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 at B. The component is represented as a rigid bar AB (length L = 1.5 m, weight W = 4.5 kN) with roller supports at A and B (neglect friction). When not in use, the machine component is supported by a single wire (diameter d = 3.5 mm) with one end attached at A and the other end supported at C (see Fig. 1-40). The wire is made of a copper alloy; the stress-strain relationship for the wire is
\sigma(\varepsilon)=\frac{124,000 \varepsilon}{1+240 \varepsilon} \quad 0 \leq \varepsilon \leq 0.03 \quad(\sigma \text { in MPa })(a) Plot a stress-strain diagram for the material; what is the modulus of elasticity E (GPa)? What is the 0.2% offset yield stress (MPa)?
(b) Find the tensile force T (kN) in the wire.
(c) Find the normal axial strain ε and elongation δ (mm) of the wire.
(d) Find the permanent set of the wire if all forces are removed .
(a) Plot a stress-strain diagram for the material. What is the modulus of elasticity? What is the 0.2% offset yield stress (MPa)?
Plot the function σ(ε) for strain values between 0 and 0.03 (Fig. 1-41).
The stress at strain ε = 0.03 is 454 MPa .
\begin{gathered}\sigma(\varepsilon)=\frac{124,000 \varepsilon}{1+240 \varepsilon} \quad \varepsilon=0,0.001, \ldots, 0.03 \\\sigma(0)=0 \quad \sigma(0.03)=454 {MPa}\end{gathered}The slope of the tangent to the stress-strain curve at strain ε = 0 is the modulus of elasticity E (see Fig. 1-42). Take the derivative of σ(ε) to get the slope of the tangent to the σ(ε) curve, and evaluate the derivative at strain ε = 0 to find E.
\begin{gathered}E(\varepsilon)=\frac{d}{d \varepsilon} \sigma(\varepsilon) \rightarrow \frac{124,000}{(240 \varepsilon+1)^{2}} \\E=E(0) \quad E=124,000 {MPa}=124 {GPa}\end{gathered}Next, find an expression for the yield strain εy, the point at which the 0.2% offset line crosses the stress-strain curve (see Fig. 1-42). Substitute the expression for \varepsilon_{y} into the σ(ε) expression and then solve for yield stress σ(\varepsilon_{y}) = \sigma_{y} :
\varepsilon_{y}=0.002+\frac{\sigma_{y}}{E} \text { and } \sigma\left(\varepsilon_{y}\right)=\sigma_{y} \text { or } \sigma_{y}=\frac{124,000 \varepsilon_{y}}{1+240 \varepsilon_{y}}Rearranging the equation in terms of \sigma_{y} gives
\sigma_{y}^{2}+\left(\frac{E}{500}\right) \sigma_{y}-\frac{E^{2}}{120000}=0Solving this quadratic equation for the 0.2% offset yield stress \sigma_{y} gives \sigma_{y} = 255 MPa.
The yield strain can be computed as
\varepsilon_{y}=0.002+\frac{\sigma_{y}}{E({GPa})}=4.056 \times 10^{-3}(b) Use statics to find the tensile force T (kN) in the wire; recall that bar weight W = 4.5 kN.
Find the angle between the x axis and cable attachment position at C:
Sum the moments about A to obtain one equation and one unknown.
The reaction B_{x} acts to the left:
Next, sum the forces in the x direction to find the cable force T_{C}:
T_{c}=\frac{-B_{x}}{\cos \left(\alpha_{c}\right)} T_{c}=3.2 {kN}(c) Find the normal axial strain ε and elongation δ (mm) of the wire.
Compute the normal stress then find the associated strain from stress-strain plot (or from the σ (ε) equation). The wire elongation is strain times wire length.
The wire diameter, cross-sectional area, and length are
We can now compute the stress and strain in the wire and the elonga-tion of the wire.
\sigma_{C}=\frac{T_{C}}{A}=333 {MPa}Note that the stress in the wire exceeds the 0.2% offset yield stress of 255 MPa. The corresponding normal strain can be found from the σ(ε) plot or by rearranging the σ(ε) equation to give
\varepsilon(\sigma)=\frac{\sigma}{124000-240 \sigma}Then , \varepsilon\left(\sigma_{C}\right)=\varepsilon_{C} \quad \text { or } \quad \varepsilon_{C}=\frac{\sigma_{C}}{124 {GPa}-240 \sigma_{C}}=7.556 \times 10^{-3}
Finally, the wire elongation is
\delta_{c}=\varepsilon_{c} L_{c}=9.68 {mm}(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 unloading line BC in Fig. 1-43 (see also Fig. 1-36b). The elastic recovery strain can be computed as
Hence, the residual strain is the difference between the total strain \left(\varepsilon_{C}\right) and the elastic recovery strain \left(\varepsilon_{er}\right) as
\varepsilon_{\text {res }}=\varepsilon_{C}-\varepsilon_{\text {er }}=7.166 \times 10^{-3}Finally, the permanent set of the wire is the product of the residual strain and the length of the wire:
P_{\text {set }}=\varepsilon_{\text {rec }} L_{r}=9.184 {mm}
