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Over-reinforced beam sectional analysis

   

Added on  2023-06-03

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Structural Engineering
Over-reinforced beam sectional analysis_1

Question
Over-reinforced beam (cross section: 110 mm by 110 mm, bar diameter is 24 mm)
• Determine moment to cause cracking (Mcrack) from sectional analysis. Use both the conservative
coefficient of 0.6 as well as the realistic coefficient of 0.85 (where f'c = 40 MPa)
𝑀𝑀𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 𝑓𝑓𝑐𝑐𝑐𝑐 ∗ 𝐼𝐼𝑔𝑔
𝑦𝑦
𝑓𝑓𝑐𝑐𝑐𝑐 = 0.6 ∗ 0.85 ∗ �𝑓𝑓𝑐𝑐′
𝑓𝑓𝑐𝑐𝑐𝑐 = 0.6 ∗ 0.85 ∗ √40
𝑓𝑓𝑐𝑐𝑐𝑐 = 3.225 𝑀𝑀𝑀𝑀𝑐𝑐
50.403
Over-reinforced beam sectional analysis_2

𝑓𝑓𝑐𝑐′ = 40 𝑀𝑀𝑀𝑀𝑐𝑐
𝑓𝑓𝑦𝑦𝑦𝑦 = 550 𝑀𝑀𝑀𝑀𝑐𝑐
𝐸𝐸𝑦𝑦 = 200 𝐺𝐺𝑀𝑀𝑐𝑐
𝐸𝐸𝑐𝑐 = 4500 ∗ �𝑓𝑓𝑐𝑐′
𝐸𝐸𝑐𝑐 = 4500 ∗ √40
𝐸𝐸𝑐𝑐 = 28.46 𝐺𝐺𝑀𝑀𝑐𝑐
Modular ratio n = 𝐸𝐸𝑦𝑦
𝐸𝐸𝑐𝑐
𝑛𝑛 = 200
28.46 = 7.027
Area of steel rod 𝐴𝐴𝑆𝑆 = 𝜋𝜋
4 ∗ 𝑑𝑑2
𝐴𝐴𝑆𝑆 = 452.3 𝑚𝑚𝑚𝑚2
Area transform 𝐴𝐴 𝑡𝑡 = (𝑛𝑛 − 1) ∗ 𝐴𝐴𝑆𝑆
𝐴𝐴 𝑡𝑡 = (7.027 − 1) ∗ 452.3
𝐴𝐴 𝑡𝑡 = 2726.14 𝑚𝑚𝑚𝑚2
Calculate y
𝑦𝑦 = 110∗110∗110
2 +2726.14∗30
110∗110+2726.14
𝑦𝑦 = 50.403 𝑚𝑚𝑚𝑚
Calculate M.O.I
𝐼𝐼𝑔𝑔 = 110 ∗ 110 3
12
𝐼𝐼𝑔𝑔 = 12.5 ∗ 106 𝑚𝑚𝑚𝑚4
Calculate 𝑀𝑀𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
𝑀𝑀𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 𝑓𝑓𝑐𝑐𝑐𝑐 ∗ 𝐼𝐼𝑔𝑔
𝑦𝑦
𝑀𝑀𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 3.225 ∗ 12.5 ∗ 106
50.403
Over-reinforced beam sectional analysis_3

𝑀𝑀𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 3.225 ∗ 12.5 ∗ 106
50.403
𝑴𝑴𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄 = 𝟖𝟖𝟖𝟖𝟖𝟖 𝑵𝑵𝑵𝑵
• Determine the ultimate moment (Mult) from sectional analysis (where fsy = 550 MPa, bar is 10 mm
diam.) • Compare these theoretically derived values to the recorded load vs deflection graph. You will
need to convert your Mcrack and Mult values into equivalent load value
Solution
From similar triangles property, stress in concrete
𝑓𝑓𝑐𝑐
3.225 = 59.59
50.403
𝑓𝑓𝑐𝑐 = 3.812 𝑀𝑀𝑀𝑀𝑐𝑐
Stress in steel
𝑓𝑓𝑦𝑦
𝑛𝑛
3.225 = 50.403 − 30
50.403
𝑓𝑓𝑦𝑦 = 11.44 𝑀𝑀𝑀𝑀𝑐𝑐
Forces act on steel and concrete
𝐹𝐹𝑐𝑐 = 1
2 ∗ 3.812 ∗ 59.59 ∗ 110
𝑭𝑭 𝒄𝒄 = 𝟏𝟏𝟏𝟏. 𝟒𝟒𝟒𝟒 𝑲𝑲𝑵𝑵
Force in steel
𝐹𝐹𝑦𝑦 = 11.44 ∗ 452.3
𝑭𝑭 𝒔𝒔 = 𝟓𝟓. 𝟏𝟏𝟏𝟏𝟒𝟒𝑲𝑲𝑵𝑵
Critical Force
𝐹𝐹𝑐𝑐𝑐𝑐 = 1
2 ∗ 3.225 ∗ 50.403 ∗ 110
𝑭𝑭 𝒄𝒄𝒄𝒄 = 𝟖𝟖. 𝟒𝟒𝟒𝟒 𝑲𝑲𝑵𝑵
Over-reinforced beam sectional analysis_4

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