Assignment No 4 Ch 7.2 Tension

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Boise State University *

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Mechanical Engineering

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Dec 6, 2023

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DESIGN OF WOOD STRUCTURES ASSIGNMENT No. 4, Chapt 7.2 Design of Tension Members Problem No. 1 For the roof truss elevation shown in the figure below, determine if a 2 x 10 spruce-pine- fir No. 1 member is adequate for the bottom chord, assuming that the roof dead load is 15 psf, the snow load is 45 psf, and the ceiling dead load is 10 psf of the horizontal plan area. Trusses are spaced 24-in. on center. Assume normal temperature and dry service conditions apply and that the members are connected with a single row of 3/4-in.- diameter bolts with 1/16-in. oversized holes. The joints of the truss are assumed to be pinned and the all gravity loads can be assumed to be applied as concentrated loads at the joints of the truss. Roof Dead Load = 15 psf; Snow load = 45 psf Ceiling Dead Load = 10 psf 9 ft 3 ft 6 ft 6 ft 3 ft 9 ft 1/4
Problem No. 2 A 5-1/8 x 15 DF combination 5 glulam is used as the tension member in a large roof truss. A single row of 1 in.-diameter bolts occurs at the net section of the member. It can be assumed that the hole diameters are 1/8 in. larger than the bolt size. Loads are a combination of dead and snow. Joints are assumed to be pin-connected. . Determine the factored axial capacity of the member assuming: a) Dry conditions (MC 16%). b) Wet conditions (MC > 16%). Problem No. 3 The truss shown in the figure below has a 2 x 4 lower chord of Select Structural Spruce- Pine-Fir (South). Trusses are space 24 in o.c. The uniform dead load on a horizontal plane is 20 psf, while the snow load is 55 psf. There is no reduction of area for fasteners and the truss is to be untreated and used in a dry normal building environment. Verify the design in tension of the 2 x 4 lower chord. 2/4
Problem No. 4 Design the tension chord for the roof diaphragm of the warehouse building shown below. The warehouse is 50 ft x 140 ft x 20 ft high. The roof is to be constructed of joists spaced at 2 ft on centre supported on the top of glulam beams spaced at 10 ft on centre supported by wood stud walls. The specified (unfactored) wind pressure on the long (140 ft) side of the building is 24 psf. All framing lumber is S-P-F No. 2 grade and is used in a normal building environment. The roof diaphragm can be assumed to act as a simply supported beam and its moment can be calculated from simple beam theory. Assume that each chord member is made of two 2 x members splice together. Assume that a splice may occur at any location along the 140 ft length of the chord member. 20 ft 50 ft 10 ft 14 @ 10 ft Joists at 2 ft on center 3/4
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Problem No. 5 The elevation of a stud wall 16 feet long by 10 feet tall in an industrial building is shown below. The wall is made of 2 x Hem Fir No. 3 studs spaced at 16 in. on center and of ½” thick plywood. The unfactored design lateral force caused by wind and the unfactored dead load resisted by the stud wall are also shown in the figure. For design purposes, it is assumed that the overturning moment due to the wind load is resisted by two reactions only at both end studs, while the gravity load is resisted in compression by all the studs based on their tributary width. The moisture content is less than 19% and normal temperatures apply. a) Calculate the factored tension force T u to be used for the design of the end studs. Note that because the dead load causes compression (i.e. 0.5 C D ) and the wind load causes tension (i.e. T w ) in the end studs, the overturning load combination (0.9D+1.0W) must be considered. b) Determine the required width of the 2 x end studs to resist the factored tension force calculated above. 4/4 16 ft Wind Load P w = 6.4 kips Dead Load w D = 250 lbs/ft Dead Load Reactions Wind Load Reactions 16 in typ. 10 ft T w C w C D 0.5C D 0.5C D