5.1 FIGURE 5.1 below shows a shaped lamina - NSC Civil Technology Civil Services - Question 5 - 2017 - Paper 1

Next, calculate the coordinates of the centroids of each area:

1. The centroid of Area A is located at:

   • ( x_A = \frac{30}{2} = 15 , \text{mm} )
   • ( y_A = \frac{30}{2} = 15 , \text{mm} )

2. The centroid of Area B is located at:

   • ( x_B = 15 + (\frac{30}{2}) = 30 , \text{mm} )
   • ( y_B = (60 + \frac{45}{3}) = 67.5 , \text{mm} )

Calculate the moments about the A-A line for both areas:

1. Moment from Area A:

   $M_A = A_1 \times y_A = 1800 \times 15 = 27000 \, \text{mm}^3$

2. Moment from Area B:

   $M_B = A_2 \times y_B = 675 \times 67.5 = 45562.5 \, \text{mm}^3$

Now, use the following formulas to get the coordinates of the centroid:

1. Horizontal coordinate of the centroid (( x )):

   $x = \frac{M_A + M_B}{A_{total}} = \frac{27000 + 45562.5}{2475} \approx 29.21 \, \text{mm}$

2. Vertical coordinate of the centroid (( y )) can be calculated similarly.