Answer to Question #242268 in Mechanical Engineering for Mom

Question #242268

1. Determine the Moment of inertia for the singly-symmetrical cross-section 

shown in figure. (5)

2. The thin-walled beam section shown in Figure is subjected to a bending 

moment M applied in a negative sense. Find the position of the neutral 

  • axis and the maximum direct stress in the section.
1
Expert's answer
2021-09-26T11:56:59-0400

1)


Iˉ=112bd3\bar I=\frac{1}{12}bd^{3}

I1xˉ=1122463=432mm4\bar {I_{1_x}}=\frac{1}{12}*24*6^{3}=432mm^{4}

I2xˉ=1128483=73728mm4\bar {I_{2_x}}=\frac{1}{12}*8*48^{3}=73728mm^{4}

I3xˉ=1124863=864mm4\bar {I_{3_x}}=\frac{1}{12}*48*6^{3}=864mm^{4}

I1x=I1xˉ+Ah2I_{1_x}=\bar {I_{1_x}}+Ah^{2}

=432+246(24+3)2=105408mm4=432+24*6*(24+3)^{2}=105408mm^{4}

I3x=I3xˉ+Ah2I_{3_x}=\bar {I_{3_x}}+Ah^{2}

=864+486(24+3)2=210816mm4=864+48*6*(24+3)^{2}=210816mm^{4}

Ix=(105408+73728+210816)mm4I_x=(105408+73728+210816)mm^{4}

=390103mm4=390*10^{3}mm^{4}


2)



Position of the neutral axis from the base of the section is,

y=A1y1+A2y2+A3y3A1+A2+A3y=\frac{A_1y_1+A_2y_2+A_3y_3}{A_1+A_2+A_3}

=att2+2at(aSin60°+t)+at(2aSin60°+2t)at+2at+at=\frac{a*t*\frac{t}{2}+2a*t*(aSin60°+t)+a*t*(2aSin60°+2t)}{a*t+2a*t+a*t}

=4.5t+3a4=\frac{4.5t+\sqrt{\smash[b]{3}}a}{4}


Maximum direct stress in the section,

σ=MyI\sigma =\frac{My}{I}

I=(I1+A1d12)+(I2+A2d22)+(I3+A3d32)

I=(at312+at(a3+t2)2)+(112tSin60(2aSin60)3+2at0)I=(\frac{at^{3}}{12}+at*(\frac{a\sqrt{\smash[b]{3}}+t}{2})^{2})+(\frac{1}{12}*\frac{t}{Sin 60}(2aSin60)^{3}+2at*0)

+(at312+at(a3+t2)2)+(\frac{at^{3}}{12}+at*(\frac{a\sqrt{\smash[b]{3}}+t}{2})^{2})

=7at3+24a3t+123a2t212=\frac{7at^{3}+24a^{3}t+12\sqrt{\smash[b]{3}}a^{2}t^{2}}{12}

\therefore Substituting the above values,

σ=M(2aSin60°+2t2)7at3+24a3t+123a2t212\sigma =\frac{M*(\frac{2aSin60°+2t}{2})}{\frac{7at^{3}+24a^{3}t+12\sqrt{\smash[b]{3}}a^{2}t^{2}}{12}}

=6M(a3+2t7at3+24a3t+123a2t2=\frac{6M*(a\sqrt{\smash[b]{3}}+2t}{7at^{3}+24a^{3}t+12\sqrt{\smash[b]{3}}a^{2}t^{2}}


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