1. By Dr. Attaullah Shah Swedish College of Engineering and Technology Wah Cantt. Reinforced Concrete Design-3 Flexural Design of Beams

2. Working Stress Design Method

3. Design of beam by Working Stress method −Design a simply supported beam of span 20 feet carrying a live load of 1.5 K/ft and dead load of 2 K/ft. the material stresses are fc=1300 psi and fs=22300 psi. −Es =29x10 6 psi and Ec= 36x10 5 psi −W=D+L = 1.5+2 = 3.5K/ft −Mmax= wl 2 /8 = 3.5*20*20/8= 175 ft-K=175*12=2100 in-K −The sizes of the beams: − The modular ratio n = Es/Ec= 9x10 6 /36x10 5= 8 − = = 0.318 j= 1-k/3 = 1-0.318/3 = 0.894 − ; bd 2 = M/0.5f c k j = 2100*1000/(0.5*1300*0.318*0.894=11364 in 3 Using d/b = 1.5 we get d=1.5b or b(1.5b) 2 = 11364 or 2.25b 3 = 11364 or b = 17.1 in. We may take b = 18 in Now 18d 2 = 11364 or d 2 = 11364/18 or d= 25 say 12 in Over all depth = 25+1.5 in ( cover)+0.5= 127 in ( beam size 18 inX27in) Design of steel As = M/fsjd = 2100*1000/22300*0.894*25 = 4.21 in 2 Using # 8 bars No of bars = 4.21/0.87 = 5 bars

6. Safety Margins

7. ACI Provisions: -Design strength > Required strength -Su = 1.2 DL + 1.6 LL

11. For f c ’ < 4000 psi ß= 0.85 For f c ’ > 4000 psi ß = 0.85 - 0.05 (fc’-4000)/1000 0.65> ß>0.85

13. Under reinforced Reinforced Concrete beams −The compression failure occurs due to crushing of concrete, which is sudden and give no warning. −The tension failure is caused by yielding of steel bars. −Due to yielding of concrete beams having large deflections, the cracks can be observed and preventive measures can be taken. −The steel ratio is taken less than balanced steel to ensure ductile failure − 200/fy

14. Strength Reduction Factor ϕ

16. Flexural Design of Rectangular beam −Design a rectangular simply supported beam subjected to computed dead load of 1.27 k/ft and live load of 2.15 k/ft. fc’ = 4000 psi and fy=60,000 psi.

18. Example: 3.7 No further iteration is required. Use As=1.49in 2 2#8 bars (No.25) bars will be used Check whether steel ratio is less than maximum allowed by the ACI Code.

19. Example 3.8

20. Use of Design Aids − We can directly use design the beam by using design charts. − The optimum reinforcement is selected. − Set the required strength Mu equal to the design strength ϕ Mn − Mn = ϕ Rbd 2 − Using Design Table A.4 select the appropriate reinforcement ratio between ρ min and ρ max Often ρ = 0.60 ρ max is more economical. − From Table A.5 for specified material strengths and selected reinforcement, find flexural resistance factor R then − bd 2 = Mn / ϕ R − Select the appropriate width and depth of beam. ( d=2-3b) − Calculate the areas of steel As= ρbd − Select the number and size of bars ( prefer larger sizes) − Alternatively we can assume the values of b and d and determine R =Mn/ ϕ bd 2 − Using Table A.5 find the reinforcement ratio ρ< ρmax that will provide the required value of R − As= ρbd

25. Table A.5 a Flexural Resistance Factor

29. Assignment 2. −Group 1 −Group 2: −Group 3: