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SSC JE Year-wise Average Questions Analysis
Number of Questions Number of Questions Number of Questions Number of Questions Average Number of
Subjects
(2016) (2018) (2019) (2020) Questions
Building Material 17 16 12 14 14.75
Estimation 10 9.5 10 7.5 9.25
SOM 3 5.5 3 4 3.875

Structural Analysis 4 6.5 3 3 4.125
RCC 9 7 12 10.5 9.625
Steel 5 7 4 3.5 4.875
Geotech 9 7 9 11 9
Transportation 9 9.5 11 7.5 9.25
Survey 9 9 9 11.5 9.625
Hydrology & Irrigation 9 7.5 10 10 9.125
Fluid Mechanics 9 6 6 9 7.5
Environmental 8 9.5 11 8.5 9.25
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Shubham Aggarwal
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Design of Concrete Structures

List of Major Indian Standard Codes Relating to Reinforced Concrete
1. IS 269 : 2015 : Specification for 33 ,43 & 53 Grade Ordinary Portland Cement
2. IS 432 (Part-1) : 1982 : Specification for Mild Steel and Medium Tensile steel
bars for concrete reinforcement.
3. IS 1893 : 2016 : Earthquake Resistance Design of Structure
4. IS 3370 : Design of Water Tank
5. IS 1343 : Prestressed Concrete
6. IS 383 : Requirements for Aggregates
7. IS 875 : Design Load for Buildings & Structures
8. IS 516 : Methods of Test for Strength of Concrete
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IS 456 : 2000
Plain and Reinforced Concrete
Depending upon the Grade of concrete , Various Grades are used:
Minimum Grade of Concrete used for

1. PCC is M15
2. RCC is M20
Q. Among the following which is the most

important code in civil engineering?
A. IS 875
Code Use
B. IS 456
C. IS 383 IS 875 Design loads (other than earthquake) for
buildings and structures
D. IS 269
IS 456 Design of PCC & RCC
IS 383 Requirements for Aggregates
Answer: B
IS 269 Specification for 33 ,43 & 53 Grade Ordinary
Self Designed Portland Cement
Q. Which of the following Indian Standard

Codes provides conservatively Wind loads
for buildings and structures?
A. IS 875 (part-4)-1987 IS 875 : 1987 Design loads (other than earthquake) for
buildings and structures
B. IS 875 (part-3)-1987
C. IS 875 (part-2)-1987 Part -1 Dead Load
D. IS 875 (part-1)-1987 Part - 2 Imposed Load

Part – 3 Wind Load
Answer: B
Part - 4 Snow Load
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Choose the CORRECT statements that explain the

properties of concrete.
1. It has the ability to gain strength over time
2. It is one of the most durable materials in existence
3. It has half the lifespan of other construction materials
4. It resists rotting, rusting and burning
कंक्रीट के गुणों की व्याख्या करने वाले सही कथनों का चयन कीजिए।

1. इसमें समय के साथ मिबत ू ी हाससल करने की क्षमता होती ह
2. यह अजततत्व में सबसे टटकाऊ (ड्यूरेबल) मटे ररयल्स में से एक ह
3. इसमें अन्य ननमााण सामग्री का आधा िीवनकाल (हाफ लाईफतपेन)
ह
4. यह सड़ने, िंग लगने और िलने का प्रनतरोध करता ह
1) Only 1, 2 & 3 statements / केवल 1, 2 और 3 कथन

4) All 1, 2, 3, & 4 statements / सभी 1, 2, 3 और 4 कथन

Grades of Concrete
Grade of IS 456 : 2000 IS 456 : 2013

Concrete
Ordinary 3 3
(M10 – M20) (M10 – M20)
Standard 7 8
(M25 – M55) (M25 – M60)
High Strength 5 8
(M60 – M80) (M65 – M100)

Size of Aggregates
Type of Work Size
Most Work 20 mm
No Restriction to the ≥ 40 mm
flow of concrete
Thin Sections with 10 mm
closely spaced R/f

Water (Permissible limits for solids shall be as liven in Table 1.)
Type of Solids Tested As per Maximum Permissible Limit
Organic Solids IS 3025 (Part 18) 200 mg/L

Inorganic Solids IS 3025 (Part 18) 3000mg/L
Sulphates (as SO3) IS 3025 (Part 24) 400mg/L
Chlorides (as Cl) IS 3025 (Part 32) 2000mg/L for Concrete not
containing embedded steel
500mg/L for reinforced concrete

work
Suspended Matter IS 3025 (Part 17) 2000mg/L

Water
Average 28 Days Compressive Strength of at least three 150mm concrete
cubes prepared with water proposed to be used shall not be less than 90%
of the Average of the Strength of 3 similar concrete cubes prepared with
distilled water.
Strength
Not Less
than 90%
Distilled Proposed
Distilled Water
Water Water
Cubes
Question: Compressive Strength of Cubes prepared with Self Designed

the help of Proposed water in comparison to cubes Answer: C
prepared with distilled water should not fall under the
category for which Compressive strength is?
a) more than 90%
b) more than 95%
c) less than 90%
d) less than 95%

Water
The pH of Water Shall not be less than 6.
Sea Water
Mixing or curing of concrete with sea water is not
recommended because of presence of harmful salts
in sea water.
Question: As Per IS 456 : 2000, the organic

content of water used for making concrete
should not be more than:
a) 200mg/L Type of Solids Maximum Permissible
b) 250mg/L Limit
c) 1000mg/L Organic Solids 200 mg/L
d) 2000mg/L Inorganic Solids 3000mg/L
Answer: A
Question: The pH of Water Shall not be less than

a) 6
b) 9
c) 11
d) 13
Answer: A
Table 3 : Environmental Exposure Conditions

Exposure Cases Min.
Condition Grade
Mild • A concrete surface protected from Rain M20
• Concrete surface exposed to rain (but not exposed to severe rain) M25
Moderate • Concrete in Submerged condition (in Normal water)
• Concrete in costal environment but protected from Saturated Salty Air.
• Concrete Exposed to Severe Rain M30

• Concrete Surface subjected to Alternate Wetting and Drying (Normal
Severe Water)
• Concrete Surface in Coastal Area (Exposed to saturated Salty Air).
Exposure Cases Minm

Condition Grade
Very Severe • Concrete surface subjected to Sea Water Spray. M35
• Exposed to Corrosive Fumes
• Exposed to Severe Freezing
Extreme • Concrete Surface in Tidal Zone M40
• Concrete Surface exposed to harmful chemicals.


Design of Concrete Structures (Table 5)

Concrete
Environmental Exposure Conditions

Q. What is the minimum grade of concrete in sea-water

or which is exposed directly along the sea-coast in
the case of plain concrete as per IS code 456-2000 ?
A. M30 Exposure PCC RCC
B. M20 Conditions Min. Grade Min. Grade
C. M40 Mild --- M20
D. M5 Moderate M15 M25
Answer: B Severe M20 M30
Very Severe M20 M35
Extreme M25 M40

Question: The Minimum Grade of Concrete for

Moderate Environmental Exposure condition in RCC
Work should be
a) M25
b) M30
c) M15
d) M20
Answer: A
Question: The Minimum Quantity of Cement content

needed in 1m3 of a reinforcement concrete which is
buried under non-aggresive soil is (in Kg).
a) 350
b) 200
c) 250
d) 300
Answer: D
Self Designed

Characteristic Compressive Strength
Characteristic strength of concrete is the value
of strength below which not more than 5% of
the test results are expected to fall.
Target Mean Strength of Concrete (fm)

The value strength which can deviate from
mean value of strength.
Target strength, fm = fck + 1.65 σ
Question: Characteristic strength of concrete is the

value of strength below which not more than ____ of
the test results are expected to fall
a) 5%
b) 10%
c) 90%
d) 95%
Answer: A
Question: What is target mean strength (N/mm2) of the

M30 grade concrete, if the standard deviation is 5.0?
a) 21.75
b) 30
c) 38.25
d) 40.25
Answer: C

Target Mean Strength of Concrete (fm)
Target strength, fm = fck + 1.65 σ
Grade M10/ M15 M20/M25 M30 & Above
σ 3.5 4.0 5.0
where σ is the standard deviation.

The standard deviation σ can be initially based on prior experience, and
later determined from trial results.
Question: What would be the value of standard

deviation for M45 grade of Concrete ?
a) 3.5
Grade M10/ M15 M20/M25 M30 & Above
b) 4
σ 3.5 4.0 5.0
c) 4.5
d) 5
Answer: D

Nominal Mix (Used for M20 or Lower Grades)
• This a Perspective Concrete Mix used for Simpler
Works.
• Used for Relatively Unimportant Works
• All Ingredients are prescribed & proportions are
specified.
• Any Specific Need of Designer is not needed.
• It is mainly used for Grades of M20 or Lower.
• NO Quality Control.
• NO Entrapped Air is Considered.

Nominal Mix (Used for M20 or Lower Grades)
• All Ingredients are prescribed & proportions
are specified.
Grade of Concrete Specified Proportion
M5 1 : 5 : 10
M10 1:3:6
M15 1:2:4
M20 1 : 1.5 : 3
M25 1 : 1 :2
Question: Nominal Mix of Concrete is preferred to

be used for Grades ?
a) Higher than M10
b) Lower than M20
c) Higher than M25
d) Lower than M25
Answer: B
Question: Mix proportion of Concrete is

preferred to be used for M20 Grade is
a) 1 : 5 : 6
b) 1 : 3 : 6
c) 1 : 2 : 4
d) 1 : 1.5 : 3
Answer: D
Question: What is the maximum quantity of water

per bag of cement if M20 grade of nominal mix of
concrete is prepared
a) 23 Kg
b) 25 kg
c) 30 kg
d) 38 kg
Answer: C
Self Designed
Question: What is the ratio of F.A. / C.A. W.r.t.

total quantity of dry aggregates per bag of cement
in a nominal mix of concrete
a) 1 : 1
Size of Aggregate Ratio FA/CA of Mix
b) 1 : 1.5
10 mm 1 : 1.5
c) 1 : 2
20 mm 1:2
d) 1 : 2.5 40 mm 1 : 2.5
Answer: C
Self Designed
Question: As per IS 456 in Reinforced Concrete Work, the

maximum cement content to used in case of M20 Grade
of concrete (Mild Exposure Conditions) in Kg/m3
a) 300
b) 320
c) 340
d) None
Answer: A
Self Designed
In a certain project, the minimum specified cement content is

370 kg/cum. Due to a failure in the aggregate crushing and
screening plant, the project manager is left with only 40 mm
aggregate and sand. How will it affect the minimum cement
content requirement?
एक ननजचचत पररयोिना में, न्यन ू तम ननटदा ष्ट सीमें ट सामग्री 370

ककग्रा/कम ह। समुच्चय संदलन और छानन संयंत्र में ववफलता के कारण,
पररयोिना प्रबंधक के पास केवल 10 सममी समच् ु चय और रे त बची ह।
यह न्यूनतम सीमें ट सामग्री की आवचयकता को कसे प्रभाववत करे गा?
1) The cement content will have to be reduced by

40 kg/cum. / सीमें ट की मात्रा को 40 ककग्रा/घन कम करना होगा।
2) The cement content will have to be increased by
40 kg/cum. / सीमें ट की मात्रा को 40 ककग्रा/कम बढाना होगा।
3) The cement content will have to be reduced by
30 kg/cum. / सीमें ट की मात्रा को 30 ककग्रा/घन कम करना होगा।
4) The cement content will have to be increased by
30 kg/cum. / सीमें ट की मात्रा को 30 ककग्रा/कम बढाना होगा।
Question: As per IS 456 : 2000, Accuracy of

measuring equipment for water shall be within
a) ±2% Material Accuracy
b) ±3%
Cement ±2%
c) ±5%
d) ±8% Aggregate ±3%
Admixture ±3%
Answer: B
Water ±3%
Question: As per IS 456 : 2000, In the absence of

exact data. only in the case of nominal mixes ,the
amount of surface water by mass in case of moist
sand may be
a) 2.5
b) 5.0
c) 7.5
d) 10
Answer: A
Self Designed



Question: As per IS 456 : 2000, What should be

the minimum striking period in case of soffits
formwork to slab constructed with Portland
cement
a) 16 – 24 Hours
b) 3 days
c) 7 days
d) 10 days
Answer: B
Self Designed
Question: As per IS 456 : 2000, Forms shall not be

released until the concrete has achieved a
strength of at least _____ the stress to which the
concrete may be subjected at the time of removal
of formwork.
a) Same
b) Twice
c) Thrice
d) Does not depend on strength
Answer: B
Self Designed

Placing of Reinforcement
Unless otherwise specified by engineer-in-charge, the

reinforcement shall be placed within which of the following
tolerances?
िब तक प्रभारी असभयंता द्वारा अन्यथा ननटदा ष्ट नहीं ककया

िाता ह, प्रबलन को ननम्नसलखित में से ककस सहनशीलता के
भीतर रिा िाएगा?
1) For effective depth, 200 mm or less ±10 mm / प्रभावी

गहराई के सलए, 200 mm या उससे कम ± 10 mm
2) For effective depth, more than 200 mm ±10 mm / प्रभावी
गहराई के सलए, 200 mm से अधधक ± 10 mm
3) Both (A) and (B) / (A) और (B) दोनों
4) None of the above / इनमे से कोई भी नहीं
Question : Placing The maximum Permissible Free Fall

For Placing of concrete may be taken as
a) 1m
b) 1.8m
c) 2m
d) 3m
Answer: A
Self Designed
Nominal Cover
Nominal Cover
Nominal cover is the design depth of
concrete cover to all steel reinforcements,
d
including links
D
Effective Cover
The distance from outer surface of
Effective Cover
concrete to centre of main reinforcement.
Nominal Cover
In a singly reinforced beam, the effective

depth is measured from its compression
edge to _____
एकल प्रबसलत (reinforced) बीम में , प्रभावी गहराई
को इसके संपीड़न ककनारे से _____ तक मापा िाता
ह
1) Tensile edge or Neutral axis of the

beam. / बीम के तन्य ककनारे या तटतथ अक्ष
2) Tensile reinforcement / तन्यता सदृ ु ढीकरण
3) Longitudinal central axis / अनुदर्धया केंद्रीय
अक्ष
4) More than one of the above / उपरोक्त में
से एक से अधधक
Nominal cover to RCC column of size 200 ×

200 mm using 12 mm diameter longitudinal
bars is _____
12 mm व्यास के लोधगटूडिनल बार का उपयोग

करके 200 × 200 mm आकार के RCC कॉलम का
नोसमनल कवर _____ का होता ह।
1) 25 mm
2) 15 mm
3) 40 mm
4) None of the above / उपरोक्त में से कोई नहीं

Nominal Cover
Question : Additional cover thickness is Exposure Minnimum

reinforced concrete members totally Condition Nominal Cover
immersed in sea water is: (mm)
A. 25mm Mild 20
B. 30 mm Moderate 30
C. 35 mm Severe 45
D. 40 mm Very Severe 50
Answer: A Extreme 75
Self Designed
Question : As per IS 456, For Severe Exposure

Conditions reduction in Nominal cover can be done
5mm , When grade of Concrete :
A. > M20
B. < M25
C. < M35
D. > M35
Answer: D
Self Designed

Nominal Cover
According to IS 456: 2000, What is the

minimum thickness of slab
under severe exposure condition, considering
durability and fire resistance conditions:
IS 456: 2000 के अनुसार, तथानयत्व और अजनन

प्रनतरोध जतथनतयों पर ववचार करते हुए, गंभीर िोखिम
की जतथनत में तलब की न्यूनतम मोटाई क्या ह?
1) 110 mm
2) 120 mm
3) 130 mm
4) 140 mm
Question : For Unit Fire Resistance in

concrete , Nominal Cover required is
20mm except in case of :
A. Simply Supported Beam
B. Continuous Slab
C. Ribbed Slabs
D. Columns
Answer: D
Self Designed
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Working Stress Method Limit State Method

(Modular Ratio Method)
Based on Linear Elastic Theory Based on Actual Stress-Strain Curve of
Concrete & Steel
Stress Based Method Strain Based Method
FOS is applied to Stress Only Load Factor are used on Service Load
Designed Sections are Uneconomical Designed Sections are Economical
Large Sections are Obtained Comparatively Thinner Sections are
obtained
Material Strength is not fully Utilized Material strength is fully utilized in the
designed member
Water Tank, Chimneys, etc
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Working Stress Method of Design

Assumptions:
1. Plane Section Remains Plane Before Bending and After
Bending.
2. 2. All the Tensile Stresses are handled by
Reinforcement, the concrete in tension zone is
neglected.
3. The Stress - Strain Relationship for concrete and steel
within Working Load are Linear.
Permissible Stresses in Concrete

Grade Tensile Compressive
Characteristic Bending Bending

Strength Tensile Compressive
fck fcr σcbc
M15 2.71 5
M20 3.31 7
M25 3.50 8.5
M30 3.83 10
M35 4.14 11.5


Assumptions:
4. The Value of Modular Ratio (m)
M15 M20 M25 M30 M35

σcbc 5 7 8.5 10 11.5
m 18.67 13.33 10.98 9.33 8.12

Question :
Calculate the Modular ratio for M20 grade of concrete in 28
days :
1. Short Term
2. IS Code Provision
3. Long Term
Which one of the following statements is NOT correct in reinforced

concrete design?
निम्िलिखित कथिों में से कौि-सा कथि प्रबलित कंक्रीट डिज़ाइि में सही
िहीं है?
1) In the cracked section, concrete below the neutral axis is

neglected in calculations / दराररत अिुभागों में तटस्थ अक्ष से िीचे के
कंक्रीट को गणिाओं में िजरअंदाज ककया जाता है ।
2) When section is subjected to external loading, resisting
moment is developed due to compression in concrete and tension
in steel / जब अिुभाग बाहरी भारण के अधीि होता है , तो प्रनतरोधी आघूणण
कंक्रीट में संपीिि और इस्पात में तिाव के कारण ववकलसत होता है ।
3) In the cracked section, the steel area below the neutral axis is
converted into equivalent concrete area / दराररत अिुभागों में तटस्थ
अक्ष से िीचे के इस्पात क्षेत्र को समकक्ष कंक्रीट क्षेत्र में पररवनतणत ककया जाता
है ।
4) The neutral axis depth does not depend on the modular
ratio / तटस्थ अक्ष की गहराई मॉिुिर अिप ु ात पर निभणर िहीं करता है ।

Modular Ratio (m)
AS AC
Stress in steel = m × stress in concrete
P
Actual Depth of Neutral Axis
Equating Moment of Area
Compressive = Tensile
Working Stress Method Design

Moment of Resistance (MOR) from Compressive Side
Ca Moment of Resistance
B = Compressive Force × Lever Arm
Compressive Force
Xa/3
Compressive C = B. Xa. Ca /2
d
d - Xa/3
D Tensile Force
Lever Arm Moment of Resistance
T = ta . Ast = Tensile Force × Lever Arm
ta / m

Under Reinforced Section
Properties :
1. UR Section fails due to failure of steel as steel
reaches it s limit first.
2. Steel is Ductile material , so before failure of
section , it gives Indications of failure.
3. Concrete is in Under Stressed Condition
This Kind of Section Is Preferred.

Question:
In working stress method of design, failure in
under reinforced beam is
(a) Brittle
(b) Ductile
(c) Can be Any
(d) Can’t Say

Balanced Section
Properties :
1. Balanced Section fails due to failure of steel &
Concrete Simultaneously.
2. Because they reach their permissible limit at
same time.
3. It also shows Ductile Behaviour at the time of
Failure.
This Kind of Section Is Preferred for Economy &

Safety Purposes.

Over Reinforced Section
Properties :
1. OR Section fails due to failure of CONCRETE as
concrete reaches it permissible value first.
2. Steel is in under stressed condition.
3. Concrete is a brittle material, so nature of
failure will brittle
4. At the time of failure, section will fail suddenly
without any indications of failure
This Kind of Section Is not Preferred for Economy &

Safety Purposes.

Value of K
Grade of Steel K
Fe 250 0.40
Fe 415 0.289
Fe 500 0.253
Question: In working stress method of design, the

critical value of depth of neutral axis for Fe500 is:
(a) 0.23 d
(b) 0.25 d
(c) 0.40 d
(d) 0.53 d
If σcbc is permissible compressive stress of

centre in bending, then modular ratio is
given by -
यदद σcbc बंकि के तहत, कंक्रीट का अिम ु ेय संपीड़ि

प्रनतबि है , तब मॉड्यि
ू र अिप
ु ात ________ द्वारा
ददया जाता है ।
𝟐𝟖𝟎
1)
𝟑𝝈𝒄𝒃𝒄
𝟐𝟖𝟎
2)
𝟒𝝈𝒄𝒃𝒄
𝟑𝟖𝟎
3)
𝟐𝝈𝒄𝒃𝒄
4) 19
If the permissible stress in steel in tension is

140 N /𝒎𝒎𝟐 , then the depth of neutral axis
for a single reinforced rectangular balanced
section will be
यदद तिाव में स्टीि में अिुमेय प्रनतबि 140 N /

𝒎𝒎𝟐 है , तो एकि प्रबलित आयताकार संतलु ित
िंि के लिए तटस्थ अक्ष की गहराई ककतिी होगी?
1) 0.45d
2) 0.30d
3) 0.40d
4) 0.35d
Consider the below statements with respect to balanced failure is RCC structures and
identify the correct answer.
Statement A: Balanced failure is expected to occur when the compressive strain in the
extreme fibre of the concrete reaches the ultimate strain and the tensile strain at the level of
centroid of the steel reaches the yield strain.
Statement B: Balanced failure is expected to occur by the simultaneous initiation of
crushing of concrete and yielding of steel.
संतलु ित भंग RCC संरचिाओं के संबंध में िीचे ददए गए कथिों पर ववचार करें और सही उत्तर की पहचाि
करें ।
कथि A: संतलु ित भंग तब होिे की उम्मीद है जब कंक्रीट के छोर तंतु में संपीड़ि ववकृनत चरम ववकृनत तक
पहुंच जाता है और इस्पात के केन्द्रक के स्तर पर तन्द्यता ववकृनत िब्धध ववकृनत तक पहुंच जाता है ।
कथि B: कंक्रीट संदिि और इस्पात िब्धध की एक साथ शरु ु आत से संतुलित भंग होिे की उम्मीद है ।
1) Both statements are correct. / दोिों कथि सही हैं।

2) Both statements are incorrect. / दोिों कथि गित हैं।
3) Statement A is correct, and statement B is incorrect. / कथि A सही है , और कथि B गित
है ।
4) Statement B is correct, and statement A is incorrect. / कथि B सही है , और कथि A गित
है ।

Design a Singly Reinforced Beam Section
Case 1 : When Size of Beam is Not Given
Steps 1: Using Compression Side


Case 2 : When Size of Beam is Given
Case a : When BM < MORbal

Design Under Reinforced Beam
Case a : When BM > MORbal

Design Over Reinforced Beam
Doubly Reinforced Beam

Modular Ratio (m) Tension M15 M20 M25 M30 M35
On Tension Side σcbc 5 7 8.5 10 11.5

m 18.6 13.33 10.9 9.33 8.12
Compression M15 M20 M25 M30 M35

σcbc 5 7 8.5 10 11.5
On Compression Side
= 1.5m 1.5m 27.9 19.99 16.35 13.9 12.18

Factor of Safety
Permissible Stress = Characteristic Compressive Strength
Factor of Safety
Concrete : σcbc = fck / 3
Steel : σst = fy / 1.8


HYSD Bars
Steel that does not show a defined yield point, yield strain is
read as that value of stress, for which permanent deformation
of (0.2% strain) called Proof Strain is Obtained.
Total Strain at Yield Point = Plastic Strain + Elastic Strain
Max. Strain at Yield Strain = 0.002 + fy/Es

Limit State Method Design

HYSD Bars
The maximum strain in tension reinforcement at the time of
collapse shall not be less than
Total Strain at Yield Point = Plastic Strain + Elastic Strain
Yield Strain < 0.002 + 0.87fy/Es




Question:
As per WSM Method, Calculate the Value of Maximum Strain
for Fe415 grade of steel
a) 0.0035
b) 0.0022
c) 0.0040
d) 0.0045
The Limit State Method (LSM) of Design

Limit State is a state of 'about to collapse' beyond which, the structure is
not of any practical use.
In LSM, two types of limits are defined which are:
1. Limit State of Collapse
1. Flexure
2. Compression
3. Shear
4. Torsion
2. Limit State of Serviceability
1. Deflection
2. Cracking
3. Corrosion
4. Excessive Vibration
Design Stress-Strain Curve for Concrete IS 456 : 2000
Maximum Strength in Characteristic Curve is 0.67Fck

for Concrete structures.
For LSM, a partial factor for concrete = 1.5 is applied to

characteristic strength of concrete.
Thus the Design Stress of concrete is obtained by

dividing the Characteristic strength by partial factor of
safety.
Design Stress-Strain Curve for Reinforcing Steel IS 456 : 2000
Factored Load = Working Load × Load F.O.S
Fyd = fy/ PFOS
= fy / 1.15
= 0.87fy
Limit State of Collapse Limit State of Serviceability
Load Dead Live Load Wind Dead Live Wind

Combination Load Load Load Load Load
DL + LL 1.5 1.5 - 1.0 1.0 -
DL + WL 0.9 / - 1.5 1.0 - 1.0

1.5
DL + LL + WL 1.2 1.2 1.2 1.0 0.8 0.8
Q. When assessing the strength of a structure

as per the limit state of collapse, the value
of partial safety factor for steel is taken as
A. 2.0
B. 1.5
C. 1.15
D. 1.00
Limit State Method of Design

Assumptions:
1. Plane Section Remains Plane Before Bending and After
Bending.
2. The Strain Relationship for steel is Linear (even at the
time of failure).
3. The Maximum Strain in concrete at the outermost
compression fibre is taken as 0.0035 in Bending.
4. An Acceptable Stress- Strain curve is shown in Figure
21 of IS 456 : 2000
5. All the Tensile Stresses are handled by Reinforcement,
All the Tensile Strength of Concrete is Ignored.
Question:
The shape of idealized stress-strain curve for concrete
as prescribed by IS 456-2000 is
(a) Rectangular
(b) Parabolic
(c) Rectangular- Parabolic
(d) None of these

Assumptions: B
6. The Maximum Strain in Tension
Reinforcement at the time of collapse shall Compressive
not be less than d
D
Statement A: Rupture of one or more critical sections

is addressed in limit state of serviceability.
Statement B: Deflection in RCC sections is addressed
in limit state of serviceability.
कथि A: सेवा योग्यता की सीमा ब्स्थनत में एक या अधधक

क्रांनतक िण्िों का ववदारण संबोधधत ककया जाता है ।
कथि B: RCC िण्िों में ववक्षेपण को सेवा योग्यता की सीमा
ब्स्थनत में संबोधधत ककया जाता है ।
1) Statement B is correct and A is incorrect / कथि B

सही है और A गित है
2) Both statements are correct / दोिों कथि सही हैं
3) Both statements are incorrect / दोिों कथि गित हैं
4) Statement A is correct and B is incorrect / कथि A
सही है और B गित है
Statement A: Rupture of one or more critical sections

is addressed in limit state of serviceability.
Statement B: Deflection in RCC sections is addressed
in limit state of serviceability.
कथि A: सेवा योग्यता की सीमा ब्स्थनत में एक या अधधक

क्रांनतक िण्िों का ववदारण संबोधधत ककया जाता है ।
कथि B: RCC िण्िों में ववक्षेपण को सेवा योग्यता की सीमा
ब्स्थनत में संबोधधत ककया जाता है ।
1) Statement B is correct and A is incorrect / कथि B

सही है और A गित है
2) Both statements are correct / दोिों कथि सही हैं
3) Both statements are incorrect / दोिों कथि गित हैं
4) Statement A is correct and B is incorrect / कथि A
सही है और B गित है
Limit State Method Design Grade of Steel k
For RCC Section Strain Diagram Fe250 0.53
B 0.0035 Fe415 0.48

Fe500 0.46
Compressive Xulim
d
D
d - Xulim
Question:
In limit state design, the limiting value of
depth of neutral axis for Fe500 is:
(a) 0.44 d
Grade of Steel k
(b) 0.46 d
Fe250 0.53
(c) 0.48 d
Fe415 0.48
(d) 0.53 d
Fe500 0.46
Question:
Partial safety factors for concrete and steel respectively
may be taken as:
a) 1.5 and 1.15
b) 1.5 and 1.78
c) 3 and 1.78
d) 3 and 1.2

RCC Section Strain Diagram Stress Diagram
B 0.0035 0.45fck CR = 0.36 fck. Xu. B
y1
3/7Xu Rectangular C1 = 0.193 fck. Xu. B
Xulim y2
Compressive CR
Parabolic d C2 = 0.171 fck. Xu. B
4/7Xu
d - Xulim
0.87fy

Actual Depth of Neutral Axis
CR = 0.36 fck. Xu. B
B 0.45fck
3/7Xu Rectangular y = 0.42Xu

Compressive Xulim
4/7Xu Parabolic d
Lever Arm
d - 0.42Xu
T = 0.87 fy. Ast
0.87fy

Moment of Resistance
CR = 0.36 fck. Xu. B
B 0.45fck
3/7Xu Rectangular y = 0.42Xu

Compressive Xulim
4/7Xu Parabolic d
Lever Arm
T = 0.87 fy. Ast
d - 0.42Xu
0.87fy
Q1. Consider the following statements concerning both the

working stress design and ultimate strength design of
reinforced concrete:
1. Plane section before bending remains plane after
bending
2. The tensile strength of concrete is ignored. Which of
these statements is/are correct?
A. 1 alone
B. 2 alone
C. Both 1 and 2
D. Neither 1 nor 2
Q20. Partial safety factor for concrete and steel

are 1.5 are 1.15 respectively, because
A. concrete is heterogeneous while steel is
homogeneous
B. the control on the quality of concrete is
not as good as that of steel
C. concrete is weak in tension
D. woids in concrete are 0.5 % while those
in steel are 0.15 %

Under Reinforced Section
Properties :
At the time of failure
a) Strain in concrete = 0.0035
b) Strain in steel >>
Xu < Xulim
This compression failure of concrete in URS is
when strain is increasing due to further increase
of strain in steel beyond
This is called Secondary Compression Failure

Q. If fcu and fy are cube compressive strength of

concrete and yieid stress of steel respectively and
Es is the modulus of elasticity of steel for all grades
of concrete, the ultimate flexural strain in concrete
can be taken as
A. 0.002
fcu
B. 1000
C. 0.0035
fy
D. + 0.002
1.15Es

Over Reinforced Section
Properties :
b) Strain in steel <
Xu > Xulim
There is not further scope of increasing the
strain & concrete will suddenly fail beyond
strain of 0.0035
This is called Primary Compression Failure


Limiting Section Xu = Xulim
Properties :
b) Strain in steel >
Xu = Xulim , Maximum Depth of N.A. Allowed
Ast = Ast lim , Maximum Area of Steel Allowed
Mu = Mu lim , Limiting Ultimate Moment of Resistance


Limiting Section Xu = Xulim
Mu = Mu lim , Limiting Ultimate Moment of Resistance
Working MOR (Max.)





Fe 250 Fe 415 Fe 500
K = [700/(1100+ 0.87fy)] 0.53 0.48 0.46
J = (1 – 0.42k) 0.78 0.80 0.81
Q = 0.36fck.k.j 0.148fck 0.138fck 0.134fck
M20 2.96 2.76 2.68

Alternate Design Formula

Valid for Under Reinforced Section & Balanced Section only
In singly reinforced sections, when the

section is under-reinforced, the relation
between depth of neutral axis (xu) and the
limiting value of depth of neutral axis
(xu' max) is:
एकि प्रबलित अिुभाग में , जब अिुभाग अध-प्रबलित

होता है , तो उदासीि अक्ष (xu) की गहराई
और उदासीि अक्ष (xu' max) की गहराई के सीलमत
माि के बीच संबंध होता है :
1) xu = xu' max
2) xu < xu' max
3) xu > xu' max
4) none of the above / उपरोक्त में से कोई िहीं
Effective Width of Flange (Bf)

As Per IS Code
1. For Isolated T-Beam
2. For Isolated L-Beam

IS 456 : 2000 Based Provsions

Effective Span
1. Simply Supported Beam / Slab
Beam
Effective Span
Leff = Lo + d
Leff = Lo + w
IS 456 : 2000 Based Provsions

Effective Span
2. Continuous Beam
When width of support,
Effective Span is Same as Simply Supported Beam
Leff = Lo + d
Leff = Lo + w
Discount Code
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Test Series
Notes
Both in English &
Hindi Medium


A continuous slab of clear span 5 m and

effective depth is 150 mm and supported on
300 mm. What is the value of effective span?
एक 5 m स्पष्ट फैिाव और 150 mm की प्रभावी

गहराई का संतत स्िैब है और वह 300 mm पर
आिब्म्बत है । प्रभावी फैिाव का माि क्या होगा?
1) 5000 mm
2) 5075 mm
3) 5300 mm
4) 5150 mm
Question : The final deflection due to all including

effects of temperature, creep and shrinkage measured
from as - cast level of the supports of floors, roofs and
all other horizontal members of reinforced concrete
should not normally exceed
A. Span / 350
B. Span / 250
C. (Span / 350) or 20 mm whichever is less
D. (5 / 348) of span
IS 456 : 2000 Based Provisions

Deflection Control
(Only Valid for Beam / 1-Way Slab)
For Span Length < 10m


Deflection Control
(Only Valid for Beam / 1-Way Slab)
For Span Length > 10m

IS 456 : 2000 Based Provisions Type of Slab ‘Z’ Value
Deflection Control Mild HYSD / TMT /

(Only Valid for 2- Way Slab) Steel CTD Bars
For Span Length < 3.5m Simply Supported Slab 35 28

Load < 3kN/m2 Continuous Slab 40 32
For two-way continuous slab of shorter span ≤ 3.5

m with HYSD reinforcement, the span to overall
depth ratio is taken as _____to satisfy the vertical
deflection limit for loading class up to 3 KN/m2
HYSD प्रबिि के साथ छोटे फैिाव ≤ 3.5 m के दो-तरफा

निरं तर स्िैब के लिए, 3 KN/m2 तक भारण वगण के लिए
िंबवत ववक्षेपण सीमा को पूरा करिे के लिए फैिाव और
समग्र गहराई के अिप ु ात की अवधध को _____ के रूप में
लिया जाता है ।
1) 26
2) 28
3) 32
4) 35
IS 456 : 2000 Based Provisions Unsupported Length of Beam (L1)

Check for Lateral Stability Simply Supported Beam
(Only Valid for Beams)
Or
Continuous Beam
Beam
Span
IS 456 : 2000 Based Provisions Unsupported Length of Beam (L1)

Check for Lateral Stability
Cantilever Beam
(Only Valid for Beams)
Beam
Span

B
Beams
Minimum Tension Reinforcement Compressive
d
D
Grade of Steel Fe250 Fe 415 Fe 500
Minimum 0.34% 0.205% 0.17%

Reinforcement

B
Beams
Minimum Compression Reinforcement
Compressive
= NIL d
D
Maximum Tension Reinforcement
= 4% of C/S Area
Maximum Compression Reinforcement =
4% of C/S Area

Maximum Diameter of Main Bar
Slab
Minimum Tension Reinforcement
Grade of Steel Minimum Value
Fe250 0.15% of C/S Area

Other Bars 0.12% of C/S Area

Slab
Maximum Spacing of Main Bar
= 3d or 300mm
Maximum Spacing of Distribution Bar

= 5d or 450mm

Slab
Minimum Spacing in Reinforcement

In a non-cellular, non-ribbed, flat slab, the

spacing of reinforcing bars shall be limited
to a maximum of m times the thickness of
the slab, where m is:
एक गैर-कोशमय, गैर-पशक ुण , समति स्िैब में ,

प्रबलित सिािों की दरू ी स्िैब की मोटाई के
अधधकतम मीटर तक सीलमत होगी, जहां m
____ है ।
1) 2.50
2) 2.00
3) 1.50
4) 1.00
The high strength deformed bar

reinforcement in either direction in slabs
shall not be less than p percent of the total
cross-sectional area, where p is:
स्िैब में ककसी भी ददशा में उच्च सामर्थयण ववकृत बार

प्रबिि कुि अिुप्रस्थ काट क्षेत्रफि के p प्रनतशत से
कम िहीं होिा चादहए, जहां p ______ है ।
1) 0.18
2) 0.10
3) 0.15
4) 0.12
R.C.C. Section
Bending stress is the reason of shear stress
All the combination of multiple stress are very difficult 45⁰ 90⁰
to analyse.
So, From Research we go to Know that there can be
three reasons of failure :
1. Diagonal Compression Failure
(Crushing Failure due to large shear force )
2. Diagonal Tension Failure
(Cracks are developed at 45⁰ due to large shear force )
3. Bending Shear Failure
(Cracks are developed at 90⁰ due to large Bending Moment)
Question :
The maximum shear stress (q) in concrete of a
reinforced cement concrete beam is
a)Shear force / (Lever arm × Width)
b)Lever arm / (Shear force × Width)
c)Width / (Lever arm × shear force)
d)None of these
Design of R.C.C. Section as Per Shear

Nominal Shear Stress (𝝉𝒗)
It is the Average Shear Stress Developed in the Section
suggested by I.S. Code
If 𝝉v > 𝝉cmax The Beam size shall be changed


Design Shear Strength of Concrete (𝝉c)
It is the shear strength of Concrete which includes shear
stress resisted by :
1. Concrete
2. Main Reinforcement
The Value of 𝝉c depends up-on :

1. Grade of Concrete
2. Percentage of Tension Reinforcement Provided

Maximum Permissible Shear Stress with
Shear Reinforcement (𝝉cmax)
The Shear Stress developed should not be greater
than Maximum shear stress in the beam.
The Value 𝝉cmax of depends up-on Grade of Concrete.
It is the Maximum Value of 𝝉v

If 𝝉v > 𝝉cmax The Beam size shall be Re-Designed

Maximum Permissible Shear Stress with
Shear Reinforcement (𝝉cmax)
M20 M25 M30 M35 M40 &
Above
LSM 2.8 3.1 3.5 3.7 4.0
𝝉cmax is kept to avoid the diagonal compression failure

of concrete.

Design Steps to Follow :
Step 7 : Design of Shear Reinforcement

a) Vertical Shear Reinforcement
LSM
WSM Spacing (Sv)

Sv – Spacing
Vsu – Strength of Shear Reinforcement
Asu – Area of Shear Reinforcement

Step 8 : Minimum Shear Reinforcement
If 𝝉v < 𝝉c
At all location in a beam shear reinforcement shall not
be less than minimum shear reinforcement

Minimum Shear Reinforcement must be
provided due to
1. To Resist Principal Tension

2. To avoid crack development due to
a) Sudden Increase in Load
a) Creep or Shrinkage
b) Principal Tension
3. To improve ductility of member
4. To improve dowel action of main
reinforcement.

Step 9 : Maximum Spacing
Vertical Shear R/f

0.75d or 300 mm whichever is less
Inclined Shear R/f

d or 300 mm Whichever is less
Bond in Reinforced Concrete

Bond in R/f concrete is the adhesion between the
R/f steel bar and Surrounding Concrete.
Due to Bond only it is possible to Transfer axial

force(Tensile / Compressive)
Due to bond only Plane section remains plane

section remains plane before bending & after
bending is valid
Note: If bond is not present this assumption is
invalid
Bond in Reinforced Concrete
NOTE:
Deformed Steel bars are better than Plain bars with
respect to Bond Stress due to Mechanical Inter-
locking due to surface ribs on deformed bars.
Bond Stress
Bond Stress is the shear stress developed between
steel & concrete due to
1. Adhesion
2. Mechanical Bond
3. Friction
Whenever there is tendency of movement of steel

over concrete surface, The bond stress developed is
of two types :
a) Flexural Bond
b) Anchorage Bond
Flexural Bond 1 2
M M + dM
T T + dT
dx
ꞇbd
dx
Flexural Bond
Maximum Permissible Bond Stress (ꞇbd)
For Mild Steel R/f (In Tension )
M20 M25 M30 M35 M40
LSM 1.2 1.4 1.5 1.7 1.9
Other type of Bars (HYSD / TMT Bars) ꞇbd

Increase the value by 60%
For Bars in Compression dx
Increase the value by 25%
Flexural Bond 1 2
If (ꞇbd)developed > (ꞇbd)permissible M M + dM

T T + dT
To make the reinforcement safe for bond
Keep same Amount of Steel (Area) dx
But 1 2
1. Reduce the diameter of Bar
2. Increase the Number of Bars
ꞇbd
dx
Question :
Which of the following option is the best way
to have proper bond between the steel &
surrounding concrete
a) Use small number of large diameter bars
b) Use large number of large diameter bars
c) Use shear stirrups
d) All of the Above
Consider the following statement regarding the factors affecting strength.

1. Larger diameter bars possess more bond strength as compared to smaller ones.
2. Bundling of bars reduces the bond strength.
3. A rise in temperature reduces the bond strength.
4. For higher strength of concrete, the increase in bond strength becomes
exponential.
5. HYSD bars possess more bond strength as compared to plain bars.
Which of the above statement are correct?
क्षमता को प्रभाववत करिे वािे कारकों के संबंध में निम्िलिखित कथिों पर ववचार करें ।
1. छोटे व्यास की ति ु िा में बड़े व्यास के बार में अधधक आबंध क्षमता होती है ।
2. बार के बंिि आबंध क्षमता को कम कर दे ते हैं।
तापमाि में वद्ृ धध से आबंध क्षमता कम हो जाती है ।
3. कंक्रीट के उच्च सामर्थयण के लिए, आबंध क्षमता में वद्ृ धध घातीय हो जाती है ।
4. HYSD बार में सरि बार की तुििा में अधधक आबंध क्षमता होती है ।
उपरोक्त में से कौि सा कथि सही है?
1) 2,3,4 2) 1,2,4,5
3) 2,3,5 4) 1,3,4,5
Consider the following statement regarding the factors affecting strength.

1. Larger diameter bars possess more bond strength as compared to smaller ones.
2. Bundling of bars reduces the bond strength.
3. A rise in temperature reduces the bond strength.
4. For higher strength of concrete, the increase in bond strength becomes
exponential.
5. HYSD bars possess more bond strength as compared to plain bars.
Which of the above statement are correct?
क्षमता को प्रभाववत करिे वािे कारकों के संबंध में निम्िलिखित कथिों पर ववचार करें ।
1. छोटे व्यास की ति ु िा में बड़े व्यास के बार में अधधक आबंध क्षमता होती है ।
2. बार के बंिि आबंध क्षमता को कम कर दे ते हैं।
तापमाि में वद्ृ धध से आबंध क्षमता कम हो जाती है ।
3. कंक्रीट के उच्च सामर्थयण के लिए, आबंध क्षमता में वद्ृ धध घातीय हो जाती है ।
4. HYSD बार में सरि बार की तुििा में अधधक आबंध क्षमता होती है ।
उपरोक्त में से कौि सा कथि सही है?
1) 2,3,4 2) 1,2,4,5
3) 2,3,5 4) 1,3,4,5
Anchorage / Development Bond

This Anchorage/development bond gets developed
at the extreme cut-off end of the bar subjected to
tensile force/compressive force.
Strength in bond = Strength in Tension
P
Ld
Anchoring Shear Reinforcement

B.L ≮ 8d B.L ≮ 6d B.L ≮ 4d
90⁰ Bend 135⁰ Bend 180⁰ Bend

For Bundled Bars For Bundled Bars , Development length of

bundled bars shall be that of individual bar
increased by :
Bars in Increase in Ld
2 Bars 10%
3 Bars 20%
4 Bars 33%
Two Bars in Three Bars in Four Bars in
Contact, 10% Contact, 20% Contact, 33% This increase in Ld is made because of
Increase in Increase in Increase in reduction in the interface area between the
Development Development Development steel & surrounding concrete.
Length Length Length
Use of Development Bond

Whenever a reinforcement is to be continued and
the available length is shorted than required.
Lap Splicing is Done
Ld
1. For R/f of diameter > 36mm , Lap Shall not be
used.
2. When Bars of 2 different diameters are lapped,
the lap length is calculated as per smaller
diameter.
3. For Bars in Compression,
Minimum Lap length ≮ 24d
Design of Column
Column is a vertical structural member that carry loads
mainly in compression.
It might transfer loads from a ceiling, floor slab, roof slab,
or from a beam, to a floor or foundations.
Y
X X D
Y
B
Slenderness Ratio =
D
If only one Effective Length is given

B
𝐋𝐞𝐟𝐟
Maximum Slenderness Ratio =
𝐋𝐞𝐚𝐬𝐭 𝐋𝐚𝐭𝐞𝐫𝐚𝐥 𝐃𝐢𝐦𝐞𝐧𝐬𝐢𝐨𝐧
𝐋𝐞𝐟𝐟
=
𝐁
Pedestal
Y
X X D
Y
B
Short Column
Long Column
Any One or Both should be satisfied

Effective Length of Column (Leff)

Degree of End Restraint of Symbol Theoretical Value of effective Recommended Value of
Compression Members Length Effective Length
(1) (2) (3) (4)

Effectively held in position
and restrained against
rotation in both ends
0.50L 0.65L
Effectively held in position at

both ends, restrained against
rotation at one end
0.70L 0.80L
Effective Length of Column (Leff)

(1) (2) (3) (4)

at both ends, but not
restrained against
1.00L 1.00L
rotation

and restrained against
rotation at one end, and at
1.00 L 1.20 L
the other restrained
against rotation but not
held in position

(1) (2) (3) (4)

Effectively held in position and
restrained against rotation at one 2.0 L 2.0 L
end but not held in position nor
restrained against rotation at the
other end
IS 456 : 2000 Provision
Y Minor Axis
Major Axis X X D
Y
B

Minimum Eccentricity(emin)
All columns shall be designed for minimum
eccentricity of
Maximum Eccentricity(emax)
= 5% of Least Lateral Dimension
Y
Design of Column (by WSM Method)
Asc
D
Axial Load Carrying Capacity of a RCC Column Ac
P = Ac σcc + Asc σsc

B
The effect of minimum eccentricity is considered in above
formula
Ac = A – Asc
Ac = BD – Asc
Asc = Area of steel
Y Minor Axis
Design of Column (by WSM Method)
Major Axis X X D
Axial Load Carrying Capacity of a RCC Column
Permissible Stress in Concrete(σcc)
Y
M15 M20 M25 M30 M35
B
σcc 4 5 6 8 9
Permissible Stress in Concrete(σsc)

Fe250 Fe415 / Fe500
σsc 130N/mm2 190 N/mm2

(Valid For Both WSM & LSM)
a) Main Reinforcement
1. Minimum Percentage of Steel = 0.8% of Total Area
2. Maximum Percentage of Steel
• If Bars are Lapped = 4% of Total Area
• If Bars are NOT Lapped = 6% of Total Area



a) Minimum Number of Bars

1. Rectangular Column = 4
2. Circular Column = 6

a) Minimum Diameter of Main Bar = 12mm

b) Maximum Spacing of Main Bar = 300mm
≯ 300mm ≯ 300mm
≯ 300mm
According to IS 456 : 2000, which of the

following equations is CORRECT for a one
way slab considering its ratio of longer span
(Ly) to shorter span (Lx)?
IS 456 : 2000 के अिुसार, िम्बे स्पेि (Ly) से छोटे

स्पेि (Lx) के अिुपात को दे िते हुए वि-वे स्िैब के
लिए कौि सा समीकरण सही है?
1) Ly/Lx >= 2
2) Ly/Lx < 2
3) Ly/Lx > 3
4) Ly/Lx <= 3
(a) Rectangular column (i) 4

(b) Circular column (ii) 5
(c) Octagonal column (iii) 6
(iv) 8
(a) आयताकार स्तंभ (i) 4

(b) वत्त
ृ ाकार स्तंभ (ii) 5
(c) अष्टकोणीय स्तंभ (iii) 6
(iv) 8
1) a - i, b - ii, c - iii
2) a - i, b - iii, c - iv
3) a - iv, b - iii, c - i
4) a - iii, b - i, c - iv

Separate Rings
a) Diameter of Lateral Ties

≯ 300mm
1. d=
2. 6 mm
} Maximum Value

Separate Rings ≯ 300mm
a) Spacing of Lateral Ties

1. Least Lateral Dimension
2. 16ϕ
3. 300 m
}
Minimum Value
Which of the following statements regarding transverse

reinforcement in the form of lateral ties provided in the
column is true?
कॉिम में ददए गए पार्शवण बंधकों के रूप में अिुप्रस्थ प्रबिि के संबंध
में निम्िलिखित में से कौि सा कथि सत्य है?
1) To resist any tension due to bending of column,

horizontal loads, eccentric loads or moments. / काॅिम के
बंकि, क्षैनतज भार, उत्केन्द्री भारों या आघूणों के कारण ककसी भी
तिाव का प्रनतरोध करिे के लिए
2) To reduce the size of column. / काॅिम का आकार कम
करिे के लिए
3) To resist longitudinal splitting of the column due to
development of transverse tension. / अिुप्रस्थ तिाव के
ववकास के कारण काॅिम के अिुदैर्धयण ववभाजि का प्रनतरोध करिे के
लिए
4) To increase the load carrying capacity. / भार वहि क्षमता
में वद्
ृ धध करिे के लिए
According to IS 456 : 2000, which of the following statements is

true with respect to reinforcement provided in reinforced
concrete columns?
IS 456: 2000 के अिुसार, प्रबलित कंक्रीट काॅिम में प्रदाि ककए गए

प्रबिि के संबंध में निम्िलिखित में से कौि सा कथि सत्य है ?
1) Spacing of lateral ties cannot be more than 16 times the

diameters of the tie bar. / पार्शवण टाई का अंतर टाई बार के व्यास से 16
गुिा से अधधक िहीं हो सकता है
2) Columns with circular sections are provided with transverse
reinforcement of helical type only. / वत्त ृ ाकार िंिों वािे काॅिमों को
केवि कंु िलििी प्रकार के अिुप्रस्थ प्रबिि के साथ प्रदाि ककया जाता है
3) Longitudinal reinforcement bar need not be in contact with
lateral ties. / अिुदैर्धयण प्रबिि बार को पार्शवण टाई के संपकण में होिे की
आवर्शयकता िहीं है
4) Maximum longitudinal reinforcement in an axially loaded short
column is 6% of the gross cross-sectional area. / अक्षीय रूप से
भाररत ककए गए छोटे कॉिम में अधधकतम अिद ु ै र्धयण प्रबिि सकि अिप्र
ु स्थ
काट क्षेत्र का 6% है
Consider the following statement regarding reinforcement provision as

per IS 456,
1. Minimum reinforcement for short column is 0.08% of gross area.
2. Minimum reinforcement for pedestal is 0.15 % of gross area.
3. The ratio of minimum reinforcement in vertical and horizontal
directions of an RC wall is 3:5
4. Maximum reinforcement in the column is 4% of the gross area when
splicing is not done.
IS 456 के अिुसार प्रबिि प्रावधाि के संबंध में निम्िलिखित कथि पर ववचार करें ,
1. िघु स्तम्भ के लिए न्द्यि
ू तम प्रबिि सकि क्षेत्र का 0.08% है ।
2. स्तम्भ आधार के लिए न्द्यूितम प्रबिि सकि क्षेत्र का 0.15% है ।
3. एक RC दीवार की ऊर्धवाणधर और क्षैनतज ददशाओं में न्द्यूितम प्रबिि का अिुपात
3:5 है ।
4. जब ब्स्लिलसंग िहीं की जाती है तो स्तम्भ में अधधकतम प्रबिि सकि क्षेत्र का
4% होता है ।
1) 1, 2 2) 2, 3
3) 3, 4 4) 1, 4
Y
Design of Column (by LSM Method)
Asc
But as per IS Code, All columns shall be designed for
minimum eccentricity . D
Ac
So previous expression has to be modified for emin
B
Axial Load Carrying Capacity of a Short Column Ac = A – Asc
P = 0.40fck.Ac + 0.67fy.Asc Ac = BD – Asc
Asc = Area of steel
For a Long Column

X X D
If Slenderness Ratio is Y
Y
Load Carrying Capacity B
P = Cr[0.40fck.Ac + 0.67fy.Asc ]
Where, Cr = for Symmetrical Section
Cr = for Un-Symmetrical Section


Design of Circular Long Column with Helical Reinforcement
1. Load Carrying Capacity is increased by 5%
2. Ductility of column is Increased
Load Carrying
P = 1.05Cr [0.40fck.Ac + 0.67fy.Asc ]
Capacity
Cr =
Dcore = (D – 2 × Clear Cover) = (D – 2 × 40) = (D – 80)

Pitch is Found of Formula
Pitch
L
• One Way Slab −  2
B
(Supported only two edges)
L
• Two way Slab −  2
B
(Supported on all edges)
Note: Max. B.M. occurred at the shorter span.

For Bending Moment:

The critical section for the bending moment is at
the face of the column.
For Shear stress:

a) One Way Shear
The critical section for one-way shear is at a
distance d from the face of the column
b) Two-Way Shear or Punching Shear

The critical section for two-way shear is at a
distance d/2 from the face of the column.
Stress resultant Location

P. Bending moment 1. at the face of column
Q. One way shear 2. at d/2 from face of column
R. Punching shear 3. • at d from face of column
प्रनतबि का पररणाम स्थाि

P. बंकि आघणू ण 1. कॉिम के फिक पर
Q. एक तरह से अपरूपण 2. स्तंभ के फिक से d/2 पर
R. नछरण अपरूपण 3. स्तंभ के फिक से d पर
1) P - 1, Q - 2, R - 2
2) P - 3, Q - 1, R - 2
3) P - 1, Q - 3, R - 2
4) P - 1, Q - 2, R - 3
Pick up the incorrect statement from the following.
निम्िलिखित में से असत्य कथि को चुनिए।
1) In the heel slab of retaining wall, reinforcement is

provided at the bottom of the slab / ररटे निंग दीवाि की
हीि स्िैब में स्िैब की निचिी सतह पर प्रबिि ददया जाता है
2) In the stem of retaining wall, reinforcement is
provided near the earth side / ररटे निंग दीवाि के तिे में
लमट्टी की तरफ के िजदीक प्रबिि ददया जाता है
3) In the toe slab of retaining wall, reinforcement is
पंजा स्िैब में स्िैब की निचिी सतह पर प्रबिि ददया जाता है
4) None of the above / उपरोक्त में से कोई िहीं
Retaining wall:
The retaining wall is used to retain earth and resist
the lateral pressure of soil at a place with a sudden
change in elevation.
1. Cantilever type retaining wall is used for heights

up to 6m.
2. Cantilever type retaining wall has the following
components:
1. Stem
2. Toe slab
3. Heel slab
Pick up the incorrect statement from the following.
निम्िलिखित में से असत्य कथि को चुनिए।
1) In the heel slab of retaining wall, reinforcement is

हीि स्िैब में स्िैब की निचिी सतह पर प्रबिि ददया जाता है
2) In the stem of retaining wall, reinforcement is
provided near the earth side / ररटे निंग दीवाि के तिे में
लमट्टी की तरफ के िजदीक प्रबिि ददया जाता है
3) In the toe slab of retaining wall, reinforcement is
पंजा स्िैब में स्िैब की निचिी सतह पर प्रबिि ददया जाता है
4) None of the above / उपरोक्त में से कोई िहीं
In an RCC column, if As = 1000 mm2, Ac =

10000 mm2, σc = 5N/mm2 and m = 20, then
load on the column is
RCC काॅिम में ,यदद As = 1000 mm2, Ac =

10000 mm2, σc = 5N/mm2 और m = 20 है ,तो
काॅिम पर भार ककतिा है?
1) 75 kN
2) 145 kN
3) 150 kN
4) 1005 KN
When HYSD bars are used in place of mild

steel bars in a beam, the bond strength
जब HYSD बार का प्रयोग बीम में मद ृ ु (माइल्ि)

स्टीि बासण के स्थाि पर ककया जाता है, तब योजक
शब्क्त
1) become zero / शून्द्य हो जाता हे

2) decreases / कम हो जाता है
3) increases / बढ़ जाता है
4) does not change / िहीं बदिता है
As per IS 456: 2000, the unsupported length (l) of column

whose one end is unrestrained in any given plane shall
NOT exceed (Given: b = width of the column cross-section
and D = depth of cross-section measured in the plane
under consideration)
(Symbols and notations carry their usual meaning)
IS 456: 2000 के अिुसार, ककसी भी ददए गए लिेि में कॉिम की

अिसलपोटे ि िें थ (I) ब्जसका एक छोर असंयलमत है , निम्ि में से
ककससे अधधक िहीं होगा? (जब ददया गया है : b = कॉिम के क्रॉस
सेक्शि की चौड़ाई और D = ववचाराधीि लिेि में मापी गई क्रॉस
सेक्शि की गहराई)
(प्रतीक और संकेत अपिे सामान्द्य अथण में ददए गए हैं)
1) (100 b) / D
2) (100 b3) / D
3) (100 D2) / b
4) (100 b2) / D
How much shall be the development length of

25 mm diameter deformed reinforcing bar of grade
Fe415 in tension?
(Given that design bond stress in limit state method
for deformed bars in tension = 2.24 N/mm2)
तिाव में ग्रेि Fe415 के 25 mm व्यास ववकृत प्रबिि छड़ की

ववकास िंबाई ककतिी होगी?
(ददया गया है कक तिाव में ववकृत छड़ों के लिए सीलमत ब्स्थनत
ववधध में डिजाइि आबंध प्रनतबि = 2.24 N/mm2)
1) 762.9 mm
2) 1220.6 mm
3) 1007.39 mm
4) 849.5 mm
AS per IS 456 ∶ 2000, the lap length of steel

reinforcement (rebar) in compression shall be equal
to _________, but not less than 24 times the diameter
of rebar.
IS 456 ∶ 2000 के अिुसार, संपीड़ि में इस्पात प्रबिि (रीबर)
की िैप िंबाई __________ के बराबर होगी, िेककि रीबर के
व्यास के 24 गुणा से कम िहीं होगी।
1) 12 times the diameter of rebar / ररबर के

व्यास के 12 गुिा
2) the development length on tension / तिाव
पर ववकास की िंबाई
3) 16 times the diameter of rebar / रीबर के
व्यास के 16 गुिा
4) the development length in compression / संपीड़ि
में ववकास की िंबाई


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AE - JE Exams

Covers CBT-1 & CBT-2

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Q. Among the following which is the most

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Question: The Minimum Grade of Concrete for

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Question: Characteristic strength of concrete is the

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Question: Nominal Mix of Concrete is preferred to

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