Question

The cross-section of a steel T-beam is shown in the figure where all dimensions are in mm.

(Figure not to scale)
The plastic section modulus of the given cross-section is \(\times 10^4\) mm\(^3\) (in integer).

Hint:

Find the plastic neutral axis by splitting the section into two equal areas, then take the sum of each area times its distance from that axis (or half the area times the distance between the two centroids).

Answer 1

Correct Answer: 12

Step 1: Understand what plastic section modulus means.
For a fully plastic cross-section, the stress is uniform and equal to the yield stress everywhere above the plastic neutral axis (PNA), and equal and opposite below it. This condition holds only when the PNA divides the cross-section into two parts of EQUAL area, since the tension and compression forces (stress times area) must balance to give zero net axial force.
The plastic section modulus is then
\[ Z_p = A_1 \bar{y}_1 + A_2 \bar{y}_2 \]
where \(A_1, A_2\) are the two equal areas above and below the PNA, and \(\bar{y}_1, \bar{y}_2\) are the distances of their centroids from the PNA.

Step 2: Compute the area of the flange and the web.
Flange: width \(100\) mm, thickness \(20\) mm:
\[ A_{flange} = 100\times20 = 2000\ \text{mm}^2 \]
Web: width \(20\) mm, height \(100\) mm:
\[ A_{web} = 20\times100 = 2000\ \text{mm}^2 \]
Total area: \(A = 2000+2000 = 4000\ \text{mm}^2\).

Step 3: Locate the plastic neutral axis (PNA).
Half of the total area is \(4000/2 = 2000\ \text{mm}^2\), which is exactly equal to the flange area. So the PNA lies exactly at the junction between the flange and the web: the flange alone (2000 mm\(^2\)) is one half, and the web alone (2000 mm\(^2\)) is the other half.

Step 4: Find the centroid distance of each half from the PNA.
The flange, of thickness \(20\) mm, sits directly above the PNA, so its own centroid is at half its thickness from the PNA:
\[ \bar{y}_1 = \frac{20}{2} = 10\ \text{mm} \]
The web, of height \(100\) mm, hangs directly below the PNA, so its centroid is at half its height from the PNA:
\[ \bar{y}_2 = \frac{100}{2} = 50\ \text{mm} \]

Step 5: Compute the plastic section modulus.
\[ Z_p = A_{flange}\bar{y}_1 + A_{web}\bar{y}_2 = 2000\times10 + 2000\times50 \]
\[ Z_p = 20000+100000 = 120000\ \text{mm}^3 \]
Expressed in units of \(10^4\ \text{mm}^3\):
\[ Z_p = 12\times10^4\ \text{mm}^3 \]

Final Answer:
\[ \boxed{Z_p = 12\times10^4\ \text{mm}^3} \]