codes_CDN.pdf

7/30/2019 codes_CDN.pdf

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6.5.5.7 National Building Code of Canada

This option calculates the seismic response according to the National Building Code of Canada (NBC) - 1995,Chapter 4.1.9 and Commentary J.

The calculations are carried out according to the procedure outlined in Commentary J, paragraph 44.

The program uses the spectrum in Figure J-3 and Table 4.1.9.1.A of Commentary J to calculate Vdyn, thedynamic base shear.

The dynamic responses may be scaled by the ratio V/Vdyn, where:

V = (Ve/R)U 4.1.9.1(4)

Ve = v S I F W 4.1.9.1(5)

where:

R = Force modification factor (4.1.9.1-8 and Table 4.1.9.1.B)

U = Level of protection factor = 0.6 (4.1.9.1-4)

v = Zonal velocity ratio

S = Seismic response factor (Table 4.1.9.1.A)

I = Seismic importance factor (4.1.9.1-10)F = Foundation factor (4.1.9.1-11 and Table 4.1.9.1.C)

W = Sustained load (Appendix A)

Define the required factors:

Codes

Select one of the codes displayed in the pull-down menu.

Number of modes

Specify the minimum number of mode shapes to be used in the seismic analysis. (Higher mode shapes usuallyinfluence the results only slightly in standard models).

Note that the program also calculates the significant number of modes according to Commentary J, item 44(d),and uses the maximum of the 'significant' no. of modes and the value entered here.


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Foundation factor (F)

Specify the Foundation factor, F, according to Table 4.1.9.1.C.

Importance factor (I)

Specify the seismic importance factor according to Section 4.1.9.1- (10).

Note: to limit the product (S I) as specified in 4.1.9.3 (3a), modify v, the zonal velocity ratio, accordingly.

Force modification factor (R)

Specify the force modification factor, R, according to Table 4.1.9.1.B.

Zonal velocity ratio (v)

Specify the zonal velocity ratio, normalized to a spectral velocity of 1.0 m/s. Refer to Appendix C of the NBC.

Note:

The value ofv can be modified to account for the Special Provisions in Section 4.1.9.3:

to increase the value ofV by 50% for buildings more than 60 m. in height, as specified in 4.1.9.3 (2),

increase v, the zonal velocity ratio, by 50%. to limit the product (S I) as specified in 4.1.9.3 (3a), modify v accordingly.

Za/Zv

Specify the ratio ofZa/Zv required to calculate S, the seismic response factor, according to Table 4.1.9.1.A.Refer to Appendix C of the NBC.

Scaling

No scalingThe program calculates the seismic responses from the dynamic base shear. The Seismic response factor,

S, is based on the fundamental period, T, calculated by the program for each mode shape.

ScalingThe program calculates the seismic responses from the dynamic base shear and modifies them by the ratio

V/Vdyn according to Commentary J, section 44(e), where V is the Minimum lateral seismic force calculatedaccording to the approximate method outlined in 4.1.9.1 (4)-(5).

The Seismic response factor, S, required for the calculation ofV, is based on the fundamental period, T,specified by the user in the dialog box; the value should be calculated from 4.1.9.1 (7).


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A7.17 CSA S16-01

A7.17.1 CSA-S16-01 - Strength of steel

The program allows for design in all grades of steel - a different grade may be assigned to each member.

Fy and Fu for the various shapes and thicknesses are taken from Tables 6.1 to 6.3 in Part Six of the Handbookof Steel Construction.

For user-defined steel types:

Fy (MPa) up to 230 230 to 260 260-300 300-350 350-400 400-480 > 480Fu (Mpa) 380 410 450 480 520 590 Fy + 100

A7.17.2 CSA-S16-01 - Classification of sections

The program strictly adheres to the width-thickness ratios listed in Table 1 for the various section types todetermine section classification and design of members in compression and bending.

for biaxial bending, the program classifies the section according to "minor axis", which always governs.

in a "Combined beam", the program calculates the classification for each segment and uses the worst case. in a tapered beam, the program calculates the classification at each end and uses the worst case.

A7.17.3 CSA-S16-01 - Shear

The design shear strength is calculated from the equations in Chapter 13.4.1.1. The program checks that theactual shear force is less than the factored shear resistance , i.e.

Vr= AwFs

where:

= 0.9

Fs is determined as follows:

the web shear buckling coefficient kv is always assumed equal to 5.34.


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the web area, Aw, is determined according to the following table:

For tapered sections, the program calculates Aw and the design shear strength at 20 intervals along themember length.

A7.17.4 CSA-S16-01 - Axial force - tension

Tension capacity is the minimum of the values calculated from equations 13.2 (a)(i) and 13.2(a)(ii), where the

effective area Ane, is calculated from the tension area reduction factor defined by the user (a default value may

be specified).

The program checks that the axial tension force is less than factored tensile resistance Tr, taken as the leastof the following:

Tr= Ag Fy

and Tr= 0.85 Ane Fu

where:

= 0.9Ag = section area

Ane = net area, i.e. the gross area multiplied by the "tension area reduction factor" defined by the user.

Fu = specified minimum tensile strength according to Tables 6.3, Part Six of the "Handbook of Steel

Construction". (If the user selects a steel grade by entering a value forFy, the program calculates a

value forFu as explained in Strength of Steel).


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A7.17.5 CSA-S16-01 - Axial force - compression

The program checks that the axial compression force is less than the factored axial compression resistance,

Cr, calculated according to the equations in section 13.3.

Class 1, 2 or 3 sections, doubly symmetric, conforming to the requirements of Clause 11 are calculated

according to Clause 13.3.1:

n = 2.24 for WWF shapes and for class H RHS and Pipe sections

n = 1.34 for all other shapes

Sections not covered by Clause 13.3.1:

The factored compressive resistence Cr is calculated according to Clause 13.3.2, based on computed

values forFex, Fey and Fez.

Class 4 Pipe sections

The effective area is calculated according to CSA S136-94 (Cold-formed sections), section 6.6.5, as follows:

Ae = A0 + (Fy/2Fp) (A - A0)

where:

A = unreduced cross-sectional area

Fe = 55E/(KL/r)55Fp = 0.833 Fe

All other Class 4 sections

The reduction factorQs is calculated in accordance with the AISC - LRFD Code.

The program may be instructed to calculate kx and ky, the effective length factors, according to theAlignment Chart in Figure C1 (Commentary) or the values may be input directly by the designer (a defaultvalue of 1.00 is assumed by the program for all members).

The program calculates the slenderness, l, between support points in both local axis directions. Theprogram compares the maximum slenderness to the user-defined limits for compression or tensionaccording to the sign of the axial force in the member.

For double angles and double channels, the program calculates the fastener interval such that theslenderness ratio of the individual component does not exceed the governing slenderness ratio of the built-up section. A minimum of one fastener is assumed.


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For tapered members or combined beams with different properties, the program calculates the exact Euler

buckling load for the member and then finds an equivalent length l1 for a member with the minimum area

which gives the same Euler buckling load. The minimum area and the length l1 are used in all theequations.

A7.17.6 CSA-S16-01 - Bending - laterally supported members

The program checks that the actual moment is less than the factored moment resistance , i.e.

Class 1 and Class 2 sections

M Mr= Zpl fy

Class 3 sections

M Mr= S fy

Class 4 Pipe sections

M Mr= S Fcr

where Fcris calculated in accordance with CSA S136-94 (Cold-formed sections), section 6.4.3.5(c):

Fcr= 0.328 E/ (D/t)

All other Class 4 sections

Reduced section properties are calculated according to section 13.5 (c) of the Code.

Note:The program calculates all result values at 1/10 of span intervals and at points of intermediate supports. Fortapered sections or combined beams with different properties, the program uses the actual section at eachpoint.

A7.17.7 CSA-S16-01 - Bending - laterally unsupported members

The program calculates all result values at: - 1/10 of span intervals, and at points of intermediate supports.

The designer can specify the exact location of intermediate supports for each member.

Lateral-torsional buckling is calculated individually for each segment between intermediate supports, andseparately along the top and bottom flanges; the calculation is done separately for positive moments (supportson bottom flange only are considered) and for negative moments (supports on top flange only are considered).

For tapered members or combined members, the program calculates the lateral-torsional buckling at 20intervals along the member length according to the section at each interval and uses the minimum capacity.

The values for the factored moment resistance, Mr, are determined from Code equations in clause 13.6 , as

follows: doubly-symmetric Class 1 and Class 2 sections:

where:


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L = unbraced portion of the beam

2 = 1.75 + 1.05k + 0.3k55 2.5= 1.00 for cantilevers

k = ratio of end moments, as explained in the Code

doubly symmetric Class 3 and Class 4 sections and channels:

where:

Mu = as defined in (a).

RHS sections: Cw = 0

Pipes: not checked

monosymmetric shapes (T, double angles) Mu is calculated as specified in the AISC-LRFD Code:

angles: calculated as supported only.


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A7.17.8 CSA-S16-01 - Combined bending and axial force

Class 1 sections of I-shaped members (clause 13.8.2)

= 0.6 + 0.4y

All other members, including I shapes other than Class 1 (clause 13.8.3)

The capacity of the member is checked for the following cases, according to clause 13.8.2:

cross-sectional strength

U1x = U1y = 1.0

Cr is based on the value of = 0.0

Mrx, Mry= as defined in clause 13.5, without lateral-torsional buckling

overall member strength

Cris based on the value of k =1.00

Mrx, Mry =as defined in clause 13.5, without lateral-torsional buckling

lateral torsional buckling strength

Cr is based on the value of k entered by the userMrx = as defined in clause 13.6, taking into account lateral-torsional buckling

Mry = as defined in clause 13.5, without lateral-torsional buckling

where:

1 is taken as defined in clause 13.8.5 and all other terms are as defined in clause 13.8.2.

A7.17.9 CSA-S16-01 - Deflections

As the deflection check is a Serviceability check, the program uses the user-defined load combinations forcalculating deflections, but does not multiply the elastic deflections by the load factors.

When checking the maximum deflection along the span of the member, the program ignores the deflection ofthe end joints, except in cantilevers where the maximum deflection is calculated at the free end.

Maximum allowable deflections per member must be entered by the designer (a default value may bespecified).

Note: The deflections calculated in the results are based on the section input in STRAPgeometry. When checking

a different section, the Postprocessor modifies the deflection value by Inew/Iold, where:

Inew = moment-of-inertia of section being checked.

Iold = moment-of-inertia ofSTRAPgeometry property.


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When checking deflections of a "combined" beam, the program uses the length of the entire combined beamand ignores any possible deflection support at the combined nodes. If such supports exists, the allowabledeflection parameter should be modified accordingly.

the program ignores intermediate supports (buckling and lateral-torsional) when checking deflections.

A7.17.10 Joists

The program can select standard joists when the American steel table is specified. For a detailed explanationon the method of calculation for joists, refer to A7.2.

A7.17.11 CSA-S16-01- Torsion

Refer to A7.1.

A7.17.12 CSA-S16 Composite beams

This section details the method used by the program to select steel sections for composite beams accordingto CAN/CSA-S16-01-2005 - Limit States Design of Steel Structures..

The user specifies the topping dimensions, properties and reinforcement, parameters that specify the type ofshear connection, details on short term vs. long term loading, etc. The program then selects the lightest steelsection that provides the required composite section capacity (the topping dimensions are not modified by theprogram during the steel beam selection process).

The program differentiates between areas of positive (sagging) and negative (hogging) moment: positive moments: designed as composite sections negative moments: designed as non-composite steel sections according to CSA S16-01 (with the addition

of reinforcement in the slab, if specified).

A7.17.12.1 Materials

Structural steel:

The structural steel type and grade is specified by the user and the program determines Fyaccording toTable 6-3 in the CISC "Handbook of Steel Construction"

Concrete:

The nominal concrete strength f'c is specified by the user. The program assumes that the topping is

stressed to a uniform compression of 0.6 (0.85 f'c) (17.9.3).

A7.17.12.2 Section types

The program can design the following section types:

A7.17.12.3 Classification

Negative moments:Classification according to Section 11.1 and Table 1.


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Positive moments:Classification according to Table 1:- flange: compression flange assumed supported along entire length and so classification is not

calculated.- web:

where:

h = clear depth of the web

tw = web thickness ( = 2tw for Type B sections)

R = -Fc/Rv

the program assumes that -1 < R < 0

Fc = the compressive force in the concrete flange

Rv = d tw Fy

d = clear depth of webFy = design strength of the steel beam (see above)

A7.17.12.4 Shear - vertical (Section 13.4.1.1)

The steel beam is designed to resist the entire vertical shear force. Refer to A7.17.3 - CSA-S16-01 - Shear.

A7.17.12.5 Shear - longitudinal

Input:

The user specifies the following input in the Default/Parameters option: capacity of a single shear connector

actual number of connectors per beam in the positive moment region, or program to calculate number

of connectors required in the positive moment region for full shear connection.

If the user specifies the number of connectors in the positive moment region and the number is less than thenumber required to develop the positive moment capacity of the section, the program designates the beam ashaving a "partial shear connection" and reduces the Bending capacity accordingly (see below).

The number of connectors in negative moment regions cannot be specified. If reinforcement in the topping isspecified, the program assumes full shear connection in the negative moment region; hence the user shouldnot specify topping reinforcement if the required number of connectors cannot be provided.

Output:

topping As =0 : Total number of shear connectors required in the positive moment region

topping As specified: Total number of shear connectors required in the positive and negative momentregions

If multiple load cases are defined, detailed results for all load cases must be displayed to determine the requireddistribution of shear connectors.


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A7.17.12.6 Bending

Positive moment

The moment capacity of the section for positive moment is calculated according to Section 13.5:

Class 1: plastic moment capacity - Mr= Z FyClass 2: plastic moment capacity - Mr= Z FyClass 3: elastic moment capacity - Mr= S FyClass 4: Mr= Se Fy according to Section 13.5c

The calculation ofZ, the plastic section modulus, assumes that:

the topping width is reduced by the factor 0.6 0.85 f'c/Fy topping area in tension (below the 'plastic neutral axis') is ignored for partial shear connections (see above), the program calculates the theoretical topping height that provides

a full shear connection and uses this reduced width to calculate the plastic section modulus.

The calculation ofS, the elastic section modulus, assumes that: topping area in tension is ignored

the topping width is reduced by the factorEs/Ec, where

The program carries out the following design checks:

where:

M1 = moment due to load applied to steel (non-composite) beam

M2 = moment due to short-term load

M3 = moment due to long-term load

Mcs = moment capacity of steel beam only, calculated according to 13.5

Mc1, Mc2, Mc3 = moment capacity corresponding to M1, M2 and M3, respectively, calculated as explained

above. For Class 1 or Class 2 sections (plastic moment capacity), Mc1= Mc2= Mc3

Note that both the individual moment capacities and a weighted average capacity are displayed in the detailedresults.

Negative moment

The Default/Parameters option allow the user to define topping reinforcement that will increase the negativemoment capacity.

The moment capacity of the section for negative moment is calculated assuming: no topping reinforcement: steel section capacity of non-composite sections

topping reinforcement: the area of the reinforcement is added to the section when calculating S:


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Reinforcement should not be specified if the shear connectors required for full shear connection cannot beprovided in the region of negative moments.

The program carries out the following design checks:

where:

M = M1 + M2 + M3

Mc= capacity of section with reinforcement

M1, M2, M3, Mcs = as defined above

A7.17.12.7 Lateral-torsional buckling

The program assumes that the top flange is continuously supported by the concrete topping. The LTB capacityof the bottom flange is calculated as for a non-composite sections. Refer to A7.17.7 - CSA-S16-01 - Lateral-torsional buckling.

A7.17.12.8 Deflections

The program calculates the total span deflection d as follows:

= (M1) + (M2) + (M3)

where:

M1, M2, M3 = as defined above (bending)

(M1) : calculated using the moment of inertia of the steel beam only(M2) = the moment-of-inertia is calculated assuming that the topping width is reduced according to a modular

ratio n = Ec/Es

(M3) = the moment-of-inertia is calculated assuming that the topping width is reduced according to a modular

ratio n = k Ec/Es, where k is specified by the User in the Default/Parameters option.

The increased deflection arising from partially composite beams is calculated using the effective moment ofinertia from equation C-I3-6:

where:

Is = moment of inertia of the structural steel section

It = transformedmoment of inertia of the fully composite uncracked section

p = fraction of full shear connection.

A7.17.12.9 Axial force

The Default/Parameters option allow the user to specify one of the following design options: ignore axial force axial force taken by steel beam only. Refer to A7.17.5 - CSA-S16-01 - Axial - compression.

A7.17.12.10 Combined bending and axial force

Design check according to equations for non-composite sections. Refer to A7.17.8 - CSA-S16-01 - Combinedforces.


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A7.17.13 CSA S16-01 - Composite columns

The program calculates the axial capacity of composite columns according to section 18.2.2 - "CompressiveResistance".

Note that the concrete is ignored when calculating the flexural capacity of the section.

The factored compressive resistance, Crc, is taken as:

All rectangular sections and for circular sections with height-to-diameter ratio > 25:

=' = 1.0

All other circular sections:

Encased sections:


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A7.32 CSA - S136-94

Code clauses in the table and the manual refer to CSA S136-94 Cold-Formed Steel Structural Members -Structures (Design), December 1994.

The program designs the following section shapes:

and additional User-defined shapes.

Sections are added to the steel section library using the Files / Utilities / Create/edit a steel sections table optionin the STRAPmain menu.

A7.32.1 CSA-S136 - Effective section properties

There are two methods available in the program for calculating the effective section properties. The methodis specified by the user in the Exit/Options-Cold formed option in the Steel postprocessor. The methods are: Use actual calculated stresses

the effective section properties are based on the actual compressive stress f. Use stresses defined in the specification

the program uses a compressive stress f specified in the Code, i.e., the compressive stress is independentof the actual load.

Use actual calculated stresses

The program calculates the effective properties using an iterative procedure:

The program calculates the reduced section properties assuming that the stress (compression or tension)

at the extreme fibre is: f = Fy = P/A +M/S

where: P = actual axial force

M = moment required so that f =Fy (i.e. M > actual moment)

A,S = area, elastic modulus of the full section

The reduced properties are used to check bending with and without LTB.

If the section is not symmetric about the relevant axis, the program calculates the reduction twice, once with

+M and again with -M; the greater reduction is used.

No iterations are performed if the centre-of-gravity is in the same location for the full and reduced sections.An additional iteration using the new location is carried out if the centre-of-gravity moves.

For biaxial bending, the program determines the maximum of Mx/Sx and My/Sy (where Mx and My are the

actual moments) and calculates the reduced section assuming uniaxial bending in the correspondingdirection, as described above.


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The program recalculates the stress at the extreme fibre:f = P/A + M/S

where: P,M = actual axial force, moment

A,S = reduced area, elastic modulus calculated in step (a).

The stress is used to recalculate the effective section properties; the new properties are used for the axialcompression check.

Deflections:The stress is calculated again, without factors of safety:

f = Fy = P/A +M/S


A,S = reduced area, elastic modulus calculated after step (b)

For each iteration the program determines:

stress gradient for each element of the section, for calculation ofk (refer to Code 5.6.2.8) whether each element of the section is unstiffened, stiffened or stiffened with an edge stiffener.

The first and last segments of a section are assumed to be stiffeners (lips) if the length of the segment is lessthan 28 mm; otherwise they are assumed to be unstiffened elements.

The program assumes that the following elements have intermediate stiffeners if the projection is greater than28 mm (1.102 in):

The program calculates the reduced properties according to Section B2.2 for sections with a hole in the web

(if a value is specified forDiameter of hole in web). The specified area is removed from all web elements inthe section; note that all stiffened elements are assumed to be webs (i.e. elements not classified as unstiffenedor as stiffened with an edge stiffener).

Use Stresses Defined in the Specification

The program uses stresses defined in the Code (based on the section capacity) for calculating the effectivesection properties and ignores the stresses resulting from the actual loads:

Bending - without LTB

f = Fy at the extreme fibre. If the section is not symmetric about the relevant axis, the program calculatesthe reduction twice, once with the extreme fibre in compression and again in tension; the greater reductionis used. The program makes four iterations for each side in compression.

Bending - with LTB

The reduced elastic modulus Sc is calculated based on f = Fc in the compression fibre (4 iterations). Referto Code item 6.4.3.

Deflections

f= the actual compressive stress in the extreme fibre = M/S + P/A


A,S = area, elastic modulus of the full section.

If this stress causes reduction in some elements of the section, then the effective properties of the sectionare recalculated, which affects the actual stress, f. This stress is used in the next iteration to calculate theeffective properties (4 iterations).

Axial force

f = Fa (No iterations). Refer to Code item 6.6 for determination of Fa.

Axial force for combined check

The calculation requires Cr= nominal axial strength determined with f = Fy (no iterations). Refer to Codeitem 6.7.


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7.32.2 CSA-S136 - Strength of Steel

The program allows for design in all grades of steel - a different grade may be assigned to each member.

Fy and Fufor various shapes and thicknesses are taken from Tables 6.1 to 6.3 in Part 6 of the CSA-S16.1"Handbook of Steel Construction".

For user-defined steel types:

Fy (mPa) up to 230: 230 to 260: 260 to 300: 300 to 350: 350 to 400 400 to 480: > 480:

Fu (mPa) 380 410 450 480 520 590 Fy + 100

The program does not calculate the increase in yield strength due to cold forming according to Section 5.2.2,even if the option is selected by the user (the option is valid only for the AISI Codes).

7.32.3 CSA-S136 - Shear

The program checks that the actual shear force is less than the allowable shear force, i.e.

Vf Vr

where:

kv = 5.34t = web thickness

h = depth of the flat portion of the web

= 0.9 - resistance factor for shear

For sections with a number of segments parallel to the direction of shear, the program calculates Vrfor each

segment and uses the sum (lips are not considered as segments).

For combined shear and moment, refer to A7.31.4 - Bending - without LTB.

7.32.4 CSA-S136 - Bending - without LTB

The program checks that the applied major axis moment is less than the allowable moment calculated

according to Section 6.4.1.1 and 6.4.2.1 of the Code, i.e.

where:

Mf= moment (factored loads)

Fy = design yield stress

Sc = elastic section modulus of the effective section about the relevant axis calculated with the extreme fibre

at Fy (compression or tension).

= 0.90 = resistance factor for bending


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Note: The program calculates all result values at: - 1/10 of span intervals, and at points of intermediate supports. When computing the moment-of-inertia and elastic section modulus of flexural members, the program uses

the effective section properties calculated according to Section 5.6.2.

Combined moment and shear:

The program checks combined moment and shear according to Section 6.4.6:

where:

Mr= factored moment resistance without LTB.

Vr = factored shear resistance.

= 0.9 (resistance factor)

7.32.5 CSA-S136 - Lateral Buckling Strength

The program calculates all result values at: - 1/10 of span intervals, and at points of intermediate supports.

The designer can specify the exact location of intermediate supports for each member.

Lateral-torsional buckling is calculated individually for each segment between intermediate supports, andseparately along the top and bottom flanges; the calculation is done separately for positive moments (supportson bottom flange only are considered) and for negative moments (supports on top flange only are considered).

For laterally unbraced segments subject to lateral buckling, the factored load moment shall not exceed thefactored moment resistance calculated according to 6.4.1.1:

where:

Mf= factored load momentSc = elastic section modulus of the effective section, with f = Fc in the extreme compression fibre

Fc = compressive stress limit calculated according to 6.4.3.

= resistance factor of safety for bending = 0.9.

Note:

symmetric section:Fb is calculated according to 6.4.3.2.

unsymmetric section: Fb is calculated according to 6.4.3.3.

The sign ofCs is specified by the user according to the direction of the flange. Sections where the principal axes directions are significantly rotated from the local axes directions:

Fb = 0.5 Fb

Values forCb are calculated as:

Cb = 1/ 2.5

where:

= 0.6 + 0.4M1/M2 for members bent in single curvature= 0.6 - 0.4M1/M2 for members bent in double curvature

M1/M2 = ratio of smaller to larger moments at opposite ends of the unbraced length in the plane of bending

The program assumes Cb = 1 when: for cantilevers

for members subject to both axial load and bending (the axial load is assumed to be zero in the Cb

calculation when the effective stress f 0.1Fy).


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Note: For sections with unstiffened flanges the factored moment resistance is limited according to 6.4.4. For hollow pipe sections the factored moment resistance is calculated according to 6.4.1.1 and 6.4.3.5.

7.32.6 CSA-S136 - Axial force - Tension

The program checks that the applied tension force does not exceed the design strength calculated accordingto 6.3.1:

Tf/Tr 1.0

where:

Tf = axial force due to factored loads

Tr = factored axial resistance, equal to the lesser of:

Ag Fy, fu An FuAg = gross section area

An = net section area

Fy = design yield strength

Fu = tensile strength

= 0.90u = 0.75

7.32.7 CSA-S136 - Axial force - Compression

The program checks that the compression force does not exceed the design strength calculated according to6.6.1.3, i.e::

where:

Cf = factored axial compressive force

Cr = AeFa

Ae = effective area at the stress Fa, calculated according to 5.6.2a = 0.9 or 0.75 according to 6.2

Fa is determined according to 6.6.1.3:

where:

Fy = tensile yield strength

Fp = critical elastic buckling stress, determined according to 6.6.2 and 6.6.3

For unsymmetric sections the program solves the following equation from the AISI Cold Formed Steel Design

Manual - Part III, Supplementary Information - Section 4 (instead of calculating Fst according to 6.6.3.1):


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Note: Torsional buckling check - 6.6.3.1

- The program first searches for the longest segment between axial supports and calculates ex for thissegment. It then searches for the longest segment defined for LTB (i.e. between +z and/or -z supports)

that overlaps (even partially) the critical axial force segment and calculates t.- xo, yo - the distances from the centroid to the shear centre along the local x,y axes of the section.

For sections with unstiffened flanges the factored compression resistance is limited according to 6.6.3.2. For hollow pipe sections the factored compression resistance is calculated according 6.6.1.3 with Ae

calculated according to 6.6.5.

7.32.8 CSA-S136 - Combined axial force & bending

Compression and Bending

Combined compression and bending moments satisfy the interaction equations in Section 6.7.1:

where all values are calculated as explained in 6.7.1.

The effective section properties for axial force and moment are calculated according to the methods outinedin B.2.7 - Axial force - compression and B.2.4 - Bending - without LTB.

Tension and Bending

Combined tension and bending moments satisfy the interaction equations in Section 6.3.2:

where all values are calculated as explained in 6.3.2 and 6.7.1.

Note that when selecting a section the program also checks bending without the axial force. The equation forcombined tension and bending is less conservative than bending alone; if a section is inadequate for bending,then combined tension and bending will not be checked.

7.32.9 CSA-S136 - Deflections

The deflection is calculated based on section properties determined according to Section 5.6.2.

The stress in compressive elements is calculated using the specified loads and the effective section properties.


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8.4 CSA A23.3-94

8.4.1 CSA A23.3-94 - General

The longitudinal reinforcement calculation is based on the design assumptions outlined in 10.1; the followingrectangular stress block is used.

where

1 = 0.85 - 0.0015 f'c > 0.67 (10-1)1 = 0.97 - 0.0025 f'c > 0.67 (10-2)c = 0.60 ( 8.4.2)s = 0.85

The modulus of elasticity:

Concrete: mPa (8-7)Steel: Es = 200,000 mPa ( 8.5.4.1)

The stress in reinforcement is calculated as Es * strain, but not greater than fy.

8.4.2 CSA A23.3-94 - Beams

The beam design procedure includes: calculation of moment and shear envelopes calculation of redistributed moments and reduced shear (option) calculation of reinforcement steel areas calculation of stirrups with variable spacing or bent-up bars with constant stirrups

Note that axial forces and torsional moments are ignored by the program.

8.4.2.1 Moment redistribution (optional)

Moments in continuous beams are redistributed as permitted in 9.2.4, according to the following guidelines: The support moments in the envelope are reduced up to the maximum percentage specified by the user,

but not less than the minimum percentage specified.

The maximum span moments in the envelope remain constant or are decreased (unless the minimumredistribution requirement forces an increase in the span moment, which will generally occur in exteriorspans with stiff columns).

The shear forces in the spans are adjusted so as to maintain equilibrium of forces and moments. For beams with columns, the moment transferred by the beam into the column before and after redistribution

is constant. This prevents redistribution in the columns and maintains equilibrium in loading cases withhorizontal loads.

Note that the program checks that the redistribution percentage does not exceed the allowable

30 - 50c/d < 20% ( 9.2.4)

afterthe reinforcement is calculated, and displays warnings if required.


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8.4.2.2 Shear Reduction (optional)

The shear stress at the face of the support is reduced to the value at a distance 'd' from the face, as specifiedin 11.3.2.

8.4.2.3 Longitudinal Reinforcement

Minimum reinforcement is specified by the user as:

(10-4)

where bt = width of the tension zoneor

1.33 As,required ( 10.5.1.3)

Maximum reinforcement is limited to 4%.

Rectangular beams:

Referring to the above figure, the reinforcement area is calculated as follows:

As = Mu/{s fy (d - 0.51c)}

where:

if > max, compression reinforcement As' is provided, where: = As / bd max = rb

where:Mr = fs fy As,max (d - 0.5 amax)

As = As,max + As'

amax = As,max fy fs/ a1 fc f'c b

As,max = max b d

Tee beams:

compression block entirely in flange: designed as rectangular beams.

compression block in web:


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As = Asf+ Asw

where:

Asf = Cf/ fysAsw = Muw / {s fy (d - 0.5 aw)}Cf = 1 f'c ( bf- bw) tfcMuw = Mu - Mnfs Mnf = s Cf (d - 0.5 tf)

if As > As,max , compression reinforcement As' is provided, where:

As,max = max bfdmax = (bw / bf) (b + f) for beams without redistribution

f= Asf/ bw d

where:Mr = fs fy As,max (d - 0.5 amax)

As = As,max + As'

amax = As,max fys / 1c f'c b

8.4.2.4 Shear Reinforcement

Stirrups only (variable spacing):

The program selects the stirrup diameter and spacing so that

(11-4, 11-11)

where:

(11-6)

c= 0.60

and subject to the following limitations:

(11-5)

(11-1)

smax = min ( 600 mm, 0.7d )


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but when , then -

smax = min ( 300 mm, 0.35d ) (11.2.11)

The program ignores axial forces.

8.4.2.5 Torsion

Torsion reinforcement is calculated according to Section 11 of the Code.

The torsion force is always assumed by the program to be carried by the rectangular web of the beam:

The following terms are used:

Torsion is calculated only when Tf> Tcr/4, where:

The critical section for torsion calculation is at a distance 'd' from the face of the support if the user specifies

Shear reduction

The program calculates the torsion reinforcement as follows:

Transverse reinforcement (stirrups)

Longitudinal reinforcement:

Combined shear and torsion: the program checks the following interaction diagram:

Minimum torsion reinforcement: refer to shear requirements

8.4.2.6 Deflections

The program checks the deflections at two stages: immediate live load deflections = (ai)L

immediate live load deflection + long term deflection under sustained loads = (ai)L + (at)sus


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Immediate live load deflection:

(ai)L = (ai)D+L - (ai)D

where:

(ai)L = immediate deflection due to live loads

(ai)D+L = immediate deflection due to dead and live loads

(ai)D = immediate deflection due to dead loads only

The deflections are calculated from the loads using Ec and Ie, where

Ec = 4500 f'c (mPa)Ie = effective moment of inertia

=

Ig = gross moment-of-inertia of concrete section, without reinforcement

Icr = moment-of-inertia of cracked concrete section, with reinforcement

Mcr = cracking moment = frIg/yt, where fr = 0.6 f'cMa = maximum service load span moment for the relevant loads

Ma is calculated separately for the (Dead) and (Dead+Live) cases, hence different values ofIe are used when

calculating (ai)Dand (ai)D+L (displayed as Ie,d and Ie,d+l in the output tables).

Long term deflection:

The total long term deflection calculated by the program includes the immediate elastic deflection due to liveloads plus the long term deflections due to sustained loads, where sustained loads are defined as all deadloads along with the live loads that act permanently on the beam.

(at)sust = (ai)sust

where:

= S/(1+50')As'= span top reinforcement

As = maximum span bottom reinforcement

' = calculated from the compression steel = taken from CSA-A23.3 - 9.8.2.5

Mmax is calculated separately for the sustained loads, hence a different values ofIe is used when calculating

(at)sust(displayed as Ie,sust in the output tables).


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8.4.3 CSA A23.3-94 - Columns

The axial load strength, Prmax, is limited to: (10.10.4)

- 0.80 Pro for all section type, except:

- 0.85 Pro round sections.

Pro = 1c f'c (Ag - As) + s fy As (10-10)

For uniaxial bending, these assumptions can be represented as:

8.4.3.1 Reinforcement

minimum: the program limits the longitudinal reinforcement to no less than 0.01 times the gross area, Ag( 10.9.1). However, if the section is larger than required (i.e. capacity factor > 1.00), theprogram reduces Ag up to 50% when calculating the capacity and the reinforcementpercentage ( 10.10.5). If the effective area is reduced, a warning is displayed.

maximum: Reinforcement > 8%: the program displays a warning that the reinforcement exceeds theallowable ( 10.9.2).

Reinforcement > 10%: the program stops the design and displays a "No solution" warning.

8.4.3.2 Slenderness

For all columns (braced and unbraced)

k lu/r < 15 = short column.

k lu/r > 15 = slender column.

where:

k : defined by the user

lu = clear column length = member length less height of support at column end.

r = radius of inertia

This is a conservative assumption; however the moment magnifier will be small for columns that are marginallyslender.

Note: k lu/r is calculated separately for both directions. If the column is slender in one direction and short inthe other, the column is considered as slender and additional moments are added automatically (ifrequested by the user).

8.4.3.3 Short columns:

No additional moments or minimum eccentricity is calculated.


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8.4.3.4 Slender columns:

The user may define the moment magnifier in each of the two major directions or specify that the programcalculate the value automatically.

The program calculates the magnified factored moment Mc as follows:

braced columns:Mc = Cm

unbraced columns:

Mc = M2

where:

= 1/(1-Pf/mPc) Equation (10-16)m = 0.75Pc = EI/(klu) Equation (10-17)EI = {(Ec Ig/5) + Es Ise}/(1+d) (10-18)d : braced columns: maximum ratio of factored dead load to factored axial load for all defined load

combinations. If the user does not identify the dead load cases , the program

assumes d = 0.4

unbraced columns: d = 0 (10.14.1)Cm = 0.6M2 + 0.4M1 > 0.4 M2 for columns not subject to transverse loading

= maximum moment for columns subject to transverse loading

M2 = larger factored end moment

M1 = smaller factored end moment

Referring to the literature, the magnified moment represents an additional moment that has an approximately

parabolic shape, i.e equal to zero at the ends and to ( -1)M at the maximum, which is assumed by the Codeto occur in the middle 20% of the span (0.6M2 + 0.4M1).

The following figure shows an example for a slender column consisting of one STRAPmember, braced in theM2 direction and unbraced in the M3 direction. The design moments at top/middle/bottom are superimposedon the moment diagrams.

8.4.3.5 Load conditions

The program calculates design moments at three locations in each column - top, middle and bottom. The

middle moment is calculated from the equation 0.6M2 + 0.4M1 and the result is increased by the momentmagnifier in the case of slender columns.

If the column is subject to transverse loads, the middle moment is taken as the maximum moment along the

span, but not less than 0.6M2 + 0.4M1


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The three design moments are calculated separately for the M2 and M3 moments. The program then checksthe capacity for all three locations, i.e.

top: Ptop M2,top M3,topmiddle: max(Ptop,Pbot) M2,mid M3,midbottom: Pbot M2,bot M3,bot

Plane frames:

The program adds additional moments to the M3 moments, and create separate load conditions with theminimum M2 moments, where both positive and negative values are generated (important for unsymmetricsections). For example:

M2i M3i M2 M3

0.0 15.7 0.0 15.7

-7.3 0.0

7.3 0.0

Space frame:The program checks for design moments about both axes; if the moments equal zero about one axis, theprogram designs the column as for plane frames. The program adds additional moments simultaneously

about both axes.

In many cases, two or more STRAPmembers may be combined to form a column. In addition, the supportlocations in the M2 and M3 directions may not be identical.

The program searches for supports at the nodes in either direction and defines "design spans". Each"design span" is calculated separately. For example:

In each design span, the program creates the combinations of M2 and M3 moments at top/middle/bottomof the span. The object is to ensure that the program creates a load situation that includes the maximummoments.

The program calculates M for the combined span. It then checks whether one of the ends is in the middle

third of the combined span. If yes, the program uses M at that end and at the middle. If both ends areoutside the middle third, the program uses the actual moments.

Referring to the example of the above figure, it calculates the load situations as follows:

Span A:

Design

Location

Design moment

M2 M3

Top Mtop MiMiddle Mi MiBottom Mbottom Mbottom


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8.4.3.6 Minimum eccentricity for slender columns

emin = 15 + 0.03h mm.

8.4.3.7 Tension

Columns with a tension axial force are always designed as short columns with = 0.90.

The capacity of a column in pure tension is equal to As fy.

The capacity of columns with combined tension and bending is calculated identically to that of column withcombined compression and bending.

8.4.3.8 Shear

The program searches for the load case with (V - Vc)max

For non-seismic load cases Vc is calculated according to ACI 318-99 Equation (11-4):

where

bwd = gross area for round sections.

Nu/Ag = N/mm2

Stirrups are detailed with uniform spacing according to the user specified parameters.

Minimum diameter: 0.3*(longitudinal bar diameter) for longitudinal bars 30

10 mm for longitudinal bars 35

Maximum spacing: minimum of - minimum dimension, 16*(longitudinal bar diameter), 48*stirrup diameter(*0.75 for fc > 50)

8.4.4 CSA A23.3 - Walls

The program calculates the minimum eccentricity and slender wall additional moments for weak axis bendingaccording to the methods used for columns. A wall is considered slender when klu/r > 15

Reinforcement details - No seismic loads minimum reinforcement area = 0.015 * Awall maximum bar spacing = minimum (3*b, 500 mm.)

minimum transverse reinforcement area: h = 0.002where b = wall thickness


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8.4.5 CSA A23.3 - Seismic - Beams

The following frame classification is used in the following explanation:

Ductile = Ductile frame membersCSA A23.3 - section 21.3

Nominal = Building members requiring nominal ductilityCSA A23.3 - section 21.9

Longitudinal Reinforcement Minimum reinforcement (21.3.2.1)

Ductile/Nominal: as in non-seismic

Maximum reinforcement

Ductile: < 0.25 (21.3.2.1)

Nominal: < 0.75 b

Ductile: Positive moment strength at joint face > 1/2 Negative moment strength (21.3.2.2)Nominal: Positive moment strength at joint face > 1/3 Negative moment strength (21.9.2.1.1)

Ductile: Moment strength (positive or negative) at any location along the span > 1/4 maximum momentstrength at the joint (21.3.2.2)

Nominal: Moment strength (positive or negative) at any location along the span > 1/5 maximum momentstrength at the joint (21.9.2.1.1)

Transverse reinforcement Hoops required over length (hinge region):

Ductile: the member depth beyond section where flexural yielding may occur, but not less than 2d from theface of the support (21.3.3.2)

Nominal: 2d from the face of the support (21.9.2.1.2)

Maximum hoop spacing in hinge region:

Ductile: min{d/4, 24s, 300 mm} (21.3.3.3)

Note that the 8L requirement cannot be checked by the program.Nominal: same as "Ductile" (21.9.2.1.2)

Maximum hoop spacing outside hinge region:Ductile: d/2 (21.3.3.5)Nominal: same as "Ductile" (29.9.2.1.3)Not: No special requirement

8.4.6 CSA A23.3 - Seismic - Columns

The following frame classification is used in the following explanation:

Ductile = Ductile frame membersCSA A23.3 - section 21.4

Nominal = Building members requiring nominal ductilityCSA A23.3 - section 21.9

Longitudinal Reinforcement Flexural strength at joint:

Ductile: Mrc > 1.1 Mnb (21.4.2 - Eq. 21-1)Nominal: No special requirement

Reinforcement ratio:

Ductile: 0.01 < g < 0.06 (21.4.3.1)Nominal: No special requirement


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Transverse reinforcement

Vc is calculated according to 21.7.3.1b.

Hoops (spiral or circular) required over length lo:Ductile: max {d, span/6, 450 mm} (21.4.4.5)Nominal: same as "Ductile" (21.9.2.2.2)

Maximum spacing within lo:Ductile: min (h/4, 6L, 100 mm) (21.4.4.3)

Nominal: min (h/2, 8L, 24s, 300 mm) (21.9.2.2.2, 7.6.5) Maximum spacing outside lo:

Ductile: min (6L, 150 mm) (21.4.4.6)

Nominal: min (h, 16L, b, 48s) (7.6.5)

8.4.7 CSA A23.3 - Seismic - Walls

Reinforcement details - Seismic loads

Concentrated reinforcement at the ends of walls:- minimum area = max(0.001 *Awall , 0.002*Aconc), except -

minimum area = max(0.002 *Awall , 0.002*Aconc) in plastic hinge region- Lconc:

lw/20, but not less than b, except -lw/10, but not less than b nor more than 3b, in plastic hinge regionThe maximum Lconc (at either end) is 0.4 * wall length

Note:- Concentrated reinforcement is not designed if the wall length< 4 * thickness- For "Nominal ductile frames", Lconc = 0.005 Aconc in the plastic hinge region and no concentrated

reinforcement is provided outside the plastic hinge region.

Distributed reinforcement:- minimum area = 0.25% * Adist- maximum spacing = 450 mm, except -

maximum spacing = 300 mm in plastic hinge region

Shear- Plastic hinge region:

Ductile frame: Vc = 0Nominal ductile frame: 0.5Vc

- outside Plastic hinge region:The shear is calculated according to the Simplified method in Code section 11.3


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