Reinforced Concrete Design-3
Flexural Design of Beams
By
Dr. Attaullah Shah
Swedish College of Engineering and Technology
Wah Cantt.
Working Stress Design Method
Design of beam by Working Stress method
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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.
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Es =29x106 psi and Ec= 36x105 psi
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W=D+L = 1.5+2 = 3.5K/ft
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Mmax= wl2/8 = 3.5*20*20/8= 175 ft-K=175*12=2100 in-K
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The sizes of the beams:
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The modular ratio n = Es/Ec= 9x106 /36x105= 8
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=
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= 0.318
j= 1-k/3 = 1-0.318/3 = 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 = 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 in2
Using
# 8 bars No of bars = 4.21/0.87 = 5 bars
Safety Margins
ACI Provisions:
- Design strength > Required strength
- Su = 1.2 DL + 1.6 LL
For fc’ < 4000 psi
For fc’ > 4000 psi
0.65> ß>0.85
ß= 0.85
ß = 0.85 - 0.05 (fc’-4000)/1000
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
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200/fy
Strength Reduction Factor ϕ
Flexural Design of Rectangular beam
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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.
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.
Example 3.8
Use of Design Aids
− We can directly use design the beam by using design charts.
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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
Table A.5 a Flexural Resistance Factor
Assignment 2.
− Group 1
− Group 2:
− Group 3: