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I don't understand how exactly can a beam as described can support a 650kN-m ultimate moment. The only allowed bars are 28mm∅ bars. Consider 40mm side covering and bar spacing to be greater or equal to 28mm, the only acceptable choice is 4-28mm∅ bars. Is there a way to make it doubly reinforced or is the question my professor gave me wrong? fc'=30MPa and fy=410MPa

On another consideration, the load was not mentioned as ultimate moment. If I treat it as a nominal moment, it still isn't manageable. Is there something I'm missing?

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  • $\begingroup$ What are the units KN-m? $\endgroup$
    – Solar Mike
    Commented May 4 at 14:01
  • $\begingroup$ Are you asking what kN-m is? If you are, that's kilonewton meter. @SolarMike $\endgroup$
    – Ndracus
    Commented May 4 at 14:03
  • $\begingroup$ So if thew N is defined as kgm/s^2, then you have (kgm/s^2)-m perhaps? or did you mean kn/m? $\endgroup$
    – Solar Mike
    Commented May 4 at 14:06
  • $\begingroup$ It's a moment. It's Force multiplied to the Moment arm. $\endgroup$
    – Ndracus
    Commented May 4 at 14:10
  • $\begingroup$ What is a USD beam? $\endgroup$
    – Forward Ed
    Commented May 9 at 17:30

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I usually do the dirty trick of dividing the moment by 0.9 effective depth to get a rough estimate of compression or tension force.

$$\frac{650kNm}{0.54m}=1,203,700N \approx120,000kg $$ $$\frac{120,000}{4500kg/cm^2}=26cm^2 \ steel\ area$$ $$26/(1.4^2 *\pi)=4.33\ say \ 5 \ 28mm \ bars$$ This is the estimated unfactored loads, then I would consider load combinations and refine this. But assuming J as 0.90 of effective depth is a good estimate!

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