝑷𝟐𝟎𝟎𝟎 = 𝟕𝟒𝟏𝟒𝟔𝟕𝟏𝟔. 𝟔
⇒ 𝑷𝟐𝟎𝟎𝟎 = 𝟕𝟒𝟏𝟒𝟔𝟕𝟏𝟕

GOI Manual Recommends..
1. Arithmetic Increase Method is
used for old cities, where growth
rate is constant
2. For new and younger cities, we will
use geometric Progression method
3. Whenever there is negative rate of
increase, incremental increase
method is used
4. Incremental Increase Method
generally gives values in between
Arithmetic progression method
and Geometric Progression
Method

Water Demand
5. Fire Demand
• For a total amount of water consumption,
for a city of 50 Lacs population, it hardly
amounts to 1 LPCD, but this water should
be easily available and kept always stored
in service reservoirs

5. Fire Demand
When population exceeds 50,000 the water required for fire demand can be
computed using the empirical formula:
𝑸 = 𝒒𝒖𝒂𝒏𝒕𝒊𝒕𝒚 𝒐𝒇 𝒘𝒂𝒕𝒆𝒓 𝒊𝒏 𝑳𝒊𝒕𝒓𝒆𝒔 𝒑𝒆𝒓 𝒎𝒊𝒏𝒖𝒕𝒆
𝑷 = 𝒑𝒐𝒑𝒖𝒍𝒂𝒕𝒊𝒐𝒏 𝒊𝒏 𝒕𝒉𝒐𝒖𝒔𝒂𝒏𝒅𝒔
1. 𝑲𝒖𝒊𝒄𝒉𝒍𝒊𝒏𝒈′
𝒔 𝑭𝒐𝒓𝒎𝒖𝒍𝒂 𝒇𝒐𝒓 𝒇𝒊𝒓𝒆 𝑫𝒆𝒎𝒂𝒏𝒅 𝑸 = 𝟑𝟏𝟖𝟐 𝑷
2. 𝑭𝒓𝒆𝒆𝒎𝒂𝒏′
𝒔 𝑭𝒐𝒓𝒎𝒖𝒍𝒂 𝑸 = 𝟏𝟏𝟑𝟔
𝑷
𝟏𝟎
+ 𝟏𝟎
3. 𝑵𝒂𝒕𝒊𝒐𝒏𝒂𝒍 𝑩𝒐𝒂𝒓𝒅 𝒐𝒇 𝑭𝒊𝒓𝒆 𝑼𝒏𝒅𝒆𝒓 𝑾𝒓𝒊𝒕𝒆𝒓′
𝒔 𝑭𝒐𝒓𝒎𝒖𝒍𝒂
a) When Population is ≤ 𝟐, 𝟎𝟎, 𝟎𝟎𝟎 𝑸 = 𝟒𝟔𝟑𝟕 𝑷(𝟏 − 𝟎. 𝟎𝟏 𝑷)
b) When Population is > 𝟐, 𝟎𝟎, 𝟎𝟎𝟎 𝑸 = 𝟓𝟒𝟔𝟎𝟎 𝒍/𝒎𝒊𝒏
4. 𝑩𝒖𝒔𝒕𝒐𝒏′𝒔 𝑭𝒐𝒓𝒎𝒖𝒍𝒂 𝑸 = 𝟓𝟔𝟔𝟑 𝑷

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Que 4. Compute the fire demand for a city having population of 140 000 as per
all formulae.
PRACTICE QUESTIONS
PRACTICE QUESTIONS
1. 𝑲𝒖𝒊𝒄𝒉𝒍𝒊𝒏𝒈′𝒔 𝑭𝒐𝒓𝒎𝒖𝒍𝒂 𝒇𝒐𝒓 𝒇𝒊𝒓𝒆 𝑫𝒆𝒎𝒂𝒏𝒅 𝑸 = 𝟑𝟏𝟖𝟐 𝑷
2. 𝑭𝒓𝒆𝒆𝒎𝒂𝒏′
𝒔 𝑭𝒐𝒓𝒎𝒖𝒍𝒂 𝑸 = 𝟏𝟏𝟑𝟔 𝟏𝟎 + 𝟏𝟎
3. 𝑵𝒂𝒕𝒊𝒐𝒏𝒂𝒍 𝑩𝒐𝒂𝒓𝒅 𝒐𝒇 𝑭𝒊𝒓𝒆 𝑼𝒏𝒅𝒆𝒓 𝑾𝒓𝒊𝒕𝒆𝒓′𝒔 𝑭𝒐𝒓𝒎𝒖𝒍𝒂
a) When Population is ≤ 𝟐, 𝟎𝟎, 𝟎𝟎𝟎 𝑸 = 𝟒𝟔𝟑𝟕 𝑷(𝟏 − 𝟎. 𝟎𝟏 𝑷)
b) When Population is > 𝟐, 𝟎𝟎, 𝟎𝟎𝟎 𝑸 = 𝟓𝟒𝟔𝟎𝟎 𝒍/𝒎𝒊𝒏
4. 𝑩𝒖𝒔𝒕𝒐𝒏′
𝒔 𝑭𝒐𝒓𝒎𝒖𝒍𝒂 𝑸 = 𝟓𝟔𝟔𝟑 𝑷

Que 4. Compute the fire demand for a city having population of 140 000 as per
all formulae.
PRACTICE QUESTIONS
PRACTICE QUESTIONS

1. Maximum Daily Demand
Various types of Demand for Design
= 𝟏. 𝟖 × 𝑨𝒗𝒆𝒓𝒂𝒈𝒆 𝑫𝒂𝒊𝒍𝒚 𝒄𝒐𝒏𝒔𝒖𝒎𝒑𝒕𝒊𝒐𝒏
= 𝒉𝒐𝒖𝒓𝒍𝒚 𝒗𝒂𝒓𝒊𝒂𝒕𝒊𝒐𝒏 𝒇𝒂𝒄𝒕𝒐𝒓 ×
𝒎𝒂𝒙 𝒅𝒂𝒊𝒍𝒚 𝒄𝒐𝒏𝒔𝒖𝒎𝒑𝒕𝒊𝒐𝒏
𝟐𝟒
2. Maximum Hourly Demand of Maximum Day/Peak Demand
= 𝟏. 𝟓 ×
𝟏. 𝟖 𝒒
𝟐𝟒
= 𝟐. 𝟕 ×
𝒒
𝟐𝟒
= 𝟏. 𝟖 𝒒
= 1.5 x Maximum daily demand/24
= 1.5 x (1.8 x average daily demand)/24
= 2.7 x average daily demand/24
= 2.7 x annual average hourly demand

3. Coincident Draft/Demand/Supply
Various types of Demand for Design
= 𝒎𝒂𝒙𝒊𝒎𝒖𝒎 𝒅𝒂𝒊𝒍𝒚 𝒅𝒆𝒎𝒂𝒏𝒅 + 𝒇𝒊𝒓𝒆 𝒅𝒆𝒎𝒂𝒏𝒅
= 𝒎𝒂𝒙𝒊𝒎𝒖𝒎 𝒐𝒇 𝑪𝒐𝒊𝒏𝒄𝒊𝒅𝒆𝒏𝒕 𝒅𝒓𝒂𝒇𝒕, 𝒎𝒂𝒙𝒊𝒎𝒖𝒎 𝒉𝒐𝒖𝒓𝒍𝒚 𝒄𝒐𝒏𝒔𝒖𝒎𝒑𝒕𝒊𝒐𝒏
4. Total Draft

2𝐻2 + 𝑂2 → 2𝐻2𝑂
Stoichiometric Principles
1 𝑚𝑜𝑙𝑒 ≡ 𝑊𝑒𝑖𝑔ℎ𝑡 𝑖𝑛 𝑔𝑟𝑎𝑚 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑 𝑏𝑦 𝑡ℎ𝑒 𝑐ℎ𝑒𝑚𝑖𝑐𝑎𝑙 𝑓𝑜𝑟𝑚𝑢𝑙𝑎
4𝑔 32𝑔 36𝑔
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑀𝑜𝑙𝑒𝑠 =
𝐺𝑖𝑣𝑒𝑛 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓𝐶𝑜𝑚𝑝𝑜𝑢𝑛𝑑
𝑀𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑊𝑒𝑖𝑔ℎ𝑡
𝐸𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑊𝑒𝑖𝑔ℎ𝑡 =
𝑀𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑊𝑒𝑖𝑔ℎ𝑡
𝑉𝑎𝑙𝑒𝑛𝑐𝑦
⇒ 𝐸𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑊𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝐶𝑎2+ =
40
2
⇒ 𝐸𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑊𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝐶𝑎𝐶𝑂3 =
100
2

Stoichiometric Principles
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑔𝑟𝑎𝑚 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 =
=
𝑔𝑖𝑣𝑒𝑛 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑜𝑚𝑝𝑜𝑢𝑛𝑑
𝐸𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑊𝑒𝑖𝑔ℎ𝑡
𝐶𝑎𝐶𝑂3 → 𝐶𝑎2+ + 𝐶𝑂3
2−
1 ∶ 1 ∶ 1
1 𝑔𝑟𝑎𝑚 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑜𝑓 𝑎𝑛𝑦𝑡ℎ𝑖𝑛𝑔 𝑟𝑒𝑎𝑐𝑡𝑠 𝑤𝑖𝑡ℎ 1𝑔𝑚 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑜𝑓
𝑎𝑛𝑦 𝑜𝑡ℎ𝑒𝑟 𝑡ℎ𝑖𝑛𝑔 𝑡𝑜 𝑔𝑖𝑣𝑒 1𝑔𝑚 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑜𝑓 𝑟𝑒𝑠𝑢𝑙𝑡𝑎𝑛𝑡
𝑁𝑜𝑟𝑚𝑎𝑙𝑖𝑡𝑦 =
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑔𝑟𝑎𝑚 𝐸𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡𝑠
𝐿𝑖𝑡𝑟𝑒𝑠 𝑜𝑓 𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛

2. Chemical Parameters
iii. Alkalinity
• For Alkalinity Measurement, 0.02N 𝑯𝟐𝑺𝑶𝟒 is used in titration. 1 ml of acid
i.e. 0.02N 𝑯𝟐𝑺𝑶𝟒 gives 1 mg/L value of Alkalinity expressed as 𝑪𝒂𝑪𝑶𝟑
0.02
1000
×
100
2
× 1000 = 1
𝑚𝑙
𝑚𝑙
𝑒𝑞 𝑤𝑡 𝑜𝑓 𝐶𝑎𝐶𝑂3
𝑁𝑜𝑟𝑚𝑎𝑙𝑖𝑡𝑦 =
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑔𝑟𝑎𝑚 𝐸𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡𝑠
𝐿𝑖𝑡𝑟𝑒𝑠 𝑜𝑓 𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛

Que . Water contains 210g of 𝐶𝑂3
2−
, 122g 𝐻𝐶𝑂3− and 68g of 𝑂𝐻−. What is the
total alkalinity of water expressed as 𝐶𝑎𝐶𝑂3?
PRACTICE QUESTIONS
PRACTICE QUESTIONS

Que . Water contains 210g of 𝐶𝑂3
2−
, 122g 𝐻𝐶𝑂3
−
and 68g of 𝑂𝐻−. What is the
total alkalinity of water expressed as 𝐶𝑎𝐶𝑂3?
PRACTICE QUESTIONS
PRACTICE QUESTIONS

2. Chemical Parameters
iv. Hardness
• The Hardness is expressed as 𝑪𝒂𝑪𝑶𝟑 equivalent of Calcium and
Magnesium present in water
𝐻𝑎𝑟𝑑𝑛𝑒𝑠𝑠 𝑎𝑠 𝑪𝒂𝑪𝑶𝟑 𝒎𝒈/𝒍 =
[𝑾𝒕 𝒐𝒇 𝑪𝒂𝟐+]
[𝑬𝒒 𝑾𝒕 𝒐𝒇 𝑪𝒂𝟐+]
× [𝒆𝒒 𝒘𝒕 𝒐𝒇 𝑪𝒂𝑪𝑶𝟑]
+
[𝑾𝒕 𝒐𝒇 𝑴𝒈𝟐+]
[𝑬𝒒 𝑾𝒕 𝒐𝒇 𝑴𝒈𝟐+]
× [𝒆𝒒 𝒘𝒕 𝒐𝒇 𝑪𝒂𝑪𝑶𝟑]
⇒ 𝐸𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑊𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝐶𝑎2+ =
40
2
⇒ 𝐸𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑊𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝐶𝑎𝐶𝑂3 =
100
2
⇒ 𝐸𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑊𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑀𝑔2+ =
24
2
Hardness as CaCO3 in mg/L Degree of Hardness
0-55 Soft Water
56-100 Slightly Hard
101-200 Moderately Hard
201-500 Very Hard

Alkalinity and Hardness
𝐶𝑎𝑟𝑏𝑜𝑛𝑎𝑡𝑒 𝐻𝑎𝑟𝑑𝑛𝑒𝑠𝑠 = 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑜𝑓{𝑇𝑜𝑡𝑎𝑙 𝐻𝑎𝑟𝑑𝑛𝑒𝑠𝑠, 𝐴𝑙𝑘𝑎𝑙𝑖𝑛𝑖𝑡𝑦}

Alkalinity and Hardness
𝐶𝑎𝑟𝑏𝑜𝑛𝑎𝑡𝑒 𝐻𝑎𝑟𝑑𝑛𝑒𝑠𝑠 = 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑜𝑓{𝑇𝑜𝑡𝑎𝑙 𝐻𝑎𝑟𝑑𝑛𝑒𝑠𝑠, 𝐴𝑙𝑘𝑎𝑙𝑖𝑛𝑖𝑡𝑦}
320 120

3. Sedimentation
Design of Sedimentation tank
VH= horizontal velocity or flow velocity
Vs= settling velocity
Time of horizontal flow =
𝑳
𝑽𝑯
=
𝑳
𝑸
𝑩𝑯
=
𝑳𝑩𝑯
𝑸
𝐃𝐞𝒕𝒆𝒏𝒕𝒊𝒐𝒏 𝒕𝒊𝒎𝒆 𝑫𝑻 =
𝑽𝒐𝒍𝒖𝒎𝒆
𝑸
L
H
VH
Vs
Time of Falling through height H => 𝐭 =
𝑯
𝑽𝑺
If first assumption is valid, t=DT
𝑫𝑻 =
𝑳𝑩𝑯
𝑸
= 𝒕 =
𝑯
𝑽𝑺
=> 𝑽𝑺 =
𝑸
𝑩𝑳
Actual settling velocity 𝑽𝒔 =
𝒈 𝑮𝒔
−𝟏 𝒅𝟐
𝟏𝟖𝝂
𝑺𝒖𝒓𝒇𝒂𝒄𝒆 𝒐𝒗𝒆𝒓𝒇𝒍𝒐𝒘 𝒓𝒂𝒕𝒆 𝑽𝟎 =
𝑸
𝑩𝑳
Surface overflow rate can be thought of a settling
velocity of that particle which if introduced at
the top most point at inlet will reach the bottom
most point at outlet. This Vs is shown by another
symbol V0

234
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238
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240
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243
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7. Biochemical Oxygen Demand (BOD)
• 𝐵𝑂𝐷 = 𝐷𝑂𝑖 − 𝐷𝑂𝑓 × 𝐷. 𝐹 .
𝐷𝑖𝑙𝑢𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟 𝐷𝐹 =
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑑𝑖𝑙𝑢𝑡𝑒𝑑 𝑠𝑜𝑚𝑝𝑙𝑒
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑢𝑛𝑑𝑖𝑙𝑢𝑡𝑒𝑑 𝑠𝑎𝑚𝑝𝑙𝑒 𝑡𝑎𝑘𝑒𝑛

Que 1. If initial DO = 5 mg/L, final DO = 2 mg/L.
5ml of sample is mixed to form 100 ml of diluted sample.
Find BOD.

Que If initial DO = 5 mg/L, final DO = 2 mg/L.
5ml of sample is mixed to form 100 ml of diluted sample.
Find BOD.
5ml of sample is mixed with 95ml of aerated water
𝐵𝑂𝐷 = (5 − 2) ×
100
5
⇒𝐵𝑂𝐷 = 60𝑚𝑔/𝐿

Que 2. If initial DO = 5 mg/L, final DO =0 mg/L at the end of 5
days.
5ml of sample is mixed with 95 ml of aerated water. Find
BOD after 5 days.

Que 2. If initial DO = 5 mg/L, final DO =0 mg/L at the end of 5
days.
5ml of sample is mixed with 95 ml of aerated water. Find
BOD after 5 days.
𝐶𝑎𝑛 𝑛𝑜𝑡 𝑏𝑒 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒𝑑 𝑎𝑠 𝑤𝑒 𝑑𝑜 𝑛𝑜𝑡 𝑘𝑛𝑜𝑤𝑛 𝑤ℎ𝑒𝑛 𝑖𝑡 𝑏𝑒𝑐𝑎𝑚𝑒 0

Let Lt = amount of organic matter present at any time t
t = time in days
K=rate constant (unit=per day) or deoxygenation constant
L0 = maximum amount of organic matter present
𝒅 𝑳𝒕
𝒅𝒕
= −𝑲 𝑳𝒕
=>
𝒅 𝑳𝒕
𝑳𝒕
= −𝑲 𝒅𝒕
=> 𝒍𝒏 𝑳𝒕 𝑳𝟎
𝑳𝒕
= −𝑲 𝒕𝟐 − 𝒕𝟏
=> 𝑳𝒕 = 𝑳𝟎 𝒆−𝒌𝒕
𝑩𝑶𝑫𝒕 = 𝑳𝒐 − 𝑳𝒕
= 𝑳𝟎 − 𝑳𝟎 𝒆−𝒌𝒕
𝑩𝑶𝑫𝒕 = 𝑳𝟎 𝟏 − 𝒆−𝒌𝒕

• Deoxygenation constant for given system depends on type of
impurities present in waste water.
• For sample impurities exp-sugar, deoxygenation constant will be
more and for complex impurities like phenol, toulene , aldehydes,
ketones, etc K will be less.
• In general, deoxygenation constant can be under base e or base 10
𝑩𝑶𝑫𝒕 = 𝑳𝟎 𝟏 − 𝒆−𝒌𝒕
• When it is not given in question whether base e or base 10, then we
will use base e.
𝑲𝑫
𝑻°𝒄
= 𝑲𝑫
𝟐𝟎°𝒄
𝟏. 𝟎𝟒𝟕 𝑻−𝟐𝟎°
This equation is called as Vanthoff’s Aeheniers Equation

Population Equivalent
• It indicates strength of industrial waste water for estimating
the treatment required at the municipal treatment plant
• Average BOD of domestic sewage is 80g/capita/day
• 𝑃𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 =
𝑡𝑜𝑡𝑎𝑙 𝐵𝑂𝐷5
𝑜𝑓 𝑡ℎ𝑒 𝑖𝑛𝑑𝑢𝑠𝑡𝑟𝑦 𝑖𝑛 𝑘𝑔/𝑑𝑎𝑦
0.08 𝑘𝑔/𝑑𝑎𝑦

ENVIRONMENTAL ENGG-2 CIVIL ENGINEERING
𝟗𝟓 ×
𝟏𝟎𝟔
𝒍/𝒅𝒂𝒚
𝟑𝟎𝟎𝒎𝒈/𝒍 𝟕𝟓𝒎𝒈/𝒅𝒂𝒚

× 𝟏𝟎𝟔
× 𝟏𝟎𝟔

ENVIRONMENTAL ENGG-2 CIVIL ENGINEERING
𝟗𝟓 ×
𝟏𝟎𝟔
𝒍/𝒅𝒂𝒚
𝟑𝟎𝟎𝒎𝒈/𝒍 𝟕𝟓𝒎𝒈/𝒅𝒂𝒚

× 𝟏𝟎𝟔
× 𝟏𝟎𝟔
𝑃𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝐸𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑜𝑓 𝑐𝑖𝑡𝑦 =
𝐵𝑂𝐷 𝑝𝑟𝑜𝑑𝑢𝑐𝑒𝑑 𝑝𝑒𝑟 𝑑𝑎𝑦 𝑖𝑛 𝑐𝑖𝑡𝑦
𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝐵𝑂𝐷 𝑝𝑒𝑟 𝑐𝑎𝑝𝑖𝑡𝑎𝑙
=
95 × 106 × 300
75
= 380 × 106

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