CIEG 407 Homework 6 Due Date: Monday 11/26/07
1. From the sketch shown below, design the interior beams for the floor plan shown. Determine the beam size using:
a. sawn lumber and
b. microlam beam material.
Determine the load at the end of the beam(s) that must be supported by the column.
Givens:
c. floor live load = 40 psf
d. floor dead load = 10 psf
e. one concentrated load in middle of beam of 2000 lb.
f. sawn lumber specs are E = 1.4x106 psi, Fb = 1100 psi, Fv = 90 psi, Fcompperpendicular = 350 psi
g. microlam beam material specs are E = 1.9x106 psi, Fb = 2500 psi, Fv = 125 psi, Fcompperpendicular = 500 psi, select size from Microlam Design Properties tables on web site
h. deflection limit = L/360
i. size specs are:
- 1 - 2x10: S = 21.39 in3 , I = 98.93 in4
- 1 - 2x12: S = 31.64 in3 , I = 178.0 in4
Answer: M = wL2/8 + P*L/2= (40+10)*12*102/8 + 1000*5 = 7500 + 5000 = 12,500 ft-lb
For sawn lumber: fb = M/S = 1100 psi = 12,500*12 in/ft/S
Solve for S = 81.82 in3
If 2x10, number = 136.36/21.39 = 6.38 or use 7 – 2x10
If 2x12, number = 136.36/31.64 = 4.3 or use 5 – 2x12
Check shear on 2x12s: fv = 4000 lb/(5*1.5*11.5) = 46.4 psi < 90 psi OK
Check Fcp on 2x12s = fcp = 4000 lb/(5*1.5*1.5) = 356 psi > 350 psi Not OK, increase bearing distance to 2”, re-check = 4000/(5*1.5*2) = 267 psi < 350 psi OK
Check deflection: add deflection for uniform load + concentrated load
Δ = 5wL4/384EI + PL3/48EI
Calculated I for 5 – 2x12 = (5*1.5)*11.53/12 = 950.54 in4
Δ = 1728*5*600*104/(384*1.4x106*950.54) + 1728*2000*103/(48*1.4x106*950.54)
Δ = .102 + .054 = .156 in < L/360 = 0.33 in. OK
For microlam: use 3 – 1.75”x9.25” microlams each with 5600 ft-lb moment capacity so 3x5600 = 16,800 ft-lb > 12,500 reqd OK
Shear allowable using microlams is: 3x3075 lb = 9225 lb > 4000 lb OK
Stresscp = 750 psi > 4000/(3*1.75*1.5) = 750 > 510 psi OK
Load at end of beam on a column = 8000 lb
2. Determine the joist size required to support the following condition:
a. Joist spacing is 16” o.c.
b. Joist allowable stresses and E are same as problem 2
c. Joist is sawn lumber
d. LL = 60 psf, DL = 10 psf
e. Joist cantilevers a wall 4 ft.
f. Joist span is 12 ft.
g. Allowable deflection = L/360
Answer: w = (60+10)*1.33 ft = 93.1 #/ft
R1 = 496.5 #
R2 = 993.1 #
M1 = (w/8*L2)*(L+a)2*(L-a)2 = (93.1/8*144)*(162)*(82) = 1324.1 ft-lb
M2 = wa2/2 = 93.1*16/2 = 744.8 ft-lb
Fb = M/S so reqd. S = M/fb = 1324.1*12/1100 = 14.4 in3
Reqd. depth t = ((S*6)/1.5)1/2 so min t = 7.6 in which means 2x8 N.G., use 2x10
Check shear: fv = 1.5*R1 or 993.1/(1.5*9.5) = 104.5 psi > Fv of 90 psi N.G.
Increase depth to 11.5 in for 2x12 so fv = 86.4 psi < 90 OK
2x12 is required for joist
Fcperpend = 993.1/(1.5*1.5) = 441.4 psi > 350 psi N.G.
Back calculate reqd bearing and must have 1.89 in, so require 2 in of bearing
Check Δ at location of max moment which is located at M1 location = 5.33 ft from left end
Δ = (wx/24EIL)*(L4-2L2x2+Lx3-2a2L2+2a2x2) where x = 5.33 ft
Δ = .128 in < L/360 = 12*12/360 = 0.4 in. OK
3. Determine the minimum sawn roof rafter size and spacing to support 30 psf roof live load and 20 psf roof dead load when the span is 18 ft and the roof is a low slope. The sawn lumber allowable stresses and E are the same as problem 2.
Answer: 2x12 @ 16” o.c. with fb = 1100 psi
4. A floor diaphragm is 45 ft long and 24 ft deep. The edge of the floor diaphragm collects wind load from 2 floors and must resist 275 lb/ft of wind pressure along the 45 ft long wall.
Find: 1. The number of 16d nails that must be used in the splices of the top wall plate when the allowable shear of one 16d nail is 140 lb
2. The length of the splice if pairs of 16d nails are spaced 4” o.c.
3. The nailing requirement for the floor for the 275 lb/ft wind pressure if the floor material is APA Rated Sheathing Sturd-I-Floor panels, the minimum width of the framing is 2” , and the minimum thickness of the floor sheathing is 19/32” (see Table 2 – APA Diaphragms and Shear Walls Guide on web site)
Answer:
a) Moment for the diaphragm = wL2/8 = 275*452/8 = 69609 ft-lb
T = M/b = 69609/24 = 2900 lb
No. of nails reqd = 2900/140 = 20.7 nails so use 22 nails
b) space 4” o.c., so 11 sets of nails x 4” – 44” say 4’ splice
c) 10d @ 6” o.c.
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