${}_w=9\mathrm{.}8k\frac{N}{m^3}.E_{\text{s}\text{l}\text{e}\text{a}\text{l}}=210$GPa.$E_{\text{e}\text{o}\text{n}\text{c}}=25$GPa$\mathrm{.}1$ inch $=25\mathrm{.}4mm$

Fig. $1$ Site $1-$ Soil Profile $-($Groundwater $1s$ at $2m$ depth$)$: HP$12$x$53$ Steel Pile $($driven$).$ Pile length $=15m$

$43$ Group Pile

Single Pile

Sand: $2m$ thick.

${}_{\text{m}\text{o}\text{i}\text{s}\text{t}}=18k\frac{N}{m^3}$

$C_u=70$kPa at top of clay.

$C_u$ increases by $3kP\frac{a}{m}$ depth

${}_{\text{s}\text{a}\text{l}}=18\mathrm{.}5k\frac{N}{m^3},$

Dense Silty sand

${}_{\text{s}\text{a}\text{t}}=19\mathrm{.}6k\frac{N}{m^3},$

${}^{\text{'}}=40$deg,$e^{\text{'}}=0$

Pile is loaded in compression.

Determine:

a$)$ ultimate axial capacity $Q_{ultsingle}(kN)$ of a single pile $=?$

b$)$ Axial elastic shortening $($in $mm)$ of a single pile loaded to $PN,=?$

Report your answer in terms of letter $P.$

c$)$ ultimate group capacity spacing $=1m)$ using an appropriate group efficiency factor $=?$ Ignore 'block' failure.

d$)$ ultimate tip resistance Qulutip$-$Block $($kN$)$ of the group, if you were to consider 'block' failure $=?$

e$)$ skin friction $)$ on the sides of the 'block' at $10m$ depth $=?$

f$)$ the magnitude of shaft resistance contributed by the Dense Silty Sand to the total ultimate shaft friction $(kN)=?$ for a $23m$ long single pile.

$f_s=K_s{}_v^{\text{'}}tan()$; If there is a range for $,$ use the average value.

Values of $$ : Steel $-0\mathrm{.}6700\mathrm{.}83{}^{\text{'}}$; Concrete $-0\mathrm{.}9001\mathrm{.}0{}^{\text{'}}$; Timber $-0\mathrm{.}8001\mathrm{.}0{}^{\text{'}}$

Adhesion $f_s={}^{**}{C}_u$;$,=0\mathrm{.}21+0\mathrm{.}26\frac{p_v}{C_u}<mn>1</mn><mn>.</mn><mn>0</mn>$;$,p_u=101$kPa,

$K_s=12$ for sand for driven piles. Lower end of $K_s$ applies to low displacement piles.

End bearing in sand $=q{N}_q{A}_p.$ End bearing in clay $=9C_u{A}_p.$

Group End Bering in Clay $=C_4^{**}{N}_c^{**}{B}_g^{**}{L}_g$;$N_c=5[1+L(5B_2)]**[1+B_b(5L_2)]<mn>9</mn>$;$L=$ Pile length $=0\mathrm{.}5+0\mathrm{.}05\frac{s}{d}$;$=1$ for $\frac{s}{d}>5.Q=N{Q}_{single.}$

pile diameter.

$=\frac{2}{3}$ for uniform $f_s$ distribution and $=\frac{1}{2}$ for triangular $f_s$ distribution.

$C_5=(0\mathrm{.}93+0\mathrm{.}16(\frac{L}{d}{)}^{0\mathrm{.}5})C_t,L=$ pile length, $d=$ pile diameter. $C_t=0\mathrm{.}02.S_{\text{g}\text{r}\text{o}\text{u}\text{p}}=S_{single}(\frac{B_8}{d_1}{)}^{0\mathrm{.}5}$;

$\backslash$table$[[{}^{\text{'}}$ deg,$30,35,40,45],[N_q,27,48,95,190]]$