Explanation Given information: The semicircular arch is subjected to a uniform distributed load along its axis of w 0 per unit length. Assumption: The normal force is positive if it creates tension, a positive shear force will cause the segment on which it acts to rotate clockwise, and a positive bending moment will tend to bend the segment on which it acts in a concave upward manner. Loadings that are opposite to these are considered negative. Consider the force indicating the right side as positive and the left side as negative, in horizontal components of forces. Consider the force indicating upside as positive and downside as negative, in vertical components of forces. Consider the clockwise moment as negative and the anticlockwise moment as positive, wherever applicable. The method of sections can be used to determine the internal loadings. Small strip: Show the free body diagram of the small strip as in Figure (1). Using Figure (1), Resultant force along the x -direction: Calculate the resultant force in the x -direction as follows: F R x = ∫ 0 θ w 0 sin θ ( r d θ ) = r w 0 ( − cos θ ) | 0 θ = r w 0 ( 1 − cos θ ) (I) Resultant force along the y-direction: Calculate the resultant force in the y -direction as follows: F R y = ∫ 0 θ w 0 cos θ ( r d θ ) = r w 0 sin θ | 0 θ = r w 0 sin θ (II) Resultant moment about point O : M R o = ∫ 0 θ w 0 ( r d θ ) r = w 0 r 2 ( θ ) | 0 θ = w 0 r 2 θ (III) Semi-circular arch at θ = 45 ° : Show the free body diagram of the semi-circular arch at θ = 45 ° as in Figure (2). Using Figure (2), Along the horizontal direction: Determine the shear force by resolving the horizontal component of forces. ∑ F x = 0 − V + F R x cos θ − F R y sin θ = 0 (IV) Along the vertical direction: Determine the normal force by resolving the vertical component of forces. ∑ F y = 0 N + F R y cos θ + F R x sin θ = 0 (V) Moment about point O : Determine the moment by taking the moment about point O . ∑ M O = 0 − M + M R o + N ( r ) = 0 (VI) Conclusion: Substitute r w 0 ( 1 − cos θ ) for F R x , 45 ° for θ , and r w 0 sin θ for F R y in Equation (IV). − V + ( r w 0 ( 1 − cos θ ) ) cos θ − ( r w 0 sin θ ) sin θ = 0 − V + ( r w 0 ( 1 − cos 45 ° ) ) cos 45 ° − ( r w 0 sin 45 ° ) sin 45 ° = 0 − V + 0

Engineering Mechanics: Statics and Study Pack (13th Edition)

13th Edition

ISBN: 9780133027990

Author: Russell C. Hibbeler

Publisher: Prentice Hall

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Chapter 7.1, Problem 39P

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The internal normal force, shear force, and moment in the arch at θ=45°.

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Chapter 7.1, Problem 38P

Chapter 7.1, Problem 40P

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Chapter 7 Solutions

Engineering Mechanics: Statics and Study Pack (13th Edition)

Ch. 7.1 - Determine the normal force, shear force, and...Ch. 7.1 - Determine the normal force, shear force, and...Ch. 7.1 - Determine the normal force, shear force, and...Ch. 7.1 - Determine the normal force, shear force, and...Ch. 7.1 - Determine the normal force, shear force, and...Ch. 7.1 - Assume A is pinned and B is a roller. Prob. F7-6 Ch. 7.1 - Assume the support at B is a roller. Point C is...Ch. 7.1 - Determine the shear force and moment at points C...Ch. 7.1 - If the suspended load has a weight of 2 kN and a...Ch. 7.1 - Prob. 4P

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The internal normal force, shear force, and moment in the arch at θ = 45 ° .

The internal normal force, shear force, and moment in the arch at θ=45°.

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Chapter 7 Solutions