ULTIMATE CAPACITY AT MIDSPAN

β_{1} = 0.85 (f o r {f `}_{c} < 4000 p s i)

Art. 8.16.2.7

A * s = 38 \times 0.153 {i n}^{2} = 5.814 {i n}^{2}

Y_cg = 5.03 in

P * = \frac{A * s}{b d}

Art. 9.1.2

where:

d = 80.50 in- 5.03 in = 75.47 in

P * = \frac{5.814 {i n}^{2}}{(102.0 i n) (75.47 i n)}

f_{s u}^{*} = {f `}_{s} (1 - (\frac{r^{*}}{β_{1}}) (P^{*}) (\frac{{f `}_{s}}{{f `}_{c}})) = 270 k s i (1 - (\frac{0.28}{0.850}) 0.000755 (\frac{270 k s i}{3.5 k s i})) = 264.8 k s i

Eq.9-17

a = \frac{A_{s}^{*} \times f_{s u}^{*}}{0.85 \times {f `}_{c} b} = \frac{5.814 \times 264.8}{0.85 \times 3.5 \times 102} = 5.07 i n

M_u-req'd = (1.3)[1320.8 k-ft. + 297.6 k-ft. + 1441.3 k-ft. + 266.7 k-ft. + (1.67) (1667.9 kft)]= 7943.9 k-ft

Since the neutral axis lies within the flange,

M_{u - p r o v ` d} = ϕ ((A_{s}^{*}) (f_{s u}^{*}) (d) (1 - (- \frac{0.6 P^{*} f_{s u}^{*}}{{f `}_{c}})))

Eq.9-13

where:

Φ = 1.0 for flexural members Art. 9.14

Δ_{f i n a l} = Δ_{e r e c t i o n} \times 2.30 = 2.30 (- 1.338) = - 3.078 i n

M_{u - p r o v ` d} = ϕ ((5.714 {i n}^{2}) (264.8 k s i) (75.47 i n) (1 - 0.6 \times 0.000755 \times \frac{264.8 k s i}{3.5 k s i})) = 112219 k - i n = 9351.6 k - f t