## Solution to Problem 213 Axial Deformation

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## Solution to Problem 212 Axial Deformation

**Problem 212**

The rigid bar ABC shown in Fig. P-212 is hinged at A and supported by a steel rod at B. Determine the largest load P that can be applied at C if the stress in the steel rod is limited to 30 ksi and the vertical movement of end C must not exceed 0.10 in.

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## Solution to Problem 211 Axial Deformation

**Problem 211**

A bronze bar is fastened between a steel bar and an aluminum bar as shown in Fig. p-211. Axial loads are applied at the positions indicated. Find the largest value of P that will not exceed an overall deformation of 3.0 mm, or the following stresses: 140 MPa in the steel, 120 MPa in the bronze, and 80 MPa in the aluminum. Assume that the assembly is suitably braced to prevent buckling. Use E_{st} = 200 GPa, E_{al} = 70 GPa, and E_{br} = 83 GPa.

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## Solution to Problem 210 Axial Deformation

**Problem 210**

Solve Prob. 209 if the points of application of the 6000-lb and the 4000-lb forces are interchanged.

**Solution 210**

P_{1} = 4000 lb compression

P_{2} = 11000 lb compression

P_{3} = 6000 lb compression

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## Solution to Problem 209 Axial Deformation

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## Solution to Problem 208 Axial Deformation

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## Solution to Problem 207 Axial Deformation

**Problem 207**

A steel wire 30 ft long, hanging vertically, supports a load of 500 lb. Neglecting the weight of the wire, determine the required diameter if the stress is not to exceed 20 ksi and the total elongation is not to exceed 0.20 in. Assume E = 29 × 10^{6} psi.

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## Solution to Problem 206 Axial Deformation

**Problem 206**

A steel rod having a cross-sectional area of 300 mm^{2} and a length of 150 m is suspended vertically from one end. It supports a tensile load of 20 kN at the lower end. If the unit mass of steel is 7850 kg/m^{3} and E = 200 × 10^{3} MN/m^{2}, find the total elongation of the rod.

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## Solution to Problem 205 Axial Deformation

**Problem 205**

A uniform bar of length L, cross-sectional area A, and unit mass ρ is suspended vertically from one end. Show that its total elongation is δ = ρgL^{2}/2E. If the total mass of the bar is M, show also that δ = MgL/2AE.

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## Solution to Problem 204 Stress-strain Diagram

**Problem 204**

The following data were obtained during a tension test of an aluminum alloy. The initial diameter of the test specimen was 0.505 in. and the gage length was 2.0 in.

Load (lb) | Elongation (in.) | Load (lb) | Elongation (in.) |

0 | 0 | 14 000 | 0.020 |

2 310 | 0.00220 | 14 400 | 0.025 |

4 640 | 0.00440 | 14 500 | 0.060 |

6 950 | 0.00660 | 14 600 | 0.080 |

9 290 | 0.00880 | 14 800 | 0.100 |

11 600 | 0.0110 | 14 600 | 0.120 |

12 600 | 0.0150 | 13 600 | Fracture |

Plot the stress-strain diagram and determine the following mechanical properties: (a) proportional limit; (b) modulus of elasticity; (c) yield point; (d) yield strength at 0.2% offset; (e) ultimate strength; and (f) rupture strength.

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