Steel Bar Elongation 8Ffcdb

1. **State the problem:** We need to find the total elongation (change in length) of a steel bar subjected to three different downward loads applied at different points along its length. 2. **Given data:** - Loads: $P_1 = 22250$ N at the lower end (0 m), $P_2 = 13500$ N at $0.35$ m above the lower end, $P_3 = 9000$ N at $1$ m above the lower end. - Modulus of elasticity: $E = 30 \times 10^{6}$ psi. - Cross-sectional area: $A = 6.5$ cm$^{2}$. 3. **Convert units:** - Convert $E$ from psi to pascals (Pa): $$1\ \text{psi} = 6894.76\ \text{Pa}$$ $$E = 30 \times 10^{6} \times 6894.76 = 2.0684 \times 10^{11} \ \text{Pa}$$ - Convert area from cm$^{2}$ to m$^{2}$: $$A = 6.5 \ \text{cm}^2 = 6.5 \times 10^{-4} \ \text{m}^2$$ 4. **Understand the loading:** The bar is loaded at three points, so the internal axial force varies along the length. The elongation is the sum of elongations of segments between loads. 5. **Calculate axial force in each segment:** - Segment 1: from $0$ m to $0.35$ m (top to $0.35$ m above lower end) Forces below $0.35$ m: $P_1 + P_2 + P_3 = 22250 + 13500 + 9000 = 44750$ N - Segment 2: from $0.35$ m to $1$ m Forces below $1$ m: $P_1 + P_2 = 22250 + 13500 = 35750$ N - Segment 3: from $1$ m to the top (assuming bar length is at least $1$ m) Forces below top: $P_1 = 22250$ N 6. **Calculate elongation for each segment:** Elongation formula: $$\Delta L = \frac{PL}{AE}$$ - Segment 1 length: $L_1 = 0.35$ m, force $P_1 = 44750$ N $$\Delta L_1 = \frac{44750 \times 0.35}{6.5 \times 10^{-4} \times 2.0684 \times 10^{11}}$$ - Segment 2 length: $L_2 = 1 - 0.35 = 0.65$ m, force $P_2 = 35750$ N $$\Delta L_2 = \frac{35750 \times 0.65}{6.5 \times 10^{-4} \times 2.0684 \times 10^{11}}$$ - Segment 3 length: $L_3 = $ (assuming bar length is $>1$ m, but no length given beyond 1 m, so no segment 3 elongation) 7. **Calculate each elongation:** - $$\Delta L_1 = \frac{44750 \times 0.35}{6.5 \times 10^{-4} \times 2.0684 \times 10^{11}} = \frac{15662.5}{1.34446 \times 10^{8}} = 1.1649 \times 10^{-4} \ \text{m}$$ - $$\Delta L_2 = \frac{35750 \times 0.65}{6.5 \times 10^{-4} \times 2.0684 \times 10^{11}} = \frac{23237.5}{1.34446 \times 10^{8}} = 1.728 \times 10^{-4} \ \text{m}$$ 8. **Total elongation:** $$\Delta L = \Delta L_1 + \Delta L_2 = 1.1649 \times 10^{-4} + 1.728 \times 10^{-4} = 2.893 \times 10^{-4} \ \text{m}$$ 9. **Convert elongation to inches:** $$1 \ \text{m} = 39.3701 \ \text{in}$$ $$\Delta L = 2.893 \times 10^{-4} \times 39.3701 = 0.01139 \ \text{in}$$ **Final answers:** - Total change in length $= 2.89 \times 10^{-4}$ m - Total change in length $= 0.0114$ in