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and the required nominal moment strength is

Mu 14,905,800

2. Choice of preliminary section. Assuming a depth of 0.6 in./ft of span as a reasonable practical guideline, we can have a trial section depth h = 0.6 x 65 ffi 40 in. (102 cm). Then assume a mild partial steel 4#6 = 4 x 0.44 = 1.76 in.2 (11.4 cm2). The empirical area ofthe concrete segment in compression up to the neutral axis can be empirically evaluated from

Mn 16,562,000 2 2

Assume on trial basis a total area of the beam section as 3A' = 3 x 121.8 « 362.0 in.2

Assume a flange width of 18 in. Then the average flange thickness = 121.8/18 ffi 7.0 in. (178 mm). So suppose the web bw = 6 in. (152 mm), to be subsequently verified for shear requirements. An empirical expression for the area of the prestressing steel can be expressed as

Number of 2-in. stress-relieved wire strands = 2.13/0.153 = 13.9. So try thirteen 2-in. tendons Aps = 13 x 0 .153 = 1. 99 in.2 (12 . 8 cm2)

### Try the section in Figure 8.8 for analysis.

3. Calculate the stress fps in the prestressing tendon at nominal strength using the strain-compatibility approach. The geometrical properties of the trial section are very close to the assumed dimensions for the depth h and the top flange width b. Hence, use the following data for the purpose of the example:

= 15 in. at midspan e2 = 225 in.2 e2/r2 = 225/187.5= 1.20

187.5 in.2

Ec = 57,000V/5ÔÔÔ = 4.03 x 106 psi (27.8 x 103 MPa) Eps = 28 x 106 psi (193 x 103 MPa)

The maximum allowable compressive strain ec at failure = 0.003 in./in. Assume that the effective prestress at service load is fpe = 155,000 psi (1,069 MPa). A.

The increase in prestressing steel strain as the concrete is decompressed by the increased external load (see Figure 8.6 and Equation 8.37b) is given as e2 edecomp . ~ (1 ^ " AcEc y

B. Assume that the stress fps — 205,000 psi as a first trial. Suppose the neutral axis inside the flange is verified on the basis of hf — 3 + 41 + 3j/2 — 9.25 in. then, from Equation 8.37c

_ Apsfps + Asfy _ 1.99 x 205,000 + 1.76 x 60,000 a — 0.85fc'b — 0.85 x 5,000 x 18

Hence, the equivalent compressive block is inside the flange and the section has to be treated as rectangular. Accordingly, for 5000 psi concrete b1 — 0.85 — 0.05 — 0.80

d — 40 — (1.5 + 2 in. for stirrups +in. for bar) ffi 37.6 in.

From Equation 4.37c, the increment of strain due to overload to the ultimate is f d — c\ f37.6 — 8.39\ e3 — eci-1 — 0.0031-—-1 — 0.0104 in./in. :> 0.005 in./in. OK.

The total strain is

— 0.0055 + 0.0004 + 0.0104 — 0.0163 in./in. From the stress-strain diagram in Figure 8.1, the fps corresponding to eps — 0.0163 is 230,000 psi. Second trial for fps value. Assume fps — 229,000 psi

a =-= 7.34 in., consider section as a rectangular beam

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