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A doubly reinforced beam of width 350 mm and effective depth 900 mm, with cover to compression steel is 50 mm. Reinforcement in tension zone is 5 bars of 20 mm dia and 2 bars of 20 mm dia in compression
Fe 415 grade steel and
M20 concrete are used
The depth of Neutral axis, if compressive strain in compression steel is 0.0028, is
  • a)
    135 mm    
  • b)
    145 mm    
  • c)
    150 mm    
  • d)
    122 rnm
Correct answer is option 'A'. Can you explain this answer?
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Given data:
Width of beam (b) = 350 mm
Effective depth of beam (d) = 900 mm
Cover to compression steel = 50 mm
Number of tension steel bars (n) = 5
Diameter of tension steel bars (diameter) = 20 mm
Number of compression steel bars = 2
Grade of steel (fy) = 415 N/mm²
Grade of concrete (fck) = 20 N/mm²
Compressive strain in compression steel (εsc) = 0.0028

Formula used:
Depth of neutral axis (xu) = (0.87fyAst - 0.48fckbxd)/(0.87fyAst + 0.48fckbxd)
Ast = (π/4) x diameter² x n (for tension steel)

Calculation:
Ast = (π/4) x (20)² x 5 = 1570 mm²
Area of compression steel (Asc) = (π/4) x (20)² x 2 = 628 mm²
Area of concrete (Aconc) = b x d = 350 x 900 = 315000 mm²

Total area of steel (Ast + Asc) = 1570 + 628 = 2198 mm²
Let xu be the depth of neutral axis
Substituting the given values in the formula, we get
xu = (0.87 x 415 x 2198 - 0.48 x 20 x 350 x 900 x 10^-3)/(0.87 x 415 x 2198 + 0.48 x 20 x 350 x 900 x 10^-3)
xu = 135 mm (approx)

Therefore, the depth of neutral axis is 135 mm. The correct answer is option A.
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A doubly reinforced beam of width 350 mm and effective depth 900 mm, with cover to compression steel is 50 mm. Reinforcement in tension zone is 5 bars of 20 mm dia and 2 bars of 20 mm dia in compressionFe 415 grade steel andM20 concrete are usedThe depth of Neutral axis, if compressive strain in compression steel is 0.0028, isa)135 mm b)145 mm c)150 mm d)122 rnmCorrect answer is option 'A'. Can you explain this answer?
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