Question

Problem 2 A reinforced concrete beam (see fig.) is acted on by a positive bending moment of M - 160 kNm. Steel reinforcement consists of 4 bars of 28-mm diameter. The modulus of elasticity for the concrete is E = 25 GPa while that of the steel is E, = 200 GPa. (a) Find the maximum stresses in steel and concrete. (b) If allowable stresses for concrete and steel are = 9.2 MPa and 0 -135 MPa, respectively, what is the maximum permissible positive bending moment? (c) What is the required area of steel reinforcement, As, if a balanced condition must be achieved? What is the allowable positive bending moment? (Recall that in a balanced design, both steel and concrete reach allowable stress values simultaneously under the design moment). E = 25 GPa E, = 200 GPa 625 mm 4 bars @28-mm n -8 M - 160 kN.m Pac = 9.2 MPa as = 135 MPa b = 300 mm d = 625 mm do - 28 mm A, - 45d,2 - 2.463 x 10–3 m? -ooo of - 300 mm

Answer 1

Solution Given that 2 2s Стра , Е.- оо еро- n = Es = 8 ; m=160 kv.m Tac = 9.2 mpa i Tas = 135 mpa b=300 d=625 į ds = 28mm Ag = 2.463 x 103 m a) maximum stress in steel and Concrete: first take moments of transformed locate Nuetral axis (PA). section to → by. n As (dy) =0 y =) (**+ 2 af 20) - (ny) 228..283 mm. moment of inertia of transformed Section Isa bus + As(tyr IN 300x228.28+8(2.463x10°) (625-228-28)

> 4.2918103 m Stress in Concrete; oes my - 160 x 228.28 4.291x103 F 8.51 mpa Stren in steel ag = namedy) Top - 8x180x (625-228.28) 4.29 14203 . = 118.3 mpa h maximum permissible positive moment, m=160kn.m. Mc= I. = 9.2 x 4-2914163 - 228.28 = 172.9 KN-m. mg = as To - 135x4.291x103 hld-y) X10 = 182.5kom L 8 (625-228.28) alle Concrele governy immax = mc = 1729 KN-m) e) Balanced design required area & steel, b=300mm, de 625mm. Equate moment capacity of Concrete Esteel then solve fory,

a noac = 8x 9.2 Tas 735 0.54518 - y = 220.517 mm use required y in 1st moments equation to find required As by (4) - n As (day) =0 As 300 (220.517) (220.517) -8 As (625-220.51) solving for in Az = 2254 mm Sneanom od ovilicon Valle we Recompute IT Dy a byla nas (dy) "7 30 = 300 220-513 + 8X2254x (625 - 220.51) = 4.023x103 mm. (6)

Now P = allowable positive bending moments are equal for balanced design. Mc = "ac IT _ 1167.8 Kuom] (izon me I rastr t Knim. n (dy) Drago! in As=2254 min & Mallow = 167.8 kNm.