Can you explain how they got their numbers for the the Ig table XAMPLE $2$GIVEN:MATERIALS: NW CONCRETE $4000$psiGR GO STEELSECTION W$/$ REINFORCEMENTREQ'Ddetermine cracking moment Mar with tue aci $318$ Method $($GROSS section$)$SOLUTION

 

 

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= M M_= , cr C C f=7.52f7.5(1.0) 4000 = 4744 r For I, break section in parts Ig A \table[[PART, /bar (y),(in), A(in), ^ar (y),(in), A,(/bar (y;) - (/bar (y))) (in),1,(^) = M M_= , cr C C f=7.52f7.5(1.0) 4000 = 4744 r For I, break section in parts Ig A \table[[PART, /bar (y),(in), A(in), ^ar (y),(in), A,(/bar (y;) - (/bar (y))) (in),1,(^)