(a) Explanation Given information: The elastic modulus ( E ) is 200 GPa . The section of the beam is W150 × 24 . Calculation: Refer Appendix C, “Properties of Rolled steel shapes”. The moment of inertia ( I ) for the given section is 13.4 × 10 6 mm 4 or 13.4 × 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 × 2.4 ) + ( 30 × 0.8 ) + ( 20 × 2.4 × 2.4 2 ) = 0 R A = 34 kN Determine the reaction of the support by considering the vertical equilibrium condition: ∑ F y = 0 R A + R B = 30 + ( 20 × 2.4 ) R A + R B = 78 Substitute 34 kN for R A . R B + 34 = 78 R B = 34 kN Show the moment ( M E I ) diagram for the span AB of the given loading beam as in Figure 2. Calculate the moment ( M 1 ) using the formula: M 1 = R A × 1.6 Substitute 34 kN for R A . M 1 = 34 × 1.6 = 54.4 kN ⋅ m Calculate the ratio of M 1 E I using the relation: Ratio = M 1 E I Substitute 54.4 kN ⋅ m for M 1 . M 1 E I = 54.4 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 .6m for b 1 and 54.4 kN ⋅ m E I for h 1 . A 1 = 1 2 × 1.6 × ( 54 .4kN ⋅ m E I ) = 43.52 E I kN ⋅ m 2 Calculate the moment ( M 2 ) using the formula: M 2 = R A × 2.4 Substitute 34 kN for R A . M 2 = 34 × 2.4 = 81.6 kN ⋅ m Calculate the ratio of M 2 E I using the relation: Ratio = M 2 E I Substitute 81 .6kN ⋅ m for M 2 . M 2 E I = 81 .6kN ⋅ m E I Calculate the area ( A 2 ) due to the moment ( M 2 ) using the formula: A 1 + A 2 = 1 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 2 .4m for b 2 and 81.6 E I kN ⋅ m for h 2 . A 2 = 1 2 × 2.4 × ( 81.6 E I kN ⋅ m ) = 97.92 E I kN ⋅ m 2 Calculate the moment ( M 3 ) using the formula: M 3 = − 20 × 1.6 × 1.6 2 = − 25.6 kN ⋅ m Calculate the ratio of M 3 E I : Ratio = M 3 E I Substitute − 25.6 kN ⋅ m for M 3 . M 3 E I = − 25.6 kN ⋅ m E I Calculate the area ( A 3 ) due to the moment ( M 3 ) using the formula: A 3 = 1 3 b 3 h 3 Here, b 3 is the width of the parabola in area ( A 3 ) and h 3 is the height of the parabola in area ( A 3 ) . Substitute 1 .6m for b 3 and − 25.6 kN ⋅ m E I for h 3 . A 3 = 1 3 × 1.6 × ( − 25.6 kN ⋅ m E I ) = − 13.6533 E I kN ⋅ m 2 Calculate the moment ( M 4 ) using the formula: M 4 = − 20 × 2 (b) To determine Find the deflection ( y D ) at point D .

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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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 130P

(a)

To determine

Find the slope (θA) at point A.

(b)

To determine

Find the deflection (yD) at point D.

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Chapter 9.6, Problem 129P

Chapter 9.6, Problem 131P

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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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Find the slope ( θ A ) at point A .

Solution Summary: The author calculates the moment of inertia (I) for the given section of the beam by considering the vertical equilibrium condition.