1)
Iˉ=121bd3
I1xˉ=121∗24∗63=432mm4
I2xˉ=121∗8∗483=73728mm4
I3xˉ=121∗48∗63=864mm4
I1x=I1xˉ+Ah2
=432+24∗6∗(24+3)2=105408mm4
I3x=I3xˉ+Ah2
=864+48∗6∗(24+3)2=210816mm4
Ix=(105408+73728+210816)mm4
=390∗103mm4
2)
Position of the neutral axis from the base of the section is,
y=A1+A2+A3A1y1+A2y2+A3y3
=a∗t+2a∗t+a∗ta∗t∗2t+2a∗t∗(aSin60°+t)+a∗t∗(2aSin60°+2t)
=44.5t+3a
Maximum direct stress in the section,
σ=IMy
I=(I1+A1d12)+(I2+A2d22)+(I3+A3d32)
I=(12at3+at∗(2a3+t)2)+(121∗Sin60t(2aSin60)3+2at∗0)
+(12at3+at∗(2a3+t)2)
=127at3+24a3t+123a2t2
∴ Substituting the above values,
σ=127at3+24a3t+123a2t2M∗(22aSin60°+2t)
=7at3+24a3t+123a2t26M∗(a3+2t
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