SERVICEABILITY

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Engineering

Montgomery College Rockville Campus

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SERVICEABILITY 1. A rectangle beam has 3 longitudinal rebars against flexure. The maximum center-to-center spacing of the longitudinal reinforcement can be limited to control cracking in flexural members. It shall be calculated using the following equation per ACI Code. 𝑠 = 15 ( 40,000 40,000 ) – 2.5 𝑐𝑐 ≤ 12 ( ) 𝑓𝑠 𝑓𝑠 (a) Sketch the state of the cross section that needs to be considered to calculate 𝑓𝑠 . I.e., the cracked part of cross section, the equivalent transformed section, the variation of strain and stress shall be sketched. See the example shown below for a fully elastic and linear cross section. Figure. Cross section with no cracks. No cracks → Equivalent transformed section → Strain → Stress 2 (b) Calculate the maximum center-to-center spacing if 𝑓𝑠 is approximated as 𝑓𝑠 = 3 𝑓𝑦 . The clear cover 𝑐𝑐 is 2.5 in., and Grade 80 bars are to be used. 2. The effective moment of inertia shall be computed per ACI Code using the equation below. 𝐼𝑒 = 𝐼𝑐𝑟 (2⁄3)𝑀𝑐𝑟 2 𝐼𝑐𝑟 1 − ( ) (1 − 𝑀𝑎 𝐼𝑔 ) ≤ 𝐼𝑔 As did for problem 1, sketch all the cross sections that need to be considered to calculate 𝐼𝑒 . For each sketched cross section, the related quantities in the formula shall be explicitly stated. Assume a rectangle cross section having longitudinal rebars at the bottom part. The area of steel shall be 𝐴𝑠 . Note that each cross section shall be presented in terms of cracked/uncracked region, equivalent transformed section, strain, and stress variations.
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