1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
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

A
46 mm
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
35 mm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
30 mm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
50 mm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is A 46 mmGiven 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

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Primary Consolidation Settlement
SOIL MECHANICS
Watch in App
Explore more
Join BYJU'S Learning Program