Question:

Consider a reinforced concrete beam section of 300 mm width and 700 mm depth. The beam is reinforced with the tension steel of 2000 mm² area at an effective cover of 50 mm. Concrete in the tension zone is assumed to be cracked. Assume the modular ratio of 12 and Young's modulus of 200 GPa for steel. When the extreme fibre in the compression zone undergoes the strain of 0.0004 due to the applied bending moment, the stress in the steel (in MPa) is ..... (rounded off to the nearest integer).

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When solving for stress in reinforced concrete beams, remember to use the strain values derived from the bending moment and the modular ratio to compute the stress in steel.
Updated On: Apr 3, 2025
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Solution and Explanation

Given Data:

  • Modular ratio, m=12 m = 12
  • Elastic modulus of steel, Es=200 E_s = 200 GPa
  • Steel reinforcement area, Ast=2000 A_{st} = 2000 mm²
  • Depth of the section, d=650 d = 650 mm

Step 1: Finding the Neutral Axis Depth

Using the given equation:

300xa22=mAst(dxa) \frac{300 x_a^2}{2} = m A_{st} (d - x_a)

Substituting the values:

150xa2+12×2000×xa12×2000×650=0 150 x_a^2 + 12 \times 2000 \times x_a - 12 \times 2000 \times 650 = 0

Solving for xa x_a :

xa=252.26mm x_a = 252.26 \, \text{mm}

Step 2: Strain in Steel

From the strain diagram:

ϵst=0.0004(650252.26)252.26 \epsilon_{st} = \frac{0.0004(650 - 252.26)}{252.26}

Evaluating:

ϵst=6.306×104 \epsilon_{st} = 6.306 \times 10^{-4}

Step 3: Stress in Steel

Using Hooke's law:

σst=ϵst×Es \sigma_{st} = \epsilon_{st} \times E_s

Substituting values:

σst=(6.306×104)×(200×103) \sigma_{st} = (6.306 \times 10^{-4}) \times (200 \times 10^3)

σst=126.136MPa126MPa(rounded) \sigma_{st} = 126.136 \, \text{MPa} \approx 126 \, \text{MPa} (\text{rounded})

Final Answer:

Correct Answer: 126 MPa.

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