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A multistory building with a basement is to be constructed. The top $4 \mathrm{~m}$ consists of loose silt, below which dense sand layer is present up to a great depth. Ground water table is at the surface. The foundation consists of the basement slab of $6 \mathrm{~m}$ width which will rest on the top of dense sand as shown in the figure. For dense sand, saturated unit weight $=20 \mathrm{kN} / \mathrm{m}^{3},$ and bearing capacity factors $\mathrm{N}_{\mathrm{q}}=40$ and $\mathrm{N}_{\gamma}=45 .$ For loose silt, saturated unit weight $=18 \mathrm{kN} / \mathrm{m}^{3}, \mathrm{~N}_{\mathrm{q}}=15$ and $\mathrm{N}_{\gamma}=20 .$ Effective cohesion $\mathrm{c}^{\prime}$ is zero for both soils. Unit weight of water is $10 \mathrm{kN} / \mathrm{m}^{3}$. Neglect shape factor and depth factor. Average elastic modulus $E$ and Poisson's ratio $\mu$ of dense sand is $60 \times 10^{3} \mathrm{kN} / \mathrm{m}^{2}$ and 0.3 respectively.

Using factor of safety $=3,$ the net safe bearing capacity (in $\mathrm{kN} / \mathrm{m}^{2}$ ) of the foundation is:
(A) 610
(B) 320
(C) 983
(D) 693



Question ID - 155919 | SaraNextGen Top Answer

A multistory building with a basement is to be constructed. The top $4 \mathrm{~m}$ consists of loose silt, below which dense sand layer is present up to a great depth. Ground water table is at the surface. The foundation consists of the basement slab of $6 \mathrm{~m}$ width which will rest on the top of dense sand as shown in the figure. For dense sand, saturated unit weight $=20 \mathrm{kN} / \mathrm{m}^{3},$ and bearing capacity factors $\mathrm{N}_{\mathrm{q}}=40$ and $\mathrm{N}_{\gamma}=45 .$ For loose silt, saturated unit weight $=18 \mathrm{kN} / \mathrm{m}^{3}, \mathrm{~N}_{\mathrm{q}}=15$ and $\mathrm{N}_{\gamma}=20 .$ Effective cohesion $\mathrm{c}^{\prime}$ is zero for both soils. Unit weight of water is $10 \mathrm{kN} / \mathrm{m}^{3}$. Neglect shape factor and depth factor. Average elastic modulus $E$ and Poisson's ratio $\mu$ of dense sand is $60 \times 10^{3} \mathrm{kN} / \mathrm{m}^{2}$ and 0.3 respectively.

Using factor of safety $=3,$ the net safe bearing capacity (in $\mathrm{kN} / \mathrm{m}^{2}$ ) of the foundation is:
(A) 610
(B) 320
(C) 983
(D) 693

1 Answer
127 votes
Answer Key / Explanation : (A, B, C, D) -

610, 320, 983, 693

127 votes


127