Design a 12 ft simply supported one-way slab to carry a service load of 100 psf. Use [tex]f_c = 3 \text{ ksi}[/tex] and [tex]f_y = 60 \text{ ksi}[/tex].

Solution:

1. Estimate Self-Weight:

- Minimum depth = \( \frac{1}{20} \) (Table 7.3.3.1)
- Depth = \( \frac{12}{20} = 0.6 \) ft
- Use a 7.5 in. thick slab
- Self-weight = \( \frac{7.5 \times 12 \times 150}{144} = 94 \) lb/ft

2. Calculate Required Strength, \( M \):

- \( M_o = \frac{94 \times 12}{8} = 1692 \) lb-ft/ft
- \( M_s = \frac{100 \times 12 \times 4}{8} = 1800 \) lb-ft/ft
- \( M_u = 1.2M_o + 1.6M_s = 1.2(1692) + 1.6(1800) = 4910 \) lb-ft/ft

3. Find Effective Depth:

- \( d = h - 1 = 7.5 - 1 = 6.5 \) in

The effective depth, represented as 'd', is then calculated to be 6.5 inches. The subject is related to Engineering, specifically the design of a simply supported one-way slab.

The problem involves the design of an engineering structure, specifically a simply supported one-way slab. The solution provided involves estimating the self-weight and required strength of the slab, using given formulas. This includes calculating the slab's minimum depth, self-weight, required strength, and effective depth.

The calculations provided in the question are used to estimate the self-weight and required strength of the slab. The minimum depth of the slab is calculated to be 0.6 feet.

A thickness of 7.5 inches is used for the slab, resulting in a self-weight of 94 lb/ft.

The required strength, denoted as M, is calculated to be 1.2 times the self-weight plus 1.6 times the service load, yielding a total required strength of 4.91 k-ft/Ft.

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