Question: d)\tA rectangular concrete pier cross-section of 500mm \times 700mm and 2.10m long carries an axial compressive load of 2.4 MN. Determine the stress in the concrete at the base of the pier also the amount of shortening that will occur in the pier. Density of concrete = 2500\nkg/m3; Young's modulus = 13 kN/mm2
1. **State the problem:** We have a concrete pier with cross-section dimensions $500\ \text{mm} \times 700\ \text{mm}$ and length $2.10\ \text{m}$. It carries an axial compressive load of $2.4\ \text{MN}$. We need to find the stress at the base and the shortening of the pier.
2. **Given data:**
- Cross-section width $b = 500\ \text{mm}$
- Cross-section height $h = 700\ \text{mm}$
- Length $L = 2.10\ \text{m} = 2100\ \text{mm}$
- Load $P = 2.4\ \text{MN} = 2400000\ \text{N}$
- Young's modulus $E = 13\ \text{kN/mm}^2 = 13000\ \text{N/mm}^2$
3. **Formula for stress:**
$$\sigma = \frac{P}{A}$$
where $A$ is the cross-sectional area.
4. **Calculate cross-sectional area:**
$$A = b \times h = 500 \times 700 = 350000\ \text{mm}^2$$
5. **Calculate stress:**
$$\sigma = \frac{2400000}{350000} = \frac{2400000}{350000}$$
Intermediate step showing cancellation:
$$\sigma = \frac{\cancel{2400000}}{\cancel{350000}} = 6.8571\ \text{N/mm}^2$$
6. **Formula for shortening:**
$$\Delta L = \frac{PL}{AE}$$
7. **Calculate shortening:**
$$\Delta L = \frac{2400000 \times 2100}{350000 \times 13000}$$
Intermediate step showing cancellation:
$$\Delta L = \frac{\cancel{2400000} \times 2100}{\cancel{350000} \times 13000} = \frac{2400000 \times 2100}{350000 \times 13000}$$
Simplify numerator and denominator:
$$\Delta L = \frac{2400000 \times 2100}{350000 \times 13000} = 1.1057\ \text{mm}$$
8. **Final answers:**
- Stress at base: $6.86\ \text{N/mm}^2$
- Shortening of pier: $1.11\ \text{mm}$
These results mean the concrete experiences a compressive stress of about $6.86\ \text{MPa}$ and shortens by approximately $1.11\ \text{mm}$ under the given load.