NTA IM 014 TIP 02 SIP Design Examples Cover 2010-10- · PDF fileNTA IM 014 TIP 02 SIP Design...

NTA, INC. • 305 NORTH OAKLAND AVENUE • P.O. BOX 490 • NAPPANEE, INDIANA 46550 PHONE: 574-773-7975 WEB: WWW.NTAINC.COM FAX: 574-773-2732 NTA IM 014 TIP 02 SIP Design Examples Cover 2010-10-13b.doc

NTA IM 14 TIP 02

SIP PANEL DESIGN EXAMPLES USING NTA IM 14 TIP 02 SIP DESIGN GUIDE

AND LISTING REPORT DATA

INTRODUCTION It is intended that this document be used in conjunction with competent engineering design, accurate fabrication, and adequate supervision of construction. NTA, Inc. does not assume any responsibility for error or omissions in this document, nor for engineering design, plans or construction prepared from it. It shall be the final responsibility of the designer to relate design assumptions to the reference design values and to make design adjustments appropriate to the end use. A summary of the notation used in this document is provided on the last page.

1 Maximum Allowable Transverse Uniform Load

2 Cladding Wall Panel Under Transverse Wind Load

3 Cladding Wall Panel Under Transverse Wind Load, Shear Strength Contribution of Fasteners Considered

4 Roof Panel Under Transverse Load

5 Maximum Allowable Axial Load

6 Maximum Allowable Axial Load Considering Increased Design Eccentricity

7 Exterior Wall Subjected to Axial, Transverse and Racking Loads

COMMENTS, QUESTIONS AND ERROR REPORTING All efforts have been made to ensure the accuracy of this document; however, if errors are found please contact Eric Tompos, P.E., S.E. via email at [email protected].

NTA, INC. • 305 NORTH OAKLAND AVENUE • P.O. BOX 490 • NAPPANEE, INDIANA 46550 PHONE: 574-773-7975 WEB: WWW.NTAINC.COM FAX: 574-773-2732 NTA IM 014 TIP 02 SIP Design Examples Cover 2010-10-13b.doc

NTA IM 14 TIP 02

NOTATION

Except where otherwise noted, the symbols used in this document have the following meanings: ∆ = Total deflection due to transverse load (in.) ∆LT = Total immediate deflection due to the long-term component of the design load (in.) ∆b = Deflection due to bending (in.) ∆c = Deflection of core under concentrated load applied to facing (in.)

∆i = Total immediate deflection due to application of a single design load acting alone (in.) ∆s = Deflection due to shear (in.) ∆2nd = Total immediate deflection considering secondary (P-delta) effects (in.) Α = Total cross sectional area of facings (in.2/ft) Av = Shear area of panel. For symmetric panels ( )chAv += 6 (in.2/ft)

c = Core thickness (in.) Ce = Eccentric load factor, Section CFv = Size factor for shear, Section 4.4.3 Cv = Shear support correction factor

e = Load eccentricity, measured as the distance from the centroid of the section to the line of action of the applied load (in.)

Eb = SIP modulus of elasticity under transverse bending (psi) Ec = Elastic modulus of core under compressive load (psi) Ef = Elastic modulus of facing under compressive load (psi) Fc = Allowable facing compressive stress (psi) Ft = Allowable facing tensile stress (psi) Fv = Allowable shear stress (through thickness) (psi) Fvip = Allowable shear load (in-plane) (plf)

G = SIP shear modulus (psi) h = Overall SIP thickness (in.) ho = Reference SIP thickness for size correction factors (in.) I = SIP moment of inertia (in.4/ft) If = Facing moment of inertia (in.4/ft) Kcr = Time dependent deformation (creep) factor for a specific load type, Section A3.5.3 L = Span length (ft) Lv = Shear span length (ft) m = Shear size factor exponent M = Applied moment (in.-lbf/ft) P = Applied axial or concentrated load (lbf/ft.) Pcr = Allowable axial load (lbf/ft) r Radius of gyration (in.) S = SIP section modulus for flexure under transverse loads (in.3/ft) V = Applied shear force (through thickness) (lbf) Vip = Applied shear force (in-plane) plf w = Uniform transverse load (psf) yc = Distance from the centroid to the extreme compression fiber (in.) β = Parameter of relative stiffness

Support Configuration:Support Spacing, L = 12.0 ft ctc

Bearing Width, l b = 1.5 in.

SIP Section Properties: SIP Material Properties:Overall Thickness, h = 4.625 in. Facing Tensile Strength, F t = 495 psi

Core Thickness, c = 3.75 in. Facing Compressive Strength, F c = 580 psiFacing Area, A f = 10.5 in.2 SIP Bending Modulus, E b = 658800 psiShear Area, A v = 50.3 in.2 SIP Shear Modulus, G = 405 psi

Moment of Inertia, I = 46.0 in.4 SIP Shear Strength, F v = 5 psiSection Modulus, S = 19.9 in.3 Shear Reference Depth, h o = 4.625 in.

Shear Depth Exponent, m = 0.86

Design Calculations:Section Properties

Shear AreaOK

OK

Moment of InertiaOK

OK

Section ModulusOK

OK

DESIGN EXAMPLE 1: MAXIMUM ALLOWABLE TRANSVERSE UNIFORM LOAD

Considering the SIP section properties and material properties listed below, calculate the tabulated maximum allowable uniform load for a 4.625-in. thick (overall) SIP panel having a 12-ft span under test support conditions. Consider deflection limits of L/180, L/240 and L/360. Assume a solid end member is provided to obviate crushing at the supports.

While the section propertes are typically tabulated in the report, the sections properties are calculated herein to explain the relatedassumptions. All properties herein are based on the assumption that the facings carry all of the flexural stress and the core carriesall of the shear stress.

The shear 'area' is calculated considering the horizontalshear stress occuring at the core-to-facing interface. Theequation shown assumes a symetric SIP. The values ismultiplied by 12 to provide the result on a per foot basis.

The moment of inertia is calculated considering the facingsonly using the parallel axis theorem. The stiffness of thefacings about their own centroid is neglected. The equationshown assumes a symetric SIP.

NTA SIP Design Guide 2009 Example 1 2011-10-11.xlsPage 1 of 3

10/14/2011 1:46 PM

Flexural StrengthAllowable Moment (tension/compression)

Solve for Uniform LoadOK

OK

Shear StrengthAllowable Shear

Solve for Uniform Load

Size Adjustment FactorOK Design Guide equation 4.4.3.

Support Adjustment FactorC v = 1.00

Allowable Shear StrengthOK Design Guide equation 4.4.2.

OK

Solve for Uniform LoadOK

Design Guide section 4.4.4.2. Bearing is provided on facingopposite the applied load; therefore C v =1.0.

Design Guide equation 4.3.1a. The allowable facing tensile and compressive stress must be considered, whichever is lesser.

Design Guide equation 4.4.5.(b). Shear taken at h away from theface of the support as shown in Design Guide section 4.4.5(a).Note that the span used for deflection calculations is taken as thecenter-to-center spacing between supports whereas the span forshear strength is taken as the clear span between bearing points (3-in. less).

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Deflection at Mid-SpanCalculated Deflection

Solve for Uniform Load

OK

Considering Various Deflection Limits

∆lim 180 240 360w (psf) 30.8 23.1 15.4

Overall Result

∆lim 180 240 360w (psf) 30.8 23.1 15.4

Design Guide equation 4.5.2a. Assumes a simply supported,uniformly load member.

The uniform load calculated here is expressed as uniformload per deflection limit ratio.

The allowable value corresponds to the smallest value consideringall limit states. The calculated values match the values provided inthe SIPA code report (NTA SIPA120908-10)

Allowable Uniform Transverse Load

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Support Configuration: LoadsSupport Spacing, L = 10.0 ft ctc Transverse Wind Load, w = 20.0 psf (C&C)

Deflection Limit, L/ 180



Moment of Inertia, I = 96.5 in.4 SIP Shear Strength, F v = 5.0 psiSection Modulus, S = 29.7 in.3 Shear Reference Depth, h o = 4.5 in.


Design Calculations:Flexural StrengthApplied Moment

OK

Allowable Moment (tension)

OKStress Ratio = 0.20 OK

Allowable Moment (compression)


Design Guide equation 4.3.1a. For illustration purposes, the moment based on the compressive strength of the facing is calculated.

DESIGN EXAMPLE 2: CLADDING WALL PANEL UNDER TRANSVERSE WIND LOAD

Verify the adequacy of the 6.5” SIP panel below using the SIP properties from NTA listing report SIPA120908-10. The wall is simply supported between spline supports (zero bearing condition) and subjected to 20 psf transverse wind load (C&C). The strong-axis of the panel is oriented perpendicular to the supports. Consider a deflection limit of L/180.

The applied moment is calculated considering a one foot width of panel. Note: because per-foot units are assumed related units are dropped.

Design Guide equation 4.3.1a. By inspection tension moment governs over compression moment (Ft<Fc).

NTA SIP Design Guide 2009 Example 2 2011-10-11.xls Page 1 of 210/14/2011 1:47 PM

Shear StrengthApplied Shear

OK



Allowable Shear Strength

OKOK

Stress Ratio = 0.94 OK

Deflection at Mid-SpanDeflection Limit

OK

Calculated Deflection

OK

OK

OK

Deflection Ratio = 0.18 OK

Overall ResultThe panel is adequate for the application, shear strength of the core governs the design (stress ratio = 0.94)

Design Guide equation 4.5.2a. Applied wind pressure is reduced inaccordance with IBC, Table 1604.3, Footnote ‘f’. Accordingly, thewind load is taken as 0.7 times the C&C loads for the purpose ofdetermine deflection limits.

The panel has adequate stiffnes to meet the specified deflection limit.

Design Guide equation 4.4.5.(b). Shear taken at the face of support.

Design Guide equation 4.4.2. The core has adequate shear strength for support conditions provided.

Design Guide section 4.4.4.2. Bearing is not provided on facingopposite the applied load; therefore Cv<1.0. From listing reportTable 4, footnote 1, Cv=0.4.

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Support Configuration: LoadsSupport Spacing, L = 10.0 ft ctc Transverse Wind Load, w = 30.0 psf (C&C)

Deflection Limit, L/ 180





Design Calculations:Flexural StrengthApplied Moment

OK




OK




OKOK

Stress Ratio = 1.41 NG

Maximum Allowable Shear Strength

OKOK


DESIGN EXAMPLE 3: CLADDING WALL UNDER TRANSVERSE WIND LOAD,

Verify the adequacy of the SIP described in Design Example 2, except the transverse wind pressure is increased to 30 psf transverse wind load (C&C).


SHEAR STRENGTH CAPACITY OF FASTENERS CONSIDERED




Design Guide equation 4.4.2. The core of the SIP alone is inadequate. Must consider fasteners in shear strength calculation.

Design Guide equation 4.4.2. Assume C v = 1.0 and determine maximum core shear strength. Refer to Design Example 1 for derivation of C Fv .


Required Connection Strength

OK

Allowable Connection Strength

Consider 0.131" x 2.5" (8d) nails at 6" ocFacing-to-Plate, each side, top-and-bottom

W' = fastener withdrawal or pull-through strength, whichever is lessorCs = spacing adjustment factor, calculated as 12 divided by the typical fastener spacingCP =

Cc =

NDS Withdrawal Strength, W' = 67.2 lbfCs = 2.0CP = 0.5Cc = 0.88

OK

Overall Assembly Shear Strength

OK



OK


OK

OK

OK


Overall ResultThe panel is adequate for the application provided that the facing-to-plate connection is as specified, shear strength of the core governs the design (stress ratio = 0.91)


Consider the adequacy of a typical minimum SIP-to-plate connection. First the withdrawal strength from the plate must be calculated using the NDS (2005 Edition). Fastener pull-through must also be considered; however, testing has shown that pull-through does not occur for the proposed connection. The NDS does not provide guidance on designing such connections, based on experiental observation several considerations need to be made to ensure the performance of the connection. These considerations are summarized in the equation provided.


Connection must be design for the difference between the applied load and the shear strength considering the actual support conditions, Cv = 0.4.

prying adjustment factor, from statics based on the solid member width and where the fastener is installed within that width. For fasteners installed in the center of the solid block the value may conservatively be taken as 0.5

sheathing continuity factor, from statics of a beam continuous over multiple supports, calculated as the inverse of the maximum reaction coefficient. May be taken as 0.88 for 4 or more fasteners installed within the width of a single SIP

For the proposed connection the adjustment factors are as shown.

The total strength of the assembly is simply the summation of the core shear strength and the shear strength contribution of the fasteners. It must be noted that the shear strength cannot exceed the maximum shear strength calculated when Cv = 1.0.

'WCCCV cpsf =

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Support Configuration: Loads:Support Spacing, L = 12.0 ft ctc Transverse Wind, WC&C = 25.0 psf (C&C)

Bearing Width, l b = 1.5 in. W = 18.0 psf (MWFRS)Deflection Limit, L/ 240 (live load only) Dead Load, D = 10.0 psfDeflection Limit, L/ 180 (total load) Roof Live, Lr = 20.0 psf

Snow, S = 30.0 psf




Shear Depth Exponent, m = 0.86Core Compressive Modulus, E c = 360 psiFacing Bending Stiffness, E f I f = 78000 lbf-in.2

Core Compressive Strength, F cc = 14.0 psiAngle of Dispersion, k = 0.055

Design CalculationLoad Cases

1 D+Lr2 D+S3 D+0.75(S+Wp) [WFRS]4 0.6D+Wn [C&C]5 D+Wp [C&C]

DESIGN EXAMPLE 4: ROOF PANEL UNDER TRANSVERSE LOAD

Verify the adequacy of the 12.25” SIP panel below using the SIP properties from NTA listing report SIPA120908-10. Assume the panel is simply supported between supports. No structural spline is provided over the supports. The panel is subjected to the load listed below. The strong-axis of the panel is oriented perpendicular to the supports. Consider a total load deflection limit of L/180 and a live load deflection limit of L/240. Assume 1.5-inches of bearing is provided at each end.

Load combinations are taken from ASCE 7-05. In accordance with ASCE 7, C&C wind loads are used when acting alone and MWFRS wind loads are used when wind loads are combined with other transient loads. The load cases D+0.75(S+Lr) and D+0.75(Lr+Wp [MWFRS]) are not considered based on judgment considering that roof live load, Lr, can not coincide with a design snow or wind event.


Flexural StrengthApplied Moment

OK

OK

Allowable Moment (tension/compression)



OK

OK



Allowable Shear StrengthDesign Guide equation 4.4.2.

OK


Bearing StrengthApplied Reaction

OK

Allowable ReactionOK

OK


Design Guide section 4.4.4.2. Bearing is provided on facingopposite the applied load; therefore C v =1.0.

The applied moment is calculated considering a one foot width of panel. Note: because per-foot units are assumed related units are dropped. By inspection D+S governs (50 psf) for strength design.

Design Guide equation 4.3.1a. By inspection tension moment governs over compression moment (F t <F c ) .

Design Guide equation 4.4.5.(a). Shear taken at a distance h from the face of support.

These provisions for bearing strength are not in the 2010 Design Guide, but will appear in subsequent revisions.


Deflection at Mid-SpanDeflection Limit (live load only)

OK

Deflection Limit (total load)OK


OK

OK

OK

LoadType Kcr

W 1.0D 4.0Lr 1.0S 3.0

Load Case1 1.0W (C&C)2 1.0Lr3 3.0S

Maximum ∆ LL

1 4.0D+1.0Lr2 4.0D+3.0S3 4.0D+0.75(3.0S+1.0Wp)4 1.0x0.6D+1.0Wn (C&C)5 4.0D+1.0Wn (C&C)

Maximum ∆TL

Ratio ∆ LL = 0.85 OK

Ratio ∆ TL = 0.93 OK

0.3280.740

Panel has adequate long-term stiffness to meet specified deflection limit.

0.3420.7400.6870.066

0.1140.5130.513

Long-TermDeflection

(in.)0.100

0.0570.1140.171

Applied C&C wind pressure is reduced in accordance with IBC, Table 1604.3, Footnote ‘f’. Accordingly, the wind load is taken as 0.7 times the C&C loads for the purpose of determine deflection limits.

(in.)Deflection

2518

0.1000.103

S

Load Type (psf)Magnitude

102030

W (C&C)W (MWFRS)

DLr

Design Guide equation 4.5.2a. Determine the deflection fora unit load (w = 1.0). The deflection for each loading maythen be determined by multiplying the load magnitude by thisresult.

Deflection for each load type, calculated above, is multiplied by the creep coefficient, K cr , to determine the total deflection for each load case. Where deflection is counteracting wind loads (case 4) the creep coefficient is taken as unity. It must be noted that the Design Guide assigns K cr = 1.0 to roof live load, L r , where this live load is a short duration construction load. Snow laods and promenade roofs subjected to occupancy live loads should be designed for K cr

= 3.0.

The Kcr factors are from the Design Guide Table 1. The Kcr value is the fractional deflection ratio, or the ratio of the long-term to inmmediate deflection.

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Bearing Deflection at SupportBearing Deflection

Calculate BetaOK

Solve for Bearing StiffnessOK

Calculate Bearing DeflectionMagnitude

(psf)2518102030

Load Case1 4.0D+1.0Lr2 4.0D+3.0S3 4.0D+0.75(3.0S+1.0Wp)5 4.0D+1.0Wn (C&C)

Maximum ∆bearing

∆ max = 0.125 in.

Ratio ∆ TL = 3.07 NG

Overall Result

60120

0.384

Panel is NOT adequate, bearing deflection governs (deflection ratio=3.07). Panel may be used if solid spline provided at each end to prevent crushing. It should also be noted that if the panel is continuous across the support the breaing deflections will be reduced by 50%

0.3840.3570.192

Comparing the calculated bearing deflection with the allowable bearing deflection of 1/8-inch the panel has inadequate bearing strength.

Bearing deflections are calculated considering creep using the total load cases as before except Case 4 has been omitted since the loads are counteracting.

Long-TermDeflection

(in.)0.177

LrS 180

0.0590.089

Load TypeW (C&C)

W (MWFRS)D

Reaction (lbf)150108

Deflection Similar to midspan deflections, the deflection of each load individually is determined. Unlike the calculation for shear, the load used in the bearing check corresponds to the support reaction. In this example the support reaction is equal to wL/2.

(in.)0.074

No solid structural spline is provided at the end of the panel, as aresult, Design Guide section 4.6.2.2 applies. With simple supports,the concentrated load is applied at the end of the panel (section4.6.2.2 Case A), as a result, equation 4.6.2.2a applies.

If the bending stiffness of the facing is not provided the the SIP manufacturer, bending stiffness values for OSB are provided in APA D510 Panel Design Specification. Table 4A of this document gives the bending stiffness.

The bearing stiffness is calculated in pounds per inch.

0.0530.030

34 βffc IE

P=∆


Support Configuration:Wall Height, L = 12.0 ft ctc




Radius of Gyration, r = 3.03 in. Shear Depth Exponent, m = 0.86Extreme Fiber Distance, y c = 3.25 in.

DESIGN EXAMPLE 5: MAXIMUM ALLOWABLE AXIAL LOAD

Considering Table 5 of NTA listing report SIPA120908-10, verify the tabulated maximum allowable axial load for a 6.5” SIP panel having a 12-ft height.


Design Calculations:Axial Strength Under Eccentric LoadsEccentric Load Factor

Minimum Eccentricity

OK

Seed Value

OK

Calculate Eccentricity Factor

OK

OK

Calculate Axial Load

OK

IterateTrial Axial Eccentricity Calcualted

Load Factor Axial LoadIteration (lbf) Ec (lbf)

1 3045 0.627 38182 3818 0.595 36223 3622 0.603 36744 3674 0.601 36605 3660 0.602 36646 3664 0.601 36637 3663 0.601 3663

Global Buckling Strength

OK

OK

Overall Result

P calc = 3663 lbfP report = 3660 lbf

A recommended seed value for the iteration process is 1/2 the axial strength of the facings.

Using the seed value, the eccentricity factor can be calculated.

From the Design Guide section 5.1.1 the minimum eccentricity, e, is taken as h/6

Design Guide equation 5.1.1b. Because the axial load, P, is a term in the eccentricity factor equation an iterative processes is required to find the maximum value.

Comparing the resulting axial load to the seed value, the loads are not equal; therefore, additional iterations are required.

Repeat the calculation process using the previous result for the subsequent calculation. As shown in the table, the trial value converges with the calculated value after seven iterations.

Design Guide equation 5.1.2. The critical buckling load can be solved directly.

The resulting maximum allowable load is the lesser of the two limit states considered. The values in the SIPA report are rounded down to the nearest 10 lbf.

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Support Configuration:Wall Height, L = 12.0 ft ctc

Design Eccentricity, e = 3.25 in.




Radius of Gyration, r = 3.03 in. Shear Depth Exponent, m = 0.86Extreme Fiber Distance, y c = 3.25 in.

DESIGN EXAMPLE 6: MAXIMUM ALLOWABLE AXIAL LOAD

Determine the maximum allowable axial load for a 6.5” SIP panel having a 12-ft height where the supported level is hung from the side of the SIP panel (i.e. load eccentricity, e = h/2).

CONSIDERING INCREASED DESIGN ECCENTRICITY


Design Calculations:Axial Strength Under Eccentric LoadsEccentric Load Factor

Minimum Eccentricity

OK

Seed Value

OK

Calculate Eccentricity Factor

OK

OK

Calculate Axial Load

OK

IterateTrial Axial Eccentricity Calcualted

Load Factor Axial LoadIteration (lbf) Ec (lbf)

1 3045 0.359 21862 2186 0.391 23823 2382 0.384 23384 2338 0.386 23485 2348 0.385 23466 2346 0.385 23467 2346 0.385 2346

Global Buckling Strength

OK

OK

Overall ResultThe resulting maximum allowable considering balloon framing is 36% less than the axially loaded capacity.

P calc = 2346 lbfP report = 3660 lbf

From the Design Guide section 5.1.1 the minimum eccentricity, e, is taken as h/6. This is less than the design eccentricity so the design eccentricity will be used.

Design Guide equation 5.1.1b. Because the axial load, P, is a term in the eccentricity factor equation an iterative processes is required to find the maximum value.

Comparing the resulting axial load to the seed value, the loads are not equal; therefore, additional iterations are required.

Repeat the calculation process using the previous result for the subsequent calculation. As shown in the table, the trial value converges with the calculated value after seven iterations.

Design Guide equation 5.1.2. The critical buckling load can be solved directly. The result is identical to the value found in Example 5.

A recommended seed value for the iteration process is 1/2 the axial strength of the facings.

Using the seed value, the eccentricity factor can be calculated.

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Support Configuration: LoadsWall Height, L = 12.0 ft ctc Transverse Wind Load, w = 25.0 psf (C&C)

Deflection Limit, L/ 180 w = 18.0 psf (Roof, Zone 2E & 3E)w = 18.0 psf (Wall, Zone 1E & 4E)w = 12.0 psf (Wall, Zone 5 & 6)

Dead, D = 10.0 psfRoof Live, Lr = 20.0 psf

Snow, S = 40.0 psfFloor Live, L = 40.0 psf

Racking Shear, V = 210.0 plf




Radius of Gyration, r = 3.03 in. Shear Depth Exponent, m = 0.86Extreme Fiber Distance, yc = 3.25 in. Assembly Racking Capacity, F vip = 380 plf

DESIGN EXAMPLE 7: EXTERIOR WALL SUBJECTED TO

Verify the adequacy of the 6.5” SIP panel below using the SIP properties from NTA listing report SIPA120908-10. Assume the panel is simply supported between supports. The panel is subjected combined axial, transverse and racking loads as listed below. The wall is 12-ft tall and has a tributary roof and floor width of 14-ft. The strong-axis of the panel is oriented perpendicular to the supports. Consider a total load deflection limit of L/180.

AXIAL, TRANSVERSE AND RACKING LOADS


Loads and Load CombinationsAxial LoadsLoads from Roof (e = 0.0-in.)

Tributary Width, T = 14.0 ftD = 140 plfLr = 280 plfS = 560 plf

W = 252 plf (MWFRS)

Loads from Floor (e = 3.25-in.)Tributary Width, T = 14.0 ft

D = 140 plfL = 560 plf

Condition 1: Axial Load OnlyLoadCase

1 D2 D+L3 D+Lr4 D+S5 D+0.75(L+Lr)6 D+0.75(L+S+Wp)

Axial Load Net EccentricityLoad LoadCase P

D(roof) 140D(floor) 140L(floor) 560

Σ 840

OK

OK

Uniform axial loads are calculated by multiplying the applied uniform loads by the tributary width. The design eccentricities associated with the loads are also noted. From accepted engineering practice, the load eccentricities are applied at the top of a column (floor load) with the bottom of the column assumed to bear concentrically (roof load).

Load Effect(plf)

840

1309

In this example six design conditions must be checked for an optimal design; each check considers different combinations of loads. The six cases arise from the load combinations provided in ASCE, the code requirement to assess the structural element for the worst loading condition, and the orientation of the element with respect to the applied wind loads.

280

910840560

Conservative (worst-case) assumptions can be made to reduce the number of load combinations; however, because SIPs are generally subjected to combined loads, such assumptions produce very conservative designs. The load cases shown in this example are not all inclusive and generally more load cases must be considered.

The design eccentricity considering all loads applied to the panel can be calculated by taking the weighted average of the worst case condition for eccentricity. In this example, the worst case eccentricity occurs when the full floor load (L+D) acts concurrently with roof dead only. Conservatively, the maximum eccentricity could be used.-- 2275

3.25

e x P(in.-lbf)

04551820

Eccentricity, e(in.)

03.25


Condition 2: Transverse Load OnlyW = 25.0 psf (C&C)

Condition 3: Combined Axial and Transverse (max bending)LoadCase

Transverse1 W 1,4 (MWFRS)

Concurrent Axial1 D2 D+Wp

Condition 4: Combined Axial and Transverse (max axial)LoadCase

Transverse1 0.75W 1,4 (MWFRS)

Concurrent Axial1 D+0.75(L+Lr)2 D+0.75(L+S+Wp)

Condition 5: Combined Axial, Transverse and Racking (max bending)LoadCase

Racking1 V

Concurrent Transverse1 W 5,6 (MWFRS)

Concurrent Axial1 D

Condition 6: Combined Axial, Transverse and Racking (max axial)LoadCase

Racking1 0.75V

Concurrent Transverse1 0.75W 5,6 (MWFRS)

Concurrent Axial1 D+0.75(L+Lr)2 D+0.75(L+S+Wp)

The transverse load combination checks the wall subjected to transverse C&C wind loads only. Generally, the wind pressure in this case corresponds to the maximum C&C pressure.

LoadEffect

The maximum bending case occurs when the only transient load is wind. The wind pressure considered is generally the maximum of the windward or leeward MWFRS wind pressure.

In this case all transient loads are applied and all transient loads are reduced by 0.75 as permitted by ASCE 7.

LoadEffect

The racking load cases are very similar to the previous two combined load cases except, the wind pressure acting concurrently with the other loads can be reduced to the maximum of Zones 5 & 6 (ASCE 7, Method 2) because the shearwall must be oriented parallel to the transverse pressures used in the previous load cases. This procedure, while rigorous, maximizes the available racking strength of the panel.

The maximum axial case occurs when all transient loads are applied—including wind load. In this case, all transient loads are taken times 0.75 as permitted in ASCE 7. The wind pressure considered is generally the maximum of the windward or leeward MWFRS wind pressure.

LoadEffect

12.0

13.5

158

280

210.0

532

9101309

LoadEffect

18.0

280

1309910

9.0


Design CalculationsCondition 1: Axial Load OnlyEccentric load factor

Design Guide equation 5.1.1b.

OK

OK

Axial Strength

OK


Buckling Strength Design Guide equation 5.1.2.

OK



Condition 2: Transverse Load OnlyFlexural StrengthApplied Moment

OK




OK




OKOK


Maximum Allowable Shear Strength

OKOK


Required Connection Strength

OK


Design Guide equation 4.4.2. The core of the SIP alone is inadequate. Must consider fasteners in shear strength calculation.

Design Guide equation 4.4.2. Assume C v = 1.0 and determine maximum core shear strength. Refer to Design Example 1 for derivation of C Fv .

Connection must be design for the difference between the applied load and the shear strength considering the actual support conditions, Cv = 0.4.





Allowable Connection Strength

Consider 0.131" x 2.5" (8d) nails at 6" ocFacing-to-Plate, each side, top-and-bottom

W' = fastener withdrawal or pull-through strength, whichever is lessorCs = spacing adjustment factor, calculated as 12 divided by the typical fastener spacingCP =

Cc =

NDS Withdrawal Strength, W' = 67.2 lbfCs = 2.0CP = 0.5Cc = 0.88

OK

Overall Assembly Shear Strength

OK



OK


OK

OK

OK


prying adjustment factor, from statics based on the solid member width and where the fastener is installed within that width. For fasteners installed in the center of the solid block the value may conservatively be taken as 0.5

sheathing continuity factor, from statics of a beam continuous over multiple supports, calculated as the inverse of the maximum reaction coefficient. May be taken as 0.88 for 4 or more fasteners installed within the width of a single SIP

For the proposed connection the adjustment factors are as shown.

The total strength of the assembly is simply the summation of the core shear strength and the shear strength contribution of the fasteners. It must be noted that the shear strength cannot exceed the maximum shear strength calculated when Cv = 1.0.


Consider the adequacy of a typical minimum SIP-to-plate connection. First the withdrawal strength from the plate must be calculated using the NDS (2005 Edition). Fastener pull-through must also be considered; however, testing has shown that pull-through does not occur for the proposed connection. The NDS does not provide guidance on designing such connections, based on experiental observation several considerations need to be made to ensure the performance of the connection. These considerations are summarized in the equation provided.


'WCCCV cpsf =

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Condition 3: Combined Axial and Transverse (max bending)Eccentric load factor

OK

OK

Axial Strength

OK


Buckling Strength


Applied Moment

OK


OK

OK

OK

Total Deflection (considering P- δ )Design Guide equation 4.5.2b.

OK

Total Moment (considering P- δ )

OK

OK

Alloable Moment (compression)

OK

Combined Loads

OK


Using the same buckling capacity as calculated for the axial only case.

Design Guide equation 5.1.1b. Because the applied axial load has changed, the eccentric load factor also changes.

Design Guide equation 4.5.2a.

Design Guide equation 7.1.1.c. Note that the first term in the moment equation is simply 1/8 times 12 which results from converting feet to inches.

Design Guide equation 4.3.1b. While tension governs flexural bending strength, the interaction equation is concerned with combined compressive loads (tension is counteracting), as a result the maximum compressive moment must be calculated.Design Guide equation 7.1.1a. Equations 7.1.1b does not govern by inspection.

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Condition 4: Combined Axial and Transverse (max axial)Eccentric load factor

OK

OK

Axial Strength

OK


Buckling Strength


Applied Moment

OK


OK

OK

OK


OK


OK

OK


OK

Combined Loads

OK







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Condition 5: Combined Axial, Transverse and Racking (max bending)Eccentric load factor

OK

OK

Axial Strength

OK


Buckling Strength


Applied Moment

OK


OK

OK

OK


OK


OK

OK


OK

Combined Loads

OK







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Condition 6: Combined Axial, Transverse and Racking (max axial)Eccentric load factor

OK

OK

Axial Strength

OK


Buckling Strength


Applied Moment

OK


OK

OK

OK


OK


OK

OK


OK

Combined Loads

OK







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Full Loads/No Load Cases: Combined Axial, Transverse and RackingEccentric load factor

OK

OK

Axial Strength

OK


Buckling Strength


Applied Moment

OK


OK

OK

OK

Total Deflection (considering P- δ )

OK


OK

OK


OK

Combined Loads

OK


A worst case design which applies all loads at full magnitude, without detailed consideration of ASCE 7 load combinations, is considered in this section for comparison with a more detailed analysis.

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Overall Result

Design ConditionCondition 1: Axial Load Only

Condition 2: Transverse Load OnlyCondition 3: Combined Axial and Transverse (max bending)

Condition 4: Combined Axial and Transverse (max axial)Condition 5: Combined Axial, Transverse and Racking (max bending)

Condition 6: Combined Axial, Transverse and Racking (max axial)Full Loads/No Load Cases: Combined Axial, Transverse and Racking

0.650.801.001.65

Stress Ratio0.460.910.41

The wall panel under combined loads is adequate (maximum combined stress ratio = 1.0). While the combined load procedure may seem tedious, a worst case design (applying all loads at full magnitude without detailed consideration of ASCE 7 load combinations) would have resulted in a stress ratio of 1.65 under combined loads, which would required a thicker panel or may necessitate an alternate framing method.


NTA IM 014 TIP 02 SIP Design Examples Cover 2010-10- · PDF fileNTA IM 014 TIP 02 SIP Design...

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