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Determine the shape factor of the cross section.

Step-by-step

The moment of inertia of the cross section about the neutral axis is

I ={ \frac { 1 }{ 2 } }(3)(9^{3}) +{ \frac { 1 }{ 2 } } (6) (3^{3}) = 195.75 in^{4}

Here, { { \sigma }_{ max }={\sigma }_{ Y }} and c = 4.5 in .Then

{ { \sigma }_{ max }=\frac { { M }c }{ I } ;\quad \quad \quad \quad { \sigma }_{ Y }=\frac { { M }_{ Y }\left( 4.5 \right) }{ 195.75 } }

 

M_{Y} = 43.5 { \sigma }_{ Y }

Referring to the stress block shown in Fig. a,

T_{1} = C_{1} = 3(3){ \sigma }_{ Y }= 9 { \sigma }_{ Y }

 

T_{2} = C_{2}= 1.5(9){ \sigma }_{ Y } = 13.5 { \sigma }_{ Y }

Thus,

M_{P} = T_{1}(6) + T_{2}(1.5)

 

= 9{ \sigma }_{ Y }(6) + 13.5{ \sigma }_{ Y }(1.5) = 74.25 { \sigma }_{ Y }

 

{ k=\frac { { M }_{ p } }{ { M }_{ Y } } =\frac { 74.25{ \sigma }_{ Y } }{ 43.5{ \sigma }_{ Y } } =1.71 }

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