The bolt shown in the figure is made of 2014-T6 aluminum alloy (Young’s modulus 10.6×10³ ksi) and is tightened so it compresses a cylindrical tube made of 1004-T61 magnesium alloy (Young’s modulus 6.48×10 ksi). The tube has an outer radius of 1/2 in. and it is assumed that both the inner radius of the tube and the radius of the bolt are 1/4 in. The washers at the top and bottom of the tube are considered to be rigid and have a negligible thickness. Initially the nut is hand-tightened (until snug-fit). Then, by using a wrench, the nut is further tightened 1/2 turn. If the bolt has 20 threads per inch (means 20 turns to advance 1 inch), determine the stress developed in the bolt.
The Correct Answer and Explanation is :
To solve the problem, we need to calculate the stress in the bolt due to the additional tightening of the nut. Here’s the step-by-step solution:
1. Determine the axial deformation caused by the nut tightening:
- Threads per inch: ( 20 \, \text{threads/in} )
- Advancement per turn: ( \frac{1}{20} \, \text{in/turn} )
- For ( \frac{1}{2} ) turn:
[
\Delta_{\text{advance}} = \frac{1}{2} \times \frac{1}{20} = 0.025 \, \text{in}.
]
The nut tightening compresses the magnesium tube and stretches the aluminum bolt by the same amount (( \Delta_{\text{advance}} )) since the washers are rigid.
2. Relate the deformation to stresses:
The deformation is shared between the bolt and the tube, and the total elongation/compression is given by:
[
\Delta_{\text{advance}} = \Delta_{\text{bolt}} + \Delta_{\text{tube}}.
]
Bolt deformation:
[
\Delta_{\text{bolt}} = \frac{\sigma_b L_b}{E_b},
]
where:
- ( \sigma_b ): stress in the bolt,
- ( L_b ): length of the bolt,
- ( E_b ): Young’s modulus of the bolt.
Tube deformation:
[
\Delta_{\text{tube}} = \frac{\sigma_t L_t}{E_t},
]
where:
- ( \sigma_t ): stress in the tube,
- ( L_t ): length of the tube,
- ( E_t ): Young’s modulus of the tube.
3. Relate stresses in the bolt and the tube:
Using the cross-sectional areas:
[
A_b = \pi \left( \frac{1}{4} \right)^2 = \frac{\pi}{16} \, \text{in}^2,
]
[
A_t = \pi \left[ \left( \frac{1}{2} \right)^2 – \left( \frac{1}{4} \right)^2 \right] = \frac{3\pi}{16} \, \text{in}^2.
]
Force equilibrium requires:
[
F_b = F_t, \quad \text{or } \sigma_b A_b = \sigma_t A_t.
]
So,
[
\sigma_t = \frac{\sigma_b A_b}{A_t} = \frac{\sigma_b \cdot \frac{\pi}{16}}{\frac{3\pi}{16}} = \frac{\sigma_b}{3}.
]
4. Solve for bolt stress:
Substitute deformations into ( \Delta_{\text{advance}} ):
[
\Delta_{\text{advance}} = \frac{\sigma_b L_b}{E_b} + \frac{\sigma_t L_t}{E_t}.
]
Replace ( \sigma_t ) with ( \frac{\sigma_b}{3} ):
[
0.025 = \frac{\sigma_b L_b}{E_b} + \frac{\frac{\sigma_b}{3} L_t}{E_t}.
]
Simplify:
[
0.025 = \sigma_b \left( \frac{L_b}{E_b} + \frac{L_t}{3E_t} \right).
]
Assume ( L_b = L_t ) (simplification, as they are not provided):
[
0.025 = \sigma_b \left( \frac{1}{10.6 \times 10^3} + \frac{1}{3 \times 6.48 \times 10^3} \right).
]
Numerical calculation:
[
0.025 = \sigma_b \left( 9.43 \times 10^{-5} + 5.14 \times 10^{-5} \right),
]
[
0.025 = \sigma_b \times 1.457 \times 10^{-4}.
]
Solve for ( \sigma_b ):
[
\sigma_b = \frac{0.025}{1.457 \times 10^{-4}} \approx 171.6 \, \text{ksi}.
]
Final Answer:
The stress developed in the bolt is approximately 171.6 ksi.
Explanation:
The problem combines concepts of axial deformation, stress-strain relationships, and compatibility of deformations. The stress in the bolt and tube is linked through equilibrium and geometry. Using the given material properties and dimensions, the final bolt stress reflects the force required to maintain the deformation imposed by the nut tightening. The aluminum alloy bolt, being stiffer, bears more stress compared to the magnesium alloy tube.