how on earth do i get to the e inequality using these values? zt $=1\mathrm{.}23$x$10^8,$ and idk what alpha or beta are they arent givennMid$-$span section design moment at transfer and SLS:

$e=400mm$

$d=660\mathrm{.}71mm$

Beam self weight $=14k\frac{N}{m}$

$M_t=700kNm(momentat$ transfer $(dueto$ self weight$))$

$M_s=1800kNm(momentatSLS(dueto$ unfactored self weight $+$ imposed loads$))$

The following four inequalities are used to establish the safe zone of the prestressing tendon force

and corresponding eccentricity:

The four inequalities controlling stresses at transfer and service conditions:

$(\frac{P_O}{A}-\frac{P_{O}e}{z_t}+\frac{M_t}{z_t}){}_{tt}$

$(\frac{P_O}{A}+\frac{P_{o}e}{z_b}-\frac{M_t}{z_b}){}_{ct}$

$(\frac{P_O}{A}-\frac{P_{o}e}{z_t}+\frac{M_s}{z_t}){}_{cs}$

$(\frac{P_O}{A}+\frac{P_{o}e}{z_b}-\frac{M_s}{z_b}){}_{ts}$

Substituting various values in the above four inequalities produces:

$z_t=122748858\mathrm{.}4m{m}^3$

$z_b=59342163\mathrm{.}36m{m}^3$

$e219\mathrm{.}194+12\mathrm{.}57(\frac{{10}^8}{P_o})$

$e-105\mathrm{.}97+23\mathrm{.}6(\frac{{10}^8}{P_o})$

$e219\mathrm{.}194-32\mathrm{.}73(\frac{{10}^8}{P_o})$

$e-105\mathrm{.}97+19\mathrm{.}27(\frac{{10}^8}{P_o})$

The students need to plot the above $4$ inequalities to a reasonable scale and identify the safe region

for Po and e$.$