Let's break down and answer each part step by step. --- ### (c) **New Arrangement Illustration and Labelling** #### **Given/Changes:** - Two extra strain gauges (Gauge Factor = 2.1) are added in the middle of the cantilever. - The force (0.6 N) direction is now upward (opposite). - Thickness and width of the cantilever are increased by 10% from their original values. - **Task:** Illustrate and label the new arrangement of the strain gauges on the cantilever (show positions and new dimensions). #### **Step 1: Original Arrangement (Assumed)** Assume the original cantilever had 4 strain gauges: - 2 at the fixed end (top and bottom) - 2 at the free end (top and bottom) #### **Step 2: New Arrangement** - Add 2 extra strain gauges **in the middle** (one on top surface, one on bottom surface). #### **Step 3: New Dimensions** If original width = \( w_0 \) and thickness = \( t_0 \), then: - New width = \( w = w_0 \times 1.10 \) - New thickness = \( t = t_0 \times 1.10 \) #### **Step 4: Illustration (Text Description)** ``` |---------------------------------------------| | | | [SG1] [SG3] [SG5] Top surface | | | [SG2] [SG4] [SG6] Bottom surface | | |---------------------------------------------| ^Fixed end Middle Free end ``` - **SG1, SG2:** Near fixed end (top/bottom) - **SG3, SG4:** Middle (top/bottom) -- *newly added* - **SG5, SG6:** Near free end (top/bottom) **Label dimensions**: - Length = \( L \) - Width = \( w = 1.1w_0 \) - Thickness = \( t = 1.1t_0 \) --- ### (d) **Find Strains and Resistances** #### **Step 1: Strain at Each Gauge Location** The strain \( \epsilon \) in a cantilever beam under load \( F \) at a distance \( x \) from the fixed end (for a rectangular cross-section) is: \[ \epsilon(x) = \frac{6F(L-x)}{Ewt^2} \] where: - \( F \) = applied force (0.6 N) - \( L \) = length of cantilever - \( x \) = distance from fixed end to gauge position - \( E \) = Young's modulus (needs to be given or assumed) - \( w \) = width (new, \( 1.1w_0 \)) - \( t \) = thickness (new, \( 1.1t_0 \)) #### **Step 2: Positions** Assume: - SG1, SG2: \( x_1 = 0 \) (fixed end) - SG3, SG4: \( x_2 = L/2 \) (middle) - SG5, SG6: \( x_3 = L \) (free end) Now, calculate strains: - At fixed end: \( \epsilon_1 = \frac{6F(L-0)}{Ewt^2} = \frac{6FL}{Ewt^2} \) - At middle: \( \epsilon_2 = \frac{6F(L-L/2)}{Ewt^2} = \frac{6F(L/2)}{Ewt^2} = \frac{3FL}{Ewt^2} \) - At free end: \( \epsilon_3 = 0 \) (since \( L-x = 0 \)) #### **Step 3: Gauge Factor and Resistance Change** Gauge factor \( GF \) relates strain to relative change in resistance: \[ \frac{\Delta R}{R} = GF \cdot \epsilon \] - **Assume original resistance \( R_0 \) for all gauges.** #### **Step 4: Calculate Resistance for Each Gauge** \[ R = R_0 + \Delta R = R_0 (1 + GF \cdot \epsilon) \] - **At fixed end (SG1, SG2):** \[ \epsilon_1 = \frac{6FL}{Ewt^2} \] \[ R_{SG1,2} = R_0 (1 + 2.1 \cdot \epsilon_1) \] - **At middle (SG3, SG4):** \[ \epsilon_2 = \frac{3FL}{Ewt^2} \] \[ R_{SG3,4} = R_0 (1 + 2.1 \cdot \epsilon_2) \] - **At free end (SG5, SG6):** \[ \epsilon_3 = 0 \] \[ R_{SG5,6} = R_0 \] --- ### **Final Answer:** #### (c) **Arrangement Illustration** - Six strain gauges: 2 at fixed end (top/bottom), 2 at middle (top/bottom, newly added), 2 at free end (top/bottom). - Dimensions: width and thickness increased by 10% (\( w = 1.1w_0, t = 1.1t_0 \)). - SG positions: \( x_1 = 0 \), \( x_2 = L/2 \), \( x_3 = L \). #### (d) **Strains & Resistances** - **Strain at SG1/SG2:** \( \epsilon_1 = \frac{6FL}{Ewt^2} \) - **Strain at SG3/SG4:** \( \epsilon_2 = \frac{3FL}{Ewt^2} \) - **Strain at SG5/SG6:** \( \epsilon_3 = 0 \) - **Resistance:** - \( R_{SG1,2} = R_0 (1 + 2.1 \cdot \epsilon_1) \) - \( R_{SG3,4} = R_0 (1 + 2.1 \cdot \epsilon_2) \) - \( R_{SG5,6} = R_0 \) > **To get numerical answers, you need values for \( R_0, w_0, t_0, L, E \). Plug in the numbers as needed.** --- If you provide original dimensions or material properties, I can compute the actual values!