Chapter 8 – Columns (Compression Members)

8.1 Effective Length and Column Classification

A column transfers axial load and bending moments to the members below. Even a nominally concentric load has some eccentricity because of construction tolerances, imperfect alignment and lateral displacement. Design therefore requires a consistent set of axial force, moments about both axes, restraint conditions and effective lengths.

Unsupported length is the actual length between lateral restraints. Effective length l_e = kl is the equivalent buckling length and depends on rotational restraint and whether the frame can sway. It is not automatically the storey height, and it can differ in the two principal directions.

Idealized restraint	Common IS 456 effective-length factor	Condition
Both ends held in position and restrained against rotation	0.65	Use the matching non-sway idealization.
Both ends held; rotation restrained at one end	0.80	The other end can rotate.
Both ends held; neither end rotationally restrained	1.00	Idealized pin-ended case.
One end fixed, the other free	2.00	Cantilever idealization.

These design values must not be confused with ideal Euler-theory factors such as 0.5 for a perfectly fixed-fixed elastic strut. A real frame may require Annex E or a second-order analysis instead of a simple tabulated idealization. An unbraced frame must not be assigned a braced factor merely because its beam-column connections are monolithic.

For a rectangular section with depth D for bending about x and breadth b for bending about y, check l_ex/D and l_ey/b. Both must satisfy the short-column criterion (less than 12 in IS 456); otherwise account for slenderness in the affected direction. A column can be short about its major axis and slender about its minor axis.

Example: the 2080 Baishakh column is 400 × 500 mm, 4.0 m unsupported, held at both ends and rotationally restrained at one end. With the matching braced factor 0.8, l_e = 3.2 m. The ratios are 3200/500 = 6.4 and 3200/400 = 8.0, so it is short about both axes.

Minimum Eccentricity

For each relevant direction, e_min = l/500 + D/30, subject to a minimum of 20 mm; use the unsupported length and the dimension in that bending direction. The corresponding moment is P_ue_min. Check the code's treatment of minimum eccentricity in biaxial design; it is not permission to ignore a larger moment already obtained from analysis.

For the 4.0 m, 400 × 500 mm example, the two values are 21.33 mm and 24.67 mm. Pair each with its corresponding axis. If a question gives only effective lengths, do not silently substitute them for the missing unsupported length in this expression.

8.2 Interaction Diagrams for Columns 2080

PAST EXAM QUESTIONS:

Define the interaction diagram with its features and neat sketches. Discuss the modes of failure for compression members in eccentric compression within the interaction diagram. [6] (2080 Bhadra, Q3a)

Q1. Define the interaction diagram with its features and neat sketches. Discuss the modes of failure for compression members in eccentric compression within the interaction diagram. [6] (2080 Bhadra, Q3a)

A column carries an axial load P together with a moment M = P·e. An interaction diagram is the section-resistance envelope of compatible axial-load and moment pairs. A design point inside the appropriate design envelope satisfies that section-capacity check; it does not by itself establish adequate slenderness, anchorage, confinement or overall frame stability.

Key features:

Top (P_uz, pure axial): the maximum squash load, M = 0.
Balanced point: the limiting concrete compression strain and the relevant tension-steel yield condition occur together. It separates characteristic tension- and compression-controlled behaviour; it is not a universal assertion that the moment is exactly the maximum possible value for every section model.
Bottom (pure moment): P = 0, the section acts as a beam.
Above the balanced point → compression-controlled (brittle) failure — adding axial load reduces the moment capacity; below → tension-controlled (ductile) failure — adding axial load increases the moment capacity.

Schematic P–M interaction envelope, not a design chart: axial compression, the balanced region and pure bending are different compatible strain states.

Choose a neutral-axis position and a compatible limiting strain distribution.
Calculate the concrete compression resultant and its position from the design stress block.
Calculate strain and design stress in each reinforcement layer; some bars may be in compression and others in tension.
Subtract the concrete displaced by compression bars if the concrete area was included in the gross block.
Add axial forces to obtain P and take moments about the section centroid to obtain M.
Repeat for other neutral-axis positions to trace the resistance envelope.

Where the entire section is compressed, the allowable strain distribution is different from the pure-flexure case; uniform axial compression uses the 0.002 limiting strain. Extending a 0.0035 flexural triangle beyond the section without the relevant compression-strain condition gives an incorrect high-load interaction curve.

═══ 8.3 ═══

Key features:

Top (P_uz, pure axial): the maximum squash load, M = 0.
Balanced point: the limiting concrete compression strain and the relevant tension-steel yield condition occur together. It separates characteristic tension- and compression-controlled behaviour; it is not a universal assertion that the moment is exactly the maximum possible value for every section model.
Bottom (pure moment): P = 0, the section acts as a beam.
Above the balanced point → compression-controlled (brittle) failure — adding axial load reduces the moment capacity; below → tension-controlled (ductile) failure — adding axial load increases the moment capacity.

Schematic P–M interaction envelope, not a design chart: axial compression, the balanced region and pure bending are different compatible strain states.

How an Interaction Curve Is Constructed

Choose a neutral-axis position and a compatible limiting strain distribution.
Calculate the concrete compression resultant and its position from the design stress block.
Calculate strain and design stress in each reinforcement layer; some bars may be in compression and others in tension.
Subtract the concrete displaced by compression bars if the concrete area was included in the gross block.
Add axial forces to obtain P and take moments about the section centroid to obtain M.
Repeat for other neutral-axis positions to trace the resistance envelope.

8.3 Design of Short Columns (Uniaxial & Biaxial) 2081208020782076207520742073207220712069

PAST EXAM QUESTIONS: COMPLETE SHORT-COLUMN QUESTION LIST

Design a reinforced-concrete column of cross-section 400 mm × 500 mm with an unsupported length of 4.0 m. Both ends are effectively held in position, and one end is restrained against rotation. The column is subjected to a factored axial load of 1600 kN and factored bending moments of 150 kN·m and 50 kN·m. Use M25 concrete and Fe500 steel. [10] (2080 Baishakh, Q2b)
Design the longitudinal and transverse reinforcement in a rectangular reinforced-concrete column of size 350 mm × 550 mm. The column is subjected to a factored axial load of 900 kN and a factored bending moment of 200 kN·m about its major axis. Its unsupported length is 3 m and its effective length is 3.6 m. Use M25 concrete and Fe415 steel. [10] (2081 Baishakh, Q4a)
Determine the longitudinal and transverse reinforcement of a reinforced-concrete column subjected to a factored axial load of 1440 kN and factored bending moments M_ux = 195 kN·m about the major axis and M_uy = 180 kN·m about the minor axis. The column has a cross-section of 350 mm × 350 mm and an unsupported length of 3.50 m. Adopt M20 concrete and Fe500 TMT steel. Also provide the ductile detailing of the transverse reinforcement. [12] (2076 Chaitra, Q2b) → also 11.5
Determine the longitudinal and transverse reinforcement in a short reinforced-concrete column subjected to a factored axial load of 1400 kN and factored bending moments M_ux = 200 kN·m and M_uy = 110 kN·m. The column has a cross-section of 300 mm × 400 mm and an unsupported length of 3 m. Adopt M20 concrete and Fe500 TMT steel. Also provide the ductile detailing of the transverse reinforcement. [10] (2076 Ashwin, Q4a) → also 11.5
Determine the longitudinal and transverse reinforcement in a biaxially loaded short square column 400 mm × 400 mm, ultimate factored axial load 800 kN inclusive of live load at an eccentricity of 80 mm in both X and Y directions. Use M20 concrete and Fe415 steel. [10] (2078 Bhadra, Q3a)
Design a short rectangular column of size 350 mm × 500 mm and unsupported length 3.30 m subjected to a factored axial load of 1500 kN and factored moments 130 kN-m and 80 kN-m about major and minor axes. Adopt M30 concrete and Fe500 steel. Sketch the reinforcement details. [14] (2075 Chaitra, Q3a)
Determine the longitudinal and transverse reinforcement in a biaxially loaded column with a cross-section of 500 mm × 600 mm and an unsupported length of 3.10 m. The column carries a factored axial load P_u = 1300 kN and factored bending moments M_ux = 125 kN·m and M_uy = 200 kN·m. Use M25 concrete and Fe500 steel, with the longitudinal reinforcement distributed on all four sides. Sketch the reinforcement details. [15] (2075 Ashwin, Q4a)
Design a short reinforced-concrete column of size 300 mm × 450 mm with an unsupported length of 3.0 m. The column is subjected to a factored axial load of 1550 kN and factored bending moments M_ux = 130 kN·m and M_uy = 90 kN·m. Provide ductile detailing for the transverse reinforcement. [14] (2074 Chaitra, Q2b) → also 11.5
An RC column of size 35 cm × 40 cm and unsupported length 3.10 m is subjected to a factored axial load of 1500 kN and factored biaxial moments M_ux = 125 kN·m and M_uy = 88 kN·m. Its ends are effectively held in position but not restrained against rotation. Design longitudinal and transverse reinforcement and sketch the details. Use M25 concrete and Fe500 steel. [15] (2074 Ashwin, Q3b)
Design the longitudinal reinforcement for a short column 400 mm × 500 mm subjected to P_u = 1600 kN, M_ux = 20.0 kN-m and M_uy = 150 kN-m. Use M25 concrete and Fe415 steel. [15] (2073 Shrawan, Q3a) Scan annotation: an added handwritten note gives unsupported length 3 m; distinguish this from the printed statement.
Design the longitudinal and transverse reinforcement for a short column of size 35 cm × 45 cm subjected to a factored axial load of 1800 kN, M_ux = 175 kN·m and M_uy = 105 kN·m. Reinforcement is distributed equally on two sides. Use M25 concrete and Fe500 steel; unsupported length is 3.1 m. [15] (2072 Chaitra, Q3a)
Determine the longitudinal and transverse reinforcement in a short rectangular column subjected to a factored axial load of 2000 kN and factored bending moments M_ux = 190 kN·m about the major axis and M_uy = 95 kN·m about the minor axis. The column has a cross-section of 300 mm × 500 mm and an unsupported length of 3 m. Adopt M30 concrete and Fe500 steel. [16] (2072 Kartik, Q3b)
Design a short rectangular column of size 450 mm × 300 mm and unsupported length 3 m subjected to an ultimate axial load of 1500 kN and ultimate moments 150 kNm and 80 kNm about major and minor axes. Adopt M30 concrete and Fe500 steel. Sketch the final design. [14] (2071 Chaitra, Q1c)
A column of 4 m length with both ends fixed and effectively held is subjected to a design axial load of 1000 kN and factored bending moment of 100 kN-m. Design the rectangular column with its longitudinal and transverse reinforcement. [14] (2069 Chaitra, Q1b)
Design a square RC column that has to carry an ultimate factored load of 800 kN inclusive of live load at an eccentricity of 80 mm in both X and Y directions. Use M20 concrete and Fe415 steel. [8] (2067 Ashadh, Q5c)
Design a column member to transfer a factored load P_u = 1350 kN, M_ux = 150 kN·m and M_uy = 100 kN·m, with an unsupported length of 3.4 m. Use M25 concrete and Fe500 steel. Distribute reinforcement equally on all four sides and assume that the column is fully restrained at both supports against displacement and rotation. [12] (2082 Baishakh, Q4b)

Q1. Design a reinforced-concrete column of cross-section 400 mm × 500 mm with an unsupported length of 4.0 m. Both ends are effectively held in position, and one end is restrained against rotation. The column is subjected to a factored axial load of 1600 kN and factored bending moments of 150 kN·m and 50 kN·m. Use M25 concrete and Fe500 steel. [10] (2080 Baishakh, Q2b)

1. Design basis. Use the stated braced end condition with effective-length factor 0.8 in both directions. Use b = 400 mm, D = 500 mm, M25 concrete and Fe500 steel. The factored axial load is 1600 kN. For this calculation x-bending acts over depth D and y-bending over breadth b. Use all four faces for the longitudinal reinforcement, as specified or explicitly assumed.

2. Eccentricity. With unsupported length 4 m, the ordinary minimum eccentricities are max(l/500 + D/30, 20) = 24.67 mm and max(l/500 + b/30, 20) = 21.33 mm. This study conservatively checks the moment envelope with at least the corresponding minimum moment in both directions.

3. Slenderness and design moments. Effective lengths are 3.2 m about x and 3.2 m about y, giving ratios 6.400 and 8.000. Where the ratio reaches the slender range, use unreduced additional moments P_uD(l_ex/D)²/2000 and P_ub(l_ey/b)²/2000: 0.000 and 0.000 kN·m. Using the larger relevant end moment conservatively gives design moments 150.000 and 50.000 kN·m. Do not add these effects again if a second-order analysis already includes them.

4. Trial reinforcement. Consider 8 bars of 16 mm diameter: A_sc = 1608.50 mm², or 0.804% of gross area. Effective cover to the bar centres is 56 mm, from 40 mm nominal cover, the selected ties and half the main-bar diameter. The candidate passes the adopted clear bar-spacing check.

5. Uniaxial capacities at the same axial load. Solve section equilibrium in each direction using a linear strain diagram, the parabolic-rectangular concrete law and the idealized steel design curve. Subtract the concrete displaced by compression bars. For a neutral axis outside the section, use the compatible all-compression strain limit rather than extending a 0.0035 triangle without modification. The neutral axes are x_ux = 372.658 mm and x_uy = 296.158 mm; the corresponding capacities are M_ux1 = 191.283 kN·m and M_uy1 = 147.601 kN·m.

These values integrate the continuous concrete design curve and are calculated section-model results, not claimed SP 16 chart readings. Small differences from the rounded 0.36-resultant/0.42-centroid hand calculation are expected. An examination solution using SP 16 should select the matching steel grade, face arrangement and cover ratios and record the corresponding readings.

6. Biaxial interaction. P_uz = 0.45f_ck(A_g − A_sc) + 0.75f_yA_sc = 2835.090 kN. Thus P_u/P_uz = 0.56436, and the code interpolation gives α_n = 1.60726.

(150.000/191.283)^1.60726 + (50.000/147.601)^1.60726 = 0.8521

7. Ordinary transverse reinforcement. Use at least 8 mm ties at 200 mm centres, with cross-ties/overlapping hoops as needed to restrain intermediate longitudinal bars. Check the minimum of 16 times the smallest main-bar diameter, the least lateral dimension and 300 mm. A single outer hoop is not sufficient if it leaves intermediate bars unrestrained. Detail laps and anchorage separately.

Conclusion: the trial passes the calculated biaxial section interaction under the stated assumptions. The result is conditional on the effective lengths, minimum eccentricity, bar arrangement and cover used; member shear, joint/foundation transfer and the final construction detailing must also satisfy the governing provisions.

Question-specific trial longitudinal arrangement. The schematic outer hoop is not a complete cross-tie detail; use the transverse requirements in the answer.

Q2. Design the longitudinal and transverse reinforcement in a rectangular reinforced-concrete column of size 350 mm × 550 mm. The column is subjected to a factored axial load of 900 kN and a factored bending moment of 200 kN·m about its major axis. Its unsupported length is 3 m and its effective length is 3.6 m. Use M25 concrete and Fe415 steel. [10] (2081 Baishakh, Q4a)

1. Design basis. Use the given 3.6 m effective length in both directions for this study check; the supplied large moment acts about the major axis. Use b = 350 mm, D = 550 mm, M25 concrete and Fe415 steel. The factored axial load is 900 kN. For this calculation x-bending acts over depth D and y-bending over breadth b. Use all four faces for the longitudinal reinforcement, as specified or explicitly assumed.

2. Eccentricity. With unsupported length 3 m, the ordinary minimum eccentricities are max(l/500 + D/30, 20) = 24.33 mm and max(l/500 + b/30, 20) = 20.00 mm. This study conservatively checks the moment envelope with at least the corresponding minimum moment in both directions.

3. Slenderness and design moments. Effective lengths are 3.6 m about x and 3.6 m about y, giving ratios 6.545 and 10.286. Where the ratio reaches the slender range, use unreduced additional moments P_uD(l_ex/D)²/2000 and P_ub(l_ey/b)²/2000: 0.000 and 0.000 kN·m. Using the larger relevant end moment conservatively gives design moments 200.000 and 18.000 kN·m. Do not add these effects again if a second-order analysis already includes them.

4. Trial reinforcement. Consider 8 bars of 16 mm diameter: A_sc = 1608.50 mm², or 0.836% of gross area. Effective cover to the bar centres is 56 mm, from 40 mm nominal cover, the selected ties and half the main-bar diameter. The candidate passes the adopted clear bar-spacing check.

5. Uniaxial capacities at the same axial load. Solve section equilibrium in each direction using a linear strain diagram, the parabolic-rectangular concrete law and the idealized steel design curve. Subtract the concrete displaced by compression bars. For a neutral axis outside the section, use the compatible all-compression strain limit rather than extending a 0.0035 triangle without modification. The neutral axes are x_ux = 283.573 mm and x_uy = 180.196 mm; the corresponding capacities are M_ux1 = 231.711 kN·m and M_uy1 = 137.388 kN·m.

6. Biaxial interaction. P_uz = 0.45f_ck(A_g − A_sc) + 0.75f_yA_sc = 2648.174 kN. Thus P_u/P_uz = 0.33986, and the code interpolation gives α_n = 1.23309.

(200.000/231.711)^1.23309 + (18.000/137.388)^1.23309 = 0.9156

Question-specific trial longitudinal arrangement. The schematic outer hoop is not a complete cross-tie detail; use the transverse requirements in the answer.

Q3. Determine the longitudinal and transverse reinforcement of a reinforced-concrete column subjected to a factored axial load of 1440 kN and factored bending moments M_ux = 195 kN·m about the major axis and M_uy = 180 kN·m about the minor axis. The column has a cross-section of 350 mm × 350 mm and an unsupported length of 3.50 m. Adopt M20 concrete and Fe500 TMT steel. Also provide the ductile detailing of the transverse reinforcement. [12] (2076 Chaitra, Q2b)

1. Design basis. Assume a non-sway effective-length factor of 1.0 where restraints are not specified. Use b = 350 mm, D = 350 mm, M20 concrete and Fe500 steel. The factored axial load is 1440 kN. For this calculation x-bending acts over depth D and y-bending over breadth b. Use all four faces for the longitudinal reinforcement, as specified or explicitly assumed.

2. Eccentricity. With unsupported length 3.5 m, the ordinary minimum eccentricities are max(l/500 + D/30, 20) = 20.00 mm and max(l/500 + b/30, 20) = 20.00 mm. This study conservatively checks the moment envelope with at least the corresponding minimum moment in both directions.

3. Slenderness and design moments. Effective lengths are 3.5 m about x and 3.5 m about y, giving ratios 10.000 and 10.000. Where the ratio reaches the slender range, use unreduced additional moments P_uD(l_ex/D)²/2000 and P_ub(l_ey/b)²/2000: 0.000 and 0.000 kN·m. Using the larger relevant end moment conservatively gives design moments 195.000 and 180.000 kN·m. Do not add these effects again if a second-order analysis already includes them.

4. Trial reinforcement. Consider 8 bars of 32 mm diameter: A_sc = 6433.98 mm², or 5.252% of gross area. Effective cover to the bar centres is 64 mm, from 40 mm nominal cover, the selected ties and half the main-bar diameter. The candidate passes the adopted clear bar-spacing check.

5. Uniaxial capacities at the same axial load. Solve section equilibrium in each direction using a linear strain diagram, the parabolic-rectangular concrete law and the idealized steel design curve. Subtract the concrete displaced by compression bars. For a neutral axis outside the section, use the compatible all-compression strain limit rather than extending a 0.0035 triangle without modification. The neutral axes are x_ux = 233.830 mm and x_uy = 233.830 mm; the corresponding capacities are M_ux1 = 193.483 kN·m and M_uy1 = 193.483 kN·m.

6. Biaxial interaction. P_uz = 0.45f_ck(A_g − A_sc) + 0.75f_yA_sc = 3457.337 kN. Thus P_u/P_uz = 0.41651, and the code interpolation gives α_n = 1.36084.

(195.000/193.483)^1.36084 + (180.000/193.483)^1.36084 = 1.9171

The checked arrangement is not adequate: the interaction sum exceeds 1.0. None of the evaluated practical single-bar perimeter arrangements within the ordinary 6% upper limit satisfied the adopted actions and spacing assumptions. This is not a mathematical proof that every possible cage is infeasible, but it does rule out accepting these trials. Revisit section size, arrangement and the justified analysis assumptions; do not reduce the given loads or quote an invented safe chart capacity.

Question-specific trial longitudinal arrangement. The schematic outer hoop is not a complete cross-tie detail; use the transverse requirements in the answer.

Q4. Determine the longitudinal and transverse reinforcement in a short reinforced-concrete column subjected to a factored axial load of 1400 kN and factored bending moments M_ux = 200 kN·m and M_uy = 110 kN·m. The column has a cross-section of 300 mm × 400 mm and an unsupported length of 3 m. Adopt M20 concrete and Fe500 TMT steel. Also provide the ductile detailing of the transverse reinforcement. [10] (2076 Ashwin, Q4a)

1. Design basis. Retain the question's short-column description and take effective length equal to unsupported length for the trial. Use b = 300 mm, D = 400 mm, M20 concrete and Fe500 steel. The factored axial load is 1400 kN. For this calculation x-bending acts over depth D and y-bending over breadth b. Use all four faces for the longitudinal reinforcement, as specified or explicitly assumed.

2. Eccentricity. With unsupported length 3 m, the ordinary minimum eccentricities are max(l/500 + D/30, 20) = 20.00 mm and max(l/500 + b/30, 20) = 20.00 mm. This study conservatively checks the moment envelope with at least the corresponding minimum moment in both directions.

3. Slenderness and design moments. Effective lengths are 3 m about x and 3 m about y, giving ratios 7.500 and 10.000. Where the ratio reaches the slender range, use unreduced additional moments P_uD(l_ex/D)²/2000 and P_ub(l_ey/b)²/2000: 0.000 and 0.000 kN·m. Using the larger relevant end moment conservatively gives design moments 200.000 and 110.000 kN·m. Do not add these effects again if a second-order analysis already includes them.

4. Trial reinforcement. Consider 8 bars of 32 mm diameter: A_sc = 6433.98 mm², or 5.362% of gross area. Effective cover to the bar centres is 64 mm, from 40 mm nominal cover, the selected ties and half the main-bar diameter. The candidate passes the adopted clear bar-spacing check.

5. Uniaxial capacities at the same axial load. Solve section equilibrium in each direction using a linear strain diagram, the parabolic-rectangular concrete law and the idealized steel design curve. Subtract the concrete displaced by compression bars. For a neutral axis outside the section, use the compatible all-compression strain limit rather than extending a 0.0035 triangle without modification. The neutral axes are x_ux = 268.535 mm and x_uy = 195.468 mm; the corresponding capacities are M_ux1 = 240.338 kN·m and M_uy1 = 149.141 kN·m.

6. Biaxial interaction. P_uz = 0.45f_ck(A_g − A_sc) + 0.75f_yA_sc = 3434.837 kN. Thus P_u/P_uz = 0.40759, and the code interpolation gives α_n = 1.34598.

(200.000/240.338)^1.34598 + (110.000/149.141)^1.34598 = 1.4447

Question-specific trial longitudinal arrangement. The schematic outer hoop is not a complete cross-tie detail; use the transverse requirements in the answer.

Q5. Determine the longitudinal and transverse reinforcement in a biaxially loaded short square column 400 mm × 400 mm, ultimate factored axial load 800 kN inclusive of live load at an eccentricity of 80 mm in both X and Y directions. Use M20 concrete and Fe415 steel. [10] (2078 Bhadra, Q3a)

The paper gives a short 400 × 400 mm M20/Fe415 column with P_u = 800 kN and eccentricity 80 mm in both directions. Therefore M_ux = M_uy = 64 kN·m.

Trial arrangement: eight 16 mm bars distributed on four faces, one at each corner and one at the midpoint of each face. With 40 mm nominal cover, 8 mm ties and 16 mm main bars, effective cover is 56 mm. A_sc = 8π16²/4 = 1608.50 mm², not the 1280 mm² minimum-steel area.

A_c = 160000 − 1608.50 = 158391.50 mm²
P_uz = [0.45 × 20 × 158391.50 + 0.75 × 415 × 1608.50]/1000
= 1926.17 kN
P_u/P_uz = 0.41533; α_n = 1.35889

As an alternative to inventing a chart reading, perform a strain-compatible section calculation. For bending about either axis the bar layers are three bars at 56 mm, two at 200 mm and three at 344 mm. Using the IS design concrete block, the idealized Fe415 design curve and the displaced-concrete correction, axial equilibrium at 800 kN gives x_u approximately 246.48 mm.

Force component	Compression positive (kN)
Concrete compression block	709.87
Net top-layer steel contribution	206.08
Net middle-layer steel contribution	51.11
Bottom-layer tension	−167.05

The unrounded forces sum to 800 kN. Taking moments about the section centroid gives M_ux1 = M_uy1 approximately 122.22 kN·m. Hence:

(64/122.22)^1.35889 + (64/122.22)^1.35889 = 0.830 < 1.0

The trial passes this approximate code-based biaxial section check. Eight 16 mm bars provide about 1.01% longitudinal steel. Ordinary 8 mm ties at 200 mm centres satisfy the basic diameter/spacing bounds; provide cross-ties or a suitable hoop arrangement to restrain intermediate bars. This is not a seismic-confinement check, and it relies on the paper's short-column classification and the stated cover/bar arrangement. In an SP 16 solution, replace the strain-derived capacities with documented chart readings for the matching section.

Question-specific trial longitudinal arrangement. The schematic outer hoop is not a complete cross-tie detail; use the transverse requirements in the answer.

Q6. Design a short rectangular column of size 350 mm × 500 mm and unsupported length 3.30 m subjected to a factored axial load of 1500 kN and factored moments 130 kN-m and 80 kN-m about major and minor axes. Adopt M30 concrete and Fe500 steel. Sketch the reinforcement details. [14] (2075 Chaitra, Q3a)

1. Design basis. Assume effective length equal to unsupported length for the stated short column. Use b = 350 mm, D = 500 mm, M30 concrete and Fe500 steel. The factored axial load is 1500 kN. For this calculation x-bending acts over depth D and y-bending over breadth b. Use all four faces for the longitudinal reinforcement, as specified or explicitly assumed.

2. Eccentricity. With unsupported length 3.3 m, the ordinary minimum eccentricities are max(l/500 + D/30, 20) = 23.27 mm and max(l/500 + b/30, 20) = 20.00 mm. This study conservatively checks the moment envelope with at least the corresponding minimum moment in both directions.

3. Slenderness and design moments. Effective lengths are 3.3 m about x and 3.3 m about y, giving ratios 6.600 and 9.429. Where the ratio reaches the slender range, use unreduced additional moments P_uD(l_ex/D)²/2000 and P_ub(l_ey/b)²/2000: 0.000 and 0.000 kN·m. Using the larger relevant end moment conservatively gives design moments 130.000 and 80.000 kN·m. Do not add these effects again if a second-order analysis already includes them.

4. Trial reinforcement. Consider 4 bars of 22 mm diameter: A_sc = 1520.53 mm², or 0.869% of gross area. Effective cover to the bar centres is 59 mm, from 40 mm nominal cover, the selected ties and half the main-bar diameter. The candidate passes the adopted clear bar-spacing check.

5. Uniaxial capacities at the same axial load. Solve section equilibrium in each direction using a linear strain diagram, the parabolic-rectangular concrete law and the idealized steel design curve. Subtract the concrete displaced by compression bars. For a neutral axis outside the section, use the compatible all-compression strain limit rather than extending a 0.0035 triangle without modification. The neutral axes are x_ux = 350.724 mm and x_uy = 241.637 mm; the corresponding capacities are M_ux1 = 223.300 kN·m and M_uy1 = 144.897 kN·m.

6. Biaxial interaction. P_uz = 0.45f_ck(A_g − A_sc) + 0.75f_yA_sc = 2912.172 kN. Thus P_u/P_uz = 0.51508, and the code interpolation gives α_n = 1.52513.

(130.000/223.300)^1.52513 + (80.000/144.897)^1.52513 = 0.8424

Question-specific trial longitudinal arrangement. The schematic outer hoop is not a complete cross-tie detail; use the transverse requirements in the answer.

Q7. Determine the longitudinal and transverse reinforcement in a biaxially loaded column with a cross-section of 500 mm × 600 mm and an unsupported length of 3.10 m. The column carries a factored axial load P_u = 1300 kN and factored bending moments M_ux = 125 kN·m and M_uy = 200 kN·m. Use M25 concrete and Fe500 steel, with the longitudinal reinforcement distributed on all four sides. Sketch the reinforcement details. [15] (2075 Ashwin, Q4a)

1. Design basis. Use reinforcement on all four faces and assume effective length equal to unsupported length. Use b = 500 mm, D = 600 mm, M25 concrete and Fe500 steel. The factored axial load is 1300 kN. For this calculation x-bending acts over depth D and y-bending over breadth b. Use all four faces for the longitudinal reinforcement, as specified or explicitly assumed.

2. Eccentricity. With unsupported length 3.1 m, the ordinary minimum eccentricities are max(l/500 + D/30, 20) = 26.20 mm and max(l/500 + b/30, 20) = 22.87 mm. This study conservatively checks the moment envelope with at least the corresponding minimum moment in both directions.

3. Slenderness and design moments. Effective lengths are 3.1 m about x and 3.1 m about y, giving ratios 5.167 and 6.200. Where the ratio reaches the slender range, use unreduced additional moments P_uD(l_ex/D)²/2000 and P_ub(l_ey/b)²/2000: 0.000 and 0.000 kN·m. Using the larger relevant end moment conservatively gives design moments 125.000 and 200.000 kN·m. Do not add these effects again if a second-order analysis already includes them.

4. Trial reinforcement. Consider 12 bars of 16 mm diameter: A_sc = 2412.74 mm², or 0.804% of gross area. Effective cover to the bar centres is 56 mm, from 40 mm nominal cover, the selected ties and half the main-bar diameter. The candidate passes the adopted clear bar-spacing check.

5. Uniaxial capacities at the same axial load. Solve section equilibrium in each direction using a linear strain diagram, the parabolic-rectangular concrete law and the idealized steel design curve. Subtract the concrete displaced by compression bars. For a neutral axis outside the section, use the compatible all-compression strain limit rather than extending a 0.0035 triangle without modification. The neutral axes are x_ux = 294.036 mm and x_uy = 245.092 mm; the corresponding capacities are M_ux1 = 409.875 kN·m and M_uy1 = 333.054 kN·m.

6. Biaxial interaction. P_uz = 0.45f_ck(A_g − A_sc) + 0.75f_yA_sc = 4252.635 kN. Thus P_u/P_uz = 0.30569, and the code interpolation gives α_n = 1.17615.

(125.000/409.875)^1.17615 + (200.000/333.054)^1.17615 = 0.7963

Question-specific trial longitudinal arrangement. The schematic outer hoop is not a complete cross-tie detail; use the transverse requirements in the answer.

Q8. Design a short reinforced-concrete column of size 300 mm × 450 mm with an unsupported length of 3.0 m. The column is subjected to a factored axial load of 1550 kN and factored bending moments M_ux = 130 kN·m and M_uy = 90 kN·m. Provide ductile detailing for the transverse reinforcement. [14] (2074 Chaitra, Q2b)

1. Design basis. The statement omits material grades; adopt M25 and Fe500 for this study design. Assume effective length equal to unsupported length. Use b = 300 mm, D = 450 mm, M25 concrete and Fe500 steel. The factored axial load is 1550 kN. For this calculation x-bending acts over depth D and y-bending over breadth b. Use all four faces for the longitudinal reinforcement, as specified or explicitly assumed.

2. Eccentricity. With unsupported length 3 m, the ordinary minimum eccentricities are max(l/500 + D/30, 20) = 21.00 mm and max(l/500 + b/30, 20) = 20.00 mm. This study conservatively checks the moment envelope with at least the corresponding minimum moment in both directions.

3. Slenderness and design moments. Effective lengths are 3 m about x and 3 m about y, giving ratios 6.667 and 10.000. Where the ratio reaches the slender range, use unreduced additional moments P_uD(l_ex/D)²/2000 and P_ub(l_ey/b)²/2000: 0.000 and 0.000 kN·m. Using the larger relevant end moment conservatively gives design moments 130.000 and 90.000 kN·m. Do not add these effects again if a second-order analysis already includes them.

4. Trial reinforcement. Consider 4 bars of 36 mm diameter: A_sc = 4071.50 mm², or 3.016% of gross area. Effective cover to the bar centres is 68 mm, from 40 mm nominal cover, the selected ties and half the main-bar diameter. The candidate passes the adopted clear bar-spacing check.

5. Uniaxial capacities at the same axial load. Solve section equilibrium in each direction using a linear strain diagram, the parabolic-rectangular concrete law and the idealized steel design curve. Subtract the concrete displaced by compression bars. For a neutral axis outside the section, use the compatible all-compression strain limit rather than extending a 0.0035 triangle without modification. The neutral axes are x_ux = 337.977 mm and x_uy = 215.283 mm; the corresponding capacities are M_ux1 = 235.202 kN·m and M_uy1 = 126.403 kN·m.

6. Biaxial interaction. P_uz = 0.45f_ck(A_g − A_sc) + 0.75f_yA_sc = 2999.760 kN. Thus P_u/P_uz = 0.51671, and the code interpolation gives α_n = 1.52785.

(130.000/235.202)^1.52785 + (90.000/126.403)^1.52785 = 0.9993

7. Ordinary transverse reinforcement. Use at least 10 mm ties at 200 mm centres, with cross-ties/overlapping hoops as needed to restrain intermediate longitudinal bars. Check the minimum of 16 times the smallest main-bar diameter, the least lateral dimension and 300 mm. A single outer hoop is not sufficient if it leaves intermediate bars unrestrained. Detail laps and anchorage separately.

8. Ductile transverse detail requested in the paper. An end-zone length is at least max(D, l/6, 450) = 500.0 mm. A conservative preliminary close-hoop spacing is 70 mm. With the core measured outside the hoops, A_k = 220 × 370 mm². Comparing the classical 0.18sh(f_ck/f_yh)(A_g/A_k − 1) expression and its 0.09 lower bound gives required directional leg areas approximately 109.96 and 184.93 mm² at this spacing, using f_yh = 415 MPa.

Choose an actual closed-hoop/cross-tie system providing those directional areas and restraining the bars. Check all adopted IS 13920 edition/amendment requirements, seismic hooks, capacity-based shear, joint confinement and permitted lap zones. This preliminary confinement calculation is not a substitute for those checks.

Question-specific trial longitudinal arrangement. The schematic outer hoop is not a complete cross-tie detail; use the transverse requirements in the answer.

Q9. An RC column of size 35 cm × 40 cm and unsupported length 3.10 m is subjected to a factored axial load of 1500 kN and factored biaxial moments M_ux = 125 kN·m and M_uy = 88 kN·m. Its ends are effectively held in position but not restrained against rotation. Design longitudinal and transverse reinforcement and sketch the details. Use M25 concrete and Fe500 steel. [15] (2074 Ashwin, Q3b)

1. Design basis. Ends are held but free to rotate, so use the idealized effective-length factor 1.0. Use b = 350 mm, D = 400 mm, M25 concrete and Fe500 steel. The factored axial load is 1500 kN. For this calculation x-bending acts over depth D and y-bending over breadth b. Use all four faces for the longitudinal reinforcement, as specified or explicitly assumed.

2. Eccentricity. With unsupported length 3.1 m, the ordinary minimum eccentricities are max(l/500 + D/30, 20) = 20.00 mm and max(l/500 + b/30, 20) = 20.00 mm. This study conservatively checks the moment envelope with at least the corresponding minimum moment in both directions.

3. Slenderness and design moments. Effective lengths are 3.1 m about x and 3.1 m about y, giving ratios 7.750 and 8.857. Where the ratio reaches the slender range, use unreduced additional moments P_uD(l_ex/D)²/2000 and P_ub(l_ey/b)²/2000: 0.000 and 0.000 kN·m. Using the larger relevant end moment conservatively gives design moments 125.000 and 88.000 kN·m. Do not add these effects again if a second-order analysis already includes them.

4. Trial reinforcement. Consider 20 bars of 16 mm diameter: A_sc = 4021.24 mm², or 2.872% of gross area. Effective cover to the bar centres is 56 mm, from 40 mm nominal cover, the selected ties and half the main-bar diameter. The candidate passes the adopted clear bar-spacing check.

5. Uniaxial capacities at the same axial load. Solve section equilibrium in each direction using a linear strain diagram, the parabolic-rectangular concrete law and the idealized steel design curve. Subtract the concrete displaced by compression bars. For a neutral axis outside the section, use the compatible all-compression strain limit rather than extending a 0.0035 triangle without modification. The neutral axes are x_ux = 282.771 mm and x_uy = 245.339 mm; the corresponding capacities are M_ux1 = 184.721 kN·m and M_uy1 = 153.668 kN·m.

6. Biaxial interaction. P_uz = 0.45f_ck(A_g − A_sc) + 0.75f_yA_sc = 3037.726 kN. Thus P_u/P_uz = 0.49379, and the code interpolation gives α_n = 1.48965.

(125.000/184.721)^1.48965 + (88.000/153.668)^1.48965 = 0.9948

Question-specific trial longitudinal arrangement. The schematic outer hoop is not a complete cross-tie detail; use the transverse requirements in the answer.

Q10. Design the longitudinal reinforcement for a short column 400 mm × 500 mm subjected to P_u = 1600 kN, M_ux = 20.0 kN-m and M_uy = 150 kN-m. Use M25 concrete and Fe415 steel. [15] (2073 Shrawan, Q3a) Scan annotation: an added handwritten note gives unsupported length 3 m; distinguish this from the printed statement.

1. Design basis. Use the scan's handwritten unsupported-length annotation of 3 m as an explicit study assumption. Retain the printed 20.0 kN-m moment. Use b = 400 mm, D = 500 mm, M25 concrete and Fe415 steel. The factored axial load is 1600 kN. For this calculation x-bending acts over depth D and y-bending over breadth b. Use all four faces for the longitudinal reinforcement, as specified or explicitly assumed.

2. Eccentricity. With unsupported length 3 m, the ordinary minimum eccentricities are max(l/500 + D/30, 20) = 22.67 mm and max(l/500 + b/30, 20) = 20.00 mm. This study conservatively checks the moment envelope with at least the corresponding minimum moment in both directions.

3. Slenderness and design moments. Effective lengths are 3 m about x and 3 m about y, giving ratios 6.000 and 7.500. Where the ratio reaches the slender range, use unreduced additional moments P_uD(l_ex/D)²/2000 and P_ub(l_ey/b)²/2000: 0.000 and 0.000 kN·m. Using the larger relevant end moment conservatively gives design moments 36.267 and 150.000 kN·m. Do not add these effects again if a second-order analysis already includes them.

4. Trial reinforcement. Consider 4 bars of 25 mm diameter: A_sc = 1963.50 mm², or 0.982% of gross area. Effective cover to the bar centres is 60.5 mm, from 40 mm nominal cover, the selected ties and half the main-bar diameter. The candidate passes the adopted clear bar-spacing check.

5. Uniaxial capacities at the same axial load. Solve section equilibrium in each direction using a linear strain diagram, the parabolic-rectangular concrete law and the idealized steel design curve. Subtract the concrete displaced by compression bars. For a neutral axis outside the section, use the compatible all-compression strain limit rather than extending a 0.0035 triangle without modification. The neutral axes are x_ux = 379.899 mm and x_uy = 300.227 mm; the corresponding capacities are M_ux1 = 210.295 kN·m and M_uy1 = 161.044 kN·m.

6. Biaxial interaction. P_uz = 0.45f_ck(A_g − A_sc) + 0.75f_yA_sc = 2839.049 kN. Thus P_u/P_uz = 0.56357, and the code interpolation gives α_n = 1.60595.

(36.267/210.295)^1.60595 + (150.000/161.044)^1.60595 = 0.9516

Question-specific trial longitudinal arrangement. The schematic outer hoop is not a complete cross-tie detail; use the transverse requirements in the answer.

Q11. Design the longitudinal and transverse reinforcement for a short column of size 35 cm × 45 cm subjected to a factored axial load of 1800 kN, M_ux = 175 kN·m and M_uy = 105 kN·m. Reinforcement is distributed equally on two sides. Use M25 concrete and Fe500 steel; unsupported length is 3.1 m. [15] (2072 Chaitra, Q3a)

1. Design basis. Use the specified two-face reinforcement arrangement and the source-confirmed 350 by 450 mm size. Assume effective length equal to unsupported length. Use b = 350 mm, D = 450 mm, M25 concrete and Fe500 steel. The factored axial load is 1800 kN. For this calculation x-bending acts over depth D and y-bending over breadth b. Use two opposite faces for the longitudinal reinforcement, as specified or explicitly assumed.

2. Eccentricity. With unsupported length 3.1 m, the ordinary minimum eccentricities are max(l/500 + D/30, 20) = 21.20 mm and max(l/500 + b/30, 20) = 20.00 mm. This study conservatively checks the moment envelope with at least the corresponding minimum moment in both directions.

3. Slenderness and design moments. Effective lengths are 3.1 m about x and 3.1 m about y, giving ratios 6.889 and 8.857. Where the ratio reaches the slender range, use unreduced additional moments P_uD(l_ex/D)²/2000 and P_ub(l_ey/b)²/2000: 0.000 and 0.000 kN·m. Using the larger relevant end moment conservatively gives design moments 175.000 and 105.000 kN·m. Do not add these effects again if a second-order analysis already includes them.

4. Trial reinforcement. Consider 6 bars of 32 mm diameter: A_sc = 4825.49 mm², or 3.064% of gross area. Effective cover to the bar centres is 64 mm, from 40 mm nominal cover, the selected ties and half the main-bar diameter. The candidate passes the adopted clear bar-spacing check.

5. Uniaxial capacities at the same axial load. Solve section equilibrium in each direction using a linear strain diagram, the parabolic-rectangular concrete law and the idealized steel design curve. Subtract the concrete displaced by compression bars. For a neutral axis outside the section, use the compatible all-compression strain limit rather than extending a 0.0035 triangle without modification. The neutral axes are x_ux = 337.626 mm and x_uy = 248.979 mm; the corresponding capacities are M_ux1 = 286.186 kN·m and M_uy1 = 161.349 kN·m.

6. Biaxial interaction. P_uz = 0.45f_ck(A_g − A_sc) + 0.75f_yA_sc = 3527.146 kN. Thus P_u/P_uz = 0.51033, and the code interpolation gives α_n = 1.51721.

(175.000/286.186)^1.51721 + (105.000/161.349)^1.51721 = 0.9952

Question-specific trial longitudinal arrangement. The schematic outer hoop is not a complete cross-tie detail; use the transverse requirements in the answer.

Q12. Determine the longitudinal and transverse reinforcement in a short rectangular column subjected to a factored axial load of 2000 kN and factored bending moments M_ux = 190 kN·m about the major axis and M_uy = 95 kN·m about the minor axis. The column has a cross-section of 300 mm × 500 mm and an unsupported length of 3 m. Adopt M30 concrete and Fe500 steel. [16] (2072 Kartik, Q3b)

1. Design basis. Assume non-sway effective lengths equal to the unsupported length. Use b = 300 mm, D = 500 mm, M30 concrete and Fe500 steel. The factored axial load is 2000 kN. For this calculation x-bending acts over depth D and y-bending over breadth b. Use all four faces for the longitudinal reinforcement, as specified or explicitly assumed.

3. Slenderness and design moments. Effective lengths are 3 m about x and 3 m about y, giving ratios 6.000 and 10.000. Where the ratio reaches the slender range, use unreduced additional moments P_uD(l_ex/D)²/2000 and P_ub(l_ey/b)²/2000: 0.000 and 0.000 kN·m. Using the larger relevant end moment conservatively gives design moments 190.000 and 95.000 kN·m. Do not add these effects again if a second-order analysis already includes them.

4. Trial reinforcement. Consider 16 bars of 20 mm diameter: A_sc = 5026.55 mm², or 3.351% of gross area. Effective cover to the bar centres is 58 mm, from 40 mm nominal cover, the selected ties and half the main-bar diameter. The candidate passes the adopted clear bar-spacing check.

5. Uniaxial capacities at the same axial load. Solve section equilibrium in each direction using a linear strain diagram, the parabolic-rectangular concrete law and the idealized steel design curve. Subtract the concrete displaced by compression bars. For a neutral axis outside the section, use the compatible all-compression strain limit rather than extending a 0.0035 triangle without modification. The neutral axes are x_ux = 368.203 mm and x_uy = 213.821 mm; the corresponding capacities are M_ux1 = 299.946 kN·m and M_uy1 = 148.757 kN·m.

6. Biaxial interaction. P_uz = 0.45f_ck(A_g − A_sc) + 0.75f_yA_sc = 3842.097 kN. Thus P_u/P_uz = 0.52055, and the code interpolation gives α_n = 1.53425.

(190.000/299.946)^1.53425 + (95.000/148.757)^1.53425 = 0.9989

Question-specific trial longitudinal arrangement. The schematic outer hoop is not a complete cross-tie detail; use the transverse requirements in the answer.

Q13. Design a short rectangular column of size 450 mm × 300 mm and unsupported length 3 m subjected to an ultimate axial load of 1500 kN and ultimate moments 150 kNm and 80 kNm about major and minor axes. Adopt M30 concrete and Fe500 steel. Sketch the final design. [14] (2071 Chaitra, Q1c)

1. Design basis. Take the 450 mm dimension as the major-axis depth and assume effective length equal to the stated unsupported length. Use b = 300 mm, D = 450 mm, M30 concrete and Fe500 steel. The factored axial load is 1500 kN. For this calculation x-bending acts over depth D and y-bending over breadth b. Use all four faces for the longitudinal reinforcement, as specified or explicitly assumed.

3. Slenderness and design moments. Effective lengths are 3 m about x and 3 m about y, giving ratios 6.667 and 10.000. Where the ratio reaches the slender range, use unreduced additional moments P_uD(l_ex/D)²/2000 and P_ub(l_ey/b)²/2000: 0.000 and 0.000 kN·m. Using the larger relevant end moment conservatively gives design moments 150.000 and 80.000 kN·m. Do not add these effects again if a second-order analysis already includes them.

4. Trial reinforcement. Consider 8 bars of 25 mm diameter: A_sc = 3926.99 mm², or 2.909% of gross area. Effective cover to the bar centres is 60.5 mm, from 40 mm nominal cover, the selected ties and half the main-bar diameter. The candidate passes the adopted clear bar-spacing check.

5. Uniaxial capacities at the same axial load. Solve section equilibrium in each direction using a linear strain diagram, the parabolic-rectangular concrete law and the idealized steel design curve. Subtract the concrete displaced by compression bars. For a neutral axis outside the section, use the compatible all-compression strain limit rather than extending a 0.0035 triangle without modification. The neutral axes are x_ux = 308.577 mm and x_uy = 199.823 mm; the corresponding capacities are M_ux1 = 238.593 kN·m and M_uy1 = 134.325 kN·m.

6. Biaxial interaction. P_uz = 0.45f_ck(A_g − A_sc) + 0.75f_yA_sc = 3242.107 kN. Thus P_u/P_uz = 0.46266, and the code interpolation gives α_n = 1.43777.

(150.000/238.593)^1.43777 + (80.000/134.325)^1.43777 = 0.9878

Question-specific trial longitudinal arrangement. The schematic outer hoop is not a complete cross-tie detail; use the transverse requirements in the answer.

Q14. A column of 4 m length with both ends fixed and effectively held is subjected to a design axial load of 1000 kN and factored bending moment of 100 kN-m. Design the rectangular column with its longitudinal and transverse reinforcement. [14] (2069 Chaitra, Q1b)

1. Design basis. The section and grades are to be selected: adopt 400 by 500 mm, M25 and Fe415. Both ends are fixed and held, so use the IS idealized factor 0.65. Use b = 400 mm, D = 500 mm, M25 concrete and Fe415 steel. The factored axial load is 1000 kN. For this calculation x-bending acts over depth D and y-bending over breadth b. Use all four faces for the longitudinal reinforcement, as specified or explicitly assumed.

3. Slenderness and design moments. Effective lengths are 2.6 m about x and 2.6 m about y, giving ratios 5.200 and 6.500. Where the ratio reaches the slender range, use unreduced additional moments P_uD(l_ex/D)²/2000 and P_ub(l_ey/b)²/2000: 0.000 and 0.000 kN·m. Using the larger relevant end moment conservatively gives design moments 100.000 and 21.333 kN·m. Do not add these effects again if a second-order analysis already includes them.

5. Uniaxial capacities at the same axial load. Solve section equilibrium in each direction using a linear strain diagram, the parabolic-rectangular concrete law and the idealized steel design curve. Subtract the concrete displaced by compression bars. For a neutral axis outside the section, use the compatible all-compression strain limit rather than extending a 0.0035 triangle without modification. The neutral axes are x_ux = 270.613 mm and x_uy = 216.279 mm; the corresponding capacities are M_ux1 = 213.577 kN·m and M_uy1 = 165.313 kN·m.

6. Biaxial interaction. P_uz = 0.45f_ck(A_g − A_sc) + 0.75f_yA_sc = 2732.549 kN. Thus P_u/P_uz = 0.36596, and the code interpolation gives α_n = 1.27660.

(100.000/213.577)^1.27660 + (21.333/165.313)^1.27660 = 0.4528

Question-specific trial longitudinal arrangement. The schematic outer hoop is not a complete cross-tie detail; use the transverse requirements in the answer.

Q15. Design a square RC column that has to carry an ultimate factored load of 800 kN inclusive of live load at an eccentricity of 80 mm in both X and Y directions. Use M20 concrete and Fe415 steel. [8] (2067 Ashadh, Q5c)

1. Design basis. Select a 400 mm square section. The 80 mm eccentricity gives 800 times 0.080 = 64 kN-m in each direction. Assume 3 m unsupported/effective length; verify the actual restraints before design use. Use b = 400 mm, D = 400 mm, M20 concrete and Fe415 steel. The factored axial load is 800 kN. For this calculation x-bending acts over depth D and y-bending over breadth b. Use all four faces for the longitudinal reinforcement, as specified or explicitly assumed.

3. Slenderness and design moments. Effective lengths are 3 m about x and 3 m about y, giving ratios 7.500 and 7.500. Where the ratio reaches the slender range, use unreduced additional moments P_uD(l_ex/D)²/2000 and P_ub(l_ey/b)²/2000: 0.000 and 0.000 kN·m. Using the larger relevant end moment conservatively gives design moments 64.000 and 64.000 kN·m. Do not add these effects again if a second-order analysis already includes them.

4. Trial reinforcement. Consider 4 bars of 22 mm diameter: A_sc = 1520.53 mm², or 0.950% of gross area. Effective cover to the bar centres is 59 mm, from 40 mm nominal cover, the selected ties and half the main-bar diameter. The candidate passes the adopted clear bar-spacing check.

5. Uniaxial capacities at the same axial load. Solve section equilibrium in each direction using a linear strain diagram, the parabolic-rectangular concrete law and the idealized steel design curve. Subtract the concrete displaced by compression bars. For a neutral axis outside the section, use the compatible all-compression strain limit rather than extending a 0.0035 triangle without modification. The neutral axes are x_ux = 252.140 mm and x_uy = 252.140 mm; the corresponding capacities are M_ux1 = 132.276 kN·m and M_uy1 = 132.276 kN·m.

6. Biaxial interaction. P_uz = 0.45f_ck(A_g − A_sc) + 0.75f_yA_sc = 1899.580 kN. Thus P_u/P_uz = 0.42115, and the code interpolation gives α_n = 1.36858.

(64.000/132.276)^1.36858 + (64.000/132.276)^1.36858 = 0.7405

Question-specific trial longitudinal arrangement. The schematic outer hoop is not a complete cross-tie detail; use the transverse requirements in the answer.

Q16. Design a column member to transfer a factored load P_u = 1350 kN, M_ux = 150 kN·m and M_uy = 100 kN·m, with an unsupported length of 3.4 m. Use M25 concrete and Fe500 steel. Distribute reinforcement equally on all four sides and assume that the column is fully restrained at both supports against displacement and rotation. [12] (2082 Baishakh, Q4b)

1. Design basis. Select a 500 mm square trial because the section size is not prescribed. Fully held and rotationally restrained ends give the idealized factor 0.65. Use b = 500 mm, D = 500 mm, M25 concrete and Fe500 steel. The factored axial load is 1350 kN. For this calculation x-bending acts over depth D and y-bending over breadth b. Use all four faces for the longitudinal reinforcement, as specified or explicitly assumed.

2. Eccentricity. With unsupported length 3.4 m, the ordinary minimum eccentricities are max(l/500 + D/30, 20) = 23.47 mm and max(l/500 + b/30, 20) = 23.47 mm. This study conservatively checks the moment envelope with at least the corresponding minimum moment in both directions.

3. Slenderness and design moments. Effective lengths are 2.21 m about x and 2.21 m about y, giving ratios 4.420 and 4.420. Where the ratio reaches the slender range, use unreduced additional moments P_uD(l_ex/D)²/2000 and P_ub(l_ey/b)²/2000: 0.000 and 0.000 kN·m. Using the larger relevant end moment conservatively gives design moments 150.000 and 100.000 kN·m. Do not add these effects again if a second-order analysis already includes them.

4. Trial reinforcement. Consider 12 bars of 16 mm diameter: A_sc = 2412.74 mm², or 0.965% of gross area. Effective cover to the bar centres is 56 mm, from 40 mm nominal cover, the selected ties and half the main-bar diameter. The candidate passes the adopted clear bar-spacing check.

5. Uniaxial capacities at the same axial load. Solve section equilibrium in each direction using a linear strain diagram, the parabolic-rectangular concrete law and the idealized steel design curve. Subtract the concrete displaced by compression bars. For a neutral axis outside the section, use the compatible all-compression strain limit rather than extending a 0.0035 triangle without modification. The neutral axes are x_ux = 281.157 mm and x_uy = 281.157 mm; the corresponding capacities are M_ux1 = 298.418 kN·m and M_uy1 = 298.418 kN·m.

6. Biaxial interaction. P_uz = 0.45f_ck(A_g − A_sc) + 0.75f_yA_sc = 3690.135 kN. Thus P_u/P_uz = 0.36584, and the code interpolation gives α_n = 1.27640.

(150.000/298.418)^1.27640 + (100.000/298.418)^1.27640 = 0.6633

Question-specific trial longitudinal arrangement. The schematic outer hoop is not a complete cross-tie detail; use the transverse requirements in the answer.

Design method (SP-16 biaxial)

First distinguish two axial expressions. For an eligible short tied column with sufficiently small eccentricity, the axial design expression is P_u = 0.4f_ckA_c + 0.67f_yA_sc. Its use is subject to the minimum-eccentricity restrictions. The separate quantity P_uz = 0.45f_ckA_c + 0.75f_yA_sc appears in the biaxial interaction procedure. Substituting P_uz as an unrestricted axial design capacity overestimates the permitted short-column resistance.

Squash load: P_uz = 0.45f_ckA_c + 0.75f_yA_sc (assume a steel ratio p = 0.8–4%, A_sc = p·bD100).
Uniaxial capacities M_ux1, M_uy1 for the given P_uf_ckbD and d′D from the SP-16 interaction charts.
Exponent α_n = 1.0 at P_uP_uz = 0.2 → 2.0 at 0.8 (linear between).
Biaxial check: (M_uxM_ux1)^α_n + (M_uyM_uy1)^α_n ≤ 1.0 — if not, increase steel or size.
Provide lateral ties: diameter at least one-quarter of the largest longitudinal bar and at least 6 mm; ordinary spacing not greater than the least of 16 times the smallest longitudinal-bar diameter, the least lateral dimension and 300 mm. Seismic confinement is a separate, often stricter check.

Reading SP 16 Correctly

Choose a chart for the correct concrete/steel grades, reinforcement arrangement and d′/D ratio. A chart for steel on two faces is not a chart for bars distributed on four faces. Determine the normalized axial load P_u/(f_ckbD), the steel parameter required by that chart, and the normalized moment M/(f_ckbD²). Interpolate between the actual plotted curves and record the values read.

For bending about the other axis, interchange the section dimensions and recompute the effective-cover ratio and the moment normalization. Both uniaxial capacities in the biaxial expression must correspond to the same applied axial load, not the two independent maximum moments on their curves.

The interaction exponent is α_n = 1 for P_u/P_uz ≤ 0.2 and 2 for ratios ≥ 0.8. Between them, α_n = 1 + [(P_u/P_uz) − 0.2]/0.6. Recalculate it after changing the steel area; do not retain the trial percentage after selecting actual bars.

8.4 Design of Long / Slender Columns 20822081208020792065

PAST EXAM QUESTIONS:

Design a column of size 650 mm × 450 mm subjected to a factored axial load of 1600 kN. At the top end, the factored bending moments are 90 kN·m about the major axis and 55 kN·m about the minor axis. At the bottom end, the corresponding moments are 35 kN·m and 30 kN·m, respectively. The effective lengths about the major and minor axes are 7 m and 6.5 m, respectively, and the column bends in double curvature. Use M30 concrete and Fe500 TMT steel. Prepare the construction drawings. [12+2] (2082 Bhadra, Q4a)
Design a slender reinforced-concrete column of cross-section 400 mm × 500 mm. Its effective lengths are L_ex = 7.0 m and L_ey = 6.0 m, and it carries a factored axial load of 1600 kN. The factored bending moments at the top are M_ux = 180 kN·m and M_uy = 110 kN·m. At the bottom, the moments are M_ux = 100 kN·m and M_uy = 50 kN·m. The column bends in single curvature. Use M20 concrete and Fe415 steel. [12] (2081 Bhadra, Q4b)
Design a braced rectangular RC column of clear height 4.0 m, cross-section 450 mm × 350 mm, subjected to a design axial load of 700 kN and design bending moments of 100 kN-m (major) and 80 kN-m (minor). The column is held effectively at both ends and restrained against rotation at one end. Use M25 concrete and Fe415 steel. [12] (2080 Bhadra, Q3b)
Design an unbraced rectangular RC column of clear height 6.0 m, cross-section 400 mm × 350 mm, subjected to a design axial load of 600 kN and design bending moments of 100 kNm (major) and 50 kNm (minor). Consider M20 concrete and Fe415 steel. [14] (2079 Bhadra, Q3b)
Write down the design steps for the design of a RC long column of an unbraced frame. [10] (2065 Kartik, Q5b)

Q1. Design a column of size 650 mm × 450 mm subjected to a factored axial load of 1600 kN. At the top end, the factored bending moments are 90 kN·m about the major axis and 55 kN·m about the minor axis. At the bottom end, the corresponding moments are 35 kN·m and 30 kN·m, respectively. The effective lengths about the major and minor axes are 7 m and 6.5 m, respectively, and the column bends in double curvature. Use M30 concrete and Fe500 TMT steel. Prepare the construction drawings. [12+2] (2082 Bhadra, Q4a)

1. Design basis. Use the supplied effective lengths. A conservative design retains the larger absolute end moments and unreduced additional moments rather than assuming a favourable double-curvature reduction without the full frame information. Use b = 450 mm, D = 650 mm, M30 concrete and Fe500 steel. The factored axial load is 1600 kN. For this calculation x-bending acts over depth D and y-bending over breadth b. Use all four faces for the longitudinal reinforcement, as specified or explicitly assumed.

2. Missing restraint information. Unsupported height is not supplied. The calculation below includes the absolute 20 mm eccentricity floor, but l/500 plus the dimensional term must be checked when the actual height is established. This missing information prevents an unconditional complete-member approval.

3. Slenderness and design moments. Effective lengths are 7 m about x and 6.5 m about y, giving ratios 10.769 and 14.444. Where the ratio reaches the slender range, use unreduced additional moments P_uD(l_ex/D)²/2000 and P_ub(l_ey/b)²/2000: 0.000 and 75.111 kN·m. Using the larger relevant end moment conservatively gives design moments 90.000 and 130.111 kN·m. Do not add these effects again if a second-order analysis already includes them.

4. Trial reinforcement. Consider 12 bars of 16 mm diameter: A_sc = 2412.74 mm², or 0.825% of gross area. Effective cover to the bar centres is 56 mm, from 40 mm nominal cover, the selected ties and half the main-bar diameter. The candidate passes the adopted clear bar-spacing check.

5. Uniaxial capacities at the same axial load. Solve section equilibrium in each direction using a linear strain diagram, the parabolic-rectangular concrete law and the idealized steel design curve. Subtract the concrete displaced by compression bars. For a neutral axis outside the section, use the compatible all-compression strain limit rather than extending a 0.0035 triangle without modification. The neutral axes are x_ux = 329.318 mm and x_uy = 227.881 mm; the corresponding capacities are M_ux1 = 492.317 kN·m and M_uy1 = 325.067 kN·m.

6. Biaxial interaction. P_uz = 0.45f_ck(A_g − A_sc) + 0.75f_yA_sc = 4820.957 kN. Thus P_u/P_uz = 0.33188, and the code interpolation gives α_n = 1.21981.

(90.000/492.317)^1.21981 + (130.111/325.067)^1.21981 = 0.4531

Question-specific trial longitudinal arrangement. The schematic outer hoop is not a complete cross-tie detail; use the transverse requirements in the answer.

Q2. Design a slender reinforced-concrete column of cross-section 400 mm × 500 mm. Its effective lengths are L_ex = 7.0 m and L_ey = 6.0 m, and it carries a factored axial load of 1600 kN. The factored bending moments at the top are M_ux = 180 kN·m and M_uy = 110 kN·m. At the bottom, the moments are M_ux = 100 kN·m and M_uy = 50 kN·m. The column bends in single curvature. Use M20 concrete and Fe415 steel. [12] (2081 Bhadra, Q4b)

1. Design basis. Use the given effective lengths. Retain the larger single-curvature end moments and unreduced additional moments, checking the end sections as well. Use b = 400 mm, D = 500 mm, M20 concrete and Fe415 steel. The factored axial load is 1600 kN. For this calculation x-bending acts over depth D and y-bending over breadth b. Use all four faces for the longitudinal reinforcement, as specified or explicitly assumed.

3. Slenderness and design moments. Effective lengths are 7 m about x and 6 m about y, giving ratios 14.000 and 15.000. Where the ratio reaches the slender range, use unreduced additional moments P_uD(l_ex/D)²/2000 and P_ub(l_ey/b)²/2000: 78.400 and 72.000 kN·m. Using the larger relevant end moment conservatively gives design moments 258.400 and 182.000 kN·m. Do not add these effects again if a second-order analysis already includes them.

4. Trial reinforcement. Consider 24 bars of 22 mm diameter: A_sc = 9123.19 mm², or 4.562% of gross area. Effective cover to the bar centres is 59 mm, from 40 mm nominal cover, the selected ties and half the main-bar diameter. The candidate passes the adopted clear bar-spacing check.

5. Uniaxial capacities at the same axial load. Solve section equilibrium in each direction using a linear strain diagram, the parabolic-rectangular concrete law and the idealized steel design curve. Subtract the concrete displaced by compression bars. For a neutral axis outside the section, use the compatible all-compression strain limit rather than extending a 0.0035 triangle without modification. The neutral axes are x_ux = 318.568 mm and x_uy = 250.473 mm; the corresponding capacities are M_ux1 = 476.642 kN·m and M_uy1 = 350.108 kN·m.

6. Biaxial interaction. P_uz = 0.45f_ck(A_g − A_sc) + 0.75f_yA_sc = 4557.483 kN. Thus P_u/P_uz = 0.35107, and the code interpolation gives α_n = 1.25178.

(258.400/476.642)^1.25178 + (182.000/350.108)^1.25178 = 0.9056

Congestion warning: this trial exceeds 4% longitudinal steel. The ordinary numerical upper limit does not ensure buildability, particularly at lap zones. Prefer a larger section or a properly qualified coupler/arrangement and recheck clear spacing and confinement.

Question-specific trial longitudinal arrangement. The schematic outer hoop is not a complete cross-tie detail; use the transverse requirements in the answer.

Q3. Design a braced rectangular RC column of clear height 4.0 m, cross-section 450 mm × 350 mm, subjected to a design axial load of 700 kN and design bending moments of 100 kN-m (major) and 80 kN-m (minor). The column is held effectively at both ends and restrained against rotation at one end. Use M25 concrete and Fe415 steel. [12] (2080 Bhadra, Q3b)

1. Design basis. The stated braced restraint corresponds to effective-length factor 0.8. Despite its location in the long-column question group, classify this actual member using both calculated ratios. Use b = 350 mm, D = 450 mm, M25 concrete and Fe415 steel. The factored axial load is 700 kN. For this calculation x-bending acts over depth D and y-bending over breadth b. Use all four faces for the longitudinal reinforcement, as specified or explicitly assumed.

2. Eccentricity. With unsupported length 4 m, the ordinary minimum eccentricities are max(l/500 + D/30, 20) = 23.00 mm and max(l/500 + b/30, 20) = 20.00 mm. This study conservatively checks the moment envelope with at least the corresponding minimum moment in both directions.

3. Slenderness and design moments. Effective lengths are 3.2 m about x and 3.2 m about y, giving ratios 7.111 and 9.143. Where the ratio reaches the slender range, use unreduced additional moments P_uD(l_ex/D)²/2000 and P_ub(l_ey/b)²/2000: 0.000 and 0.000 kN·m. Using the larger relevant end moment conservatively gives design moments 100.000 and 80.000 kN·m. Do not add these effects again if a second-order analysis already includes them.

4. Trial reinforcement. Consider 4 bars of 25 mm diameter: A_sc = 1963.50 mm², or 1.247% of gross area. Effective cover to the bar centres is 60.5 mm, from 40 mm nominal cover, the selected ties and half the main-bar diameter. The candidate passes the adopted clear bar-spacing check.

5. Uniaxial capacities at the same axial load. Solve section equilibrium in each direction using a linear strain diagram, the parabolic-rectangular concrete law and the idealized steel design curve. Subtract the concrete displaced by compression bars. For a neutral axis outside the section, use the compatible all-compression strain limit rather than extending a 0.0035 triangle without modification. The neutral axes are x_ux = 225.034 mm and x_uy = 175.024 mm; the corresponding capacities are M_ux1 = 203.608 kN·m and M_uy1 = 147.474 kN·m.

6. Biaxial interaction. P_uz = 0.45f_ck(A_g − A_sc) + 0.75f_yA_sc = 2360.924 kN. Thus P_u/P_uz = 0.29649, and the code interpolation gives α_n = 1.16082.

(100.000/203.608)^1.16082 + (80.000/147.474)^1.16082 = 0.9297

Question-specific trial longitudinal arrangement. The schematic outer hoop is not a complete cross-tie detail; use the transverse requirements in the answer.

Q4. Design an unbraced rectangular RC column of clear height 6.0 m, cross-section 400 mm × 350 mm, subjected to a design axial load of 600 kN and design bending moments of 100 kNm (major) and 50 kNm (minor). Consider M20 concrete and Fe415 steel. [14] (2079 Bhadra, Q3b)

1. Design basis. The unbraced frame stiffness is not supplied. Use an explicitly assumed effective-length factor of 1.5 in each direction and unreduced additional moments for this illustrative calculation. A real sway analysis or justified effective-length factors are needed before accepting the design. Use b = 350 mm, D = 400 mm, M20 concrete and Fe415 steel. The factored axial load is 600 kN. For this calculation x-bending acts over depth D and y-bending over breadth b. Use all four faces for the longitudinal reinforcement, as specified or explicitly assumed.

2. Eccentricity. With unsupported length 6 m, the ordinary minimum eccentricities are max(l/500 + D/30, 20) = 25.33 mm and max(l/500 + b/30, 20) = 23.67 mm. This study conservatively checks the moment envelope with at least the corresponding minimum moment in both directions.

3. Slenderness and design moments. Effective lengths are 9 m about x and 9 m about y, giving ratios 22.500 and 25.714. Where the ratio reaches the slender range, use unreduced additional moments P_uD(l_ex/D)²/2000 and P_ub(l_ey/b)²/2000: 60.750 and 69.429 kN·m. Using the larger relevant end moment conservatively gives design moments 160.750 and 119.429 kN·m. Do not add these effects again if a second-order analysis already includes them.

4. Trial reinforcement. Consider 16 bars of 25 mm diameter: A_sc = 7853.98 mm², or 5.610% of gross area. Effective cover to the bar centres is 60.5 mm, from 40 mm nominal cover, the selected ties and half the main-bar diameter. The candidate passes the adopted clear bar-spacing check.

5. Uniaxial capacities at the same axial load. Solve section equilibrium in each direction using a linear strain diagram, the parabolic-rectangular concrete law and the idealized steel design curve. Subtract the concrete displaced by compression bars. For a neutral axis outside the section, use the compatible all-compression strain limit rather than extending a 0.0035 triangle without modification. The neutral axes are x_ux = 208.217 mm and x_uy = 182.071 mm; the corresponding capacities are M_ux1 = 321.050 kN·m and M_uy1 = 262.364 kN·m.

6. Biaxial interaction. P_uz = 0.45f_ck(A_g − A_sc) + 0.75f_yA_sc = 3633.866 kN. Thus P_u/P_uz = 0.16511, and the code interpolation gives α_n = 1.00000.

(160.750/321.050)^1.00000 + (119.429/262.364)^1.00000 = 0.9559

Question-specific trial longitudinal arrangement. The schematic outer hoop is not a complete cross-tie detail; use the transverse requirements in the answer.

Q5. Write down the design steps for the design of a RC long column of an unbraced frame. [10] (2065 Kartik, Q5b)

A slender column develops additional bending because axial load acts through its lateral displacement. Global frame sway produces a P–Δ effect, while curvature within a member produces a P–δ effect. The short-column classification is not restricted to braced frames; bracing affects effective length and analysis. A slender member may reach material failure after second-order amplification, so "all slender columns fail by Euler buckling" is too crude.

Where the IS 456 simplified procedure is applicable and second-order effects have not already been included, the unreduced additional moments are:

M_ax = P_u·D2000(l_exD)² (× a reduction factor k = P_uz − P_uP_uz − P_bal ≤ 1)

M_ay = (P_ub/2000)(l_ey/b)²

Take P_bal from the appropriate balanced condition, and apply a reduction only where its code conditions are satisfied. Taking k = 1 is an unreduced conservative study choice. The reduction factor k is not the effective-length factor used earlier. Do not add the simplified moments again to an analysis that already includes the same second-order effects.

Record end moments with a consistent sign convention. In single curvature the bending diagram has the same curvature sense along the member; double curvature includes a point of contraflexure. For the applicable braced-column procedure, an equivalent initial moment uses 0.6M₂ + 0.4M₁, not less than 0.4M₂, with M₂ the larger end moment and the code's relative sign for M₁. Check the end sections as well as the amplified interior section. Never use this interpolation as a universal sway-frame rule.

═══ 8.5 ═══

═══ 8.4 ═══

Complete design sequence: establish sway restraints and stiffness, obtain compatible factored actions, select a trial section and bar arrangement, check both slenderness ratios and minimum eccentricity, include second-order effects once, and test biaxial interaction. Iterate the size/reinforcement as necessary. Finish with shear, tie/confinement design, permitted lap locations, joint anchorage and construction drawings. A braced-column end-moment interpolation cannot be used automatically for an unbraced frame.

Where the IS 456 simplified procedure is applicable and second-order effects have not already been included, the unreduced additional moments are:

M_ax = P_u·D2000(l_exD)² (× a reduction factor k = P_uz − P_uP_uz − P_bal ≤ 1)

M_ay = (P_ub/2000)(l_ey/b)²

Single and Double Curvature

8.5 Reinforcement Detailing / Capacity Check 2065

PAST EXAM QUESTIONS:

A reinforced-concrete column of cross-section 600 mm × 400 mm is reinforced with 8 longitudinal bars of 28 mm diameter. It is subjected to an axial load P_u = 2000 kN and bending moments M_ux = 350 kN·m and M_uy = 50 kN·m. Check whether the column is safe for this combination of actions. Use M20 concrete and Fe415 steel, and take the effective cover as 60 mm. [10] (2065 Kartik, Q1b)

Q1. A reinforced-concrete column of cross-section 600 mm × 400 mm is reinforced with 8 longitudinal bars of 28 mm diameter. It is subjected to an axial load P_u = 2000 kN and bending moments M_ux = 350 kN·m and M_uy = 50 kN·m. Check whether the column is safe for this combination of actions. Use M20 concrete and Fe415 steel, and take the effective cover as 60 mm. [10] (2065 Kartik, Q1b)

1. Design basis. Check the supplied eight 28 mm bars, assumed distributed at corners and face midpoints. Use the given 60 mm effective cover. Restraint data are absent, so this is not a full slenderness check. Use b = 400 mm, D = 600 mm, M20 concrete and Fe415 steel. The factored axial load is 2000 kN. For this calculation x-bending acts over depth D and y-bending over breadth b. Use all four faces for the longitudinal reinforcement, as specified or explicitly assumed.

3. Scope of this check. Effective lengths are not available, so the following is a section-capacity check using moments 350.000 and 50.000 kN·m. Establish whether the actual member is short or slender before adopting the result for the column.

4. Provided reinforcement. Check 8 bars of 28 mm diameter: A_sc = 4926.02 mm², or 2.053% of gross area. Effective cover to the bar centres is 60 mm as supplied. The candidate passes the adopted clear bar-spacing check.

5. Uniaxial capacities at the same axial load. Solve section equilibrium in each direction using a linear strain diagram, the parabolic-rectangular concrete law and the idealized steel design curve. Subtract the concrete displaced by compression bars. For a neutral axis outside the section, use the compatible all-compression strain limit rather than extending a 0.0035 triangle without modification. The neutral axes are x_ux = 455.618 mm and x_uy = 296.427 mm; the corresponding capacities are M_ux1 = 356.045 kN·m and M_uy1 = 213.913 kN·m.

6. Biaxial interaction. P_uz = 0.45f_ck(A_g − A_sc) + 0.75f_yA_sc = 3648.889 kN. Thus P_u/P_uz = 0.54811, and the code interpolation gives α_n = 1.58019.

(350.000/356.045)^1.58019 + (50.000/213.913)^1.58019 = 1.0739

The checked arrangement is not adequate: the interaction sum exceeds 1.0. The supplied reinforcement therefore fails this study section check. Revisit section size, arrangement and the justified analysis assumptions; do not reduce the given loads or quote an invented safe chart capacity.

Question-specific trial longitudinal arrangement. The schematic outer hoop is not a complete cross-tie detail; use the transverse requirements in the answer.

To check an existing column: compute P_uz from the provided steel, read M_ux1 and M_uy1 from SP-16 for the actual P_uf_ckbD, find α_n from P_uP_uz, then verify (M_uxM_ux1)^α_n + (M_uyM_uy1)^α_n ≤ 1.0. Detailing: longitudinal steel 0.8–6% of the gross area (≥ 4 bars, ≥ 12 mm), lateral ties as in 8.3, and for earthquake resistance special confining hoops over a length l₀ at each end (see Ch 11).

Construction and Detailing Requirements

Use at least four longitudinal bars in a rectangular column and six in a circular column; longitudinal bars are at least 12 mm diameter.
The ordinary gross-area limits are 0.8% minimum and 6% maximum steel. Congestion and lap zones often justify a substantially lower practical maximum; the numerical maximum is not a target.
Provide the required nominal cover to the outside of the ties, ordinarily at least 40 mm for these column sizes and not less than the relevant bar-diameter requirement. Derive effective cover to the main-bar centre separately.
Restrain longitudinal bars against buckling with closed ties and cross-ties as required. A large outer rectangular tie alone may not restrain all intermediate bars.
Show splice locations, lap/coupler details, starter anchorage, normal tie zones and special confinement zones on the elevation and section.
When a question asks for ductile transverse detailing, perform the confinement and capacity-design checks in Chapter 11; passing the ordinary tie-spacing rule is insufficient.

Completion criterion: an assumed percentage of steel is only a trial. A completed answer records the actual bar area, cover ratios, the two uniaxial capacities at the specified axial load, the interaction exponent, the utilization, slenderness treatment and the transverse detail. If no feasible bar arrangement satisfies the fixed size, state that limitation instead of inventing a chart capacity.