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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 mv varies over the depth of clay as
mv=(20+1.6Z2)×105m2/kN
where 'Z' is distance from top of clay layer
The consolidation settlement of clay stratum in mm is


A
46 mm
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B
35 mm
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C
30 mm
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D
50 mm
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Solution

The correct option is A 46 mm
Given
mV=(20+1.6Z2)×105
ΔH=H.mV.Δσ
Let us assume a thickness of dz for which MV remain constant
form 2 V : 1H dispersion
Actual procedure :
Strip loading :



Δσ=Q[B+(2+Z)]×1

Δσ=2002.5+2+Z=2004.5+Z

ΔH=50(20+1.6Z2)×105×2004.5+z×dz

put 4.5+Z=xZ=x4.5

dZ=dx
ΔH=9.54.520+1.6(x4.5)2x×200×105×dx

ΔH=2×1039.54.520+1.6[x2+4.522×4.5×x]xdx

=2×1039.54.5[20x+1.6x+32.4x14.4]dx

=2×103[52.4[logex]9.54.514.4[9.54.5]+1.62[9.524.52]]

=2×103[52.4(loge9.5loge4.5)72+56]

ΔH=2×103×[39.1572+56]

=2×103×23.15

=46.3×103m

ΔH=46.3mm46mm

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