2DVERIFY

RISA-2DRapid Interactive Structural Analysis - 2 Dimensional

Verification Problems

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Table of Contents

Page i

IV) Verification

Verification Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-1Verification Problem 1 (Truss) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-2Verification Problem 2 (Large Model Cantilever) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-3Verification Problem 3 (Thermal) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-4Verification Problem 4 (Two Story Frame) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-6Verification Problem 5 (Large Frame) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-8Verification Problem 6 (Plates) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-10Verification Problem 7 (Dynamics) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-12Verification Problem 8 (Dynamics and Response Spectrum Analysis) . . . . . . . . . . . . . . . . . 1-14Verification Problem 9 (AISC Code Checks, ASD and LRFD) . . . . . . . . . . . . . . . . . . . . . . 1-16Verification Problem 10 (NDS Timber Code Checks) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1-28Verification Problem 11 (Tapered WF Analysis and ASD Code Checks) . . . . . . . . . . . . . . 1-34

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Verification OverviewWe at RISA Technologies maintain a library of dozens of test problems usedto validate the computational aspects of RISA-2D. In this section we presenta representative sample of these test problems for your review.

These test problems should not necessarily be used as design examples; insome cases the input and assumptions we use in the test problems may notmatch what a design engineer would do in a real world situation; the inputfor these test problems was formulated to test RISA-2Ds performance, notnecessarily to show how certain structures should be modeled.

The RISA-2D solutions for each of these problems are compared to eitherhand calculations or solutions from other well established programs. By"well established" we mean programs that have been in general use for manyyears, such as the Berkeley SAPIV program. This original SAPIV program isstill the basis for several commercial programs currently on the market (butnot RISA-2D).

The reasoning is if two or more independently developed programs that usetheoretically sound solution methods arrive at the same results for the sameproblem, those results are correct. The likelihood that both programs willgive the same wrong answers is considered extremely remote.

If discrepancies occur between the RISA-2D and SAPIV results duringRISA-2D testing, we don't automatically assume SAPIV is correct.Additional testing and hand calculations are used to verify which program, ifeither, is correct. There have been instances where the SAPIV results haveproven to be erroneous!

The data for each of these verification problems is provided. The files areVPROB1.R2D for problem 1, VPROB2.R2D for problem 2, etc. When youinstall RISA-2D these datafiles are copied into the /RISA directory, so whenyou go to the Files (Alt-F) screen, you will see them listed. If you want to runany of these problems yourself, just read in the appropriate datafile and haveat it.

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Verification Problem 1Description This problem is a typical truss model. The members are pinned at both ends,

thus they behave as truss elements. This particular problem is presented inthe text Structural Analysis and Design, by Ketter, Lee and Prawel. It canbe found on page 171 as example 3.7. The text lists "Q" as the loadmagnitude and "a" as the panel width. For this solution "Q" is taken as 10MT and "a" is taken as 2 meters (continental metric units).

This problem provides a comparison of the stiffness method used in RISA-2D with the joint equilibrium method used in the text. The joint equilibriummethod may be used to solve statically determinate structures only, while thestiffness method can solve either determinate or indeterminate models.

Model Sketch

Validation The axial force results calculated by RISA-2D are comparedMethod with the axial force results presented in the text. In the text the results are

listed as multiples of "Q".

Comparison Axial Force Comparison

Member RISA-2D Text

1 - 2 39.13 39.13

8 - 9 -23.75 -23.75

3 - 9 5.59 5.59

4 - 10 11.18 11.18

As can be seen, the results match exactly. Note the text lists tension aspositive, compression as negative.

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Verification Problem 2Description This model is simply a cantilever with a vertical load applied at the end. The

cantilever is 999 feet in length, modeled using a series of 999 beams, each 1ft in length. This problem tests the numerical accuracy of RISA-2D, and alsothe program limits for number of joints. Any significant precision errorswould show up dramatically in a model like this.

Model Sketch

Validation The RISA-2D solution will be compared with the theoreticalMethod displacement and rotation for a cantilever with a load at its end.

The equations for the displacement and rotation are:

> = P * L^3 / (3 * E * I) = P * L^2 / (2 * E * I)

For this model, the following values were used:P = -1 KL = 999 ( 11,988" )E = 100,000 KsiI = 10,000 in^4

So the theoretical solution values are:

> = -574.274 inches= -.07186 radian

Comparison Cantilever Solution

Value RISA-2D Theoretical

Displacement -574.270 -574.274

Rotation -.07186 -.07186

As can be seen, the results match almost exactly.

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Verification Problem 3Description This model is used to test the thermal force calculations in RISA-2D. The

model is a five member cantilever with a spring in the local x direction at thefree end. As the model is loaded thermally the spring causes some, but notall, of the thermal expansion to be resisted.

Thermal loads cause structural behavior somewhat different from otherloads. For gravity loads, displacements induce stress, but for thermalloading displacements cause stress to be relieved. For example, a free endcantilever that undergoes a thermal loading would expand without resistanceand thus see no stress. Conversely, a fixed-fixed member that undergoes thesame thermal loading would see a stress increase with no displacements.

This model uses a spring to provide partial resistance to the thermal load.This is realistic in that members generally would have only partial resistanceto thermal effects.

Model Sketch

Validation This model is validated by the use of hand calculations. TheMethod theoretically exact solution may be calculated for comparison with the RISA-

2D result. Following are those calculations.

Property Values:

Area(A) = 50 cmYoung's Modulus(E) = 70,000 MPaThermal Load (>T) = 300ECoef Therm Exp.( ) = .000012 cm/cm ECSpring (K) = 500 KN/cmLength (L) = 10 Meters

The unrestrained thermal expansion (>free) is:

>free = * >T * L

The general equation for the displacement of a member due to an axial load(>axial) is:

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>axial = P * L / (A * E)

Well call the actual displacement of the member ">actual". Now, well say"P" is the force in the spring, so:

P = >actual * K

So, using these formulations, the following is true:

>actual * K * L / (A * E) = >free - >actual

In other words, the "resisted expansion" of the member is the thermalexpansion that is not allowed to occur because of the spring and is equal to>free - >actual. Think of it as the spring force pushing the member end backthis resisted expansion distance.

This leads to the equation for the actual displacement:

>actual = * >T * L / ( 1 + (K*L)/(A*E) )

The force in the member is:

Force = (>free - >actual) * A * E / L

So for the given property values,

>actual = 1.482 cmForce = 741.2 KN

Comparison Thermal Results Comparison

Solution Method Displacement Force

Exact 1.482 741.2

RISA-2D 1.482 741.2

As can be seen, the results match exactly.

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Verification Problem 4Description This problem is a small but typical moment frame, with uniform distributed

dead and live loads and lateral nodal loads used to approximate seismiceffects. The columns are W14X61s and the beams are W18X50s.

This model was chosen because it represents the type of model typicallysolved by hand. It can be solved relatively easily using moment distribution,though a hand solution is time consuming. The hand solution gives shearsand moments for the members.

Model Sketch

Validation The validation of this solution is against a hand solution using the moment Method distribution method. The effects of sideway are considered in the moment

distribution solution for the seismic lateral loading case. Note that there areroundoff errors in the moment distribution solution and any small differencescan be attributed to this.

Comparison The moment distribution calculations for the dead load analysis will belisted, and a table comparing the dead load and seismic results is provided.

K for 1-2 is : 640/(17.25 *12) = 3.09K for 2-3 is : 640/(18 *12) = 2.96K for 3-4 is : 800/(25 *12) = 2.67K for 2-5 is : 800/(25 *12) = 2.67K for 5-4 is : 640/(18 *12) = 2.96

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K for 6-5 is : 640/(17.25 *12) = 3.09

Moment Distribution Calculations ( Dead Load Only)1 2 3 4 5 6

1-2 2-1 2-3 2-5 3-2 3-4 4-3 4-5 5-2 5-4 5-6 6-50. .35 .34 .31 .53 .47 .47 .53 .31 .34 .35 0.

0. 0. 8.65 0. 8.65 -8.65 0. -8.65 0. 0. 0.-3.03 -2.94 -2.68 -4.58 -4.07 4.07 4.58 2.68 2.94 3.03

-1.52 0. -2.29 1.34 -1.47 2.04 2.04 1.47 -1.34 2.29 0. 1.52.33 .32 .29 -.30 -.27 .27 .30 .29 -.32 -.33

.17 -.15 -.15 .16 .14 -.14 -.16 .15 .15 0. -.17.11 .10 .10 -.16 -.14 .14 .16 -.1 -.1 -.11

-1.35 -2.59 -4.96 7.55 -6.35 6.35 -6.35 6.35 -7.55 4.96 2.59 1.35

Comparison of Beam End Forces

Beam Load Force RISA-2D Mom. Dist Diff.2 - 5 Dead I, Shear 2.07 2.08 0.5%

Dead I, Moment 7.54 7.55 0.1%Dead J, Shear 2.08 2.08 0.0%Dead J, Moment -7.54 -7.55 0.1%

3 - 4 Dead I, Shear 2.07 2.08 0.5%Dead I, Moment 6.32 6.35 0.5%Dead J, Shear 2.08 2.08 0.0%Dead J, Moment -6.32 -6.35 0.5%

5 - 4 Seismic I, Shear 8.14 8.2 0.7%Seismic I, Moment 63.33 64.2 1.4%Seismic J, Shear -8.14 -8.2 0.7%Seismic J, Moment 83.27 82.5 0.9%

2 - 5 Seismic I, Shear -11.04 -11.1 0.5%Seismic I, Moment -138.05 -139.1 0.8%Seismic J, Shear 11.04 11.1 0.5%Seismic J, Moment -137.87 -139.1 0.9%

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Verification Problem 5Description This problem is a relatively large frame with a high degree of load and

member complexity. All types of beam and node loads are included, thebeam loads being; point loads, point moments, partial length trapezoidalloads, thermal loads, and self weight. All possible combinations of beamaxial, shear, and moment releases are included. Several sets of beamproperties are used. The loads are separated into four basic load cases andcombined into five load combinations.

This model is fairly complicated, so instead of attempting a modeldescription the user is advised to read in and look at the VPROB6.R2D filefor any details of this model.

This problem is primarily a test of the overall static analysis capabilities ofRISA-2D. This problem also tests that complex loadings, in concert withcomplex model geometries, are handled properly.

Model Sketch

Validation The validation for this problem is a comparison against the same modelMethod solved with an established and industry accepted program. For this model,

the results will be compared against the SAP IV program. The SAP IVprogram is a widely accepted program based on the finite element method,which was developed at the University of California at Berkeley. Selectedjoint displacements and member end forces are the quantities that will becompared.

Note that since RISA-2D displays left hand section forces, the signs of the J-end results must be reversed to give equivalent J-end member forces. I-end

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left hand section forces are the same as the I-end member end forces.

Comparison As can be seen from the tabulated values below, the only differences are dueto the fact that RISA-2D rounds displacements to 3 places after the decimal.

Comparison of Nodal Displacements

Node No. LoadComb.

Dir. RISA-2DResults

SAP IVResults

% Diff.

8 1 X 23.579 23.57944 0.016 1 Y .223 .22297 0.04 2 X .826 .82625 0.016 3 .00267 .00267 0.09 4 X -3.145 -3.14479 0.08 5 X 16.473 16.47341 0.012 5 Y .294 .29412 0.024 5 -.00647 -.00647 0.0

Comparison of Member End Forces

Memb.Num.

Nodes LoadComb.

End,Dir.

RISA-2DResult

SAP IVResult

% Diff.

11 13 - 21 1 J,x -25.88 -25.88 0.0J,y 18.00 18.00 0.0J, -54.00 -54.00 0.0

14 16 - 24 2 I,x -.77 -.77 0.0I,y 0.00 0.00 0.0I, -126.40 -126.40 0.0

7 8 - 16 3 I,x .95 .95 0.0I,y 1.88 1.88 0.0I, 6.29 6.29 0.0

15 12 - 19 4 I,x 15.25 15.25 0.0I,y 0.00 0.00 0.0I, 0.00 0.00 0.0

8 10 - 18 5 I,x -23.99 -23.99 0.0I,y -.88 -.88 0.0

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I, -20.20 -20.20 0.0

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Verification Problem 6Description This problem is used to test the plate elements for in-plane membrane action.

The model is of two cantilever beams, the first modeled using a mesh offinite elements and the second modeled using a rectangular beam. The loadis applied at the free end of the cantilever.

Note that more example problems for plate elements are given in the PlateData section of the General Reference.

Model Sketch

Validation This model is validated by comparing the deflection at the cantilevers freeMethod end between the beam and plate model. The results will also be checked

against hand calculations. Following are the theoretical calculations toobtain the tip deformation.

Property Values:

Beam Depth (D) = 60 in.Beam Width (B) = 6 in.Area (A) = 360 in.2Length (L) = 30 ft. (360 in.)

Youngs Modulus (E) = 4000 KsiShear Modulus (G) = 1539 Ksi

Membrane Load applied at the free end (Pm) = 5000 Kips

Moment of Inertia for the Membrane Load (Im) = 108,000 in.4

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Therefore, for the given property values:

The free end deflection due to the Membrane load is:

m3/ 3

(,%

3/$*

LQ

Comparison Free End Deflection Comparison, Plates vs. Beam vs. Theory

Loading Plates Beam Theory

Membrane 180.17 in. 183.90 in. 183.89 in.

As can be seen, the results match closely.

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Verification Problem 7Description This problem is designed to test the dynamic analysis capability of RISA-2D.

The first 10 frequencies for a simply supported beam, modeled as a series of50 individual beam elements, are calculated.

Model Sketch

Validation The frequencies calculated by RISA-2D will be compared to Method the "exact" frequencies presented by the text Formulas for Natural

Frequency and Mode Shape, by Dr. Robert D. Blevins (1979, VanNostrand Reinhold).

The equation presented by Blevins for the transverse frequencies is:

Fi = [ i / 2 L ] * (EI/m)

The equation presented by Blevins for the longitudinal frequencies is:

Fi = [ i / 2 L ] * (E/)

where: = i*

m = mass per unit length = mass densityi = Frequency Number (i=1,2,3,...)

For our model,

E = 30,000 ksiI = 20,000 in4m = .10783 slugs/in

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= .00074885 slugs/in3

Comparison Frequency Comparison

Freq No. Blevins RISA-2D

1 .64328 .643

2 2.573 2.573

3 5.7895 5.789

4 10.292 10.292

5 16.082 16.082

6 23.158 23.158

7 31.521 31.52

8 41.170 41.168

9 41.699* 41.692*

10 52.106 52.101

* This is the first longitudinal frequency.

As can be seen, the results match very closely.

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Verification Problem 8Description This problem tests the dynamic analysis and response spectrum analysis

(RSA) features of RISA-2D. The model is a 10 story building. Thecalculation of mode shapes and frequencies is followed by an RSA using the1940 El Centra earthquake spectra. The SRSS combination method will beused to combine the modes.

All the Y-direction degrees of freedom have been restrained using very stiffsprings. This will make the building vibrate laterally as an idealized shearbuilding.

The members use heavier sections at the bottom and lighter sections at thetop. The mass used for the dynamic analysis is only the self weight of themembers.

Model Sketch

Validation This problem will be validated against the ALGOR FEA program. The Method model used for this problem is identical to the Algor verification problem

VE02004. The ALGOR program is primarily verified versus the originalSAP IV program developed at U.C. Berkeley.

The nodal displacements for selected nodes and the bending moments forselected members will be compared. The frequency and modal participationresults are not shown here because if the nodal displacements match, theintermediate results must also match.

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Comparison The compared values are shown below. As can be seen, the results matchvery closely.

Comparison of Nodal Displacements (in.)Node Algor RISA-2D % Diff.

3 .260 .260 0.0

7 .856 .856 0.0

11 1.411 1.411 0.0

15 1.924 1.924 0.0

21 2.393 2.392 0.04

Comparison of Bending Moments (K-ft)Member Algor RISA-2D % Diff.

1 159.42 159.27 0.09

7 84.08 84.10 0.02

13 64.74 64.72 0.03

19 46.06 45.96 0.22

25 19.62 19.54 0.41

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Verification Problem 9

Description This verification model is a 2 bay, 2 story frame. The model is comprised ofWF, WT, Channel, Pipe, Single Angle, Double Angle and Structural Tubemembers.

This problem is used to verify the stress and steel code check calculations inRISA-2D. Both ASD and LRFD codes will be checked.

Model Sketch

Validation Following are the hand calculations for various members for variousMethod load combinations. The steel codes used are the AISC ASD 9th edition, and

AISC LRFD 2nd edition. At least one member of each type is validated.

As part of the validation, the member stresses also are calculated. The forcesfor each member come from the RISA-2D analysis. The "1.333" factor is theresult of the Wind/Seismic increase in allowable stresses activated for LoadCombination 3. This increase applies to the ASD code checks only.

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ASD Following are the ASD hand calcs:

Member 2, Load Comb 1

Shape is W8X24

A = 7.08I = 82.8, S = 20.88ry = 1.61, rz = 3.42L = 10 (120")Stress Calculations:

fa = 7.05/7.08 = 1.0 Ksifb = 6.67(12)/20.88 = 3.83 Ksicompact section

K_in = K_out = 1.0, No sidesway (K*L/ry)out = 149.1 , (K*L/rz)in = 35.1, slender Fa = 6.7 Ksi ( Eq. E2-2 )Lcomp = Lb = 20Lc = 6.85 , ( Lb > Lc ) Fb = 16.39 Ksi ( Eq. F1-8 )Cm = .57 Fez = 121.2

Eqn H1-1 : .282Eqn H1-2 : .280Eqn H1-3 : .382 (so dont use)


Shape is an 8 Diameter Pipe, Standard Weight

A = 8.4I = 72.5, S = 16.8r = 2.94L = 10' (120")Stress Calculations:

fa =. 71/8.4 = .08 Ksifb = 11.09(12)/16.8 = 7.92 Ksicompact section

K_in = K_out = 1.0, No sidesway

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(K*L/r) = 40.82 nonslender Fa = 19.152 Ksi (Table 3, p. 5-119, Cc = 126.1)Fb = 23.76 Ksi ( Eq. F3-1 )Cm = .85 Fez = 89.62



Shape is C10x30

A = 8.82I = 103, S = 20.6ry = .669, rz = 3.42d/Af = 7.55

L = 10 (120")Stress Calculations:

fa = 0.0 Ksifb = 8.94(12)/20.6 = 5.21 Ksicompact section

Lcomp = Lb = 10Lc = 3.2 , ( Lb > Lc ) Fb = 6.62 Ksi ( Eq. F1-8 )Cm = .85 Fez = 121.2

Eqn H1-1 : .669Eqn H1-2 : .787Eqn H1-3 : .787


Shape is a Tube 6X6X3/8

Note that Fy for tube steel is 44 Ksi.

A = 8.08I = 41.6, S = 13.9

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ry = 2.27, rz = 2.27L = 10 (120")Stress Calculations:

fa = 10.65/8.08 = 1.32 Ksifb = 23.22(12)/13.9 = 20.05 Ksicompact section

Lb_in = 10, Lb_out = 20K_in = K_out = 1.0, No sidesway (K*L/ry)out = 105.73 , (K*L/rz)in = 52.86, nonslender Fa = 13.068(1.333) = 17.42 Ksi (Table 3, p. 5-119, Cc = 114.06)Fb = .66Fy = 29.04(1.333) = 38.71 Ksi ( Eq. F3-1 )Cm = .30 Fez = 53.44



Shape is WT4X9

A = 2.63ry = 1.23, rz = 1.14L = 14.14 (169.7")Stress Calculations:

fa = 13.68/2.63 = 5.2 Ksi

compact section

K_in = K_out = 1.0, No sidesway (K*L/rz)in = 148.84, slender Fa = 6.74(1.333) = 8.98 Ksi ( Eq. E2-2 )Eqn H1-1 : .579Eqn H1-2 : .181Eqn H1-3 : skip by inspection (same as H1-1)


Shape is LL4x4x1/4 thick X 3/8 spacing

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A = 3.88ry = 1.79, rz = 1.25L = 14.14 (169.7")Stress Calculations:

fa = 13.65/3.88 = -3.52 Ksi (Tension)compact section

Ft = .6Fy(1.333) = 28.79 Ksi Eqn H2-1 : .122


Shape is L5x5x7/16

A = 4.18rz-z = .986 ( This is the radius of gyration about the weakest axis )L = 10' (120")Stress Calculations:

fa = 3.79/4.18 = .907 Ksi

compact section

K_in = K_out = 1.0, No sidesway (K*L/rz) = 121.7, nonslender Fa = 10.04 Ksi (Table 3, p. 5-119, Cc = 126.1)UC : .090

Note that unity check for single angles is based upon the axial load only andEuler buckling is considered about the weakest (z-z) axis.

Comparison ASD Unity Check Comparisons

Member L.C. RISA-2D Hand Calc

2 1 .283 .282

Verification

ASD Unity Check Comparisons


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7 1 .337 .337

6 1 .787 .787

4 3 .556 .555

8 3 .581 .579

9 3 .122 .122

10 1 .090 .090

As can be seen, the results match very closely. The hand calculations usedtables 3 and 8 to calculate Fa and Fe, and usually only carried 2 places to theright of the decimal. This and other roundoff differences cause any slightdiscrepancies.

No examples of using Appendix B were shown, however the program willinclude these provisions if a member is non-compact or slender. ( Note thatseveral of the members were declared slender based on their Kl/r ratiobeing greater than Cc. This slender is different than the sectioncompactness criteria )

For the above members in several instances, equation H1-3 produced a largerunity check than H1-1 or H1-2 and this value was NOT used. This isbecause H1-3 is allowed in lieu of H1-1, or H1-2. Since H1-1 and H1-2 arethe more accurate expressions for the combined stresses, a lower unity checkfrom H1-1 or H1-2 can be used in place of H1-3.

LRFD Following are the LRFD hand calcs:

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Shape is W8X24

A = 7.08I = 82.8, S = 20.88Z = 23.2ry = 1.61, rz = 3.42L = 10 (120")Strength Calculations:

Pu = 9.87 KipsMu = 9.3 ft-Kips

compact section

K_in = K_out = 1.0, No sidesway (K*L/ry)out = 149.1 c = 1.672 > 1.5

Fcr = 11.29 Ksi ( Eqn. E2-3 )Pn = Ag*Fcr = 79.94 Kips

Lcomp = Lb = 20Lp = 6.71 , ( Lb > Lp ) Lr = 24.4 , (Lb < Lr )Mp = 69.6 ft-Kips Mr = 45.2 ft-Kips

Mn = 51.3 ft-Kips

b = .90, c = .85

Eqn H1-1b : .274 (compression)


Shape is an 8 Diameter Pipe, Standard Weight

A = 8.4I = 72.5, S = 16.8Z = 22.2r = 2.94L = 10' (120")Strength Calculations:

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compact section

K_in = K_out = 1.0, No sidesway (K*L/r) = 40.82 c = .458 < 1.5


Mn = Mp = 66.6 ft-Kips

b = .90, c = .85



Shape is C10x30

A = 8.82Iz = 103, Sz = 20.6Iy = 3.94Z = 26.6ry = .669, rz = 3.42d/Af = 7.55J = 1.23Cw = 79.3X1 = 6401X2 = 1.8 x 10-4

L = 10 (120")Strength Calculations:


compact section

Lcomp = Lb = 20Lp = 3.2 , ( Lb > Lp ) Lr = 19.7 , (Lb > Lr )Mn = Mcr = 43.89 ft-Kips

b = .90, c = .85


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Shape is a Tube 6X6X3/8

Note that Fy for tube steel is 44 Ksi.

A = 8.08I = 41.6, S = 13.9Z = 16.8ry = 2.27, rz = 2.27L = 10' (120")Strength Calculations:


compact section

Lb_in = 10', Lb_out = 20'K_in = K_out = 1.0, No sidesway (K*L/ry)out = 105.73 , (K*L/rz)in = 52.86c = 1.311 < 1.5


Mn = Mp = 61.6 ft-Kips

b = .90, c = .85



Shape is WT4X9

A = 2.63ry = 1.23, rz = 1.14L = 14.14' (169.7")Strength Calculations:


compact section

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K_in = K_out = 1.0, No sidesway (K*L/rz)in = 148.84 c = 1.669 > 1.5


b = .90, c = .85

Eqn H1-1a : .696 (compression)


Shape is LL4x4x1/4 thick X 3/8 spacing

A = 3.88ry = 1.79, rz = 1.25L = 14.14' (169.7")Strength Calculations:

Pu = -17.94 Kips (Tension)Mu = 0.0 ft-Kips

compact section

Pn = Fy*Ag = 139.7 Kips (Eqn. D1-1)

t = .90

Eqn H1-1b : .071 (tension)


Shape is L5x5x7/16

A = 4.18rz-z = .986 ( This is the radius of gyration about the weakest axis )L = 10' (120")Strength Calculations:


compact section

Verification

Page 1-27

K_in = K_out = 1.0, No sidesway (K*L/rz) = 121.7c = 1.365 < 1.5


UC : .105

Note that unity check for single angles is based upon the axial load only andEuler buckling is considered about the weakest (z-z) axis.

Comparison LRFD Unity Check Comparisons

Member L.C. Hand Calc RISA-2D

2 5 .274 .275

7 5 .261 .261

Verification

LRFD Unity Check Comparisons

Member L.C. Hand Calc RISA-2D

Page 1-28

6 5 .317 .318

4 7 .544 .544

8 8 .696 .699

9 8 .071 .071

10 8 .105 .105

As can be seen, the results match very closely. The hand calculations usuallyonly carried 2 places to the right of the decimal. This and other roundoffdifferences cause any slight discrepancies.

No examples of using Appendix B were shown, however the program willinclude these provisions if a member is non-compact or slender.

Note that all load combinations used for the LRFD problems included a P-Delta analysis, which is required to satisfy the requirements of Chapter C. See the P-Delta Analysis section of the General Reference for moreinformation on this.

Verification

Page 1-29

Verification Problem 10Description This problem tests the NDS Timber Code checking. The model is a 2 bay

portal frame, with one bay braced. The model is loaded with combinationsof Dead Load, Live Load, and Lateral (Wind) Load. A different CD (LoadDuration) factor is used for each combination.

Model Sketch

Validation Following are the hand calculations for various members for variousMethod load combinations. All code check calculations and wood properties are

from the 1991 NDS and 1991 NDS Supplement. Several different situationscommonly encountered in timber design are shown here, such as columns,beams, and combined beam/column members.

The member stresses ( axial, bending, and shear ) will also be calculated aspart of the verification.

Only NDS adjustment factors that are non-unity will be shown in thecalculations.

Member 1, Load Comb 3 (DL + LL + Wind)NDS Wood Shape is DFLARN1D_6X8

This is a Douglas Fir Larch, Dense No. 1, Post & Timber. DesignProperties from Table 4D of the NDS Supplement : (All values in Psi)

Fb = 1400 Ft = 950 Fv = 85 Fc = 1200 E = 1,700,000

Verification

Page 1-30

Section Properties

b=5.5, d=7.5A = 41.25 in2, S = 51.6 in3

Stress Calculations:

ft = 3980 Lbs / 41.25 in2 = 96.5 psifb = 2.4 kip-ft (12)(1000)/ 51.6 in3 = 558.6 psifv = 1.5 * (1200 Lbs) / 41.25 in2 = 43.6 psiEffective Lengths

K_in = K_out = 1.0, no sideswayLb_in = Le1 = 96 ; Lb_out = :Le2 = 48 Lu = 48, Lu/d = 6.4; Le_bend = 2.06(Lu) = 98.88 Le1/d = 12.8 Le2/b = 8.72

Applicable NDS Adjustment FactorsFb' CD = 1.6Ft' CD = 1.6Fv' CD = 1.6Fc' CD = 1.6; CM = 0.91E'

Ft' = 950(1.6) = 1520 psiFb* = 1400(1.6) = 2240 psiEqn 3.9-1 : .313

Also Check ShearFv' = 85(1.6) = 136 psiShear check = .321

Member 2, Load Comb 2 (DL + LL)NDS Wood Shape is HEMFRSEL_R6

This is a Hem Fir, Select Structural, Post & Timber Round shape. DesignProperties from Table 4D of the NDS Supplement : (All values in Psi)

Fb = 1200 Ft =800 Fv = 70 Fc = 975 E = 1,300,000

Section Properties

diameter = 6, Actual A = 28.27 in2 , S_actual = 21.21in3Equivalent square member properties d_equiv = 5.32, I_equiv = 66.75 in4


Verification

Page 1-31

fc = 5480 Lbs / 28.27 in2 = 193.8 psi

Effective Lengths

K_in = K_out = 1.0, no sideswayLb_in = Le1 = 96 ; Lb_out = :Le2 = 48 Le1/d = 18.04 Le2/b = 9.02

Fce = 1197.7 psi ( Kce = .3, c = 0.85 )Fc* = 975 psi

Applicable NDS Adjustment FactorsFb' Cf = 1.18Ft'Fv'Fc' Cp = .789E'

Fc' = 950(0.789) = 768.9 psiEqn 3.9-3 : .064

Member 3, Load Comb 3 (DL + LL + Wind)NDS Wood Shape is YLPOPNO1_2X6

This is a Yellow Poplar, No. 1, Dimension Lumber. Design Properties fromTable 4A of the NDS Supplement : (All values in Psi)

Fb = 725 Ft = 425 Fv = 75 Fc = 725 E = 1,400,000

Section Properties

b = 1.5, d = 5.5A = 8.25 in2, S = 2.06 in3


fc = 2110 Lbs / 8.25 in2 = 255.8 psifb = 0.75 kip-ft (12)(1000)/ 2.06 in3 = 4363.6 psifv = 1.5 * (375 Lbs) / 8.25 in2 = 68.2 psiEffective Lengths

K_in = K_out = 1.0, no sideswayLb_in = Le1 = 24 ; Lb_out = :Le2 = 24 Le1/d = 4.36 Le2/b = 16.0

Note : CL is 1.0 since d < b ( weak axis bending )Fce1 = 22094 psi ( Kce = .3 )Fce2 = 1640.6 psi (Kce = .3 )Fc* = 1276 psi

Verification

Page 1-32

c = 0.80

Applicable NDS Adjustment FactorsFb CD = 1.6 CF = 1.3 Cfu = 1.15Ft CD = 1.6 CF = 1.3Fv CD = 1.6Fc CD = 1.6; CF = 1.1 Cp = .770E

Fc = 725(1.6)(1.1) (.770) = 982.5 psiFb2 = 725(1.6)(1.3)(1.15) = 1734.2 psiEqn 3.9-3 : 3.048

Also Check ShearFv = 75(1.6) = 120 psiShear check = .568

Member 5, Load Comb 1 (DL Only)NDS Wood Shape is DFLANNO2_2X14

This is a Douglas Fir Larch (North), No. 2, Dimension Lumber. DesignProperties from Table 4A of the NDS Supplement : (All values in Psi)

Fb = 825 Ft =500 Fv = 95 Fc = 1350 E = 1,600,000

Section Properties

b = 1.5, d = 13.25A = 19.88 in2, S = 43.89 in3


fb = 5.82 kip-ft (12)(1000)/ 43.89 in3 = 1591.3 psifv = 1.5 * (3030 Lbs) / 19.88 in2 = 228.6 psiEffective Lengths

K_in = K_out = 1.0, no sideswayLe_bend = 60

RB = 18.8Fbe = 1982.8 psiFb* = 768.5 psi

Applicable NDS Adjustment FactorsFb' CD = 0.9 CF = 0.9 CL = 0.971 Cr = 1.15Ft' CD = 0.9 CF = 0.9

Verification

Page 1-33

Fv CD = 0.9Fc CD = 0.9; CF = 0.9E

Fb = 825(0.9)(0.9)(0.971)(1.115) = 746.2 psiEqn 3.9-3 : 2.132

Also Check ShearFv = 95(.9) = 85.8 psiShear check = 2.674

Member 6, Load Comb 3 (DL + LL + Wind)NDS Wood Shape is SOPINCON_4X4

This is a Southern Pine, Construction Grade, Dimension Lumber. DesignProperties from Table 4B of the NDS Supplement : (All values in Psi)

Fb = 1100 Ft = 625 Fv = 100 Fc = 1800 E = 1,500,000

Section Properties

b=3.5, d=3.5A = 12.25 in2


fc = 1377 Lbs / 12.25 in2 = 112.4 psi

Effective Lengths

K_in = K_out = 1.0, no sideswayLb_in = Le1 = Lb_out = Le2 = 153.7 Le1/d = Le2/b = 43.92

Fce = 233.3 psi ( Kce = .3)Fc* = 2880 psi

Applicable NDS Adjustment FactorsFb' CD = 1.6Ft' CD = 1.6Fv' CD = 1.6Fc' CD = 1.6; Cp = .080E'

Fc' = 1800(1.6)(.08) = 230.4 psiEqn 3.9-3 : .238

Verification

Page 1-34

Comparison NDS Timber Unity Check Comparisons


1 3 .313 .313

2 2 .064 .064

3 3 3.048 3.048

5 1 2.131 2.132

6 3 .240 .238

As can be seen, the results match very closely. Numerical roundoff is thecause of any slight differences.

Verification

Page 1-35

Verification Problem 11Description This problem is used to test the tapered WF sections. A typical single bay

with a sloped roof will be analyzed using tapered WF sections for thecolumns and beams. Loading will consist of vertical member projected loads, lateral member distributed loads, and member point loads. Gravity selfweight will also be applied.

Model Sketch

Validation The frame analyzed with the tapered WF sections will be compared to a Method similar frame, which is modeled with 14 piecewise prismatic sections for

each tapered WF member in the original frame. Since each tapered WFmember is modeled internally as a 14 member piecewise prismaticmember, the results should match very closely. Selected joint deflections, reactions, and member section forces will be compared. The ASD codechecks on the tapered WF sections will be compared to hand calculationsusing the ASD 9th edition steel code.

Comparison Comparison of Joint DeflectionsTapered WF Section Equivalent "piecewise" sections

Node Direction Deflection Node Direction Deflection

2 X -0.877 20 X -0.877

3 Y -3.002 21 Y -3.002

4 X 0.290 49 X 0.290The joint deflections were checked at the top left corner, peak, and top rightcorner respectively. The results match exactly.

Verification

Page 1-36

Comparison of Base ReactionsTapered WF Section Equivalent "piecewise" sections

Node X Y MZ Node X Y MZ

1 5.66 18.53 0 6 5.66 18.53 0

5 -10.86 17.09 41.75 35 -10.86 17.09 41.75

The reactions were checked at the two base nodes. A moment only appearson one side because the other side had a moment release at the bottom of thecolumn. As can be seen, the results match exactly.

Comparison of Member Section ForcesTapered WF Section Equivalent "piecewise" sections

Member SectionLoc.

LocalDirection

Value Member SectionLoc.

LocalDirection

Value

1 5 Mz 108.63 18 5 Mz 108.63

1 1 x 18.53 5 1 x 18.53

2 5 y 15.92 32 5 y 15.92

2 5 Mz -108.63 32 5 Mz -108.63

2 1 Mz 30.97 19 1 Mz 30.97

3 1 Mz -30.97 47 1 Mz -30.97

3 5 Mz 99.78 60 5 Mz 99.78

3 5 y -14.50 60 5 y -14.50

4 5 Mz -99.78 46 5 Mz -99.78

4 1 x 17.09 33 1 x 17.09

The section forces were checked at the base of the columns, at the cornerjoints, and at the peak. As can be seen, the results match exactly.

The properties of the Tapered WF section used for this analysis are shownbelow:

Tapered WF Properties

Index

Page 1-37

Taper Start Taper EndTotal Depth 7 14

Web Thickness 0.25 0.25Flange Width 6 6

Flange Thickness 0.375 0.375Area 6.062 7.812Iyy 13.51 13.52Izz 54.52 257.36rT 1.492 1.315

The ASD code check for member 2 follows.

By inspection of the member stresses, design of the J end will control. The Jend is the large end of the tapered member.

Stress Calculations: ( All stress calcs done at the large end of the member )fa = 13.238 kips / 7.812 = 1.695 Ksi (Compression)fbz = 108.63 kip-ft (12)(14/2)/257.36 = 35.46 Ksicompact section

sidesway for z-z axis onlyL = 20.4' (244.8) Lbyy = 12, Lbzz = 244.8Lcomp = 12Ky = Kz = 1.0 Note Kz will taken as 1.0 for simplicity, however it can calculated

via the figures in the Appendix F Commentary in the ASD 9thedition, pg. 5-191.

Cc = 126.1Ky*Lbyy/roy = 8.04 roy = radius of gyration at the small end = sqrt ( Iyy / A ) Kz*Lbzz/roz = 81.63 roz = radius of gyration at the small end = sqrt ( Izz / A ) Fa = 15.17 Ksi (Eqn. A-F7-2)Eqns F1-1, F1-6, F1-7, and F1-8 were checked, however only A-F7-4 will beshown. This equation usually governs the design of tapered members.

B is taken as 1.0 for simplicity, however this is to be determined from sectionAF7.4. hs = 1.1405; hw = 1.011; Fs = 281.83; Fw = 2571.1

Fbz = 23.94 > .6*Fy; Fbz = 21.6 Ksi ( Eqn A-F7-4)Eqn A-F7-14 : .1.753

The unity check obtained from RISA-2D is also 1.753, so the results matchexactly.

2DVERIFY

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