Question

An inverted T-shaped concrete beam (B1) in the figure, with centroidal axis \(X - X\), is subjected to an effective prestressing force of 1000 kN acting at the bottom kern point of the beam cross-section. Also consider an identical concrete beam (B2) with the same grade of concrete but without any prestressing force. \includegraphics[width=0.5\linewidth]{87image.png} The additional cracking moment (in kN.m) that can be carried by beam B1 in comparison to beam B2 is __ (rounded off to the nearest integer).

Hint:

In concrete structures, the additional cracking moment due to prestressing is calculated by multiplying the prestressing force by its eccentricity with respect to the centroid. This represents how the prestressing force introduces bending in the section.

Answer 1

Step 1: Calculate the distance from the centroid to the bottom fiber. Given the dimensions, the centroidal axis \(X - X\) is located \( \frac{H}{3} \) from the bottom. The total height of the beam is \( H \), hence the distance to the bottom fiber is \( \frac{2H}{3} \). Step 2: Determine the location of the prestressing force. The prestressing force is applied at 100 mm (0.1 m) from the bottom, or \( \frac{2H}{3} - 0.1 \) m from the centroid. Step 3: Calculate the eccentricity \( e \) of the prestressing force with respect to the centroid. \[ e = \left(\frac{2H}{3} - \frac{H}{3}\right) - 0.1 = \frac{H}{3} - 0.1 \] Step 4: Calculate the additional moment due to prestressing (\( M_p \)). \[ M_p = P \times e = 1000 \text{ kN} \times \left(\frac{H}{3} - 0.1\right) \] Assuming \( H = 0.45 \) m (since \(\frac{H}{3}\) = 0.15 m), \[ M_p = 1000 \times (0.15 - 0.1) = 1000 \times 0.05 = 50 \text{ kN.m} \] Step 5: Compute the total additional moment at cracking for B1. The effective depth \( d \) for the prestressing force action is typically \( \frac{2H}{3} \). \[ M_c = \frac{M_p \times d}{e} = \frac{50 \times \frac{2 \times 0.45}{3}}{0.05} \] \[ M_c = \frac{50 \times 0.3}{0.05} = 300 \text{ kN.m} \]