- Beams Analysis And Design For Serviceability Example 3 3 Elastic Section Analysis Doubly Reinforced Rectangular Beam T 1 (170.97 KiB) Viewed 48 times
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Beams Analysis and design for serviceability EXAMPLE 3.3 ELASTIC SECTION ANALYSIS - DOUBLY REINFORCED RECTANGULAR BEAM The doubly reinforced rectangular section of Figure 3.12 contains 2N24 bars for top reinforcement and 4N24 bars for bottom reinforcement. Determine the neutral axis depth dy, and the concrete and steel stresses induced by a positive bending moment of 80 kNm, given that the section is cracked but still in the elastic range, and that the elastic moduli are: Ec = 30100 MPa. E = 200x100 MPa. Eso = &old - 60)/d, (3.34) Est = {(390 - d.)/d (3.35) Substituting Equations 3.34 and 3.35 into Equations 3.32 and 3.33, respec- tively, and noting that EC - ET = 0, gives: 0 + 119.6d- 33490 = 0 (3.36) + Solving for the positive root of Equation 3.36 gives dn = 133 mm. 300 60 2N24 450 390 Asc= 900 mm2 Ast=1800 mm It is worthy of note that if solving the doubly reinforced problem of Example 3.3 by the modular ratio method we would find that dn = 136 mm. The difference in the results comes about from the way that holes in the com- pression concrete due to the reinforcing steel are treated in each method. In the modular ratio method the holes are deducted in the conversion of the steel to an equivalent area of concrete via the -1 in the (n-1) term applied to the area of compression steel. This is done as it is relatively easy, and convenient, to do so. In the application of the method of equilibrium and compatibility, however, removing the holes from the compression concrete adds undesirable complexity and, thus, the holes are usually ignored. The relative error induced by this assumption is small. 4N24 Figure 3.12 Doubly reinforced rectangular beam: Example 3.3 Calculating the stresses in the concrete and reinforcing steel for an applied moment of 80 kNm by taking moments about the tensile reinforcement: SOLUTION The equilibrium and compatibility conditions for the section are given in Figure 3.11. Forces on the cross-section (Equations 3.24 to 3.26): 300 Cc = 30100 x 2 ex € dn= 4.515x10°£,dn N (3.31) M= C (d-4/3)+C (d-dsc) = (3.37) 80x10º = 346C+ 330C, Nmm From Equations 3.31, 3.32 and 3.34: C. = 600x10'e, N C = 98.8x10°€. N (3.38) and substituting Equation 3.38 into Equation 3.37 and solving gives € = 333x10 X (3.32) C = 200x10' x 900 x 8 = 180.0x10°c N To = 200x10° x 1800 x£x= 360.0x10'est sc N (3.33) Compatibility (Equations 3.27 and 3.28):