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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 ′m′v varies over the depth of clay as

mv=(20+1.6Z2)×10−5m2/kN

where 'Z' is distance from top of clay layer

The consolidation settlement of clay stratum in mm is

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Solution

The correct option is **A** 46 mm

Given

mV=(20+1.6Z2)×10−5

∴ΔH=H.mV.Δσ′

Let us assume a thickness of dz for which ′M′V 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)×10−5×2004.5+z×dz

put 4.5+Z=x⇒Z=x−4.5

dZ=dx

ΔH=∫9.54.520+1.6(x−4.5)2x×200×10−5×dx

ΔH=2×10−3∫9.54.520+1.6[x2+4.52−2×4.5×x]xdx

=2×10−3∫9.54.5[20x+1.6x+32.4x−14.4]dx

=2×10−3[52.4[logex]9.54.5−14.4[9.5−4.5]+1.62[9.52−4.52]]

=2×10−3[52.4(loge9.5−loge4.5)−72+56]

ΔH=2×10−3×[39.15−72+56]

=2×10−3×23.15

=46.3×10−3m

ΔH=46.3mm≈46mm

Given

mV=(20+1.6Z2)×10−5

∴ΔH=H.mV.Δσ′

Let us assume a thickness of dz for which ′M′V remain constant

form 2 V : 1H dispersion

Strip loading :

Δσ′=Q[B+(2+Z)]×1

Δσ′=2002.5+2+Z=2004.5+Z

ΔH=∫50(20+1.6Z2)×10−5×2004.5+z×dz

put 4.5+Z=x⇒Z=x−4.5

dZ=dx

ΔH=∫9.54.520+1.6(x−4.5)2x×200×10−5×dx

ΔH=2×10−3∫9.54.520+1.6[x2+4.52−2×4.5×x]xdx

=2×10−3∫9.54.5[20x+1.6x+32.4x−14.4]dx

=2×10−3[52.4[logex]9.54.5−14.4[9.5−4.5]+1.62[9.52−4.52]]

=2×10−3[52.4(loge9.5−loge4.5)−72+56]

ΔH=2×10−3×[39.15−72+56]

=2×10−3×23.15

=46.3×10−3m

ΔH=46.3mm≈46mm

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