Industrial Truss are triangular formation of metal sections, usually used to span large lengths in space instead of solid girders. The external load apply mostly axial forces on the members in a truss. Depending upon how the force is applied.
Design of an Industrial Truss
- Total Span of the Truss = 6@20 = 120 , Total Height of the Truss = 24 , Spacing = 30,
- Pitch Angle = tan 1 (24/60) = 21.80
- Dead Loads: Roofing = 2 psf, Purlins = 1.5 psf, Sagrods + Bracings = 1 psf
- Basic Wind Speed V = 120 mph
- Material Properties: f c = 3 ksi, fy = 40 ksi
Design of Purlins and Sagrods
Purlin length = Spacing of truss = 30
Load Combination and Biaxial Bending:
It is clear from the preceding analyses that leeward side is more critical for wind loading. Therefore, the combination of dead load and leeward wind load provides the governing design condition for purlins.
Calculation of Point Dead Load, Wind Load and Truss Analysis
In this section, the uniformly distributed loads from the roof as well as from the wind are concentrated on the truss joints for subsequent analyses.
Load Combination and Design of Truss Members
The member forces and support reactions calculated earlier for point dead loads and wind loads are combined in this section to obtain the design member forces and support reactions.
Combination of Dead Load and Wind Load:
The calculation for the design forces is best carried out in a tabular form as shown below.
Design Concept of Truss Members:
The design of truss members is carried out using the following equations.
Members under tension:
Rather than designing for all the members individually, only one section will be chosen for all the bottom cord members, one section for all the top cord members and one section for all the other (vertical and diagonal; i.e., ‘web’) members.
Since the design truss has a long span, design sections will be chosen from double angle sections rather than single angle sections, which are chosen for smaller trusses.
Design of Bottom Cord Members:
For the bottom cord members (L0L1~ L5L6), the maximum tensile force = 30.69 kips and the maximum compressive force = 60.24 kips. Since all the bottom cord members are of equal length (i.e., if effective length factor k is taken = 1, then Le = 20 ft = 240 inch), the maximum forces are taken as the design forces.
Although the design compressive force is much larger and is likely to be governing condition in this case, the sections are designed for both tension and compression for completeness of design.
Design of Vertical and Diagonal Members:
For the vertical and diagonal members (U1L1~ U5L5, U1L2~ U5L4), the maximum tensile force =20.88 kips and the maximum compressive force = 21.86 kips. The maximum tension acts on U2L3 and U4L3 while the maximum compression acts on U3L3, which is 24 ft long. Although U2L3 and U4L3 are slightly longer (i.e., 25.61 ft long), the maximum compressive forces on them are much smaller (i.e., 7.86 kips). Therefore the effective length of the design members is taken as Le = 24 ft = 288 inch. The sections are designed for both tension and compression although it is likely to be governed by compression.
Design of Bracings and Connections
The truss members designed in the previous section are supported against out-of-plane loads by several bracings, joined to each other by welded plate connections and connected to column or wall supports. This section discusses the design of these so-called ‘non-structural members’.
Design of Bracings:
The bracings connect joints of two successive trusses in order to provide structural support
against out-of-plane loadings. The bracing system used for the truss illustrated here is shown
below, consisting of three types of bracings; i.e.,
(i) Bottom cord bracings connect the corresponding bottom joints (e.g., L0 with L0)
(ii) Top cord bracings connect the bottom joints (e.g., L0) with top joints (e.g., U2) diagonally
(iii) Vertical bracings connect the bottom joints (e.g., L3) with top joints (e.g., U3) vertically
Since the structural analysis of the bracing system is complicated, the design follows simplified guidelines, according to which the slenderness ratio (Le/rmin) 400 for bracings under tension and 300 for bracings under compression. In the absence of accurate calculations, the more conservative second criterion (i.e., rmin Le/300) is chosen for design here. The design (using single equal angles) is best carried out in a tabular form as shown below.
Design of Connections:
The joints provided here are actually gusset plates joining two or more members with welded connections. The design is based on the following material and structural properties.
Allowable shear stress fv = 0.3 fy = 0.3 40 = 12 ksi
The gusset plate should be designed to resist the maximum possible design load condition. In this case, however, the thickness of the plate is approximately estimated based on the maximum axial force 64.63 kips. If allowable tensile stress = 0.5 fy = 0.5 40 = 20 ksi, and plate thickness = 0.25 , then the maximum required width of the plate = 64.63/(20 0.25) 13 . This should not be too large, based on the weld-lengths calculated subsequently.
- Thickness of gusset plate is chosen tentatively as = 0.25 and thickness of weld t = 0.25
- The length of weld, Lw = P/(fv 0.707t) = P/(12 0.707 0.25) = P/2.12
Design of Anchorage and Support
The truss is supported by reinforced concrete columns and footings, their reactions having been calculated earlier for point dead load and wind loads. The connections between the truss and support are designed in this section for the combined design loads.
Combination of Support Reactions from Dead Load and Wind Load:
The calculation for the design support reactions is carried out in the following tabular form.
Therefore the design conditions are summarized as follows
- Compressive force = 14.73 kips
- Tensile Force = 29.29 kips, Shear Force = 3.34 kips
Design of Base plate and Anchorage:
Since the truss is supported on base plates on concrete pedestals supported by masonry columns, the design in this study deals mainly with the connections between the truss and the columns.
The column forces are nominal, therefore a 10 20 masonry column is chosen.
The maximum tensile stress on the column = 29.29/(10 20) = 0.146 ksi, which is within the allowable limit (Tensile strength 300 psi).
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