Explanation Given information: The allowable normal stress of the timber beam is σ a l l = 1750 psi . Design of beam: Show the free-body diagram of the beam as in Figure 1. Determine the vertical reaction at point D by taking moment at point A . ∑ M A = 0 − ( 1 2 × 1.5 × 3 ) × 2 − ( 1.5 × 6 ) × 6 − ( 1 2 × 1.5 × 3 ) × 10 + D y ( 12 ) = 0 − 4.5 − 54 − 22.5 + 12 D y = 0 D y = 6.75 kips Determine the vertical reaction at point A by resolving the vertical component of forces. ∑ F y = 0 A y − ( 1 2 × 1.5 × 3 ) − ( 1.5 × 6 ) − ( 1 2 × 1.5 × 3 ) + D y = 0 A y − 2.25 − 9 − 2.25 + 6.75 = 0 A y = 6.75 kips Shear force: Show the calculation of shear force as follows; V A = 6.75 kips V B − V A = − 1 2 × 1.5 × 3 V B = − 2.25 + V A = − 2.25 + 6.75 = 4.5 kips V C − V B = − 1.5 × 6 V C = − 9 + V B = − 9 + 4.5 = − 4.5 kips V D − V C = − 1 2 × 1.5 × 3 V D = − 2.25 + V C = − 2.25 − 4.5 = − 6.75 kips Show the calculated shear force values as in Table 1. Location ( x ) ft Shear force ( V ) kips A 6.75 B 4.5 C –4.5 D –6.75 Plot the shear force diagram as in Figure 2. Bending moment: Show the calculation of the bending moment as follows; M A = 0 M B − M A = 4

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 5, Problem 158RP

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Chapter 5 Solutions

Mechanics of Materials, 7th Edition

Ch. 5.1 - 5.1 through 5.6 For the beam and loading shown,...Ch. 5.1 - 5.1 through 5.6 For the beam and loading shown,...Ch. 5.1 - 5.1 through 5.6 For the beam and loading shown,...Ch. 5.1 - 5.1 through 5.6 For the beam and loading shown,...Ch. 5.1 - 5.1 through 5.6 For the beam and loading shown,...Ch. 5.1 - 5.1 through 5.6 For the beam and loading shown,...Ch. 5.1 - 5.7 and 5.8 Draw the shear and bending-moment...Ch. 5.1 - 5.7 and 5.8 Draw the shear and bending-moment...Ch. 5.1 - 5.9 and 5.10 Draw the shear and bending-moment...Ch. 5.1 - 5.9 and 5.10 Draw the shear and bending-moment...

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