Let's solve part (c) step-by-step: --- ### **(c) Determine the spring constants of the SMA spring at high and low temperatures** --- #### **Given Data:** | Parameter | Value | |---|---| | Load, P | 6 N | | Shear stress in SMA at high temp, \( \tau_{hot} \) | 100 MPa (from problem statement) | | Shear stress in SMA at low temp, \( \tau_{cold} \) | 15 MPa (from problem statement) | | Shear modulus of SMA, \( G \) | 27 GPa = \( 27 \times 10^3 \) MPa | | Shear modulus, \( G \) | 27,000 MPa | --- ### **Step 1: Relate shear stress to shear strain** The shear modulus \(G\) relates shear stress \( \tau \) and shear strain \( \gamma \): \[ \tau = G \times \gamma \] --- ### **Step 2: Find shear strain at high and low temperatures** Using the relation: \[ \gamma = \frac{\tau}{G} \] - **At high temperature:** \[ \gamma_{hot} = \frac{100\, \text{MPa}}{27,000\, \text{MPa}} = \frac{100}{27,000} \approx 3.70 \times 10^{-3} \] - **At low temperature:** \[ \gamma_{cold} = \frac{15\, \text{MPa}}{27,000\, \text{MPa}} = \frac{15}{27,000} \approx 5.56 \times 10^{-4} \] --- ### **Step 3: Find shear strain in terms of shear displacement** Shear strain \( \gamma \) is also related to shear displacement \( \delta \) and shear plane length \( l_s \): \[ \gamma = \frac{\delta}{l_s} \] where \( l_s \) is the shear plane length (we'll consider the shear plane length as the original shear length, which is the length of the SMA spring, \(l_s\)). --- ### **Step 4: Find shear displacement \( \delta \) due to the load** - The shear displacement \( \delta \) relates to the shear strain: \[ \delta_{hot} = \gamma_{hot} \times l_s \] \[ \delta_{cold} = \gamma_{cold} \times l_s \] --- ### **Step 5: Relate shear force and shear displacement to find spring constant** - The shear force \( F \) related to shear displacement: \[ F = k_s \times \delta \] - Where \( k_s \) is the shear spring constant. - Given the shear force: \[ F = P = 6\, \text{N} \] - Therefore: \[ k_s = \frac{F}{\delta} \] - Since \( \delta = \gamma \times l_s \), then: \[ k_s = \frac{F}{\gamma \times l_s} \] --- ### **Step 6: Final calculation of shear spring constants** - **At high temperature:** \[ k_{s,hot} = \frac{6}{\gamma_{hot} \times l_s} \] \[ k_{s,hot} = \frac{6}{(3.70 \times 10^{-3}) \times l_s} \] - **At low temperature:** \[ k_{s,cold} = \frac{6}{(5.56 \times 10^{-4}) \times l_s} \] --- ### **Note:** To find the numerical values of the spring constants, **you need the shear plane length \( l_s \)** (which should be given or assumed). If \( l_s \) is not provided, the answer is expressed in terms of \( l_s \). --- ### **Final expressions:** \[ \boxed{ \begin{aligned} k_{s,hot} &= \frac{6}{(3.70 \times 10^{-3}) \times l_s} \\ k_{s,cold} &= \frac{6}{(5.56 \times 10^{-4}) \times l_s} \end{aligned} } \] This completes the calculation of the shear spring constants at high and low temperatures. --- **Let me know if you have the shear plane length \( l_s \), and I can help compute the exact numerical values!**