Explanation Given information: The ultimate strength of the beam is σ U = 60 MPa . The uniformly distributed dead load is w D = 0.75 kips/ft . The concentrated live loads P 1 = 24 kips ; P 2 = 6 kips . Calculation: Maximum moment due to dead load: Show the free-body diagram of the beam AB as in Figure 1. Determine the vertical reaction at point B by taking moment at point A . ∑ M A = 0 B y ( 48 ) − ( 0.75 × 48 ) × 48 2 = 0 48 B y − 864 = 0 B y = 18 kips Determine the vertical reaction at point A by resolving the vertical component of forces. ∑ F y = 0 A y − ( 0.75 × 48 ) + B y = 0 A y − 36 + 18 = 0 A y = 18 kips Shear force: Show the calculation of shear force as follows; V A = 18 kips V B − V A = − 0.75 × 48 V B = − 36 + V A = − 36 + 18 = − 18 kips Show the calculated shear force values as in Table 1. Location ( x ) ft Shear force ( V ) kips A 18 B –18 Plot the shear force diagram as in Figure 2. Due to symmetry of the beam the maximum bending moment occurs at the mid-point. The maximum bending moment occurs at a distance of 8 m from the left end of the beam. Bending moment: Show the calculation of the bending moment as follows; M A = 0 M max − M A = 1 2 × 18 × 24 M max = 216 + M A = 216 + 0 = 216 kips-ft M B = 0 Show the calculated bending moment values as in Table 2. Location ( x ) ft Bending moment ( M ) kips-ft A 0 Max BM 216 C 0 Plot the bending moment diagram as in Figure 3. Refer to the Figure 3; The maximum bending moment due to dead load in the beam is M D = 216 kips-ft . Maximum moment due to live load: The distance u is; u = a P 2 2 ( P 1 + P 2 ) Substitute 14 ft for a , 6 kips for P 2 , 24 kips for P 1 . u = 14 × 6 2 ( 24 + 6 ) = 1.4 ft Find the distance of x ; x = L 2 − u = 48 2 − 1.4 = 22.6 ft Show the free-body diagram of the beam AB as in Figure 4. Determine the vertical reaction at point B by taking moment at point A . ∑ M A = 0 B y ( 48 ) − 24 ( 22.6 ) − 6 ( 36.6 ) = 0 48 B y − 542.4 − 219.6 = 0 B y = 15.875 kips Determine the vertical reaction at point A by resolving the vertical component of forces. ∑ F y = 0 A y − 24 − 6 + B y = 0 A y − 30 + 15.875 = 0 A y = 14.125 kips Shear force: Show the calculation of shear force as follows; V A = 14

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 5.3, Problem 96P

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The most economical wide-flange shape for the beam.

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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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The most economical wide-flange shape for the beam.

Solution Summary: The author explains the most economical wide-flange shape for the beam. The uniformly distributed dead load is w_D=0.75kips/ft

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The most economical wide-flange shape for the beam.

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Chapter 5 Solutions