# Info

Total compressive stress = 0.849 + 0.469 = 1.381 ksi This is less than 1.8 ksi. Therefore, design is OK. Check the design against positive moment of 14.33 kip-ft:

Check the design against positive moment of 14.33 kip-ft: This is less than 0.380 ksi. Therefore, end bay design is OK.

7.3.6.2.2. Example 2: Post-Tensioned Continuous Beam. Refer to Fig. 7.42 for dimensions and loading. Determine flange width of beam using the criteria given in ACI 318-99.

The flange width bf is the least of:

2. Web width + 16 x (flange thickness)

3. Web width + ^ clear distance to next web

Therefore Section properties:

I = 16,650 in.4 Y = 7.69 in. St = 2637 in.8 Sb = 2166 in.3 A = 1065 in.2

 Dead load of 7 y in slab = 94 psf Mech. & elec. = 6 psf Ceiling = 6 psf Partitions = 20 psf Additional dead load 615 x 60 x 150 13 5 due to beam self wt 144 x 30 = 14 psf Total dead load = 140 psf Live load at owner's request = 80 psf D + L = 220 psf

Uniform load per ft of beam = 0.220 x 30 = 6.6 klf. The resulting service load moments are shown in Fig. 7.43. As before we design for the moments at the face of supports.

Interior Span. Calculate through tendons by using interior span moment of 427 kip-ft at the inside face of third column (Fig. 7.43).

Assume h1 = 11.5 in., h2 = 2.5 in., Lx = 2.5 ft, and L2 = 12.5 ft. Refer to Fig. 7.41 for notations.

The height of inflection point

3 15

 Wp = 3.5 klf 