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Structural Concrete - Hassoun

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17.12 Equivalent Frame Method 687<br />

Figure 17.40 Example 17.13: Equivalent frame method— final bending moments and<br />

shear forces. (Slabs can be designed for the negative moments at the face of the<br />

columns as shown.)<br />

of the bending moment diagram, considering the variation of the moment of inertia along the<br />

span, is<br />

Total area (A m )=A 1 + A 2 + 2A 3<br />

Fixed − end moment coefficient = A m<br />

A a l 2 1<br />

where A a for the slab is 24.72, as calculated in step 3:<br />

k ′′ 1300<br />

=<br />

24.72(25) = 0.084<br />

2<br />

= 2 × 23.33(78.1 − 10)+23.33 × 10<br />

3<br />

( )<br />

1<br />

+ 2<br />

2 × 0.83 × 10 (0.84) =1300

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