Correct Answer - Option 2 :
Dynamic shear modulus of the soil
Cyclic plate load test:
Plate Load Test is a field test for determining the ultimate load carrying capacity of soil and the maximum settlement under an applied load.
The plate load test basically consists of loading a steel plate placed at the foundation level and recording the settlements corresponding to each load increment.
The load applied is gradually increased till the plate starts to sink at a rapid rate. The total value of load on the plate in such a stage divided by the area of the steel plate gives the value of the ultimate bearing capacity of soil.
\({\rm{q}} = \frac{{{\rm{load\;on\;the\;plate}}}}{{{\rm{area\;of\;the\;plate}}}}= \frac{{{\rm{Q}}}}{{{\rm{A}}}}\)
After calculating the bearing capacity of soil, the Shear modulus (G) is calculated as follows
\({{\rm{C}}_{\rm{z}}} = \frac{{\rm{q}}}{{{{\rm{S}}_{\rm{e}}}}} = 1.13\frac{{\rm{E}}}{{1 - {{\rm{\mu }}^2}}}\frac{1}{{\sqrt {\rm{A}} }}\)
where,
Cz = Subgrade modulus
E = Modulus of elasticity
μ = Poisson’s ratio
A = Area of plate
\({\rm{G}} = \frac{{\rm{E}}}{{2\left( {1 + {\rm{\mu }}} \right)}}\)
\({{\rm{C}}_{\rm{z}}} = \frac{{2.26{\rm{G}}\left( {1 + {\rm{\mu }}} \right)}}{{1 - {{\rm{\mu }}^2}}}\frac{1}{{\sqrt {\rm{A}} }}\)
\(\therefore {\rm{G}} = \frac{{\left( {1 - {\rm{\mu }}} \right){{\rm{C}}_{\rm{Z}}}\sqrt {\rm{A}} }}{{2.26}}\)
Hence the cyclic plate load tests can be adopted to determine the Dynamic shear modulus of the soil
