A freely-floating rectangular barge of length 200 m is divided into five equal compartments. In light-weight condition, the weight and buoyancy are uniformly distributed along the length of the barge. Assume \( g = 9.81 \, {m/s}^2 \). If 500 tonne of liquid cargo is added to each of the two end compartments as shown in the figure, then the maximum bending moment is {98.10 MN·m(rounded off to two decimal places).
Show Hint
For symmetric loading, the maximum bending moment typically occurs at the midspan of the barge, especially when the loading is concentrated at the ends.
Step 1: Calculate the Weight of Cargo in Each Compartment The weight of the cargo in each of the end compartments is calculated as: \[ W = m \cdot g = 500 \, {tonne} \times 9.81 \, {m/s}^2 \] where \( 1 \, {tonne} = 10^3 \, {kg} \). Therefore: \[ W = 500 \times 10^3 \times 9.81 = 4.905 \times 10^6 \, {N} \] Thus, the weight of cargo in each of compartments 1 and 5 is \( 4.905 \times 10^6 \, {N} \). Step 2: Maximum Bending Moment Calculation The maximum bending moment occurs at the center of the barge due to the point loads at the ends. The formula for the maximum bending moment for a beam with point loads at both ends is: \[ M_{{max}} = \frac{W \cdot L}{4} \] where \( W \) is the total load acting on the barge and \( L \) is the length of the barge. The total weight acting on the barge is: \[ W_{{total}} = 2 \times 4.905 \times 10^6 \, {N} = 9.81 \times 10^6 \, {N} \] Now, we calculate the maximum bending moment: \[ M_{{max}} = \frac{9.81 \times 10^6 \, {N} \times 200 \, {m}}{4} \] \[ M_{{max}} = 4.905 \times 10^8 \, {N-m} \] Finally, we convert it into MN-m (Mega Newton meter): \[ M_{{max}} = 490.5 \, {MN-m} \] Thus, the maximum bending moment is \( \boxed{490.5 \, {MN-m}} \).
