2015/2018 NDS Example Problems...2020/04/16 · 2015/2018 NDS Example Problems Member Designs and Connection Basics (DES 221) Lori Koch, P.E. Manager, Educational Outreach American

2015/2018 NDS Example ProblemsMember Designs and Connection Basics (DES 221)

Lori Koch, P.E.Manager, Educational OutreachAmerican Wood Council

1

2 0 1 5 / 2 0 1 8 N D S E x a m p l e P r o b l e m s

This presentation is protected by US and International Copyright laws. Reproduction, distribution, display and use of the presentation without written permission of AWC is

prohibited.

© American Wood Council 2020

COPYRIGHT MATERIALS

2


This course is registered with AIA CES for continuing professional education. As such, it does not include content that may be deemed or construed to be an approval or endorsement by the AIA of any material of construction or any method or manner ofhandling, using, distributing, or dealing in any material or product.

Questions related to specific materials, methods, and services will be addressed at the conclusion of this presentation.

The American Wood Council is a Registered Provider with the American Institute of Architects Continuing Education Systems (AIA/CES), Provider #50111237.

Credit(s) earned on completion of this course will be reported to AIA CES for AIA members. Certificates of Completion for both AIA members and non-AIA members are available upon request.

3


LEARNING OBJECTIVESUpon completion, participants will be better able to identify:

Discuss application of NDS design provisions for beams, columns, connections, and calculating design values.

Documentation

Apply the reference design values from the NDS Supplement.

Resources

Explain the design value adjustment factors from the NDS and NDS Supplement.

Adjustment Factors

Recognize the updated design provisions for connections new to the 2018 NDS.

Updated Provisions

1 3

2 4

4


POLLING QUESTION

1. What is your profession?a) Architectb) Engineerc) Code Officiald) Fire Servicee) Builder/Product Manufacturer/Other

5


OUTLINE

• Example Problems Document• Member Design Examples• Connection Design Examples

• Shear• Lateral

6


DOCUMENTS

• 2018 NDS

• 2018 NDS Supplement

• Examples document – NEW!

• https://awc.org/codes-standards/publications/nds-2018

7

https://awc.org/codes-standards/publications/nds-2018

E1.2a - Simply Supported Beam Capacity Check (ASD)

A Select Structural Douglas Fir-Larch (DF-L) nominal 4X16 beam on a 20 ft span supports a hoist located at thecenter of the span. Determine the maximum allowable load on the hoist (including its weight) based onbending. Assume normal load duration. The beam is supported on a 2x4 top plate. Lateral support is providedonly at the ends of the member and the ends are considered pinned.

Check beam's capacity to resist shear stress from maximum (moment controlled) load; determine deflectionfrom maximum load and check bearing capacity.

Notes: Load cases used in this example have been simplified for clarity. Refer to NDS Section 1.4.4 for requirementson load combinations.

8


NDS SUPPLEMENT

• Unadjusted design values

9

E1.2a - Simply Supported Beam Capacity Check (ASD)

A Select Structural Douglas Fir-Larch (DF-L) nominal 4X16 beam on a 20 ft span supports a hoist located at thecenter of the span. Determine the maximum allowable load on the hoist (including its weight) based onbending. Assume normal load duration. The beam is supported on a 2x4 top plate. Lateral support is providedonly at the ends of the member and the ends are considered pinned.

Check beam's capacity to resist shear stress from maximum (moment controlled) load; determine deflectionfrom maximum load and check bearing capacity.

Notes: Load cases used in this example have been simplified for clarity. Refer to NDS Section 1.4.4 for requirementson load combinations.

Reference and Adjusted Design Values for 4x16 Select Structural DF-L (size adjusted 4x12 values)

Fb 1500 psi E 1900000 psi Emin 690000 psi (Table 4A)

Fc⊥ 625 psi Fv 180 psi

CD 1.0 CM 1.0 Ct 1.0

Cfu 1.0 Cr 1.0 Ci 1.0

CT 1.0 CF 1.0 (Table 4A 14" and wider)

E' E CM Ct Ci E'min Emin CM Ct Ci CT

E' 1900000 psi E'min 690000 psi

Member dimensions and properties

l 20 ft b 3.5 in d 15.25 in wbearing 3.5 in (width of bearing)

Ib d312

Ag b d Sb d26

Ag 53.38 in2

S 135.66 in3 I 1034 in4

Beam Stability Factor

F'b* Fb CD CM Ct CF Ci Cr F'b* is adjusted bending design value with all adjustmentfactors except the beam stability factor CL and flat use factorCfu applied.

F'b* 1500 psi

lu 12inft l lu 240 in Laterally unsupported length

lud

15.7 lu/d >7 (Table 3.3.3)

le 1.37 lu 3 d le 375 in (Table 3.3.3)

10

RBle d

b2 RB 21.6 Slenderness ratio for bending (3.3-5)

FbE1.20 E'min

RB2

FbE 1776 psi Critical bucking design value for bending (3.7.1)

CL

1FbEF'b*

1.9

1FbEF'b*

1.9

2 FbEF'b*

0.95

CL 0.876

F'b F'b* Cfu CL F'b 1313 psi Adjusted bending design value with all adjustment factors

Determine Maximum Moment Allowed on Beam

Maximum total moment is the adjusted bending design value F'b times the section modulus S

Mmax F'bS

12inft

Mmax 14849 ftꞏlbf

Determine Maximum Hoist Load P

Maximum hoist load P is determined from subtracting moment due to beam weight from the maximum total momentallowed on the beam and solving for hoist load P. Load P creates a moment on beam length L of PL/4. Assumedensity of beam material is 37.5 lbs/ft3 (110% of tabulated of the specific gravity G for Southern Pine).

ρ 37.5lbf

ft3 wbeamweight ρ

b

12inft

d

12inft

wbeamweight 13.9 plf Note:wbeamweight is self weight of beam

Mbeamweight is moment due to selfweight Mallow is maximum allowablemoment due to applied hoist load

Mbeamweightwbeamweight l( )

2

8 Mbeamweight 695 ftꞏlbf

Mallow Mmax Mbeamweight Mallow 14154 ftꞏlbf

P 4Mallow

l

Result: The total allowable concentrated moment-limited midspan load (hoist plus payload) is P 2831 lbf

11

Check Beam's Capacity to Resist Shear from Maximum (bending controlled) Load

VP2

V 1415 lbf

fv3 V2 b d

fv 40 psi

F'v Fv CD CM Ct Ci F'v 180 psi fv

E1.5b - Compression Member Analysis (LRFD)

A No 2 Spruce Pine Fir (SPF) nominal 2X6 interior bearing stud, 91.5 inches long, sheathed on both sides withgypsum board, carries dead load and snow load from the roof (assume load combination 1.2D + 1.6S, λ=0.8).Determine CP and the allowable compression parallel to grain design value (Fc') for the stud. Assume studs areplaced 16" on center and top and bottom plates are the same grade and species. Determine axial loads basedon buckling and bearing limit states.

13

E1.5b - Compression Member Analysis (LRFD)

A No 2 Spruce Pine Fir (SPF) nominal 2X6 interior bearing stud, 91.5 inches long, sheathed on both sides withgypsum board, carries dead load and snow load from the roof (assume load combination 1.2D + 1.6S, λ=0.8).Determine CP and the allowable compression parallel to grain design value (Fc') for the stud. Assume studs areplaced 16" on center and top and bottom plates are the same grade and species. Determine axial loads basedon buckling and bearing limit states.

Reference and Adjusted Design Values for No. 2 SPF 2x6

Fc 1150 psi Emin 510000 psi Fc⊥ 425 psi (NDS Table 4A)

λ 0.8 CM 1.0 Ct 1.0 CF 1.1 Ci 1.0 CT 1.0 (NDS Table 4.3.1 and Appendix N)

KFc 2.40 ϕc 0.9 KFEmin 1.76 ϕEmin 0.85

Kc⊥ 1.67 ϕc⊥ 0.9

E'min Emin CM Ct Ci CT KFEmin ϕEmin E'min 762960 psi

F'c⊥ Fc⊥ CM Ct Ci Kc⊥ ϕc⊥ F'c⊥ 638.775 psi

Member length and properties

l 91.5 in b 1.5 in d 5.5 in

Column Stability FactorFc* Fc CM Ct CF Ci KFc ϕc λ Fc* is adjusted bending design value with all adjustment

factors except the column stability factor CP ,Fc* 2186 psi

le2 0 le1 l Effective lengths of compression member in planes of lateralsupport. Strong axis buckling controls

le2b

0le1d

16.636

F'c Fc* CP F'c 1537 psi F'c is adjusted compression design value with all adjustment

factors.

Determine Axial Loads Based on Buckling and Bearing

PBuckling b d F'c PBuckling 12683 lbf

PBearing b d F'c⊥ PBearing 5270 lbf Perpendicular to grain bearing

PBearing2 b d Fc* PBearing2 18034 lbf Parallel to grain bearing

Note: Bearing area factor (Cb) can be used to increase the bearing controlled load on interior studs. The bearingfactor for the 1-1/2 bearing length measured parallel to grain is 1.25 (NDS Equation 3.10-2 and Table 3.10.4) Cb 1.25

PBearingIncreased b d F'c⊥ Cb PBearingIncreased 6587 lbf Controlling Value

Note: With a 3:1 snow to dead load ratio, this translates to 3,294 lbs snow load and 1098 lbs dead load.

15


POLLING QUESTION

2. The aspect ratio (length divided by depth) of a compression member needs to be less than 50.

a) Trueb) False

16

E1.6 - Combined Bending and Axial Tension Loading of a Truss Chord Member (ASD)

A No. 2 Hem-Fir nominal 2x8 is considered for use as the bottom chord of a 24-ft roof truss (12 ft between panelpoints). The chord will be subject to a uniform dead load of 8 psf as well as tension forces (assuming pinnedconnections) of 880 lb from roof wind loads (WL), 880 lb from roof live (RLL) and 1420 lb from dead loads (DL).Trusses are to be spaced 4 ft on center. Framing will have a 19% (max) moisture content. Check the adequacyof the bottom chord member for bending and tension for the appropriate load cases.

Note: Load cases used in this example have been simplified for clarity. Refer to NDS Section 1.4.4 forrequirements on load combinations.

17

E1.6 - Combined Bending and Axial Tension Loading of a Truss Chord Member (ASD)

A No. 2 Hem-Fir nominal 2x8 is considered for use as the bottom chord of a 24-ft roof truss (12 ft between panelpoints). The chord will be subject to a uniform dead load of 8 psf as well as tension forces (assuming pinnedconnections) of 880 lb from roof wind loads (WL), 880 lb from roof live (RLL) and 1420 lb from dead loads (DL).Trusses are to be spaced 4 ft on center. Framing will have a 19% (max) moisture content. Check the adequacyof the bottom chord member for bending and tension for the appropriate load cases.

Note: Load cases used in this example have been simplified for clarity. Refer to NDS Section 1.4.4 forrequirements on load combinations.

Reference and Adjusted Design Values for No. 2 Hem-Fir 2x8

Fb 850 psi E 1300000 psi Emin 470000 psi (NDS Supplement Table 4A)

Ft 525 psi CM 1.0 Ct 1.0 CF 1.2 (NDS Table 4.3.1)

Cfu 1.0 Ci 1.0 Cr 1.0

CT 1.0

E' E CM Ct Ci E'min Emin CM Ct Ci CT

E' 1300000 psi E'min 470000 psi

Member length and properties

l 12 ft b 1.5 in d 7.25 in

Ag b d Sb d26

Ag 10.875 in2

S 13.141 in3

Applied Loads

wD 8lbf

ft2 wtrib 4 ft Twind 880 lbf TLive 880 lbf TDead 1420 lbf

18

Load Case 1: DL + RLL + WL

CD 1.6 NDS Appendix B Section B.2 (non-mandatory)

Tension

Ft' Ft CD CM Ct CF Ci Adjusted tension parallel to grain design value for shortduration loads (NDS 2.3.1 and 4.3.1)

Ft' 1008 psi

T1 Twind TLive TDead Subscripts refer to Load Case

T1 3180 lbf

ft1T1Ag

Tensile stress in bottom chord

ft1 292 psi Ft' 1008 psi Actual tension stress is less than adjusted tension parallel todesign value. OK (NDS 3.8.1)

BendingF'b* is adjusted bending design value with all adjustmentfactors except the beam stability factor CL and flat use factorCfu applied. The following calculations determine the beamstabilty factor CL:

F'b* Fb CD CM Ct CF Ci Cr

F'b* 1632 psi

Determine Beam Stability Factor CL (NDS 3.3.3)

lu 12inft l lu 144 in Laterally unsupported length

lud

19.9 lu/d >7 (NDS Table 3.3.3)

le 1.63 lu 3 d le 256.5 in (NDS Table 3.3.3)

RBle d

b2 RB 28.75 RB < 50 OK (NDS 3.3.3.7)

FbE1.20 E'min

RB2

FbE 682 psi (NDS 3.3.3.6)

CL

1FbEF'b*

1.9

1FbEF'b*

1.9

2 FbEF'b*

0.95 (NDS Equation 3.3-6)

CL 0.404 Resulting beam stability factor CL.

19

F'b F'b* CL Cfu F'b is the fully adjusted bending design value with alladjustment factors including the beam stability factor CLand flat use factor applied F'b 660 psi

F'b** F'b Since Cv does not apply to solid sawn lumber, F'b** isequal to F'b

F'b** 660 psi

Bending resulting from dead loadMmax

wD wtrib l2 12inft

8

Mmax 6912 inꞏlbf

fbMmaxS

fb 526 psi

fb 526 psi F'b 660 psi Ok. Actual bending stress fb does not exceed adjustedbending design value F'b

Combined Bending and Axial Tension ft1Ft'

fbF'b*

0.61

Load Case 2: DL+RLL

CD 1.25 Appendix B Section B.2 (non-mandatory). Roof Live Load isa construction load.

Tension

Ft' Ft CD CM Ct CF Ci Adjusted tension parallel to grain design value for shortduration loads (NDS 2.3.1 and 4.3.1)

Ft' 787.5 psi

T2 TLive TDead

T2 2300 lbf

ft2T2Ag

ft2 211 psi Ft' 787 psi


F'b* 1275 psi

CL

1FbEF'b*

1.9

1FbEF'b*

1.9

2 FbEF'b*

0.95

CL 0.509

F'b F'b* CL Cfu

F'b 649 psi

F'b** F'b

F'b** 649 psi

fb 526 psi F'b 649 psi

Combined Bending and Axial Tension ft2Ft'

fbF'b*

0.68

Load Case 3: DL only

CD 0.9

Tension

Ft' Ft CD CF CM Ci Ct

Ft' 567 psi

T3 TDead

T3 1420 lbf

ft3T3Ag

ft3 131 psi Ft' 567 psi

Bending


F'b* 918 psi

CL

1FbEF'b*

1.9

1FbEF'b*

1.9

2 FbEF'b*

0.95

CL 0.674

F'b F'b* CL Cfu

F'b 619 psi

F'b** F'b

F'b** 619 psi

fb 526 psi F'b 619 psi

Combined Bending and Axial Tension

ft3Ft'

fbF'b*

0.8

E2.1b - Fastener Uplift Capacity - Roof Sheathing Ring Shank Nail in 5/8" WSP (2018 NDS Only)Using 2018 NDS section 12.2, calculate the Allowable Stress Design (ASD) reference withdrawaldesign value in pounds (capacity) and head pull-through design value in pounds (capacity) of a0.131" diameter, 3" long roof sheathing ring shank (RSRS-05) nail in the narrow face of a DouglasFir-Larch nominal 2x6 with a 5/8 in. thick Douglas-Fir Wood Structural Panel (plywood ororiented strand board) side member. Assume all adjustment factors are unity. Main member: Douglas Fir-Larch (DFL) 2x6 (G = 0.5) Side member: 5/8 in. thick Wood Structural Panel (WSP) (G = 0.5) Fastener Dimensions: Dash No. 05 (NDS Table L6) Length = 3 in. Diameter = 0.131 in. Head diameter = 0.281 in. TL = 1.5 in.

23

E2.1b - Fastener Uplift Capacity - Roof Sheathing Ring Shank Nail in 5/8" WSP (2018 NDS Only)Using 2018 NDS section 12.2, calculate the Allowable Stress Design (ASD) reference withdrawaldesign value in pounds (capacity) and head pull-through design value in pounds (capacity) of a0.131" diameter, 3" long roof sheathing ring shank (RSRS-05) nail in the narrow face of a DouglasFir-Larch nominal 2x6 with a 5/8 in. thick Douglas-Fir Wood Structural Panel (plywood ororiented strand board) side member. Assume all adjustment factors are unity. Main member: Douglas Fir-Larch (DFL) 2x6 (G = 0.5) Side member: 5/8 in. thick Wood Structural Panel (WSP) (G = 0.5) Fastener Dimensions: Dash No. 05 (NDS Table L6) Length = 3 in. Diameter = 0.131 in. Head diameter = 0.281 in. TL = 1.5 in.

D 0.131 Fastener diameter (in.)

DH 0.281 Fastener head diameter (in.)

TL 1.5 Deformed Shank Length (in.)

tns 0.625 Net Side Member thickness (in.)

G 0.5 Specific gravity, main and side members (NDS Table 12.3.3A and 12.3.3B)

Checking Fastener Withdrawal

W 1800 G2 D NDS Equation 12.2-5

W 59 Reference withdrawal design value. Compare to NDS Table 12.2E, W =59 lbs/in

W TL 88 Reference withdrawal design value based on deformed shank fastenerpenetration (TL) in main member (lbs)

Checking Fastener Head Pull-Through

tns 0.625

2.5DH 0.703 2.5DH greater than tns, so NDS Equation 12.2-6a applies

WH 690 π DH G2

tns NDS Equation 12.2-6a

WH 95 Head pull-through design value (lbs). Compare to NDS Table 12.2F, WH = 95 lbs

Fastener head pull-through design value of 95 lbs is greater than withdrawal design value of 88 lbs;withdrawal controls design capacity. See NDS Table 11.3.1 for application of additional adjustmentfactors for connections based on end use conditions.

24

E2.3 - Withdrawal Design Value - Lag Screw

Using 2015/2018 NDS provisions (NDS 12.2) calculate the Allowable Stress Design (ASD)withdrawal design value of a lag screw in the connection below. Assume all adjustment factorsare unity. Main member: Southern Pine Nominal 6x (Actual thickness = 5.5 in.) (G = 0.55) (NDS Table 12.3.3A) Side member: Southern Pine Nominal 2x (Actual thickness = 1.5 in.) (G = 0.55) (NDS Table 12.3.3A) Fastener Dimensions: 1/2 in. diameter lag screw (NDS Table L2) Length = 4 in. Tip Length = 0.3125 in.

25

E2.3 - Withdrawal Design Value - Lag ScrewUsing 2015/2018 NDS provisions (NDS 12.2) calculate the Allowable Stress Design (ASD)withdrawal design value of a lag screw in the connection below. Assume all adjustment factorsare unity. Main member: Southern Pine Nominal 6x (Actual thickness = 5.5 in.) (G = 0.55) (NDS Table 12.3.3A) Side member: Southern Pine Nominal 2x (Actual thickness = 1.5 in.) (G = 0.55) (NDS Table 12.3.3A) Fastener Dimensions: 1/2 in. diameter lag screw (NDS Table L2) Length = 4 in. Tip Length = 0.3125 in.

D 0.5 Fastener diameter (in.)

tip 0.3125 Fastener tapered tip length (in.)

G 0.55 Specific gravity (NDS Table 12.3.3A)

L 4 Lag screw length (in.)

Ls 1.5 Side Member thickness (in.)

pt L Ls tip Lag screw penetration into main member (in.)

Note: Per Table L2, the unthreaded body length, S = 1.5".Therefore, the threaded portion begins at the wood-to-woodinterface. Note pt also matches Table L2 dimension T-E = 2-3/16".Longer lag screws have different thread length dimensionsrequiring evaluation to determine actual thread penetration.

pt 2.188

W 1800 G

32

D

34

NDS Equation 12.2-1

W 436.6 Compare to NDS Reference Withdrawal Design Value Table12.2A, W = 437 lbs/in.

W pt 955 Withdrawal design value based on main member penetration(lbs)

See NDS Table 11.3.1 for application of additional adjustment factors for connections based on enduse conditions.

26


POLLING QUESTION

3. Fastener head pull-through analysis is required per the 2018 NDS only.a) Trueb) False

27

E2.2 - Single Common Nail Lateral Design Value - Single Shear Wood-to-wood Connection Using the 2015/2018 NDS yield limit equations in section 12.3, determine the AllowableStress Design (ASD) reference lateral design value of a single shear connection with thefollowing configuration. Assume all adjustment factors are unity.

Main member Nominal 3x Douglas Fir-Larch (Actual thickness = 2.5 in.) (G = 0.5) (NDS Table 12.3.3A)

Side member Nominal 1x Douglas Fir-Larch (Actual thickness = 0.75 in.) (G = 0.5) (NDS Table 12.3.3A) Fastener Dimensions: 10d Common Nail (NDS Table L4) D = 0.148 in. Length = 3 in.

28

E2.2 - Single Common Nail Lateral Design Value - Single Shear Wood-to-wood Connection Using the 2015/2018 NDS yield limit equations in section 12.3, determine the AllowableStress Design (ASD) reference lateral design value of a single shear connection with thefollowing configuration. Assume all adjustment factors are unity.

Main member Nominal 3x Douglas Fir-Larch (Actual thickness = 2.5 in.) (G = 0.5) (NDS Table 12.3.3A)

Side member Nominal 1x Douglas Fir-Larch (Actual thickness = 0.75 in.) (G = 0.5) (NDS Table 12.3.3A) Fastener Dimensions: 10d Common Nail (NDS Table L4) D = 0.148 in. Length = 3 in.

Define parameters:

Fem 4650 Main member Dowel Bearing Strength (NDS Table 12.3.3) (psi)

Fes 4650 Side member Dowel Bearing Strength (NDS Table 12.3.3) (psi)

ReFemFes

Re 1

Fyb 90000 Fastener dowel bending yield strength (psi) (NDS Table I1)

D 0.148 Nail Diameter (in.)

Tip 2 D Length of tapered fastener tip (in.) (NDS 12.3.5.3b)

Ls 0.75 Side member Dowel Bearing Length (in.) (NDS 12.3.5)

Lm 3 LsTip2

Main member Dowel Bearing Length (in.) (NDS 12.3.5.3)

Lm 2.1

2015 NDS 12.1.6.5 (2018 NDS 12.1.6.4) Requires minimum main member penetrationequal to 6D, Lm > 0.89 in.

Rd 2.2 Reduction Term (NDS Table 12.3.1B)

29

Calculate k 1 , k 2 , and k 3 (NDS Table 12.3.1A)

RtLmLs

Rt 2.803

k1Re 2 Re

2 1 Rt Rt

2

Rt

2 Re3

Re 1 Rt

1 Re

k1 0.935

k2 1 2 1 Re 2 Fyb 1 2 Re D2

3 Fem Lm2

k2 1.047

k3 12 1 Re

Re

2 Fyb 2 Re D23 Fem Ls

2

k3 1.347

Yield Mode Calculations (NDS Table 12.3.1A)

Mode Im

ZImD Lm Fem

Rd

ZIm 658 Yield Mode Im Solution (lbs)

Mode Is

ZIsD Ls Fes

Rd

ZIs 235 Yield Mode Is Solution (lbs)

30

Mode II

ZIIk1 D Ls Fes

Rd

ZII 219 Yield Mode II Solution (lbs)

Mode IIIm

ZIIImk2 D Lm Fem

1 2 Re Rd

ZIIIm 230 Yield Mode IIIm Solution (lbs)

Mode IIIs

ZIIIsk3 D Ls Fem

2 Re Rd

ZIIIs 105 Yield Mode IIIs Solution (lbs)

Mode IV

ZIVD2

Rd

2 Fem Fyb

3 1 Re

ZIV 118 Yield Mode IV Solution (lbs)

Zdist

ZIm

ZIs

ZII

ZIIIm

ZIIIs

ZIV

Zdist

658

235

219

230

105

118

Creating an array with all Yield Mode Solutions

31

Z min Zdist

Z 105 Minimum value of all Yield Modes provides Z-reference lateraldesign value (lbs). Mode IIIs controls. Compare to NDS Table 12Nvalue = 105 lbs. See NDS Table 11.3.1 for application ofadditional adjustment factors for connections based on end useconditions.

32


POLLING QUESTION

4. The new equations for fastener head pull‐through are based on which of the following.a) Fastener head diameterb) Species and weight of side memberc) Net side member thicknessd) All of the above.e) a. and c.

33

T h i s p r e s e n t a t i o n i s p r o t e c t e d b y U S a n d I n t e r n a t i o n a l C o p y r i g h t l a w s . R e p r o d u c t i o n , d i s t r i b u t i o n , d i s p l a y a n d u s e o f t h e p r e s e n t a t i o n w i t h o u t w r i t t e n p e r m i s s i o n o f A m e r i c a n W o o d C o u n c i l ( A W C ) i s p r o h i b i t e d . © A m e r i c a n W o o d C o u n c i l 2 0 2 0

in [email protected] | www.awc.org

34

DES221 FINAL slides with POLLSDES 221 Examples2015/2018 NDS Example ProblemsCopyright MaterialsSlide Number 3Learning ObjectivesPolling QuestionOUTLINEDocuments

Combined ExamplesMathcad - E1.2a - Solid Sawn Beam ASD

DES 221 Examples Table4aDES221 FINAL slides with POLLSCombined ExamplesMathcad - E1.5b - Column Capacity Analysis LRFD

DES 221 ExamplesPolling Question

Combined ExamplesMathcad - E1.6 - Combined Bending and Axial TensionMathcad - E2.1b - Ring Shank Nail WithdrawalMathcad - E2.3 - Lag Screw Withdrawal

DES 221 ExamplesPolling Question

Combined ExamplesMathcad - E2.2 - Single Nail Single Shear

DES 221 Example poll4DES 221 ExamplesSlide Number 10