DETENSIONING

Verify stresses at 2.083 feet from beam end after group 6 has been released. Strand eccentricity at beam end for groups 1-6:

36.60 - \frac{((2) (2) + (2) (66.50) + 2 (64.5) + (2) (62.5) + (2) (60.50))}{12} = 17.483 i n

Strand eccentricity at depression point for groups 1-6:

e = 36.60 - \frac{((4) (2.50) + (2) (10.50) + 2 (8.50) + (2) (6.50) + (2) (4.50))}{12} = 30.767 i n

Strand eccentricity at 2.083 for groups 1-6:

e = - 17.483 i n - \frac{((30.767 i n + 17.483 i n) (2.083 f t))}{(0.50 f t + 45.90 f t)} = - 15.317 i n

$P = \frac{12}{38} (1076.2 k) = 339.7 k$

$\Pr e s t r e s s - t o p = 339.7 k (\frac{1}{767.0 {i n}^{2}} + \frac{15.40 i n}{15420.7 {i n}^{3}}) = 0.7821 k s i = 782.1 k s i$

M_{s e l f - w e i g h t} = \frac{(799.0 p l f) (2.083 f t) (116.00 f t - 2.083 f t)}{2} = 94797 l b - f t

S e l f - W e i g h t - t o p = \frac{(94794 l b - f t) (12 f t / f t)}{15420.7 {i n}^{3}} = 73.8 p s i

Total Stress at top = 782.1 psi + 73.8 psi = 855.9 psi

Prestress

$P = \frac{12}{38} (1076.2 k) = 339.7 k$

$\Pr e s t r e s s - t o p = 339.7 k (\frac{1}{767.0 {i n}^{2}} + \frac{15.40 i n}{15420.7 {i n}^{3}}) = 0.7821 k s i = 782.1 k s i$

M_{s e l f - w e i g h t} = \frac{(799.0 p l f) (2.083 f t) (116.00 f t - 2.083 f t)}{2} = 94797 l b - f t

S e l f - W e i g h t - t o p = \frac{(94794 l b - f t) (12 f t / f t)}{15420.7 {i n}^{3}} = 73.8 p s i

Total Stress at top = 782.1 psi + 73.8 psi = 855.9 psi