Example 03: Moment Capacity of a Timber Beam Reinforced with Steel and Aluminum Strips

$y_1 = \frac{1}{2}(20) = 10 ~ \text{mm}$
 

Wood
$A_2 = 80(150) = 12\,000 ~ \text{mm}^2$

$y_2 = 20 + \frac{1}{2}(150) = 95 ~ \text{mm}$
 

Transformed aluminum
$A_3 = 560(50) = 28\,000 ~ \text{mm}^2$

$y_3 = 20 + 150 + \frac{1}{2}(50) = 195 ~ \text{mm}$
 

Total area
$A = A_1 + A_2 + A_3 = 32\,000 + 12\,000 + 28\,000$

$A = 72\,000 ~ \text{mm}^2$

$72\,000\bar{y} = 32\,000(10) + 12\,000(95) + 28\,000(195)$

$\bar{y} = 96.11 ~ \text{mm}$

$I_{NA_1} = \dfrac{1600(20^3)}{12} + 32\,000(\bar{y} - 10)^2 = 238\,350\,617.3 ~ \text{mm}^4$

$I_{NA_2} = \dfrac{80(150^3)}{12} + 12\,000(\bar{y} - 95)^2 = 22\,514\,814.81 ~ \text{mm}^4$

$I_{NA_3} = \dfrac{560(50^3)}{12} + 28\,000(195 - \bar{y})^2 = 279\,645\,679 ~ \text{mm}^4$
 

$I_{NA} = 238\,350\,617.3 + 22\,514\,814.81 + 279\,645\,679$

$I_{NA} = 540\,511\,111.11 ~ \text{mm}^4$

Based on steel
$\dfrac{120}{n_{st}} = \dfrac{M(\bar{y})(1000^2)}{I_{NA}}$

$M = 33.74 ~ \text{kN}\cdot\text{m}$
 

Based on Wood
$10 = \dfrac{M(\bar{y} - 20)(1000^2)}{I_{NA}}$

$M = 71.02 ~ \text{kN}\cdot\text{m}$
 

Based on Aluminum
$\dfrac{80}{n_{al}} = \dfrac{M(220 - \bar{y})(1000^2)}{I_{NA}}$

$M = 49.9 ~ \text{kN}\cdot\text{m}$