Question

A strip footing of 2.5 m wide located at a depth of 1 m below G.L as shown in figure below is assumed to distribute the load over the soil stratum at 2 V : 1 H dispersion. If the value of ${}^{\prime }{m}_v^{\prime }$ varies over the depth of clay as
$m_v=(20+1.6Z^2)\times {10}^{-5}{m}^2/kN$
where 'Z' is distance from top of clay layer
The consolidation settlement of clay stratum in mm is

• A. 46 mm
• B. 35 mm
• C. 30 mm
• D. 50 mm

Answer 1

The correct option is A 46 mm

Given
$m_V=(20+1.6Z^2)\times {10}^{-5}$
$\therefore \Delta H=H.m_V.\Delta {\sigma }^{\prime }$
Let us assume a thickness of dz for which ${}^{\prime }{M}_V^{\prime }$ remain constant
form 2 V : 1H dispersion
Actual procedure :
Strip loading :


$\Delta {\sigma }^{\prime }=\frac{Q}{[B+(2+Z\left)\right]\times 1}$

$\Delta {\sigma }^{\prime }=\frac{200}{2.5+2+Z}=\frac{200}{4.5+Z}$

$\Delta H={\int }_0^5(20+1.6Z^2)\times {10}^{-5}\times \frac{200}{4.5+z}\times dz$

put $4.5+Z=x\Rightarrow Z=x-4.5$

$dZ=dx$
$$\Delta H={\int }_{4.5}^{9.5}\frac{20+1.6(x-4.5{)}^2}{x}\times 200\times {10}^{-5}\times dx$$

$$\Delta H=2\times {10}^{-3}{\int }_{4.5}^{9.5}\frac{20+1.6[x^2+{4.5}^2-2\times 4.5\times x]}{x}dx$$

$$=2\times {10}^{-3}{\int }_{4.5}^{9.5}[\frac{20}{x}+1.6x+\frac{32.4}{x}-14.4]dx$$

$=2\times {10}^{-3}[52.4{\left[lo{g}_{e}x\right]}_{4.5}^{9.5}-14.4[9.5-4.5]+\frac{1.6}{2}[{9.5}^2-{4.5}^2]]$

$=2\times {10}^{-3}\left[52.4\right(lo{g}_{e}9.5-lo{g}_{e}4.5)-72+56]$

$\Delta H=2\times {10}^{-3}\times [39.15-72+56]$

$=2\times {10}^{-3}\times 23.15$

$=46.3\times {10}^{-3}m$

$\Delta H=46.3mm\approx 46mm$