Structural Design- II CEA 3180

Structural Design- II CEA 3180

Dr. Moonis Zaheer

Course Objectives

• Develop concepts of analysis and design of reinforced concrete elements not covered under 1^st course of the subject (CEA3110).

• To introduce analysis and design of structural members like

Compression members, footings, continuous beams and portal frames.

• To introduce students the significance of structural construction with steel and develop basic concept of analysis and design of steel

structures.

Course contents

• Unit 1: Design of Compression members, short column, Column with uni-axial & bi-axial bending, long column, use of design charts,

reinforcement Detailing.

• Unit 2 : Introduction design of foundation: wall footing, Isolated and Combined footing for columns Reinforcement Detailing.

• Unit 3: Analysis & Design of Continuous beams & portal frames &

their Detailing.

• Unit 4: Design of riveted & welding connections (simple cases only) Tension & Compression members, beams & Plate girders.

Introduction to grillage foundation, Trusses.

Text and Reference Books:

• Jain, A. K. “Reinforced concrete- limit State Design” NCB, Roorkee, India

• Pillai and Menon “Reinforced Concrete Design”, TMH, New Delhi, India.

• Verghese, P. C. “Advanced Reinforced Concrete Design” PHI, Delhi, India

• IS: 456 - 2000. Code of Practice for Plain and Reinforced Concrete, BIS, New Delhi, India.

• SP: 16 - Design Aids for Reinforced Concrete to IS: 456, BIS, New Delhi, India.

• Kazmi, S. M. A. and Jindal, R.S. “Design of Steel Structures” PHI, New Delhi, India.

• Arya and Ajmani “Design of Steel Structures”, NCB, Roorkee, India.

• Ramamrutham, S. “Design of Steel Structures” Dhanpat Rai, Delhi, India.

• IS: 800-2000 - Code of Practice for General Construction in Steel, BIS, New Delhi, India.

Columns

• Structural element subjected to axial compressive forces is called compression member.

 Columns- when a compression member is vertical

 Struts- when a compression member is horizontal or inclined (concrete trusses)

 Shear walls

• A column that springs from a beam is referred to as floating column.

• Compression members that supports deck in bridges are called as piers.

• Upright slender members (circular) and subjected to predominant B.M and nominal compression are called poles, pillars or posts.

Definitions according to code

• Clause 25.1.1 – Column or strut is a compression member, effective length of which exceeds three times the least lateral dimension.

• Clause 26.5.3.1 h – Pedestal is the compression member, the effective length of which does not exceed three times the least lateral dimension 𝑏.

• The other horizontal dimension 𝐷 shall not exceed four times of 𝑏.

• Location?

• At the base of columns to transfer the load of columns to a footing, pile cap, or mat.

Definitions

Pedestal Column

Wall

• Wall is a vertical compression member whose effective height 𝐻 to thickness 𝑡 (least lateral dimension) shall not exceed 30 (CL. 32.2.3 of IS 456).

• The larger horizontal dimension i.e., the length of the wall 𝐿 is more than 4𝑡.

Why steel is provided in compression members

• Very few members are truly axially loaded

• Steel is essential for resisting any local bending that may occur

• Part of total load is carried by steel with its much greater

compressive strength

Based on cross section

 Rectangular

 Square

 Circular

 Hexagonal

 T, L, or + shapes

Classification of columns

Based on type of reinforcement

Tied columns- most common (square, rectangular)

Columns with helical

reinforcement- Architects choice

Composite column

• The main longitudinal reinforcement of the composite columns consists of structural steel sections or pipes with or without longitudinal bars.

• Axially loaded columns – all inside columns

• Axially loaded column with uniaxial bending – all side columns

• Axially loaded column with biaxial bending – all corner columns

Based on loading

Columns based on location

Euler buckling load

• 𝑃_𝑐𝑟 = ^{𝜋2𝐸𝐼}

𝐿²_𝑒

• 𝜎_𝑐𝑟 = ^𝑃𝑐𝑟

𝐴 = ^{𝜋2𝐸𝐼}

𝐿²_𝑒𝐴

• 𝜎_𝑐𝑟 = _𝐿^𝜋2𝐸

𝑒 𝑟 ²

• where

• 𝑟 = ^𝐼

𝐴

Slenderness ratio

• The factor 𝑙_𝑒 𝑟 is called as the slenderness ratio.

• It provides a measure of probability of column buckling.

• short columns fail by crushing with the material reaching its ultimate strength.

• Long columns fails in buckling

Based on slenderness ratio

a) Short column

b) Long or slender column

• IS 456 stipulates the slenderness ratio as the ratio of its effective length 𝑙_𝑒 to its least lateral dimension.

• The effective length 𝑙_𝑒is different from the unsupported length.

• The rectangular RC column of cross-sectional dimensions b and D shall have two effective lengths in the two directions of b and D.

• Slenderness ratio about the major axis = 𝑙_𝑒𝑥 𝐷

• Slenderness ratio about the minor axis = 𝑙_𝑒𝑦 𝑏

• CL. 25.1.2 of IS 456 stipulates the following:

• A compression member may be considered as short when both the slenderness ratios 𝑙_𝑒𝑥 𝐷 and 𝑙_𝑒𝑦 𝑏 are less than 12.

• When both the slenderness ratios 𝑙_𝑒𝑥 𝐷 and 𝑙_𝑒𝑦 𝑏 are more than 12, it is considered as long or slender compression member.

• Accordingly, CL. 25.3, the unsupported length between end restraints shall not exceed 60 times the least lateral dimension of a column (max).

Based on slenderness ratio

Braced and unbraced columns

• It is desirable that the columns do not have to resist any horizontal loads due to wind/ EQ.

• This can be achieved by bracing the columns of water tank or tall buildings.

• Lateral tie members for the columns of water tank or shear walls of tall buildings resist the horizontal forces and these columns are called braced columns.

• The bracings can be in one or more directions depending on the directions of the lateral loads.

• The vertical distance between the points of inflection of the column in the buckled configuration is termed as effective length 𝑙_𝑒.

• The effective length is different from the unsupported length 𝐿 of the member.

• The relation between the effective and unsupported lengths of any compression member is 𝑙_𝑒 = 𝑘𝐿

• where 𝑘 is the ratio of effective to the unsupported lengths.

• Clause 25.2 of IS 456 stipulates the effective lengths of compression members (vide Annex E of IS 456). This parameter is needed in classifying and designing the compression members.

Effective length (CL. 25.2)

Code requirements for reinforcement and detailing

• Longitudinal reinforcement Clause 26.5.3.1

• (a) Min.≮ 0.8 %

• Max.≯ 6 % (4% is actually recommended).

• (b) Number of longitudinal bars

• Min.-In a rectangular column = 4

• In circular columns = 6

• Minimum bar dia. =12 mm

• (c) Columns having helical reinforcement shall have at least six longitudinal bars within and in contact with the helical reinforcement.

The bars shall be placed equidistant around its inner circumference.

• (d) Spacing of longitudinal bars ≯ 300mm measured along the periphery of the column

• (e) In pedestals, nominal reinforcement = 0.15% of the cross- sectional area shall be provided.

• Transverse Reinforcement Clause 26.5.3.2

• All longitudinal reinforcement in a compression member must be enclosed within transverse reinforcement, comprising either lateral ties (with internal angles not exceeding 135˚) or spirals.

• The pitch of transverse reinforcement shall not be more than least of the following:

• i) The least lateral dimension

• ii) Sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied.

• iii) 300 mm

• The diameter lateral ties shall not be less than ¼ of diameter of largest longitudinal bar and in no case less than 6 mm.

Code requirements for reinforcement and

detailing

Helical reinforcement

• Pitch – Helical reinforcement shall be regular formation with the turns of the helix spaced evenly and its ends shall be anchored properly by providing one and a half extra turns of the spiral bar.

• Pitch < = min (75 mm, core diameter/6)

• Pitch = > max (25 mm, 3 x diameter of bar forming the helix)

• (b) Longitudinal bars spaced at a maximum distance of 48 times the diameter of the tie shall be tied by single tie and additional open ties for in between longitudinal bars

Clause 26.5.3.2 stipulates the guidelines of the arrangement of transverse reinforcement. The salient points are:

• (a) Transverse reinforcement shall only go round corner and alternate bars if the longitudinal bars are

not spaced more than 75 mm on either side

• (c) For bars placed in more than one row (i) transverse reinforcement is provided for the outer-most row in accordance with (a) above, and (ii) no bar of the inner row is closer to the nearest compression face than three times the diameter of the largest bar in the inner row.

• (d) For long. bars arranged in a group such that they are not in contact and each group is adequately tied as per (a), (b) or (c) above, as appropriate, the transverse reinforcement as a whole may be provided assuming that each group is a single longitudinal bar for determining the pitch and diameter of the transverse reinforcement. The diameter of such transverse reinforcement should not, however, exceed 20 mm

Thumb Rules

• Width of column should not be less than width of beam.

• Orientation of column should be such that depth of column (larger dimension) shall be in the plane of maximum bending.

• To avoid projection of column outside from a wall, long dimension is placed along the wall.

• Size of column may be changed (if desired) after every three floors.

• Size of such columns be reduced at upper floors by the following amounts

1. Width = by 50 mm 2. Depth = by 100 mm

• Change of shape of column from circular to rectangular/ square may be possible.

• Minimum size of column 1. Circular: 230 mm (9”)

2. Square: 230 x 230 mm (9” x 9”)

3. Rectangular: 230 x 300 mm (9” x 12”)

4. Boundary wall column: 230 x 300 mm (9” x 12”)

• Due to ease of fixing of precast panels, L-shape columns are generally provided on corner of buildings.

• Expansion joint:

1. 30 – 50 mm (concrete structures) 2. 50 – 60 mm (steel structures)

• Conditions for providing expansion joint 1. New building adjoining existing building

2. Intersection of wings of L, T or U shaped buildings 3. Low building abutting high building

4. Different types of foundation in structure (building on raft or isolated footing)

• Settlement joint is provided when two portions of the same builg. have large difference in height.

• Commonly used module sizes (basic module) 1. 1.2 m (golden module)

2. 1.5 m

3. 1.8 m (used in hospitals, where it is required to move machinery/beds in wards)

• Following module sizes are considered as good (architectural module/ design module)

1. 4.8 m (4 x 1.2m) 2. 6.0 m (5 x 1.2m) 3. 7.2 m (6 x 1.2m) 4. 8.4 m (7 x 1.2m)

• Module size less than 3.6 m are not good.

• Structural module may be same or twice of architectural module/design module.

e.g, if architectural module = 4.8 m (size of any space), then structural module = 4.8 x 2 = 9.6 m (size governed by position of columns).

• Location of lifts:

1. Maximum spacing = 70 m (30-50 m is desirable) 2. Near the entrance along with stairs

3. Lifts should be in multiples (not single) and should be symmetrical as far as possible along the length of building.

• Minimum eccentricity

• Clause 25.4- All columns shall be designed for minimum eccentricity

• 𝑒_{min 𝑥} = 𝑚𝑖𝑛 𝐿 500 + 𝐷 30 > 20 𝑚𝑚

• 𝑒_{min 𝑦} = 𝑚𝑖𝑛 𝐿 500 + 𝑏 30 > 20 𝑚𝑚

• Assumptions

• Assumptions for the design of compression members (CL. 39.1 of IS 456).

• (i) The maximum compressive strain in concrete in axial compression is taken as 0.002.

• Maximum compressive strain at the highly compressed extreme fibre in concrete subjected to axial compression and bending and when there is no tension on the section shall be 0.0035 minus 0.75 times the strain at the least compressed extreme fibre.

• (ii) Other assumptions for the design of limit state of collapse in flexure are also applicable here.

• Design of short column under axial loading

• Maximum design strength of concrete = 0.446𝑓_𝑐𝑘 (for 𝜖 = 0.002 − 0.0035)

• Maximum design strength of steel = 0.87𝑓_𝑦

• 𝑒_{x 𝑚𝑖𝑛} = 𝐿 500 + 𝐷 30 > 20 𝑚𝑚

• 𝑜𝑟 12𝐷 500 + 𝐷 30 = 0.057𝐷 ≈ 0.05𝐷

• 𝑒_{y 𝑚𝑖𝑛} = 𝐿 500 + 𝑏 30 > 20 𝑚𝑚 𝑜𝑟 0.05𝑏

• To account for minimum eccentricity, design strength of concrete and steel are further reduced by 10%.

• Therefore, design strength of concrete for short column is taken as

• = 0.9 × 0.446𝑓_𝑐𝑘 = 0.4𝑓_𝑐𝑘

• For Fe 250, the design strength at which the strain is 0.002 is 𝑓_𝑦 1.15 = 0.87𝑓_𝑦

• For Fe 415,

• 0.9𝑓_𝑦𝑑 + 0.05𝑓_𝑦𝑑 0.002−0.00192

0.00226−0.00192 = 0.908𝑓_𝑦𝑑 = 0.908 1.15 = 0.789𝑓_𝑦

• For Fe 500,

• 0.85𝑓_𝑦𝑑 + 0.05𝑓_𝑦𝑑 0.002−0.00195

0.00226−0.00195 = 0.859𝑓_𝑦𝑑 = 0.859 1.15 = 0.746𝑓_𝑦

• To take care of minimum eccentricity, reduce design strength by 10%

• Therefore, for Fe 250 = 0.9 × 0.87𝑓_𝑦 = 0.783𝑓_𝑦

• For Fe 415 = 0.9 × 0.789𝑓_𝑦 = 0.710𝑓_𝑦

• For Fe 500 = 0.9 × 0.746𝑓_𝑦 = 0.671𝑓_𝑦

• Accordingly, CL 39.3 of IS 456 stipulates 0.67𝑓_𝑦 as the design strength for all grades of steel for design of short columns

• Consequently,

• Design strength of concrete = 0.446𝑓_𝑐𝑘

• Design strength of steel = 0.67𝑓_𝑦

• Therefore, the design strength of a short column is obtainable as,

• 𝑃_𝑢 = 𝑃_𝑐 + 𝑃_𝑠

• = 𝑓_𝑐𝑐 𝐴_𝑐 + 𝑓_𝑠𝑐𝐴_𝑠𝑐

• 𝑃_𝑢 = 0.4𝑓_𝑐𝑘𝐴_𝑐 + 0.67𝑓_𝑦𝐴_𝑠𝑐

• where

• 𝑃_𝑢 = factored axial load on the member,

• 𝑓_𝑐𝑘= characteristic compressive strength of the concrete,

• 𝐴_𝑐= area of concrete,

• 𝑓_𝑦 = characteristic strength of the compression reinforcement, and

• 𝐴_𝑠𝑐 = area of longitudinal reinforcement for columns.

• ^𝐴𝑠𝑐

𝐴_𝑔 × 100 = 𝜌 ⇒ 𝐴_𝑠𝑐 = ^𝜌𝐴𝑔

100

• 𝐴_𝑔 = 𝐴_𝑐 + 𝐴_𝑠𝑐

• ⇒ 𝐴_𝑐 = 𝐴_𝑔 − 𝐴_𝑠𝑐 = 𝐴_𝑔 − ^𝜌𝐴𝑔

100 = 𝐴_𝑔 1 − ^𝜌

100

• 𝑃_𝑢 = 0.4𝑓_𝑐𝑘𝐴_𝑔 1 − ^𝜌

100 + 0.67𝑓_𝑦 ^𝜌𝐴𝑔

100

•

• ^𝑃𝑢

𝐴_𝑔 = 0.4𝑓_𝑐𝑘 1 − ^𝜌

100 + 0.67𝑓_𝑦 ^𝜌

100

•

• _𝐴^𝑃𝑢

𝑔 = 0.4𝑓_𝑐𝑘 − 0.4𝑓_{𝑐𝑘 100}^𝜌 + 0.67𝑓_{𝑦 100}^𝜌

• ^𝑃𝑢

𝐴_𝑔 = 0.4𝑓_𝑐𝑘 + ^𝜌

100 0.67𝑓_𝑦 − 0.4𝑓_𝑐𝑘

• 𝜌 = 0.8 − 4% (𝑑𝑖𝑟𝑒𝑐𝑡 𝑐𝑜𝑚𝑝𝑢𝑡𝑎𝑡𝑖𝑜𝑛 𝑚𝑒𝑡ℎ𝑜𝑑)

Example: Design the reinforcement in a column of size 400 mm x 600 mm subjected to an axial load of 2000 kN under service dead load and live load. The column has an unsupported length of 4.0 m and effectively held in position and restrained against rotation in both ends. Use M 25 concrete and Fe 415 steel.

• Step 1: To check if the column is short or slender

• Given 𝐿 = 4000 mm, 𝑏 = 400 mm and 𝐷 = 600 mm.

• Service load (working load) = 2000 kN

• Factored load = 1.5 × 2000 = 3000 𝐾𝑁

• Table 28 of IS 456 =𝑙_𝑒𝑥 = 𝑙_𝑒𝑦 = 0.65𝐿 = 0.65 × 4000 = 2600 mm. So, we have

• 𝑙_𝑒𝑥 𝐷 = 2600 600 = 4.33 < 12

• 𝑙_𝑒𝑦 𝑏 = 2600 400 = 6.5 < 12

• Hence, it is a short column.

• Step 2: Minimum eccentricity

• 𝑒_{x 𝑚𝑖𝑛} = 𝑔𝑟𝑒𝑎𝑡𝑒𝑟 𝑜𝑓 𝐿 500 + 𝐷 30 𝑎𝑛𝑑 20 𝑚𝑚

• 4000 500 + 600 30 = 8 + 20 = 28 𝑚𝑚 𝑎𝑛𝑑 20 𝑚𝑚 = 28 𝑚𝑚

• 𝑒_{y 𝑚𝑖𝑛} = 𝑔𝑟𝑒𝑎𝑡𝑒𝑟 𝑜𝑓 𝐿 500 + 𝑏 30 𝑎𝑛𝑑 20 𝑚𝑚

• 4000 500 + 400 30 = 8 + 13.33 = 21.33 𝑎𝑛𝑑 20 𝑚𝑚 = 21.33 𝑚𝑚

• 0.05𝐷 = 0.05 × 600 = 30 𝑚𝑚 > 28 𝑚𝑚 = 𝑒_{𝑥 𝑚𝑖𝑛}

• 0.05𝑏 = 0.05 × 400 = 20 𝑚𝑚 = 20 𝑚𝑚 = 𝑒_{𝑦 𝑚𝑖𝑛}

• Hence, the equation given in CL.39.3 of IS 456 is applicable for the design here.

• Step 3: Area of steel

• We know that

• 𝑃_𝑢 = 0.4𝑓_𝑐𝑘𝐴_𝑐 + 0.67𝑓_𝑦𝐴_𝑠𝑐 or

• 𝑃_𝑢 = 0.4𝑓_𝑐𝑘𝐴_𝑔 + 0.67𝑓_𝑦 − 0.4𝑓_𝑐𝑘 𝐴_𝑠𝑐 or 𝑃_𝑢 = 0.4𝑓_𝑐𝑘{𝐴_𝑔−𝐴_𝑠𝑐} + 0.67𝑓_𝑦𝐴_𝑠𝑐

• 3000 10³ = 0.4 25 400 600 − 𝐴_𝑠𝑐 + 0.67 415 𝐴_𝑠𝑐

• 𝑤ℎ𝑖𝑐ℎ 𝑔𝑖𝑣𝑒𝑠,

• 𝐴_𝑠𝑐 = 2238.39 𝑚𝑚²

• Provide 6-20 mm diameter and 2-16 mm diameter bars giving 2287 mm² (>2238.39 mm²) and 𝜌 = 0.953%, which is more than minimum percentage of 0.8 and less than maximum percentage of 4.0. Hence, o.k.

• Step 4: Lateral ties

• The diameter of transverse reinforcement (lateral ties) is determined from CL.26.5.3.2 C-2 of IS 456 as not less than

• ∅ 4 and

• (ii) 6 mm.

• Here,∅ = largest bar diameter used as longitudinal

reinforcement = 20 mm. So, the diameter of bars used as lateral ties = 6 mm.

• The pitch of lateral ties, as per CL.26.5.3.2 C-1 of IS 456, should be not more than the least of

• (i) The least lateral dimension of the column = 400 mm

• (ii) Sixteen times the smallest diameter of longitudinal reinforcement bar to be tied = 16 (16) = 256 mm

• (iii) 300 mm

• Adopt pitch of lateral ties = 250 mm

Example 2: Design the column of Problem 1 employing the chart of SP-16.

• Solution 2: Steps 1 and 2 are the same as those of Problem 1.

• Step 3: Area of steel

• 𝑃_𝑢 𝐴_𝑔 = 3000 10³ 600 × 400 = 12.5 𝑀𝑃𝑎

• From the lower section of Chart 25 of SP-16, we get 𝑝 = 0.95% when 𝑃_𝑢 𝐴_𝑔 =12.5 N/mm²and

concrete grade is M 25. This gives 𝐴_𝑠𝑐 = 0.95(400) (600)/100 =2288 mm². The results of both the

problems are in good agreement. Marginally

higher value of A_sc while using the chart is due to parallax error while reading the value from the chart. Here also, 6-20 mm diameter bars + 2-16 mm diameter bars (A_sc provided = 2287 mm²) is o.k., though it is 1 mm² less.

• Step 4 is the same as that of Problem 1. Figure showing the reinforcing bars (longitudinal and transverse reinforcement) of this problem (same column as that of Problem 1).

Spiral columns

• Columns with helical reinforcement take more load than that of tied columns due to additional strength of spirals.

• CL. 39.4 recommends a multiplying factor of 1.05 regarding the strength of such columns.

• The code further recommends that the ratio of volume of helical reinforcement to the volume of core shall not be less than

• 0.36 (𝐴_𝑔 𝐴_𝑐 − 1) 𝑓_𝑐𝑘 𝑓_𝑦 as per CL. 39.4.1.

• The governing equation of the spiral columns may be written as

• 𝑃_𝑢 = 1.05(0.4𝑓_𝑐𝑘𝐴_𝑐 + 0.67𝑓_𝑦𝐴_𝑠𝑐)

Pitch of helical reinforcement

• Volume of helical reinforcement in one loop = 𝜋 𝐷_𝑐 − 𝜑_𝑠𝑝 𝑎_𝑠𝑝

• Volume of core= 𝜋 4 𝐷_𝑐². 𝑝

• where 𝐷_𝑐 =diameter of the core

• 𝜑_𝑠𝑝 =Diameter of the spiral reinforcement

• 𝑎_𝑠𝑝 =Area of cross section of spiral reinforcement

• 𝑝 =Pitch of the spiral reinforcement

• To satisfy the condition of CL 39.4.1 of IS 456, i.e ratio of volume of helical reinforcement to the volume of core shall not be less than 0.36 𝐴_𝑔 𝐴_𝑐 − 1 𝑓_𝑐𝑘 𝑓_𝑦

• 𝜋 𝐷_𝑐 − 𝜑_𝑠𝑝 𝑎_𝑠𝑝 𝜋 4 𝐷_𝑐²𝑝 ≥ 0.36 𝐴_𝑔 𝐴_𝑐 − 1 𝑓_𝑐𝑘 𝑓_𝑦

• which finally gives

• 𝑝 ≤ 11.1 𝐷_𝑐 − 𝜑_𝑠𝑝 𝑎_𝑠𝑝𝑓_𝑦/ 𝐷² − 𝐷_𝑐² 𝑓_𝑐𝑘

Example: Design a circular column of 400 mm diameter with helical reinforcement subjected to an axial load of 1500 kN under service dead and live load. The column has an unsupported length of 3 m effectively held in position at both ends but not restrained against rotation. Use M 25 concrete and Fe 415 steel.

• Step 1: To check if the column is short or slender

• Given 𝐿 = 3000 mm, 𝐷 = 400 mm. Table 28 of Annexure E of IS 456 gives effective length

• =𝑙_𝑒 = 𝐿= 3000 mm. So, we have

• 𝑙_𝑒 𝐷 = 3000 400 = 7.5 < 12

• Hence, it is a short column.

• Gross area of column = ^𝜋₄ 400 ² = 125663.7 𝑚𝑚²

• Step 2: Minimum eccentricity

• 𝑒 _𝑚𝑖𝑛 = 𝐿 500 + 𝑏 30 (> 20 𝑚𝑚)

• 𝑒 _𝑚𝑖𝑛 = 3000 500 + 400 30 = 19.33 𝑚𝑚 (≈ 20 𝑚𝑚)

• 0.05𝐷 = 0.05 × 400 = 20 𝑚𝑚

• As per CL. 39.3 of IS 456, 𝑒_𝑚𝑖𝑛 should not exceed 0.05𝐷. Here, both the eccentricities are the same. So, we can use the equation in that clause of IS 456.

• Step 3: Area of steel

• 𝑃_𝑢 = 1.05(0.4𝑓_𝑐𝑘𝐴_𝑐 + 0.67𝑓_𝑦𝐴_𝑠𝑐)

• 𝐴_𝑐 = 𝐴_𝑔 − 𝐴_𝑠𝑐 = 125663.7 − 𝐴_𝑠𝑐

• 1.5 × 1500 10³ = 1.05{0.4 25 125663.7 − 𝐴_𝑠𝑐 + 0.67 415 𝐴_𝑠𝑐

• 𝑤ℎ𝑖𝑐ℎ 𝑔𝑖𝑣𝑒𝑠,

• 𝐴_𝑠𝑐 = 3304.29 𝑚𝑚² (2.64% of gross area)

• Provide 11-20 mm diameter bars = 3455𝑚𝑚² as longitudinal reinforcement giving 𝜌 = 2.75%, this 𝜌 is between 0.8 (min.) and of 4 (max.) per cent. Hence, o.k.

• Diameter of helical reinforcement (CL.26.5.3.2 d-2) shall be not less than greater of

• (i) one-fourth of the diameter of largest longitudinal bar, and (ii) 6 mm.

• Therefore, with 20 mm diameter bars as longitudinal reinforcement, the diameter of helical reinforcement = 6 mm.

• 𝑝 ≤ 11.1 𝐷_𝑐 − 𝜑_𝑠𝑝 𝑎_𝑠𝑝𝑓_𝑦/ 𝐷² − 𝐷_𝑐² 𝑓_𝑐𝑘

• 𝐷_𝑐 = 400 − 40 − 40 = 320 𝑚𝑚

• 𝜑_𝑠𝑝 = 6 𝑚𝑚

• 𝑎_𝑠𝑝 = 28 𝑚𝑚²

• 𝐷 = 400 𝑚𝑚

• 𝑝 ≤ 11.1 320 − 6 28 415 400² − 320² 25 ≤ 28.125𝑚𝑚

• As per CL.26.5.3.2 d-1, the maximum pitch is the lesser of 75 mm and 320/6 = 53.34 mm and the minimum pitch is greater of 25 mm and 3(6) = 18 mm. We adopt pitch = 25 mm So, provide 6 mm bars @ 25 mm pitch forming the helix.

• Checking of CL. 39.4.1 of IS 456

• To satisfy the condition of CL 39.4.1 of IS 456, we have

• 𝜋 𝐷_𝑐 − 𝜑_𝑠𝑝 𝑎_𝑠𝑝 𝜋 4 𝐷_𝑐²𝑝 ≥ 0.36 𝐴_𝑔 𝐴_𝑐 − 1 𝑓_𝑐𝑘 𝑓_𝑦

• Volume of helical reinforcement in one loop

= 𝜋 𝐷_𝑐 − 𝜑_𝑠𝑝 𝑎_𝑠𝑝

• = 𝜋 320 − 6 28 = 27621𝑚𝑚³

• Volume of core in one loop = 𝜋 4 𝐷_𝑐². 𝑝

• = 𝜋 4 𝐷_𝑐². 𝑝 = 𝜋 4 320 ². 25 = 2011619.2 mm³

• Their ratio = 27621/2010619.2 = 0.0137307

• 0.36 (𝐴_𝑔 𝐴_𝑐 − 1) 𝑓_𝑐𝑘 𝑓_𝑦 = 0.012198795

• Thus, it is seen that the above ratio is not less than 0.36 (𝐴_𝑔 𝐴_𝑐 − 1) 𝑓_𝑐𝑘 𝑓_𝑦 ok