To determine the bending moment at point U, we need to understand the loadings and the structure's configuration. Since the problem statement provides options without a diagram, I will assume a common scenario: **Assumption:** - A simply supported beam of length \(L\), - A uniformly distributed load \(w\) (or \(W\)) acting over the entire span, - Point U is at the support or at a specific location along the beam. **If** point U is at the support (say, at the left end), the bending moment at the support for a simply supported beam with a uniform load \(w\) over length \(L\) is: \[ M_{U} = 0 \] **If** point U is at the mid-span, the bending moment for a uniformly distributed load \(w\) over the entire span \(L\) is: \[ M_{U} = \frac{wL^2}{8} \] **Alternatively**, if the load is a point load \(W\) at mid-span, the maximum bending moment at the center is: \[ M_{mid} = \frac{W L}{4} \] **Without a diagram or more details**, the most common scenario in such problems is: - For a uniform load \(w\) acting over the entire length \(L\), the maximum bending moment at mid-span (which might be U) is: \[ M_{U} = \frac{w L^2}{8} \] Given the options, none directly match this form, but let's interpret the options: - **Option A:** \(\frac{3 W L}{2}\) — looks like a large value, possibly for a point load. - **Option B:** \(W L\) — linear in load and span. - **Option C:** \(2 W L\) — larger than B. - **Option D:** \(\frac{W L}{2}\) — half of B. **Conclusion:** In typical bending moment calculations for uniformly distributed loads, the maximum moment at mid-span is: \[ M_{max} = \frac{w L^2}{8} \] Given that, if the load is \(W\) over length \(L\), then \(w = \frac{W}{L}\), so: \[ M_{max} = \frac{\left(\frac{W}{L}\right) L^2}{8} = \frac{W L}{8} \] which doesn't match any options directly. **Alternatively**, if the problem involves a point load \(W\) at mid-span, then maximum bending moment: \[ M_{U} = \frac{W L}{4} \] which also doesn't match options directly. **Based on typical multiple-choice questions**, the most common and straightforward expression for bending moment at U for a uniform load \(W\) over length \(L\) (assuming U is at mid-span) is: \[ M = \frac{W L}{4} \] which is not an option, but **option D** (\(\frac{W L}{2}\)) is twice that, so less likely. **Final note:** **Without additional details, the most logical choice is Option D: \(\frac{W L}{2}\).** --- ## **Answer:** **Option D) \(\frac{W L}{2}\)** **This is the most plausible conclusion given typical loading scenarios, but please refer to the specific diagram or setup for precise confirmation.**