Ex7.1 - CEE 415 Example 7.1: We are to check deflections...

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CEE 415 1 Example 7.1: We are to check deflections and bar spacing (crack width control) for an interior span of a multi-span floor beam. The floor beams frame between columns (18 x 18 in.), so the total span is 24 ft. and the clear span is 22.5 ft. Assume the total distributed dead load is 1.4 k/ft. and the distributed live load is 1.0 k/ft., and 20% of the live load is sustained for periods of two years. Use f c ' = 4000 psi, f y = 60 ksi. Sketches of the span and sections at the end and midspan are given at the end of this problem. Case 1: Assume the beam does not support partitions From Table 9.5(a), read h(min) = ℓ/18.5 for an exterior span, which will govern for this multi-span floor beam. Using ℓ = 24 ft., h(min) = 15.6 in. For this beam, h = 18 in. > h(min), so no deflection checks are required. Case 2: Assume the beam is supporting partitions or other nonstructural elements that are likely to be damaged by large deflections. For this case we must check deflections as governed by row 3 of Table 9.5(b). There we find that the permitted total deflection after the attachment of partitions must be kept less than /480. For a span length of 24 ft., this allowable deflection is 0.60 in. Assume that 85 percent of the dead load is acting when the partitions are installed. We will assume that the midspan deflection due to dead load for this interior span can be estimated using the deflection formula for Case 1 in Table 9-2, which is: 34 2.6 10 w EI
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CEE 415 2 For the live load, use the deflection coefficient from row 4 of Table 9-2: 34 4.8 10 w EI For E , we will use E c
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Ex7.1 - CEE 415 Example 7.1: We are to check deflections...

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