CHECK OF STRENGTH LIMIT STATE

Check of Flexural Resistance

Total ultimate bending moment for Strength I limit state:

M_u= 1.25(DC) + 1.5(DW) + 1.75(LL+IM)

Therefore, ultimate bending moment at midspan is

M_u = 1.25(1097.7 + 40.5 + 93.8) + 1.5(158.6) + 1.75(792.6)= 3164.9 ft-kips

Average stress in prestressing steel when $f_{p e} \geq 0.5 f_{p u}$

f_{p s} = f_{p u} (1 - k \frac{c}{d_{p}})

where:

f_ps = average stress in prestressing steel, ksi

k = 0.28 (for low relaxation strands) [LRFD Table C5.7.3.1.1-1]

c = distance from extreme compression fiber to the neutral axis, in

Assume Rectangular Section:

c = \frac{A_{p s} f_{p u} + A_{s} f_{y} - {A `}_{s} {f `}_{y}}{0.85 {f `}_{c} β_{1} b_{w} + k A_{p s} \frac{f_{p u}}{d_{p}}}

where:

A_ps = area of prestressing steel, in²

A_s = area of mild steel tension reinforcement = 0 in²

A`_s = area of compression reinforcement = 0 in²

β₁=stress factor of compression block

= 0.85 f o r {f `}_{c} \leq 4.0

= 0.85 - 0.05 (f ` - 4.0) = 0.85 - 0.05 (5.5 - 4.0) = 0.775 \geq 0.65 f o r {f `}_{c} > 4.0

b = width of compression flange = 48 in.

a = depth of the equivalent stress block =β₁c

d_p = distance from extreme compression fiber to the centroid of the prestressing tendons = h - y_bs

h_f= depth of compression flange = 5.5 in.

b_w = width of web = 2(5) = 10 in.

Therefore, area of prestressing steel, A_ps = 27 × 0.153 = 4.131 in²

For the 27 bottom strands, the distance between the center of gravity of the strands and the bottom fiber of the beam, y_bs, is

y_{b s} = \frac{23 \times 3 \times 0.75 + 4 \times 4 \times 0.75}{27 \times 0.75} = 2.30 i n

Thus, d_p = 42 - 2.30 = 39.70 in.

c = \frac{4.131 \times 270 + 0 - 0}{0.85 \times 5.5 \times 0.775 \times 10 + 0.28 \times 4.131 \times \frac{270}{39.70}} = 6.1 i n

Therefore, assumption of Rectangular Section is correct.

a = β_{1} c = 0.775 \times 6.1 = 4.8 i n < h_{f} = 5.5

Therefore, average stress in prestressing steel:

f_{p s} = 270 (1 - 0.28 \frac{6.1}{39.70}) = 258.4 k s i

Nominal flexural resistance, M_n: [LRFD Art. 5.7.3.2.2]

M_{n} = A_{p s} f_{p s} (d - \frac{a}{2}) = \frac{4.131 \times 258.4 \times (39.70 - \frac{4.8}{2})}{12} = 3319.9 k i p s - f t

Factored flexural resistance, M_r:

M_{r} = ϕ M_{n}

Where:

As per 2006 Interims, the c/dt value will be used in computing the resistance factor

d_t- the distance from the extreme compression fiber to the extreme tension steel element:

M_r = 3319.9 ft-kips > M_u = 3164.9 kips-ft