example roof truss analysis
TRANSCRIPT
CE 331, Fall 2010 Example: Roof Truss Analysis 1 / 6
In this example, a parallel‐chord steel roof truss is analyzed for typical dead and roof live loads. The photo below shows a truss girder (painted gray) supporting the roof of a gymnasium.
Figure 1. Truss girders (gray) supporting bar joists (white) supporting metal roof deck for a gymnasium
The truss girder in the photo is supported by columns (not seen in Figure 1) and supports bar joists at the panel points (chord connections) and midway between the panel points. A similar truss girder is analyzed in this example, except that the bar joists are located at the panel points only. Information about truss girder members is presented below. Table 1. Truss girder components.
Type Member Shape Available Strength (φ Pn)
Chords WT 6 x 20 160 k (compression)
Diagonals LL 2.5 x 2.0 x 3/16 73 k (tension)
Verticals LL 2.5 x 2.5 x 3/16 43 k (compression)
The total weight of truss girder (self weight) is 4.05 k, and the bar joists weigh 9 plf. Other roof components are listed below. Roof & Ceiling:
20 ga metal deck Waterproof membrane with gravel 1” thick Perlite insulating roof boards Heating & cooling ductwork Steel suspended ceiling Acoustic Fiber Board
CE 331, Fall 2010 Example: Roof Truss Analysis 2 / 6
8 @ 10’
Plan View
Front Elevation View
6’
bar joist
metal decking
Side
Elevatio
n of Roo
f Framing
8 @ 10’
6’
3 @ 25’
truss girder
bar joists
truss girder
column
3 @ 25’
Example Roof Truss Analysis 3 / 6
Stability & Determinacy
assume that truss is externally statically determinate for gravity loads
Num_Forces = 33 + 3 = 36Num_Eqns = 18 x 2 = 36
therefore stable & determinate
Dead LoadRoof & Ceiling Wt: weight, psf
20 ga metal deck 2.5Waterproof membrane with gravel 5.5Fiberglass insulation 0.7Heating & cooling ductwork 4Steel suspended ceiling 2Acoustis Fiber Board 1
Total 15.7 psf use 16 psf
Structural Model of Truss
truss girder self wt 4.05 k = 4.05 k / ( 80 ft x 25 ft ) = 2.03 psf18.03 psf
bar joist wt 9 plf
PDint (dead load at an interior panel point)
= 18.025 psf x 25 ft x10 ft = 4.51 k due roof, ceiling wt & truss girder= 9 plf x 25 ft = 0.225 k due purlin wt
4.73 kPDext (dead load at an exterior panel point)
= 18.025 psf x 25 ft x 10/2 ft = 2.25 k due roof, ceiling wt & truss girder= 9 plf x 25 ft = 0.225 k due purlin wt
2.48 k
7 @ 4.73 k2.48 k 2.48 k
Structural Model of Truss
Dead Loads on Truss Girder
Example Roof Truss Analysis 4 / 6
Live LoadRoof live load = Lr = (20 psf) R1 0.6 <= R1 <= 1.0
R1 = 1.2 ‐ 0.001 At
At = 25 ft x 10 ft/panel x 8 panels = 2000 sf
R1 = 1.2 ‐ 0.001 x 2000 sf = 0.60
Lr = 20 psf x 0.6 = 12 psf
PLrint = 12 psf x 25 ft x10 ft = 3.00 k due roof live load
PLrext = 12 psf x 25 ft x 10/2 ft = 1.50 k due roof live load
7 @ 3 k1.5 k 1.5 k
Factored Load
Pu = 1.2 PD + 1.6 PLr
Live Loads on Truss Girder
Pu
PU_int = 1.2 (4.73 k ) + 1.6 (3 k) = 10.476 k
PU_ext = 1.2 (2.48 k ) + 1.6 (1.5 k) = 5.376 k
7 @ 10.476 k5.376 k 5.376 k
Live Loads on Truss Girder
Factored Loads on Truss Girder
Example Roof Truss Analysis 5 / 6
Maximum Chord Compressive Force
Draw deflected shape of loaded truss. Identify chord with max. compressive force.
The top "fibers" of the beam are in compression, and
the fibers in the middle of the beam have the maximum compression.
Therefore, the top chord in the middle of the truss has the max. compressive force.
Calculate the force in the top chord of Panel #4
4 @ 10.476 k5 376 k
C
T
5.376 k
5
R = [7 ( 10.476 k) + 2 ( 5.376 k) ] / 2 = 42.042 k
Σ M about Pt 5 = 0:(f_top) ( 6 ft ) ‐ (42.042 k ‐ 5.376 k ) ( 4 x 10 ft) + (3 x 10.476 k) (20 ft) = 0
f_top 139.7 k in panels at midspan
Check the strength of the chords
factored force in member (Pu) <? Available strength (φc Pn)
Pu 139.7 k
φC Pn 160 k OK
f_top
R
C
T
Example Roof Truss Analysis 6 / 6
Maximum Diagonal Tensile Force
Looking at the parallel‐chord truss as if it were a beam, the max. shear occurs near the supports
analagous beam (assume load is uniformly distributed along beam)
shear
bendingmoment
Therefore cut the truss in the first panel to calculate max diagonal forceTherefore, cut the truss in the first panel to calculate max. diagonal force
5.38 k
6 ft 11.66 ft
10 ft
42.04 k
Σ FV = 0: 42.042 k ‐ 5.376 k ‐ 6 / 11.66 x f_diag
f_diag 71.3 k in end panels
Check the strength of the diagonalsTu 71.3 k
φT Pn 73 k OK
f_diag