Explanation Calculation: Show the free body diagram of the beam as in Figure 1. Calculate the vertical reaction at point A by taking moment about point B . ∑ M B = 0 − R A L + P × ( L 4 + L 4 ) − P × ( L 4 ) = 0 − R A L + P L 2 − P L 4 = 0 R A = P L 4 L R A = P 4 Calculate the vertical reaction at point B by resolving the vertical component of forces. ∑ F y = 0 P 4 − P + P + R B = 0 R B = − P 4 Calculate the moment due to the reaction at A : M 1 = P 4 × L 2 = P L 8 Calculate the M 1 E I value; Substitute P L 8 for M 1 . M 1 E I = P L 8 E I = P L 8 E I Show the M 1 E I diagram in below Figure 2. Calculate the area ( a ) using the relation: A 1 = 1 2 ( L A D ) ( M 1 E I ) Substitute L 2 for L A D and ( P L 8 E I ) for ( M 1 E I ) . A 1 = 1 2 ( L 2 ) ( P L 8 E I ) = P L 2 32 E I Calculate the moment due to the point load: M 2 = ( P 4 × ( L 2 + L 4 ) ) − ( P × L 4 ) = − P L 16 E I Calculate the M 2 E I value; Substitute − P L 16 E I for M 2 . M 2 E I = − P L 16 E I E I = − P L 16 E I Show the ( M 2 E I ) diagram in below Figure 3. Calculate the position of point of contra flexure as below: P 4 ( L 2 + x ) − P × x = 0 P L 8 + P x 4 − P x = 0 P L 8 − 3 P x 4 = 0 − 3 P x 4 = − P L 8 x = P L 8 × 4 3 P x = L 6 Calculate the area ( A 2 ) using the relation: A 2 = 1 2 ( x ) ( M 1 E I ) Substitute L 6 for x and P L 8 E I for ( M 1 E I ) . A 2 = 1 2 ( L 6 ) ( P L 8 E I ) = P L 2 96 E I Calculate the area ( A 3 ) using the relation: A 3 = 1 2 ( L D E − x ) ( M 2 E I ) Substitute L 6 for x and − P L 16 E I for ( M 2 E I ) . A 3 = 1 2 ( L 4 − L 6 ) ( − P L 16 E I ) = − P L 2 384 E I Calculate the moment due to reaction at point B: M 3 = − P 4 × L 4 = − P L 16 Calculate the M 3 E I value; Substitute − P L 16 for M 3 . ( M 3 E I ) = − P L 16 E I = − P L 16 E I Show the ( M 3 E I ) diagram in below Figure 4. Calculate the area ( A 4 ) using the relation: A 4 = 1 2 ( L E B ) ( M 3 E I ) Substitute ( L 4 ) for L E B and − P L 16 E I for ( M 3 E I ) . A 4 = 1 2 ( L 4 ) ( − P L 16 E I ) = − P L 2 128 E I Calculate the tangential deviation of B with respect to A using the relation: t B / A = A 1 ( ( 1 3 × L A D ) + L D E + L E B ) + A 2 ( 2 3 x + ( L D E − x ) + L E B ) + A 3 ( 1 3 ( L D E − x ) + L E B ) + A 4 ( 2 3 L E B ) Substitute P L 2 32 E I for A 1 , P L 2 96 E I for A 2 , ( − P L 2 384 E I ) for A 3 , ( − P L 2 128 E I ) for A 4 , ( L 6 ) for x , ( L 2 ) for L A D , ( L 4 ) for L D E , and ( L 4 ) for L E B

7th Edition

ISBN: 9780073398235

Author: Ferdinand P. Beer, E. Russell Johnston Jr., John T. DeWolf, David F. Mazurek

Publisher: McGraw-Hill Education

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Chapter 9.6, Problem 141P

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The magnitude (yK) and location of the largest downward deflection (xK).

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Chapter 9.6, Problem 140P

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Chapter 9 Solutions

Mechanics of Materials, 7th Edition

Ch. 9.2 - In the following problems assume that the flexural...Ch. 9.2 - In the following problems assume that the flexural...Ch. 9.2 - In the following problems assume that the flexural...Ch. 9.2 - 9.1 through 9.4 For the loading shown, determine...Ch. 9.2 - 9.5 and 9.6 For the cantilever beam and loading...Ch. 9.2 - 9.5 and 9.6 For the cantilever beam and loading...Ch. 9.2 - For the beam and loading shown, determine (a) the...Ch. 9.2 - For the beam and loading shown, determine (a) the...Ch. 9.2 - Knowing that beam .AB is a W10 33 rolled shape...Ch. 9.2 - Knowing that beam AB is an S200 34 roiled shape...

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The magnitude ( y K ) and location of the largest downward deflection ( x K ) .

Solution Summary: The author calculates the vertical reaction at point A by taking moment about point B.

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The magnitude (yK) and location of the largest downward deflection (xK).

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