Explanation Assumption: The normal force is positive if it creates tension, a positive shear force acting on the segment causes it to rotate clockwise, and a positive bending moment acting on the segment will cause it to bend and concave upwards. 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 the horizontal components of forces. Consider the force indicating upward as positive and downward as negative, in the vertical components of forces. Consider the clockwise moment as negative and the anti-clockwise moment 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 the moment about the point A . ∑ M A = 0 B y ( 6 ) − 1 2 × 12 × 3 × 2 3 × 3 − 12 × 3 × ( 3 2 + 3 ) = 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 − 1 2 × 12 × 3 − 12 × 3 = 0 (II) Along the horizontal direction: Determine the horizontal reaction at point B by resolving the horizontal component of forces. ∑ F x = 0 B x = 0 Conclusion: Solve Equation (I). 6 B y − 36 − 36 × 4.5 = 0 6 B y = 198 B y = 33 kN Substitute 33 kN for B y in Equation (II). A y + 33 − 18 − 36 = 0 A y = 21 kN Determine the shear force and moment: Region ( 0 ≤ x < 3 m ) : Show the free body diagram of the region ( 0 ≤ x < 3 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 − 1 2 × 12 3 × x × x = 0 (III) Moment about the section: Determine the bending moment at a distance ‘ x ’ from left end A by taking the moment at the section. ∑ M x = 0 M − A y ( x ) + 1 2 × 12 3 × x × x × 1 3 × x = 0 (IV) Conclusion: Substitute 21 kN for A y in Equation (III). 21 − V − 2 x 2 = 0 V = 21 − 2 x 2 (V) Substitute 21 kN for A y in Equation (IV). M − 21 ( x ) + 2 x 3 3 = 0 M = 21 x − 2 x 3 3 (VI) Region ( 3 m < x ≤ 6 m ) : Show the free body diagram of the region ( 3 m < x ≤ 6 m ) as in Figure 3. Using Figure (3), Along the vertical direction: Determine the shear force at the section by resolving the vertical component of forces. ∑ F y = 0 A y − V − 1 2 × 12 × 3 − 12 ( x − 3 ) = 0 (VII) Moment about the section: Determine the bending moment at a distance ‘ x ’ from left end A by taking the moment at the section

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ISBN: 9780137514724

Author: HIBBELER

Publisher: PEARSON

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Bending Moment

In general, we can define the bending moment as the reaction induced when a certain amount of load or force is applied to an element which makes it bend.

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Chapter 7, Problem 65P

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Chapter 7, Problem 64P

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

Solution Summary: The author explains the shear force and moment diagrams for the beam, and the method of sections to determine the internal loadings.

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