A 0.5-in.-diameter steel [E = 30,000 ksi] bolt (1) is placed in a copper tube (2), as shown in Figure P5.47. The copper [E = 16,000 ksi] tube has an outside diameter of 1.00 in., a wall thickness of 0.125 in., and a length of L = 8.0 in. Rigid washers, each with a thickness of t = 0.125 in., cap the ends of the copper tube. The bolt has 20 threads per inch. This means that each time the nut is turned one complete revolution, the nut advances 0.05 in. (i.e., 1/20 in.). The nut is handtightened on the bolt until the bolt, nut, washers, and tube are just snug, meaning that all slack has been removed from the assembly but no stress has yet been induced. What stresses are produced in the bolt and in the tube if the nut is tightened an additional quarter turn past the snug-tight condition?
Question 5.47: A 0.5-in.-diameter steel [E = 30,000 ksi] bolt (1) is placed...
Equilibrium: The force induced in the bolt by advancing the nut will be balanced by the force created in the tube; therefore:
F_{1}=-F_{2} (a)
Geometry-of-deformation relationship: When the nut is turned on the bolt, the bolt is shortened by that same amount. However, the resistance exerted by the tube causes the bolt to elongate and the tube to shorten. This relationship can be expressed by:
\delta_{1}-\Delta_{\text {nut }}=\delta_{2}In words, this relationship says that the elongation of the bolt plus the shortening of the bolt caused by the nut advance (i.e., a negative deformation) must equal the deformation of the tube. When the nut is turned an additional quarter turn past the snug-tight condition, the nut has advanced
(0.25 \text { turns }) \frac{1 \text { in. }}{20 \text { turns }}=0.0125 \text { in. }Therefore, the geometry-of-deformation equation can be written as
\delta_{1}-0.0125 in .=\delta_{2} (b)
Force-Deformation Relationships:
\delta_{1}=\frac{F_{1} L_{1}}{A_{1} E_{1}} \quad \delta_{2}=\frac{F_{2} L_{2}}{A_{2} E_{2}} (c)
Compatibility Equation:
Substitute Eqs. (c) in Eq. (b) to derive the compatibility equation.
\frac{F_{1} L_{1}}{A_{1} E_{1}}-0.0125 in .=\frac{F_{2} L_{2}}{A_{2} E_{2}} (d)
Solve the equations:
Substitute Eq. (a) into Eq. (b) to obtain:
Numeric values:
\begin{array}{ll}A_{1}=\frac{\pi}{4}(0.5 in .)^{2}=0.196350 in .^{2} & A_{2}=\frac{\pi}{4}\left[(1.0 in .)^{2}-(0.75 in .)^{2}\right]=0.343612 in. ^{2} \\L_{1}=L+2 t=8.25 in . & L_{2}=8.00 in . \\E_{1}=30,000 ksi & E_{2}=16,000 ksi\end{array}Substitute these values into Eq. (e) and solve for F_{1}:
F_{1}=\frac{0.0125 in .}{\frac{L_{1}}{A_{1} E_{1}}+\frac{L_{2}}{A_{2} E_{2}}}=\frac{0.0125 in .}{\frac{8.25 in .}{\left(0.196350 in. ^{2}\right)(30,000 ksi )}+\frac{8.00 in .}{\left(0.343612 in .^{2}\right)(16,000 ksi )}}=4.3772 kipsThe force in the tube is the same magnitude; however, it is compression:
F_{2}=-4.3772 \text { kips }Normal stresses:
The normal stress in the bolt is:
The normal stress in the tube is:
\sigma_{2}=\frac{-4.3772 kips }{0.343612 in .^{2}}=-12.7389 ksi =12.74 ksi ( C )