Example 24.
2—Investigation of Stresses at Prestress Transfer and at Service Load
For the simply supported double-tee considered in Example 24.1, check all permissible concrete stresses im-
mediately after prestress transfer and at service load assuming the unit is used for roof framing. Use losses
computed in Example 24.1.
live load = 40 psf
roof load = 20 psf 10' - 0"
dead load = 47 psf = 468 plf 2"
5 3/4"
span = 48 ft
fci′ = 3500 psi
fc′ = 5000 psi
8 - 0.5 in. diameter low-relaxation strands .. .. 24"
.. ..
Aps = 8 (0.153 in.2) = 1.224 in.2 ..
..
..
..
.. ..
e = 9.77 in. (all strands straight) . . 8"
fpu = 270,000 psi 5' - 0"
fpy = 0.90fpu
3 3/4"
jacking stress = 0.74fpu = 200 ksi
stress after transfer = 193 ksi
force after transfer = Pp = 1.224 3 193 = 236 kips
Section Properties
Ac = 449 in.2
Ic = 22,469 in.4
yb = 17.77 in.
yt = 6.23 in.
V/S = 1.35 in.
Code
Calculations and Discussion Reference
1. Calculate permissible stresses in concrete. 18.4
At prestress transfer (before time-dependent losses): 18.4.1
Compression: 0.60 fci′ = 0.60(3500) = 2100 psi
Compression at the ends = 0.70› = 0.70(3500) = 2450 psi
Tension: 6 fci′ = 355 psi (at ends of simply supported members; otherwise 3 fci′ )
At service load (after allowance for all prestress losses): 18.4.2
Compression: 0.45 fc′ = 2250 psi - Due to sustained loads
Compression: 0.60 fc′ = 3000 psi - Due to total loads
Tension: 12 fc′ = 849 psi 18.3.3(b)
2. Calculate service load moments at midspan:
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Code
Example 24.2 (cont’d) Calculations and Discussion Reference
w d l2 0.468 × 48 2
Md = = = 134.8 ft-kips (beam dead load)
8 8
w ds l2 0.02 × 10 × 48 2
Mds = = = 57.6 ft-kips (roof dead load)
8 8
Msus = Md + Mds = 134.8 + 57.6 = 192.4 ft-kips (sustained load)
w l l2 0.04 × 10 × 48 2
M l = = = 115.2 ft-kips (live load)
8 8
Mtot = Md + Mds + Ml = 134.8 + 57.6 + 115.2 = 307.6 ft-kips (total load)
3. Calculate service load moments at transfer point
Assume transfer point located at 50db = 25 in. from end of beam. Assume distance from 11.4.3
end of beam to center of support is 4 in. Therefore, x = 25 - 4 = 21 in. = 1.75 ft.
wdx 0.468 × 1.75
Md = (l - x) = (48 - 1.75) = 18.9 ft-kips (beam dead load)
2 2
Additional moment calculations at this location are unnecessary because conditions im-
mediately after release govern at this location.
4. Calculate extreme fiber stresses by “linear elastic theory” which leads to the following
well known formulas:
P Pey t My t
ft = - +
A I I
P Pey b My b
fb = + -
A I I
where, from Example 24.1
P = Pp = 236 kips (immediately after transfer)
P = Pe = 219 kips (at service load)
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Example 24.2 (cont’d) Calculations and Discussion Reference
Table 24-6 Stresses in Concrete Immediately after Prestress Transfer (psi)
At Assumed Transfer Point At Mid Span
Top Bottom Top Bottom
Pp/A +526 +526 +526 +526
Ppey/I -639 +1824 -639 +1824
Mdy/I +63 -180 +448 -1279
Total -50 (O.K.) +2170 (O.K.) +335 (O.K.) +1071 (O.K.)
Permissible -355 +2450 +2100 +2100
Compression (+)
Tension (-)
Table 24-7 Stresses in Concrete at Service Loads (psi)
At Midspan – Sustained Loads At Midspan – Total Loads
Top Bottom Top Bottom
Pe/A +488 +488 +488 +488
Peey/I -594 +1695 -594 +1695
My/I +640 -1826 +1023 -2919
Total +534 (O.K.) +357 (O.K.) +917 (O.K.) -736 (O.K.)
Permissible +2250 +2250 +3000 -849
Compression (+)
Tension (-)
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