# 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.

# 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.

# 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.

# 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.

# Solution to Problem 203 Stress-strain Diagram

**Problem 203**

The following data were recorded during the tensile test of a 14-mm-diameter mild steel rod. The gage length was 50 mm.

Load (N) | Elongation (mm) | Load (N) | Elongation (mm) |

0 | 0 | 46 200 | 1.25 |

6 310 | 0.010 | 52 400 | 2.50 |

12 600 | 0.020 | 58 500 | 4.50 |

18 800 | 0.030 | 68 000 | 7.50 |

25 100 | 0.040 | 59 000 | 12.5 |

31 300 | 0.050 | 67 800 | 15.5 |

37 900 | 0.060 | 65 000 | 20.0 |

40 100 | 0.163 | 65 500 | Fracture |

41 600 | 0.433 |

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

# Solution to Problem 142 Pressure Vessel

**Problem 142**

A pipe carrying steam at 3.5 MPa has an outside diameter of 450 mm and a wall thickness of 10 mm. A gasket is inserted between the flange at one end of the pipe and a flat plate used to cap the end. How many 40-mm-diameter bolts must be used to hold the cap on if the allowable stress in the bolts is 80 MPa, of which 55 MPa is the initial stress? What circumferential stress is developed in the pipe? Why is it necessary to tighten the bolt initially, and what will happen if the steam pressure should cause the stress in the bolts to be twice the value of the initial stress?

# Solution to Problem 141 Pressure Vessel

**Problem 141**

The tank shown in Fig. P-141 is fabricated from 1/8-in steel plate. Calculate the maximum longitudinal and circumferential stress caused by an internal pressure of 125 psi.

# Solution to Problem 140 Pressure Vessel

**Problem 140**

At what angular velocity will the stress of the rotating steel ring equal 150 MPa if its mean radius is 220 mm? The density of steel 7.85 Mg/m^{3}.

**Solution 140**

$CF = M \omega^2 \bar x$

$M = \rho V = \rho A \pi R$

$x = 2R / \pi$

Thus,

$CF = \rho A \pi R \omega^2 (2R / \pi)$