RCC STRUCTURES DESIGN QUIZ

In a combined footing if shear stress does not exceed 5 kg/cm², the nominal stirrups provided are

8 legged

6 legged

10 legged

12 legged

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A foundation rests on

Both B and C

Sub-grade

Foundation soil

Base of the foundation

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A pre-stressed concrete member is preferred because

Its dimensions are not decided from the diagonal tensile stress

Removal of cracks in the members due to shrinkage

Large size of long beams carrying large shear force need not be adopted

All listed here

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If Ac, Asc and A are areas of concrete, longitudinal steel and section of a R.C.C. column and m and σc are the modular ratio and maximum stress in the configuration of concrete, the strength of column is

σc[A + (m - 1)Asc]

σcAc + m σcAsc

σc(A - Asc) + m σcAsc

All listed here

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If K is a constant depending upon the ratio of the width of the slab to its effective span l, x is the distance of the concentrated load from the nearer support, bw is the width of the area of contact of the concentrated load measured parallel to the supported edge, the effective width of the slab be is

Kx (1 + x/l) + bw

Kx (1 - x/l) + bw

All listed here

K/x (1 + x/d) + bw

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As per IS : 1343, total shrinkage for a pre-tensioned beam, is

3.0 × 10⁻²

3.5 × 10⁻⁵

3.0 × 10⁻³

3.0 × 10⁻⁵

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If K is a constant depending upon the ratio of the width of the slab to its effective span l, x is the distance of the concentrated load from the nearer support, bw is the width of the area of contact of the concentrated load measured parallel to the supported edge, the effective width of the slab be is

Kx (1 + x/l) + bw

All listed here

K/x (1 + x/d) + bw

Kx (1 - x/l) + bw

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Total pressure on the vertical face of a retaining wall of height ‘h’ per unit run exerted by the retained earth weighing ‘w’ per unit volume, is

wh² [(1 - sin φ)/2(1 + sin φ)]

wh² [(1 - sin φ)/3(1 + sin φ)]

wh [(1 - sin φ)/(1 + sin φ)]

wh² [(1 - sin φ)/(1 + sin φ)]

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If the length of a combined footing for two columns l meters apart is L and the projection on the left side of the exterior column is x, then the projection y on the right side of the exterior column, in order to have a uniformly distributed load, is (where x̅ is the distance of centre of gravity of column loads).

y = L/2 + (l - x̅)

y = L - (l - x̅)

y = L/2 - (l + x̅)

y = L/2 - (l - x̅)

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A pile of length ‘L’ carrying a uniformly distributed load ‘W’ per metre length is suspended at the centre and from other two points 0.15 L from either end ; the maximum hogging moment will be

WL²/60

WL²/90

WL²/30

WL²/15

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