I=112(bh3−b1h13)=340.69⋅106 mm4,I=\frac 1{12}(bh^3-b_1h_1^3)=340.69\cdot 10^6~mm^4,I=121(bh3−b1h13)=340.69⋅106 mm4,
2Fs=VQI, ⟹ smax=2FIVQ,\frac{2F}s=\frac{VQ}I,\implies s_{max}=\frac{2FI}{VQ},s2F=IVQ,⟹smax=VQ2FI,
a)
Q=Apdp=bth−t2=680⋅103 mm3,Q=A_pd_p=bt\frac{h-t}2=680\cdot 10^3~mm^3,Q=Apdp=bt2h−t=680⋅103 mm3,
sA=78.3 mm,s_A=78.3~mm,sA=78.3 mm,
b)
Q=Afdf=(b−2t)th−t2=544⋅103 mm3,Q=A_fd_f=(b-2t)t\frac{h-t}2=544\cdot 10^3~mm^3,Q=Afdf=(b−2t)t2h−t=544⋅103 mm3,
sB=97.9 mm,s_B=97.9~mm,sB=97.9 mm,
c)
sB>sAs_B>s_AsB>sA beam B is more efficient.
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