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

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 =29x106 psi and Ec= 36x105 psi
W=D+L = = 3.5K/ft
Mmax= wl2/8 = 3.5*20*20/8= 175 ft-K=175*12=2100 in-K
The sizes of the beams:
The modular ratio n = Es/Ec= 9x106 /36x105= 8
= = j= 1-k/3 = /3 = 0.894
; bd2 = M/0.5fc 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 = or 2.25b 3 = or b = 17.1 in. We may take b = 18 in Now 18d 2 = or d 2 = 11364/18 or d= 25 say 12 in
Over all depth = in ( cover)+0.5= 127 in ( beam size 18 inX27in)
Design of steel As = M/fsjd = 2100*1000/22300*0.894*25 = 4.21 in2
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 LL

11. For fc’ < 4000 psi ß= 0.85
For fc’ > 4000 psi ß = (fc’-4000)/ > 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 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=1.49in2 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 ϕMn
Mn = ϕRbd2
Using Design Table A.4 select the appropriate reinforcement ratio between ρmin and ρmax Often ρ = 0.60 ρmax is more economical.
From Table A.5 for specified material strengths and selected reinforcement, find flexural resistance factor R then
bd2 = Mn /ϕR
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/ ϕbd2
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: