Your Answer : (Not Answered)

Correct Answer : D

Lateral ties are provided to resist buckling of longitudinal steel bars.

The longitudinal bars of a column are slender and heavily compressed. Without restraint they would bow outward and burst the concrete cover. Lateral ties hold each bar in position at close intervals, preventing this buckling, and they also confine the core concrete.

Columns are designed mainly for axial load with some moment, which the longitudinal steel and concrete resist. Shear in ordinary columns is small, and ties are not sized for bending or shear, so the other options are wrong.

Your Answer : (Not Answered)

Correct Answer : B

The basic values are 20 and 26 respectively.

IS 456:2000, clause 23.2.1, gives basic span to effective depth ratios for spans up to 10 m: 7 for a cantilever, 20 for a simply supported beam and 26 for a continuous beam. These are then modified for the amount of tension and compression steel and for longer spans.

A continuous beam is stiffer than a simply supported one, so it is allowed a larger ratio, which rules out 26 and 20. The 7 in the other options is the cantilever value.

Your Answer : (Not Answered)

Correct Answer : B

The effective depth is D − d_t.

Effective depth is the distance from the extreme compression face to the centroid of the tension reinforcement. The effective cover to the tension steel, d_t, is the distance from the tension face to that centroid, so subtracting it from the overall depth D gives the effective depth.

The compression steel does not enter the definition, so subtracting d_c, alone, together with d_t, or as part of an average, gives a different dimension. D − d_c is the depth of the compression steel's centroid measured from the tension face.

Your Answer : (Not Answered)

Correct Answer : A

Side face reinforcement is required when the depth exceeds 750 mm.

IS 456:2000, clause 26.5.1.3, requires side face reinforcement where the depth of the web of a beam exceeds 750 mm. It amounts to 0.1 percent of the web area, distributed equally on both faces, at a spacing not more than 300 mm or the web thickness, whichever is less. It controls shrinkage and temperature cracking on the deep sides of the beam.

450, 500 and 600 mm are not limits in the code.

Your Answer : (Not Answered)

Correct Answer : B

The minimum factor of safety against overturning is 1.5.

A retaining wall is checked so that the moment resisting overturning about the toe is comfortably larger than the moment of the earth pressure. IS 456:2000 (clause 20.1) requires the restoring moment, counting only 0.9 times the dead load, to be at least 1.4 times the overturning moment, which works out to about 1.4/0.9 = 1.55, and design texts take the minimum as 1.5.

2.0, 2.5 and 3.0 are more conservative values a designer may choose, but none is the minimum the question asks for.

Your Answer : (Not Answered)

Correct Answer : D

Main steel goes at the top in the heel slab and at the bottom in the toe slab.

The heel slab, under the retained earth, carries the heavy downward weight of the backfill, which exceeds the upward soil pressure beneath it. It bends as a cantilever from the stem with tension on its top face, so its main bars are placed at the top.

The toe slab, in front of the stem, has little load on it but receives the largest upward soil pressure. It bends upward as a cantilever, with tension on the bottom face, so its main bars are at the bottom. The other three options put the steel on the compression face of one or both slabs.

Your Answer : (Not Answered)

Correct Answer : D

Punching shear is mostly related to the isolated square footing.

Punching shear, or two-way shear, is the tendency of a column to punch through the slab or footing that carries it, along a truncated pyramid around the column. It is checked on a perimeter at d/2 from the column face, and it usually governs the depth of an isolated column footing, of which the square footing is the standard case.

In a long rectangular footing one-way beam shear often governs instead. A combined footing, including the trapezoidal form, carries two columns and is designed largely like a beam, so one-way shear and bending dominate.

Your Answer : (Not Answered)

Correct Answer : D

The design bond stress in compression is 25% more.

IS 456:2000, clause 26.2.1.1, gives the design bond stress for plain bars in tension, and says that for bars in compression these values are to be increased by 25 percent. A bar in compression also bears on the concrete at its end, which helps its anchorage.

60% more is the increase allowed for deformed bars over plain bars, a different rule. Bond in compression is not reduced, so the 'less' options are wrong.

Your Answer : (Not Answered)

Correct Answer : C

The most economical way is using smaller diameter bars in greater numbers.

Bond resistance depends on the surface area of steel in contact with concrete. For the same total area of steel, many small bars have a much larger total perimeter than a few large ones (perimeter grows with diameter, area with its square), so the bond stress per unit length falls without adding any steel.

Vertical stirrups resist shear, not bond. Increasing the beam depth costs more concrete and steel. Using larger bars in fewer numbers reduces the total perimeter and makes bond worse.

Your Answer : (Not Answered)

Correct Answer : B

The shear capacity of the stirrups is \(1.16 (f_{y} A_{sv})\).

For vertical stirrups, IS 456 gives the shear resisted as:

\[ \begin{aligned} V_{us} &= \frac{0.87 f_y A_{sv} d}{s_v} \\ &= \frac{0.87 f_y A_{sv} d}{0.75 d} = 1.16 f_y A_{sv} \end{aligned} \]

The effective depth cancels. The other values do not combine the factor 0.87 with the ratio \(d/s_v = 1/0.75\); for example, \(1/0.75\) alone would give 1.33.

Your Answer : (Not Answered)

Correct Answer : C

The assumption is used in both working stress and limit state method of design.

The assumption that plane sections before bending remain plane after bending, which makes strain proportional to distance from the neutral axis, is one of the basic assumptions of IS 456 both for the working stress method (Annex B) and for the limit state method of design for flexure (clause 38.1).

Each of the other options names the older ultimate load method. The assumption holds there too, but the option that covers both methods of the current code, working stress and limit state, is the complete answer, and saying 'only' the ultimate method is plainly wrong.

Your Answer : (Not Answered)

Correct Answer : C

The minimum reinforcement for Fe-250 is 0.15%.

IS 456:2000, clause 26.5.2.1, requires slab reinforcement in either direction of at least 0.15 percent of the gross cross-sectional area when mild steel (Fe 250) is used, reduced to 0.12 percent for high strength deformed bars.

0.12% is the near miss, the figure for HYSD bars. 0.08% and 0.04% are below any minimum in the code.

Your Answer : (Not Answered)

Correct Answer : C

The thickened part of the slab over a column is the drop panel.

In a flat slab the slab rests directly on the columns, so shear and negative moment are concentrated around each column. A drop is a local thickening of the slab itself over the column, which increases the depth there to resist punching shear and hogging moment.

A column head, also called a capital, is the flared top of the column, not part of the slab. 'Beam head and panel' is not a flat slab term; a flat slab has no beams at all.

Your Answer : (Not Answered)

Correct Answer : C

The nominal shear stress is \(\tau_{v}\) = 0.189 MPa.

Effective span is the lesser of clear span plus d (4.165 m) and centre-to-centre of supports (4.23 m), so 4.165 m. Loads per square metre:

\[ \begin{aligned} w &= 0.2 \times 25 + 1 + 4 = 10\ \text{kN/m}^2 \\ w_u &= 1.5 \times 10 = 15\ \text{kN/m}^2 \end{aligned} \]

For a 1 m wide strip:

\[ \begin{aligned} V_u &= \frac{15 \times 4.165}{2} = 31.24\ \text{kN} \\ \tau_v &= \frac{31.24 \times 10^3}{1000 \times 165} = 0.189\ \text{MPa} \end{aligned} \]

Leaving out the factor 1.5 gives about 0.126 MPa, and other options come from wrong spans or loads.

Your Answer : (Not Answered)

Correct Answer : D

The minimum reinforcement for Fe 250 in a slab is 0.15% of its gross sectional area.

IS 456 (clause 26.5.2.1) requires the reinforcement in either direction in slabs to be at least 0.15 percent of the gross cross-sectional area when mild steel (Fe 250) is used. The limit is reduced to 0.12 percent for high strength deformed bars, which are stronger.

0.12 percent is the HYSD value, the usual trap, and 0.10 and 0.14 percent are not code values.

Your Answer : (Not Answered)

Correct Answer : C

The soil can safely bear 200 kN/m\(^2\), so the width of footing for 800 kN/m is \(800/200 = 4\) m. The wall is 1 m wide, so each side of the footing projects \((4 - 1)/2 = 1.5\) m beyond the wall face. The maximum bending moment is at the wall face:

The diagonal shear is taken at a distance equal to the effective depth (0.5 m) from the face, so the overhang is 1.5 - 0.5 = 1.0 m:

The other options give different moments and shears from the same data and do not follow this procedure.