Subject reinforced concrete design solve p02 ### P02: Determining Modular Ratio ‘n’ and Modulus of Rupture ‘fr’In this exercise, we will determine the modular ratio ‘n’ and the modulus of rupture ‘fr’ for each case outlined in P01.#### Modulus of Rupture (\( f_r \))The modulus of rupture (\( f_r \)) can be calculated using the formulae given for different units:1. **For psi units:** \[ f_r = 7.5 \lambda \sqrt{f'_c} \text{ psi} \]2. **For \( \text{N/mm}^2 \) units:** \[ f_r = 0.62 \lambda \sqrt{f'_c} \, \text{N/mm}^2 \]#### Modular Ratio (\( n \))The modular ratio (\( n \)) can be obtained from the formulas below for given values of the elastic moduli of steel and concrete:1. **For ksi units:** \[ n = \frac{29000 \, (\text{ksi})}{E_c \, (\text{ksi})} \]2. **For MPa units:** \[ n = \frac{200000 \, (\text{MPa})}{E_c \, (\text{MPa})} \]By using these formulas, you can calculate both the modular ratio and modulus of rupture for each specific scenario described in problem P01. Ensure the appropriate units are used for each calculation to get accurate results.**Note:** \( \lambda \) is a modification factor, and \( f'_c \) represents the compressive strength of the concrete. \( E_c \) is the modulus of elasticity of concrete. **Calculation of Modulus of Elasticity \( E_c \)**### Problem Statement:Calculate the modulus of elasticity \( E_c \) for each case shown in the table below:### Formulas:- For units in psi:\[ E_c = 33 w^{1.5} \sqrt{f'_c} \, \text{psi} \]- For units in MPa:\[ E_c = 0.043 w^{1.5} \sqrt{f'_c} \, \text{MPa} \]### Table 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 |**Explanation:**- The first formula is used when density (\( w \)) is given in pounds per cubic foot (pcf) and \( f'_c \) is in psi.- The second formula is used when density (\( w \)) is given in kilograms per cubic meter (kg/m³) and \( f'_c \) is in MPa.This table includes various combinations of densities and compressive strengths (\( f'_c \)). Use the relevant formula based on the units provided to calculate the modulus of elasticity for each case.For educational purposes, students would apply these formulas to the values provided in the table, performing appropriate unit conversions where necessary to ensure consistency within each formula's application context.

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P02: Determine the modular ratio 'n' and the modulus of rapture 'fr' for each case in P01 for each case shown in P01. f₁=7.5f, psi n= 29000 (ksi) E. (ksi) 11- f₁=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=7.5f, psi n= 29000 (ksi) E. (ksi) 11- f₁=0.624/f' N/mm² 200000 (MPa) E. (MPa)

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Question

Subject reinforced concrete design solve p02

### P02: Determining Modular Ratio ‘n’ and Modulus of Rupture ‘fr’

In this exercise, we will determine the modular ratio ‘n’ and the modulus of rupture ‘fr’ for each case outlined in P01.

#### Modulus of Rupture (\( f_r \))

The modulus of rupture (\( f_r \)) can be calculated using the formulae given for different units:

1. **For psi units:**
\[
f_r = 7.5 \lambda \sqrt{f'_c} \text{ psi}
\]

2. **For \( \text{N/mm}^2 \) units:**
\[
f_r = 0.62 \lambda \sqrt{f'_c} \, \text{N/mm}^2
\]

#### Modular Ratio (\( n \))

The modular ratio (\( n \)) can be obtained from the formulas below for given values of the elastic moduli of steel and concrete:

1. **For ksi units:**
\[
n = \frac{29000 \, (\text{ksi})}{E_c \, (\text{ksi})}
\]

2. **For MPa units:**
\[
n = \frac{200000 \, (\text{MPa})}{E_c \, (\text{MPa})}
\]

By using these formulas, you can calculate both the modular ratio and modulus of rupture for each specific scenario described in problem P01. Ensure the appropriate units are used for each calculation to get accurate results.

**Note:** \( \lambda \) is a modification factor, and \( f'_c \) represents the compressive strength of the concrete. \( E_c \) is the modulus of elasticity of concrete.

**Calculation of Modulus of Elasticity \( E_c \)**

### Problem Statement:
Calculate the modulus of elasticity \( E_c \) for each case shown in the table below:

### Formulas:
- For units in psi:
\[ E_c = 33 w^{1.5} \sqrt{f'_c} \, \text{psi} \]
- For units in MPa:
\[ E_c = 0.043 w^{1.5} \sqrt{f'_c} \, \text{MPa} \]

### Table 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 |

**Explanation:**
- The first formula is used when density (\( w \)) is given in pounds per cubic foot (pcf) and \( f'_c \) is in psi.
- The second formula is used when density (\( w \)) is given in kilograms per cubic meter (kg/m³) and \( f'_c \) is in MPa.

This table includes various combinations of densities and compressive strengths (\( f'_c \)). Use the relevant formula based on the units provided to calculate the modulus of elasticity for each case.

For educational purposes, students would apply these formulas to the values provided in the table, performing appropriate unit conversions where necessary to ensure consistency within each formula's application context.

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