Table of Contents
Appendix II: Design

— David Wagner 2007/12/03 16:16

Purlins

The controlling load combination is 1.2D + 1.6(Lr or S) + (L or 0.8W) =53 psf, λ=0.8.

Roof joists are 37 No. 1 DF-L, 2×6 at 16 in. o.c., directly support the roof sheathing attached to brace them continuously, and are considered simply-supported on each beam. The overhangs are small enough to be neglected.

SymbolReference Design ValueCMCtCLCFCfuCiCrCPCTCbKFφλAdjusted Design Value
Fb 1000 1 1 1 1.3 1 1 1.15 – – – 2.16/0.85 0.85 0.8 2583
Ft 675 1 1 – 1.3 – 1 – – – – 2.16/0.80 0.80 0.8 1516
Fv 180 1 1 – – – 1 – – – – 2.16/0.75 0.75 0.8 311
Fc⊥ 625 1 1 – – – 1 – – – 1.19 1.875/0.90 0.90 0.8 1116
Fc 1500 1 1 – 1.1 – 1 – 1 – – 2.16/0.90 0.90 0.8 2851
E 1.7e6 1 1 – – – 1 – – – – – – – 1.7e6
Emin 0.62e6 1 1 – – – 1 – – 1 – 1.5/0.85 0.85 – 0.93e6

p=53 psf=0.368 psi; w=0.368*16=5.9 lb/in @16” oc; l≈ 24/2 = 12 ft = 144 in; S=7.563.

  • f_b le phi F prime_b
  • {M/S} le phi F_b prod{n}{}{C_n}K_F phi_b lambda
  • {{5.9*144^2}/{8*7.563}} le 0.85*1000*1.3*1.15*{2.16/0.85}*0.85*0.8
  • 2022 le 2196✔

Note these 2×6@16” are 1 psf lighter than the 2×10@16” purlins originally assumed in the 53 psf dead load.

Roof Beams

The central roof beam spans the entire length, and is simply supported on the ends. In the middle it bears on an extension of the central CMU partition wall. The front beam is identical, though it also bears on the columns around the windows.

p=53-1.2*1=52 psf=0.361 psi; w=0.361*12*12=52 lb/in; l≈ 39*12 = 468 in.

  • f_b le phi F prime_b
  • {M/S} le phi F_b prod{n}{}{C_n}K_F phi_b lambda
  • →S {F_b prod{n}{}{C_n}} ge M/{phi K_F lambda} = {{52*468^2*0.85} / {8*0.85*2.16*0.8}} = 823875
  • For 20F-1.5E glulam, {F_b prod{n}{}{C_n}} approx2000 psi, S >= 412.

Try 6-3/4×22. Sx = 544.5; Ix = 5990. Weight ≈ 0.5*62.43*6.75*22/144=32.2 lb/ft = 2.7 lb/in → p=52+2.7=55 lb/in

  • Southern pine: C_v = root{20}{{21*12*5.125}/{L b d}} = root{20}{{1291.5}/{39*6.75*22}} = root{20}{{1291.5}/{39*6.75*22}} = root{20}{0.223} = 0.93 le 1
SymbolReference Design ValueCMCtCLCVCfuCcCPCbKFφλAdjusted Design Value
Fbx+ 2000 1 1 1 0.93 1 1 – – 2.16/0.85 0.85 0.8 3214
Fbx- 1100 1 1 1 0.93 1 1 – – 2.16/0.85 0.85 0.8 1768
Ft 725 1 1 – – – – – – 2.16/0.80 0.80 0.8 1253
Fvx 210 1 1 – – – – – – 2.16/0.75 0.75 0.8 363
Fc⊥x 425 1 1 – – – – – 1 1.875/0.90 0.90 0.8 638
Fc 925 1 1 – – – – 1 – 2.16/0.90 0.90 0.8 1598
Ex 1.5e6 1 1 – – – – – – – – – 1.5e6
Ex,min 0.78e6 1 1 – – – – – – 1.5/0.85 0.85 – 1.17e6
  • {M/S} le phi F_b prod{n}{}{C_n}K_F phi_b lambda
  • {{55*468^2}/{8*544.5}} le 0.85*2000* prod{n}{}{C_n}{2.16/0.85} *0.8
  • 2765 le 3214✔

Deflection, beams not supporting the ceiling. (The ceiling is attached to the rafters; maximum deflection is l/180 = 2.6.)

  • w_{LT}=23; w_{ST}=32
  • Delta_{LT} = {5w_{LT}l^4}/{304EI} = {5*23*468^4}/{304*1.5x10^6*5990}=2.0 in
  • Delta_{ST} = {32/23}Delta_{LT}=2.8 in
  • Delta_T = K_{cr}Delta_{LT} + Delta_{ST}=1.5*2 +2.8=5.8 in
  • 5.8 > 2.6✘

Use a support in the middle by extending the center partition wall so it is the same shape as the end walls and bears the beam at the middle of the span. The attachment to this column does not transfer any loading other than vertical bearing, so conservatively consider the beam supported only on the ends, as is.

  • Delta_{LT} = {5w_{LT}l^4}/{304EI} = {5*23*234^4}/{304*1.5x10^6*5990}=0.13 in
  • Delta_{ST} = {32/23}Delta_{LT}=0.18 in
  • Delta_T = K_{cr}Delta_{LT} + Delta_{ST}=1.5*0.13 +0.18=0.38 in
  • 0.38 ≤ 1.3✔

Halving the span may introduce negative bending at the support (depending on the connection) of at most a quarter the full-span stress.

  • f_{bx-} lt {1/4}2765 =691 << 1768✔

Note, although these beams are overdesigned for this smaller span, the width is chosen in part to fit nicely, with their connections, into a standard 8” gap in the CMU walls. Reducing the depth to as small as 12” only cuts the beam cost $1,000 out of the $24,000 project cost. Call the extra depth, “architectural”. Also note the beams are to be delivered inexpensively in 24' lengths. Information on connection design is not specified here, other than the condition that the beam connection to the middle wall induces nothing but a vertical bearing load on the CMU wall section below.

Window Columns

Window columns bear the load from the front roof beam. The columns are fully grouted and laterally supported to the glass block windows as required by ASCE 5-05 §7.3.3.1.

The largest tributary area is 9×6=54 sf, for a load of 52x9x6 +32.2×9 = 2808+300=3100 lb. fm=3100/7.625^2 = 53 psi « f'm for any masonry system.

Windows

The windows are glass block, isolated from in-plane loads [ASCE 5-05 §7.3.1].

The beam above the windows deflects no more than l/600 = 9×12/600=0.18.

  • w_{LT}=13; w_{ST}=16; l=9×12=108 in.
  • Delta_{LT} = {5w_{LT}l^4}/{304EI} = {5*13*108^4}/{304*1.5x10^6*5990}=0.0032 in
  • Delta_{ST} = {16/13}Delta_{LT}=0.0040 in
  • Delta_T = K_{cr}Delta_{LT} + Delta_{ST}=1.5*0.0032 +0.0040=0.0088 in
  • 0.0088 < 0.18✔

Walls

The greatest in-plane shear is produced by wind on the front of the structure and is resisted by the central partition. The force may be considered as a zone C pressure at the center of the wall, at the top of the partition, pplied as a point load at the far end of a cantilever. The maximum in-plane shear occurs at the base of this wall. Analysis is by allowable stress design, ASCE 5-05 §2.2.

  • V=1.6*21.8*20*20=13952 lb

Consider face-shell bedding, ungrouted.

  • f_v={VQ}/{I_n b} = {3V}/{2 b d}= {3*13952}/{2*2*1.25*24.2*12} =29 psi
  • For the weakest 1000 psi masonry system and considering only the masonry load: f_v le f prime_v le [1.5sqrt{1000}, 120, 37+0.45*0.9*120*20/144]≡29 le [47,120,44]✔

∴Any CMU system compliant with ASCE 5-05 is sufficient to resist shear in this structure.

The greatest bending occurs near the middle of an end (gable) wall at close to the mean height of this wall. Wind loads are transferred horizontally because the walls are stiffer horizontally. This occurs because simple supports at 24' produce more stiffness than a 16' cantilever equivalent to a simple span length of 32'.

  • w=1.6*21.8*16=558 lb/ft = 46.5 lb/in over a span of 24 ft = 288 in.
  • There is no axial load parallel to the bedding.

Consider face-shell bedding, ungrouted.

  • S=b*7.625^2 - b*5.125^2= b(58.14-26.27)=31.9b
  • f_b={M/S}={{6wl^2}/{8*31.9b}}={{46.5*288^2}}/{8*31.9*288} =52.5 psi > 50 (Type M or S PCL)
  • Grouting required: < {52.5-50}/{80-50}={1/12}=1 void of every 12, one void every 8'.

Strictly speaking, slightly less grouting would be required since filling voids increases the section modulus somewhat. Consider full bedding, ungrouted.

  • S=b*7.625^2 - {{2*5.125}/16}b*5.125^2= b(58.14-16.83)=41.3b
  • f_b={M/S}={{6wl^2}/{8*41.3b}}={{46.5*288^2}}/{8*41.3*288} =40.5 psi < 50 (Type M or S PCL)✔
  • No grouting required.

Bond Beams

Bond beams and collar beams are not required.

Discussion


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