Review Module – RCD Beam – Ultimate Strength Design

ULTIMATE STRENGTH DESIGN: BEAMS

NSCP 2015: 3. Over-reinforced design – concrete yields first than steel
Load Factors: U = 1.2 D + 1.6 L

SINGLY REINFORCED BEAMS

Beams reinforced for tension only

Strain Diagram of Beam Section:

𝜀𝑐 = 0.003; 𝜀𝑡 < 𝜀𝑦 (𝑓𝑠 = 𝜀𝑡 𝐸𝑠 ) < 𝑓𝑦

“STEEL DOES NOT YIELD”

MODES OF DESIGN 600(𝑑 − 𝑐)

𝑓𝑠 = 𝜀𝑡 𝐸𝑠 =
1. Balanced design – concrete and steel yield simultaneously 𝑐
STEEL RATIO:
𝜀𝑐 = 0.003; 𝜀𝑡 = 𝜀𝑦 𝐴𝑠
𝜌𝑚𝑖𝑛 ≤ [𝜌𝑎𝑐𝑡 = ] ≤ 𝜌𝑚𝑎𝑥
𝑏𝑑
𝑓𝑠 = 𝜀𝑡 𝐸𝑠 = 𝑓𝑦
Minimum Steel Ratio: (the bigger governs)
𝑓𝑦 1.4
𝜀𝑦 = 𝜌𝑚𝑖𝑛 =
𝐸𝑠 𝑓𝑦

“STEEL YIELDS” 0.25√𝑓𝑐 ′

𝜌𝑚𝑖𝑛 =
𝑓𝑦
General Formula for steel ratio, 𝝆:
0.85 𝑓𝑐′ 𝛽 𝑐
2. Under-Reinforced design – steel yields first than concrete 𝜌= [ ]
𝑓𝑦 𝑑

0.85 𝑓𝑐′ 𝛽 600

𝜌𝑏𝑎𝑙 = [ ]
𝑓𝑦 600 + 𝑓𝑦

𝛽 = 0.85 if 𝑓𝑐′ ≤ 28 𝑀𝑃𝑎

0.05
𝛽 = 0.85 − [𝑓𝑐′ − 28] ≥ 0.65 if 𝑓𝑐′ > 2 8 𝑀𝑃𝑎
7

Maximum Steel Ratio:

0.85 𝑓𝑐′ 𝛽 3
𝜌𝑚𝑎𝑥 = [ ]
𝑓𝑦 7
Based on 𝜀𝑡 = 0.004

BENDING MOMENT
Nominal:
𝑎 𝑎
𝑀𝑁 = 𝐶 (𝑑 − ) = 𝑇 (𝑑 − )
𝜀𝑐 = 0.003; 𝜀𝑡 > 𝜀𝑦 (𝑓𝑠 = 𝜀𝑡 𝐸𝑠 ) > 𝑓𝑦 2 2
𝑎 = 𝛽𝑐
“STEEL YIELDS”
𝑎 = 𝑑𝑒𝑝𝑡ℎ 𝑜𝑓 𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛 𝑏𝑙𝑜𝑐𝑘
𝑐 = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑓𝑟𝑜𝑚 𝑛𝑒𝑢𝑡𝑟𝑎𝑙 𝑎𝑥𝑖𝑠 𝑡𝑜 𝑡ℎ𝑒 𝑜𝑢𝑡𝑒𝑟𝑚𝑜𝑠𝑡
𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛 𝑓𝑖𝑏𝑒
 Ultimate:
𝑀𝑈 = ∅𝑀𝑁

Variation of ∅ with net tensile strain in extreme tension

reinforcement, 𝜀𝑡

Reduction Factor, ∅:

1. Determine the positive nominal moment capacity of the beam.

PAST CE BOARD PROBLEMS:
2. Determine the negative nominal moment capacity of the beam.
SITUATION: A 12m simply supported beam is provided by an additional
support at midspan. The beam has a width of b = 300mm and a total Additional question:
depth, h = 450mm. It is reinforced with 4 – 25mm dia. at the tension side
and 2 – 25mm dia. at the compression side with 70mm cover to centroid 3. Determine the maximum uniform load that the beam could support
of reinforcements. fc’ = 30 MPa, fy = 415 MPa. based on the positive ultimate moment capacity.
Use NSCP 2015 PROVISIONS

1. Determine the depth of compression block.

SITUATION:
2. Determine the nominal bending moment. Beam section is b = 300mm, h = 450 mm. Effective depth is 380 mm.
Compressive strength of concrete fc’ = 30 MPa, steel strength fy = 415
3. Determine the total factored uniform load including the beam’s MPa. The beam is simply supported on a span of 5m and carries the
weight. following loads:

Superimposed dead load = 16 kN/m

live load = 14 kN/m
SITUATION: [MAY 2022]
The beam shown has the following data: U = 1.2D + 1.6L

Beam: b x h = 350mm x 550mm 1. What is the maximum factored moment at ultimate condition?
Slab thickness = 100mm
Reinforcement at the top of the beam: 3 – 28mm∅ 2. Find the number of 16mm dia. bars required if the design moment at
Reinforcement at the bottom of the beam: 5 – 28mm∅ ultimate load is 200 kN-m.
Web reinforcement diameter = 12mm
Clear concrete cover = 40mm 3. Find the number of 16mm dia. bars required if the ultimate
concentrated load at the midspan is 50kN.
fc’ = 28 MPa
fy = 415 MPa
𝜌𝑏𝑎𝑙 = 0.0288

The beam having a length of 12m is simply supported at the left end at a
point 3m from the right end of the beam.
DOUBLY REINFORCED BEAMS
Beams reinforced for tension and compression PAST CE BOARD PROBLEMS

Strain Diagram of Beam Section: SITUATION: A simply – supported beam is reinforced with 4 – 28 mm 
at the bottom and 2 – 28 mm  at the top of the beam. Steel covering to
centroid of reinforcement is 70 mm at the top and bottom of the beam.
The beam has a total depth of 400 mm and a width of 300 mm.
fc’ = 30 MPa, fy = 415 MPa.

1. Determine the depth of compression block.

2. Determine the design strength of the beam.

3. Determine the live load at the midspan in addition to a DL = 20 kN/m

including the weight of the beam if it has a span of 6 m.
Use U = 1.2D + 1.6L

𝐴′𝑠 = 𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛 𝑏𝑎𝑟𝑠

𝐴𝑠 = 𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑏𝑎𝑟𝑠 SITUATION: A reinforced concrete beam having a width of 300 mm and
an effective depth of 400mm is reinforced with a total tensile area of 2500
STEEL RATIO: mm2. fc’= 21MPa fy = 415 MPa.
𝝆𝒂𝒄𝒕 > 𝝆𝒎𝒂𝒙
Cover to the centroid of reinforcements = 70mm
𝝆𝒂𝒄𝒕 = 𝝆𝑺𝑹𝑩 + 𝝆′
1. Determine the depth of compression block.
′
𝑨′𝒔
𝝆 =
𝒃𝒅 2. Determine the ultimate moment capacity.

𝟔𝟎𝟎 (𝒄 − 𝒅′ ) 3. Determine the compression reinforcement

𝒇′𝒔 =
𝒄

BENDING MOMENT SITUATION. A fixed ended beam with width “b” = 400mm, total depth “h’
Nominal: = 500mm, is subjected to factored design forces resulting to factored
𝑴𝑵𝑫𝑹𝑩 = 𝑴𝑵𝟏 + 𝑴𝑵𝟐 moment at the fixed ends 440 kN-m.

𝑴𝑵𝑫𝑹𝑩 = 𝑛𝑜𝑚𝑖𝑛𝑎𝑙 𝑚𝑜𝑚𝑒𝑛𝑡 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑜𝑓 𝑡ℎ𝑒 fc’= 20.7 MPa

𝑑𝑜𝑢𝑏𝑙𝑦 𝑟𝑒𝑖𝑛𝑓𝑜𝑟𝑐𝑒𝑑 𝑏𝑒𝑎𝑚 fy = 415 MPa
Concrete cover to the centroid of steel = 65 mm
𝑴𝑵𝟏 = 𝑛𝑜𝑚𝑖𝑛𝑎𝑙 𝑚𝑜𝑚𝑒𝑛𝑡 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑜𝑓 𝑡ℎ𝑒
𝑠𝑖𝑛𝑔𝑙𝑦 𝑟𝑒𝑖𝑛𝑓𝑜𝑟𝑐𝑒𝑑 𝑏𝑒𝑎𝑚 𝑜𝑟 1. Determine the ultimate moment capacity of concrete.
𝑡ℎ𝑒 𝑛𝑜𝑚𝑖𝑛𝑎𝑙 𝑚𝑜𝑚𝑒𝑛𝑡 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑜𝑓 𝑐𝑜𝑛𝑐𝑟𝑒𝑡𝑒
2. Determine the tension reinforcement required for factored moment.
𝑴𝑵𝟐 = 𝑛𝑜𝑚𝑖𝑛𝑎𝑙 𝑚𝑜𝑚𝑒𝑛𝑡 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑜𝑓
𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛 𝑏𝑎𝑟 3. Determine the compression reinforcement required for factored
moment
Ultimate:
𝑴𝑼𝑫𝑹𝑩 = ∅ 𝑴𝑵𝑫𝑹𝑩

𝑴𝑼𝑫𝑹𝑩 = 𝑢𝑙𝑡𝑖𝑚𝑎𝑡𝑒 𝑚𝑜𝑚𝑒𝑛𝑡 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑜𝑓 𝑡ℎ𝑒

𝑑𝑜𝑢𝑏𝑙𝑦 𝑟𝑒𝑖𝑛𝑓𝑜𝑟𝑐𝑒𝑑 𝑏𝑒𝑎𝑚