BPSC AE Paper 5 (Civil Engineering) 2019 Official Paper (Previous Year Paper)

Concept:

Initially the horizontal thrust is towards right at A.

When support B sinks.

∑ M_B = 0

V_A × 1 – H_A × δ  = 0

$\Rightarrow {H_A} = \frac{{{V_A} \cdot l}}{\delta }$

So, H_A is acting towards left

So overall horizontal thrust will decrease.

Mistake Points When the support sinks/settles the horizontal thrust acts in the opposite direction, hence the horizontal thrust gets decreased. 

Alternate Method:

The equation for horizontal thrust in a 2-hinged parabolic arch is

$H; = ;\frac{{\smallint \frac{{{M_x}ydx}}{{EI}}; + ;\alpha tl}}{{\smallint \frac{{{y^2}dx}}{{E{I_C}}}; + ;\frac{l}{{AE}}; + ;k}}$

Where A = Horizontal thrust

E = Modulus of elasticity of the arch material

I = moment of inertia of Arch cross-section

A = Area of cross-section of the arch

α = Coefficient of thermal expansion

t = Difference of temperature

l = length of arch, k = yielding of supports

Here there is no clear mention of temperature Change, (t = 0),

so the horizontal thrust will decrease as the support sinks, i.e k has some value.

And H inversely proportional to k

Concept:

Suppose 4 loads P1, P2, P3, P4 move in a series. Maximum bending moment will occur under P2 when the centre of span is midway between the centre of gravity of the load system R and the wheel load P2.

A maximum bending moment under the wheel load is always below some point load. To find maximum bending moment below a point load, the point load should be placed so that the resultant load and the point load are equidistant from the mid-span.

To find absolute bending moment following steps are followed:

Step 1: Find the resultant of all the loads.

Step 2: Place the individual load as per the condition of absolute maximum bending moment and compute the bending moment under the concentrated load.

Step 3: The maximum of all the possible cases will give the absolute maximum bending moment.

Here the maximum bending moment below the load is obtained.

Important point:

The influence line diagram represents the variation of the stress function at a specified point in a member as a concentrated load moves over the member.

To find the value of stress function, the value of the ordinate of ILD at the location of point load is multiplied by the value of point load.

Coefficient of consolidation is the rate at which saturated soil undergoes consolidation. It can be only measured in laboratory.

Time factor is given by ${T_v} = \frac{{{C_v}t}}{{{d^2}}}$.

From this equation, the unit of coefficient of consolidation comes out to be cm²/sec.

Concept:

Dry density: The dry density is the mass of solids per unit area of total volume of the soil mass.

${\rho _d} = \frac{{{M_d}}}{V}$

Saturated density: When the soil mass is saturated its bulk density is called saturated density. Thus, saturated density is the ratio of the total mass of the saturated sample to its total volume.

Submerged density: The submerged density is the submerged mass of soil solids per unit of total volume V of the soil mass.

Submerged density (ρ’) = ρ_sat - ρ_w

In saturated soil, since voids are completely filled with water hence mass of soil is very high.

In wet soil, voids are partially filled with water hence mass of wet soil is less than saturated

In dry soil, voids are filled with air only hence mass of dry soil is less than the wet & saturated.

∴ Saturated density (saturated soil mass) > wet density (partially saturated soil mass) > dry density (dry soil mass)

When a soil in water is submerged, buoyant force reduces the weight of soil or mass of soil due to which submerged density becomes very less.

So the order will be

Saturated density > wet density > dry density > submerged density

Tricks to remember:

1. ${\gamma_t} = \left( {\frac{{G + Se}}{{1 + e}}} \right){\gamma_w}$

2. ${\gamma_{sat}} = \left( {\frac{{G + e}}{{1 + e}}} \right){\gamma_w}$

3. ${\gamma_d} = \frac{{G{\gamma_w}}}{{1 + e}} = \frac{{{\gamma_t}}}{{1 + w}} = \frac{{\left( {1 - {\eta _a}} \right)G{\gamma_w}}}{{1 + wG}}$

4. ${\gamma'} = \frac{{\left( {G - 1} \right)}}{{1 + e}}{\gamma_w}$

Where G = Specific gravity of soil

E = void ratio

S = degree of saturation

From the above four formula the order can be decided mathematically.

For all the densities the Denominator is constant.

From the numerator values the order can be decided for a constant value of G and e.

1. Open Drive Thin-Walled Sampler:

• This is a basic type of sampler. The tube is advanced into the soil by applying weight and turning it. It can provide relatively undisturbed samples of fine sands, silts, and clays.
• The advantage of this sampler is its simplicity and low cost. However, in saturated conditions, it can cause considerable disturbance to the soil structure.

2. Standard Split-Spoon Sampler:
3. This is the most common type of soil sampler and is used in Standard Penetration Testing (SPT).
4. It consists of a split tube that can be opened lengthwise to retrieve the sample. It's driven into the soil by a hammer and provides a disturbed sample, which is suitable for identification purposes and providing a measure of density or strength in coarser soils. In finer, saturated soils, it can cause significant disturbance.
5. Stationary Piston Sampler:
6. This sampler works under a principle that minimizes the disturbance of the soil sample.
7. The piston is held stationary at the bottom of the sample tube while the tube is advanced into the soil. This way, the soil enters the tube with minimal disturbance.
8. It's especially effective in conditions where the soil is soft and saturated, as it can minimize the remoulding of the soil and maintain the structure of the sample.
9. Rotary Sampler:

• This method is used to obtain soil samples from hard, stiff, or rocky soils that cannot be sampled using the traditional sampling methods.
• A drill bit is used to break down the soil and bring it to the surface.
• However, this method is unsuitable for soft, saturated soils as it greatly disturbs the soil structure and is likely to cause soil liquefaction.

Concept:

The velocity potential function may be defined as a scalar function of space and time such that its derivative with respect to any direction gives the fluid velocity in that direction.

Velocity potential function (φ) satisfies the Laplace equation.

So $\frac{{{\partial ^2}\phi }}{{\partial {x^2}}} + \frac{{{\partial ^2}\phi }}{{\partial {y^2}}} = 0$ 

The solution of the above equation gives two sets of curves known as equipotential lines and stream lines (or flow lines), mutually orthogonal to each other as shown in figure.

The equipotential lines represent contours of equal head (potential).

The direction of seepage is always perpendicular to the equipotential lines.

The path along which the individual particles of water seep through the soil are called streamlines or flow lines.

Concept:

Permeability is defined as the property of a porous material that permits the passage or seepage of water (or other fluids) through its interconnecting voids.

Factors affecting the permeability of soil are

1. Grain size
2. Properties of the pore fluid.
3. Voids ratio of the soil.
4. The structural arrangement of the soil particles.
5. Entrapped air and foreign-matter
6. Adsorbed water in clayey soils

The structural arrangement of the particle may vary, at the same voids ratio, depending upon the method of deposition or compacting the soil mass. The structure may be entirely different for a disturbed sample as compared to an undisturbed sample which may process stratification. The effect of structural disturbance on permeability is much pronounced in fine-grained soils.

A sample of clay and a sample of sand has the same specific gravity and void ratio. Their permeabilities would differ because the size ranges of their voids would be different.

Concept:

In spring analogy, the grain structure represents the spring while the voids filled with water represent the cylinder. The valve opening is represented by the permeability of the soil mass, and the rate of load transfer from water to soil depends upon the permeability and the boundary conditions (i.e. the drainage faces available).

The pressure that builds up in pore water due to load increment on the soil is termed excess pore pressure or excess hydrodynamic pressure u, because it is in excess of the initial pressure in water under static condition. The excess hydrostatic pressure forces the water to drain out of the voids.

As water starts escaping from the voids, the excess hydrostatic pressure in water gets gradually dissipated and the pressure increment is shifted as an increase in effective pressure on the soil solids and the soil mass decreases in volume.

So when the drainage layer is provided at top the pressure of water at top will be less and pore water pressure will be maximum at the bottom of the soil layer.

Concept:

Time required for consolidation is given by,

$t = \frac{{{d^2}}}{{{c_v}}}{T_v}$

Where, t = time required for consolidation

d = length of drainage path

e_v = co-eff of consolidation

T_v = time factor

${T_v} = \frac{\pi }{4}{\left( {\frac{{v\% }}{{100}}} \right)^2}$ when U < 60%

T_v = 1.7813 – 0.9332 log₁₀ (100 – U%) when U > 60%

Where U = degree of consolidation

Calculation:

Thickness of clay layer = 10 m

Sand layer is on both top & bottom.

So, Two-way drainage will occur.

% of consolidation = 50%

Coefficient of consolidation (c_v) = 0.002 cm² / s = 0.002 × 10^-4 m² / s

For 50% consolidation, T_v = 0.197

$t = \frac{{{{\left( 5 \right)}^2} \times 0.197}}{{0.002 \times {{10}^{ - 4}}}}\sec = 2462.5 \times {10^4}\sec = \frac{{2462.5 \times {{10}^4}}}{{24 \times 60 \times 60}}$

t ≈ 284 days

Note:

T_v is constant for a given degree of consolidation and given boundary conditions.

The angle of friction between soil particle depends upon the shape, size, roughness, grading, and packaging of particles.

For Sand, particles size and shape roughly remain the same and thus it cannot be used to estimate the angle of friction of sandy soil.

However, the angle of internal friction would be very high if the sand particle is compactly dense which is indicated by the value of the density index.

Density index is the ratio of the difference between the void ratios of a cohesionless soil in its loosest state and existing natural state to the difference between its void ratio in the loosest and densest states.

$Density;Index;/;Relative;Density; = \frac{{{e_{max}} - {e_{natural;}}}}{{{e_{max}} - ;{e_{min;}}}};$

The higher value of the relative index indicates a higher angle of internal friction.

Sheet pile:

A sheet pile wall consists of a series of sheet piles driven side by side into the ground, thus forming a continuous vertical wall for the purpose of retaining an earth bank.

They are commonly used for waterfront structures temporary construction and lightweight construction where sub-soil is poor for supporting a retaining wall.

A sheet pile wall may be of three types.

1. Cantilever sheet piling
2. Anchored sheet pilling
3. Braced sheeting.

Certain configurations of sheet pilling are termed as bulkheads or cofferdams. A bulkhead is a sheet pile retaining wall of water front, backed up ground.

A cantilever sheet pile wall or bulkhead derives its stability entirely from the lateral resistance of the soil into which it is driven.

The bulkhead is adequately embedded into the soil below the dredge into so that a driven line of sheeting acts as wide cantilever beam in resisting the lateral earth pressures developed above the dredge line. A cantilever bulkhead is used for moderate height only.

Concept:

Lateral earth pressure is the pressure that soil exerts against a structure in a sideways, mainly horizontal direction. Since most open cuts are excavated in stages within the boundaries of sheet pile walls or walls consisting of soldier piles and laggings and since struts are inserted progressively as the excavation proceeds, the walls are likely to deform (as shown in the figure).

Little inward movement can occur at the top of the cut after the first strut is inserted. As the earth pressure magnitude increases due to an increase in effective stress while moving from top to bottom, so deflection of a sheet pile in a braced cut increases from top to bottom.

Concept:

Degree of consolidation (U) $= \frac{{{\rm{\Delta }}h}}{{{\rm{\Delta }}H}} \times 100$ 

Δh = settlement of soil after time ‘t’.

ΔH = Ultimate consolidation settlement when U = 100%

Δh will depend on two way or one-way drainage condition

Time factor [T_V]

${T_V} = \frac{\pi }{4}{\left( U \right)^2},;when;U \le 60\%$

T_V = (-0.9332 log₁₀ (1 - U) – 0.0851) when U > 60%

${T_V} = \frac{{{C_v}t}}{{{d^2}}}$

Where,

d = length of drainage path

$d = \frac{{{H}}}{2}\left[ {Two;way;drainage} \right]$

d = H [one way drainage]

H = thickness of clay layer

t = time required for degree of consolidation equal to (U%)

C_V = coefficient of consolidation​

∴ t ∝ H² and t ∝ 1/C_v

Concept:

Correlation Between SPT and CPT

Meyerhof (1965) presented comparative data between SPT and CPT. For fine or silty medium loose to medium dense sands, he presents the correlation as

q_c = 0.4 N MN/m²

His findings are as given in Table 9.6.

Table 9.6 Approximate relationship between relative density of fine sand, the SPT, the static cone resistance and the angle of internal fraction (Meyerhof, 1965)

Calculation:

The value of static cone penetration test (q_c) = 8 MPa and

The N value is 20.

q_c = 8 MPa

So the type of soil is medium dense sand.

Concept:

Shear stress of a soil sample is given by,

τ = c + σ tan ϕ

Where, τ = shear strength of soil

σ = normal stress of soil

c = cohesion

ϕ = angle of internal friction or shearing resistance

Calculation:

Dry sand sample, So, C = 0

σ = τ = 120 kN / m²

τ = 0 + σ tan ϕ ⇒ 120 = 120 tan ϕ

⇒ tan ϕ = 1

⇒ ϕ = 45°

Concept:

In an unconfined compression test, the compressive stress is calculated on the basis of changed C / S area A₂ at failure.

${A_2} = \frac{V}{{{L_1} - {\rm{\Delta }}L}} = \frac{{{A_1}}}{{1 - \frac{{{\rm{\Delta }}L}}{L_1 }}} = \frac{{{A_1}}}{{1 - ϵ}}$

as, Volume = Area × Length

Where, V = initial vol. of the specimen

L₁ = initial length of the specimen

ΔL = change in length at failure

${ϵ} = \frac{{{\rm{\Delta }}L}}{{{L_1}}} =$ strain at failure.

Calculation:

Given,

 Initial C/S A₁ = 15 cm²

ϵ = 25% = 0.25

${A_2} = \frac{{15}}{{1 - 0.25}} = \frac{{15}}{{0.75}} = 20;c{m^2}$

∴ Corrected area = 20 cm²

Concept:

Time factor, \({{\rm{\tau }}{\rm{v}}} = \frac{{{{\rm{c}}{\rm{v}}}{\rm{t}}}}{{{{\rm{d}}^2}}}\)

Where,

cv = coefficient of consolidation, d = length of drainage, and t = time taken for consolidation

${{\rm{\tau }}_{\rm{v}}} = \frac{{\rm{\pi }}}{4} \times {{\rm{u}}^2}$

(Is u ≤ 60%) where u is degree of consolidation

Now:

$\frac{{\rm{\pi }}}{4}{{\rm{u}}^2} = \frac{{{{\rm{c}}_{\rm{v}}}{\rm{t}}}}{{{{\rm{d}}^2}}}$

$\Rightarrow {\rm{u}} = \sqrt {\frac{4}{{\rm{\pi }}}\frac{{{{\rm{c}}_{\rm{v}}}{\rm{t}}}}{{{{\rm{d}}^2}}}}$

Calculation:

For a given cv, and d ⇒ u ∝ √t

$\frac{{{{\rm{u}}_1}}}{{{{\rm{u}}_2}}} = \sqrt {\frac{{{{\rm{t}}_1}}}{{{{\rm{t}}_2}}}} \Rightarrow \frac{{30}}{{60}} = \sqrt {\frac{{100}}{{{{\rm{t}}_2}}}}$

⇒ t2 = 400 days

∴ Time taken for 60% consolidation is 400 days.

Concept:

Using the relation

${\sigma _1} = {\sigma _3}{\tan ^2}\left( {45^\circ + \frac{\phi }{2}} \right) + 2c\tan \left( {45^\circ + \frac{\phi }{2}} \right)$

Where,

σ1 = Major principal stress

σ3 = all round cell pressure

ϕ = Angle of internal friction of the soil

c = cohesion of the soil

Calculation:

For a sandy soil

C = 0

Now,

$\begin{array}{l} {\sigma _1} = {\sigma _3}{\tan ^2}\left( {45^\circ + \frac{\phi }{2}} \right) + 2c\tan \left( {45^\circ + \frac{\phi }{2}} \right)\ 50 = {\sigma _3}{\tan ^2}\left( {45^\circ + \frac{{30}}{2}} \right) + 0 \end{array}$

50 = σ₃ tan² (60°)

$\frac{{50}}{{{{\tan }^2}\left( {60^\circ } \right)}} = {\sigma _3}$

= 16.66 kN/m².

Concept:

Failure condition for an infinite slope of cohesionless soil is represented in above figure.

$FOS = \frac{{{\tau _f}}}{2} = \frac{{\tan \phi }}{{\tan i}}$

ϕ = angle of shearing resistance

i = slope angle

Calculation:

FOS = 1.723

ϕ = 30°

So, $\Rightarrow 1.723 = \frac{{\tan 30^\circ }}{{\tan i}}$

$\Rightarrow \tan i = \frac{{\tan 30^\circ }}{{1.723}} = 0.335$

⇒ i = tan^-1 (0.335)

Concept:

When assessing the strength of a structure or structural member for the limit state of collapse, the values of partial safety factor, γ_m should be taken as 1.15 for steel. Thus, in the limit state method, the design stress for steel reinforcement is given by,

fy/γ_m = fy/1.15 = 0.87fy

According to IS 456:2000 for design purpose, the compressive strength of concrete in the structure shall be assumed to be 0.67 times the characteristic strength of concrete in the cube and partial safety factor, γ_mc = 1.5 shall be applied in addition to this. Thus, the design stress in concrete is given by,

0.67f_ck / γ_mc=0.67f_ck /1.5=0.446 f_ck

The partial safety factors for loads, as per IS 456 : 2000 are given in table below:

(* This value is to be considered when stability against overturning or stress reversal is critical)

Concept:

Creep is the plastic permanent deformation of a structure under constant load for a very long period of time.

It can occur as a result of long-term exposure to high levels of stress that are still below the yield strength of the material. Creep is more severe in materials that are subjected to heat for long periods and generally increases as they near their melting point.

It occurs due to dead load only.

Factors affecting creep

1. Type of loading
2. Magnitude of load
3. Time
4. Temperature

If temperature is more, creep is more and if the temperature reaches to half of melting point temperature the creep is intolerable. Such a temperature is defined as homologous temperature.

Concept:

The bond between concrete and steel restrains shrinkage. Therefore, the shrinkage curvature is influenced by the area of tension and compression reinforcement equally.

So when there is only tension reinforcement is provided at the bottom end the curvature due to shrinkage deflection will be same as the curvature of the reinforcement.

As in a simply supported beam the beam deflects downward due to loading the reinforcement will also deflect downward.

So the deflection due to shrinkage stresses in a simply supported beam having reinforcement only at the bottom ends will occur in downward direction.

Important Points:

Shrinkage deflection can be eliminated by making tension steel equal to compression steel.

Concept:

Domes are self-supporting structures they take the form of an arch distributing external loads around the sides and down to the foundations. They are tightly compacted by gravity and any external loads are carried by the compressive forces that develop internally.

Meridional forces act in a longitudinal direction of a dome and increase from crown to base in magnitude. Meridional forces are due to the weight of the masonry and applied loads.

Hoop forces act in a latitudinal direction of a dome on the lateral faces of the voussoirs, and contribute significantly to the stability of domes and their construction.  For a dome with uniform axisymmetric loading, hoop forces from adjacent slices are equal in magnitude and act on the same plane.

In a spherical dome subjected to concentrated lead at crown or uniformly distributed load, the meridional force is always compressive.

Concept:

Prestressed concrete is basically concrete in which internal stresses of a suitable magnitude and distribution are introduced so that the stresses resulting from external loads are counteracted to a desired degree. In reinforced concrete members, the pre-stress is commonly introduced by tensioning the steel reinforcement.

Generally, due to external loading the bottom part of a beam is in tension and strength of concrete is less in tension.

So to counteract this problem the beams are generally pre-stressed at lower portion so that they will be in compression initially and after loading the tensile stress can be counteracted by the pre-stressing cables.

Traverse reinforcement/ties are provided in columns for the following reasons:

i) To prevent the premature buckling of individual bars

ii) To confine the concrete in the ‘core’, thus improving ductility and strength

iii) To provide resistance against shear and torsion is required

iv) To hold the longitudinal bars in position during constructions

∴ One of the purposes of the provision of lateral ties in columns is to avoid buckling of longitudinal bars.

Explanation:

Shear reinforcements in the form of vertical stirrups and or bent up longitudinal bars.

1. Side face reinforcement in the web of the beam provided when  the depth of the web in a beam exceeds 750 mm
2. It shall not be less than 0.1% of the web area
3. It should be distributed equally on two faces at a spacing not exceeding 300 mm or web thickness, whichever is less.

Deep beams are structural elements loaded as simple beams in which a significant amount of the load is carried to the supports by a compression force combining the load and the reaction. As a result, the strain distribution is no longer considered linear, and the shear deformations become significant when compared to pure flexure.

In view of ample shear strength, deep beams are primarily recommended as transfer girders. These members transfer loads through-loading face to supports in the transverse direction. The deep horizontal members predominantly fail in shear rather than flexure. These beams are characterized with small span-to-depth ratio. Pile caps, corbel, brackets, foundation walls and off-shore structures are few examples of RC deep beams.

According to IS 456-2000 a beam shall be deemed to be a deep beam when the ratio of effective span-to-overall depth, l/D is less than:

1. 2.0, for simply supported beam; and

2. 2.5, for a continuous beam.

Though different codes define deep beams in different clear span-to-depth ratios, as a general rule deep beams are recognized by their relatively small span-to-depth ratio.

Hence, Deep beams are designed for bending moment and checked for shear.

IS 456:2000 page 51.

Vertical Bunker:

The bunkers are large size shallow bins to store grains, coal, and cement. In bunkers, the plane of rupture intersects the free surface of the stored material. Generally, steel bunkers are used to store coal at power plants and loco-running sheds. Generally, these are square or rectangular shaped.

Maximum bending moment at corner or edge = pl2/12

Maximum bending moment at center = pl2/24

Minimum clear cover requirement:

1. Slab = 20 mm
2. Beam = 25 mm
3. Column = 40 mm
4. Footings = 50 mm

For Slab:

For Mild exposure – 20 mm, and For Moderate exposure – 30 mm

However, if the diameter of the bar does not exceed 12 mm, or cover may be reduced by 5 mm. Thus for main reinforcement up to 12 mm diameter bar and for mild exposure, the nominal cover is 15 mm.

For Column:

For a longitudinal reinforcing bar in a column, concrete cover not less than 40 mm not less than the diameter of such a bar should be provided. In the case of columns of the minimum dimension of 20 cm or under, whose reinforcing bars do not exceed 12 mm, the concrete cover of 25 mm to be used for reinforcement.

Concept:

Spacing between the bars for 1 m(1000 mm) width:

${s} = \frac{a}{{{A_{st}}}} \times 1000$

Where, a = Area of one bar = $\frac{\pi \ \times \ ϕ^2}{4}$

ϕ = Diameter of bar

s = spacing of bars

A_st = Area of total main reinforcement

Calculations:

Given, 

Case 1: when ϕ = 10 mm then spacing(s₁) = 10 cm = 100 mm

Case 2: when ϕ = 12 mm then spacing(s₂) = ?

Case 1:

When the main reinforcement of an RC slab consists of 10 mm bars at 10 cm spacing, A_st will be

${s_1} = \frac{a}{{{A_{st}}}} \times 1000$

$\Rightarrow {A_{st}} = \frac{{\frac{{\pi \times 10^2 }}{4}}}{{100}} \times 1000 = 785.4;m{m^2}$

Case 2:

⇒ If 10 mm bars is to be replaced by 12 mm bars, then the spacing of 12 mm bars

As the Area of the main reinforcement will be the same so A_st = 785.4 mm²

${s_2} = \frac{a}{{{A_{st}}}} \times 1000$

$\Rightarrow ;{s_2} = \frac{{\frac{{\pi \times 12^2}}{4}}}{{785.4}} \times 1000 = 14.40;cm$

Shortcut Trick

$S_2 \ ={\left( {\frac{{{d_2}}}{{{d_1}}}} \right)^2} \times {S_1} = {\left( {\frac{{12}}{{10}}} \right)^2} \times 10 = 14.4;cm$

Specifications of Pitch of rivet:

Pitch is the distance between two adjacent rows of rivet parallel to the direction of application of force. The minimum pitch will be 2.5 times the gross diameter.

The minimum pitch should not be less than 2.5 times the nominal diameter of the rivet.

As a thumb rule pitch equal to 3 times the nominal diameter of the rivet is adopted.

The maximum pitch shall not exceed 32 times the thickness of the thinner outside plate or 300 mm whichever is less.

Concept:

Spandrel beam:

In steel or concrete structures, the spandrel beam is the exterior beam that stretches horizontally from one column to another column. These are also known as edge beam.

Spandrel beams are provided on each floor which helps distinguish floor levels in high-rise buildings. These are used to withstand the load of the peripheral walls, in some cases, also roof loads of a building because masonry walls generally cannot carry self-weight and slab weight entirely.

Lintel beam:

A lintel is a beam placed across the openings like doors, windows etc. in buildings to support the load from the structure above. The width of lintel beam is equal to the width of wall, and the ends of it is built into the wall.

Header beam:

A header in the construction and engineering world is a beam over an opening that disperses the structural load to the outside of the opening to keep structural integrity.

Otherwise, the load from the roof, floor, or whatever is above the opening can lead to stress that can cause cracks, shifting, or even worse.

Concept:

Compression members fail under buckling.

Buckling of a member depends upon its slenderness ratio.

Slenderness ratio is defined as the ratio of the effective length of a column to the least radius of gyration of its cross section.

S.R = (KL/r) where KL = Effective length of the column

r = Radius of gyration

Radius of gyration of a body about an axis of rotation is defined as the radial distance to a point which would have a moment of inertia the same as the body's actual distribution of mass, if the total mass of the body were concentrated.

The channels or angles in the compression chords of the steel truss girder bridges are turned outward in order to increase the radius of gyration so that the slenderness ratio will decrease and the beam will become safe in buckling.

Buckling strength:

A certain portion of the beam at supports acts as a column to transfer the load from the beam to the support. Hence, under this compressive force, the web may buckle. This may happen under a concentrated load on the beam also. The load dispersion angle is taken as 45° with the horizontal.

Bearing area is calculated at the root of the web by dispersion at a slope of 1:2.5.

∴ The angle dispersion of a concentration load on the flange to the web plate of a steel beam is 45° with the horizontal.

Transverse Stiffeners are provided to increase buckling resistance of the web due to shear, It is provided vertically along the span.

Horizontal Stiffener / Longitudinal Stiffener is designed to prevent web buckling due to bending compression.

End Bearing Stiffeners are provided at the supports & Load Bearing Stiffeners are provided at the points of concentrated loads.

Important Point:

if $\frac{d}{{{t_W}}}$< 67ϵ ⇒ unstiffened girder can be designed i.e. No girder required.

If 85 ϵ < $\frac{d}{{{t_W}}}$< 200 ϵ ⇒ Vertical stiffener (C1 and C2 ) may be provided.

If 200 ϵ < $\frac{d}{{{t_W}}}$ < 250 ϵ ⇒ Vertical stiffener along with longitudinal stiffener at 0.2 d may be provided.

If 250 ϵ < $\frac{d}{{{t_W}}}$ < 345 ϵ ⇒ Vertical stiffeners along with two longitudinal stiffeners at 0.2 d and 0.5 d respectively may be provided.

Calculation:

$\frac{d}{{{t_w}}} = \frac{{1000}}{6} = 166.67$

As $\frac{d}{{{t_W}}} > 85$ and $\frac{d}{{{t_w}}} < 200$

Therefore, intermediate vertical stiffeners are to be provided.

The absolute maximum bending moment will occur at the centre when the load is spread equally on either side of the centre.

$\begin{array}{l} {M_{max}}; = ;\frac{{wa}}{4}\left( {L - \frac{a}{2}} \right)\ ; = ;\frac{{4 \times 5}}{4} \times \left( {20 - \frac{5}{2}} \right) \end{array}$

= 5 × 17.5 = 87.5 t-m at centre

Concept:

As loading is UDL and the shape of the arch is parabolic so there will be no moment in the arch rib.

But as the temperature increases in a two hinged arch (degree of indeterminacy one), the horizontal thrust will increase.

Moment due to horizontal thrust is – Py.

So maximum bending moment will be at crown as crown has highest value of y.

In particular, if the 2-hinge arch has a parabolic shape and it is subjected to a uniform horizontally distributed vertical load, then only compressive forces will be resisted by the arch. Under these conditions, the arch shape is called a funicular arch because no bending or shear forces occur within the arch.

Bending moment, Mx = 0

Radial shear, Sx = 0

Normal thrust, Nx ≠ 0

Important Point:

Concept:

Where

M_F = fixed moment.

Drawing M/EI diagram for (I) & (II).

The slope at fixed and will be zero.

So using moment area theorem, we have

Area of $\frac{M}{EI}$ diagram upto L/2 = 0

\(\Rightarrow \left( \frac{1}{2}\times \frac{L}{4}\times \frac{PL}{8EI} \right)+\left( \frac{PL}{16EI}+\frac{PL}{8EI} \right)\times \frac{L}{8}+\frac{{{M}{F}}}{EI}\times \frac{L}{4}+\frac{{{M}{F}}\times L}{2EI\times 4}\)

\(\Rightarrow \frac{P{{L}^{2}}}{64EI}+\frac{3P{{L}^{2}}}{128EI}+\frac{3{{M}_{F}}\times L}{8EI}=0\)

\(\Rightarrow \frac{5P{{L}^{2}}}{128EJ}=\frac{-3{{M}_{F}}\times L}{8EJ}\)

$\Rightarrow {{M}_{F}}=\frac{-5PL}{48~EI}$

(-ve sign indicates the assumed direction is opposite to the actual moment)

Concept:

The influence line diagram for thrust in parabolic arch:

The horizontal thrust due to a concentrated load W at a distance x from end A is given by:

${\rm{H}} = \frac{5}{8}\frac{{\rm{W}}}{{{{\rm{L}}^3}{\rm{h}}}}{\rm{x}}\left( {{\rm{L}} - {\rm{x}}} \right)\left( {{{\rm{L}}^2} + {\rm{xL}} - {{\rm{x}}^2}} \right)$

Which is a 4° parabolic equation.

Now, to draw influence line diagram for horizontal thrust, W = 1 kN

${\rm{H}} = \frac{5}{{8{\rm{h}}{{\rm{L}}^3}}}{\rm{x}}\left( {{\rm{L}} - {\rm{x}}} \right)\left( {{{\rm{L}}^2} + {\rm{xL}} - {{\rm{x}}^2}} \right)$

$\Rightarrow {\rm{H}} = \frac{5}{{8{\rm{h}}}}{\rm{x}}\left( {1 - \frac{{2{{\rm{x}}^2}}}{{{{\rm{L}}^2}}} + \frac{{{{\rm{x}}^3}}}{{{{\rm{L}}^3}}}} \right)$

Concept:

For a 3 hinged arch

ILD for beam moment in arch.

Bending moment in an arch is due to the combined action of beam moment & moment due to horizontal thrust.

ILD for BM:

For absolute maximum BM:

Let at a dist. x from left end, maximum +ve BM occurs

For maximum +ve BM, AB should be minimum

Using similar triangle property,

$AB = \frac{{2{x^2}}}{{{l^2}}}\left( {l - x} \right)$

$\left( {BM} \right) = \frac{{x\left( {l - x} \right)}}{x} - \frac{{2{x^2}}}{{{l^2}}}\left( {l - x} \right) = \frac{{2{x^3} - 3{x^3}l + x{l^2}}}{{{l^2}}}$

For (BM) to be maximum, $\frac{{dz}}{{dx}} = 0$

⇒ 6x² – 6xl + l² = 0

Solving this, x = 0.211 l (from both ends)

Components of roof trusses are:

1. Upper chord members: The uppermost line of members which extend from one support to the other through the apex is called the upper chord.
2. Bottom chord members: The lowermost line of members which extend from one support to the other through the apex is called the lower chord.
3. Web members or diagonals: The top and the bottom chord members are connected by vertical or diagonal members called web members.

In trusses simply supported at the ends, the members in the top chord are subjected to compression and the members of the bottom chord are subjected to tension. Tension members are called ties while compression members are called struts.

A roof truss will slightly deflect under a vertical or moving load. The upper chord will be under compression while the lower chord is under tension. Diagonals can either be compression or tension members depending on their inclination. Therefore, diagonals are truss members the stress depends upon whether the load is moving on the top chord or bottom chord.

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:

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:

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 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-

Conjugate Beam Theorem-

Theorem 1: The slope at a point in the real beam is numerically equal to the shear at the corresponding point in the conjugate beam.

Theorem 2: The displacement of a point in the real beam is numerically equal to the moment at the corresponding point in the conjugate beam.

Column analogous method:

The column analogous method is the force method used to analyze the indeterminate structure.

In the method of column analogy, the actual structure is considered under the action of applied loads and the redundant acting simultaneously. The load on the top of the analogous column is usually the B.M.D. due to applied loads on simple spans and therefore the reaction to this applied load is the B.M.D. due to redundant on simple spans.

It is based on the analogy between moments at ends and pressure at the edges of the short column. If both ends of the beam are fixed then the end moments are called fixed end moments.

An analogous column will be a short column with cross-section dimensions L and 1/EI.

Concept:

Calculation:

Simply supported beam deflects by 5 mm

Concentrated load = 10 kN

$\delta = \frac{{W{L^3}}}{{48EI}}$

$\delta = \frac{{10 \times {{\left( L \right)}^3}}}{{48EI}} = \frac{{10 \times {{\left( L \right)}^3}}}{{48 \times E \times \frac{{b{d^3}}}{{12}}}} = 5;mm$     …i)

For a 1/10 model, ${L_M} = \frac{L}{{10}},;{b_m} = \frac{b}{{10}},;{d_m} = \frac{d}{{10}}$

\(\underbrace {{\delta m}}{\left( {due;to;1;kN} \right)} = \frac{{P \times L_m^3}}{{48E{I_m}}} = \frac{{1 \times {{\left( {\frac{L}{{10}}} \right)}^3}}}{{48E \times \frac{b}{{10}} \times {{\left( {\frac{d}{{10}}} \right)}^3} \times \frac{1}{{12}}}}\)

$= \frac{{10 \times {L^3}}}{{48E\frac{{b{d^3}}}{{12}}}}$

From equation i), δ_m = 5 mm

Concept:

Flexibility: It is defined as the displacement/deformation produced due to unit load. It is inverse of stiffness.

Mathematically, f = 1/K

Stiffness: It is defined as the load required for unit deflection in spring

In spring F = -K × Δ

Where,

K = spring constant/stiffness, f = Flexibility constant,  Δ = deflection, and F = force

When Δ = 1 unit, ⇒ F = k

So, stiffness is equal to the applied load.

Castigliano’s first theorem: The first partial derivative of the total internal energy (strain energy) in a structure with respect to any particular deflection component at a point is equal to the force applied at that point and in the direction corresponding to that deflection component.

This first theorem is applicable to linearly or non-linearly elastic structures in which the temperature is constant, and the supports are unyielding.

Castigliano’s second theorem: The first partial derivative of the total internal energy in a structure with respect to the force applied at any point is equal to the deflection at the point of application of that force in the direction of its line of action.

The second theorem of Castigliano is applicable to linearly elastic structures with constant temperature and unyielding supports. Note that in the above statements, force may mean point force or a couple (moment) and displacement may mean translation or angular rotation.

∴ These theorems are applicable to any structure for which the force deformation relations are linear.