Question 1.7: A machine component slides along a horizontal bar at A and m...
A machine component slides along a horizontal bar at A and moves in a vertical slot B. The component is represented as a rigid bar AB (length L = 5 ft, weight W = 985 lb) with roller supports at A and B (neglect friction). When not in use, the machine component is supported by a single wire (diameter d = 1/8 in.) with one end attached at A and the other end supported at C (see Fig.1-43). The wire is made of a copper alloy; the stress-strain relationship for the wire is
\sigma(\varepsilon)=\frac{17,500 \varepsilon}{1+240 \varepsilon} 0 \leq \varepsilon \leq 0.03 \quad(\sigma \text { in ksi })
(a) Plot a stress-strain diagram for the material; What is the modulus of elasticity E (ksi)? What is the 0.2% offset yield stress (ksi)?
(b) Find the tensile force T (lb) in the wire.
(c) Find the normal axial strain \varepsilon and elongation \delta \text { (in.) } of the wire.
(d) Find the permanent set of the wire if all forces are removed.
Use a four-step problem-solving approach to find the modulus of elasticity, yield stress, tensile force, normal strain and elongation, and the permanent set of copper alloy wire AC.
1. Conceptualize: The copper alloy has considerable ductility but will have a stress-strain curve without a well-defined yield point. Define the yield point using an offset method as illustrated in Fig.1-35. Find the residual strain and then the permanent set of the wire, as shown in Fig.1-39.
2. Categorize: The given analytical expression for the stress-strain curve \sigma(\varepsilon) is based on measured laboratory data for the copper alloy used to manufacture this wire. Hence, the analytical expression is an approximation of the actual behavior of this material and was formulated based on test data. Analytical representations of actual stress-strain curves are often used in computer programs to model and analyze structures of different materials under applied loads of various kinds.
3. Analyze : Part (a): Plot a stress-strain diagram for the material; What is the modulus of elasticity E (ksi)? What is the 0.2% offset yield stress (ksi)?
Plot the function \sigma(\varepsilon) for strain values between 0 and 0.03 (Fig.1-44). The stress at strain \varepsilon=0.03 is 64 ksi.
\sigma(\varepsilon)=\frac{17,500 \varepsilon}{1+240 \varepsilon} \quad \varepsilon=0,0.001, \ldots, 0.03\sigma(0)=0 \quad \sigma(0.03)=64 ksi
The slope of the tangent to the stress-strain curve at strain \varepsilon=0 is the modulus of elasticity E (see Fig.1-45). Take the derivative of \sigma(\varepsilon) to get the slope of the tangent to the \sigma(\varepsilon) curve, and evaluate the derivative at strain \varepsilon=0 to find E:
E(\varepsilon)=\frac{d}{d \varepsilon} \sigma(\varepsilon)=\frac{17,500}{(240 \varepsilon+1)^{2}}E=E(0) E=17,500 ksi
Next, find an expression for the yield strain \varepsilon_{y}, the point at which the 0.2% offset line crosses the stress-strain curve (see Fig.1-45). Substitute the expression \varepsilon_{y} into the s \sigma(\varepsilon) expression and then solve for yield \sigma\left(\varepsilon_{y}\right)=\sigma_{y} :
\varepsilon_{y}=0.002+\frac{\sigma_{y}}{E} \text { and } \sigma\left(\varepsilon_{y}\right)=\sigma_{y} \quad \text { or } \quad \sigma_{y}=\frac{17,500 \varepsilon_{y}}{1+240 \varepsilon_{y}}Rearranging the equation in terms of \sigma_{y} gives
\sigma_{y}^{2}+\left(\frac{37 E}{6000}-\frac{875}{12}\right) \sigma_{y}-\frac{7 E}{48}=0Solving this quadratic equation for the 0.2% offset yield, stress \sigma_{y} gives \sigma_{y} = 36 ksi .
The yield strain is computed as
Part (b): Find the tensile force T (lb) in the wire. Recall that bar weight W = 985 lb.
Find the angle between the x-axis and cable attachment position at C:
\alpha_{C}=\arctan \left(\frac{1.5}{4}\right)=20.556^{\circ}Sum the moments about A to obtain one equation and one unknown. The reaction B_{x} acts to the left:
B_{x}=\frac{-W(2 ft )}{3 ft }=-656.667 lbNext, sum the forces in the x direction to find the cable force T_{C}:
T_{C}=\frac{-B_{x}}{\cos \left(\alpha_{C}\right)} \quad T_{C}=701 lbPart (c): Find the normal axial strain ε and elongation δ (in.) of the wire.
Compute the normal stress then find the associated strain from stress-strain plot (or from the \sigma(\varepsilon) equation). The wire elongation is strain times wire length.
The wire diameter, cross-sectional area, and length are
L_{C}=\sqrt{(4 ft )^{2}+(1.5 ft )^{2}}=4.272 ft
Now compute the stress and strain in the wire and the elongation of the wire as
\sigma_{C}=\frac{T_{C}}{A}=57.1 ksiNote that the stress in the wire exceeds the 0.2% offset yield stress of 36 ksi.
The corresponding normal strain is found from the \sigma(\varepsilon) plot or by rearranging the \sigma(\varepsilon) equation to give
Then,
\varepsilon\left(\sigma_{C}\right)=\varepsilon_{C}, \quad \text { or } \quad \varepsilon_{C}=\frac{\sigma_{C}}{17,500 ksi -240 \sigma_{ C }}=0.015Finally, the wire elongation is
\delta_{C}=\varepsilon_{C} L_{C}=0.774 inPart (d): Find the permanent set of the wire if all forces are removed.
If the load is removed from the wire, the stress in the wire will return to zero following the unloading line in Fig.1-46 (see also Fig.1-39b). The elastic recovery strain is
\varepsilon_{ er }=\frac{\sigma_{C}}{E}=3.266 \times 10^{-3}The residual strain is the difference between the total strain \left(\varepsilon_{C}\right) and the elastic recovery strain \left(\varepsilon_{er}\right)
\varepsilon_{ res }=\varepsilon_{C}-\varepsilon_{ er }=0.012Finally, the permanent set of the wire is the product of the residual strain and the length of the wire:
\varepsilon_{ res } L_{C}=0.607 in .4. Finalize: This example presents an analytical model of the stress-strain relationship for a copper alloy. The computed values of modulus of elasticity E and yield stress \sigma_{y} are consistent with values listed in Appendix I. The tensile force, normal strain and elongation, and permanent set are computed for the wire when stressed beyond the apparent yield point of the material.
