Question: Q7. Show the yield line patterns for the following slabs. [Image of yield line patterns for three…

Q7. Show the yield line patterns for the following slabs.

[Image of yield line patterns for three different slab configurations]

Q8. Design the slab panels (only S2 and S5) to carry a floor finishing load of 2 KN/m² in addition to its own weight and live load of 10 KN/m² using the yield line method. It is required to:

a) Determine the ultimate moment resistance (M) of the slab, which is to be isotopically reinforced using Yield line method.

b) Determine flexural reinforcement for the ultimate moment (M).

Use C20/25 and S400 class I work. Take thickness of slab 170mm.

[Image of slab layout with dimensions and panel labels]

A B C D 6 m 3 m 3 m 6 m

3 m 3 m 3 m 3 m 4 m

S1 S2 S3 S4 S5 S6

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Solution

To solve part a) of the problem, we will determine the ultimate moment resistance \( M \) of the slab using the Yield Line Method. Step 1: Calculate the total factored load The total load is the sum of the floor finishing load and the live load. \[ \text{Total load} = \text{Floor finishing load} + \text{Live load} \] & \[ \text{Total load} = 2\, \text{kN/m}^2 + 10\, \text{kN/m}^2 = 12\, \text{kN/m}^2 \] & Step 2: Determine the ultimate load Assuming a safety factor (\(\gamma\)) of 1.5 for the ultimate load design. \[ \text{Ultimate load} = \gamma \times \text{Total load} \] & \[ \text{Ultimate load} = 1.5 \times 12\, \text{kN/m}^2 = 18\, \text{kN/m}^2 \] & Step 3: Calculate the ultimate moment resistance \( M \) Using the Yield Line Method for a simply supported slab, the ultimate moment can be estimated as: \[ M = \frac{w \times l^2}{8} \] & Where: - \( w = 18\, \text{kN/m}^2 \) (ultimate load) - \( l = 6\, \text{m} \) (span length) \[ M = \frac{18 \times 6^2}{8} = \frac{18 \times 36}{8} = \frac{648}{8} = 81\, \text{kN}\cdot\text{m} \] & Answer for part a): The ultimate moment resistance \( M \) of the slab is 81 kN·m. --- For part b), we will determine the required flexural reinforcement for the ultimate moment \( M \). Step 1: Calculate the section properties Given: - Slab thickness \( h = 170\, \text{mm} = 0.17\, \text{m} \) - Effective depth \( d = h - \text{cover} - \frac{\phi}{2} \) (Assuming cover and bar diameter are negligible for this calculation) \[ d \approx 0.17\, \text{m} \] & Step 2: Determine the required steel area \( A_s \) Using the formula: \[ M = 0.138 \times f_y \times A_s \times d^2 \] & Where: - \( f_y = 400\, \text{N/mm}^2 = 400 \times 10^3\, \text{N/m}^2 \) - \( M = 81 \times 10^6\, \text{N}\cdot\text{mm} \) (converted to mm) \[ 81 \times 10^6 = 0.138 \times 400 \times 10^3 \times A_s \times (170)^2 \] & Step 3: Solve for \( A_s \) \[ A_s = \frac{81 \times 10^6}{0.138 \times 400 \times 10^3 \times 28900} \] & \[ A_s = \frac{81 \times 10^6}{1.596 \times 10^9} \approx 50.88\, \text{mm}^2 \] & Step 4: Select appropriate reinforcement bars Assuming use of 16 mm diameter bars: \[ \text{Area of one } 16\, \text{mm bar} = \frac{\pi}{4} \times 16^2 \approx 201\, \text{mm}^2 \] & Step 5: Determine the number of bars required \[ \text{Number of bars} = \frac{50.88}{201} \approx 0.25 \] & Since we cannot have a fraction of a bar, we round up to 1 bar. Answer for part b): A single 16 mm diameter reinforcement bar is required for the ultimate moment.