RCC Design Test 2

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

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:

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.