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).
Show 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).
Step 1: Recall a shortcut formula for the plastic modulus.
When a section is split by the plastic neutral axis (PNA) into two equal areas $A/2$ each, the plastic section modulus can be written as
\[ Z_p = \frac{A}{2}\left(\bar{y}_1+\bar{y}_2\right) \]
which just says $Z_p$ equals half the total area times the distance between the centroids of the two halves (the compression half and the tension half), since $\bar{y}_1+\bar{y}_2$ is that centroid-to-centroid distance.
Step 2: Find the total area and check where the equal-area split falls.
Flange area $= 100\times20 = 2000\text{ mm}^2$. Web area $= 20\times100=2000\text{ mm}^2$. Total area $A = 4000\text{ mm}^2$, so each half must be $2000\text{ mm}^2$.
Since the flange itself is exactly $2000\text{ mm}^2$, the flange is one whole half by itself, and the web is the other whole half by itself. The PNA sits exactly on the line separating the flange from the web.
Step 3: Find the centroid of each half, measured from the top and bottom of the section.
Centroid of the flange, measured from the top face, is at $20/2 = 10\text{ mm}$, which is $10\text{ mm}$ above the PNA.
Centroid of the web, measured from the bottom face, is at $100/2 = 50\text{ mm}$, which is $50\text{ mm}$ below the PNA.
Step 4: Find the distance between the two centroids and apply the shortcut formula.
Distance between the flange centroid and the web centroid $= 10+50 = 60\text{ mm}$, measured straight through the PNA.
\[ Z_p = \frac{A}{2}\times60 = 2000\times60 = 120000\text{ mm}^3 \]
\[ Z_p = 12\times10^4\text{ mm}^3 \]
Final Answer:
\[ \boxed{Z_p = 12\times10^4\text{ mm}^3} \]
