RCC Design Test 1

Concept:

As per IS 456 : 2000, Table 5;

Cement content should not exceed 450 kg/m³ of concrete to limit cracking due to drying shrinkage or reduce alkali aggregate reaction.

Important Point:

Concept:

When overall Depth D ≥ 700 mm, then 0.1 % of gross web area is provided as the side face reinforcement equally distributed on both faces with spacing not more than 300 mm. It is provided to take care of shrinkage crack and torsional effect due to high depth of beam.

Calculation:

D = 900 mm, and b = 300 mm

 Side face reinforcement = 0.1% of gross area = \(\frac{{0.1 \times 900 \times 300}}{{100}} = 270{\rm{\;m}}{{\rm{m}}^2}\)

This side reinforcement should be distributed equally on both faces of beam.

Side face reinforcement on one face of beam = 270/2 = 135 mm²

Mistake point:

Half of the total side face R/F is provided on each of the faces.

According to Table No. 19 of IS 456:2000,

The allowable shear stress (design shear strength) depends on both percentage of longitudinal steel provided and grade of concrete used in the beam.

Concept:

Values of partial safety factors for loads (γ _f):-

Calculation:

Hence the design load is maximum of above loads i.e. 192 kN

Mistake Point:-

The 0.9 value of DL is to be considered when stability against overturning or stress reversal is critical. Whenever there is stability in structure due to dead load we reduce the partial load factor of dead load.

Concept:

From the static equilibrium condition, equating the compressive force (C) due to concrete and tensile force (T) due to steel along the neutral axis:

C = T

C = 0.36 f_ck × B × x_{u (lim)} …... (for limiting case)

T = f_ds× A_{st (lim)}                  …... (for limiting case)

\(\therefore {{\text{A}}_{\text{s}{{\text{t}}_{\left( \text{lim} \right)}}}}=\frac{0.36\times {{\text{f}}_{\text{ck}}}\times \text{ }\!\!~\!\!\text{ B}\times {{\text{x}}_{{{\text{u}}_{\text{lim}}}}}}{{{\text{f}}_{\text{ds}}}}\)

Where,

f_ck = compressive strength of concrete, f_ds­ = design strength of steel, b = width of section and x_{u (lim)} = limiting depth of neutral axis

Given Data:

B = 300 mm, D = 650 mm, Effective cover = 50 mm, FOS = 1.5

Effective depth (d’) = D- Effective cover = 650 – 50 = 600 mm

f_ds= f_y/FOS = f_y/1.5 = 0.66­f_y

x_{u (lim)} = 0.48 × d = 0.48× 600 = 288 mm

Calculations:

\(\therefore {{\text{A}}_{\text{s}{{\text{t}}_{\left( \text{lim} \right)}}}}=\frac{0.36\times 300\times 35\times 288\text{ }\!\!~\!\!\text{ }}{\frac{415}{1.5}}=3934.84\text{ }\!\!~\!\!\text{ m}{{\text{m}}^{2}}\)

A_{st, lim} = 3935 mm²

Check for minimum reinforcement:

A_{st (min)} = 0.85 bd/f_y = 0.85 × 300 × 600/415 = 373 mm²

∵ A_{st (lim)} > A_{st (min)} ⇒ OK

Check for maximum reinforcement:

A_{st (max)} = 4% of gross area = 0.04 × B × D = 0.04 × 300 × 650 = 7800 mm²