The New and Improved Pultex ® Pultrusion Global Design Manual Chapter 4 1 Previous Page Next Page Top of Page Table of Contents Chapter 4 Load Tables for Flexural Members and Connections Beam Deflections - A pultruded beam will be designed for deflection, strength and buckling. Fiber reinforced composite beams exhibit both flexural and shear deflections. Shear deflections are most apparent when the span to depth ratios are less than 20. At short spans, the shear deflections comprise a significant portion of the actual deflections; therefore, the designer should always account for shear deflections when designing with composites. Reference Pultex ® Fiber Reinforced Polymer Structural Profiles Material Properties Sheets for the appropriate properties of the profiles for which you are utilizing in your design. Although coupon testing is a good quality control method, composite materials are not homogeneous and will exhibit different properties in the web and flange areas. Deflection predictions should be made with values based on full section testing. Please reference Appendix B per ASTM D198 for full section testing procedure. The Allowable Uniform Load Tables were calculated using physical properties that were derived from full section testing. The Load Tables are based on simply supported end conditions with uniformly distributed loads. For beam loadings and end conditions not represented in the Allowable Uniform Load Tables, reference the Beam Deflection Formula and relative design tables. The following formula was used to predict the deflections in The Allowable Uniform Load Tables: ∆ = 5wL 4 + w L 2 Where: 384 EI 8A′G A′ = kA w (mm 2 ) A w = Shear area of profile (mm 2 ) (Ref. Table 2) k = Shear coefficients (Ref. Table 2) E x = Modulus of elasticity (GPa) G = Modulus of rigidity (Shear Modulus) (GPa) I = Moment of inertia (mm 4 ) L = Length of span (m) ∆ = Deflection (mm) w = Load on the beam (N/m) Allowable Stresses Fiber reinforced composite beams exhibit compressive, flexural, and shear stresses under various load conditions. The dominating failure mode for long span flexural members is typically local buckling of the compressive flange, while short spans are dominated by in-plane shear failures. Safety Factors The allowable stresses used in The Allowable Uniform Load Tables are based on the ultimate compressive buckling, flexural and shear strengths with applied safety factors. Specifically, a “2.5” safety factor is used for local buckling and flexural stresses while a “3” safety factor is used for shear. The following shear and flexure formula were used to predict the Allowable Loads. V=f v (A w ); where f v = allowable shear stress = 31 MPa = 10.3 MPa (Equation P-4)
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The New and Improved Pultex® Pultrusion Global Design ManualChapter 4
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Chapter 4 Load Tables for Flexural Members and Connections
Beam Deflections - A pultruded beam will be designed for deflection, strength and buckling.
Fiber reinforced composite beams exhibit both flexural and shear deflections. Shear deflections are mostapparent when the span to depth ratios are less than 20. At short spans, the shear deflections comprise asignificant portion of the actual deflections; therefore, the designer should always account for sheardeflections when designing with composites. Reference Pultex® Fiber Reinforced Polymer StructuralProfiles Material Properties Sheets for the appropriate properties of the profiles for which you areutilizing in your design.
Although coupon testing is a good quality control method, composite materials are not homogeneous andwill exhibit different properties in the web and flange areas. Deflection predictions should be made withvalues based on full section testing. Please reference Appendix B per ASTM D198 for full section testingprocedure.
The Allowable Uniform Load Tables were calculated using physical properties that were derived fromfull section testing. The Load Tables are based on simply supported end conditions with uniformlydistributed loads. For beam loadings and end conditions not represented in the Allowable Uniform LoadTables, reference the Beam Deflection Formula and relative design tables.
The following formula was used to predict the deflections in The Allowable Uniform Load Tables:
∆ = 5wL4 + w L2 Where: 384 EI 8A′G A′ = kAw (mm2)
Aw = Shear area of profile (mm2) (Ref. Table 2) k = Shear coefficients (Ref. Table 2)Ex = Modulus of elasticity (GPa)G = Modulus of rigidity (Shear Modulus) (GPa)I = Moment of inertia (mm4)L = Length of span (m)∆ = Deflection (mm)w = Load on the beam (N/m)
Allowable Stresses
Fiber reinforced composite beams exhibit compressive, flexural, and shear stresses under various loadconditions. The dominating failure mode for long span flexural members is typically local buckling ofthe compressive flange, while short spans are dominated by in-plane shear failures.
Safety Factors
The allowable stresses used in The Allowable Uniform Load Tables are based on the ultimatecompressive buckling, flexural and shear strengths with applied safety factors. Specifically, a “2.5”safety factor is used for local buckling and flexural stresses while a “3” safety factor is used for shear. Thefollowing shear and flexure formula were used to predict the Allowable Loads.
Local Buckling of the Compression Flange for Wide Flange, I-Sections, Square Tube andRectangular Tube Sections
The local compression buckling strength of pultruded wide flange, I-Sections, square tubes andrectangular tubes can be determined by utilizing the following equations. The local bucking equationswere derived from University research. (Reference Step by Step Design Equations for Fiber-reinforcedPlastic Beams for Transportation Structures)Davalos,Barbero and Qiao
Where, σx is the critical stress, and p and q are constants that are defined by the coefficient ofrestraint (ζ) at the junction of the plates:
I/W sections:
Box sections:
Where:
σxcr = Critical buckling stress in (MPa)
b = Half the width of the compression flange for I/W sections (mm)b = The width of the compression flange for box sections, b=bf (mmbf = Width of the compression flange (mm)bw = Height of the section (mm)Ex = Longitudinal modulus of elasticity (GPa)Ey = Transverse modulus of elasticity (GPa)f = FlangeGxy = Modulus of rigidity (Shear Modulus) (GPa)p = Constant defined by the coefficient of restraint (ζ)q = Constant defined by the coefficient of restraint (ζ)t = Thickness of the compression flange (mm) ζ = Coefficient of restraint of the compression platesw = Web
( ) ( )( ) ( ) ( ) ( )( )[ ]fxyfxyfyfyfx
2f
2cr
x G2EpEE2qbt
12++
= νπσ
( )( ) 2
;2
;4.0
065.0025.0;5.0
004.03.0 f
wyf
fyw bb
Eb
Ebqp ==
+
+=
−
+= ζζζ
( )( ) f
wyf
fywbb
Eb
Ebqp ==
+
+=
−
+= ;;2.0
08.00.1;3.1
002.00.2 ζζζ
able ofontents
)
(Equation P-3)
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Stress Calculations of Channels
The Wide Flange-, I- and Box sections mentioned above are loaded in the plane of symmetry and bend inthe plane of loading. Channel sections, however, do not exhibit such behavior unless the loading isapplied through the shear center. In normal construction with channel members, such a loading conditionis seldom observed; therefore, the top flange of channel sections, must be adequately supported to resistrotation due to off shear center loading. The Maximum Uniform Loads in the Allowable Uniform LoadTables were calculated using working stress analysis with the assumption that the members are fullylaterally supported. Reference Equations P-4 and P-5 on page 1. (Note: CPI is currently developing localbuckling and laterally unsupported beam equations for the next update)
Lateral-Torsional Buckling
The Allowable Uniform Loads in the Allowable Uniform Load Tables are derived assuming that adequatelateral support is provided for the flexural members. The degree of lateral support for structures isdifficult to predict. Figures a.– d. represent common bracing scenarios that are considered to provideadequate lateral support. Note that the bracing intervals must be adequate. In the event that lateralsupport is not used, the designer must investigate lateral torsional buckling criteria. The AllowableUniform Load Tables contain a column titled Allowable load, laterally unsupported beam global bucklingcapacity. Please note that the global buckling load tables include a 2.5x safety factor.
For I-Sections or Wide Flange Sections, the lateral torsional buckling load for various loading conditionscan be determined by using the following equation:
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Design Equation for Lateral-Torsional Buckling
Where, for Wide Flange Sections and I-Sections
Cw = Warping constant (mm6) J = Torsion constant (mm4)Cb = Moment variation constant (Ref. Table 1) Mcr = Critical moment (N-m)Lb = Unsupported length between points that have lateral Ey = Modulus of elasticity for bending about minor axis y (Use same value as Ex, for simplicity. Values are veG = Shear modulus (GPa)K = Effective length coefficient (Ref. Table 1)Iy = Moment of Intertia about the minor axis (mm4)
Cb is a moment gradient adjuster, that depends on the type of load and for Cb can be located in Table 1.
(Equation P-1)JGIEIC
KLE
KLCM yyyw
b
y
bbcr +
=
2ππ
( )33231
wf htbtJ +=4
2y
w
IhC =
Table ofContents
restraint (mm)-y (GPa)ry similar) Ey ≈Ex
end restraint conditions. Values
The New an
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Table 1Lateral Buckling Coefficient for Various End Conditions1
Lateral Supportabout y-axis
Moment gradientadjuster (Cb)
Effective lengthcoefficient (K)
None 1.0 1.0
NoneFull
1.130.97
1.00.5
NoneFull
1.300.86
1.00.5
NoneFull
1.351.07
1.00.5
None 1.70 1.0
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Full 1.04 0.5
6
Beam Deflection Formula
Note: Reference Table 2. Shear Areas and Shear Coefficients for Various Cross Sections A', A'=kAw.
Uniform load on simple beam
( )
( )xlw
wl
wV
wlVisR
xlxlwAG
wlwlwl
x
−=
=
−=
=
+−=∆
′+=∆
=
2xM
8midpoint)(at max. M
x21
2
2 24x
8 384 5 midpoint)(at max.
Load UniformEquiv. Total
x
2
x
323
24
EI
EI
Uniform load on beam fixed at both ends
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Point load on simply supported beam
( )
2Px
21x whenM
4lPload) of point (at max. M
2PV
x4l3 48
Px21x when
AG4Pl
48lP load) of point (at max.
2P Load Uniform Equiv. Total
x
22x
3
=
<
=
=
−=
<∆
′+=∆
=
EI
EI
( )
( )22x
1
x
2
2
x
4
xl xl12wM
24lwmidpoint) (atM
12lwends) (at max. M
xwV
2wlVisR
xll 24
xwAG
wl384
lw midpoint) (at max.
3wl Load Uniform Equiv. Total
66
21
8
2
2
2
2
2
−−=
=
=
−=
=
−=∆
′+=∆
=
EI
EI
T
Point load on beam with fixed ends
( )
( )l x48P
21 x whenM
8lPends) and center (at max. M
2PV
x4l3l 48
Px21 x when
AG4lP
192lP midpoint) (at max.
P Load Uniform Equiv. Total
x
2
2
x
3
−=
<
=
=
−=
<
′+=
=
EI
EI
∆
∆
o
Point load on cantilever beam
( )
( )
( )PxM
lPend fixed at max. MPVisR
xxl3l2 6P
AGlP
3lP end free at max.
8PLoad Uniform Equiv. Total
x
323x
3
===
+−=
′+=
=
EI
EI
∆
∆
Uniform load on cantilever beam
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( )
2xM
2end) fixed(at max. M
VVisR
3 x 4x 24
2 8w end) free(at max.
4 Load UniformEquiv. tal
2
x
2
x
434
24
w
wl
wxwl
llwAG
wllwl
x
=
=
==
+−=∆
′+=∆
=
EI
EI
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Table 2
Two Concentrated Loads Equally Spaced ona Simply Supported Beam
( )
( ) ( )
( ) ( )
( ) xPa x whenMPaloads) (between max. MPVisR
ax3xl3 6Paa) - l( and a x when
xa3al3 6Pxa x when
AGPa
6a
8l
EIPa center at max.
lPa8 Load Uniform Equiv. Total
x
22x
22x
22
=<==
−−=<>
−−=<
′+
−=
=
EI
EI
∆
∆
∆
for calculating A', A' = kAw
Type Shear Area k Type Shear Area k
Rectangular Section Aw= bd 5/6
Channel Section Aw= 2bt 5/6
I or W-Section Aw= 2bt 5/6Channel Section Aw= ht 1
Note: Arrows indicate direction of shear forces k = Shear coefficient Aw = Shear areaNote: Values are approximated for simplicity. For exact shear coefficients reference Timoshenko's Beam Theory.
Cross Section Cross Section
Shear Areas and Shear Coefficients for Various Cross Sections
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Examples of Beam Selection of Pultex® Profiles used as Flexural Members
Example 1.
Design ParametersSelect a Pultex® Wide Flange Section capable of withstanding a uniform load of 1314 N/m, over a simplysupported span of 5m, with a deflection limit of L/180. The beam is laterally supported and has anassumed weight of 7.34 kg/m.SolutionRefer to the Allowable Uniform Load Tables. The load tables do not take into account the weight of thebeam; therefore, add the weight of the section to the design load. From the Allowable Uniform LoadTables, reference the 152mm x 152mm x 9.5 mm (nominal) Wide Flange Section. Locate the spancolumn and find the 5m span and look across the columns to the L/180 column. The number in the spacerepresents a uniform load of 1542 N/m. This load is more than the design load 1386 N/m (includedweight of selected beam). Therefore, the section is adequate. Select a 152 mm x 152 mm x 9.5mmWide Flange Section.
Example 2.
Design ParametersSelect a Pultex® Wide Flange Section that is simply supported and is capable of withstanding a laterallyunsupported load of 2550 N/m at a span of 6.4m with a deflection less than L/240. SolutionReference the Allowable Uniform Load Tables. Select a member size to begin the process. Locate the203mm x 203mm x 9.5mm Wide Flange Section and the span of 6.5m. Locate the Allowable load,laterally unsupported beam, global buckling capacity and locate the 6.5m span and load interface. Themaximum load is 960 N/m and is not adequate; therefore, select a larger Wide Flange Section. Select a254mm x 254mm x 12.7mm Wide Flange Section. Locate the 6.5m span and Maximum Load LaterallyUnsupported column. The maximum load is 2883 N/m with a 2x safety factor. The 2883 N/m load isgreater than the design load plus the weight of the Wide Flange Section; therefore, the 254mm x 254mmx 12.7mm beam is adequate. Scanning across the columns, notice that the 2711 N/m design load is lessthan the 3193 N/m load in the L/240 column; therefore, the deflection will be less than L/240.Select a 254mm x 254mm x 12.7mm Wide Flange Section.
Example 3.
Design ParametersDetermine the maximum allowable point load and deflection of a laterally unsupported 152mm x 152mmx 9.5mm Wide Flange Section that is simply supported at a span of 3.66mm. The beam is to be used in a10% concentration of Potassium Hydroxide.SolutionStep 1. Reference equation (P-1) for lateral-torsional buckling.
Cw = Warping constant (mm6)
J = Torsion constant (mm4)Cb = Moment gradient adjusterMcr = Critical moment (mm-N)L = Unsupported length (mm)
GJIEICKLE
KLCM yyyw
b
y
bbcr +
=
2ππ
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Ey = Modulus of elasticity for bending about minor axis (GPa) Ey ≈E
Step 2. Use equation (P-2) to predict the critical moment Mcr. Obtain the moment variation constant Cbfrom Table 1. Cb is 1.35 for the simply supported beam with no end constraints and a point load (Table 1.)L is the laterally unsupported length of 3.66m or 3660 mm.E is the modulus of elasticity (reference the Pultex® Fiber Reinforced Polymer Structural Profiles MaterialProperties Sheet) E = 27.59 GPa.G is the modulus of rigidity (Shear Modulus) (reference the Pultex® Fiber Reinforced Polymer StructuralProfiles Material Properties Sheet) G = 3.45 GPaCw is the warping constant; a value can be located in the Elements of Section in the Design Manual. Forthe 152mm x 152mm x 9.5mm Wide Flange Section, Cw = 3.218E10mm6.J is torsion constant, a value can be found in the Elements of Section in the Design Manual. For the 152mm x 152mm x 9.5mm Wide Flange Section, J = 131698mm4.Iy is the moment if inertia about the weak axis, Iy = 5542163mm4. (from Elements of Section)K is the effective length coefficient from Table 1., K = 1.
Step 3. Equate Mcr
Mcr = 15,093 N-m
For a simply supported span with a point load at mid span, the maximum moment is given byM= PL/4. Where:P = point load in (N)L = length of span (m), equals Lb in the present case.
Therefore, calculate P.
15093 N-m = P(3.66m)/4P= 16,495 N
Apply the desired safety factor. In this case, use a 2.5 safety factor.Therefore, Pallowable = 6,598N
Step 4. Calculate the allowable deflection with the 6,60N load.
From the beam deflection equations, determine the equation for the simply supported, mid-span, pointload condition.
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G = Modulus of rigidity (Shear modulus) (GPa) i.e., 3.45 GPaE = Full section modulus of elasticity (GPa), i.e., 27.59 GPaIx = Moment of inertia (mm4), i.e., 16964344mm4
Example 3 (cont’d)
A′ = kAw, i.e. 1(1,452mm2)=1,452mm2
Aw = Cross sectional area of web 1,452mm2
k = Shear Coefficient Reference Table 2. (Shear area of common cross sections), i.e.,(Table 2)
Step 5. Solve for deflection ∆.
∆ = 15.57mm or L/235
Step 6. Determine if the flexural strength is adequate.
σf = M/Sx
Where: σf = flexural stress (GPa)M = maximum moment (N - m)Sx = Section modulus (mm3)
From the Elements of Section of The New and Improved Design Manual for Pultrusion ofStandard and Custom Fiber Reinforced Polymer Structural Profiles, determine Sx for the 152mmx 152mm x 9.5mm Wide Flange Section. Sx = 222630mm3.From the Pultex® SuperStructural Profiles for Wide Flange Sections and I-Sections MaterialProperties Sheets, determine the ultimate flexural strength and apply the proper safety factor,which in the present case is 2.5.
σf = 227.6 MPa ultimate flexural strength.
(227.6 MPa/2.5) = (3.66mm/4)/222630mm3
Pflexural= 4,983 lbs.
Pflexural > Pallowable therefore, the strength is adequate.
Step 7. Calculate the Critical Buckling load and determine if it is adequate. From equation (P-2):
Where:
σxcr = Critical buckling stress in (MPa)
b = Half the width of the compression flange for I/W sections (mm)b = The width of the compression flange for box sections, b=bf (mm)bf = width of the compression flange (mm)
)mm0015.mm452,1)(GPa45.3()m66.3)(N600,6(
41
)mm16964344)(GPa59.27()m66.3)(N600,6(
481
24
3
−+=∆
( ) ( )( ) ( ) ( ) ( )( )[ ]fxyfxyfyfyfx
fcrx
GEpEEqbt
22*212
22
++
= νπσ
( )( ) 2
;2
;4.0
065.0025.0;5.0
004.03.0 f
wyf
fyw bb
Eb
Ebqp ==
+
+=
−
+= ζζζ
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bw = Height of the section (mm)Ex = Longitudinal modulus of elasticity (GPa)Ey = Transverse modulus of elasticity (GPa)
f = FlangeGxy = Modulus of rigidity (Shear Modulus) (GPa)p = Constant defined by the coefficient of restraint (ζ)q = Constant defined by the coefficient of restraint (ζ)t = Thickness of the compression flange (mm) ζ = Coefficient of restraint of the compression plates
σxcr = 156 MPa
Step 7. The allowable local buckling load is determined by evaluating the critical buckling stress tobending stress and applying the appropriate safety factor. In this case use 2.5.
Pallowable =15,198N> Pglobal buckling 6598N; therefore, global buckling governs the design.
The design is governed by Mcr Lateral Torsional Buckling (Global buckling) and is limited to6598N.
Reference the Chemical Compatibility Guide to determine the proper Pultex® Series.
Choose Pultex® 1625 Series.
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Nomenclature
∆ = Deflection (mm)∆ = 1-(νxyνyx)ζ = Coefficient of restraint of the compression platesσc = Compressive stress (GPa)σx
cr = Critical buckling stress in (GPa)νxy = Poisson’s ratio (longitudinal)νyx = Poisson’s ratio (transverse)a = Unsupported length or region over which Nx acts (length of beam) inchesAw = Shear area of profile (Table 2) (mm2)A′ = kAw, Shear coefficient x shear area of profile (mm2)b = Half the width of the compression flange for I/W sections (mm)b = The width of the compression flange for box sections, b=bf (mm)bf = Width of the compression flange (mm)bw = Height of the section (mm)Cb = Moment Variation Constant Cw = Warping Constant (mm6) D = Deflection (mm)D11, D22 = Flexural rigidity in 1, 2 and radial directions ExorEy = Modulus of elasticity of the major or minor axis(GPa)Ey local = Local transverse modulus of Elasticity (GPa)f = Flangefb = Flexural stress (GPa)fv = Shear stress (GPa) G = Shear modulus (modulus of rigidity) (GPa)Gxy = Shear modulus (GPa)h = Depth of section (mm)Ix or y = Moment of Inertia about desired axis (mm4)J = Torsion Constant (mm4)K = Effective length coefficientk = Shear coefficient (Table 2.)L = Length (mm)Lb = Unsupported length between points that have lateral restraint (mm)M = Maximum moment (m -N)Mcr = Critical Moment that causes lateral buckling (m-N)P = Point load (N)p = Constant defined by the coefficient of restraint (ζ)q = Constant defined by the coefficient of restraint (ζ)r = Radius of gyration (mm)Sx = Section modulus (mm3)t = Thickness of compression flange (mm)V = Shear Force (N)Wt. = Weight of profile in N/mWlb = Maximum load governed by critical local buckling (N/m)Wf = Maximum load governed by flexural stress (N/m)Wv = Maximum load governed by shear strength (N/m)Wlu = Maximum laterally unsupported load (N/m)L/D = Ratio of length of the span to the deflection
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Introduction to Pultex SuperStructural Profiles
Product Advantage Summary
When comparing pultruded fiber reinforced polymer composites to traditional building materials such assteel, one will notice that the strengths of the materials are generally comparable while the stiffnesscharacteristics are dissimilar. For example, the modulus of elasticity of steel is approximately 29E6 psi.,while the modulus of elasticity of a typical pultruded Wide Flange Section is 2.5-2.8E6 psi. The stiffnessdifference is 11.5 times between the two materials. In an effort to improve stiffness, Creative Pultrusions,has modified the fiber architecture of selected structural profiles. The result improved the modulus ofelasticity from 2.5-2.8E6 psi to 3.9-4.0E6 psi., an average improvement in E-Modulus of 49%. Pultex
SuperStructural profiles offer the designer the ability to design longer spans with heavier loads. Themost important advantage is a more economical design, as material and labor costs are greatly reduced.
The following example is a comparison of a standard pultruded section to a Pultex® SuperStructuralprofile.
Example 1.0Reference Creative Pultrusions former Design Guide, Volume 2, Revision 1, Allowable Uniform LoadTables, page 3-17. The allowable uniform load of a standard 6" x 6" x 3/8" Wide Flange Section at aspan of 10′ and L/D ratio of 360 is 149 lbs./ft. Referencing The New and Improved Pultex® PultrusionDesign Manual, the allowable uniform load for the same loading, span and deflection criteria is 220lbs./ft. The difference is a 48% increase in E-Modulus. The graph below demonstrates the differencebetween the Pultex SuperStructural 6" x 6" x 3/8" Wide Flange Section and a standard pultruded 6" x6" x 3/8" Wide Flange Section. The graph demonstrates the allowable uniform loads for each beam atvarious spans with the deflection limit of L/D = 360.
Comparison of Standard Structural Profiles and Pultex SuperStructural Profiles
Project example: Plating Tank Cover Design
Pultex® Supe rSt ruct ura l vs Pultex® Standard Structural Allowable Uniform Load Comparison
0
100
200
300
400
8 9 10 11 12 13 14 15Span (ft)
(Uni
form
Loa
d (lb
s./ft
48% increase in E-Modulus
Pultex® SuperStructural Profiles 6" x 6" x 3/8" Wide Flange Section Pultex® Standard Structural Profiles 6" x 6" x 3/8" Wide Flange Section
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1. Standard Structural Project Design FactorsDescription: Plating tank Design load: 80 psfMaximum deflection: L/180 or .67"Service temperature: 80°F5% concentration of Chromic Acid
Step 1. Based on the 80 psf + the 3.46 psf DL of Flowgrip, determine the allowable beam spacing.
Step 2. Reference Creative Pultrusions former Design Guide, Volume 2, Revision 1.For a 6"x6"x3/8" Wide Flange Beam, the allowable uniform load at L/180 is 298 lbs/ft.
Step 3. Determine the allowable spacing by dividing the allowable load by the design load, i.e., (298 lbs/ft)/83.46lbs/ft2= 3.57' O.C.
Step 4. Space the beams @ 3.5' O.C. (Note: beam weight excluded)
Bill of Materials for Standard Structural Project DesignItem Quantity Price $ Total
2. Pultex® SuperStructural Project Design FactorsDescription: Plating tank cover 10' x 60'Design load: 80 psfMaximum deflection: L/180 or .67"Service temperature: 80°F5% concentration of Chromic Acid
Step 1. Determine the maximum span of Flowgrip Solid Floor Panel.a. Reference page 7 of the Solutions that Work---The Most Complete Line of Grating and
Access Structure Products in the IndustryNote: The Flowgrip® Solid Floor Panel will span 60" and satisfy the above design criteria.
(The beam spacing is based on 5' O.C.)
Pultex SuperStructural Profiles Pultex® Standard Structural Profiles
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Step 2. Determine which Wide Flange Section profile will satisfy the loading condition above.a. 80 psf x 5' panel width = 400 lbs/ft live load on the beams.b. Calculate the dead load. Assume that a 6" x 6" x 3/8" Wide Flange Section profile is
sufficient.c. The weight of the 6" x 6" x 3/8" section is 4.92 lbs/ft.d. Calculate the weight of the Flowgrip 3.46 psf x 5' = 17.3 lbs/ft.
Step 3. Determine the total live load (LL) and dead load (DL) combination.a. 400 lbs/ft LL + 4.92 lbs/ft DL + 17.3 lbs/ft DL = 422.2 lbs./ft.
Step 4. Determine if the 6" x 6" x 3/8" Wide Flange Section profile is adequate.a. Reference the 6" x 6" x 3/8" Wide Flange Section in the Allowable Uniform Load Tables.b. Locate the 10' span row and look across to the l/180 deflection column.c. The Pultex SuperStructural 6" x 6" x 3/8" Wide Flange Section will hold 441 lbs/ft and
deflect less than L/180; therefore, the 6" x 6" x 3/8" Wide Flange Section profile is adequate.
Step 5. Space all beams at 5' O.C. across the 10' section of the span.
Bill of Materials for Pultex SuperStructural Project Design Item Quantity Price $ Total
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Allowable Uniform Load Tables (Metric)
76.2 x 76.2 x 6.3 Ultimate In-Plane Shear Strength (MPa) 48.3 Simply Supported beam with Uniform1500/1525/1625 Series Ultimate Flexural Strength (MPa) 227.6 Loads at various L/D ratios
E= 27.5 GPa G =3.4 GPa Ultimate Local Buckling Strength (MPa) 277.7 (N/m)Ix= 1.34E6 mm4 Sx= 3.52E4 mm3 Simply Supported with a Uniform Load Maximum Lb = .61 mAw= 4.83E2
2Wt.= 2.42 kg/lm Laterally Supported beams L/D ratio
Note: Bold numbers in the Factored Load Tables represent the governing load
101.6 x 101.6 x 6.3 Ultimate In-Plane Shear Strength (MPa) 48.3 Simply Supported beam with Uniform1500/1525/1625 Series Ultimate Flexural Strength (MPa) 227.6 Loads at various L/D ratios
E= 27.5 GPa G =3.4 GPa Ultimate Local Buckling Strength (MPa) 156.2 (N/m)Ix= 3.35E6 mm4 Sx= 6.60E4 mm3 Simply Supported with a Uniform Load Maximum Lb = .76 mAw= 6.45E2
2Wt.= 3.26 kg/lm Laterally Supported beams L/D ratio
Note: Bold numbers in the Factored Load Tables represent the governing load
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152.4 x 152.4 x 6.3 Ultimate In-Plane Shear Strength (MPa) 48.3 Simply Supported beam with Uniform1500/1525/1625 Series Ultimate Flexural Strength (MPa) 227.6 Loads at various L/D ratios
E= 27.5 GPa G =3.4 GPa Ultimate Local Buckling Strength (MPa) 69.4 (N/m)Ix= 1.19E7 mm4 Sx= 1.56E5 mm3 Simply Supported with a Uniform Load Maximum Lb = 1.07 mAw= 9.67E3
2Wt.= 4.94 kg/lm Laterally Supported beams L/D ratio
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152.4 x 152.4 x 9.5 Ultimate In-Plane Shear Strength (MPa) 48.3 Simply Supported beam with Uniform1500/1525/1625 Series Ultimate Flexural Strength (MPa) 227.6 Loads at various L/D ratios
E= 27.5 GPa G =3.4 GPa Ultimate Local Buckling Strength (MPa) 156.2 (N/m)Ix= 1.69E7 mm4 Sx= 2.22E5 mm3 Simply Supported with a Uniform Load Maximum Lb = 1.07 mAw= 1.45E3
2Wt.= 7.34 kg/lm Laterally Supported beams L/D ratio
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203.2 x 203.2 x 9.5 Ultimate In-Plane Shear Strength (MPa) 48.3 Simply Supported beam with Uniform1500/1525/1625 Series Ultimate Flexural Strength (MPa) 227.6 Loads at various L/D ratios
E= 27.5 GPa G =3.4 GPa Ultimate Local Buckling Strength (MPa) 87.9 (N/m)Ix= 4.18E7 mm4 Sx= 4.11E5 mm3 Simply Supported with a Uniform Load Maximum Lb = 1.52 mAw= 1.93E3
2Wt.= 9.85 kg/lm Laterally Supported beams L/D ratio
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203.2 x 203.2 x 12.7 Ultimate In-Plane Shear Strength (MPa) 48.3 Simply Supported beam with Uniform1500/1525/1625 Series Ultimate Flexural Strength (MPa) 227.6 Loads at various L/D ratios
E= 26.8 GPa G =3.4 GPa Ultimate Local Buckling Strength (MPa) 154.8 (N/m)Ix= 5.36E7 mm4 Sx= 5.28E5 mm3 Simply Supported with a Uniform Load Maximum Lb = 1.52 mAw= 2.58E3
2Wt.= 13.04 kg/lm Laterally Supported beams L/D ratio
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254 x 254 x 9.5 Ultimate In-Plane Shear Strength (MPa) 48.3 Simply Supported beam with Uniform1500/1525/1625 Series Ultimate Flexural Strength (MPa) 227.6 Loads at various L/D ratios
E= 27.5 GPa G =3.4 GPa Ultimate Local Buckling Strength (MPa) 56.2 (N/m)Ix= 8.34E7 mm4 Sx= 6.75E5 mm3 Simply Supported with a Uniform Load Maximum Lb = 1.83 mAw= 2.42E3
2Wt.= 12.37 kg/lm Laterally Supported beams L/D ratio
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254 x 254 x 12.7 Ultimate In-Plane Shear Strength (MPa) 48.3 Simply Supported beam with Uniform1500/1525/1625 Series Ultimate Flexural Strength (MPa) 227.6 Loads at various L/D ratios
E= 26.8 GPa G =3.4 GPa Ultimate Local Buckling Strength (MPa) 99.1 (N/m)Ix= 1.08E8 mm4 Sx= 8.5E5 mm3 Simply Supported with a Uniform Load Maximum Lb = 1.83 mAw= 3.22E3
2Wt.= 16.40 kg/lm Laterally Supported beams L/D ratio
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304.8 x 304.8 x 12.7 Ultimate In-Plane Shear Strength (MPa) 48.3 Simply Supported beam with Uniform1500/1525/1625 Series Ultimate Flexural Strength (MPa) 227.6 Loads at various L/D ratios
E= 26.8 GPa G =3.4 GPa Ultimate Local Buckling Strength (MPa) 68.8 (N/m)Ix= 1.90E8 mm4 Sx= 1.25E6 mm3 Simply Supported with a Uniform Load Maximum Lb = 2.13 mAw= 3.87E3
2Wt.= 19.75 kg/lm Laterally Supported beams L/D ratio
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76.2 x 38.1 x 6.3 Ultimate In-Plane Shear Strength (MPa) 48.3 Simply Supported beam with Uniform1500/1525/1625 Series Ultimate Flexural Strength (MPa) 227.6 Loads at various L/D ratios
E= 27.5 GPa G =3.4 GPa Ultimate Local Buckling Strength (MPa) 1,009.1 (N/m)Ix= 7.51E5 mm4 Sx= 1.97E4 mm3 Simply Supported with a Uniform Load Maximum Lb = .30 mAw= 4.84E2 mm2 Wt.= 1.58 kg/lm Laterally Supported beams L/D ratio
Note: Bold numbers in the Factored Load Tables represent the governing load
101.6 x 50.8 x 6.3 Ultimate In-Plane Shear Strength (MPa) 48.3 Simply Supported beam with Uniform1500/1525/1625 Series Ultimate Flexural Strength (MPa) 227.6 Loads at various L/D ratios
E= 27.5 GPa G =3.4 GPa Ultimate Local Buckling Strength (MPa) 567.6 (N/m)Ix= 1.89E6 mm4 Sx= 3.71E4 mm3 Simply Supported with a Uniform Load Maximum Lb = .38 mAw= 6.45E2 mm2 Wt.= 2.14 kg/lm Laterally Supported beams L/D ratio
Note: Bold numbers in the Factored Load Tables represent the governing load
Pultex® SuperStructural ProfilesI-Sections
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152.4 x 76.2 x 6.3 Ultimate In-Plane Shear Strength (MPa) 48.3 Simply Supported beam with Uniform1500/1525/1625 Series Ultimate Flexural Strength (MPa) 227.6 Loads at various L/D ratios
E= 27.5 GPa G =3.4 GPa Ultimate Local Buckling Strength (MPa) 252.3 (N/m)Ix= 6.73E6 mm4 Sx= 8.83E4 mm3 Simply Supported with a Uniform Load Maximum Lb = .53 mAw= 9.68E2 mm2 Wt.= 3.26 kg/lm Laterally Supported beams L/D ratio
Note: Bold numbers in the Factored Load Tables represent the governing load
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152.4 x 76.2 x 9.5 Ultimate In-Plane Shear Strength (MPa) 48.3 Simply Supported beam with Uniform1500/1525/1625 Series Ultimate Flexural Strength (MPa) 227.6 Loads at various L/D ratios
E= 27.5 GPa G =3.4 GPa Ultimate Local Buckling Strength (MPa) 567.6 (N/m)Ix= 9.55E6 mm4 Sx= 1.25E5 mm3 Simply Supported with a Uniform Load Maximum Lb = .53 mAw= 1.45E3 mm2 Wt.= 4.82 kg/lm Laterally Supported beams L/D ratio
Note: Bold numbers in the Factored Load Tables represent the governing load
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203.2 x 101.6 x 9.5 Ultimate In-Plane Shear Strength (MPa) 48.3 Simply Supported beam with Uniform1500/1525/1625 Series Ultimate Flexural Strength (MPa) 227.6 Loads at various L/D ratios
E= 26.9 GPa G =3.4 GPa Ultimate Local Buckling Strength (MPa) 319.3 (N/m)Ix= 2.36E7 mm4 Sx= 2.32E5 mm3 Simply Supported with a Uniform Load Maximum Lb = .76 mAw= 1.93E3 mm2 Wt.= 6.50 kg/lm Laterally Supported beams L/D ratio
Note: Bold numbers in the Factored Load Tables represent the governing load
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203.2 x 101.6 x 12.7 Ultimate In-Plane Shear Strength (MPa) 48.3 Simply Supported beam with Uniform1500/1525/1625 Series Ultimate Flexural Strength (MPa) 227.6 Loads at various L/D ratios
E= 26.9 GPa G =3.4 GPa Ultimate Local Buckling Strength (MPa) 562.7 (N/m)Ix= 3.02E7 mm4 Sx= 2.97E5 mm3 Simply Supported with a Uniform Load Maximum Lb = .76 mAw= 2.58E3 mm2 Wt.= 8.57 kg/lm Laterally Supported beams L/D ratio
Note: Bold numbers in the Factored Load Tables represent the governing load
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254 x 127 x 9.5 Ultimate In-Plane Shear Strength (MPa) 48.3 Simply Supported beam with Uniform1500/1525/1625 Series Ultimate Flexural Strength (MPa) 227.6 Loads at various L/D ratios
E= 27.5 GPa G =3.4 GPa Ultimate Local Buckling Strength (MPa) 204.3 (N/m)Ix= 4.73E7 mm4 Sx= 3.72E5 mm3 Simply Supported with a Uniform Load Maximum Lb = .91 mAw= 2.42E3 mm2 Wt.= 8.17 kg/lm Laterally Supported beams L/D ratio
Note: Bold numbers in the Factored Load Tables represent the governing load
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254 x 127 x 12.7 Ultimate In-Plane Shear Strength (MPa) 48.3 Simply Supported beam with Uniform1500/1525/1625 Series Ultimate Flexural Strength (MPa) 227.6 Loads at various L/D ratios
E= 26.9 GPa G =3.4 GPa Ultimate Local Buckling Strength (MPa) 360.1 (N/m)Ix= 6.10E7 mm4 Sx= 4.80E5 mm3 Simply Supported with a Uniform Load Maximum Lb = 1.07 mAw= 3.22E3 mm2 Wt.= 10.81 kg/lm Laterally Supported beams L/D ratio
Note: Bold numbers in the Factored Load Tables represent the governing load
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304.8 x 152.4 x 12.7 Ultimate In-Plane Shear Strength (MPa) 48.3 Simply Supported beam with Uniform1500/1525/1625 Series Ultimate Flexural Strength (MPa) 227.6 Loads at various L/D ratios
E= 26.9 GPa G =3.4 GPa Ultimate Local Buckling Strength (MPa) 250.1 (N/m)Ix= 1.08E8 mm4 Sx= 7.07E5 mm3 Simply Supported with a Uniform Load Maximum Lb = 1.07 mAw= 3.87E3 mm2 Wt.= 13.04 kg/lm Laterally Supported beams L/D ratio
Note: Bold numbers in the Factored Load Tables represent the governing load
Pultex® Standard Structural ProfilesRectangular Tubes
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50.8 x 152.3 x 3.4 Multicell Ultimate In-Plane Shear Strength (MPa) 48.31500/1525/1625 Series Ultimate Flexural Strength (MPa) 227.6 Simply Supported beam with Uniform
E= 22.1 GPa G =2.9 GPa Ultimate Local Buckling Strength (MPa) 22.2 Loads at various L/D ratios Ix= 6.71E5mm4 Sx= 3.49E5 mm3 Simply Supported with a Uniform Load (N/m)Aw= 5.54E2
2Wt.= 2.66 kg/lm Laterally Supported beams L/D ratio
Note: Bold numbers in the Factored Load Tables represent the governing load
Pultex® Standard Structural ProfilesRectangular Tubes
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177.8 x 101.6 x 6.4 Ultimate In-Plane Shear Strength (MPa) 48.31500/1525/1625 Series Ultimate Flexural Strength (MPa) 227.6 Simply Supported beam with Uniform
E= 22.1 GPa G =2.9 GPa Ultimate Local Buckling Strength (MPa) 129.4 Loads at various L/D ratios Ix= 1.40E7mm4 Sx= 1.57E5 mm3 Simply Supported with a Uniform Load (N/m)Aw= 2.26E2
2Wt.= 5.80 kg/lm Laterally Supported beams L/D ratio
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Metric Clip Connection Load Tables with Pultex® SuperStructural Angles
The following Clip Connection Load Tables were developed based on full section testing of the Pultex®
SuperStructural angles. The experimental procedure can be located in Appendix A. The load tableswere developed based on the following:
1. Experimental test results2. Room temperature (22.8°C)3. A safety factor has not been applied. All loads are ultimate unless noted.4. An ultimate bearing strength of 228 MPa5. A 4% hole deformation bearing stress of .083 GPa6. No damage to the composite materials7. No chemical exposure8. Bolted connection only; no adhesive. 9. Full section shear strength through angle heel =23.4 MPa (1500/1525 Series); 26.9 MPa (1625 Series)
Note: Clip angle connection dimensions are governed by internal dimensions of beams and externaldimensions of the flanges on the columns.
Note: The maximum slip of the connection at failure is approximately 8.9 mm without adhesive. The maximum slip of the connection at failure is approximately 2.6 mm with adhesive. Theadhesive does not affect the ultimate load capacity of the connection, but does affect the amountof slip of the connection.
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Metric Bolt Hole Bearing Capacity of Pultex® SuperStructural Beams and Columns
Tables 3a and 3b represent the bearing capacity of the flange and web sections of various beams andcolumns. The load table is based on the following:
1. The ultimate lengthwise bearing strength of the flange sections is 228 MPa.2. The ultimate crosswise bearing strength of the web section is 207 MPa.3. The 4% hole deformation longitudinal bearing strength of the flange is 83 MPa.4. The 4% hole deformation transverse bearing strength of the web is 78 MPa.5. No safety factors are applied.6. Testing performed at room temperature (23 °C)7. No adhesive is used.8. Bolt torque is not considered.
Table 3aBolt Hole Bearing Capacity of Pultex® SuperStructural Columns and Beams for 1500/1525 Series
Bolt recommendations based on ASTM A325 grade five coarse threaded boltsConnection testing performed with grade 8 oversized washers (2.5 times hole dia.)Proof strength for ASTM A325 bolt equals 75,000 psi
D
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Design Example Using the Clip Connection Charts
Task Design a connection for a 152mm x 152mm x 9.5mm Wide Flange Section beam that is clipped into a203mm x 203mm x 9.5mm Wide Flange Section column. (Reference drawing below.)
GivenDesign load on connection is a static load of 26688NThe corrosive environment requires 1625 Series.Operating temperature = 23°C.Minimum Safety Factor = 2.5
Solution:1. Reference the 1625 Series of Pultex® SuperStructural angle clip connection charts for the 114.3mm
clip angle length (depth of the 152mm x 152mm x 9.5mm section dictates the 114.3mm angle length).2. First, investigate the 9.5mm thick angles. Looking across to the Ultimate Load Before Failure
Column, the ultimate load of the 9.5mm thick clip angle connection is 13,163 lbs.3. The design load is 26688N(2.5 S.F.) = 66720N; therefore, the 9.5mm section is not adequate.4. Investigate the 12.7mm thick angles. The Ultimate load is 78,062N; therefore, the 12.7mm angle is
adequate.5. Specify the 101.6mm x 101.6mm x12.7mm angle 114.3mm long 1625 Series of Pultex
SuperStructural angle.6. Determine the bearing capacity of the two holes in the web of the 152mm x 152mm x 9.5mm Wide
Flange Section and the bearing capacity of the four holes in the column. 7. Reference the Hole Bearing Capacity of Pultex® Columns and Beams Design Chart. Determine
the row that satisfies the desired bolting condition and member sizes.8. Enter the row for the 9.5mm member thickness and two bolts in the beam. The ultimate beam web
bearing load in the transverse direction is 171,933N for the 15.9m,m diameter bolts. The design loadwith the 2.5 S.F. is 15,000 lbs.; the connection is adequate.
10. From Table 1a, the column bearing load is adequate at 158,460N.
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Moment Capacity of Pultex® SuperStructural Angles
Table 1
Moment capacity is in (N-m/m) per longitudinal inch of angle
Table 2
Moment capacity is in (N-m/m) per longitudinal inch of angle
Moment Capacity of 12.7mm Thick SuperStructural Angles1500/1525/1625 Series
0.0
1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
9.0
10.0
0 500 1000 1500 2000 2500 3000 3500 4000
Moment (N-m/m)
Deg
rees
of R
otat
ion
Ultimate Moment at Failure = 4244 N-m/m
Moment Capacity of 12.7mm Thick SuperStructural Angles1500/1525/1625 Series
0
1
2
3
0 100 200 300 400 500 600 700 800
Ultimate Moment at Failure = 1045 N-/
Deg
rees
of R
otat
ion
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Table 3
Moment capacity is in (N-m/m) per longitudinal inch of angle
Table 4
Moment capacity is in (N-m/m) per longitudinal inch of angle
Moment Capacity of 9.5mm Thick SuperStructural Angles1500/1525/1625 Series
0.0
1.0
2.0
3.0
4.0
5.0
6.0
7.0
8.0
0 200 400 600 800 1000 1200 1400 1600 1800 2000
Moment (N-m/m)
Deg
rees
of R
otat
ion
Ultimate Moment at Failure = 2251 N-m/m
Moment Capacity of 9.5mm Thick SuperStructural Angles1500/1525/1625 Series
0
1
2
0 50 100 150 200 250 300 350 400 450 500
Moment (N-m/m)
Deg
rees
of R
otat
ion
Ultimate Moment at Failure = 667 N-m/m
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Designing a Connection With a Coped Flange*Examples Illustrated in Imperial
When designing a connection with a coped flange, block shear must be checked to insure that theconnection is adequate. The two factors affecting block shear are the transverse tensile and shearstrengths of the web section of the beam. Values for transverse tensile strength and in-plane shearstrengths can be located in the Material Properties Sheets of Pultex® Fiber Reinforced Polymer StructuralProfiles and Pultex® SuperStructural Profiles for Wide Flange Sections and I-Sections.
The procedure for determining the block shear (Vblock) allowable load is as follows:
1. Determine the in-plane shear and tensile strengths of the beam profile with which you are designing. (Material Properties Sheets of Pultex®Fiber ReinforcedPolymer Structural Profiles and Pultex® SuperStructural Profiles for Wide FlangeSections and I-Sections).
2. Determine the shear length (Lv) of the bolted connection. (Reference Figure 2.)3. Determine the net tensile length (Lt) of the bolted connection. (Reference Figure 2.)4. Apply the proper safety factors to the connection. (Use “2” for demonstration.)
Determine if Vblock allowable is greater than the design load.
Example
Design a framed connection using a Pultex® SuperStructural 6" x 6" x 3/8" Wide Flange Section copedinto an Pultex® SuperStructural 8" x 8" x 3/8" beam. The holes are 9/16" to accommodate 1/2" bolts.(Reference Figure 1.)
1. Referencing the Material Properties Sheets for Pultex® SuperStructural Profiles, the in-planeshear and tensile strengths of the web section are 7,000 psi and 10,500 psi, respectively.
2. Calculate the shear length (Lv) of the Figure 1. connection, i.e., (.130 + 2 + 1.25 -9/16 - 9/32) = 2.53"
3. Calculate the net tension length (Lt) of the above connection, i.e., (1.5 –.5(9/16)) =1.219"
4. Apply the proper safety factors to the transverse shear and tensile strengths, i.e., (Allowable shear = 7,000/2 = 3,500 psi)(Allowable tension = 10,500/2 = 5,250 psi)
5. Calculate the allowable shear force for block shear Vblock i.e., (5000(1.219)(.375) + 3,500(2.53)(.375) = 5,720 lbs.
Coped Flange ConnectionFig 1. Fig 2.
6. Compare Vblock allowable to the design load.3
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Endnotes
1 Task Committee on Design of the Structural Plastics Research Council of the Technical Council
on Research of the American Society of Civil Engineers. Structural Plastics Design Manual
Vol. 2. New York: American Society of Civil Engineers, 1984, pg. 732.
2 Task Committee on Design of the Structural Plastics Research Council of the Technical Council
on Research of the American Society of Civil Engineers. Structural Plastics Design Manual
Vol. 1. New York: American Society of Civil Engineers, 1984, pg. 439.
3 Bank, L.C. “Bolted Connections for Pultruded Frame Structures.” Diss. The Catholic