The composite bar BCD in the figure is composed of an inner aluminum cylindrical core of length 2L=2 m and radius R=1 cm, and a sleeve of steel of length L=1 m and thickness R=1 cm surrounding the aluminum core in the CD section of the bar. The bar is fixed at B and at D, and a concentrated load FC=100 kN is applied at the midsection C.
0.4 mm for last problem.
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HW3_4B: 0.49
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HW3_1A: -86.8
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HW3_2c:
0.79 mm
0.71 m
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HW3_2a:
HW3_2b:
or HW3_3?
The composite bar BCD in the figure is composed of an inner aluminum cylindrical core of length 2L=2 m and radius R=1 cm, and a sleeve of steel of length L=1 m and thickness R=1 cm surrounding the aluminum core in the CD section of the bar. The bar is fixed at B and at D, and a concentrated load FC=100 kN is applied at the midsection C.
plz help
HW3___1B ?
hw3_1b?
any one knows the answer?
To find the stress and strain in the composite bar BCD, we need to consider the properties of both materials (aluminum and steel) and apply relevant formulas.
1. Find the cross-sectional area of each segment:
- Aluminum core (BC): The cross-sectional area of a cylinder is given by the formula A = π*r^2, where r is the radius. In this case, the radius is 1 cm, or 0.01 m. So, the cross-sectional area of the aluminum core is A_BC = π*(0.01 m)^2.
- Steel sleeve (CD): The cross-sectional area of a hollow cylinder is given by the difference of two circles, i.e., A = π*(R_outer^2 - R_inner^2), where R_outer is the outer radius and R_inner is the inner radius. In this case, R_outer is the same as the radius of the aluminum core, i.e., 0.01 m, and R_inner is obtained by subtracting the thickness of the steel sleeve (0.01 m) from the radius of the aluminum core. Therefore, R_inner = 0.01 m - 0.01 m = 0 m. So, the cross-sectional area of the steel sleeve is A_CD = π*(0.01 m)^2.
2. Determine the total cross-sectional area of the composite bar BCD:
The total cross-sectional area of the composite bar BCD is the sum of the aluminum core and steel sleeve areas, i.e., A_total = A_BC + A_CD.
3. Calculate the stress in each segment:
- Aluminum core (BC): The stress in an object is given by the formula stress = force / area. In this case, the force applied is FC = 100 kN, which is equal to 100,000 N. So, the stress in the aluminum core is stress_BC = FC / A_BC.
- Steel sleeve (CD): The stress in the steel sleeve is also given by the formula stress = force / area. Since the force applied at point C is still FC = 100 kN, the stress in the steel sleeve is stress_CD = FC / A_CD.
4. Find the total stress in the composite bar BCD:
The total stress in the composite bar BCD can be obtained by summing up the stresses in the aluminum core and steel sleeve, i.e., stress_total = stress_BC + stress_CD.
5. Calculate the strain in each segment:
- Aluminum core (BC): The strain in an object is given by the formula strain = change in length / original length. In this case, the change in length is L/2 (half the length of the aluminum core) and the original length is L (the length of the aluminum core). So, the strain in the aluminum core is strain_BC = (L/2) / L.
- Steel sleeve (CD): Since the steel sleeve is surrounding the aluminum core, its length remains the same as the aluminum core (L = 1 m). Therefore, the strain in the steel sleeve is zero (strain_CD = 0).
6. Find the total strain in the composite bar BCD:
The total strain in the composite bar BCD can be obtained by summing up the strains in the aluminum core and steel sleeve, i.e., strain_total = strain_BC + strain_CD.
By following these steps, you can calculate the stress and strain in the composite bar BCD.