Give step-by-step solution with explanation and final answer:
You are renovating your house and plan to convert the space above the garage into a small home gym. The floor of this room is a one
-
way

Question

Give step-by-step solution with explanation and final answer:
You are renovating your house and plan to convert the space above the garage into a small home gym. The floor of this room is a one
-
way reinforced concrete slab that spans 
1
2
 ﻿ft from the center of one wall support to the center of the opposite wall. Since the roof is supported directly by the surrounding walls, the only dead load on the slab is its own self
-
weight. Originally, the slab was used only as a light storage area, but adding gym equipment 
(
treadmills
,
 ﻿weights, etc.
)
 ﻿introduces much higher live loads. Before installing the equipment, you want to check if the slab can safely support the new usage. From the contractor who built the house you came to know that 
3
0
0
0
 ﻿psi concrete and 
4
0
 ﻿grade reinforcing steel was used. Determine the service live load capacity 
(
psf
)
 ﻿that the slab may safely carry.

VIPSolutions · Accepted Answer

# Step-by-Step Solution

## 1. **Given Data**

- **Slab span (L):** $12\, \text{ft}$
- **Concrete strength ($f'_c$):** $300\, \text{psi}$
- **Rebar yield strength ($f_y$):** $40\, \text{ksi} = 40{,}000\, \text{psi}$
- **Slab is one-way, simply supported**
- **Unknown:** *Service live load capacity* ($w_{LL,service}$) in $psf$.

## 2. **Assumptions**

- **Slab thickness ($h$):** Not specified. Assume a typical thickness for a 12 ft span slab. For residential/light commercial, **6-inch** slab is common.
- **Reinforcement ratio ($\rho$):** Assume minimum code-compliant value.
- **Unit weight of concrete ($\gamma_c$):** $150\, \text{pcf}$.

## 3. **Calculate Slab Self-Weight (Dead Load)**

$$
h = 6\, \text{in} = .5\, \text{ft}
$$
$$
w_{DL} = \gamma_c \times h = 150\, \text{pcf} \times .5\, \text{ft} = 75\, \text{psf}
$$

## 4. **Strength Design (Ultimate Limit State)**

### a. **Factored Load Combination**

$$
w_u = 1.2\, w_{DL} + 1.6\, w_{LL}
$$

### b. **Slab Flexural Capacity**

#### i. **Effective Depth ($d$)**

Assume bottom cover + bar diameter = 1.5 in  
$$
d \approx 6\, \text{in} - 1.5\, \text{in} = 4.5\, \text{in} = .375\, \text{ft}
$$

#### ii. **Minimum Reinforcement Ratio ($\rho_{min}$)**

$$
\rho_{min} = \max\left( \frac{3\sqrt{f'_c}}{f_y}, \frac{200}{f_y} \right)
$$
$$
\frac{3\sqrt{300}}{40,000} = \frac{3 \times 54.77}{40,000} \approx .0041
$$
$$
\frac{200}{40,000} = .005
$$

So,  
$$
\rho_{min} = .005
$$

#### iii. **Area of steel per foot width ($b = 12\, \text{in}$)**

$$
A_s = \rho_{min} \times b \times d = .005 \times 12\, \text{in} \times 4.5\, \text{in} = .005 \times 54 = .27\, \text{in}^2/\text{ft}
$$

#### iv. **Nominal Moment Capacity ($M_n$)**

For a singly reinforced rectangular slab (neglecting compression steel):

$$
M_n = A_s f_y \left( d - \frac{a}{2} \right)
$$
where $a = \frac{A_s f_y}{.85 f'_c b}$

Let's compute $a$:

$$
A_s = .27\, \text{in}^2/\text{ft}
$$
$$
b = 12\, \text{in}
$$
$$
a = \frac{.27 \times 40,000}{.85 \times 3,000 \times 12}
$$
$$
.85 \times 3,000 \times 12 = 30,600
$$
$$
a = \frac{10,800}{30,600} \approx .353\, \text{in}
$$

$$
d - a/2 = 4.5 - .176 = 4.324\, \text{in}
$$
$$
M_n = .27 \times 40,000 \times 4.324 = .27 \times 40,000 = 10,800
$$
$$
10,800 \times 4.324 = 46,699\, \text{in-lb}
$$

Convert to ft-lb:
$$
\frac{46,699}{12} = 3,892\, \text{ft-lb}
$$

Convert to ft-kip:
$$
\frac{3,892}{1,000} = 3.89\, \text{ft-kip}
$$

#### v. **Strength Reduction Factor ($\phi$)**

For flexure, $\phi = .9$:

$$
M_u = \phi M_n = .9 \times 3.89 = 3.50\, \text{ft-kip} = 3,500\, \text{ft-lb}
$$

## 5. **Relate Moment to Uniform Load**

Maximum moment for simply supported slab (1 ft width):

$$
M_u = \frac{w_u L^2}{8}
$$
$$
w_u = \frac{8 M_u}{L^2}
$$

Plug in values ($L = 12\, \text{ft}$):

$$
w_u = \frac{8 \times 3.50}{12^2} = \frac{28}{144} = .194\, \text{kips/ft}
$$
$$
.194\, \text{kips/ft} \times 1,000 = 194\, \text{lb/ft}
$$

This is the **factored load** the slab can carry, per foot width.

## 6. **Subtract Dead Load to Find Live Load Capacity**

Recall:
$$
w_u = 1.2\, w_{DL} + 1.6\, w_{LL}
$$
$$
194 = 1.2 \times 75 + 1.6\, w_{LL}
$$
$$
1.2 \times 75 = 90
$$
$$
194 = 90 + 1.6\, w_{LL}
$$
$$
1.6\, w_{LL} = 104
$$
$$
w_{LL} = \frac{104}{1.6} = 65\, \text{psf}
$$

## 7. **Final Answer**

### **Service Live Load Capacity**

$$
\boxed{65\, \text{psf}}
$$

---

## **Summary Table**

| Parameter          | Value    |
|--------------------|----------|
| Span, $L$        | 12 ft    |
| Slab thickness, $h$ | 6 in   |
| Concrete strength, $f'_c$ | 3,000 psi |
| Steel yield, $f_y$   | 40,000 psi |
| Dead load, $w_{DL}$ | 75 psf |
| Service live load capacity, $w_{LL}$ | **65 psf** |

---

## **Conclusion**

- The **maximum service live load** the slab may safely carry for gym equipment use is **65 psf**.
- If the actual slab thickness or reinforcement is greater, the capacity may be higher.
- For heavy gym equipment, check if this is adequate; free weights and racks may exceed this load locally.
- For a more accurate answer, verify the actual slab thickness and reinforcement details.

---

**Image:**  
*Typical one-way slab section*  
![Typical one-way slab section with reinforcement](https://i.imgur.com/7QFgVnM.png)
*Alt text: Diagram showing a typical one-way slab with bottom reinforcement, slab thickness, and span indicated*

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