Flanged Section Shear, Bond, and Torsion: GATE Civil Engineering Notes by Rajat Johari Sir

Flanged beams, including T-beams and L-beams, play a vital role in reinforced concrete structures by efficiently resisting compressive stresses. Here, GATE Civil Engineering Notes by Rajat Johari Sir explain the structural analysis of flanged beams in a clear and exam-focused manner. It begins with the concept and importance of the effective width of the flange (Bf) as per design codes.

The discussion then moves step by step through the analysis of T-beams, explaining how to locate the neutral axis and how it affects stress distribution. Finally, the blog covers the calculation of the moment of resistance, making it highly useful for GATE and other competitive exam preparation.

Flanged Beams: T-Beams and L-Beams

Flanged beams are structural elements featuring a flange (a wide, flat part) and a web (a narrow, vertical part). The two primary types of flanged beams are:

1. T-Beam: This beam has a symmetrical flange on both sides of the web.

2. L-Beam: This beam has a flange on only one side of the web.

The analysis and design of both T-Beams and L-Beams are critical.

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Effective Width of Flange (Bf)

The effective width of the flange (Bf) is the portion of the slab or flange assumed to be effective in resisting compressive stresses. For design and analysis, we consider a reduced width, not the total physical width of the flange.

Terminology for a T-Beam:

• B: Overall or total width of the flange.

• Bf: Effective width of the flange.

• Df: Depth of the flange.

• Bw: Width of the web.

• d: Effective depth of the beam (from the top compression fiber to the centroid of the tension reinforcement).

The flange is the horizontal part of the beam, and the web is the vertical part.

Check: GATE Civil Engineering Notes

1. For Isolated T-Beams and L-Beams

An isolated beam is one not cast monolithically with a connecting slab.

• Isolated T-Beam:
  The effective width of the flange is given by the formula:
  Bf = Bw + [ L₀ / ( (L₀ / B) + 4 ) ]

• Isolated L-Beam:
  An L-Beam can be thought of as half of a T-Beam, with a flange on only one side. The formula adjusts accordingly:
  Bf = Bw + [ 0.5 * L₀ / ( (L₀ / B) + 4 ) ]

Definition of L₀:

L₀ is defined as the distance between the points of contraflexure (points where the bending moment is zero). Determining L₀ for Different Support Conditions:

2. For Beams Cast Monolithically with a Slab (Continuous Beams)

When a beam and slab are cast together, the slab acts as the flange for the beam.

• Monolithic T-Beam:
  The effective width of the flange, Bf, is the minimum of the following two values:

1. Bf = Bw + L₁/2 + L₂/2 (where L₁ and L₂ are the clear distances to the adjacent parallel beams on either side).

2. Bf = Bw + L₀/6 + 6*Df

• Monolithic L-Beam:
  The effective width of the flange, Bf, is the minimum of the following two values:

1. Bf = Bw + L₃/2 (where L₃ is the clear distance to the adjacent parallel beam).

2. Bf = Bw + L₀/12 + 3*Df

Analysis of T-Beams

The analysis of a T-Beam involves a systematic, step-by-step procedure.

Step 1: Determine the Limiting Depth of the Neutral Axis (xu,lim)

This value is identical to that used for singly and doubly reinforced sections and depends only on the grade of steel (Fy).

xu,lim = k * d

Where 'k' is a constant based on the steel grade:

• Fe 250: k = 0.53

• Fe 415: k = 0.48

• Fe 500: k = 0.46

Step 2: Determine the Actual Depth of the Neutral Axis (xu)

To find the actual depth of the neutral axis, we equate the total compressive force (C) to the total tensile force (T).

C = T

However, the calculation depends on the location of the neutral axis. There are two primary possibilities:

1. The neutral axis lies within the flange (xu ≤ Df).

2. The neutral axis lies within the web (xu > Df).

Case 1: Neutral Axis Lies Within the Flange (xu ≤ Df)

Assumption: Assume the neutral axis lies within the flange (xu ≤ Df).

• Analysis: If the neutral axis is in the flange, the concrete in compression has a rectangular shape with width Bf. The analysis becomes identical to that of a singly reinforced rectangular beam of width Bf.

• Force Calculation:

• Compressive Force (C): C = 0.36 * fck * Bf * xu

• Tensile Force (T): T = 0.87 * fy * Ast

• Calculating xu:
  Equating C and T:
  xu = (0.87 * fy * Ast) / (0.36 * fck * Bf)

• Verification of Assumption:

• If the calculated xu ≤ Df, the assumption is correct.

• If the calculated xu > Df, the assumption is incorrect, and the neutral axis must lie in the web (proceed to Case 2).

Case 2: Neutral Axis Lies Within the Web (xu > Df)

When the neutral axis is in the web, the compressive stress block is no longer a simple rectangle. This case is further divided into two sub-cases based on the depth of the flange (Df).

Sub-Cases for NA in Web

Methodology for Case 2 Analysis:

To simplify the analysis, the T-beam is conceptually divided into two components:

1. A rectangular web section of width Bw.

2. Two rectangular flange sections with a combined width of (Bf - Bw).

The total compressive force and moment of resistance are the sum of the contributions from the web and the flange.

Sub-Case 2a: Neutral Axis in Web (xu > Df) and Df ≤ (3/7)xu

• Force Calculation:

• Compressive Force in Web (Cw): Cw = 0.36 * fck * Bw * xu

• Compressive Force in Flange (Cf): Cf = 0.45 * fck * (Bf - Bw) * Df

• Total Compressive Force (C): C = Cw + Cf

• Total Tensile Force (T): T = 0.87 * fy * Ast

• Calculating xu:
  Equate total compression to total tension (C = T):
  0.36*fck*Bw*xu + 0.45*fck*(Bf - Bw)*Df = 0.87*fy*Ast
  Solve this equation for xu.

• Verification: Check if the calculated xu and given Df satisfy xu > Df and Df ≤ (3/7)xu. If so, the assumption is correct.

Sub-Case 2b: Neutral Axis in Web (xu > Df) and Df > (3/7)xu

• IS Code Recommendation: When the stress distribution over the flange is non-uniform, the IS code simplifies the calculation by introducing an equivalent uniform stress block.

• Instead of using the actual flange depth Df, an equivalent depth (yf) is used, over which the stress is assumed to be uniform at 0.45 fck.

• Equivalent Depth (yf):
  yf = 0.15 * xu + 0.65 * Df
  (Note: This yf value should not be greater than Df).

• Force Calculation:

• Compressive Force in Web (Cw): Cw = 0.36 * fck * Bw * xu (remains the same)

• Compressive Force in Flange (Cf): Cf = 0.45 * fck * (Bf - Bw) * yf

• Total Compressive Force (C): C = Cw + Cf

• Total Tensile Force (T): T = 0.87 * fy * Ast

• Calculating xu:
  Equate total compression to total tension (C = T):
  0.36*fck*Bw*xu + 0.45*fck*(Bf - Bw)*yf = 0.87*fy*Ast
  Substitute the expression for yf into this equation and solve for xu. This will be an iterative process or require solving a quadratic equation.

Step 3: Determine the Type of Section and Moment of Resistance (Mu)

1. Compare xu and xu,lim:

• If xu < xu,lim: The section is Under-Reinforced.

• If xu = xu,lim: The section is Balanced / Limiting.

• If xu > xu,lim: The section is Over-Reinforced.

2. Calculate the Moment of Resistance (Mu):
   Mu can be calculated from either the compression side or the tension side. It is the sum of moments from the web and flange components.
   Mu = C * (Lever Arm) = T * (Lever Arm)

Formulas for Moment of Resistance (Mu)

Important Note on Over-Reinforced Sections:

• Do not design or analyze over-reinforced sections in Limit State Method as they lead to brittle, sudden failure.

• If a section is found to be over-reinforced (xu > xu,lim), its moment of resistance must be calculated as a limiting section.

• To do this, use xu = xu,lim in the moment of resistance formula.

• NEVER use the tension-side formula to calculate Mu for an over-reinforced section. The provided Ast is greater than Ast,lim, which would give an unsafe, overestimated moment capacity. Always use the compression-side formula with xu = xu,lim.

Design Formulas for Steel Area (Ast)

If the section dimensions and material properties are known, and the goal is to find the required area of steel (Ast), the force equilibrium equations can be rearranged. This is typically done by separating the steel required for the web (Ast,w) and the flange (Ast,f).

• Steel for Web (Ast,w):
  Corresponds to the compressive force in the web.
  0.87 * fy * Ast,w = 0.36 * fck * Bw * xu

• Steel for Flange (Ast,f): Corresponds to the compressive force in the flange.

• If Df ≤ (3/7)xu: 0.87 * fy * Ast,f = 0.45 * fck * (Bf - Bw) * Df

• If Df > (3/7)xu: 0.87 * fy * Ast,f = 0.45 * fck * (Bf - Bw) * yf

The total steel required is Ast = Ast,w + Ast,f.

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