A copper rod that is 5,2 m long is subjected to a tensile load of 850 kg. The rod elongates by 1,25 mm. 7.1.1 Calculate the tensile stress on the rod. 7.1.2 Calculate the strain on the rod.

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Asked on August 2, 2026Physics

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.

Accepted Answer · 2026-08-02T07:03:34.091Z

Here are the solutions to the questions: QUESTION 7: STRESS, STRAIN AND YOUNG'S MODULUS 7.1.1 Calculate the tensile stress on the rod. Step 1: Calculate the tensile force F. The mass m = 850 kg. The acceleration due to gravity g = 9.8 m/s^2. F = mg = 850 kg × 9.8 m/s^2 = 8330 N Step 2: State the formula for tensile stress. Tensile stress = (F)/(A), where F is the tensile force and A is the cross-sectional area of the rod. Note: The cross-sectional area of the copper rod is not provided in the question. Therefore, a numerical value for the tensile stress cannot be calculated. The tensile force is 8330 N. 7.1.2 Calculate the strain on the rod. Step 1: Identify the original length and the elongation. Original length L = 5.2 m. Elongation L = 1.25 mm = 0.00125 m. Step 2: Calculate the strain using the formula = ( L)/(L). = 0.00125 m5.2 m ≈ 0.00024038 The strain on the rod is 0.000240. 7.2.1 Draw a stress strain graph for these values. Step 1: Calculate the original cross-sectional area A_0. Original diameter D_0 = 16 mm = 0.016 m. A_0 = ( D_0^2)/(4) = (0.016 m)^24 = × 0.000256 m^24 = 0.000064 m^2 ≈ 2.0106 × 10^-4 m^2 Step 2: Calculate the stress ( = LoadA_0) and strain ( = ExtensionL_0) for each data point. Gauge length L_0 = 80 mm = 0.080 m. | Load (kN) | Extension (mm) | Stress ( = Load/A_0) (MPa) | Strain ( = Extension/L_0) | |:---------:|:--------------:|:---------------------------------------:|:---------------------------------------:| | 0 | 0 | 0 | 0 | | 15 | 0.0182 | (15 × 10^3)/(2.0106 × 10^-4) ≈ 74.60 | (0.0182)/(80) = 0.0002275 | | 30 | 0.0363 | (30 × 10^3)/(2.0106 × 10^-4) ≈ 149.21 | (0.0363)/(80) = 0.00045375 | | 45 | 0.0547 | (45 × 10^3)/(2.0106 × 10^-4) ≈ 223.91 | (0.0547)/(80) = 0.00068375 | | 60 | 0.0724 | (60 × 10^3)/(2.0106 × 10^-4) ≈ 298.42 | (0.0724)/(80) = 0.000905 | | 75 | 0.0910 | (75 × 10^3)/(2.0106 × 10^-4) ≈ 373.02 | (0.0910)/(80) = 0.0011375 | Step 3: Describe how to draw the graph. Plot the calculated stress values on the y-axis and the strain values on the x-axis. Use the given scale: 10 mm = 10 MPa for the stress axis and 10 mm = 50 × 10^-6 for the strain axis. The points should form a roughly linear relationship, representing the elastic region of the material. (As an AI, I cannot physically draw the graph, but the table above provides the necessary data points for plotting.) 7.2.2 Determine Young's modulus of elasticity with the aid of the graph. Step 1: Young's modulus E is the slope of the linear portion of the stress-strain graph. We can calculate it using any point in the elastic region (e.g., the last data point before yielding, assuming the given data is within the elastic range). Using the last data point: = 373.02 MPa and = 0.0011375. E = ()/() = 373.02 × 10^6 Pa0.0011375 E ≈ 3.279 × 10^11 Pa = 327.9 GPa Young's modulus of elasticity is 327.9 GPa. 7.2.3 Determine the percentage reduction in area. Step 1: Calculate the original area A_0 and the final area A_f. Original diameter D_0 = 16 mm. Neck diameter at fracture D_f = 11.52 mm. A_0 = ( D_0^2)/(4) = (16 mm)^24 = 64 mm^2 A_f = ( D_f^2)/(4) = (11.52 mm)^24 = × 132.7104 mm^24 = 33.1776 mm^2 Step 2: Calculate the percentage reduction in area. Percentage reduction in area = (A_0 - A_f)/(A_0) × 100\% Percentage reduction in area = 64 mm^2 - 33.1776 mm^264 mm^2 × 100\% Percentage reduction in area = (64 - 33.1776)/(64) × 100\% Percentage reduction in area = (30.8224)/(64) × 100\% = 0.4816 × 100\% = 48.16\% The percentage reduction in area is 48.16\%. 7.3.1 Calculate the normal stress in the tube. Step 1: Calculate the cross-sectional area A of the hollow tube. Outside diameter D_o = 50 mm = 0.050 m. Inside diameter D_i = 30 mm = 0.030 m. A = ()/(4)(D_o^2 - D_i^2) = ()/(4)((0.050 m)^2 - (0.030 m)^2) A = ()/(4)(0.0025 m^2 - 0.0009 m^2) = ()/(4)(0.0016 m^2) = 0.0004 m^2 ≈ 1.2566 × 10^-3 m^2 Step 2: Calculate the normal stress using the formula = (F)/(A). Axial tensile load F = 60 kN = 60 × 10^3 N. = 60 × 10^3 N0.0004 m^2 ≈ 47746482.9 Pa ≈ 47.75 MPa The normal stress in the tube is 47.75 MPa. 7.3.2 If the extension of the tube is 0,76 mm, calculate the strain. Step 1: Identify the original length and the extension. Original length L = 3.8 m. Extension L = 0.76 mm = 0.00076 m. Step 2: Calculate the strain using the formula = ( L)/(L). = 0.00076 m3.8 m = 0.0002 The strain is 0.0002. 7.3.3 Determine Young's modulus of the material. Step 1: Use the calculated stress from 7.3.1 and strain from 7.3.2. Stress = 47.75 MPa = 47.75 × 10^6 Pa. Strain = 0.0002. Step 2: Calculate Young's modulus E using the formula E = ()/(). E = 47.75 × 10^6 Pa0.0002 E = 238.75 × 10^9 Pa = 238.75 GPa Young's modulus of the material is 238.75 GPa. Drop the next question.

Accepted Answer · 2026-08-02T07:03:34.091Z

Here are the solutions to the questions: QUESTION 7: STRESS, STRAIN AND YOUNG'S MODULUS 7.1.1 Calculate the tensile stress on the rod. Step 1: Calculate the tensile force F. The mass m = 850 kg. The acceleration due to gravity g = 9.8 m/s^2. F = mg = 850 kg × 9.8 m/s^2 = 8330 N Step 2: State the formula for tensile stress. Tensile stress = (F)/(A), where F is the tensile force and A is the cross-sectional area of the rod. Note: The cross-sectional area of the copper rod is not provided in the question. Therefore, a numerical value for the tensile stress cannot be calculated. The tensile force is 8330 N. 7.1.2 Calculate the strain on the rod. Step 1: Identify the original length and the elongation. Original length L = 5.2 m. Elongation L = 1.25 mm = 0.00125 m. Step 2: Calculate the strain using the formula = ( L)/(L). = 0.00125 m5.2 m ≈ 0.00024038 The strain on the rod is 0.000240. 7.2.1 Draw a stress strain graph for these values. Step 1: Calculate the original cross-sectional area A_0. Original diameter D_0 = 16 mm = 0.016 m. A_0 = ( D_0^2)/(4) = (0.016 m)^24 = × 0.000256 m^24 = 0.000064 m^2 ≈ 2.0106 × 10^-4 m^2 Step 2: Calculate the stress ( = LoadA_0) and strain ( = ExtensionL_0) for each data point. Gauge length L_0 = 80 mm = 0.080 m. | Load (kN) | Extension (mm) | Stress ( = Load/A_0) (MPa) | Strain ( = Extension/L_0) | |:---------:|:--------------:|:---------------------------------------:|:---------------------------------------:| | 0 | 0 | 0 | 0 | | 15 | 0.0182 | (15 × 10^3)/(2.0106 × 10^-4) ≈ 74.60 | (0.0182)/(80) = 0.0002275 | | 30 | 0.0363 | (30 × 10^3)/(2.0106 × 10^-4) ≈ 149.21 | (0.0363)/(80) = 0.00045375 | | 45 | 0.0547 | (45 × 10^3)/(2.0106 × 10^-4) ≈ 223.91 | (0.0547)/(80) = 0.00068375 | | 60 | 0.0724 | (60 × 10^3)/(2.0106 × 10^-4) ≈ 298.42 | (0.0724)/(80) = 0.000905 | | 75 | 0.0910 | (75 × 10^3)/(2.0106 × 10^-4) ≈ 373.02 | (0.0910)/(80) = 0.0011375 | Step 3: Describe how to draw the graph. Plot the calculated stress values on the y-axis and the strain values on the x-axis. Use the given scale: 10 mm = 10 MPa for the stress axis and 10 mm = 50 × 10^-6 for the strain axis. The points should form a roughly linear relationship, representing the elastic region of the material. (As an AI, I cannot physically draw the graph, but the table above provides the necessary data points for plotting.) 7.2.2 Determine Young's modulus of elasticity with the aid of the graph. Step 1: Young's modulus E is the slope of the linear portion of the stress-strain graph. We can calculate it using any point in the elastic region (e.g., the last data point before yielding, assuming the given data is within the elastic range). Using the last data point: = 373.02 MPa and = 0.0011375. E = ()/() = 373.02 × 10^6 Pa0.0011375 E ≈ 3.279 × 10^11 Pa = 327.9 GPa Young's modulus of elasticity is 327.9 GPa. 7.2.3 Determine the percentage reduction in area. Step 1: Calculate the original area A_0 and the final area A_f. Original diameter D_0 = 16 mm. Neck diameter at fracture D_f = 11.52 mm. A_0 = ( D_0^2)/(4) = (16 mm)^24 = 64 mm^2 A_f = ( D_f^2)/(4) = (11.52 mm)^24 = × 132.7104 mm^24 = 33.1776 mm^2 Step 2: Calculate the percentage reduction in area. Percentage reduction in area = (A_0 - A_f)/(A_0) × 100\% Percentage reduction in area = 64 mm^2 - 33.1776 mm^264 mm^2 × 100\% Percentage reduction in area = (64 - 33.1776)/(64) × 100\% Percentage reduction in area = (30.8224)/(64) × 100\% = 0.4816 × 100\% = 48.16\% The percentage reduction in area is 48.16\%. 7.3.1 Calculate the normal stress in the tube. Step 1: Calculate the cross-sectional area A of the hollow tube. Outside diameter D_o = 50 mm = 0.050 m. Inside diameter D_i = 30 mm = 0.030 m. A = ()/(4)(D_o^2 - D_i^2) = ()/(4)((0.050 m)^2 - (0.030 m)^2) A = ()/(4)(0.0025 m^2 - 0.0009 m^2) = ()/(4)(0.0016 m^2) = 0.0004 m^2 ≈ 1.2566 × 10^-3 m^2 Step 2: Calculate the normal stress using the formula = (F)/(A). Axial tensile load F = 60 kN = 60 × 10^3 N. = 60 × 10^3 N0.0004 m^2 ≈ 47746482.9 Pa ≈ 47.75 MPa The normal stress in the tube is 47.75 MPa. 7.3.2 If the extension of the tube is 0,76 mm, calculate the strain. Step 1: Identify the original length and the extension. Original length L = 3.8 m. Extension L = 0.76 mm = 0.00076 m. Step 2: Calculate the strain using the formula = ( L)/(L). = 0.00076 m3.8 m = 0.0002 The strain is 0.0002. 7.3.3 Determine Young's modulus of the material. Step 1: Use the calculated stress from 7.3.1 and strain from 7.3.2. Stress = 47.75 MPa = 47.75 × 10^6 Pa. Strain = 0.0002. Step 2: Calculate Young's modulus E using the formula E = ()/(). E = 47.75 × 10^6 Pa0.0002 E = 238.75 × 10^9 Pa = 238.75 GPa Young's modulus of the material is 238.75 GPa. Drop the next question.

A copper rod that is 5,2 m long is subjected to a tensile load of 850 kg. The rod elongates by 1,25 mm. 7.1.1 Calculate the tensile stress on the rod. 7.1.2 Calculate the strain on the rod.

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A copper rod that is 5,2 m long is subjected to a tensile load of 850 kg. The rod elongates by 1,25 mm. 7.1.1 Calculate the tensile stress on the rod. 7.1.2 Calculate the strain on the rod.

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