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Explanation:

• Flat slab is a slab that directly rests on the column without any beam. Sometimes column capital or drop panels are provided to prevent punching failure of flat slab.
• The Thickened part of the flat slab over its supporting column is known as the column Drop panel.
• Its thickness is invariably higher than the thickness of the slab in a normal slab system in which beams are provided.
• Flat slabs are uneconomical and can be used for medium span only because in large span deflection is comparatively more.
• Some of the advantages of flat slab:

1. It has a better aesthetic appearance than normal slab system.
2. It has better dispersion of light than normal slab system.
3. Flat slab provides ease in providing AC ducts, Electrical ducts, etc.
4. Used in case of lesser story height.

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Explanation:

(i) There are two basic design criteria. One is the bearing capacity criteria or the shear failure criteria another is the settlement criteria. When we design the footing, it should take the load coming from the superstructure without shear cracking.

(ii) At the same time there should not be an excessive amount of settlement in the footing. So the pressure that we can apply on a soil is the minimum pressure on a soil in terms of shear criteria or in terms of the settlement criteria.

(iii) Tension failure, compression failure or bond failure hardly occur in foundation structures and are never a serious concern.

So The correct answer is Both allowable settlement and ultimate bearing capacity

Explanation:

A riveted joint may fail in the following ways

• Tearing of the plate at an edge
• Shear failure of the rivets
• Tensile failure of rivet across the row of rivets
• Crushing failure of plate
• Shear failure of plate at the marginal area.

Shearing of the plate  

A joint may fail due to shearing of the plate at an edge as shown in Fig. This can be avoided by keeping the margin, m = 1.5d, where “d” is the diameter of the rivet hole.

Confusion Points The minimum edge distance from the center of any hole to the nearest edge of plate shall not less than 1.7 times the hole diameter in case of sheared or hand flame cut edges; and 1.5 times the hole diameter in case of rolled, machine-flame cut, sawn and planned edges. Adequate edge distance (margin) is maintained to prevent the plate from failing due to tearing of plate.

Note: According to commission option (c) is correct

Concept:

Area moment of inertia is given by, I = A × k2

Where A is area of section and k is radius of gyration of the section.

For circular section, k = D/4

\({\rm{A}} = \frac{{\rm{\pi }}}{4}{{\rm{D}}^2}\)

\({\rm{k}} = \frac{{\rm{D}}}{4}\)

\(\therefore {\rm{I}} = {\rm{A}} \times {{\rm{k}}^2} = \frac{{\rm{\pi }}}{{64}}{{\rm{D}}^4}\)

Calculation:

D = 20 cm

Moment of Inertia for a circular section

\(\therefore {\rm{I}} = \frac{{\rm{\pi }}}{{64}}{{\rm{(20)}}^4}\) = 7854 cm⁴ 

Important Points

Polar Moment of Inertia (J) = 2I = \(\frac{{\rm{\pi }}}{{32}}{{\rm{D}}^4}\)

Centre of pressure:

The point of action of total hydrostatic force on a submerged surface is called the centre of pressure. It is the point at which the resultant pressure on a immersed surface acts.

Explanation:

The total energy of a flowing fluid can be represented in terms of head, which is given by -

\(\frac{p}{ρ{g}};+;\frac{V^2}{2g};+;z;\)

Piezometric head:

The sum of pressure head and datum is known as piezometric head. It is given by -

\(\frac{p}{ρ{g}}+z\)

\(where;\frac{p}{\rho{g}}=pressure;head,;\frac{V^2}{2g}=velocity;head;and;z=datum\)

∴ Piezometric head is the sum of the pressure head and datum head.

A traverse consists of series of straight lines of known length related to one another by known angles between the lines.

Traverse survey is a method of establishing control points, their position being determined by measuring the distances between the traverse stations which serve as control points and the angles subtended at the various stations by their adjacent stations.

A traverse is developed by measuring the distance and angles between points that found the boundary of a site.

Explanation:

The given beam with end moment and resultant reaction:

Taking clockwise moment as a positive and anticlockwise moment as negative

Taking the moment of the beam at A is 

Σ M_A = 0

2 × 2 - 4 - M_A = 0

M_A = 0

Explanation:

The most efficient section is one that gives maximum discharge for a given amount of excavation. 

For maximum discharge through the rectangular channels, the geometric conditions can be found by minimizing the wetted perimeter P.

\(P=B+2d\)

\(P=\frac{Bd}{d}+2d\)

\(P=\frac{A}{d}+2d\)

Now, 

\(\frac{dP}{dd}=-\frac{A}{d^2}+2=0\)

A = 2d2

Bd = 2d2

B = 2d

\(Hydraulic;mean;radius,;{y_m} = \frac{A}{P} = \frac{{Bd}}{{B + 2d}} = \frac{{2d.d}}{{2d + 2d}} = \frac{d}{2} \)

Concept:

The hydraulic mean depth or also called Hydraulic radius is defined as:

\(R = \frac{{A}}{{P}}\)

where A = cross-sectional area of the flow, P = wetted perimeter.

Wetted area,  \(A = \frac{\pi }{4}{d^2} \)

Wetted perimeter, P = πd

Hydraulic radius, \(R = \frac{A}{P}\)

Hydraulic radius, \(R=\frac{\frac{\pi d^2}{4}}{\pi d}=\frac{d}{4}\)