## Question:

Determine the shape factor for the cross section of the beam.

## Step-by-step

$I = { \frac { 1 }{ 2 } } (0.2)(0.22)^{3}-{ \frac { 1 }{ 2 } }(0.185)(0.2)^{3} = 54.133(10^{-6}) m^{4}$

$C_{1} ={ \sigma }_{ Y } (0.01)(0.2) = (0.002){ \sigma }_{ Y }$

$C_{2} ={ \sigma }_{ Y } (0.1)(0.015) = (0.0015) { \sigma }_{ Y }$

$M_{p} = 0.002 M_{p} = 0.002{ \sigma }_{ Y }(0.21) + 0.0015{ \sigma }_{ Y }(0.1) = 0.0005 { \sigma }_{ Y }$

${ \sigma }_{ Y }=\frac { M_{Y}c }{ I }$

${ { M }_{ Y }=\frac { { \sigma }_{ Y }\left( 54.133 \right) \left( { 10 }^{ -6 } \right) }{ 0.11 } =0.000492 }{ \sigma }_{ Y }$

${ k=\frac { { M }_{ p } }{ { M }_{ Y } } =\frac { 0.00057{ \sigma }_{ Y } }{ 0.000492{ \sigma }_{ Y } } =1.16 }$