RCC Design Test 1

Concept:

Flexural strength is used to determine the onset of cracking or the loading at which cracking starts.

\({f_{cr}} = 0.7\;\sqrt {{f_{ck}}} \)

It is valid for LSM only.

Calculation:

f_ck = 25

\(\therefore {f_{cr}} = 0.7 \times \sqrt {25} = 3.5\;MPa\)

Important Point:

f_ct = f_cr × 0.66

Concept:

Spacing for vertical shear stirrups:

\({S_v}\; \not > \;\left\{ {\begin{array}{*{20}{c}} {0.75\;d}\\ {or}\\ {300\;m} \end{array}} \right.\;\)  whichever is less.

Calculation:

b = 300 mm, d = 460 mm.

spacing \( \not > \;\left\{ {\begin{array}{*{20}{c}} {0.75 \times 460 = 345\;mm}\\ {or\;}\\ {300\;mm} \end{array}} \right.\) whichever is less

= 300 mm

The total shrinkage of concrete depends upon the constituents of concrete, size of the member and environmental conditions.

For a given humidity and temperature, the total shrinkage of concrete is most influenced by the total amount of water present in the concrete at the time of mixing and, to a lesser extent, by the cement content.

The total shrinkage strain is composed of two components, the autogenous shrinkage strain and the drying shrinkage strain. The value of the total shrinkage strain (ϵ_cs) is given by:

ϵ_cs = ϵ_cd + ϵ_ca

where,

ϵ_cd = drying shrinkage strain

ϵ_ca = autogenous shrinkage strain

Autogenous shrinkage: It develops during hardening of concrete in the early days after casting. It is a linear function of concrete strength. It should be considered specifically when new concrete is cast against hardened concrete.

Drying shrinkage: The drying shrinkage strain develops slowly, as it is a function of migration of the water through the hardened concrete.

Concept:

A counterfort retaining wall is a cantilever wall with counterforts, or buttresses, attached to the inside face of the wall to further resist lateral thrust.

• In the counterfort retaining wall, the stem and the base and the base slab are tied together by counterforts, at suitable intervals
• Because of the provision of counterforts, the vertical stem, as well as the heel slab, acts as a continuous slab, in contrast to the cantilevers of cantilever retaining wall
• Counterforts are firmly attached to the face slab as well as the base slab; The earth pressure acting on the face slab is transferred to the counter forts which deflect as vertical cantilevers
• The back of the rear counter forts comes in tension and their front face is under compression
• Hence the inclined (back) face of rear counter forts should be provided with main reinforcement and in front counterforts, the tension develops at the bottom face so it is provided with main reinforcement
• Counterfort retaining walls are economical for height over about 6m

∴ In a counterfort retaining wall, the main reinforcement is provided on the bottom face in front counterfort and inclined face in back counterfort.

As per Is 456:2000, the maximum strain in extreme fibre in the tension reinforcement will be:

\({\varepsilon _1} = 0.002 + \frac{{0.87{f_y}}}{{{E_s}}}\)

\({\varepsilon _1} = 0.002 + \frac{{0.87 \times 500}}{{2 \times {{10}^5}}}\)

ε₁ = 4.175 × 10^-3

The maximum strain in extreme fibre in the concrete in compression will be = 0.0035

ε₂ = 0.0035

\(\frac{{{\varepsilon _1}}}{{{\varepsilon _2}}} = \frac{{4.175 \times {{10}^{ - 3}}}}{{0.0035}} = 1.193\)

Concept:

Ultimate creep coefficient \(\left( \theta \right) = \frac{{Ultimate\;creep\;strain}}{{Elastic\;strain}}\)

Long term elastic modulus including the effect of creep is given as,

\({E_{cc}} = \frac{{{E_c}}}{{1 + \theta }}\)

Where, E_c = Short term elastic modulus \( = 5000\sqrt {{f_{ck}}} \)

Calculation:

θ = 1.6,  f_ck = 25

\(\therefore {E_c} = 5000\sqrt {{f_{ck}}} = 5000 \times \sqrt {25} = 25000\;MPa\)

Concept:

Minimum Reinforcement is given as

\(\frac{{{A_{st\;min}}}}{{bd}} = \frac{{0.85}}{{{f_y}}}\)

This is based on the criteria for providing ductility in the beam.

Calculation:

f_y = 415, b = 300 mm, d = 600 mm

\(\therefore \frac{{{A_{st\;min}}}}{{300 \times 600}} = \frac{{0.85}}{{415}}\)

⇒ A_{st min} = 368.67 mm²

Important Point:

As per IS 456:2000:

As per clause 26.2.2.1,

The anchorage value of bend shall be taken as 4 times the diameter of bar for each 45^o bend subjected to a maximum of 16 times the diameter of bar.

The anchorage value of a standard U-type hook shall be equal to 16 times the diameter of the bar.

As per clause 26.1.1,

Bars larger than 32 mm diameter shall not be bundled except in case of columns.

As per clause 26.2.5.1 (e),

When bars of two different diameters are to be spliced, the lap length shall be calculated on the basis of diameter of the smaller bar.

Concept:

As per Rankine's formula, the preliminary estimate of a minimum depth of footing is given as,

\({D_f} = \frac{q}{\gamma }{\left( {\frac{{1 - \sin \phi }}{{1 + \sin \phi }}} \right)^2}\)

Where,

q = Gross bearing capacity, γ = unit weight of soil, ϕ = Angle of friction.

Calculation:

q = 360 kN, ϕ = 30°, γ = 18 kN/m³

\(\therefore {D_f} = \frac{{360}}{{18}} \times {\left( {\frac{{1 - \sin 30^\circ }}{{1 + \sin 30^\circ }}} \right)^2} = 2.22\;m\)

Concept:

Spacing between main reinforcement or secondary reinforcement is given by following formula:

Spacing = \(\frac{1000\times {{\text{A}}_{\text{b}}}}{\text{Ast}}\)

Where, A_b = Area of one bar and A_st = Area of reinforcement required.

For given A_st: Spacing \(\propto\) A_b

Calculation:

For Main reinforcement:

Effective depth, d = 80 mm

Diameter of bar is changed from 20 mm to 28 mm

For given A_st : Spacing ∝ A_b

\(\frac{150}{\text{S}}=\text{ }\!\!~\!\!\text{ }\frac{\frac{\text{ }\!\!\pi\!\!\text{ }}{4}\times {{20}^{2}}}{\frac{\text{ }\!\!\pi\!\!\text{ }}{4\text{ }\!\!~\!\!\text{ }}\times {{28}^{2}}}\)

On solving, we get: S = 294 mm

Check for maximum spacing requirement:

S_max = min (3d, 300) = min (3 × 80, 300)

S_max = 240 mm < S not ok

So, spacing is limited to S = 240 mm

% change in spacing, \(\text{X}=\frac{240-150}{150}\times 100=60\text{ }\!\!%\!\!\text{ }\)

For Secondary Reinforcement

Effective depth, d = 80 mm

Diameter of bar is changed from 16 mm to 20 mm.

For given A_st: Spacing A_b

\(\dfrac{200}{S} = \dfrac{\dfrac{\pi}{4}\times 16^2}{\dfrac{\pi}{4}\times 20^2}\)

On solving, we get: S = 312.50 mm

Check for maximum spacing requirement:

S_max = min (5d, 450) = min (5 × 80, 450)

S_max = 340 mm > S Ok

% change in spacing,  \(\text{Y}=\frac{312.5-200}{200}\times 100=\text{ }\!\!~\!\!\text{ }56.25\text{ }\!\!%\!\!\text{ }\)

Ratio = X/Y = 60/56.25 = 1.07

Concept:

Check for lateral stability:

For lateral stability, span \( \not > \;\left\{ {\begin{array}{*{20}{c}}{25\;b}\\{or}\\{\frac{{100\;{b^2}}}{d}}\end{array}} \right.\)  whichever is less

Check for Deflection:

For deflection, \({\left( {\frac{\ell }{d}} \right)_{actual}} \not > {\left( {\frac{\ell }{d}} \right)_{basic}}\;{k_1}.{k_2}.{k_3}\)

a) k₁ depends on the area of tensile steel and stress level of tensile steel.

b) k₂ depends on the area of compression steel.

c) k₃ depends on the type of beam cross-section.

(ℓ/d)_basic values for different beams:

Cantilever = 7

Simply supported = 20

Continuous = 26

Calculation:

ℓ = 6 m = 6000 mm, b = 300 mm, d = 600 – 40 = 560 mm

For lateral stability ; span \( \not > min\left\{ {\begin{array}{*{20}{c}}{25\;b = 25 \times 300 = 7500\;mm}\\{or}\\{\frac{{100 \times {b^2}}}{d} = \frac{{100 \times {{300}^2}}}{{560}} = 16071.43\;mm}\end{array}} \right.\)

⇒ span ≯ 7500 mm

Actual span = 6000 mm ≯ 7500 mm

∴ Beam is laterally stable.

For deflection criteria;

\({\left( {\frac{\ell }{d}} \right)_{beam}} \not > 7\)

\( \Rightarrow {\left( {\frac{\ell }{d}} \right)_{actual}} = \frac{{6000}}{{560}} = 10.714\)  

Beam is not safe in deflection.

Mistake Point:

Concept:

As per IS 456: 2000

f_cm = f_ck + 1.65 × σ

Where

f_cm = mean compression strength

f_ck = characteristics compressive strength

σ = standard deviation (as per table 8 of IS 456: 2000)

\({{\rm{f}}_{{\rm{cm}}}} = \frac{{{{\rm{f}}_{{\rm{c}}1}} + {{\rm{f}}_{{\rm{c}}2}} + {{\rm{f}}_{{\rm{c}}3}}}}{3}\)

f_c1, f_c2 and f_c3 are average compressive strength of specimen 1, 2 and 3 respectively

(σ)_M20 = 4.0 (as per Table 8 of 13 456: 2000)

Calculation:

f_cm = 20 + 1.65 × 4 = 26.6 N/mm²

∴  Average compressive strength of specimen should be 26.6 N/mm²

Now,

\({{\rm{f}}_{{\rm{cm}}}} = {\rm{\;}}\frac{{{{\rm{f}}_{{\rm{c}}1}} + {{\rm{f}}_{{\rm{c}}2}} + {{\rm{f}}_{{\rm{c}}3}}}}{3}\)

\(\Rightarrow 26.6 = \frac{{28{\rm{\;}} + {\rm{\;}}24{\rm{\;}} + {\rm{\;}}{{\rm{f}}_{{\rm{c}}3}}}}{3}\)

⇒ f_c3 = 27.8 N/mm²

Calculation:

w = 0.55

Let the weight of cement be x kg

∴ Weight of water = 0.55 x

Weight of fine aggregate = 2x

Weight of coarse aggregate = 4x

In order to make 1 m³ of concrete,

Volume (cement + FA + CA + water) = 1 m³

\( \Rightarrow \frac{x}{{3.15 \times 1000}} + \frac{{2x}}{{2.65 \times 1000}} + \frac{{4x}}{{2.6 \times 1000}} + \frac{{0.55\;x}}{{1000}} = 1\)

For Working Stress Method:

i) Critical neutral axis (x_c) = k × d = 0.4 × 400 = 160 mm

ii) Actual neutral axis (x_a) = 96.76 mm     …(Given)

∴ Given section is an under reinforced section

iii) Moment of resistance as per WSM is given by:

\({{\rm{M}}_{\rm{r}}} = {{\rm{\sigma }}_{{\rm{st}}}} \times {{\rm{A}}_{{\rm{st}}}} \times \left( {{\rm{d}} - \frac{{{{\rm{x}}_{\rm{a}}}}}{3}} \right) = 230 \times {\rm{\pi }} \times {12^2} \times \left( {400 - \frac{{96.76}}{3}} \right) = {\rm{\;}}38.26{\rm{kNm}}\)

For Limit State Method:

i) Limiting depth of neutral axis = 0.48 × d = 0.48 × 400 = 192 mm

ii) Actual depth of neutral axis:

0.36 × f_ck × b × x_u = 0.87 × f_y × A_st

⇒ 0.36 × 20 × 400 × x_u = 0.87 × 415 × π × 12²

⇒ x_u = 56.71 mm

∴ Given section is an under reinforced section

iii) Moment of resistance as per LSM is given by:

M_r = 0.36 × f_ck × b × x_u × (d – 0.42 × x_u)

M_r = 0.36 × 20 × 400 × 56.71 × (400 – 0.42 × 56.71) = 61.44 kNm

Difference of bending moment (M) = M₂ – M₁ = 23. 18 kNm