UPPSC AE Civil 2013 Official Paper - I (Previous Year Paper)

Concept:

From a fundamental relationship, the density of soil is given by:

${{\rm{\gamma }}_{{\rm{bulk}}}} = \frac{{\left( {{\rm{G}} + S × e} \right) × {\gamma _w}}}{{1 + e}} = \frac{{\left( {{\rm{G}} + w × G} \right) × {\gamma _w}}}{{1 + e}} = \frac{{\left( {1 + {\rm{w}}} \right) × G × {\gamma _w}}}{{1 + e}};$      ----(i)

Where

γ = Density of soil, γw = Density of water, G = Specific gravity of soil, S = Degree of saturation, w = Water content, and e = void ratio

For the dry density of soil: Degree of saturation (S) = 0

The dry density of soil is given by ${{\rm{\gamma }}_{\rm{d}}} = \frac{{\left( {\rm{G}} \right) × {\gamma _w}}}{{1 + e}}$      ----(ii)

Calculation:

1 + e = (2.70 × 1)/1.84

⇒ e = 0.467 ≈ 0.47

In compaction test:

 

${\rm{Compactive;Energy;Ratio;}}\left( {\rm{r}} \right) = \frac{{{\rm{Compactive;effort;in;modified;compaction;test}}}}{{{\rm{Compactive;effcot;in;standart;compaction;test}}}}$

$\therefore {\rm{r}} = \frac{{4.9 \times 450 \times 5 \times 25}}{{2.6 \times 310 \times 3 \times 25}} = 4.559$

Concept:

Critical hydraulic gradient (ic): The quicksand / boiling condition occurs at a critical upward hydraulic gradient typically around 1.0 for many soils, when the seepage force just balances the buoyant weight of an element of soil.

At the critical conditions, the effective stress is equal to zero.

${i_c} = \frac{{G - 1}}{{1 + e}};{\rm{OR;}}\left( {{\rm{G}} - 1} \right) \times \left( {1 - {\rm{n}}} \right)$

Calculations:

Critical hydraulic gradient

\(\begin{array}{*{20}{c}} {{i_c} = \left( {{\rm{G}} - 1} \right) \times \left( {1 - {\rm{n}}} \right) = 1.65 \times \left( {1 - 0.45} \right) = 0.9075} \end{array}\)

i_s = i_c/4 = 0.9075/4 = 0.225

Concept:

Flow net: It is a graphical representation of two dimensional steady state flow consisting of two sets of lines – the flow and the equipotential lines.

In case of isotropic soil these lines are orthogonal to each other but need not be orthogonal in anisotropic soil.

Various methods of obtaining flow nets are:

1. Graphical method

2. Electrical flow Analogy method

3. Sand Model

4. Capillary flow Model

5. Solution of Laplace equation/Analytical method.

Flow model is not used to obtain flow nets.

Soil classification as per Indian standards:

∴ According to IS Classification, there are 18 groups of soils:

8 groups of coarse-grained, 9 group of fine-grained, and 1 of peat

From the above reference, highly compressible silty soil is designated by MH.

Explanation

Mohr’s circle for a c-ϕ soil is shown below:

Any angle α_c⁰ on the horizontal plane becomes equal to 2α_c⁰ Mohr’s circle.

Here,

σ_1f = Major Principal Stress, σ_3f = Minor Principal Stress, ϕ = Angle of internal friction of soil, and c = Cohesion.

In ΔAOB,

Φ + (180 – 2 α_c⁰) + 90° = 180°

⇒ Φ = (2 α_c⁰ – 90°)

⇒ Φ = 2 (α_c⁰ – 45°)

Important Point:

• The triaxial test is the most widely used shear strength test in which drainage can be controlled irrespective of the type of soil. Both pore water pressure and volume changes can be measured in a triaxial test.
• At the time of failure, the horizontal plane is the major principal plane and the vertical plane is the minor principal plane.
• A triaxial test is performed in two stages.

 

 

The three types of triaxial tests are as follows:

1. CD → Consolidated drained Test
2. CU → Consolidated undrained Test
3. UU → Unconsolidated undrained Test

Rankine’s Active earth Pressure theory in cohesive Soil:

$\rm{{P_a} = {K_a}\gamma Z - 2c'\sqrt {{K_a}}}$

So, at Z =0,

$\rm{{P_a} = - 2c'\sqrt {{K_a}}}$

Negative sign shows that the pressure is negative.

For the stress to reverse its nature, a point will reach where the earth pressure value will be zero,

Therefore, Pa = 0,

$\rm{{K_a}\gamma Z - 2c'\sqrt {{K_a}} = 0}$

$\rm{{Z_c} = \frac{{2c'\sqrt {{K_a}} }}{{{K_a}\gamma }} = \frac{{2c'}}{{\gamma \sqrt {{K_a}} }}}$

The above depth is known as the depth of tensile crack (Zc).

Concept:

For purely cohesive soil,

\({{\rm{s}}{\rm{n}}} = \frac{{\rm{c}}}{{{{\rm{\gamma }}{\rm{t}}}{\rm{H;}}{{\rm{f}}_{\rm{c}}}}}\)

Where,

c = cohesion, f_c = Factor of safety with respect to cohesion, and H = Depth of excavation.

Calculation:

Given:

c = 26 kN/m³, γ_t = 18 kN/m³, S_n = 0.172, and H = 6 m

Using the formula,

$0.172 = \frac{{26}}{{18 \times 6 \times {F_c}}}$

⇒ F_c = 1.4

Important Point:

Taylor’s stability number (s_n) = cos β × sin β

Where, β = slope angle

Theoretical maximum value of stability number is 0.5 and practically its value is in the range of 0.25 - 0.30.

Let the percentage/ratio of sand be 1 : x (soil: sand)

Thus,

$6 = \frac{{\left( {15 \times 1} \right) + \left( {0 \times x} \right)}}{{\left( {1 + x} \right)}}$

⇒ x = 1.5

Thus soil: sand ∷ 1 : 1.5

Thus,

${\rm{Percentage;of;sand}} = \frac{{1.5}}{{1 + 1.5}} \times 100;$= 60%

Important point:

Plasticity index is the range of water content over which soil behaves in plastic range.

Plasticity is exhibited due to clay minerals present in soil and hence clay can assimilate large quantity of water without getting converted into liquid state.

Normally plasticity index of clay is 15 - 100.

Concept:

${\rm{;}}\frac{{{\rm{\Delta H}}}}{{\rm{H}}} = \frac{{{\rm{\Delta e}}}}{{1 + {{\rm{e}}_0}}} = \frac{{{\rm{\Delta V}}}}{{\rm{V}}}$

Where,

e₀ = initial void ratio, H = Initial height of strata, V = Initial volume

Calculation:

Given:

H = 3 m, e₀ = 0.90, and e_f = 0.71

∴ Δe = 0.90 – 0.71 = 0.19

Applying the formula given above,

$\frac{{{\rm{\Delta H}}}}{3} = \frac{{0.19}}{{1 + 0.9}}$

⇒ ΔH = 0.1 × 3 = 0.3 m = 30 cm

Thus, total consolidation settlement = 30 cm

Concept:

The given pile is a friction pile hence all the load is carried through friction only.

Static pile load formula is given as-

Q_U = Q_PU + Q_f

where,

Q_U = Ultimate pile load capacity,

Q_PU = q_p × A_b ; Ultimate point bearing capacity.

Q_f = f_sA_S ; Frictional pile load capacity.

where,

f_s  = α c_u

c_u  =  Cohesion averaged over the pile length.

α = Adhesion factor.

q_p = Effective stress at the base of pile.

A_b = Base area of pile.

A_s = Surface area of pile.

Given:

α = 0.7, c_u = 4t/m², FOS = 2.5, Pile diameter (φ) = 30 cm = 0.3 m

Thus, by using static pile load formula –

Q_U = Q_PU + Q_f

⇒ Q_U = 0 + f_s­A_S

⇒ Q_U  = (α C_U)A_s

⇒ Q_U  = (0.7 × 4)× (π ∅ l)

⇒ Q_U  = (0.7 × 4)× (π × 0.3× 10)

⇒ Q_U  = 26.389 t

Thus,

Safe load = $\frac{{{{\rm{Q}}_{\rm{U}}}}}{{{\rm{FOS}}}}{\rm{;}} = {\rm{;}}\frac{{26.389}}{{2.5}}{\rm{;}} = {\rm{;}}10.55{\rm{;t}}$

Important Point:

Static pile load formula assumes that both point bearing capacity and frictional capacity develop its full value. But both do not mobiles to full capacity simultaneously as for developing full frictional resistance movement of the pile required is small but for developing point bearing moment of pile required is large. Thus when full point bearing develops friction falls below its peak value.

Adhesion factor (α) depends upon adhesion b/w soil and pile. Softer is the clay, larger is the value of α.

Concept:

Newmark’s influence chart: It is a graphical method used to compute vertical, horizontal and shear stress due to uniformly distributed load over an area of any shape or geometry, below any point that lies either below or outside the loaded area. It is based on Boussinesq’s equation.

Stress below any point is calculated as :-

${{\rm{\sigma }}_{\rm{z}}}{\rm{;}} = {\rm{;}}\frac{1}{{{\rm{m}} \times {\rm{n}}}} \times {\rm{q}} \times {\rm{N;}}$

where,

m = No. of concentric circles, n = No. of Radial lines, q = Intensity of load, and N = Equivalent number of areas covered by plan area.

The product $\left( {\frac{1}{{m \times n}}} \right)$ is called influence factor.

Given: m = 10, and n = 10

∴ Influence factor $\left( {{{\rm{I}}_{\rm{f}}}} \right){\rm{;}} = {\rm{;}}\frac{1}{{10 \times 10}}$ = 0.01

Concept:

Ultimate Bearing capacity (q_u): It is defined as the maximum gross intensity of loading that the soil can support before it fails in shear.

As per Terzaghi’s bearing capacity equation –

q_u = cN_c + qN_q + $\frac{1}{2}{\rm{B\gamma }}{{\rm{N}}_{\rm{\gamma }}}$

where,

N_c, N_q and Nγ are the dimensionless bearing capacity factors that are a function of ∅ only.

The above formula is derived for strip footing and some modifications are carried out various shapes of footing.

Modifications in Terzaghi’s Bearing capacity equation:-

a) Rectangular footing :

\({{\rm{q}}{{\rm{nu}}}}{\rm{;}} = {\rm{;}}\left( {1 + 0.3\frac{{\rm{B}}}{{\rm{L}}}} \right){\rm{C}}{{\rm{N}}{\rm{C}}} + {\rm{q}}\left( {{{\rm{N}}{\rm{q}}} - 1} \right) + \left( {1 - 0.2\frac{{\rm{B}}}{{\rm{L}}}} \right)0.5{\rm{B\gamma }}{{\rm{N}}{\rm{\gamma }}}\)

b) Square footing:

q_nu = 1.3 CN­c + q(N_q – 1) + 0.8 × 0.5 BγNγ

q_nu = 1.3 CN_c + q(N_q -1) + 0.4 BγNγ

c) Circular footing :

q_nu = 1.3 cN_c + q(N_q - 1) + 0.6 × 0.5 BγNγ

q_nu = 1.3CN_c + q(N_q -1) + 0.3BγNγ

Given:

c = 0 {∵ sandy soil} , γ _t = 18 kN/m³, N_q = 33, and Nγ = 48

Thus by using Terzaghi’s bearing capacity equation ,

Q_u = (qN_q) + \(\left( {1 - 0.2} \right) \times \left( {\frac{1}{2}{\rm{B}}{{\rm{\gamma }}{\rm{t}}}{{\rm{N}}{\rm{\gamma }}}} \right)\)

Q_u = (γ_t × D_f × N_q) + 0.8 × \(\left( {\frac{1}{2}{\rm{B}}{{\rm{\gamma }}{\rm{t}}}{{\rm{N}}{\rm{\gamma }}}} \right)\)

Q_u = (18 × 1.5 × 33) + 0.8 × (0.5 × 2.5 × 18 × 48)

Q_u = 1755 kN/m²

Thus safe load (q_s) $= \frac{{{{\rm{q}}_{\rm{u}}}}}{{{\rm{FOS}}}} = \frac{{1755}}{3}$

⇒ q_s = 585 kN

The above calculation has been done according to the options give although the calculation is incorrect as Factor of safety is applied on net ultimate bearing capacity and net gross ultimate bearing capacity.

Correct explanation should be as under

Q_nu = q(N_q -1) + 0.8 × $\left( {\frac{1}{2}{\rm{B\gamma }}{{\rm{N}}_{\rm{\gamma }}}} \right)$

⇒ q_nu = (18 × 1.5) × (33 - 1) + 0.8 × (0.5 × 2.5 × 18 × 48) = 1728 kN/m²

∴q_ns = $\frac{{{{\rm{q}}_{{\rm{nu}}}}}}{{{\rm{FOS}}}} = \frac{{1728}}{3} = 576{\rm{;kN}}/{{\rm{m}}^2}$

Hence \({{\rm{q}}{\rm{S}}} = \frac{{{{\rm{q}}{{\rm{nu}}}}}}{{{\rm{FOS}}}} + {{\rm{\gamma }}{\rm{t}}}{{\rm{D}}{\rm{f}}}\) = 576 + 18 × 1.5 = 603 kN/m²

Engineering News formula:

Q allowable$= \frac{{wH}}{{6\left( {s + c} \right)}}$

where

w = Weight of drop (kg), H = Height of free fall in ‘cm’, and S = Settlement per blow in ‘cm’

The factor of Safety is taken as 6.

‘c’ is a constant which depends on the type of hammer used.

c = 2.5 cm; For drop hammer.

c = 0.25 cm, For single-acting steam hammer

Note: On single-acting steam hammer, steam is used to lift the hammer and it is always free fall.

Another form of engg. News formula.

${{\rm{Q}}_{{\rm{kN}}}}{\rm{;}} = {\rm{;}}\frac{{166.64{\rm{E}}}}{{{\rm{S}} + 2.54}}$

E = Energy per blow in ‘KJ’

S = Settlements per blow in ‘mm’

Given:

w = 30 KN, H = 1.5 m = 1500 cm, S = 5 mm, c = 2.5 cm, and FOS = 6

As per Engineering News formula:

${{\rm{Q}}_{{\rm{allowable}}}} = \frac{{{\rm{wH}}}}{{6\left( {{\rm{S}} + {\rm{c}}} \right)}} = \frac{{30 \times \left( {1.5 \times 100} \right)}}{{6\left( {2.5{\rm{;}} + {\rm{;}}0.5} \right)}} = 250{\rm{;kN}}$

Let the thickness of first layers be H

Then the thickness of 2^nd layer = 2H

The thickness of 3^rd layer = 3H

Also,

Let the permeability of first layer be K.

Thusthe permeability of 2^nd layer = 2K

Permeability of 3^rd layer = 2 × (2k) = 4k

The formula for horizontal and vertical permeability is given as:-

${{\rm{K}}_{{\rm{eq;horizontal}}}} = \frac{{{{\rm{H}}_1}{{\rm{K}}_1} + {{\rm{H}}_2}{{\rm{k}}_2} + {\rm{H}}{{\rm{k}}_3}}}{{{{\rm{H}}_1} + {{\rm{H}}_2} + {{\rm{H}}_3}}}$

${{\rm{K}}_{{\rm{eq;vertical}}}} = \frac{{{{\rm{H}}_1} + {{\rm{H}}_2} + {{\rm{H}}_3}}}{{\left( {\frac{{{{\rm{H}}_1}}}{{{{\rm{K}}_1}}}} \right) + \left( {\frac{{{{\rm{H}}_2}}}{{{{\rm{K}}_2}}}} \right) + \left( {\frac{{{{\rm{H}}_3}}}{{{{\rm{K}}_3}}}} \right)}}$

Applying the formula given above,

\({{\rm{K}}{{\rm{eq;horizontal}}}} = {{\rm{k}}{\rm{n}}} = \frac{{\left( {{\rm{H}} \times {\rm{K}}} \right) + \left( {2{\rm{H}} \times 2{\rm{K}}} \right) + \left( {3{\rm{H}} \times 4{\rm{K}}} \right)}}{{{\rm{H}} + 2{\rm{H}} + 3{\rm{H}}}} = \frac{{17{\rm{K}}}}{6}\)

\({{\rm{K}}{{\rm{eq;vertical;}}}} = {{\rm{k}}{\rm{v}}} = \frac{{{\rm{H;}} + {\rm{;}}2{\rm{H}} + 3{\rm{H}}}}{{\left( {\frac{{\rm{H}}}{{\rm{K}}}} \right) + \left( {\frac{{2{\rm{H}}}}{{2{\rm{K}}}}} \right) + \left( {\frac{{3{\rm{H}}}}{{4{\rm{K}}}}} \right)}} = \frac{{6{\rm{H}}}}{{\left( {\frac{{2{\rm{H}}}}{{\rm{K}}} + \frac{{3{\rm{H}}}}{{4{\rm{K}}}}} \right)}} = \frac{{24}}{{11}}{\rm{k}}\)

\(\therefore {\rm{Ratio}} = \frac{{{{\rm{k}}{\rm{n}}}}}{{{{\rm{K}}{\rm{v}}}}} = \frac{{\left( {\frac{{17}}{6}{\rm{k}}} \right)}}{{\left( {\frac{{24}}{{11}}{\rm{k}}} \right)}}\)

⇒ K_n : K_v = 1.29 = 1: 0.77

Concept:

The sensitivity of the soil is given by:

$S{\rm{}} = {\rm{}}\frac{{{q_u}}}{{{q_r}}}$

qu = unconfined compressive strength of soil in an undisturbed state

qr = unconfined compressive strength of soil in the remoulded state

Calculation:

$\rm{s = \frac{40.5}{26.6}=1.52}$

Classification of soil based on Sensitivity:

Concept:

Vertical stress due to concentrated load is given as per Boussinesq’s equation.

Formula:

${{\rm{\sigma }}_{\rm{z}}}{\rm{;}} = {\rm{;}}\frac{3}{{2{\rm{\pi }}}}\frac{{\rm{Q}}}{{{{\rm{z}}^2}}}{\left( {\frac{1}{{1 + {{\left( {\frac{{\rm{r}}}{{\rm{Z}}}} \right)}^2}}}} \right)^{\frac{5}{2}}}$

where,

σ_z = Stress magnitude, q = Vertical point load, r = Radial distance, and Z = Vertical distance.

Given:

Z = 10 m, and Q = 500 kN

R = 0  {∵ directly below the load axis}

$\therefore {{\rm{\sigma }}_{\rm{z}}} = \frac{3}{{2{\rm{\pi }}}} \times \frac{{500}}{{\left( {{{10}^2}} \right)}} \times {\left( {\frac{1}{{1 + {{\left( {\frac{0}{{10}}} \right)}^2}}}} \right)^{5/2}} = 2.38{\rm{;kN}}/{{\rm{m}}^2}$

Important Point:

As per Boussinesq’s stress directly below a point load at any distance ‘z’ is given as

${{\rm{\sigma }}_{\rm{z}}}{\rm{;}} = {\rm{;}}\frac{3}{{2{\rm{\pi }}}}\frac{{\rm{Q}}}{{{{\rm{z}}^2}}}$

Difference between Boussinesq’s and Westergaard’s equation of vertical stress :

Concept:

$\rm{Seepage;Velocity;(V_s) = \frac{Discharge;Velocity;(V)}{Porosity;(n)}}$

$\rm{Porosity;(n)=\frac{e}{1+e}}$

Where, e = Void Ratio

Given:

Void ratio, e = 0.5, and Discharge velocity = 6 × 10-7 m/s

Calculation:

Porosity, $\rm{n= \frac{{\rm{e}}}{{1 + {\rm{e}}}} = \frac{{0.5}}{{1 + 0.5}} = 0.333}$ 

Seepage velocity, $Vs = \frac{V}{n} = \frac{{6 \times {{10}^{ - 7}}}}{{0.333}} = 18 \times {10^{ - 7}}m/s$

Skempton’s Pore Pressure coefficients are A and B,

Where A → depends upon the degree of saturation and over consolidation ratio

B → depends upon the degree of saturation of soil

For over-consolidated soil, the value of A is negative.

For fully saturated soil, the value of B is 1

For a Good Soil Sampler,

Area Ratio should be less than 10%

Inside Clearance should be between 0.5 to 3

Outside Clearance should be between 0 to 2

Cutting edge angle should be within 20°

∴ The area ratio shall be as low as possible to minimize the disturbance to the soil sample.

Concept:

Columns can be classified as:

1. Long Column: It fails due to buckling.
2. Short column: It fails due to crushing.
3. Intermediate column: It can fail due to both crushing and buckling.

Analysis of long column is done using Euler’s formula:

${{\rm{f}}_{{\rm{cr}}}}{\rm{;}} = {\rm{;}}\frac{{{{\rm{\pi }}^2}{\rm{E}}}}{{{{\rm{\lambda }}^2}}}$

Where,

E = Modulus of elasticity of material, and λ = slenderness Ratio

\({\rm{\lambda ;}} = {\rm{;}}\frac{{{\rm{l}}{{\rm{e}}{{\rm{ff}}}}}}{{{{\rm{r}}{{\rm{min}}}}}}\)  & I_min = A × r_min²

Where,

r_min = Minimum radius of gyration of the section, I_min = Minor principle Moment of Inertia, A = Cross-sectional Area, and L_eff­ = It depends on the end conditions of the column section.

Eulers Load is given as

\({\rm{P;}} = {\rm{;}}\frac{{{{\rm{\pi }}^2}{\rm{E}}}}{{{\rm{l}}{{\rm{eff}}}^2}}{\rm{r}}{{\rm{min}}}^2 \times {\rm{A;}} = {\rm{;}}\frac{{{{\rm{\pi }}^2}{\rm{E;Imin}}}}{{{\rm{l}}_{{\rm{eff}}}^2}}\)

Thus by Euler’s load formula,

\({\rm{Load;}} \propto {\rm{;}}{{\rm{I}}{{\rm{min}}}}{\rm{;and;Load}} \propto \frac{1}{{{{\left( {{{\rm{l}}{{\rm{eff}}}}} \right)}^2}}}\)

For same end conditions:

Effective length will remain same for every cross-section of the column.

Thus, Load α I_min.

Hence buckling load will be largest for the column with the largest value of minor principal moment of inertia.

Given:

Four different sections with variable cross-sections but similar end conditions and having the same height.

Moment of inertia is maximum for the hollow circular section because the maximum mass, in this case, lies away from the centroidal axis as compared to all other cross sections.

Hence load carrying capacity is maximum for hollow circular section.

As per IS: 800 – 2007

For column, the effective length of a column effectively held at both ends in position but not restrained in directions L.

The design value of the effective length factor for various combination is given below:

Concept:

Effective length is defined as the distance between two adjacent points of zero bending moment or contra flexure.

The value of the effective length varies according to the support condition.

According to Euler's column theory, the crippling load for a column of length L,

${P_{cr}}= ;\frac{{{\pi ^2}EI}}{{{{\left( {L_e} \right)}^2}}}$

Where Leq is the effective length of the column.

Concept:

Net Sectional area:

When tension members are attached to the gusset plate, some material of plate is removed due to bolt/rivet hole. The net area at any section is equal to the gross area minus the deductions for holes at that section.

As per IS 800,

i. For a rivet upto 25 mm diameter:

Dia of rivet hole = Rivet Dia + 1.5 mm

ii. For rivet diameter > 25 mm

Dia of rivet hole = (d + 2) mm

Given:

d = 18 mm, B = 300 mm, t = 10 mm

Net sectional area = (b – nd₀) x t

Where,

n = Number of rivet holes, d₀ = Diameter of one rivet hole, t = Thickness of plate, and b = Width of plate.

Here, d₀ = (18 + 1.5) = 19.5mm

∴ Net Sectional Area = (300 – 19.5) × 10 = 2805 mm²

Important Point:

Consideration of Net Sectional area makes conservative design.

Dia of holes in case of:

Stiffeners are classified into two types:

(a) Vertical Stiffeners

(b) Horizontal stiffeners.

a) Vertical stiffeners: These are provided at right angles to the flanges and are also known as transverse stiffeners.

These are further classified as:

(i) End stiffeners: These are provided at the ends of the girder as they receive the load from the beam and transfer it to support. The web when subjected to compressive load, may get crushed and thus the end stiffeners protects web from crushing.

(ii) Intermediate stiffeners: These are used when there is concentrated load acting on the plate girder. When the web thickness is less than (1/85 × depth of web), web may buckle due to heavy shear. These stiffeners are provided to sustain the web from buckling and avoid buckling of web plate.

b**) Horizontal stiffeners** can be classified as continuous and discontinuous horizontal stiffeners.

Continuous horizontal stiffeners are designed to take loads from the flanges whereas discontinuous stiffeners are not designed to take any load and they only prevent buckling of web.

Concept:

Shear stress distribution in I-section-

Shear stress for flange:

${\rm{\tau ;}} = {\rm{;}}\frac{{{\rm{V}}\left[ {\frac{{{{\rm{D}}^2}}}{4} - \frac{{{{\rm{d}}^2}}}{4}} \right]}}{{2{\rm{I}}}}$

${\rm{\tau ;}} = {\rm{;}}\frac{{{\rm{V}}\left[ {{\rm{B}}\left( {\frac{{\rm{D}}}{2} - {\rm{y}}} \right)} \right]\left[ {{\rm{y}} + \frac{{\frac{{\rm{D}}}{2} - {\rm{y}}}}{2}} \right]}}{{{\rm{I}} \times {\rm{B}}}}{\rm{;}} = {\rm{;}}\frac{{{\rm{V}}\left( {\frac{{{{\rm{D}}^2}}}{4} - {\rm{y}}} \right)}}{{2{\rm{I}}}}$

Shear stress for web:

${\rm{A\bar y;}} = {\rm{;}}{{\rm{A}}_1}{{\rm{\bar y}}_1} + {{\rm{A}}_2}{{\rm{\bar y}}_2}$

\(\therefore {\rm{\tau ;}} = {\rm{;}}\frac{{{\rm{V}}\left[ {{\rm{B}}\left( {\frac{{{\rm{D}} - {\rm{d}}}}{2}} \right)\left{ {\frac{{\rm{d}}}{2} + \frac{{\frac{{{\rm{D}} - {\rm{d}}}}{2}}}{2}} \right} + \left( {\frac{{\rm{d}}}{2} - {\rm{y}}} \right) + \left( {{\rm{y}} + \frac{{\frac{{\rm{d}}}{2} - {\rm{y}}}}{2}} \right)} \right]}}{{{\rm{I}} \times {\rm{t}}}}\)

${\rm{\tau ;}} = {\rm{;}}\frac{{\rm{V}}}{{2{\rm{I}}}}\left( {\frac{{{{\rm{D}}^2}}}{4} - \frac{{{{\rm{d}}^2}}}{4}} \right)\frac{{\rm{B}}}{{\rm{t}}} + \frac{{{\rm{V}}\left( {\frac{{{{\rm{D}}^2}}}{4} - \frac{{{{\rm{y}}^2}}}{4}} \right)}}{{2{\rm{I}}}}$

Note that more than 85% of shear in the I section is resisted by the web.

Thus because most of the shear stress is taken by the web of a steel section, hence shear capacity of the flange is neglected.

Important Point:

Shear stress distribution for various important sections are:

Diamond:

Circular

H/Plus Beam:

Punched section:

Butt Weld: In this joint, the parts lie in the same plane and are joined at their edges. This joint is designed if members are to be subjected to direct tension or compression.

Fillet Weld: A fillet weld is used to fill in the edges of the plates created by corner, lap and tee joints. It is adopted when two members are to be welded at 90° to each other. e.g. Single J, Single bevel, square, etc.

Slot Weld: It is used for attaching flat plates using one or more holes or slots in the top part and then filling with filler metal to fuse the two parts together.

Seam Weld: In seam welding, overlapping sheets are gripped between two wheels or roller disc electrodes and current is passed to obtain either the continuous seam. Hence pressure is continuously applied for welding the two members.

For a given amount of weld material, the fillet weld is not as strong as due to the heavy concentration of the stress.

∴ For fillet weld, the weakest plane is along the throat section.

Concept:

Impact strength: It is also called as toughness i.e. resistance against impact loading.

Toughness is the total strain energy that can be stored in a given volume of material up to fracture limit and it is equal to the area of load-deflection curve up to fracture limit.

Tensile strength: It is the resistance against pull applied in longitudinal direction on material. It is the maximum amount of stress developed in the material under tension before it fails or stress at the fracture point in stress-strain curve under tension.

Yield strength: It is defined as the strength at first fibre failure and at which the material begins to change its shape.

Fatigue strength: In case of cyclic loading, in every cycle, certain amount of strain energy is stored in the specimen which is not released due to fast rate of loading.

After large no. of such cycles, the stress accumulated exceeds the modulus of toughness and specimen fails due to fatigue.

Hence, Fatigue strength is that level of stress, below which there is very high probability that structure will not fail after large no. of repetitions (approx. 10⁶ cycles).

The design stress in case of cyclic loading is known as endurance limit.

Concept:

The Value of Effective throat thickness either depends upon K values which depend on the angle between fusion faces or depend upon the size of the weld.

Throat Thickness (Tt) = K × Size of the weld

The K values vary as mentioned in the table below

Explanation:

If the angle between fusion faces of a fillet weld is 60° - 90°, the effective throat thickness as per Indian standard is equal to $\frac{1}{{\sqrt 2 }}$ size of weld.

Butt Weld: In this joint, the parts lie in the same plane and are joined at their edges. This joint is designed if members are to be subjected to direct tension or compression.

Fillet Weld: A fillet weld is used to fill in the edges of the plates created by corner, lap and tee joints. It is adopted when two members are to be welded at 90° to each other. e.g. Single J, Single bevel, square, etc.

Slot Weld: It is used for attaching flat plates using one or more holes or slots in the top part and then filling with filler metal to fuse the two parts together.

Seam Weld: In seam welding, overlapping sheets are gripped between two wheels or roller disc electrodes and current is passed to obtain either the continuous seam. Hence pressure is continuously applied for welding the two members.

Concept:

In the force method of analysis:

Primary unknown are forces in the members, and compatibility equations are written for displacement and rotations (which are calculated by force displacement equations) in this method.

Solving these equations, redundant forces are calculated. Once the redundant forces are calculated, the remaining reactions are evaluated by equations of equilibrium.

In the displacement method of analysis:

Primary unknowns are the displacements and initially force -displacement relations are computed and subsequently equations are written satisfying the equilibrium conditions of the structure in this method. After determining the unknown displacements, the other forces are calculated satisfying the compatibility conditions and force displacement relations.

Difference between Force & Displacement Methods

Concept:

Shear stress distribution is given by-

${\rm{\tau }} = \frac{{{\rm{SA\bar y}}}}{{{\rm{I}} \cdot {\rm{b}}}}$

where,

τ = Shear stress.

S = Shear force at section where stress is required.

A = Cross-section area above or below the point at which stress is required.

y̅ = Centroidal distance of area A from neutral axis,

I = Area moment of inertia of cross-section about neutral axis, and

b = Width of section at the point where stress is required.

Calculation:

 

${\rm{A}} = \frac{1}{2} \times {\rm{b'}} \times {\rm{y}}$

${\rm{\bar y}} = \frac{{2{\rm{d}}}}{3} - \frac{{2{\rm{y}}}}{3}$

${\rm{I}} = \frac{{{\rm{b}}{{\rm{d}}^3}}}{{36}}$

Substituting the above values:

${\rm{\tau }} = \frac{{{\rm{S}} \times \left( {\frac{1}{2}{\rm{b'y}}} \right) \times \left( {\frac{{2{\rm{d}}}}{3} - \frac{{2{\rm{y}}}}{3}} \right)}}{{\frac{{{\rm{b}}{{\rm{d}}^3}}}{{36}} \times {\rm{b'}}}}{\rm{;}}$

${\rm{\tau }} = \frac{{18{\rm{Sy}}\left( {\frac{{2{\rm{d}}}}{3} - \frac{{2{\rm{y}}}}{3}} \right)}}{{{\rm{b}}{{\rm{d}}^3}}}$

Now, to maximize shear stress, τ' = 0

$\frac{{2{\rm{d}}}}{3} - \frac{{2{\rm{y}}}}{3} \times 2 = 0$

${\rm{y}} = \frac{{\rm{d}}}{2}$

Now using above equation:

${\rm{\tau }} = \frac{{18{\rm{S}} \times \frac{{\rm{d}}}{2} \times \left( {\frac{{2{\rm{d}}}}{3} - \frac{{2{\rm{d}}}}{{2 \times 3}}} \right)}}{{{\rm{b}}{{\rm{d}}^3}}}$

${\rm{\tau }} = \frac{{3{\rm{s}}}}{{{\rm{bd}}}}$  (Maximum shear stress)

Now, average shear stress (τ_avg­)

${{\rm{\tau }}_{{\rm{avg}}}} = \frac{{{\rm{Total;shear;force}}}}{{{\rm{Total;area}}}}$

${{\rm{\tau }}_{{\rm{avg}}}} = \frac{{\rm{s}}}{{\frac{{{\rm{bd}}}}{2}}}$

${{\rm{\tau }}_{{\rm{avg}}}} = \frac{{2{\rm{s}}}}{{{\rm{bd}}}}$

Now, ratio of maximum to average shear stress:

$\frac{{\rm{\tau }}}{{{{\rm{\tau }}_{{\rm{avg}}}}}} = \frac{{\frac{{3{\rm{s}}}}{{{\rm{bd}}}}}}{{\frac{{2{\rm{s}}}}{{{\rm{bd}}}}}}$

τ/τ_avg­ = 1.5

Shear stress about neutral axis = $\frac{4}{3} \times {{\rm{\tau }}_{{\rm{avg}}}} = \frac{4}{3} \times \frac{{2{\rm{S}}}}{{{\rm{bd}}}} = {\rm{;}}\frac{{8{\rm{S}}}}{{3{\rm{bd}}}}$

Concept:

Slenderness ratio: It is defined as the ratio of the effective length of the column to the radius of gyration corresponding to the minor axis. It is denoted as λ.

Given,

Cantilever beam

This is a case of one end free and the other end fixed. In this case, the effective length is twice the length of the column.

Leff = 2L and slenderness ratio= 2L/r

 

Following are the different values of effective slenderness ratios for different arrangements:

<br>

Clearly, for the cantilever column; i.e. fixed at one end and free from other ends, the value of slenderness ratio is KL/R = 2L/R.

Explanation:

As it is a Semi-Circular Beam Length = 2 R

The semi-circular fixed beam can be depicted as the horizontal beam length of ‘2R’.

The standard result for fixed end moments can be used to obtain the fixed end moments at the supports.

Fixed end moment:

\({{\rm{M}}{{\rm{AB}}}} = {{\rm{M}}{{\rm{BA}}}} = \frac{{{\rm{W}}{\ell ^2}}}{{12}} = \frac{{{\rm{W}} \times {{\left( {2{\rm{R}}} \right)}^2}}}{{12}} = \frac{{{\rm{W}} \times {{\rm{R}}^2}}}{3}\)

Important Point:

Fixed end moments developed due to various load combinations is given below:

 

Concept:

The two theorems of Castigliano's for structural analysis are as follows:

Castigliano’s 1^st theorem: The partial derivative of total strain energy of the system with respect to any particular deflection at a point is equal to the force applied at that point in the same direction as that of the deflection.

$\frac{{\partial {\rm{u}}}}{{\partial {\rm{\delta }}}} = {\rm{w}}$

Castigliano’s 2^nd theorem: The partial derivative of strain energy of the system with respect to load at any point is equal to deflection at that point.

$\frac{{\partial {\rm{u}}}}{{\partial {\rm{w}}}} = {\rm{\delta }}$

Also, the partial derivative of strain energy of the system with respect to couple at any point is equal to slope at that point.

$\frac{{\partial {\rm{u}}}}{{\partial {\rm{M}}}} = {\rm{\theta }}$

Where,

u is the strain energy of the system.

Note:

These theorems are valid in both the beams and truss, but in truss strain energy is only due to axial loads.

Concept:

Primary stresses are mainly force driven and are caused by imposed mechanical loading on the structure and excessive primary stress causes yielding and plastic deformation of the material. They are not self limiting. e.g Dead and Live load on a cantilever beam etc.

Secondary stresses are mainly displacement driven and are caused due to geometric discontinuity and stress concentration. They are self limiting and beyond a certain stage the stresses begin to decrease. e.g Thermal loads, Eccentric loads etc.

Eccentric connection: If the applied forces do not pass through the C.G of joint, it causes an additional moment about the C.G and these types of connections are called eccentric connections.

The stresses induced due to these eccentric connections are known as secondary stresses because they arise due to geometric dissimilarity.

Here eccentricity ‘e’ introduces a moment Pe and secondary stresses are generated consequently.'

Eccentric connection in lap joint

Here also secondary stresses are generated due to eccentricity in connections.

Let the vertical reactions at A and B be R_A and R_B respectively.  Similarly, the horizontal reactions at A and B are H_A and H_B respectively.

∑F_y = 0

⇒ R_A + R_B = W     --- (i)

∑M_A = 0 

⇒ W × a - R_B × 2 L = 0

$\Rightarrow {{\rm{R}}_{\rm{B}}} = \frac{{{\rm{wa}}}}{{2{\rm{L}}}}$

$\Rightarrow {{\rm{R}}_{\rm{A}}} = \left( {{\rm{w}} - \frac{{{\rm{wa}}}}{{2{\rm{L}}}}} \right)$

∑F_x = 0

⇒ H_A = H_B

∵ There is a hinge at crown C, the bending moment at C must be zero.

∑ M_C on the BC part of arch = 0

⇒ H_B × h - R_B × ℓ = 0

$\Rightarrow {{\rm{H}}_{\rm{B}}} \times {\rm{h}} - \frac{{{\rm{wa}}}}{{2\ell }} \times \ell = 0$

$\Rightarrow {{\rm{H}}_{\rm{B}}} = \frac{{{\rm{wa}}}}{{2{\rm{h}}}}$

Hence \({{\rm{H}}{\rm{A}}} = {{\rm{H}}{\rm{B}}} = \frac{{{\rm{wa}}}}{{2{\rm{h}}}}\)

Important Point:

A three hinged arch is a statically determinate structure with two supports hinged and an internal hinge provided in the body of the arch generally at the crown. Such arches are analysed using equations of static equilibrium only.

As per IS 456: 2000, clause 26.5.3.1,

Longitudinal reinforcement in a column:

Percentage of steel:

The cross-sectional area of longitudinal reinforcement shall be not less than 0.8 percent and not more than 6 percent of the gross cross-sectional area of the column.

Minimum diameter of longitudinal reinforcement bars:

The main longitudinal reinforcement bars used in column shall not be less than 12 mm in diameter.

Minimum number of longitudinal bar:

The minimum number of the longitudinal bar provided in column shall be

• four in rectangular columns
• six in circular columns

Helical reinforcement:

A reinforced concrete column having helical reinforcement shall have at least six-bar of longitudinal reinforcement within the helical reinforcement.

Spacing:

The spacing of longitudinal bars measured along the periphery of the column shall not exceed 300 mm.

Development Length:

A development length is the amount of rebar length that is needed to be embedded or projected into concrete to create desired bond strength between the two materials and also to develop required stress in steel at that section.

As per IS 456: 2000, clause 26.2.1,

The development length is given by: 

${L_d} = \frac{{\phi {\sigma _s}}}{{4{\tau _{bd}}}}$

where

ϕ = nominal diameter of the bar

σs = stress in the bar at the section considered at design load

τbd = design bond stress

Points to be remembered:

1. The development length includes anchorage values of hooks in tension reinforcement.

2. For bars other than circular cross-section, the development length should be sufficient to develop the stress in the bar by the bond.

Concept:

${f_m} = {f_{ck}} + 1.65;σ$

Where,

fm is target mean compressive strength

fck is the characteristics strength of concrete

σ is the standard deviation

The standard deviation for different grade of concrete as per Clause 9.2.4.2 IS:456-2000

Concept:

Energy loss in a hydraulic jump is given as

${{\rm{E}}_{\rm{L}}} = \frac{{{{\left( {{{\rm{y}}_2} - {{\rm{y}}_1}} \right)}^3}}}{{4{{\rm{y}}_1}{{\rm{y}}_2}}}$

Where, y₁ & y₂ are conjugate depths of flow.

Given: y₁ = 0.2 m, y₂ = 1.2 m

$\therefore {{\rm{E}}_{\rm{L}}}{\rm{;}} = {\rm{;}}\frac{{{{\left( {1.2 - 0.2} \right)}^3}}}{{4 \times 1.2 \times 0.2}}$

E_L = 1.042 m

Important Point:

Conjugate depth: The conjugate depths refer to the depth upstream and the depth downstream of the hydraulic jump whose momentum fluxes are equal for a given discharge. 

When the flow condition changes from supercritical (F_r > 1) to subcritical (F_r < 1), the result is an abrupt rise in water surface accompanied by turbulent rollers. This phenomenon is called hydraulic jump and is used to dissipate excess available energy in the flow to avoid scouring of bed.

Concept:

Any flow in open channel is classified on the basis of Froude’s number.

Froude’s number > 1 ⇒ super critical flow

Froude’s number = 1 ⇒ critical flow

Froude’s number < 1 ⇒ subcritical flow

For rectangular channel

Froude’s number $=\frac{{\rm{V}}}{{\sqrt {\frac{{\rm{g}}}{{\rm{A}}}{\rm{/T}}} }} = \frac{{\rm{V}}}{{\sqrt {{\rm{g}} \times \left( {\frac{{{\rm{By}}}}{{\rm{B}}}} \right)} }} = \frac{{\rm{V}}}{{\sqrt {{\rm{gy}}} }}$

Calculation:

Given: y = 1.6 m, and E = 2.7 m

We know, specific energy

${\rm{E}} = {\rm{y}} + \frac{{{{\rm{v}}^2}}}{{2{\rm{g}}}}$

$2.7 = 1.6 + \frac{{{{\rm{v}}^2}}}{{2 \times 9.81}}$

⇒ v = 4.65 m

Froude’s number $= \frac{{4.65}}{{\sqrt {9.81 \times 1.6} }};$= 1.17

∵ Froude’s number > 1

⇒ Super critical flow.

Froude’s number= $\frac{{\rm{v}}}{{\sqrt {{\rm{g}} \times \left( {\frac{{\rm{A}}}{{\rm{T}}}} \right)} }}$

Where,

A = Area of cross-section, and T = Top width

For a triangular channel as stated above

${{\rm{F}}_{\rm{r}}} = \frac{{\rm{v}}}{{\sqrt {\frac{{{\rm{gA}}}}{{\rm{T}}}} }}$

T = 2y + 2y = 4y

${\rm{A}} = \frac{1}{2} \times 4{\rm{y}} \times {\rm{y}} = 2{y^2}$

$\therefore {{\rm{F}}_{\rm{r}}} = \frac{{\rm{v}}}{{\sqrt {{\rm{g}} \times \frac{{2{{\rm{y}}^2}}}{{4{\rm{y}}}}} }}$

${{\rm{F}}_{\rm{r}}} = \frac{{\rm{v}}}{{\sqrt {{\rm{g}}\left( {\frac{{\rm{y}}}{2}} \right)} }}$

Concept:

According to Lacey, the design formulas to build canal is as follows:

1) silt factor ⇒ $\rm{f = 1.76\sqrt {{d_{mm}}}}$

2. Lacey's Normal scour depth ⇒ $R_r' = 1.35{\left[ {\frac{{q^2}}{{f}}} \right]^{\frac{1}{3}}}$

3. velocity of flow ⇒$V = {\left[ {\frac{{Q{f^2}}}{{140}}} \right]^{\frac{1}{6}}}$

4) hydraulic mean depth ⇒$R = \frac{{5{V^2}}}{{2f}}$

5. wetted perimeter ⇒ $P = 4.75\sqrt Q$

6) bed slope ⇒$S = \frac{{{f^{\frac{5}{3}}}}}{{3340 \times {Q^{\frac{1}{6}}}}}$

Concept:

Gravity Dam: Gravity dams are massive structure dam which is constructed of concrete or stone masonry. These dams are held by the gravity to the ground and resist the water force by its weight. As they rely on their weight, it is necessary to construct them on a solid foundation of bedrocks.

It is designed such that there is no tension anywhere in the dam as the concrete is weak in tension. It should also resist sliding and overturning to avoid any kind of failure.

1. No tension condition:

Let resultant of self-weight (W), hydrostatic force (P), and uplift force (U) passes through the outermost point of middle 3^rd portion.

Moment about B = ${\rm{W}} \times \frac{{\rm{B}}}{3} - {\rm{U}} \times \frac{{\rm{B}}}{3} - {\rm{P}} \times \frac{{\rm{H}}}{3}{\rm{;}} = {\rm{;}}0$

⇒ WB – UB = PH

\( \Rightarrow \frac{1}{2}{{\rm{B}}^2}{\rm{H}}{{\rm{\gamma }}{{\rm{concrete}}}} - \frac{1}{2}{\rm{C}}{{\rm{\gamma }}{\rm{w}}}{{\rm{B}}^2}{\rm{H}} = \frac{1}{2}{{\rm{\gamma }}_{\rm{w}}}{{\rm{H}}^3}\)

\( \Rightarrow \frac{1}{2}{{\rm{B}}^2}{\rm{H}}\left( {{{\rm{\gamma }}{\rm{w}}}} \right)\left( {{{\rm{G}}{\rm{c}}} - {\rm{C}}} \right) = \frac{1}{2}{{\rm{\gamma }}_{\rm{w}}}{{\rm{H}}^3}\)

$\Rightarrow {\rm{B}} = \frac{{\rm{H}}}{{\sqrt {{{\rm{G}}_{\rm{c}}} - {\rm{C}}} }}$

2. No Sliding Condition:

\({{\rm{\mu }}{\rm{v}}} > {\rm{\Sigma }}{{\rm{F}}{\rm{H}}}\)

⇒μ (W – U) > P

\( \Rightarrow {\rm{\mu }}\left( {\frac{1}{2}{\rm{BH}}{{\rm{\gamma }}{{\rm{concrete}}}} - \frac{1}{2}{\rm{C}}{{\rm{\gamma }}{\rm{w}}}{\rm{BH}}} \right) > \frac{1}{2}{{\rm{\gamma }}_{\rm{w}}}{{\rm{H}}^2}\)

\( \Rightarrow \frac{1}{2}{\rm{HB}}{{\rm{\gamma }}{\rm{w}}}{\rm{\mu ;}}\left( {{{\rm{G}}{\rm{c}}} - {\rm{C}}} \right) > \frac{1}{2}{{\rm{\gamma }}_{\rm{w}}}{{\rm{H}}^2}\)

$\Rightarrow {\rm{B}} > \frac{{\rm{H}}}{{{\rm{\mu }}\left( {{{\rm{G}}_{\rm{c}}} - {\rm{C}}} \right)}}$

Neglecting the uplift pressure (c) in both cases i.e. for no tension and no sliding conditions;

${\rm{B}} > \frac{{\rm{H}}}{{\sqrt {{{\rm{G}}_{\rm{c}}}} }}$      ----(i)

${\rm{B}} > \frac{{\rm{H}}}{{{\rm{\mu }}{{\rm{G}}_{\rm{c}}}}}$      ----(ii)

Hence base width should be greater of $\frac{{\rm{H}}}{{\sqrt {\rm{G}} }}{\rm{;and}}\frac{{\rm{H}}}{{{\rm{\mu G}}}}$

Concept:

Kennedy carried out his research in the upper Baridoab canal system in Punjab and introduced the term critical velocity ‘V₀’ which will keep the channel free from silting and scouring. This is given as:

${{\rm{V}}_0}{\rm{;}} = {\rm{;}}0.55{\rm{;m;}}{{\rm{y}}^{0.64}}$

where_,

y = Depth of flow in ‘m’, and m = critical velocity ratio, whose value depends on the type of soil.

m > 1 ; for coarse-grained soil.

m < 1 ; for fine grained soil.

Kennedy suggested using Kutter’s equation to find the actual velocity of flow.

Kuttler’s equation is given as:

$V; = ;\left[ {\frac{{23 + \frac{{0.00155}}{S} + \frac{1}{n}}}{{1 + \left( {\frac{{23 + 0.00155}}{S}} \right)\frac{n}{{\sqrt R }}}}} \right]\sqrt {RS}$

A lot of mathematical calculations are required in designing irrigation channels by Kennedy’s method. To avoid this a graphical solution was suggested by Garrett for Kennedy’s and Kutter’s equation.

Garret’s diagram is a graphical representation of the design of canal dimensions based on Kennedy's theory. The diagram is discharge plotted on the abscissa (X-axis), and slope of the channel on the left side of an ordinate (Primary Y-axis) and depth of the channel and critical velocity on the right side of the ordinate. (Secondary Y-axis).

Garret’s diagram is applicable only for the side slope of the channel of 0.5:1.

Concept:

Critical Flow: It is defined as the flow for which the specific energy is minimum and Froude’s number is equal to unity.

Froude’s no. (F_r) = $\sqrt {\frac{{{\rm{Inertial;force}}}}{{{\rm{gravity;force}}}}}$

Where,

Inertial force = ρ × Q × v

(ρ = Density, Q = Discharge, v = velocity)

Gravity force = ρ × g × V;

(ρ = Density, g = Gravity acceleration, V = Volume)

Now,

${{\rm{F}}_{\rm{r}}} = \sqrt {\frac{{{\rm{\rho Qv}}}}{{{\rm{\rho gV}}}}} = \sqrt {\frac{{{\rm{Qv}}}}{{{\rm{gAL}}}}} = \sqrt {\frac{{{\rm{Q}}\left( {\frac{{\rm{Q}}}{{\rm{A}}}} \right)}}{{{\rm{gA}}\left( {\frac{{\rm{A}}}{{\rm{T}}}} \right)}}} = \sqrt {\frac{{{{\rm{Q}}^2}{\rm{T}}}}{{{\rm{g}}{{\rm{A}}^3}}}}$

 (∵ A = L × T, ν = Q/A)

A = Area, L = Length, and T = Top width

Now, for a critical flow F_r = 1

$\therefore {{\rm{F}}_{{{r}}}} = 1$

$\Rightarrow \frac{{{{\rm{Q}}^2}{\rm{T}}}}{{{\rm{g}}{{\rm{A}}^3}}} = 1$

Important points:

For a critical flow:

Specific energy is minimum for a given discharge.

Discharge is maximum for a given specific energy.

Specific force is minimum for a given discharge.

Concept:

If pipe is rigid, then pressure wave will travel in the medium at a velocity C equal to the velocity of sound which is given by:

${\rm{C}} = \sqrt {\frac{{\rm{K}}}{{\rm{\rho }}}}$

Where, K = bulk modulus of fluid in MPa and ρ = density of fluid in kg/m³.

Hence, the velocity of pressure wave is proportional to 1/√ρ

Important Point:

Pressure waves are generated due to the water hammer action which is caused due to sudden closure of valve.

Due to this the momentum of water gets destroyed and hence a pressure wave is generated which will travel in opposite direction.