Question: User: 5000 lb \u2190
6000 lb \u2192 \u2190
4000 lb \u2192
3000 lb \u2192
3 ft
5 ft
4 ft
A horizontal bar/beam is shown near the top-center of the image; the left segment is labeled 6000 lb, the middle segment is unlabeled, and the right segment is labeled 4000 lb, with 5000 lb force acting to the left at the far left end and 3000 lb force acting to the right at the far right end. The dimensions below the beam are 3 ft from the left end to the first division, 5 ft between the first and second divisions, and 4 ft from the second division to the right end. position_hint=top-centerAn aluminium bar having a cross-sectional area of 0.4 𝑖𝑛2 carries the axial loads applied at the positions shown in Fig. Compute the total change in length of the bar if E = 10 \times 10^{6} psi. Assume the bar is suitably braced to prevent lateral buckling. (7 marks)
1. **Problem Statement:**
Calculate the total change in length of an aluminium bar with cross-sectional area $0.4\ \text{in}^2$ subjected to axial loads at different segments along its length. The modulus of elasticity is $E = 10 \times 10^{6}\ \text{psi}$. The bar is divided into three segments of lengths $3\ \text{ft}$, $5\ \text{ft}$, and $4\ \text{ft}$ respectively.
2. **Given Data:**
- Cross-sectional area, $A = 0.4\ \text{in}^2$
- Modulus of elasticity, $E = 10 \times 10^{6}\ \text{psi}$
- Segment lengths: $L_1 = 3\ \text{ft}$, $L_2 = 5\ \text{ft}$, $L_3 = 4\ \text{ft}$
- Forces: $F_1 = 5000\ \text{lb}$ (left, compressive), $F_2 = 6000\ \text{lb}$ (right, tensile), $F_3 = 4000\ \text{lb}$ (right, tensile), $F_4 = 3000\ \text{lb}$ (right, tensile)
3. **Convert lengths to inches:**
$$L_1 = 3 \times 12 = 36\ \text{in}$$
$$L_2 = 5 \times 12 = 60\ \text{in}$$
$$L_3 = 4 \times 12 = 48\ \text{in}$$
4. **Determine axial force in each segment:**
- Segment 1 (left end to first division): Forces acting are $5000\ \text{lb}$ left and $6000\ \text{lb}$ right. Net force on segment 1 is $F_1 = 6000 - 5000 = 1000\ \text{lb}$ tensile.
- Segment 2 (between first and second division): Forces acting are $6000\ \text{lb}$ right and $4000\ \text{lb}$ right. Net force on segment 2 is $F_2 = 4000\ \text{lb}$ tensile.
- Segment 3 (second division to right end): Forces acting are $4000\ \text{lb}$ right and $3000\ \text{lb}$ right. Net force on segment 3 is $F_3 = 3000\ \text{lb}$ tensile.
5. **Formula for axial deformation:**
$$\Delta L = \frac{F L}{A E}$$
where $F$ is axial force, $L$ is length of segment, $A$ is cross-sectional area, and $E$ is modulus of elasticity.
6. **Calculate deformation for each segment:**
$$\Delta L_1 = \frac{1000 \times 36}{0.4 \times 10 \times 10^{6}} = \frac{36000}{4 \times 10^{6}} = 0.009\ \text{in}$$
$$\Delta L_2 = \frac{4000 \times 60}{0.4 \times 10 \times 10^{6}} = \frac{240000}{4 \times 10^{6}} = 0.06\ \text{in}$$
$$\Delta L_3 = \frac{3000 \times 48}{0.4 \times 10 \times 10^{6}} = \frac{144000}{4 \times 10^{6}} = 0.036\ \text{in}$$
7. **Total change in length:**
$$\Delta L_{total} = \Delta L_1 + \Delta L_2 + \Delta L_3 = 0.009 + 0.06 + 0.036 = 0.105\ \text{in}$$
**Answer:** The total elongation of the aluminium bar is $0.105\ \text{in}$.