Explanation Given information: The diameter d of a solid steel rod is 0.75 in . The young’s modulus of steel is 29 × 10 6 psi . Calculation: Calculate the radius ( c ) of a solid steel rod using the relation: c = 1 2 d Substitute 0.75 in . for d . c = 1 2 ( 0.75 ) = 0.375 in . Calculate the moment of inertia ( I ) using the relation: I = π 4 c 4 Substitute 0.375 in . for c . I = π 4 ( 0.375 ) 4 = 0.015532 in . 4 Calculate the moment at point D due to the point load at A : M D = 150 × 4 = 600 lb ⋅ in . Calculate the moment at point E due to the point load at B : M E = − 300 × 6 = − 1800 lb ⋅ in . Calculate the M D E I value for point D ; Substitute 600 lb ⋅ in . for M D , 0.015532 in . 4 for I , and 29 × 10 6 psi for E . M D E I = 600 29 × 10 6 × 0.015532 = 1.33215 × 10 − 3 in . -1 Show the M D E I diagram in below Figure 1. Calculate the M E E I value for point D : Substitute 1800 lb ⋅ in . for M E , 0.015532 in . 4 for I , and 29 × 10 6 psi for E . M E E I = − 1800 29 × 10 6 × 0.015532 = − 3.99619 × 10 − 3 in . − 1 Show the M E E I diagram in below Figure 2. Calculate the area ( A 1 ) using the relation: A 1 = 1 2 b 1 ( M D E I ) Here, b 1 is length between point D and E Substitute 24 in . for b 1 and 1.33215 × 10 − 3 in . − 1 for ( M D E I ) . A 1 = 1 2 ( 24 ) ( 1.33215 × 10 − 3 ) = 0.0159858 Calculate the area ( A 3 ) using the relation: A 3 = 1 2 b 2 ( M D E I ) Here, b 2 is length between point A and D . Substitute 4 in . for b 2 and 1.33215 × 10 − 3 in . − 1 for ( M D E I ) . A 3 = 1 2 ( 4 ) ( 1.33215 × 10 − 3 ) = 2.6643 × 10 − 3 Calculate the area ( A 2 ) using the relation: A 2 = 1 2 b 1 ( M E E I ) Substitute 24 in . for b 1 and − 3.99619 × 10 − 3 in . − 1 for ( M E E I ) . A 2 = 1 2 ( 24 ) ( − 3.99619 × 10 − 3 ) = − 0.047954 Calculate the area ( A 4 ) using the relation: A 4 = 1 2 b 3 ( M E E I ) Here, b 3 is length between point E and B . Substitute 6 in . for b 3 and − 3.99619 × 10 − 3 in . − 1 for ( M E E I ) . A 4 = 1 2 ( 6 ) ( 3.99619 × 10 − 3 ) = 0.011989 Calculate the tangential deviation of E with respect to D using the relation: t E / D = A 1 ( 2 3 b 1 ) + A 2 ( 1 3 b 1 ) Substitute 24 in . for b 1 , 0.0159858 for A 1 , and 0.047954 for A 2 . t E / D = 0.0159858 × ( 2 3 × 24 ) + ( − 0.047954 ) × ( 1 3 × 24 ) = 0.2557728 − 0.383632 = − 0.1278592 in . Calculate the slope ( θ D ) at point D using the relation: θ D = − t E / D b 1 Substitute − 0.1278592 in . for t E / D and 24 in . for b 1 . θ D = − ( − 0.1278592 ) 24 = 5.33 × 10 − 3 rad Let point K be the location of largest downward deflection. Show the point K in ( M E I ) diagram in below Figure 3. Calculate the M 1 E I value: M 1 E I = M D E I ( 1 − u 24 ) Substitute 1.33215 × 10 − 3 in

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ISBN: 9780100257061

Author: BEER

Publisher: YUZU

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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 9.6, Problem 145P

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The largest upward deflection (yK) in span DE.

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Chapter 9.6, Problem 144P

Chapter 9.6, Problem 146P

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The largest upward deflection ( y K ) in span DE.

Solution Summary: The author calculates the radius of a solid steel rod using the relation: c=12d.

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The largest upward deflection (yK) in span DE.

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