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mal forces in structural members under different port conditions. Part A - Internal Loading Due to a Variable, Distributed Load ctural member, it is necessary to know the resultant ents within the member to ensure the material will rt the loading. The method of sections is used to al loadings. Consider the cantilever beam shown in Consider the cantilever beam and loading shown in the image below where d = 6.00 m, wR = 530 N/m, and wa = 290 N/m. (Figure 3) Determine the magnitudes of the internal loadings on the beam at point C. dings acting on the beam at B are to be make an imaginary cut at section a-a perpendicular beam and separate the beam into two segments. lings at B then become external on the free-body segment, as shown in the image below Express your answers, separated by commas, to three significant figures. > View Available Hint(s) onent NB acting perpendicular to the cut, or is of the beam, is called the normal force; the force acting tangentially to the cut, or perpendicular to eam, is called the shear force; the couple moment e bending moment. The forces and moments acting ents act in opposite directions, as required by w, and can be determined by applying the tions to the free-body diagram of either segment. vec Nc =, Vc =, Mc = N, N, N- m Submit itural member is homogeneous and is not

mal forces in structural members under different port conditions. Part A - Internal Loading Due to a Variable, Distributed Load ctural member, it is necessary to know the resultant ents within the member to ensure the material will rt the loading. The method of sections is used to al loadings. Consider the cantilever beam shown in Consider the cantilever beam and loading shown in the image below where d = 6.00 m, wR = 530 N/m, and wa = 290 N/m. (Figure 3) Determine the magnitudes of the internal loadings on the beam at point C. dings acting on the beam at B are to be make an imaginary cut at section a-a perpendicular beam and separate the beam into two segments. lings at B then become external on the free-body segment, as shown in the image below Express your answers, separated by commas, to three significant figures. > View Available Hint(s) onent NB acting perpendicular to the cut, or is of the beam, is called the normal force; the force acting tangentially to the cut, or perpendicular to eam, is called the shear force; the couple moment e bending moment. The forces and moments acting ents act in opposite directions, as required by w, and can be determined by applying the tions to the free-body diagram of either segment. vec Nc =, Vc =, Mc = N, N, N- m Submit itural member is homogeneous and is not

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
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Item 1 Learning Goal: To calculate internal forces in structural members under different loading and support conditions. Part A - Internal Loading Due to a Variable, Distributed Load To design a structural member, it is necessary to know the resultant forces and moments within the member to ensure the material will be able to support the loading. The method of sections is used to determine internal loadings. Consider the cantilever beam shown in Consider the cantilever beam and loading shown in the image below where d = 6.00 m, wp = 530 N/m, and wa = 290 N/m. (Figure 3) Determine the magnitudes of the internal loadings on the beam at point C. the image below: (Figure 1) If the internal loadings acting on the beam at B are to be determined, we make an imaginary cut at section a-a perpendicular to the axis of the beam and separate the beam into two segments. The internal loadings at B then become external on the free-body diagram of each segment, as shown in the image below (Figure 2) The force component NB acting perpendicular to the cut, or parallel to the axis of the beam, is called the normal force; the force component Vg acting tangentially to the cut, or perpendicular to the axis of the beam, is called the shear force; the couple moment Mg is called the bending moment. The forces and moments acting on the two segments act in opposite directions, as required by Newton's third law, and can be determined by applying the equilibrium equations to the free-body diagram of either segment. Express your answers, separated by commas, to three significant figures. > View Available Hint(s) Nνα ΑΣφ I1 vec ? Nc =, Vc =, Mc = N, N, N - m Submit Provided the structural member is homogeneous and is not irreversibly deformed, these loadings can be considered to act at the centroid of the member's cross-section. Part B - Internal Loading on a Semicircular Member The customary sign conventions are to say the normal force is positive if it creates tension, the shear force is positive if it causes the beam segment to rotate clockwise, and a bending moment is positive if it causes the beam to bend concave upward. Oppositely- acting loadings are considered negative. Consider the semicircular member and loading shown in the image where d = 0.750 m and F = 65.0 N. (Figure 4) Determine the magnitudes of the internal loadings on the beam at point B. Express your answers, separated by commas, to three significant figures. • View Available Hint(s) Figure < 1 of 4 > P, V AZO It vec P2 a NB =, VB =, MB = Ν.N. Ν. m В a Submit
Transcribed Image Text:Item 1 Learning Goal: To calculate internal forces in structural members under different loading and support conditions. Part A - Internal Loading Due to a Variable, Distributed Load To design a structural member, it is necessary to know the resultant forces and moments within the member to ensure the material will be able to support the loading. The method of sections is used to determine internal loadings. Consider the cantilever beam shown in Consider the cantilever beam and loading shown in the image below where d = 6.00 m, wp = 530 N/m, and wa = 290 N/m. (Figure 3) Determine the magnitudes of the internal loadings on the beam at point C. the image below: (Figure 1) If the internal loadings acting on the beam at B are to be determined, we make an imaginary cut at section a-a perpendicular to the axis of the beam and separate the beam into two segments. The internal loadings at B then become external on the free-body diagram of each segment, as shown in the image below (Figure 2) The force component NB acting perpendicular to the cut, or parallel to the axis of the beam, is called the normal force; the force component Vg acting tangentially to the cut, or perpendicular to the axis of the beam, is called the shear force; the couple moment Mg is called the bending moment. The forces and moments acting on the two segments act in opposite directions, as required by Newton's third law, and can be determined by applying the equilibrium equations to the free-body diagram of either segment. Express your answers, separated by commas, to three significant figures. > View Available Hint(s) Nνα ΑΣφ I1 vec ? Nc =, Vc =, Mc = N, N, N - m Submit Provided the structural member is homogeneous and is not irreversibly deformed, these loadings can be considered to act at the centroid of the member's cross-section. Part B - Internal Loading on a Semicircular Member The customary sign conventions are to say the normal force is positive if it creates tension, the shear force is positive if it causes the beam segment to rotate clockwise, and a bending moment is positive if it causes the beam to bend concave upward. Oppositely- acting loadings are considered negative. Consider the semicircular member and loading shown in the image where d = 0.750 m and F = 65.0 N. (Figure 4) Determine the magnitudes of the internal loadings on the beam at point B. Express your answers, separated by commas, to three significant figures. • View Available Hint(s) Figure < 1 of 4 > P, V AZO It vec P2 a NB =, VB =, MB = Ν.N. Ν. m В a Submit
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