This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
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Here are the solutions to Question 6: QUESTION 6: STRESS, STRAIN AND YOUNG'S MODULUS 6.1 A tensile force of 25 kN is applied to a rectangular bar 12 mm thick and 22 mm wide. The original length is 1.8 m and Young's modulus for the steel is 195 GPa. Given: Tensile force, F = 25 kN = 25 × 10^3 N Thickness, t = 12 mm = 0.012 m Width, w = 22 mm = 0.022 m Original length, L_0 = 1.8 m Young's modulus, E = 195 GPa = 195 × 10^9 Pa 6.1.1 The stress Step 1: Calculate the cross-sectional area (A). A = w × t A = 0.022 m × 0.012 m = 0.000264 m^2 Step 2: Calculate the stress (). = (F)/(A) = 25 × 10^3 N0.000264 m^2 ≈ 94696969.7 Pa ≈ 94.70 MPa The stress is 94.70 MPa. 6.1.2 The final length of the bar Step 1: Calculate the strain (). E = ()/() = ()/(E) = 94.69697 × 10^6 Pa195 × 10^9 Pa ≈ 0.0004856 Step 2: Calculate the change in length ( L). = ( L)/(L_0) L = L_0 L = 0.0004856 × 1.8 m L ≈ 0.000874 m Step 3: Calculate the final length (L_f). L_f = L_0 + L L_f = 1.8 m + 0.000874 m L_f = 1.800874 m The final length of the bar is 1.800874 m. 6.2 FIGURE 3 below shows a round steel rod subjected to a tensile load of 55 kN. Young's modulus for the steel is 200 GPa. Given: Tensile load, F = 55 kN = 55 × 10^3 N Young's modulus, E = 200 GPa = 200 × 10^9 Pa Section 1: Diameter D_1 = 25 mm = 0.025 m, Length L_1 = 100 mm = 0.1 m Section 2: Diameter D_2 = 50 mm = 0.050 m, Length L_2 = 100 mm = 0.1 m 6.2.1 The maximum tensile stress Step 1: Identify the section with the smallest area (Section 1) and calculate its area (A_1). A_1 = ( D_1^2)/(4) = (0.025 m)^24 A_1 = × 0.000625 m^24 ≈ 0.00049087 m^2 Step 2: Calculate the maximum tensile stress (_max). _max = (F)/(A_1) _max = 55 × 10^3 N0.00049087 m^2 _max ≈ 112040000 Pa ≈ 112.04 MPa The maximum tensile stress is 112.04 MPa. 6.2.2 The minimum stress in the rod Step 1: Identify the section with the largest area (Section 2) and calculate its area (A_2). A_2 = ( D_2^2)/(4) = (0.050 m)^24 A_2 = × 0.0025 m^24 ≈ 0.0019635 m^2 Step 2: Calculate the minimum stress (_min). _min = (F)/(A_2) _min = 55 × 10^3 N0.0019635 m^2 _min ≈ 28011713 Pa ≈ 28.01 MPa The minimum stress in the rod is 28.01 MPa. 6.2.3 The total strain The question asks for "total strain". For a composite rod, this typically refers to the total elongation divided by the total original length. Step 1: Calculate the strain and elongation for Section 1. _1 = _maxE = 112.04 × 10^6 Pa200 × 10^9 Pa ≈ 0.0005602 L_1 = _1 L_1 = 0.0005602 × 0.1 m ≈ 0.00005602 m Step 2: Calculate the strain and elongation for Section 2. _2 = _minE = 28.011713 × 10^6 Pa200 × 10^9 Pa ≈ 0.00014006 L_2 = _2 L_2 = 0.00014006 × 0.1 m ≈ 0.000014006 m Step 3: Calculate the total elongation ( L_total). L_total = L_1 + L_2 L_total = 0.00005602 m + 0.000014006 m = 0.000070026 m Step 4: Calculate the total original length (L_total). L_total = L_1 + L_2 = 0.1 m + 0.1 m = 0.2 m Step 5: Calculate the overall "total strain". Total Strain = L_totalL_total Total Strain = 0.000070026 m0.2 m ≈ 0.00035013 The total strain is 0.00035013. Drop the next question.