P02: Determine the modular ratio 'n' and the modulus of rapture 'fr' for each case in P01 for each case shown in P01. f₁=7.54√f, psi n= 29000 (ksi) E. (ksi) 11- 1₁=0.624/f' N/mm² 200000 (MPa) E, (MPa)
P02: Determine the modular ratio 'n' and the modulus of rapture 'fr' for each case in P01 for each case shown in P01. f₁=7.54√f, psi n= 29000 (ksi) E. (ksi) 11- 1₁=0.624/f' N/mm² 200000 (MPa) E, (MPa)
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Question
Answer P02 given P01
![### Problem P02
**Objective:**
Determine the modular ratio ‘n’ and the modulus of rupture ‘fr’ for each case in P01 for each case shown in P01.
#### Key Formulas
The modulus of rupture (\(f_r\)) and the modular ratio (\(n\)) are calculated using the following formulas:
**Modulus of Rupture:**
- In pounds per square inch (psi):
\[
f_r = 7.5 \cdot \lambda \cdot \sqrt{f'_c} \, \text{psi}
\]
- In Newtons per square millimeter (N/mm\(^2\)):
\[
f_r = 0.62 \cdot \lambda \cdot \sqrt{f'_c} \, \text{N/mm}^2
\]
**Modular Ratio:**
- In kilopounds per square inch (ksi):
\[
n = \frac{29000 \, (\text{ksi})}{E_c \, (\text{ksi})}
\]
- In Megapascals (MPa):
\[
n = \frac{200000 \, (\text{MPa})}{E_c \, (\text{MPa})}
\]
#### Explanation
- **\(f_r\)**: Modulus of rupture of concrete.
- **\(\lambda\)**: A modification factor that accounts for the type of aggregate used in the concrete.
- **\(f'_c\)**: Compressive strength of the concrete.
- **\(E_c\)**: Modulus of elasticity of the concrete.
- **\(n\)**: Modular ratio, defined as the ratio of the modulus of elasticity of steel to the modulus of elasticity of concrete.
Substitute the given values into the appropriate formulas to solve for \(n\) and \(f_r\) for each case.
#### Note:
- Ensure to use consistent units throughout your calculations.
- The formulas provided are standard, but it's crucial to adapt them to the specific conditions and values provided in P01.
Use the discussed relationships to derive the respective values for each specific case scenario defined in P01. This would allow for detailed structural analysis and ensure accuracy in design and application.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb816de81-d091-4434-aa0d-8a250bb5db6b%2Fa51d8f13-a080-44f9-8fba-4d1e249048a8%2F8uutud_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem P02
**Objective:**
Determine the modular ratio ‘n’ and the modulus of rupture ‘fr’ for each case in P01 for each case shown in P01.
#### Key Formulas
The modulus of rupture (\(f_r\)) and the modular ratio (\(n\)) are calculated using the following formulas:
**Modulus of Rupture:**
- In pounds per square inch (psi):
\[
f_r = 7.5 \cdot \lambda \cdot \sqrt{f'_c} \, \text{psi}
\]
- In Newtons per square millimeter (N/mm\(^2\)):
\[
f_r = 0.62 \cdot \lambda \cdot \sqrt{f'_c} \, \text{N/mm}^2
\]
**Modular Ratio:**
- In kilopounds per square inch (ksi):
\[
n = \frac{29000 \, (\text{ksi})}{E_c \, (\text{ksi})}
\]
- In Megapascals (MPa):
\[
n = \frac{200000 \, (\text{MPa})}{E_c \, (\text{MPa})}
\]
#### Explanation
- **\(f_r\)**: Modulus of rupture of concrete.
- **\(\lambda\)**: A modification factor that accounts for the type of aggregate used in the concrete.
- **\(f'_c\)**: Compressive strength of the concrete.
- **\(E_c\)**: Modulus of elasticity of the concrete.
- **\(n\)**: Modular ratio, defined as the ratio of the modulus of elasticity of steel to the modulus of elasticity of concrete.
Substitute the given values into the appropriate formulas to solve for \(n\) and \(f_r\) for each case.
#### Note:
- Ensure to use consistent units throughout your calculations.
- The formulas provided are standard, but it's crucial to adapt them to the specific conditions and values provided in P01.
Use the discussed relationships to derive the respective values for each specific case scenario defined in P01. This would allow for detailed structural analysis and ensure accuracy in design and application.
![## Calculation of the Modulus of Elasticity \( E_c \)
### Problem Statement
Calculate the modulus of elasticity \( E_c \) for each case shown in the table below.
### Formulas
For calculations in pounds per cubic foot (pcf) and pounds per square inch (psi):
\[ E_c = 33 w^{1.5} \sqrt{f'_c} \ \text{psi} \]
For calculations in kilograms per cubic meter (kg/m³) and megapascals (MPa):
\[ E_c = 0.043 w^{1.5} \sqrt{f'_c} \ \text{MPa} \]
### Given Data
| Density | \( f'_c \) |
|-----------------|----------------|
| 180 pcf | 5000 psi |
| 160 pcf | 4000 psi |
| 145 pcf | 3000 psi |
| 2550 kg/m³ | 45 MPa |
| 2400 kg/m³ | 40 MPa |
| 2300 kg/m³ | 25 MPa |
### Description of Table
The table provides the density and the compressive strength (\( f'_c \)) of concrete for various conditions. The density is given in two different units: pounds per cubic foot (pcf) and kilograms per cubic meter (kg/m³). Similarly, the compressive strength (\( f'_c \)) is given in pounds per square inch (psi) and megapascals (MPa).
Using the appropriate formulas, the modulus of elasticity (\( E_c \)) can be calculated for each scenario in psi or MPa as provided.
### Example Calculation
1. For the first row with density \( w = 180 \) pcf and \( f'_c = 5000 \) psi:
\[ E_c = 33 \times (180)^{1.5} \times \sqrt{5000} \ \text{psi} \]
Following through with the calculation steps will yield the modulus of elasticity for that specific condition.
Repeat the procedure for each row to find the corresponding \( E_c \) values in psi or MPa, as detailed by the provided formulas.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb816de81-d091-4434-aa0d-8a250bb5db6b%2Fa51d8f13-a080-44f9-8fba-4d1e249048a8%2Fkzkjxtm_processed.jpeg&w=3840&q=75)
Transcribed Image Text:## Calculation of the Modulus of Elasticity \( E_c \)
### Problem Statement
Calculate the modulus of elasticity \( E_c \) for each case shown in the table below.
### Formulas
For calculations in pounds per cubic foot (pcf) and pounds per square inch (psi):
\[ E_c = 33 w^{1.5} \sqrt{f'_c} \ \text{psi} \]
For calculations in kilograms per cubic meter (kg/m³) and megapascals (MPa):
\[ E_c = 0.043 w^{1.5} \sqrt{f'_c} \ \text{MPa} \]
### Given Data
| Density | \( f'_c \) |
|-----------------|----------------|
| 180 pcf | 5000 psi |
| 160 pcf | 4000 psi |
| 145 pcf | 3000 psi |
| 2550 kg/m³ | 45 MPa |
| 2400 kg/m³ | 40 MPa |
| 2300 kg/m³ | 25 MPa |
### Description of Table
The table provides the density and the compressive strength (\( f'_c \)) of concrete for various conditions. The density is given in two different units: pounds per cubic foot (pcf) and kilograms per cubic meter (kg/m³). Similarly, the compressive strength (\( f'_c \)) is given in pounds per square inch (psi) and megapascals (MPa).
Using the appropriate formulas, the modulus of elasticity (\( E_c \)) can be calculated for each scenario in psi or MPa as provided.
### Example Calculation
1. For the first row with density \( w = 180 \) pcf and \( f'_c = 5000 \) psi:
\[ E_c = 33 \times (180)^{1.5} \times \sqrt{5000} \ \text{psi} \]
Following through with the calculation steps will yield the modulus of elasticity for that specific condition.
Repeat the procedure for each row to find the corresponding \( E_c \) values in psi or MPa, as detailed by the provided formulas.
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