To determine the placement b of the hooks to minimize the largest moment when the concrete member is being hoisted. To draw: The shear and moment diagrams. Explanation Method adopted: Graphical method. Given information: The member has a square cross section of dimension a on each side. The specific weight of concrete is γ . Assumption: Consider that the right side force as positive and left side as negative in horizontal components of forces. Consider that the upward force as positive and downward as negative in vertical components of forces. Consider that the clockwise moment as negative and anticlockwise moment as positive. Apply the Equation of equilibrium wherever applicable. Determine the reactions: Concrete member: Show the free-body diagram of the concrete member hoisted as in Figure 1. The self-weight of the concrete is uniformly throughout the section. Write the equation to determine the self-weight. γ = W A γ = W a 2 W = γ a 2 Here, γ is specific weight of concrete, W is self-weight of the concrete block, and a is side of the square cross section. The concrete member is symmetrical about the x -axis. Along the vertical direction: Determine the vertical resultant force by resolving the vertical component of forces. ∑ F y = 0 F y + F y − γ a 2 L = 0 2 F y = γ a 2 L F y = γ a 2 L 2 Absolute minimum moment: Absolute minimum moment will occur where the positive bending moment is equal to negative bending moment. ( M max ) n e g a t i v e = ( M max ) p o s i t i v e (1) Maximum negative bending moment: The maximum negative bending moment occurs at the supports. Show the free-body diagram of the concrete member up to a distance b from left end as in Figure 2. Moment about the section: Determine the moment at the section by taking moment about the section. ∑ M x = 0 γ a 2 × b × b 2 − ( M max ) n e g a t i v e = 0 ( M max ) n e g a t i v e = γ a 2 b 2 2 Maximum positive bending moment: The maximum bending moment will occur at the mid-point of the concrete member. Show the free-body diagram of the section as in Figure 3. Moment about the section: Determine the moment at the section by taking moment about the section. ∑ M x = 0 γ a 2 × L 2 × L 4 + ( M max ) p o s i t i v e − γ a 2 × L 2 × ( L 2 − b ) = 0 γ a 2 L 2 8 + ( M max ) p o s i t i v e − γ a 2 L 2 4 + γ a 2 L b 2 = 0 ( M max ) p o s i t i v e = γ a 2 L 8 ( L − 4 b ) Substitute γ a 2 b 2 2 for ( M max ) n e g a t i v e and γ a 2 L 8 ( L − 4 b ) for ( M max ) p o s i t i v e in Equation (1). γ a 2 b 2 2 = γ a 2 L 8 ( L − 4 b ) b 2 = L 4 ( L − 4 b ) b 2 = L 2 4 − L b 4 b 2 + 4 L b − L 2 = 0 Solve the Equation. b = 0.207 L Therefore, the placement b of the hooks to minimize the largest moment when the concrete member is being hoisted is b = 0.207 L _ . Entire structure: Show the free-body diagram of the force in the concrete member as in Figure 4. Determine the shear force: The substitution of reaction values and moment included in the same calculation steps. Consider the distance x from left end. Show the calculation of shear force values as follows: Shear force: Shear force at x = 0 : V | x = 0 = 0 Shear force at x = 0.207 L : V | x = 0.207 L ( Just left ) = − γ a 2 × 0.207 L = − 0.207 γ a 2 L V | x = 0.207 L ( Just right ) = − 0.207 γ a 2 L + F y = − 0

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ISBN: 9780133254426

Author: Russell C. Hibbeler

Publisher: Prentice Hall

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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 6.2, Problem 6.46P

To determine

the placement b of the hooks to minimize the largest moment when the concrete member is being hoisted.

To draw: The shear and moment diagrams.

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Chapter 6.2, Problem 6.45P

Chapter 6.4, Problem 6.2PP

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