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Ex1: Calculate The Ultimate Moment Capacity For The Given Beam. FC' 25 Mpa Fy 400 Mpa Es 200000 Mpa

The document outlines the calculation of the ultimate moment capacity for a beam with specified material properties and dimensions. It includes the determination of various parameters such as strain, stress, and reinforcement ratios, ultimately leading to the calculation of the nominal moment capacity (Mn) and the ultimate moment capacity (Mu). The calculations indicate that the section is under-reinforced and provides the necessary equations and values used in the analysis.

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15 views2 pages

Ex1: Calculate The Ultimate Moment Capacity For The Given Beam. FC' 25 Mpa Fy 400 Mpa Es 200000 Mpa

The document outlines the calculation of the ultimate moment capacity for a beam with specified material properties and dimensions. It includes the determination of various parameters such as strain, stress, and reinforcement ratios, ultimately leading to the calculation of the nominal moment capacity (Mn) and the ultimate moment capacity (Mu). The calculations indicate that the section is under-reinforced and provides the necessary equations and values used in the analysis.

Uploaded by

orthko
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
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Ex1: Calculate the ultimate moment capacity for𝑛 the given beam.

𝜙𝑀
fc' = 25 mPa
fy= 400 mPa
Es= 200000 mPa

Strain Stress
D= 600 mm 𝐸𝑠 200000
d= 550 mm 𝑛= = = 9
𝐸𝑐 4700 25
b= 300 mm
𝜋
Ast= 3 Y 20 ⇒ 𝐴𝑠𝑡 = 3𝑥 202 = 942.8 mm²
4
𝛽1 = 0.85 17 ≤ 𝑓𝑐 ′ ≤ 28 ⇒ 𝛽1 = 0.85
𝐴𝑠𝑡 942.8
𝜌= = = 0.0057
𝑏𝑑 300𝑥550
𝑓𝑐 ′ 600
𝜌𝑚𝑎𝑥/𝑑𝑢𝑐𝑡𝑖𝑙𝑒 = 0.75(0.85 𝛽1 )
𝑓𝑦 600 + 𝑓𝑠𝑦
25 600
𝜌𝑚𝑎𝑥 = 0.75 0.85 𝑥0.85𝑥 = 0.0203
400 600 + 400
0.25 𝑓𝑐′ 1.4 1.4
𝜌𝑚𝑖𝑛 = ≥ ⇒ 0.0031 ≥ = 0.0035
𝑓𝑦 𝑓𝑦 400
Pmin ≤ P ≤ Pmax => Section is under reinforced (Ductile)
From equilibrium : C=T
′
0.85 𝑓𝑐 𝛽1 𝑐 𝑏 = 𝐴𝑠𝑡 𝑓𝑠𝑦 𝑎 = 𝛽1 𝑐 𝑐 = 𝑘𝑑
𝐴𝑠𝑡 𝑓𝑠𝑦 𝜌 𝑏𝑑 𝑓𝑠𝑦
𝑘= =
0.85 𝑓𝑐 ′ 𝛽1 𝑑 𝑏 0.85 𝑓𝑐 ′ 𝛽1 𝑑 𝑏

𝜌 𝑓𝑠𝑦 0.0057𝑥400
⇒ 𝑘= = 0.126
0.85 𝑓𝑐 ′ 𝛽1 0.85𝑥25𝑥0.85
𝑓 𝛽
𝛽1 𝑘𝑑
𝑀𝑛 = 𝑇𝑗𝑑 = 𝐶𝑗𝑑 = 𝐴𝑠𝑡 𝑓𝑠𝑦 𝑑 −
2
𝛽1 𝑘𝑑
Or 𝑀𝑛 = 0.85𝑓𝑐 ′ 𝛽1 𝑘𝑑 𝑏 𝑑 −
2
0.126𝑥0.85𝑥550
𝑀𝑛 = 942.8𝑥400𝑥 550 − 10−6 = 196.3 kN.m
2

𝑀𝑢 = 𝜙𝑀𝑛 = 0.9𝑀𝑛 = 0.9𝑥 194.3 = 176.67 kN.m

A) If 𝜌 < 𝜌𝑚𝑖𝑛 ⇒ 𝑈𝑠𝑒 𝑃𝑚𝑖𝑛

0.25 𝑓𝑐′ 1.4 𝐴𝑠𝑡


𝜌𝑚𝑖𝑛 = ≥ = ⇒ 𝐴𝑠𝑡 = 𝜌𝑚𝑖𝑛 𝑏𝑑
𝑓𝑦 𝑓𝑦 𝑏𝑑

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