Given:
- A concrete column with a cross-section of [tex][tex]$200 \times 400 \, \text{mm}$[/tex][/tex] is reinforced with 4 bars of mild steel, each with a diameter of [tex][tex]$25 \, \text{mm}$[/tex][/tex].
- The column is to carry an axial load of [tex][tex]$P = 850 \, \text{kN}$[/tex][/tex].
- The concrete contracts the same amount as the steel.

Calculate:
- The stress in the concrete and in the steel.

Given:
- [tex]E_s = 205 \, \text{kN/mm}^2[/tex]
- [tex]E_c = \frac{14 \, \text{kN/mm}^2}{850 \, \text{kN}}[/tex]

Use the formula:
[tex]\sigma = \frac{E}{A}[/tex]



Answer :

Certainly! Let's go step-by-step through the solution for the problem.

Given data:
- Width of concrete section ([tex]\(b\)[/tex]): 200 mm
- Height of concrete section ([tex]\(h\)[/tex]): 400 mm
- Diameter of steel bars ([tex]\(d\)[/tex]): 25 mm
- Axial load ([tex]\(P\)[/tex]): 850 kN (which is 850,000 N since 1 kN = 1000 N)
- Modulus of elasticity of steel ([tex]\(E_s\)[/tex]): 205,000 N/mm²
- Modulus of elasticity of concrete ([tex]\(E_c\)[/tex]): 14,000 N/mm²

### Step 1: Calculate the area of the concrete section
The area of the concrete section, [tex]\(A_c\)[/tex], is calculated by the formula:
[tex]\[ A_c = b \times h \][/tex]

Substituting the values:
[tex]\[ A_c = 200 \, \text{mm} \times 400 \, \text{mm} = 80,000 \, \text{mm}^2 \][/tex]

### Step 2: Calculate the area of the steel bars
Each steel bar has a circular cross-section. The radius of the steel bar, [tex]\(r\)[/tex], is half of the diameter:
[tex]\[ r = \frac{d}{2} = \frac{25 \, \text{mm}}{2} = 12.5 \, \text{mm} \][/tex]

The area of one steel bar, [tex]\(A_{s1}\)[/tex], is calculated by the formula for the area of a circle:
[tex]\[ A_{s1} = \pi r^2 \][/tex]
[tex]\[ A_{s1} = \pi (12.5 \, \text{mm})^2 = \pi \times 156.25 \, \text{mm}^2 \approx 490.87 \, \text{mm}^2 \][/tex]

Since there are 4 steel bars,
[tex]\[ A_s = 4 \times A_{s1} = 4 \times 490.87 \, \text{mm}^2 \approx 1,963.50 \, \text{mm}^2 \][/tex]

### Step 3: Determine the load shared by concrete and steel
The total axial load [tex]\(P\)[/tex] is shared between the concrete and steel. The load carried by the concrete, [tex]\(P_c\)[/tex], and the load carried by the steel, [tex]\(P_s\)[/tex], are calculated based on their stiffness (product of modulus of elasticity and area).

Using the ratio of the stiffness:
[tex]\[ P_c = \frac{E_c A_c}{E_c A_c + E_s A_s} \times P \][/tex]
[tex]\[ P_c = \frac{14,000 \, \text{N}/\text{mm}^2 \times 80,000 \, \text{mm}^2}{14,000 \, \text{N}/\text{mm}^2 \times 80,000 \, \text{mm}^2 + 205,000 \, \text{N}/\text{mm}^2 \times 1,963.50 \, \text{mm}^2} \times 850,000 \, \text{N} \][/tex]
[tex]\[ P_c \approx \frac{1,120,000,000 \, \text{N}}{27,178,967,000 \, \text{N}} \times 850,000 \, \text{N} \][/tex]
[tex]\[ P_c \approx 625,280.56 \, \text{N} \][/tex]

The load carried by the steel bars is the remaining load not carried by the concrete:
[tex]\[ P_s = P - P_c \][/tex]
[tex]\[ P_s = 850,000 \, \text{N} - 625,280.56 \, \text{N} \][/tex]
[tex]\[ P_s \approx 224,719.44 \, \text{N} \][/tex]

### Step 4: Calculate the stress in the concrete and steel
Stress is given by [tex]\( \sigma = \frac{P}{A} \)[/tex].

- Stress in concrete ([tex]\(\sigma_c\)[/tex]):
[tex]\[ \sigma_c = \frac{P_c}{A_c} \][/tex]
[tex]\[ \sigma_c = \frac{625,280.56 \, \text{N}}{80,000 \, \text{mm}^2} \][/tex]
[tex]\[ \sigma_c \approx 7.82 \, \text{N}/\text{mm}^2 \][/tex]

- Stress in steel ([tex]\(\sigma_s\)[/tex]):
[tex]\[ \sigma_s = \frac{P_s}{A_s} \][/tex]
[tex]\[ \sigma_s = \frac{224,719.44 \, \text{N}}{1,963.50 \, \text{mm}^2} \][/tex]
[tex]\[ \sigma_s \approx 114.45 \, \text{N}/\text{mm}^2 \][/tex]

### Conclusion:
- The area of concrete is [tex]\(80,000 \, \text{mm}^2\)[/tex].
- The total area of the steel bars is [tex]\(1,963.50 \, \text{mm}^2\)[/tex].
- The load carried by the concrete is approximately [tex]\(625,280.56 \, \text{N}\)[/tex].
- The load carried by the steel is approximately [tex]\(224,719.44 \, \text{N}\)[/tex].
- The stress in the concrete is approximately [tex]\(7.82 \, \text{N}/\text{mm}^2\)[/tex].
- The stress in the steel is approximately [tex]\(114.45 \, \text{N}/\text{mm}^2\)[/tex].

Other Questions