9. Slab Design Example

PS21rev

Post-tensioned, un-bonded one-way slab with draped tendons (midspan moments will be assumed to control)

 A = \frac{262}{4} = 65.5in^2, where 262in^2 is the area of the entire section shown above and 65.5in^2 is the shaded area.
 I = \frac{4949}{4} = 1237.25in^4, which is the moment of inertia of the shaded area
e = 10.5 - 6 = 4.5''
c = 6”
f'_c = 5000 psi
f'_{ci} = 3500 psi
f_{pu} = 270 ksi, low-relaxation (f_p = 0.75f_{pu}) 

LL = 100 psf
DSL = 10 psf
\frac{L}{h} = 37

Friction, anchor set, long-term losses

 A detailed friction analysis could be performed, as done in an earlier section, but here we will just estimate  f_{pi} as 187 ksi, and f_{pe} as 157 ksi.

 Finding A_ps

 member self-weight = 150*\frac{262in^2}{4}*\frac{1}{144} = 68 plf (per foot width)

SDL = 10 psf * 1' = 10 plf (per foot width)

LL = 100 psf * 1' = 100 plf (per foot width)

M_{max} = (0.68 + 0.010 + 0.100)*\frac{36^2}{8} = 28 ft*kips (per foor width, working stress)

e + k_t = 4.5" + \frac{S_b}{A} = 4.5 + \frac{1237.25/6}{65.5} = 7.65"

6\sqrt{f'_c} = 424psi

(1)   \begin{equation*} P_f \geq \frac{(28ft*kips*12 ) - (\frac{1237.2}{6})(0.424)}{7.65} = 32.5 kips \text{(per foot width)} \end{equation*}

Note:  this is based on bottom tension at service

A_{ps} \geq \frac{32.5 kips}{153 ksi} = 0.212 in^2 (per foot width)

 Also, for strength,

 M_{max} = [1.2*(0.068+0.01) + 1.6*0.100]\frac{36^2}{8} = 41 kip*ft (per foot width, LRFD)

A_{ps} \geq \frac{41*12}{0.77*270*12" = h} = 0.197 in^2 

Try two \frac{1}{2}'' diameter strands; A_{ps} = 2*0.153 = 0.306 in^2 (per foot width)

 Check top tension at transfer

3\sqrt{3500} = 177 psi

(2)   \begin{equation*} e \leq k_b + \frac{M_{min} + S_tf_t}{P_i}\end{equation*}

where M_{min} = (0.068) \frac{36^2}{8} = 11.1kip*ft, P_i = 183ksi*0.306 = 56 kips,

S_t = S_b = \frac{1237.25}{6} = 206.2in^3, k_b = \frac{S_t}{A} = \frac{(\frac{1237.25}{6})}{65.5} = 3.15 in

(3)   \begin{equation*} e \leq 3.15 + \frac{(11.1*12) + (206.2)(0.177)}{56} = 6.18 in > 4.5 in \therefore \text{ OK}\end{equation*}

Check bottom compression at transfer

\frac{-P_i}{A} = -855 psi

\frac{-P_ie}{S_b} = -1222 psi

\frac{M_{min}}{S_b} = 646 psi 

\Sigma = -1431 psi < 0.6f'_{ci} = 2100 psi \therefore \text{ OK}

Check top compression service

\frac{P_f}{A} = \frac{-153*0.306*1000}{65.5} = -715 psi

\frac{P_fe}{S_t} = 1022 psi

\frac{-M_{max}}{S_t} = -1629 psi

\Sigma = -1322 psi <0.45f'_c = 2250 psi \therefore \text{ OK}

Code specified minimum mild steel in slab

A_s = 0.004*\frac{65.5}{2} = 0.131in^2 (per foot width)

Choose one #4 bar per foot (d \cong d_p = 10.5")

A_s = 0.2in^2 (per foot width)

Strength 

For members with unbonded tendons:

For members with span-to-depth ratios of 35 or less,

(4)   \begin{equation*} f_{ps} = f_{pe} + 10000 + \frac{f'_c}{100\rho_p}, \text{ psi (empirical)}\end{equation*}

Else,

(5)   \begin{equation*}f_{ps} = f_{pe} + 10000 + \frac{f'_c}{300\rho_p}, \text{ psi} \end{equation*}

where \rho_p = \frac{A_{ps}}{bd_p}

 For our example,

\frac{L}{h} = 36

f_{ps} = 153000 + 10000 + \frac{5000}{300\rho_p}, \rho_p = \frac{0.306}{(12)(10.5)} = 0.0024

f_{ps} = 169.9 ksi

There are two checks:

 169.9 < 0.9(270) = 243 and <153 + 30 = 183 \therefore \text { OK}

(6)   \begin{equation*} \omega = \frac{A_s}{bd}\frac{f_y}{f'_c} = \frac{0.2}{(12)(10.5)}\frac{60}{5} = 0.019\end{equation*}

(7)   \begin{equation*} \omega_p = \frac{A_{ps}}{bd_p}\frac{f_{ps}}{f'_c} = \frac{0.306}{(12)(10.5}\frac{169.9}{5} = 0.0825 \end{equation*}

(8)   \begin{equation*} \omega_p + \frac{d}{d_p}\omega = 0.0825 + \frac{10.5}{10.5}(0.019) = 0.1 < 0.36(0.8) = 0.288 \therefore \text{ OK} \end{equation*}

(9)   \begin{equation*} \beta_1 = 0.85 - (f'_c - 4000)(0.00005) = 0.8 \geq 0.65 \end{equation*}

(10)   \begin{equation*} a = \frac{A_{ps}f_{ps}}{\gamma f'_cb} = \frac{(0.306)(169.9)}{(0.85)(5)(12)} = 1.02" < ~ 1.5" \Rightarrow \text {the n.a. is below the hollow portion}\end{equation*}

(11)   \begin{equation*} \phi M_n = 0.9[A_{ps}f_{ps}(d_p - \frac{a}{2}) + A_sf_y(d-\frac{a}{2})] = 575kip*in = 47.9kip*ft \text { (per foot width)} > 41 kip*ft \therefore \text{ OK} \end{equation*}

Check Ductility

7.5\sqrt{f'_c} = 530 psi

530 = -\frac{P_f}{A} - \frac{P_fe}{S_b} + \frac{M_{cr}}{S_b} = -715 psi - 1022 psi + \frac{M_{cr}}{206.2} \Rightarrow M_{cr} = 467 kip*in

(12)   \begin{equation*} \frac{\phi M_n}{M_{cr}} = \frac{575}{467} = 1.23 > 1.2 \therefore \text{ OK} \end{equation*}

 

Quick deflection check

(13)   \begin{equation*} \Delta_L = \frac{5}{384} \frac{0.1(36)^4*12^3}{(\frac{57000\sqrt{5000}}{1000})(1237.25)}= 0.76" < \frac{L}{360} = \frac{36*12}{360} = 1.2" \therefore \text{ OK} \end{equation*}