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: 8220100257063

Author: BEER

Publisher: YUZU

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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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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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