CEE 377 HW3 SOLUTION

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University of Washington *

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377

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Mechanical Engineering

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Dec 6, 2023

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pdf

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Homework 3 Solution Problem 1 A new downtown Seattle restaurant is framed as shown 25 ft 8 ft 12 ft 12 ft 12 ft A A B B 8 × 8 Timber Support Column (Typ.) Indoor Restaurant Area Outdoor Patio Plan View Elevation View Simple Connection (Typ.) Supporting 2 × 12 Supporting 2 × 12 Full Height Glass Window Full Height Glass Window 2 × 10 Wood Beam 2 × 10 Wood Beam The beam along line A that supports the second story may be analyzed as pin-supported at the left end and propped on a roller at the locaton of the 2 × 12 supporting beam. The dead and live loads are calculated as shown below. Using the LRFD load combinations from ASCE 7, determine the maximum positive and negative moments for use in design of the beam (assume both loads at in full) D = 1 . 5 kip / ft 25 ft 8 ft L = 0 . 8 kip / ft 25 ft 8 ft Since both beams have the same layout, we’ll start this problem by solving the moment diagram for an arbitrary uniform distributed load of w .

w 25 ft 8 ft Solving for reactions (and acknowledging that there are no forces in the x -direction), we get X M pin = 0 ⟲ + R roller (25 ft) − ( w )(33 ft)(33 ft / 2) = 0 → R roller = 21 . 78 w X F y = 0 ↑ + : R pin + R roller − w (33 ft) = 0 → R pin = 11 . 22 w Note that the units above are in feet, but once we add the dimensional load it will convert to a force. Now we can get the moment diagram. I’m going to work through the shear and moment diagram by observation from the free body diagram below. For the shear, we know (1) the reactions represent a discontinuity in the shear diagram, and (2) the distributed load represents to slope of the shear diagram. For the moment, we have zero moment at the pin support by definition, then we can use the magnitudes from the shear diagram to give us the slopes of the moment diagram. To determine the magnitudes of the moment diagram at a point x , we can take the area of the shear diagram from x = 0 to x = x . FBD 11 . 22 w 21 . 78 w 25 ft 8 ft w V ( x ) (kip) 11 . 22 w − 13 . 78 w 8 . 0 w 11 . 22 ft M ( x ) (k-ft) 62 . 9 w − 32 w So we see here that the maximum positive moment occurs 11.22 feet from the pin support and is going to be equal to 62.9 w , while the maximum negative moment occurs at the roller support with a magnitude of − 32 w . Specifically, we find that D = 1 . 5 kip / ft → M + D,max = 94 . 4 kip ft , M − D,max = − 48 kip ft D = 0 . 8 kip / ft → M + L,max = 50 . 4 kip ft , M − L,max = − 25 . 6 kip ft Finally we consider the load combinations. We only have a live and a dead load, so wwe will only need to consider two load combinations: 1 . 4 D and 1 . 2 D + 1 . 6 L . M + max : max [1 . 4 D = 1 . 4(94 . 4) = 132 . 2 kip ft , 1 . 2 D + 1 . 6 L = 1 . 2(94 . 4) + 1 . 6(50 . 4) = 193 . 9 kip ft] M − max : max [1 . 4 D = − 67 . 2 kip ft , 1 . 2 D + 1 . 6 L = − 98 . 6 kip ft] Thus, the maximum factored moments are M + max = 193 . 9 kip ft M − max = − 98 . 6 kip ft

If the live load may instead be placed either on the 25 ft or the 8 ft spans (cases 1 and 2 below), determine the maximum factored positive and negative moments. L = 0 . 8 kip / ft 25 ft 8 ft Live Load Case 1 L = 0 . 8 kip / ft 25 ft 8 ft Live Load Case 2 We are going to approach the problem the same way as we did previously by solving the moment diagram. The dead load does not change, so we need to consider the modified live load. – Live Load Case 1 X M pin = 0 ⟲ + R roller (25 ft) − (0 . 8 kip /ft )(25 ft)(25 ft / 2) = 0 → R roller = 10 kip X F y = 0 ↑ + : R pin + R roller − (0 . 8 kip / ft)(25 ft) = 0 → R pin = 10 kip FBD 10 kip 10 kip 25 ft 8 ft 0 . 8 kip / ft V ( x ) (kip) 10 − 10 12 . 5 ft M ( x ) (k-ft) 62 . 5 kip ft – Live Load Case 2 X M pin = 0 ⟲ + R roller (25 ft) − (0 . 8 kip /ft )(8 ft)(29 ft) = 0 → R roller = 7 . 42 kip X F y = 0 ↑ + : R pin + R roller − (0 . 8 kip / ft)(8 ft) = 0 → R pin = − 1 . 02 kip FBD 1 . 02 kip 7 . 42 kip 25 ft 8 ft 0 . 8 kip / ft V ( x ) (kip) − 1 . 02 6 . 4 M ( x ) (k-ft) − 25 . 6 kip ft

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