perpendicular to the axis of the beam, is called the shear force; the couple moment MB 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. 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. 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. Part B - Internal Loading on a Semicircular Member Consider the semicircular member and loading shown in the image where d = 0.730 m and F = 75.0 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) ΜΕ ΑΣΦ vec NB =, VB = MB Figure < 4 of 4 > Submit B 30° d A Provide Feedback ? N, N, N. m Next >

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perpendicular to the axis of the beam, is called the shear force;
the couple moment MB 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. 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. 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.
Part B - Internal Loading on a Semicircular Member
Consider the semicircular member and loading shown in the image where d = 0.730 m and F = 75.0 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)
ΜΕ ΑΣΦ
vec
NB =, VB = MB
Figure
<
4 of 4 >
Submit
B
30°
d
A
Provide Feedback
?
N, N, N. m
Next >
Transcribed Image Text:perpendicular to the axis of the beam, is called the shear force; the couple moment MB 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. 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. 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. Part B - Internal Loading on a Semicircular Member Consider the semicircular member and loading shown in the image where d = 0.730 m and F = 75.0 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) ΜΕ ΑΣΦ vec NB =, VB = MB Figure < 4 of 4 > Submit B 30° d A Provide Feedback ? N, N, N. m Next >
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