Explanation Given information: The length of the beam is L . EI is constant. Calculation: Convert the uniformly varying load into point load as shown below. P = 1 2 × Intensity of the loading × Length of the loading Substitute w 0 for intensity of the loading and L for length of the loading. P = 1 2 × w 0 × L = w 0 L 2 Sketch the Free Body Diagram of the beamas 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 vertical direction is Equal to zero. ∑ F y = 0 A y + B y − w 0 L 2 = 0 (1) Take moment about A is Equal to zero. ∑ M A = 0 B y × L + M A − M B − w 0 L 2 × L 3 = 0 B y L + M A − M B − w 0 L 2 6 = 0 (2) Sketch the Free Body diagram of the beam section at a distance x from B as shown in Figure 2. Refer to Figure 2. Take moment about the section as shown below. M ( x ) = B y × x − M B − 1 2 ( w 0 x L ) x × x 3 = B y x − w 0 6 L x 3 − M B 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 B y x − w 0 6 L x 3 − M B for M ( x ) in Equation (31). E I d 2 v d x 2 = B y x − w 0 6 L x 3 − M B Integrate both sides of the Equation. ∫ E I d 2 v d x 2 = ∫ ( B y x − w 0 6 L x 3 − M B ) E I d v d x = B y x 2 2 − w 0 6 L × x 4 4 − M B x + C 1 E I d v d x = B y 2 x 2 − w 0 24 L x 4 − M B x + C 1 (4) Integrate both sides of the Equation (4). ∫ E I d v d x = ∫ ( B y 2 x 2 − w 0 24 L x 4 − M B x + C 1 ) E I v = B y 2 x 3 3 − w 0 24 L x 5 5 − M B x 2 2 + C 1 x + C 2 E I v = B y 6 x 3 − w 0 120 L x 5 − M B 2 x 2 + C 1 x + C 2 (5) Apply the boundary conditions as shown below. i. The value of d v d x = 0 at x = 0 . ii. The value of v = 0 at x = 0 . iii. The value of d v d x = 0 at x = L . iv. The value of v = 0 at x = L . Apply boundary condition (i) in Equation (4). Substitute 0 for d v d x and 0 for x in Equation (4). E I ( 0 ) = B y 2 ( 0 ) 2 − w 0 24 L ( 0 ) 4 − M B ( 0 ) + C 1 C 1 = 0 Apply boundary condition (ii) in Equation (5). Substitute 0 for v , 0 for x , and 0 for C 1 in Equation (5). E I ( 0 ) = B y 6 ( 0 ) 3 − w 0 120 L ( 0 ) 5 − M B 2 ( 0 ) 2 + ( 0 ) ( 0 ) + C 2 C 2 = 0 Apply boundary condition (iii) in Equation (4)

Statics and Mechanics of Materials Plus Mastering Engineering with Pearson eText - Access Card Package (5th Edition)

5th Edition

ISBN: 9780134301006

Author: Russell C. Hibbeler

Publisher: PEARSON

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Statics and Mechanics of Materials Plus Mastering Engineering with Pearson eText - Access Card Package (5th Edition)

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Find the moment reactions at the supports A and B using the method of integration.

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