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This is subtracted from the total service load moment of 14.7 kip-ft to obtain the unbalanced moment Mub.

The flexural stresses at top and bottom are obtained by dividing Mub by the section moduli of the structure's cross section.

The minimum required compressive prestress fp is found by subtracting the maximum allowable tensile stress fa = from the calculated tensile stress. Thus the smallest required compressive stress is:

fp ft fa

= 0.567 - 380 = 0.187 ksi The prestress force is calculated by multiplying fp by the cross-sectional area: P = 0.187 x 9 x 12 = 20.20 kip-ft

Determine the equivalent load due to prestress force P by the relation

For the example problem,

P = 20.20 kip-ft, e = 5.90'' Le = 2 x 12.90 = 25.8 ft

Therefore W„ =---= 0.120 klf = 120 plf p 25.82 x 12

Comparing this with the value of 93 plf assumed at the beginning of first cycle, we find the two values are not equal. Therefore, we assume a new value and repeat the procedure until convergence is obtained.

Second Cycle

fp = 0.527 - 0.380 = 0.147 ksi P = 0.147 x 9 x 12 = 15.92 kip-ft

This is less than 99 plf assumed at the beginning of second cycle. Therefore, we assume a new value and repeat the procedure.

Third Cycle

7.22 x 12 f = fb = 162 = 0.535 ksi f„ = 0.535 - 0.380 = 0.155 ksi p

This is nearly equal to 97.7 plf assumed at the beginning of the third cycle. Therefore OK. Check compressive stress at the support:

Axial compressive stress due to post-tension =-= 155 psi

Total compressive stress = 527 + 155 = 682 psi

This is less than the allowable compressive stress of 1800 psi. Therefore, the design is satisfactory.

End Bay Design. The placement of tendons within the end bay presents a few problems. The first problem is in determining the location of the tendon over the exterior support. Placing the tendon above the neutral axis of the member results in an increase in the total tendon drape, allowing the designer to use less prestress than would otherwise be required. Raising the tendon, however, introduces an extra moment that effectively cancels out some of the benefits from the increased drape. For this reason, the tendon is usually placed at a neutral axis at exterior supports.

The second problem is in making a choice in the tendon profile: whether to use a profile with a reverse curvature over each support (see Fig. 7.45, profile 1), or over the first interior support only (see Fig. 7.45, profile 2). A profile with the reversed curvature over the first interior support only gives a greater cable drape than the first profile, suggesting a larger equivalent load with the same amount of prestress. On the other hand, the effective length Le between inflection points of profile 1 is less than that of profile 2 which suggests the opposite. To determine which profile is in fact more efficient, it is necessary to evaluate the amount of prestress for both profiles. More usually, a tendon profile with reverse curvature over both supports is 5 to 10% more efficient since the equivalent load produced is a function of the square of the effective length.

The last item addresses the extra end bay prestressing required in most situations. The exterior span in an equal span structure has the greatest moments due to support rotations. Because of this, extra prestressing is commonly added to end bays to allow efficient design of end spans. For design purposes, the extra end bay prestressing is considered to act within the end bay only. These tendons actually extend well into the adjacent span for anchorage, as shown in Fig. 7.46. Advantage can be taken of this condition by designing the through tendons using the largest moment found within the interior spans, including the moment at the interior face of the first support. The end bay prestress force is determined using the largest moment within the exterior span. The stress at the inside face of the first support is checked using the equivalent loads produced by the through tendons and the axial compression provided by both the through and added tendons. If the calculated stresses are less than the allowable values, the design is complete. If not, more stress is provided either by through tendons or added tendons or both.

The design of end bay using profiles 1 and 2 follows.

Profile 1:Reverse Curvature at the Right Support Only (Fig. 7.47b). Observe that the height of inflection point is exact if the tendon profile is symmetrical about the center of span. If it is not, as in span 1 of the example problem, sufficiently accurate value can be obtained by taking the average of the tendon inflection point at each end as follows.

Figure 7.46. Anchorage of added tendons.
Figure 7.47. Example problem 3: flat plate, tendon profiles: (a) interior span; (b) exterior span, reverse curvature at right support; (c) exterior span, reverse curvature at both supports.

Left end:

Right end:

First Cycle To show the quick convergence of the procedure, we start with a rather high value of

= 8 x 15.87 x 4.11 = 29.352 x12 = 0.050 klf = 50.0 plf

This is less than 106 plf. N.G.

Second Cycle

This is less than 91.5 psi used at the beginning of second cycle. N.G.

Third Cycle

f = fb = 162 = 0.631 ksi fp = 0.631 - 0.380 = 0.251 ksi p

This is nearly equal to 89 plf used at the beginning of third cycle. Therefore OK. Profile 2. Reverse Curvature Over Each Support (Fig. 7.47c).

Left end:

Right end:

First Cycle We start with an assumed balanced load of 0.65 DL = 92 plf.

Balanced moment Mb

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