# m4l9 - Module 4 Doubly Reinforced Beams Theory and Problems...

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Module 4 Doubly Reinforced Beams – Theory and Problems Version 2 CE IIT, Kharagpur

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Lesson 9 Doubly Reinforced Beams – Theory Version 2 CE IIT, Kharagpur
Instructional Objectives: At the end of this lesson, the student should be able to: design the amounts of compression and tensile reinforcement if the b , d , d' , f ck , f y and M u are given, and determine the moment of resistance of a beam if b , d , d' , f ck , f y , A sc and A st are given. 4.9.1 Introduction This lesson illustrates the application of the theory of doubly reinforced beams in solving the two types of problems mentioned in Lesson 8. Both the design and analysis types of problems are solved by (i) direct computation method, and (ii) using tables of SP-16. The step by step solution of the problems will help in understanding the theory of Lesson 8 and its application. 4.9.2 Numerical problems 4.9.2.1 Problem 4.1 Design a simply supported beam of effective span 8 m subjected to imposed loads of 35 kN/m. The beam dimensions and other data are: b = 300 mm, D = 700 mm, M 20 concrete, Fe 415 steel (Fig. 4.9.1). Determine f sc from d'/d as given in Table 4.2 of Lesson 8. Version 2 CE IIT, Kharagpur

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(a) Solution by direct computation method Dead load of the beam = 0.3 (0.7) (25) = 5.25 kN/m Imposed loads (given) = 35.00 kN/m Total loads = 5.25 + 35.00 = 40.25 kN/m Factored bending moment = 2 1 5 40 25 8 8 (1 5 482 96 88 wl ( . ) ( . ) ( )( ) . ) . == kNm Assuming d' = 70 mm, d = 700 - 70 = 630 mm 48 0 max . d x u, = gives x u , max = 0.48 (630) = 302.4 mm Step 1: Determination of M u , lim and A st , lim ck u, u, u, f b d d x . - d x . M 2 max max lim ) 42 0 1 ( ) ( 36 0 = (4.2) = 0.36(0.48) {1 – 0.42 (0.48)} (300) (630) 2 (20) (10 -6 ) kNm = 328.55 kNm ) 42 0 ( 87 0 max lim lim u, y u, st, x . d - f . M A = (6.8) So, 2 6 1 mm 14 1809 } 630 ) 48 0 ( 42 0 630 { ) 415 ( 87 0 Nmm ) 10 ( 55 328 . . . . . A st = = Step 2: Determination of M u2 , A sc , A st2 and A st (Please refer to Eqs. 4.1, 4.4, 4.6 and 4.7 of Lesson 8.) 2 482 96 328 55 154 41 uu u , l i m M M - M . - . . = kNm Here, d'/d = 70/630 = 0.11 From Table 4.2 of Lesson 8, by linear interpolation, we get, Version 2 CE IIT, Kharagpur
8 350 5 342 353 353 . - - f sc = = N/mm 2 2 6 2 mm 806.517 N/mm 70) - (630 )} 20 ( 446 . 0 8 . 350 { Nmm ) (10 154.41 ) ( ) ( = = = d - d' - f f M A cc sc u sc 2 2 mm 694 763 ) 415 ( ) 87 0 ( ) 92 8 8 350 ( 517 806 87 0 ) ( . . . - . . f . - f f A A y cc sc sc st = = = 2 2 1 mm 834 2572 621 783 14 1809 . . . A A A st st st = + = + = Step 3: Check for minimum and maximum tension and compression steel. (vide sec.4.8.5 of Lesson 8) (i) In compression: (a) Minimum 2 mm 420 (700) (300) 100 0.2 = = sc A (b) Maximum 2 mm 8400 (700) (300) 100 4 = = sc A Thus, 420 mm 2 < 806.517 mm 2 < 8400 mm 2 . Hence, o.k. (ii) In tension: (a) Minimum 2 mm 387.1 415 (630) (300) 0.85 85 0 = = f b d . A y st (b) Maximum 2 mm 8400 (700) (300) 100 4 = = st A Here, 387.1 mm 2 < 2572.834 mm 2 < 8400 mm 2 . Hence, o.k. Step 4: Selection of bar diameter and numbers. (i) for A sc : Provide 2-20 T + 2-12 T (= 628 + 226 = 854 mm 2 ) (ii) for A st : Provide 4-25 T + 2-20 T (= 1963 + 628 = 2591 mm 2 ) Version 2 CE IIT, Kharagpur

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It may be noted that A st is provided in two layers in order to provide adequate space for concreting around reinforcement. Also the centroid of the tensile bars is at 70 mm from bottom (Fig. 4.9.1). (b) Solution by use of table of SP-16 For this problem, 0.11.
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m4l9 - Module 4 Doubly Reinforced Beams Theory and Problems...

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