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:

The width of the flange with constant flexural stress equal to the actual flexural compressive stress which leads to the same longitudinal compressive force as due to the original stress distribution is called effective width of flange.

For Independent L-beam or T-beam:-

For L-Beam:-

\({b_f} = {b_w} + \frac{{0.5\;{\ell _0}}}{{\left( {\frac{{{\ell _0}}}{b} + 4} \right)}}\; \not > b\)

For T-Beam:-

\({b_f} = {b_w} + \frac{{{\ell _0}}}{{\left( {\frac{{{\ell _0}}}{b}} \right) + 4}}\; \not > b\)

Calculation:

b = 1500 mm, l₀ = 7500 mm, b_w = 200 mm

\({b_f} = {b_w} + \frac{{{\ell _0}}}{{\left( {\frac{{{\ell _0}}}{b}} \right) + 4}} \not > b\)

\(\therefore {b_f} = 200 + \left( {\frac{{7.5 \times 1000}}{{\frac{{7500}}{{1500}} + 4}}} \right)\) = 1033.33 mm

Important Point:

For continuous L beam or T beam:

For L – Beam:

\({b_f} = {b_w} + 3{D_f} + \frac{{{\ell _0}}}{{12}}\)

\( \not > {b_w} + \frac{{{\ell _1}}}{2}\)

For T Beam:

\({b_f} = {b_w} + 6{D_f} + \frac{{{\ell _0}}}{6}\)

\(\not > \;{b_w} + \frac{{{l_1} + {\ell _2}}}{2}\)

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.

Calculation:

d = 500 mm, t = 140 mm, M20

For under reinforced section: Steel will yield.

∴ σ_st = 140 MPa and let the stress in concrete is σ_C

Actual Neutral axis distance:-

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaWGcbWdaiaGbccacayGxdWdbiaacIcacaWGybWdamaaDaaaleaa % peGaamyyaaWdaeaapeGaaGOmaaaakiaacMcacaGGVaGaaGOmaiabg2 % da9iaad2gapaGaag41a8qacaWGbbWdamaaBaaaleaapeGaam4CaaWd % aeqaaOWdbiaadshapaGaag41a8qacaGGOaGaamizaiabgkHiTiaadI % fapaWaaSbaaSqaa8qacaWGHbaapaqabaGcpeGaaiykaaaa!4E30! B{\rm{ \times }}(x_a^2)/2 = m{\rm{ \times }}{A_s}t{\rm{ \times }}(d - {x_a})\)\(% MathType!MTEF!2!1!+- % feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaWGcbWdaiaGbccacayGxdWdbiaacIcacaWGybWdamaaDaaaleaa % peGaamyyaaWdaeaapeGaaGOmaaaakiaacMcacaGGVaGaaGOmaiabg2 % da9iaad2gapaGaag41a8qacaWGbbWdamaaBaaaleaapeGaam4CaaWd % aeqaaOWdbiaadshapaGaag41a8qacaGGOaGaamizaiabgkHiTiaadI % fapaWaaSbaaSqaa8qacaWGHbaapaqabaGcpeGaaiykaaaa!4E31! B{\rm{ \times }}(x_a^2)/2 = m{\rm{ \times }}{A_s}t{\rm{ \times }}(d - {x_a})\)\(% MathType!MTEF!2!1!+- % feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaWGcbWdaiaGbccacayGxdWdbiaacIcacaWGybWdamaaDaaaleaa % peGaamyyaaWdaeaapeGaaGOmaaaakiaacMcacaGGVaGaaGOmaiabg2 % da9iaad2gapaGaag41a8qacaWGbbWdamaaBaaaleaapeGaam4CaaWd % aeqaaOWdbiaadshapaGaag41a8qacaGGOaGaamizaiabgkHiTiaadI % fapaWaaSbaaSqaa8qacaWGHbaapaqabaGcpeGaaiykaaaa!4E31! B{\rm{ \times }}(x_a^2)/2 = m{\rm{ \times }}{A_s}t{\rm{ \times }}(d - {x_a})\)

Here B is not given and hence we cannot obtain the actual Neutral axis depth.

Also,

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qadaWcaaWdaeaapeGaamiEa8aadaWgaaWcbaWdbiaadggaa8aabeaa % aOqaa8qacaWGJbaaaiabg2da9maalaaapaqaa8qadaWcaaWdaeaape % Gaeq4Wdm3damaaBaaaleaapeGaam4Caiaadshaa8aabeaaaOqaa8qa % caWGTbaaaaWdaeaapeGaamizaiabgkHiTiaadIhapaWaaSbaaSqaa8 % qacaWGHbaapaqabaaaaaaa!4435! \frac{{{x_a}}}{c} = \frac{{\frac{{{\sigma _{st}}}}{m}}}{{d - {x_a}}}\)\(% MathType!MTEF!2!1!+- % feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qadaWcaaWdaeaapeGaamiEa8aadaWgaaWcbaWdbiaadggaa8aabeaa % aOqaa8qacaWGJbaaaiabg2da9maalaaapaqaa8qadaWcaaWdaeaape % Gaeq4Wdm3damaaBaaaleaapeGaam4Caiaadshaa8aabeaaaOqaa8qa % caWGTbaaaaWdaeaapeGaamizaiabgkHiTiaadIhapaWaaSbaaSqaa8 % qacaWGHbaapaqabaaaaaaa!4435! \frac{{{x_a}}}{c} = \frac{{\frac{{{\sigma _{st}}}}{m}}}{{d - {x_a}}}\)\(% MathType!MTEF!2!1!+- % feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qadaWcaaWdaeaapeGaamiEa8aadaWgaaWcbaWdbiaadggaa8aabeaa % aOqaa8qacaWGJbaaaiabg2da9maalaaapaqaa8qadaWcaaWdaeaape % Gaeq4Wdm3damaaBaaaleaapeGaam4Caiaadshaa8aabeaaaOqaa8qa % caWGTbaaaaWdaeaapeGaamizaiabgkHiTiaadIhapaWaaSbaaSqaa8 % qacaWGHbaapaqabaaaaaaa!4435! \frac{{{x_a}}}{c} = \frac{{\frac{{{\sigma _{st}}}}{m}}}{{d - {x_a}}}\)

Again c cannot be determined because neutral axis depth is not found.

Hence the date given is inadequate.

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

Concept:

Shear stress for Beam of varying depth:

\({\tau _v} = \frac{{{V_u} \pm \frac{{{M_u}}}{d}\tan \beta }}{{bd}}\)

Where,

V_u = Factored shear force at the section.

M_u = Factored Bending moment at the section.

β = Angle between the top and the bottom surface of the beam.

The negative sign is used when the magnitude of bending moment increases in the direction in which the depth increases

The positive sign is used when the magnitude of bending moment increases in the direction in which the depth decreases.

Minimum shear reinforcement is given by

\(0.87\;{f_y}\;{A_{sv}}\frac{d}{{{S_v}}} = 0.4\;bd\)

\({f_y} \not\succ 415\;MPa\)

Calculation:

b = 300 mm, d = 800 mm (at critical section)

\({\tau _v} = \frac{{{V_u} - \frac{{{M_u}}}{d}\tan \beta }}{{bd}} \Rightarrow \frac{{150 \times {{10}^3} - \frac{{90 \times {{10}^6}}}{{800}} \times \left( {\frac{{450}}{{3000}}} \right)}}{{300 \times 800}}\) ≈ 0.594

Percentage of tensile reinforcement at the section

\({p_t} = \frac{{{A_{st}}}}{{bd}} \times 100 = \frac{{\left( {4 \times 201} \right)}}{{300 \times 800}} \times 100 = 0.335\% \)

Adopting the value from the table given,

τ_c = 0.40

∴ τ_v - τ_c = 0.594 – 0.40 = 0.194

∵ τ_v - τ_c < 0.4 ⇒ Provide minimum shear reinforcement.

Minimum shear reinforcement is given by,

\(0.87\;{f_y}\;{A_{sv}}\frac{d}{{{S_v}}} = 0.4\;bd\)

Here, Fe 500, but f_y ⊁ 415

Two-legged shear stirrup,

∴ \({A_{sv}} = 2 \times \left( {\frac{\pi }{4} \times {8^2}} \right) = 100.53\;m{m^2}\)

Putting values above,

\(0.87 \times 415 \times 100.53 \times \frac{{800}}{{{S_v}}} = 0.4 \times 300 \times 800\)

⇒ S_v = 302.496 mm

But the maximum spacing of vertical stirrups is limited to 300 mm.

So provide 8 mm diameter bar @ 300 mm c/c.

Concept:

Moment of resistance of the doubly reinforced section is given as

MOR = 0.36 f_ck bx (d - 0.42x) + A_sc (f_sc - f_cc) (d – d’)

Where,

f_sc = stress in the compression steel.

f_cc = stress in concrete at the level of compression steel.

Calculation:

For Fe 415, x_lim = 0.48 d = 0.48 × (500 - 40)

= 220.8 mm

Strain at the level of compression steel.

\(\frac{{{\varepsilon _{sc}}}}{{{x_{{\rm{lim}}}} - d'}} = \frac{{0.0035}}{{{x_{lim}}}}\)

\(\Rightarrow \;{\varepsilon _{sc}} = \frac{{0.0035 \times \left( {220.8 - 0} \right)}}{{220.8}} = 2.865 \times {10^{ - 3}}\)

Thus from the table given, f_sc = 352 MPa

Also, for a strain greater than 0.002, f_cc = 0.45 f_ck

∴ MOR_lim = 0.36 × 20 × 250 × 220.8 (d – 0.42 × 220.8) + 900 × (352 – 0.45 × 20) (460 - 40)

= 275.62 kN-m

Concept:

Creep coefficient : (C_t)

\({C_t} = \frac{{Creep\;strain}}{{Instantaneous\;strain}}\)

\({\delta _h} = \frac{{FL}}{{AE}},\;{\varepsilon _h} = \frac{F}{{AE}}\)

∴ Creep strain \(= \frac{{{C_t}F}}{{AE}}\)

\({\delta _c} = \frac{{{C_t}FL}}{{AE}}\)

Total Deformation \( = \frac{{FL}}{{AE}} + {C_t}\frac{{FL}}{{AE}} = \frac{{FL}}{{A\left[ {\frac{E}{{1 + {C_t}}}} \right]}}\;\)

Elastic modulus of concrete \(= 5000\;\sqrt {{f_{ck}}} \)

Calculation:

Short term elastic modulus of concrete \(= 5000\;\sqrt {{f_{ck}}} = 5000 \times \sqrt {25} = 25000\;MPa\)

Total strain \(= \frac{\sigma }{E} + {C_t} \times \frac{\sigma }{E}\)     (∵ F/A = σ )

\( = \left( {1 + {C_t}} \right)\frac{\sigma }{E} = \left( {1 + 1.5} \right)\frac{{2500}}{{25000}}\;\)= 0.25