Question

If the 600-kN force acts through the centroid of the cross section, determine the location y of the centroid and the average normal stress on the cross section. Also, sketch the normal stress distribution over the cross section.

Answer 1

y_c = 130 mm

stress = 10 MPa

Explanation:

Given:

y_c = sum (A_i*y_i) / sum (A_i)

A_i : Unit Surface Area

y_i: Distance between the axis of unit surface and x-axis.

Find:

a) y_c y-coordinate of centroid.

b) Average normal stress

Solution:

- Compute 3 unit surface Areas:

A_1 = 800*80*0.5 = 12,000 m^2

A_2 = 300*120 = 36,000 m^2

A_3 = 800*80*0.5 = 12,000 m^2

- Compute 3 distance between the axes, A_i and x-axis:

y_1 = 300 / 3 = 100 mm

y_2 = 300 / 2 = 150 mm

y_3 = 300 / 3 = 100 mm

- Compute y_c:

y_c = (2*12,000*100 + 36,000*150) / (2*12,000+36,000)

y_c = 130 mm

- For normal stress

stress = F / A

stress = 600KN / (0.5*(0.4)*0.3)

stress = 600 KN / 0.06

stress = 10 MPa