A multistory building with a basement is to be constructed- Top 4 m consists of loose silt- below which dense sand layer is present upto a great depth- Ground water table is at the surface- The foundation consist of the basement slab of 6 m width which will rest on the top of dense sand as shown in the figure- For dense sand- saturated unit weight - 20 kN-m3- and bearing capacity factors Nq - 40 and N- - 45- For loose slit- saturated unit weight - 18 kN-m3- Nq- 15- N- - 20- Effective cohesion C- is zero for both soils- Unit weight of water is 10 kN- m3- Neglect shape factor and depth factor- Average elastic modulus E and Poisson-s ratio 1-m of dense sand is 60 - 103 kN-m3 and 0-3 respectively-Using factor of safety - 3- the net safe bearing capacity in kN-m2 of the foundation is

Q_u = CN_c + qN_q + 12γ BN_γ Here, nbsp; nbsp;γ = nbsp;γsub1 nbsp;and q = nbsp;γsub2 nbsp; ×4 γsub1 nbsp;= nbsp;γsat(sand) γ_w nbsp; nb ...

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Foundation Engineering

Effect of Water Table -II

A multistory ...

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A multistory building with a basement is to be constructed. Top 4 m consists of loose silt, below which dense sand layer is present upto a great depth. Ground water table is at the surface. The foundation consist of the basement slab of 6 m width which will rest on the top of dense sand as shown in the figure. For dense sand, saturated unit weight = $20 k N / m^{3}$ , and bearing capacity factors $N_{q} = 40$ and $N_{γ} = 45$ . For loose slit, saturated unit weight = $18 k N / m^{3}$ , $N_{q}$ = 15, $N_{γ}$ = 20. Effective cohesion C' is zero for both soils. Unit weight of water is $10 k N / m^{3}$ . Neglect shape factor and depth factor. Average elastic modulus E and Poisson's ratio (1/m) of dense sand is 60 $\times$ $10^{3}$ kN/ $m^{3}$ and 0.3 respectively.

Using factor of safety = 3, the net safe bearing capacity (in kN/ $m^{2}$ ) of the foundation is

876.7

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983

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Solution

The correct option is A 876.7
$q_{u} = C N_{c} + q N_{q} + \frac{1}{2} γ B N_{γ}$

Here, $γ$ = $γ_{s u b_{1}}$ and q = $γ_{s u b_{2}}$ $\times$ 4
$γ_{s u b_{1}}$ = $γ_{s a t (s a n d)}$ - $γ_{w}$
= 20-10 = 10 kN/ $m^{3}$

$γ_{s u b_{2}} = γ_{s a t (s i l t)} - γ_{w} = 18 - 10 = 8 k N / m^{3}$

$∴ q_{u} = 0 + 8 \times 4 \times 40 + 0.5 \times 10 \times 6 \times 45 = 2630 k N / m^{3}$

Since in this case, soil is not filled
$q_{n u} = q_{u}$

$q_{n s} = \frac{q_{n u}}{F} = \frac{2630}{3}$

$= 876.66 k N / m^{2}$

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Similar questions

Q. A multistory building with a basement is to be constructed. Top 4 m consists of loose silt, below which dense sand layer is present upto a great depth. ground water table is at the surface. The foundation consist of the basement slab of 6 m width which will rest on the top of dense sand as shown in the figure. For dense sand, saturated unit weight = 20 kN/

m^{3}

, and bearing capacity factors

N_{q}

= 40 and

N_{γ}

= 45. For loose slit, saturated unit weight = 18 kN/

m^{3}

N_{q}

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N_{γ}

= 20. Effect cohesion C' is zero for both soils. Unit weight of water is 10 kN/

m^{3}

. Neglect shape factor and depth factor. Average elastic modulus E and Poisson's ratio (1/m) of dense sand is 60

\times

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and 0.3 respectively.

The foundation slab is subjected to net safe bearing capacity derived in the above in the above question. Using influence factor

I_{f}

= 2.0, and neglecting embedment depth and rigidity corrections, the immediate settlement of the dense sand layer will be:

Q. The ultimate bearing capacity of a soil is 300 kN/

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Q. A square footing of 4 m. side is placed at 1 m depth in a sand deposit. The dry unit weight (

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Q. A 4 m wide strip footing is founded at a depth of 1.5 m below the ground surface in a

c - ϕ

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ϕ^{'}

{28.63}^{o}

γ_{s a t}

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γ_{b u l k}

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and

γ_{w}

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. the values of bearing capacity factors for different

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\begin{matrix} ϕ^{^{'}} & N_{c} & N_{q} & N_{γ} 15^{\circ} & 12.9 & 4.4 & 2.5 20^{\circ} & 17.7 & 7.4 & 5.0 25^{\circ} & 25.1 & 12.7 & 9.7 30^{\circ} & 37.2 & 22.5 & 19.7 \end{matrix}

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. Find the depth at which the pier is to be located such that a factor of safety 3.0 (in m).
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ϕ = 30^{0} a r e N_{c} = 30.1, N_{q} = 18.4

N_{γ} = 22.4

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The correct option is A 876.7qu=CNc+qNq+12γ BNγ Here, γ = γsub1 and q = γsub2 ×4 γsub1 = γsat(sand)- γw = 20-10 = 10 kN/ m3 γsub2=γsat(silt)−γw=18−10=8kN/m3 ∴qu=0+8×4×40+0.5×10×6×45=2630kN/m3 Since in this case, soil is not filled qnu=qu qns=qnuF=26303 =876.66kN/m2