Explanation Given information: The loading on a floor beam used in the airplane. Calculation: Sketch the Free Body Diagram of the beam as shown in Figure 1. Refer to Figure 1. Find the support reaction as shown below. Apply the Equations of Equilibrium as shown below. Summation of forces along horizontal direction is Equal to zero. ∑ F x = 0 B x = 0 Summation of forces along vertical direction is Equal to zero. ∑ F y = 0 B y + A y − w L 2 = 0 B y = w L 2 − A y (1) Take moment about B is Equal to zero. ∑ M B = 0 M B + A y × L − w L 2 ( L 4 ) = 0 M B + A y L − w L 2 8 = 0 M B = w L 2 8 − A y L (2) Take moment about the section at a distance x from A as shown below. M ( x ) = − ( − A y ) ( x − 0 ) − w 2 ( x − L 2 ) 2 = A y x − w 2 ( x − L 2 ) 2 Apply the slope and elastic curve as shown below. E I d 2 v d x 2 = M ( x ) (3) Here, EI is the flexural rigidity, d 2 v d x 2 is the second derivatives of the function v , and M ( x ) is the moment at the section x . Substitute A y x − w 2 ( x − L 2 ) 2 for M ( x ) in Equation (3). E I d 2 v d x 2 = A y x − w 2 ( x − L 2 ) 2 Integrate both sides of the Equation. ∫ E I d 2 v d x 2 = ∫ ( A y x − w 2 ( x − L 2 ) 2 ) E I d v d x = A y x 2 2 − w 6 ( x − L 2 ) 3 + C 1 (4) Integrate both sides of the Equation (4). ∫ E I d v d x = ∫ ( A y x 2 2 − w 6 ( x − L 2 ) 3 + C 1 ) E I v = A y 2 × x 3 3 − w 6 × 4 ( x − L 2 ) 4 + C 1 x + C 2 E I v = A y x 3 6 − w 24 ( x − L 2 ) 4 + C 1 x + C 2 (5) Apply the boundary conditions as shown below. i. The value of v = 0 at x = 0 . ii. The value of v = 0 at x = L . iii. The value of d v d x = 0 at x = L . Apply boundary condition (i) in Equation (5). Substitute 0 for x and 0 for v in Equation (5). E I ( 0 ) = A y ( 0 ) 3 6 − w 24 ( 0 − L 2 ) 4 + C 1 ( 0 ) + C 2 C 2 = 0 Apply boundary condition (i) in Equation (5). Substitute L for x , 0 for v , and 0 for C 2 in Equation (5). E I ( 0 ) = A y L 3 6 − w 24 ( L − L 2 ) 4 + C 1 ( L ) + 0 0 = A y L 3 6 − w L 4 384 + C 1 L C 1 = w L 3 384 − A y L 2 6 (6) Apply boundary condition (iii) in Equation (4). Substitute L for x and 0 for d v d x in Equation (4). E I ( 0 ) = A y L 2 2 − w 6 ( L − L 2 ) 3 + C 1 0 = A y L 2 2 − w L 3 48 + C 1 C 1 = w L 3 48 − A y L 2 2 (7) Equating Equations (6) and (7) as shown below

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ISBN: 2810014920922

Author: HIBBELER

Publisher: PEARSON

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Slope and Deflection

The slope in a deflected beam is described as an angle in radians made by the tangent of the section with the original axis of the beam. The deflection at any point on the axis of the beam is defined as the lateral displacement of the point before and af…

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Chapter 12.7, Problem 12.104P

To determine

The reactions at the supports A and B using discontinuity functions.

To draw: The moment diagram for the beam.

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Chapter 12.7, Problem 12.103P

Chapter 12.7, Problem 12.105P

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The reactions at the supports A and B using discontinuity functions. To draw: The moment diagram for the beam.