Answer to Question #242268 in Mechanical Engineering for Mom

Question

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.

Answer 1

1)

$\bar I=\frac{1}{12}bd^{3}$

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

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

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

$I_{1_x}=\bar {I_{1_x}}+Ah^{2}$

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

$I_{3_x}=\bar {I_{3_x}}+Ah^{2}$

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

$I_x=(105408+73728+210816)mm^{4}$

$=390*10^{3}mm^{4}$

2)

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

$y=\frac{A_1y_1+A_2y_2+A_3y_3}{A_1+A_2+A_3}$

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

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

Maximum direct stress in the section,

$\sigma =\frac{My}{I}$

I=(I₁+A₁d₁^2)+(I₂+A₂d₂²)+(I₃+A₃d₃²)

$I=(\frac{at^{3}}{12}+at*(\frac{a\sqrt{\smash[b]{3}}+t}{2})^{2})+(\frac{1}{12}*\frac{t}{Sin 60}(2aSin60)^{3}+2at*0)$

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

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

$\therefore$ Substituting the above values,

$\sigma =\frac{M*(\frac{2aSin60°+2t}{2})}{\frac{7at^{3}+24a^{3}t+12\sqrt{\smash[b]{3}}a^{2}t^{2}}{12}}$

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