## 080

Then the total strain is eps = 0.0055 + 0.0004 + 0.0093 = 0.0152 in./in. From Figure 8.1, fps = 229,000 psi (1.579 MPa), OK. Use

4. Available moment strength. From Equation 8.46, if the neutral axis were to fall within the flange

Mn = 1.99 x 229,000 ( 36.16--— J + 1. 76 x 60,000 (37 . 6 — —

= 14,806,017 + 3,583,008 = 18,389,025 in. lb (2,078 kN m) > required Mn = 16,562,000 in. lb, OK.

5. Check of the reinforcement allowable limits.

A. Min As = 0.004A, where A is the area of the segment of the concrete section between the tension face and the centroid of the entire section

Min As = 0.004 x 201.0 = 0.80 in.2 < 1.76 in.2 used, hence alright. B. Maximum reinforcement index oT = d/dp (op — o) < 0 . 36 b1< 0 . 29 for b1 = 0 . 80 Actual

18 x 36.16 x 5,000 36.16 ^18 x 37.6 x 5,000 = 0 .14 + 0 . 03 = 0 .17 < 0 .29, hence alright .

Alternatively, the ACI Code limit strain provisions as given in Figure 8.7 do not prescribe a maximum percentage of reinforcement. They require that a check be made of the strain et at the level of the extreme tensile reinforcement to determine whether the beam is in the tensile, the transition, or the compression zone for verifying the appropriate f value. In this case, for c = 9.17 and dt = 37.6 in., and from similar triangles in the strain distribution across the beam depth et = 0 .003 x(37. 6 — 9 .17)/9.17 = 0.0093 > 0 . 005

Hence, the beam is in the tensile zone of Figure 8.7, with f = 0.90 as used in the solution, and the design is alright in terms of ductility and reinforcement limits.

8.6.9 Example 6: Ultimate Limit State Design of Prestressed Beams in SI Units

Solve Example 5 using SI units. Strands are bonded. Data

Ac = 5045 cm2, Ic = 7.04 x 106 cm4 r2 = 1394 cm2 cb = 89 . 4 cm, ee = 84 . 2 cm, Sb = 78,707 cm3, wD = 11 . 9 x 103 kN/m, 1= 19 . 8 m b = 45 . 7 cm,

C = 32 . 5 cm ee = 60 . 4 cm St = 216,210 cm3 wSD = 1,459 N/m,

W = 16 .1 kN/m bw = 15 .2 cm fc' = 34.5 MPa, fpi = 1300 MPa fpu = 1860 MPa, fpy = 1580 MPa

Solution

Assume f = 0.90 to be subsequently verified. Hence, Mu = fMn = 0.9 x 106 = 14.9 x 106 kNm

1. Section properties:

Average thickness hf = 4.5 + 2(3.5) ffi 6.25 in. = 15.7 cm

Try 4 No. 20 M mild steel bars for partial prestressing (diameter = 19.5 mm, As = 300 mm2)

2. Stress fps in the prestressing steel at nominal strength and neutral axis position:

Verify neutral axis position. If outside flange, its depth has to be greater than a = Apwfps/ 0 . 85f0bw; 0.5fpu = 0.50 x 1860 = 930 MPa < 1066. Hence, one can use the ACI approximate procedure for determining fps. From Equation 8.37e.

dp — 36 .16 in . — 91. 8 cm, d — 37. 6 in. — 95 . 5 cm fpy 1580

Aps 1287

Kp bdp 457 x 918

As 1200

bd 457 x 955

Aps fps 1674

As fy 414

bd fc0 34 5

0 0