Explanation Given information: The elastic modulus ( E ) is 200 GPa . The section of the beam is W250 × 44.8 . Calculation: Refer Appendix C, “Properties of Rolled steel shapes”. The moment of inertia ( I ) for the given section is 70.8 × 10 6 mm 4 or 70.8 × 10 − 6 m 4 . Use moment area method: Show the free body diagram of the beam as in Figure 1. Determine the reaction of the support by taking moment at point B . ∑ M B = 0 − ( R A × 6 ) + ( 40 × 4.5 ) + ( 20 × 3 ) = 0 R A = 40 kN Determine the reaction of the support by considering the vertical equilibrium condition: ∑ F y = 0 R A + R B = 40 + 20 Substitute 40 kN for R A . R B + 40 = 60 R B = 20 kN Show the shear force ( V ) diagram for the given loading beam as in Figure 2. Show the moment ( M E I ) diagram for the span AB of the given loading beam as in Figure 3. Calculate the moment ( M 1 ) using the formula: M 1 = R A × 1.5 Substitute 40 kN for R A . M 1 = 40 × 1.5 = 60 kN ⋅ m Calculate the ratio of M 1 E I : Ratio = M 1 E I Substitute 60 kN ⋅ m for M 1 . M 1 E I = 60 kN ⋅ m E I Calculate the area ( A 1 ) due to the moment ( M 1 ) using the formula: A 1 = 1 2 b 1 h 1 Here, b 1 is the width of the triangle in area ( A 1 ) and h 1 is the height of the triangle in area ( A 1 ) . Substitute 1 .5m for b 1 and 60 kN ⋅ m E I for h 1 . A 1 = 1 2 × 1.5 × ( 60 kN ⋅ m E I ) = 45 E I kN ⋅ m 2 Calculate the moment ( M 2 ) using the formula: M 2 = 40 × 1.5 = 60 kN ⋅ m Calculate the ratio of M 2 E I : Ratio = M 2 E I Substitute 60kN ⋅ m for M 2 . M 2 E I = 60kN ⋅ m E I Calculate the area ( A 2 ) due to the moment ( M 2 ) using the formula: A 2 = b 2 h 2 Here, b 2 is the width of the rectangle in area ( A 2 ) and h 2 is the height of the rectangle in area ( A 2 ) . Substitute 1 .5m for b 2 and 60 E I kN ⋅ m for h 2 . A 2 = 1.5 × ( 60 E I kN ⋅ m ) = 90 E I kN ⋅ m 2 Calculate the moment ( M 3 ) using the formula: M 3 = 20 × 3 = 60kN ⋅ m Calculate the ratio of M 3 E I : Ratio = M 3 E I Substitute 60kN ⋅ m for M 3 . M 3 E I = 60kN ⋅ m E I Calculate the area ( A 3 ) due to the moment ( M 3 ) using the formula: A 3 = 1 2 b 3 h 3 Here, b 3 is the width of the triangle in area ( A 3 ) and h 3 is the height of the triangle in area ( A 3 ) . Substitute 3m for b 3 and 60 E I kN ⋅ m for h 3 . A 3 = 1 2 × 3 × ( 60 E I kN ⋅ m ) = 90 E I kN ⋅ m 2 Place the reference tangent at the point A . Show the tangent for deflection related to reference tangent as in Figure 4. Calculate the deflection at the end A related to the point B ( t B / A ) using the formula: t B / A = [ A 1 × ( 4.5 + ( 1 3 × 1.5 ) ) ] + [ A 2 × ( 3 + 1.5 2 ) ] + [ A 3 × ( 2 3 × 3 ) ] Substitute 45 E I kN ⋅ m 2 for A 1 , 90 E I kN ⋅ m 2 for A 2 , 90 E I kN ⋅ m 2 for A 3 , 200 × 10 9 GPa for E , and 70.8 × 10 − 6 m 4 for I . t B / A = [ 45 E I × ( 4.5 + ( 1 3 × 1.5 ) ) ] + [ 90 E I × ( 3 + 1

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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 143P

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

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Chapter 9.6, Problem 142P

Chapter 9.6, Problem 144P

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Chapter 9 Solutions

EBK MECHANICS OF MATERIALS

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

Solution Summary: The author calculates the moment of inertia (I) for the given section of the beam by taking moment at point B.

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

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