📖 Explanation
As per Meyerhoff's approach qu=CNc[icScdc]+qNq[qSqdr]+21βγNr[irSrdr] Given; shape factor Sc,Sq,Sr=1 depth factor dc,dv,dr=1 Assuming no inclination of loading hence iciqir=1 For ϕ=0∘
Nc=5.14Nq=1Nr=0∴qu=25×5.14×1×1×1+19×1×1×1×1=147.5kPaqν=qu−rDf=147.5−19×1=128.5kPa
qns=FOSqnu=3128.5=42.83kPa Hence, B2500≤42.83 or B≥3.416m Hence B=3.4m