Concept:

Torsional Reinforcement detailing in Slab:-

Torsional reinforcement is provided in the form of a grid or mesh both at the top and bottom of the slab. IS 456: 2000 recommends that the torsional reinforcement grid should extend beyond the edge for a distance not less than 20% of shorter span. The total area of torsional steel provided in each of the four layers should not be less than:-

1. \(0.75\;{A_{stx}}\;\)if both the meeting edges are discontinuous.

2. \(0.375\;{A_{stx}}\) if one of the two meeting edges, one in continuous and other discontinuous

Here, \({A_{stx}}\) denotes are of flexural steel required for maximum mid span moment in the short direction.

No torsional reinforcement is required if both the meeting edges are continuous.

Calculation:

A_st-x = 1500 mm²

Concept:

Lap splicing is done by overlapping the bars over a certain length that makes the transfer of force from terminating bar to the connecting bar.

The straight length of the lap should not be less than 15 times the bar diameter or 200 mm whichever is more. The lap length must at least be equal to development length.

In flexural and direct tension, minimum lap length should be 30 times the bar diameter. In compression minimum lap length should be twenty four times the bar diameter.

Concept:

Torsional moment exist along with bending moment as well as shear force, axial force and generally shear stresses are induced in the beam due to torsion.

Equilibrium torsion:- The torsion induced in an RCC member due to eccentric loading with respect to the center is called equilibrium torsion.

Equilibrium Torsion in beam 2 due to load P shown above.

Compatibility Torsion: This type of torsion is induced due to the need of the member to undergo a certain angle of  twist to maintain the compatibility conditions. The twisting moment developed is statically indeterminate.

Compatibility Torsion/ Secondary Torsion/ Statically Indeterminate Torsion in 2 and 3 due to load P shown above.

Explanation:

If torsional stiffness of member is neglected the member is designed for bending moment only and shear force is assumed to cater for compatibility torsion.

Thus the design is done only for equilibrium torsion.

Equivalent shear force (V_ue) is given by

\(\begin{array}{l}{V_{ue}} = {V_u} + \frac{{1.6{T_u}}}{B}\\{V_{ue}} = 50 + \frac{{1.6 \times 150}}{{0.300}}\end{array}\)

Concept:

Development Length \(\left( {{\ell _d}} \right) = \frac{{0.87\;{f_y}}}{{4\;{\tau _{bd}}}}\)

Also,

\({\ell _d} \le 1.3\frac{{{M_u}}}{{{V_u}}} + {\ell _0}\)             (When there is compression confinement)

Where,

M_u = Ultimate MOR at a section considering stress in all bars as 0.87 f_y

V_u = Factored shear force at the point of zero moment.

ℓ₀ = Anchorage length

Calculation:

M_u = 50 kN – m, V_u = 200 kN, ℓ₀ = 400 mm

\(\therefore {\ell _d} \le 1.3 \times \frac{{50}}{{200}} + \frac{{400}}{{1000}} = 0.725\;m\)

⇒ ℓ_d ≤ 725 mm

At limiting condition,

ℓ_d = 725 mm

Also,

\({\ell _d} \ge \frac{{0.87\;{f_y}\;\phi }}{{4\;{\tau _{bd}}}} \Rightarrow 725 \ge \frac{{0.87 \times 415 \times \phi }}{{4 \times 1.2}}\)

⇒ ϕ ≤ 9.638 mm  

Thus the maximum bar dia that can be provided is 9 mm.

Mistake Point:

The value of bond stress is given for Fe 415 bar and we need not add 60 % in the given value.

Important Point:

In case of simple support or the places where there is no compression confinement, tension R/F should be limited to a diameter, such that

\({\ell _d} \le \frac{{{M_u}}}{{{V_u}}} + {\ell _0}\)

Concept:

Area of steel required will be:

\({A_{st\left( {req} \right)}} = \frac{{0.5{f_{ck}}}}{{{f_y}}}\left( {1 - \sqrt {1 - \frac{{4.6{M_u}}}{{{f_{ck}}b{d^2}}}} } \right) \times bd\)

Where,

M_u = ultimate moment of the slab, b = width of the slab, and d = effective depth of the slab,

f_ck and f_y have usual meanings

Spacing of the bar = \(\frac{{1000}}{{{{\rm{A}}_{{\rm{st}}}}}} \times {{\rm{A}}_\phi }\)

Minimum Area of steel in slab = 0.12% of cross-sectional area of slab = 0.0012bd

Maximum spacing of main bar in slab = 3d or 300 mm whichever is lesser.

Calculation:

M_u = 37 kN-m, b = 1000 mm, and d = 400 mm

\({{\rm{A}}_{{\rm{st}}\left( {{\rm{req}}} \right)}} = \frac{{0.5 \times 25}}{{500}}\left( {1 - \sqrt {1 - \frac{{4.6 \times 37 \times {{10}^6}}}{{25 \times 1000 \times {{400}^2}}}} } \right)1000 \times 400 = 215{\rm{\;m}}{{\rm{m}}^2}/{\rm{m}}\)

But minimum Area of steel should be,

\({\rm{As}}{{\rm{t}}_{{\rm{min}}}} = \frac{{0.12}}{{100}} \times 1000 \times 400 = 480{\rm{m}}{{\rm{m}}^2}/{\rm{m}}\)

Hence the spacing of the bar should be,

\({{\rm{A}}_\phi } = \frac{{\rm{\pi }}}{4} \times 10 \times 10 = 78.54{\rm{\;m}}{{\rm{m}}^2}\)

\({\rm{Spacing}} = \frac{{1000}}{{480}} \times 78.54 = 163.62{\rm{\;mm}}\)

Check for spacing:

Maximum spacing should be lesser of 3d or 300 mm whichever is lesser.

∴ Spacing = 3 x 400 = 1200 mm or 300 mm whichever is lesser.

Hence provide 8ϕ -160 mm c/c main bar in the slab.

Concept:

Equivalent Bending Moment:

M_e1 = M_u + M_t

\({M_t} = \frac{{{T_u}}}{{1.7}}\left( {1 + \frac{D}{b}} \right)\)

Where, M_e1 = equivalent factored B.M, T_u = Factored Torsional Moment, D = Overall Depth of section, b = Width of the section

Design for Bending:

• Longitudinal Reinforcement shall be designed to resist equivalent B.M  M_e1.
• Design the cross-section either as singly reinforced or doubly reinforced depending on the value of M_e1 & cross-section size. For singly reinforced section,
  \({M_{e1}} = 0.87\;{f_y}\;Ast\;\left( {d - \frac{{{f_y}\;{A_{st}}}}{{{f_{ck}}\;b}}} \right)\)
• If |M_t| > |M_u|, i.e. the numerical value of M_t exceeds the numerical value of moment M_u, then longitudinal reinforcement shall be provided on the flexural compression side for moment M_e2.
  M_e2 = |M_t| - |M_u|
  Also, M_e2 = 0.87 f_y A_st (d – d’)
  A_st = Tensile reinforcement on the flexural compression side
• If |M_t| < |M_u|, then no need to provide extra reinforcement on the flexural compression side.

Calculation:

M_u = 200 kN-m, T_u = 150 kN-m

\(\therefore {M_t} = \frac{{{T_u}}}{{1.7}}\left( {1 + \frac{D}{b}} \right) = \frac{{150}}{{1.7}}\left( {1 + \frac{{500}}{{250}}} \right) = 264.71\;kN - m\)

Here, M_t > M_u

∴ Compression R/F is provided on the flexural compression side for moment M_e2.

M_e2 = M_t - M_u = 264.71 – 200 = 64.71 kN-m

Also, d’ = d = 25 mm.

∴ M_e2 = 0.87 f_y A_st (d – d’)

⇒ 64.71 × 10⁶ = 0.87 × 415 × A_st (475 - 25)

⇒ A_st = 398.28 mm² ≈ 400 mm²

Concept:

Development length required by limit state method is:

\({{\rm{L}}_{\rm{d}}} = \frac{{\phi \times 0.87 \times {{\rm{f}}_{\rm{y}}}}}{{4 \times {{\rm{\tau }}_{{\rm{b}}{{\rm{d}}_{{\rm{perm}}}}}}}}\)

Where

ϕ = Bar diameter, f_y = Grade of steel, and \({{\rm{\tau }}_{{\rm{b}}{{\rm{d}}_{{\rm{perm}}}}}} = {\rm{Permissible\;bond\;stress}}\) 

For bars in compression, the permissible bond stress is increased by 25% and for HYSD, CTD, TMT bars, the bond stress is additionally increased by 60%.

Calculation:

For bar A:

\({\left( {{{\rm{L}}_{\rm{d}}}} \right)_{\rm{A}}} = \frac{{{\phi _{\rm{A}}} \times 0.87 \times {{\rm{f}}_{\rm{y}}}}}{{4 \times {{\rm{\tau }}_{{\rm{b}}{{\rm{d}}_{\rm{A}}}}}}}\)

As bar A is in tension but it should be increased by 60% as the bar used is HYSD bar.

\({{\rm{\tau }}_{{\rm{b}}{{\rm{d}}_{\rm{A}}}}} = 1.4 \times 1.60 = 2.24{\rm{\;N}}/{\rm{m}}{{\rm{m}}^2}\)

Development length for bar A is given by:

\({\left( {{{\rm{L}}_{\rm{d}}}} \right)_{\rm{A}}} = \frac{{{\phi _{\rm{A}}} \times 0.87 \times 500}}{{4 \times 1.4 \times 1.60}} = 48.55 \times {\phi _{\rm{A}}} = 48.55{\rm{\;}}\phi \)

For bar B:

\({\left( {{L_d}} \right)_B} = \frac{{{\phi _B} \times 0.87 \times {f_y}}}{{4{\tau _{b{d_B}}}}}\)

As bar B is in compression therefore it should be increased by 25% apart from increased by 60% due to bar of type HYSD.

\({{\rm{\tau }}_{{\rm{b}}{{\rm{d}}_{\rm{B}}}}} = 1.4 \times 1.6 \times 1.25 = 2.8{\rm{\;N}}/{\rm{m}}{{\rm{m}}^2}\)

\({\left( {{{\rm{L}}_{\rm{d}}}} \right)_{\rm{B}}} = \frac{{\phi \times 0.87 \times 500}}{{4 \times 1.4 \times 1.6 \times 1.25}} = 38.84\;\phi \)

 (L_d)_A > (L_d)_B

Concept:

Reinforcement Requirements in One Way Slab:-

Mild steel requirement in either direction of the slab shall not be less than 0.15% of the total cross-sectional area (bD). However for HYSD bars in either direction, the reinforcement shall not be less than 0.12% of the gross cross-sectional area.

Maximum dia of reinforcing bars shall not exceed \(\frac{1}{8}\)of the total thickness of slab. { e.g. t = 80 mm ⇒ ϕ = 10 mm}

Calculation:-

b = 2000 mm, D = 200 mm

∴ Minimum percentage of mild steel = 0.15% of total cross-sectional area

\( = \frac{{0.15}}{{100}} \times 2000 \times 200 = 600\;m{m^2}\)

Concept:

The attainment of the ultimate flexural strain at one cross-section doesn’t lead to the collapse of the slab. The yield line analysis is carried out to economize the structure. It is based on the upper bound theory. Assumptions of yield line analysis are:-

(i) The reinforcing steel is fully yielded along the yield line at failure and undergoes plastic deformation.

(ii) The slab separates into segments by the yield lines. The individual segments behave elastically.

(iii) Elastic deformations are negligible compared to plastic deformations.

(iv) The Bending & Twisting moment are uniformly distributed along the yield line and have the maximum value provided by the ultimate moment capacity.

Explanation:

P is true because we have plastic deformation along yield line.

Q is false as each segment behaves elastically.

R is true and it follows from the assumption iv listed above.