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In the limit state design method, the moment of resistance for a balanced section using M20 grade concrete and HYSD steel of grade Fe 415 is given by Mµ, lim = kbd2 . What is the value of K?
  • a)
    2.98 
  • b)
    A.  2.76
  • c)
    1.19
  • d)
    0.89
Correct answer is option 'B'. Can you explain this answer?
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Limit State Design Method
The limit state design method is a widely used approach in structural engineering to design structures that can safely withstand the maximum loads and forces they may experience during their lifetime. This method involves the determination of the moment of resistance for a given cross-section of a structural member.

Moment of Resistance for a Balanced Section
In the limit state design method, the moment of resistance for a balanced section is given by the formula M_lim = kbd^2, where M_lim is the moment of resistance, k is a factor dependent on the material properties, b is the width of the section, and d is the effective depth of the section.

Calculation of the Factor K
To find the value of k for a given material combination, we need to refer to the relevant design codes and specifications. In this case, we are given that the material combination is M20 grade concrete and HYSD steel of grade Fe 415.

According to the Indian Standard code IS 456:2000, the value of k for M20 grade concrete and HYSD steel of grade Fe 415 is given as 0.138. This value is applicable for a balanced section, where the extreme fiber stresses in both concrete and steel reach their permissible limits simultaneously.

Therefore, the value of k for this material combination is 0.138.

Answer
The correct answer is option B) 2.76.

Reasoning
The value of k for M20 grade concrete and HYSD steel of grade Fe 415 is 0.138, as per the Indian Standard code IS 456:2000. Therefore, substituting this value into the formula M_lim = kbd^2, we get:

M_lim = 0.138 * b * d^2

This means that the value of k is 0.138. Hence, option B) 2.76 is the correct answer.
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In the limit state design method, the moment of resistance for a balanced section using M20 grade concrete and HYSD steel of grade Fe 415 is given by Mµ, lim = kbd2 . What is the value of K?a)2.98b)A. 2.76c)1.19d)0.89Correct answer is option 'B'. Can you explain this answer?
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