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Structural Concrete - Hassoun

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17.12 Equivalent Frame Method 683<br />

the face of the column and the edge of the support. Circular section columns must be treated<br />

as square columns with the same area.<br />

8. Sum of moments: A two-way slab floor system that satisfied the limitations of the direct design<br />

method can also be analyzed by the equivalent frame method. To ensure that both methods will<br />

produce similar results, the ACI Code, Section 8.11.6.5, states that the computed moments<br />

determined by the equivalent frame method may be reduced in such proportion that the numerical<br />

sum of the positive and average negative moments used in the design must not exceed<br />

the total statical moment, M 0 .<br />

Example 17.13<br />

Using the equivalent frame method, analyze a typical interior frame of the flat-plate floor system given<br />

in Example 17.3 in the longitudinal direction only. The floor system consists of four panels in each<br />

direction with a panel size of 25 × 20 ft. All panels are supported by a 20 × 20-in. columns, 12 ft long.<br />

The service live load is 60 psf and the service dead load is 124 psf (including the weight of the slab).<br />

Use f c ′ = 3ksiandf y = 60 ksi. Edge beams are not used. Refer to Fig. 17.38.<br />

Solution<br />

1. A slab thickness of 8.0 in. is used.<br />

2. Factored load is q u = 1.2 × 124 + 1.6 × 60 = 245 psf. The ratio of service live load to service dead<br />

load is 60/124 = 0.48 < 0.75; therefore, the frame can be analyzed with the full factored load, q u ,<br />

acting on all spans instead of pattern loading.<br />

3. Determine the slab stiffness, K s :<br />

where k is the stiffness factor and<br />

I s = l 2 h3<br />

12<br />

K s = k E cs I s<br />

l s<br />

20 × 12<br />

= (8) 3 = 10,240 in. 4<br />

12<br />

The stiffness factor can be determined by the column analogy method described in books on<br />

structural analysis. Considering the moment of inertia for the slab I s to be 1.0 as a reference, the<br />

moment of inertia between the column centerline and the face of the column is<br />

1.0<br />

(1 − c 2 ∕l 2 ) = 1.0<br />

2 [1 − 20∕(20 × 12)] = 1.19<br />

2<br />

The width of the analogous column varies with 1/I, as shown in Fig. 17.38b: (1/1.19) = 0.84:<br />

(<br />

1<br />

Slab stiffness factor k = l 1 + Mc<br />

)<br />

A a I a<br />

where<br />

A a = area of analogous column section<br />

I a = moment of inertia of analogous column<br />

M = moment due to unit load at extreme fiber of analogous column located at center of slab<br />

M = 1.0 × l 1<br />

2<br />

A a = 23.33 + 2 × (0.83 ft)(0.84) = 23.33 + 1.40 = 24.72<br />

I a = I (for slab portion of 23.33) + I (of end portion) about centerline<br />

(<br />

I a = (23.33)3 + 2(0.83)(0.84) 12.5 − 0.83 ) 2<br />

12<br />

= 1263<br />

2

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