m3l7 - Module 3 Limit State of Collapse Flexure (Theories...

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Module 3 Limit State of Collapse - Flexure (Theories and Examples) Version 2 CE IIT, Kharagpur
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Lesson 7 Numerical Problems on Singly Reinforced Rectangular Beams (Continued) Version 2 CE IIT, Kharagpur
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Instructional Objectives: At the end of this lesson, the student should be able to: Apply the principles to analyse a given cross-section of a beam with specific reinforcement to determine its moment of resistance. 3.7.1 Introduction This lesson explains the determination of moment of resistance of given singly reinforced rectangular beam sections with the help of illustrative analysis type of problem. The numerical problem is solved by (i) direct computation method, (ii) using charts of SP-16 and (iii) using tables of SP-16. Step by step solutions illustrate the procedure clearly. 3.7.2 Analysis Type of Problems It may be required to estimate the moment of resistance and hence the service load of a beam already designed earlier with specific dimensions of b , d and D and amount of steel reinforcement A st . The grades of concrete and steel are also known. In such a situation, the designer has to find out first if the beam is under-reinforced or over-reinforced. The following are the steps to be followed for such problems. 3.7.2.1 x u , max The maximum depth of the neutral axis x u , max is determined from Table 3.2 of Lesson 5 using the known value of f y . 3.7.2.2 x u The depth of the neutral axis for the particular beam is determined from Eq. 3.16 of Lesson 5 employing the known values of f y , f ck , b and A st . 3.7.2.3 M u and service imposed loads The moment of resistance M u is calculated for the three different cases as follows: (a) If x u < x u , max , the beam is under-reinforced and M u is obtained from Eq. 3.22 of Lesson 5. (b) If x u = x u , max , the M u is obtained from Eq. 3.24 of Lesson 5. Version 2 CE IIT, Kharagpur
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(c) If x u > x u , max , the beam is over-reinforced for which x u is taken as x u , max and then M u is calculated from Eq. 3.24 of Lesson 5, using x u = x u , max . With the known value of M u , which is the factored moment, the total factored load can be obtained from the boundary condition of the beam. The total service imposed loads is then determined dividing the total factored load by partial safety factor for loads (= 1.5). The service imposed loads are then obtained by subtracting the dead load of the beam from the total service loads. 3.7.3 Analysis Problems 3.2 and 3.3 Determine the service imposed loads of two simply supported beam of same effective span of 8 m (Figs. 3.7.1 and 2) and same cross-sectional dimensions, but having two different amounts of reinforcement. Both the beams are made of M 20 and Fe 415. 3.7.4
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m3l7 - Module 3 Limit State of Collapse Flexure (Theories...

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