Tension: wood — analysis. Find the capacity (safe load) of a 6x8 Hem-Fir No.1 tension element connected with two lines of 3/4"-diameter bolts. The bolt holes are drilled through the short dimension of the 6x8. Assume that only live and dead load are present.
The allowable stress, Ft = 650 psi (for Hem-Fir No.1 "posts and timbers"). There are no adjustments, so F't = 650 psi. The net area of the section, An = Ag - (bolt hole area) = 41.25 - 2(5.5 x 7/8) = 41.25 - 9.625 = 31.625 in2. The capacity (safe load), P = stress x area = F't x An = 650 x 31.625 = 20,556#.
Tension: wood — design. A roof is supported by wooden trusses spaced at 10-feet on center. The roof load consists of a dead load, D = 30 psf, a snow load, S = 20 psf, and a construction/maintenance roof load, LR = 30 psf. Design the tension chord of the truss, assuming that the governing load is applied to the top chord of the truss at the truss "nodes" only, and that the loads at these nodes can be found using tributary areas. Also assume that the tension member is a single 2x Hem-Fir No.1, and that it is spliced together at mid-span with 2 lines of 3/4"-diameter bolts.
First, find the governing load combination on the roof: we can compare the following two cases:
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(D + S) / CD = (30 + 20) / 1.15 = 43.5
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(D + LR) / CD = (30 + 30) / 1.25 = 48
The second case governs, with the governing load = (30 + 30) = 60 psf and the duration of load factor, CD = 1.25.
The tributary area for the mid-span node of a typical truss is At = 15 x 10 = 150 sq.ft. Note that the loads acting on the outside nodes can be omitted from the calculation, as they have no effect on the design of the tension chord.
The load acting on the mid-span node, P = 60(150) = 9,000#. By symmetry, each reaction = 4,500#.
We then cut a free-body diagram (FBD) through the tension chord, and show all unknown internal bar forces. Taking moments about point "C," we can solve for the unknown tension force, P (see diagram above, right):
ΣMC = -4500(15) + P(10) = 0; from which P = 6,750#.
The allowable stress for Hem-Fir No.1 (dimension lumber), Ft = 600 psi. CD = 1.25; CM = 1.0; and CF is unknown (assume 1.0 for now; check later). Therefore, the adjusted allowable stress, F't = 600(1.25) = 750 psi.
The required net area, An = load / stress = P / F't = 6750 / 750 = 9.0 in2. Adding in the bolt hole area, we get a required gross area, Ag = 9.0 + 2(1.5 x 7/8) = 9.0 + 2.625 = 11.625 in2. From Table A-4.1, we select a 2x10 with Ag = 13.88 in2. Since the size factor for a 2x10 is 1.1 (greater than the value assumed), we know that the 2x10 will work. However, it is possible that a smaller section with a larger size factor will also work: the next step is to check a 2x8 with CF = 1.2 and Ag = 10.88 in2.
The new adjusted allowable stress, F't = 600(1.25)(1.2) = 900 psi. The required net area, An = P / F't = 6750 / 900 = 7.5 in2. Adding in the bolt hole area, we get a required gross area, Ag = 7.5 + 2(1.5 x 7/8) = 9.0 + 2.625 = 10.125 in2. Since the actual gross area of a 2x8 (10.88 in2) is larger than the required area, we select the 2x8. Final answer.
Tension: steel — analysis.
a) Find the capacity of the threaded rod hangers shown in the image below. Assume 5/8"-diameter "double" rods, A-36 steel, and an allowable stress, Ft = 0.33Fu on the unthreaded part of the rod. Assume that the rods are spaced at 8-feet on center. See text, p. 63, for more information on threaded rods in tension.
NY Times Headquarters (Renzo Piano, Architect), photo by J. Ochshorn
b) As shown in the diagram below, the mezzanine span is assumed to be 20-feet. If the mezzanine supports a dead load, D = 45 psf, and a live load, L = 100 psf (with no live load reduction permitted), are the hangers acceptable?
a) For A-36 steel, Fu = 58 ksi (see Table A-3.11), so the allowable stress, Ft = 0.33(58) = 19.14 ksi. The capacity of one 5/8"-diameter threaded rod equals stress x area = Ft x πr2 = 19.14 x (π x 5/162) = 19.14 x 0.307 = 5.87 k (see discussion, page 63 of text). The capacity of the double rod is therefore 2 x 5.87 = 11.74 k.
b) The actual load on a typical double rod can be found by multiplying the unit load (D + L) by the tributary area: P = (45 + 100)(8 x 10) = 11,600# = 11.6 k. Since the actual load ≤ the allowable load (capacity), the hangers are acceptable.
Tension: steel — design. Assume that the tension force in the vertical first-floor chord of a lateral-force-resisting truss is 136 k. Select (design) the lightest acceptable W8 member to resist this tensile force. To begin, assume the tension element is connected with four lines of 5/8-diameter bolts (with 6 bolts per line), that the flange thickness is 1/4", and that the steel used is A-992. After selecting a provisional cross-section, check the section using its actual dimensions. Revise your selection if necessary and re-check.
For steel tension members, we must do the calculations twice (once each for gross area and effective net area), and design for the worst case. For A-992 steel (see Table A-3.11), Fy = 50 ksi and Fu = 65 ksi. In general, the required area = load / stress. Therefore:
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required Ag = 136 / (0.6 x 50) = 4.53 in2;
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required Ae = 136 / (0.5 x 65) = 4.185 in2;
For the second case: An = Ae / U where U (the shear lag coefficient, Table A-6.1) is unknown. Assuming U = 1.0 for now (check later), we use a provisional value of An = 4.185 in2. Assuming a flange thickness of 1/4", we can add in the bolt hole area from 4 bolts and get a required gross area, Ag = 4.185 + 4(1/4 x 3/4) = 4.935 in2.
Effective net area governs: from Table A-4.3, we select the lightest W-section with a gross area at least equal to 4.935 in2. A W8x18 has an actual flange thickness of 0.330 in. and an area of 5.26 in2. Because it's flange width (5.25 in) is not at least 2/3 of its depth (2/3 x 8.14 = 5.43 in.), we cannot use U = 0.9, but rather must use U = 0.85. See Table A-6.1 for the specific conditions that must be met. We now check the W8x18, using its actual dimensions:
Check W8x18: An = 5.26 - 4(0.330 x 3/4) = 4.27 in2. The actual effective net area, Ae = An x U = 4.27 x 0.85 = 3.63 in2. Because this is less than the required Ae = 4.185 in2, we must try the next larger section:
Check W8x21: An = 6.16 - 4(0.400 x 3/4) = 4.96 in2. Because the flange width of the W8x21 (5.27 in) is not at least 2/3 of its depth (2/3 x 8.28 = 5.52 in.), we again cannot use U = 0.9, but rather must use U = 0.85.
The actual effective net area, Ae = An x U = 4.96 x 0.85 = 4.22 in2. Because this is greater or equal to the required Ae = 4.185 in2, the W8x21 is acceptable.