DesignOfDoublyReinforcedBeams CE153
DesignOfDoublyReinforcedBeams CE153
DesignOfDoublyReinforcedBeams CE153
F e l i x V. Ga r d e , J r .
4. Fabrication ease.
𝑀𝑛 = 𝑀𝑛1 + 𝑀𝑛2
where,
𝑀𝑛1 = 𝐴′𝑠 𝑓 𝑦 (𝑑 − 𝑑 ′ )
𝑎
𝑀𝑛2 = ( 𝐴𝑠 − 𝐴′𝑠 ) 𝑓 𝑦 𝑑 −
2
and
( 𝐴𝑠 − 𝐴′𝑠 ) 𝑓 𝑦
𝑎=
0.85 𝑓𝑐′ 𝑏
Fel ix V. Ga r d e , J r . (Institute of Computing and Engineering
Reinforced
DAVAOConcrete
ORIENTAL
DesignSTATE UNIVERSITY) 9 / 39
Case I: ( 𝑓𝑠 = 𝑓𝑠′ = 𝑓 𝑦 )
Limiting values that will ensure yielding of the compression steel at failure
• minimum tensile reinforcement ratio
𝑓𝑐′ 𝑑 ′ 0.003
𝜌¯ 𝑐𝑦 = 0.85𝛽1 + 𝜌′
𝑓 𝑦 𝑑 0.003 − 𝜖 𝑦
Limiting values that will ensure yielding of the compression steel at failure
• (𝑎/𝑑 ′ )lim ratio
600𝛽1
(𝑎/𝑑 ′ )lim =
600 − 𝑓 𝑦
Strain-Compatibility Check:
𝑐 − 𝑑′
𝜀 ′𝑠 = 0.003
𝑐
𝑓𝑠′ = 𝜀 ′𝑠 𝐸 𝑠 ≤ 𝑓 𝑦
Í
Applying condition of equilibrium on internal forces, ( 𝐹𝐻 = 0)
Strain-Compatibility Check:
𝑑𝑡 − 𝑐
𝜀 𝑡 = 0.003
𝑐
𝐴𝑠 = 𝐴𝑠1 + 𝐴𝑠2
5. Analyze the beam. If (𝑎/𝑑 ′ )lim > (𝑎/𝑑 ′ )act , then 𝑓𝑠′ < 𝑓 𝑦 and the
compression steel area must be increased.
6. Revise the compression area using
′ ′ 𝑓𝑦
𝐴s,revised = 𝐴s,trial
𝑓𝑠′
where,
𝑐 − 𝑑′
𝑓𝑠′ = 600
𝑐
Fel ix V. Ga r d e , J r . (Institute of Computing and Engineering
Reinforced
DAVAOConcrete
ORIENTAL
DesignSTATE UNIVERSITY) 19 / 39
Maximum Reinforcement Ratio, 𝜌¯ max
𝐴′ 0.85 𝑓 𝑐′ 𝛽1 𝑑𝑡
𝐴𝑠 0.003
= 𝑠 + · ·
𝑏𝑑 𝑏𝑑 𝑓𝑦 𝑑 0.003 + 𝜀𝑡
Let,
𝐴𝑠
𝜌¯ = ; tension steel reinforcement ratio for doubly-reinforced
𝑏𝑑
rectangular beam
𝐴′
𝜌′ = 𝑠 ; compression steel reinforcement ratio
𝑏𝑑
1 For 𝑓 𝑠 = 𝑓 𝑦 , 𝑓 𝑠′ = 𝑓 𝑦
Then,
𝑇 = 𝐶𝑠 + 𝐶𝑐
0.85 𝑓 𝑐′ 𝛽1 𝑑𝑡
𝐴𝑠 𝑓 𝑦 = 𝐴′𝑠 𝑓 𝑦 + 0.85 𝑓 𝑐 𝑎𝑏 𝜌¯ = 𝜌′ + · ·
0.003
𝑓𝑦 𝑑 0.003 + 𝜀𝑡
𝑎 = 𝛽1 𝑐
0.003𝑑𝑡 From ACI 10.3.5, for maximum ratio, 𝜀𝑡 ≥ 0.004
𝑐=
0.003 + 𝜀𝑡 0.85 𝑓 𝑐′ 𝛽1 𝑑𝑡
0.003
𝜌max = · ·
0.003𝑑𝑡 𝑓𝑦 𝑑 0.003 + (0.004)
𝐴𝑠 𝑓 𝑦 = 𝐴′𝑠 𝑓 𝑦 + 0.85 𝑓 𝑐 𝑏𝛽1
0.003 + 𝜀𝑡
Therefore,
divide both sides by, 𝑏𝑑 𝑓 𝑦
Let,
𝐴𝑠
𝜌¯ = ; tension steel reinforcement ratio for doubly-reinforced
2 For 𝑓 𝑠 = 𝑓 𝑦 , 𝑓 𝑠′ ≠ 𝑓 𝑦 𝑏𝑑
𝑇 = 𝐶𝑠 + 𝐶𝑐 rectangular beam
𝑎 = 𝛽1 𝑐 Then,
𝑓′ 0.85 𝑓 𝑐′ 𝛽1 𝑑𝑡
0.003𝑑𝑡 0.003
𝑐= 𝜌¯ = 𝜌′ 𝑠 + · ·
0.003 + 𝜀𝑡 𝑓𝑦 𝑓𝑦 𝑑 0.003 + 𝜀𝑡
0.003𝑑𝑡 From ACI 10.3.5, for maximum ratio, 𝜀𝑡 ≥ 0.004
𝐴𝑠 𝑓 𝑦 = 𝐴′𝑠 𝑓 𝑠′ + 0.85 𝑓 𝑐 𝑏𝛽1
0.003 + 𝜀𝑡
0.85 𝑓 𝑐′ 𝛽1 𝑑𝑡
0.003
divide both sides by, 𝑏𝑑 𝑓 𝑦 𝜌max = · ·
𝑓𝑦 𝑑 0.003 + (0.004)
𝐴′ 𝑓 ′ 0.85 𝑓 𝑐′ 𝑏𝛽1 𝑑𝑡
𝐴𝑠 𝑓 𝑦 0.003 Therefore,
= 𝑠 𝑠 + · ·
𝑏𝑑 𝑓 𝑦 𝑏𝑑 𝑓 𝑦 𝑏 𝑓𝑦 𝑑 0.003 + 𝜀𝑡
𝐴′ 𝑓 ′ 0.85 𝑓 𝑐′ 𝛽1 𝑑𝑡
𝐴𝑠 0.003
= 𝑠 𝑠 + · ·
𝑏𝑑 𝑏𝑑 𝑓 𝑦 𝑓𝑦 𝑑 0.003 + 𝜀𝑡
𝑓′
𝜌¯max = 𝜌′ 𝑠 + 𝜌max
𝑓𝑦
1. Compute the design moment strength of the beam shown below if 𝑓𝑐′ = 21 MPa
and 𝑓 𝑦 = 420 MPa, Check the maximum permissible 𝐴𝑠 to ensure ductile failure.
0.85 𝑓 𝑐′ 𝛽1 𝑑𝑡
𝑎
𝑀𝑛2 = ( 𝐴𝑠 − 𝐴′𝑠 ) 𝑓 𝑦 𝑑 −
0.003
2 𝜌max = · ·
𝑓𝑦 𝑑 0.003 + (0.004)
191.176
𝑀𝑛2 = (4, 024 − 774) (420) 700 − = 825.02 kN · m 0.85(21) (0.85) 0.003
2 𝜌max = ·
420 0.003 + (0.004)
The nominal moment strength is, 𝜌max = 0.015
𝜌min = 0.0033
𝜌 = 0.014 > 𝜌min = 0.0033 Okay.
2. Compute the design moment strength of the beam shown below. If 𝑓𝑐′ = 21 MPa
and 𝑓 𝑦 = 420 MPa. Check the maximum permissible 𝐴𝑠 to ensure ductile failure?
𝜌¯max = 𝜌′ + 𝜌max
𝜌¯max = 0.013 + 0.015
𝜌¯max = 0.028
and the total tension steel reinforcement ratio is,
𝐴𝑠 4, 095
𝜌= =
𝑏𝑑 (200) (500)
𝜌 = 0.041 > 𝜌¯max = 0.028
Or,
𝑑𝑡 − 𝑐 500 − 311.486
𝜀𝑡 = 0.003 = 0.003 = 0.00182 < 0.004
𝑐 311.486
The section exceeds the maximum tension steel reinforcement ratio. To comply the requirement of ACI 10.3.5, increase the amount of 𝜌′ .
3. A rectangular beam that must carry a service dead load of 35 kN/m and a
calculated live load of 15 kN/m on a 5.5m simple span is limited in cross section
for architectural reasons to 250mm width and 500mm total depth. If 𝑓𝑐′ = 25 MPa
and 𝑓 𝑦 = 420 MPa, what steel area(s) must be provided?
69.54(5.5) 2
𝑀𝑢 = = 262.95 kN · m
8
b) Determine the required moment, 𝑀𝑢 Since, 𝜙𝑀𝑛 < 𝑀𝑢 , therefore compression reinforcements
are necessary.
𝑤𝑢 𝐿 2
𝑀𝑢 =
8
Fel ix V. Ga r d e, J r . (Institute of Computing and Engineering
Reinforced
DAVAOConcrete
ORIENTAL
DesignSTATE UNIVERSITY) 31 / 39
The total tension steel area,
𝐴𝑠 = 𝐴𝑠1 + 𝐴𝑠2
𝐴𝑠1 = 𝐴′𝑠 = 125.09 mm2
𝐴𝑠2 = 1, 763.92 mm2
𝐴𝑠 = 1, 889.01 mm2
Let, Select bars:
𝐴𝑠2 = 𝐴𝑠 = 𝜌bd Try 16 mm𝜙 for compression reinforcement.
𝐴𝑠2 = 0.0161(250) (437.5) 𝐴′𝑠 125.09
No. of bars = =
𝐴𝑠2 = 1, 763.92 mm2 𝐴16 mm𝜙 𝜋 (16) 2
𝑀2 = 𝑀𝑛 = 272.43 kN · m 4
No. of bars = 0.622 pc, say 2 pcs 16 mm𝜙
Calculate the excess moment, 𝑀1 ,
Try 20 mm𝜙 for tension reinforcement.
𝑀𝑢 = 𝜙𝑀1 + 𝜙𝑀2
𝐴𝑠 1, 889.01
𝑀𝑢 22.95 No. of bars = =
𝑀1 = − 𝑀2 = − 272.463 𝐴20 mm𝜙 𝜋 (20) 2
𝜙 0.9
4
𝑀1 = 19.702 kN · m
No. of bars = 6.013 pc, say 6 pcs 20 mm𝜙
Assume, 𝑓 𝑠′ = 𝑓 𝑦 , then
𝑇 = 𝐶𝑠 + 𝐶𝑐
𝐴𝑠 𝑓 𝑦 = 𝐴′𝑠 𝑓 𝑠′ + 0.85 𝑓 𝑐′ 𝑎𝑏
𝐴𝑠 𝑓 𝑦 = 𝐴′𝑠 ( 𝜖𝑠′ 𝐸𝑠 ) + 0.85 𝑓 𝑐′ (𝛽1 𝑐) 𝑏
𝑐 − 𝑑′
𝐴𝑠 𝑓 𝑦 = 𝐴′𝑠 0.003 𝐸𝑠 + 0.85 𝑓 𝑐′ (𝛽1 𝑐) 𝑏
Where, 𝑐
𝑑 ′ = 60 mm
𝑑 = 417.5 mm 𝑑−𝑐
𝜖𝑠 = 0.003 = 0.00569
𝑐
Assume, 𝑓 𝑠 = 𝑓 𝑠′ = 𝑓 𝑦
𝑐−𝑑 ′
( 𝐴𝑠 − 𝐴′𝑠 ) 𝑓 𝑦 (1885.96 − 402.12) (420) 𝜖𝑠′ = 0.003 = 0.00175
𝑎= = 𝑐
0.85 𝑓 𝑐′ 𝑏 0.85(25) (250)
The nominal strength is,
𝑎 = 117.231 mm; 𝑐 = 𝑎/𝛽1 = 137.92 mm
𝑀𝑛 = 𝑀𝑛1 + 𝑀𝑛2
Verify assumptions;
𝑀𝑛1 = 𝐴′𝑠 𝑓 𝑠′ (𝑑 − 𝑑 ′ )
𝑑−𝑐 𝑀𝑛1 = (405.124) (0.00175) (200, 000) (417.5 − 60)
𝜖𝑠 = 0.003 = 0.00608
𝑐
𝑀𝑛1 = 50.35 kN · m
𝑐−𝑑 ′
𝜖𝑠′ = 0.003 = 0.00169
𝑐
Fel ix V. Ga r d e, J r . (Institute of Computing and Engineering
Reinforced
DAVAOConcrete
ORIENTAL
DesignSTATE UNIVERSITY) 33 / 39
Try this section.
𝑎
𝑀𝑛2 = 0.85 𝑓 𝑐′ 𝑎𝑏 𝑑 −
2
122.51
𝑀𝑛2 = 0.85(25) (122.51) (250) 417.5 −
2
𝑀𝑛2 = 231.86 kN · m
𝑀𝑛 = 50.35 kN · m + 231.86 kN · m
𝑀𝑛 = 282.21 kN · m
where,
Determine the strength reduction factor, 𝜙
𝑓 𝑐′ = 25 MPa 𝑑𝑡 = 440 mm
𝑑𝑡 − 𝑐 𝑓 𝑦 = 420 MPa 𝐴𝑠 = 1, 885.96 mm2
𝜖𝑡 = 0.003 = 0.00616
𝑐
𝑏 = 250 mm 𝐴′𝑠 = 942.478 mm2
The section is tension-controlled, therefore 𝑑 = 417.5 mm 𝑑 ′ = 60 mm
𝑀𝑛 = 𝑀𝑛1 + 𝑀𝑛2
𝑀𝑛1 = 𝐴′𝑠 𝑓 𝑠′ (𝑑 − 𝑑 ′ )
𝑀𝑛1 = (942.478) (0.00144) (200, 000) (417.5 − 60)
𝑀𝑛1 = 96.94 kN · m
𝑎
𝑀𝑛2 = 0.85 𝑓 𝑐′ 𝑎𝑏 𝑑 −
2
𝐴′𝑠 = 2 pcs 16 mm𝜙 = 402.12 mm2
97.98
𝑀𝑛2 = 0.85(25) (97.98) (250) (250) 417.5 − 𝐴𝑠 = 6 pcs 25 mm𝜙 = 2, 945.243 mm2
2
𝑀𝑛2 = 191.82 kN · m Verify if compression steel will yield, from
𝑓 ′ 𝑑′ 0.003
Then, 𝜌¯ 𝑐𝑦 = 0.85𝛽1 𝑐 + 𝜌′ = 0.02475
𝑓 𝑦 𝑑 0.003 − 𝜖 𝑦
𝑀𝑛 = 𝑀𝑛1 + 𝑀𝑛2 = 96.94 kN · m + 191.82 kN · m 𝐴𝑠 2, 945.243
𝑀𝑛 = 288.7 kN · m 𝜌𝑠 = = = 0.0286 > 𝜌¯ 𝑐𝑦
𝑏𝑑 (250) (412.5)
Determine the strength reduction factor, therefore, 𝑓 𝑠′ = 𝑓 𝑦
𝑑𝑡 − 𝑐 440 − 115.27 Verify the design moment strength.
𝜖𝑡 = 0.003 = 0.003 = 0.00845
𝑐 115.27
Assignment: