Answer :
Answer:
To determine the loss of prestress in a prestressed concrete beam due to elastic shortening, we need to follow these steps:
1. **Calculate the initial prestressing force (P):**
\[
P = f_s \cdot A_s
\]
where \( f_s \) is the stress in the steel and \( A_s \) is the area of the prestressing steel. Since \( f_s \) is not given directly, it is typically taken as the allowable stress in the steel, which is generally 0.7 times the ultimate tensile strength. However, this value is not provided, so we will assume an initial prestress value, as the exact initial force isn't required to find the percentage loss. Let's denote it by \( P \).
2. **Calculate the modulus of elasticity of concrete (Ec):**
\[
E_c = 5000 \sqrt{f_{ck}}
\]
where \( f_{ck} \) is the characteristic strength of concrete.
\[
E_c = 5000 \sqrt{50} = 5000 \cdot 7.071 = 35355 \text{ N/mm}^2
\]
3. **Calculate the stress in concrete due to prestressing force (f_cp):**
\[
f_{cp} = \frac{P}{A_c}
\]
where \( A_c \) is the cross-sectional area of the concrete.
\[
A_c = 150 \times 10^3 \text{ mm}^2
\]
\[
f_{cp} = \frac{P}{150 \times 10^3}
\]
4. **Determine the strain in concrete (ε_c):**
\[
\varepsilon_c = \frac{f_{cp}}{E_c}
\]
5. **Calculate the loss of prestress in steel (Δf_s):**
\[
\Delta f_s = \varepsilon_c \cdot E_s
\]
where \( E_s \) is the modulus of elasticity of the prestressing steel.
\[
E_s = 200 \times 10^3 \text{ N/mm}^2
\]
\[
\Delta f_s = \left( \frac{P}{150 \times 10^3 \times 35355} \right) \cdot 200 \times 10^3
\]
6. **Calculate the percentage loss of prestress:**
\[
\text{Percentage loss} = \frac{\Delta f_s}{f_s} \times 100
\]
where \( f_s \) is the initial stress in the steel.
Now, let's go through the calculations step-by-step:
1. **Initial Prestressing Force (P):**
We are assuming an initial prestress value of \( P \).
2. **Modulus of Elasticity of Concrete (Ec):**
\[
E_c = 35355 \text{ N/mm}^2
\]
3. **Stress in Concrete Due to Prestressing Force (f_cp):**
\[
f_{cp} = \frac{P}{150 \times 10^3}
\]
4. **Strain in Concrete (ε_c):**
\[
\varepsilon_c = \frac{f_{cp}}{E_c} = \frac{\frac{P}{150 \times 10^3}}{35355} = \frac{P}{150 \times 10^3 \times 35355}
\]
5. **Loss of Prestress in Steel (Δf_s):**
\[
\Delta f_s = \left( \frac{P}{150 \times 10^3 \times 35355} \right) \cdot 200 \times 10^3 = \frac{P \cdot 200 \times 10^3}{150 \times 10^3 \times 35355} = \frac{P}{150 \times 353.55}
\]
\[
\Delta f_s = \frac{P}{53032.5}
\]
6. **Percentage Loss of Prestress:**
\[
\text{Percentage loss} = \frac{\Delta f_s}{f_s} \times 100
\]
Since \( P = f_s \cdot A_s \):
\[
f_s = \frac{P}{A_s} = \frac{P}{1500}
\]
\[
\Delta f_s = \frac{P}{53032.5}
\]
\[
\text{Percentage loss} = \frac{\frac{P}{53032.5}}{\frac{P}{1500}} \times 100 = \frac{1500}{53032.5} \times 100 = 2.83\%
\]
Therefore, the loss of prestress due to elastic shortening is approximately **2.83%** of the initial prestressing force.
