5.1 FIGURE 5.1 below shows a shaped lamina with a right-angled triangular hole - NSC Civil Technology Civil Services - Question 5 - 2016 - Paper 1

The area of part 2 can be calculated similarly:

$ext{Area}_{part2} = ext{Length} imes ext{Width} = 60 ext{ mm} imes 60 ext{ mm} = 3600 ext{ mm}^2$

For part 3, which is a right triangle, the area is found using:

ext{Area}_{part3} = rac{1}{2} imes ext{Base} imes ext{Height} = rac{1}{2} imes 15 ext{ mm} imes 30 ext{ mm} = 225 ext{ mm}^2

The total area of the lamina is the sum of the areas of all three parts:

$ext{Total Area} = ext{Area}_{part1} + ext{Area}_{part2} - ext{Area}_{part3} = 2700 ext{ mm}^2 + 3600 ext{ mm}^2 - 225 ext{ mm}^2 = 8500 ext{ mm}^2$

To find the centroid of part 3, we calculate:

x_{centroid} = rac{1}{3} imes ext{Base} = rac{1}{3} imes 15 ext{ mm} = 5 ext{ mm} ext{ (from base)} Position from A–A is 30 mm - 5 mm = 55mm.

For part 1:

x_{centroid} = rac{30}{2} = 15 ext{ mm} Position from B–B is 30 mm - 15 mm = 15 mm.

For part 3 (considering its height):

Position from B–B is 30 mm + 5 mm = 40 mm.

For part 2:

x_{centroid} = rac{60}{2} = 30 ext{ mm} Position from B–B is 60 mm + 30 mm (part width) = 45 mm.

The vector diagram requires constructing the angles using the given forces and solving for the resultant forces using graphical methods.

From the vector diagram:

• Member AE: Strut, Magnitude: 95.2 N
• Member DE: Tie, Magnitude: 47.6 N

The shear forces can be deduced using equilibrium equations. The shear force diagram will show values as calculated from the loads applied.

The bending moment diagram will display the changes in moment across the beam based on point loads and distributed loads applied.