Question 3.14: A 14-mm-diameter steel bolt, and a steel pipe with 19-mm ID ......
A 14-mm-diameter steel bolt, and a steel pipe with 19-mm ID and 25-mm OD are arranged as shown in Fig. 1. For both the steel bolt and steel pipe, E = 200 GPa. The pitch of the (single) thread is 2 mm. What stresses will be produced in the steel bolt and sleeve if the nut is tightened by \frac{1}{8} turn? (For a single thread the pitch is the distance the nut advances along the thread in one complete revolution of the nut.) Neglect the thickness of the washers.
Plan the Solution As we discussed earlier in this section, the misfit due to tightening of the nut will enter the solution through the deformation-geometry equations. Tightening the nut should put the bolt in tension and the sleeve in compression. We will use the Basic Force Method to solve this problem.
Equilibrium: We should ask the question: What free-body diagram can we draw that will enable us to relate the internal element forces to each other and to the external forces? (In this problem there are no external forces.) One answer is that a cut made just inside the washer at either end will expose the internal forces so that they can be included in the equilibrium equation. A free-body diagram of the left end is shown in Fig. 2. As always, for each of the elements we take tension to be positive.
Then, we sum forces in the axial direction.
\underrightarrow{+}\sum{F_x} = 0: F_1 + F_2 =0 Equilibrium (1)
Element Force-Deformation Behavior: Since the Basic Force Method will be used, Eq. 3.14 is the most convenient form.
e = fF, \text{where} f ≡ \frac{L}{ AE} (3.14)
e_1 = f_1F_1 Element Force-Deformation Behavior (2)
e_2 = f_2F_2where, as always, e is the \underline{\text{total elongation}} of an element and is positive when the element gets longer.
f_1 = \left(\frac{ L} {AE}\right)_1 = \frac{250 mm}{\pi(7 mm)^2(200 kN/mm^2)} = 8.12(10^{-3}) mm/kN
f_2 = \left(\frac{ L} {AE}\right)_2 = \frac{250 mm}{\pi[(12.5 mm)^2 – (9.5 mm)^2](200 kN/mm^2)}= 6.03(10^{-3}) mm/kN
Deformation Geometry: We need to assess the deformation that results when the nut is tightened. To do this, let us suppose that the head of the bolt does not move, while the right-hand washer and nut move to the left an amount \bar{δ} when the nut is tightened. Therefore, we need to relate the elongations e_1 and e_2 in Eqs. (2) to the displacement \bar{δ}.
Figure 3 is a deformation diagram showing the shortening of the sleeve by an amount \bar{δ}.
The sleeve is shortened by the amount \bar{δ} that the right-hand washer moves to the left, so
If the sleeve were to be removed and the nut advanced \frac{1}{8} turn, the working length of the bolt would be [250 mm – (\frac{1}{8})(2 mm)] = 250 mm – 0.25 mm = 249.75 mm. Thus, there is a “misfit” \bar{δ} = 0.25 mm, and we would have to stretch the bolt by 0.25 mm to restore it to the 250-mm length, from which the displacement \bar{δ} is measured. Therefore, the element elongations are related to the displacement δ by
e_1 = 0.25 mm – \bar{δ} Deformation Geometry (3)
e_2 = – \bar{δ}Solution of the Equations by the Basic Force Method: We can eliminate the displacement from Eqs. (3) to get the compatibility equation
e_1 – e_2 = 0.25 mm
Substituting the element force-deformation equations into this equation gives the following compatibility equation in terms of element forces:
f_1F_1 – f_2F_2 = 0.25 mm Compatibility in Terms of Element Forces (4)
Solving this compatibility equation and the equilibrium equation simultaneously, we get
F_1 = 17.7 kN (5)
F_2 =-17.7 kN
Finally, the element stresses are given by
\left\{\begin{matrix} σ_{\text{bolt}} ≡ σ_1 =\frac{F_1}{A_1}=\frac{17.7 kN}{\pi(7.0 mm)^2} = 114.8 MPa \\ σ_{\text{sleeve}} ≡ σ_2 =\frac{F_2}{A_2} = \frac{-17.7 kN}{\pi[(12.5 mm)^2 – (9.5 mm)^2 ]}\\= -85.2 MPa \end{matrix} \right\}Thus, the bolt has a tensile stress of 115 MPa, and the sleeve has a compressive
stress of 85 MPa as a result of tightening the nut by \frac{1}{8} turn.
Review the Solution We expected the bolt to be in tension and the sleeve to be in compression, and this is the result that we obtained. The yield strength of the steel is not stated in the problem, but assuming that it is 250 MPa or greater (see Table F.3 in Appendix F), the values of σ_{\text{bolt}} and σ_{\text{sleeve}} are reasonable.
