# Solution Given: - **Diameter of slab, \( D = 6 \) m** - **Radius, \( r = \frac{D}{2} = 3 \) m** - **Uniform distributed load, \( w = 7 \) kPa = 7 kN/m²** - **Slab is simply supported at the edge** - **Yield line pattern is radial (as shown in Figure 6(b))** - **Slab is isotropically reinforced** Let's use the **yield line theory** for a simply supported circular slab under uniform load. --- ## 1. **Work Done by External Load** The total load on the slab: \[ W_{\text{total}} = w \cdot \text{Area} \] \[ \text{Area} = \pi r^2 = \pi (3^2) = 9\pi \text{ m}^2 \] \[ W_{\text{total}} = 7 \times 9\pi = 63\pi \text{ kN} \] The **deflection at the center** (let's call it \(\delta\)) is the maximum vertical displacement. The **work done by the external load**: \[ \text{External Work} = W_{\text{total}} \times \frac{\delta}{2} \] (The average load moves through half the deflection.) --- ## 2. **Internal Work (Yield Line Mechanism)** For a **radial yield line pattern** (fan mechanism), each radial segment rotates about the support. The **internal work** done by the slab (using moment capacity \(m_u\) per unit length): \[ \text{Internal Work} = (\text{Total length of yield lines}) \times m_u \times \theta \] - **Total length of yield lines** = \(2\pi r\) (circumference of the circle) - **\(\theta\)** is the rotation at the edge For a simply supported circular slab, the **vertical displacement at the center is \(\delta\)** and the **rotation at the edge** is: \[ \theta = \frac{\delta}{r} \] So, \[ \text{Internal Work} = (2\pi r) m_u \left(\frac{\delta}{r}\right) = 2\pi m_u \delta \] --- ## 3. **Equate External and Internal Work** \[ \text{External Work} = \text{Internal Work} \] \[ 63\pi \cdot \frac{\delta}{2} = 2\pi m_u \delta \] Divide both sides by \(\delta\) (since \(\delta \neq \)): \[ \frac{63\pi}{2} = 2\pi m_u \] Divide both sides by \(2\pi\): \[ m_u = \frac{63\pi}{2 \times 2\pi} = \frac{63}{4} = 15.75 \text{ kNm/m} \] --- ## 4. **Final Answer** \[ \boxed{m_u = 15.75 \text{ kNm/m}} \] --- ### **Summary of Steps** 1. **Calculate total load:** \(W_{\text{total}} = 63\pi\) kN 2. **External work:** \(W_{\text{total}} \cdot \frac{\delta}{2}\) 3. **Internal work:** \(2\pi m_u \delta\) 4. **Equate and solve:** \(m_u = 15.75\) kNm/m --- ### **Assumption** - Slab yields along radial lines (fan mechanism). - Isotropic reinforcement, so capacity per meter is constant. --- **The required bending moment capacity for the slab is \( \boxed{15.75 \text{ kNm/m}} \).**