Explanation 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 opposite to these and are considered as negative. Consider the force indicating right side as positive and left side as negative in the horizontal components of forces. Consider that the force indicating upward is taken as a positive and downward as a negative in the vertical components of forces. Consider the clockwise movement as negative and the anticlockwise movement as positive, wherever applicable. The method of sections can be used to determine the internal loadings. Determine the reactions: Entire beam: Show the free body diagram of the entire beam as in Figure (1). Using Figure (1), Moment about the point A : Determine the vertical reaction at point B by taking moment about point A . ∑ M A = 0 B y ( 4 ) − 8 × 6 × 6 2 = 0 (I) Along the vertical direction: Determine the vertical reaction at point A by resolving the vertical component of forces. ∑ F y = 0 A y + B y − 8 × 6 = 0 (II) Along the horizontal direction: Determine the horizontal reaction at point A by resolving the horizontal component of force. ∑ F x = 0 A x = 0 Conclusion: Solve Equation (I). 4 B y − 144 = 0 B y = 36 kN Substitute 36 kN for B y in Equation (II). A y + 36 − 48 = 0 A y = 12 kN Determine the shear force and moment: Segment AB ( 0 ≤ x < 4 m ) : Show the free body diagram of the region ( 0 ≤ x < 4 m ) as in Figure (2). Using Figure (2), Along the vertical direction: Determine the shear force at the section by resolving the vertical component of forces. ∑ F y = 0 A y − V − 8 × x = 0 (III) Moment about the section: Determine the bending moment at a distance ‘ x ’ from left end A by taking moment at the section. ∑ M x = 0 M − A y ( x ) + 8 × x × x 2 = 0 (IV) Conclusion: Substitute 12 kN for A y in Equation (III). 12 − V − 8 x = 0 V = 12 − 8 x (V) Substitute 12 kN for A y in Equation (IV). M − 12 ( x ) + 4 x 2 = 0 M = 12 x − 4 x 2 (VI) Segment CB ( 4 m < x ≤ 6 m ) : Show the free body diagram of the region ( 4 m < x ≤ 6 m ) as in Figure (3)

13th Edition

ISBN: 9780132915540

Author: Russell C. Hibbeler

Publisher: Prentice Hall

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Question

Chapter 7.2, Problem 63P

To determine

To draw: The shear force and moment diagrams for the overhang beam.

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Chapter 7.2, Problem 62P

Chapter 7.2, Problem 64P

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Chapter 7 Solutions

Engineering Mechanics: Statics

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