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1 Level II: Section 04 Simplified Method (optional) Section Downloads Section Downloads Handouts & Slides can be printed Version.0 Other documents cannot be printed Course binders are available for purchase Not required Download & Print TTT II Sec 04 Slides TTT II, Sec 04 Handout 3 Level II: Section 04 Version.0 Copyright WTCA

2 Required Material ANSI/TPI -00 Commentary & Appendices ANSI/TPI Outline Member Design Plate Calculations Appendix D 5 Member Design Utilizing the TPI Simplified Method Level II: Section 04 Version.0 Copyright WTCA

3 TPI Simplified Design Method Used since 978 Closely resembles traditional design methods used before computers Uniform design loads converted to point loads and added to additional concentrated loads Primary reactions are solved by summation of moments about each support 7 Assumptions Made in Simplified Method Members are connected at their ends by smooth frictionless pins Member lengths are straight Line connecting the center of the joints at each end of the member is the line of force transmission Loads and reactions are applied only at joints 8 Simplified Method Example Span (L o ) = 30 feet Slope (S) = 5/ On-Center Spacing (SP) = feet Duration of Load (DOL) = 5% Bearing (B) = 3.5 inches Top Chord Live Load (TCLL) = 30 psf Top Chord Dead Load (TCDL) = psf Bottom Chord Live Load (BCLL) = 0 psf Bottom Chord Dead Load (BCDL) = 7 psf Total Load (TL) = 47 psf 9 Level II: Section 04 Version.0 Copyright WTCA 3

4 Determine Scarf Lengths 7.8 Stin. 4.3 Sain. S Section 0: Truss Math 4" B S t = d 0.5 = = 7.8 inches S/ 5/ S a = S t - B = = 4.3 inches Determine Panel Lengths span L i = L o (S T ) in. scarf length = 360 (7.8) 4 four panels 86. in in. = 86. in. 86. Li in. 86. in. 86. in. 80 in in. Calculation of Member Forces ASD Basic Combinations. D. D + L + F + H + T + (L r or S or R) 3. D + (W or 0.7E) + L + (L r or S or R) D + W + H D + 0.7E + H The truss is designed for each loading condition above so that it will withstand the worst-case load condition. Level II: Section 04 Version.0 Copyright WTCA 4

5 Calculation of Member Forces Worst load condition: Dead-plus-Live Load Top Chord: + 30 = 40 psf 40 psf (SP) = 40 psf ( ft) = 80 plf Bottom Chord: = 7 psf = 4 plf 3 Calculation of Member Forces REACTIONS R = R = (TL) (SP) (L o ) = (47 psf) ( ft) (30 ft) = 4 # 4 # 4 # R R 4 Calculation of Member Forces 600# 3 574# 600# 4 33# 33# # # 4# 5# 0.45# 5# 4# 5 Level II: Section 04 Version.0 Copyright WTCA 5

6 Calculation of Member Forces AXIAL FORCES: Joints #/ # + ΣF V = 0 cos(.6) x F.6 θ sin(.6) x F # F 33# F # 4# sin(.6)f - 4 = 0 F = # (C) 6 Calculation of Member Forces AXIAL FORCES: Joints #/ # + ΣF H = F 9 = 0 F 9 = 50.3# (T) cos(.6) x F = 50.3# F 50.3# 7 Calculation of Member Forces AXIAL FORCES: Joint #9 + ΣF V = F 9 = 0: F 9 = 5# (T) 5# F 9 F 9 F 9 = 50.3# 50.3# F 98 + ΣF H = F 98 = 0: F 98 = 50.3# (T) 5# 9 8 Level II: Section 04 Version.0 Copyright WTCA 6

7 Calculation of Member Forces AXIAL FORCES: Joint # 600# 793.8# F 3 sin(.6)f 3 + ΣF V = sin(.6)f sin(.6)f 3 = 0 F 8 = F # # 4.3# cos(.6)f 3 F96.49# 8 sin(.6)f 8 cos(.6)f 8 5# + ΣF H = cos(.6)f 3 - cos(.6)f 8 = 0 F 3 = F 8 Therefore, F 3 = 793.8# (C) and F 8 = 96.49# (C) 9 Calculation of Member Forces AXIAL FORCES: 574# Joint # # # 793.8# 793.8# # # + ΣF V = 0 F # F 38 = 0 F 38 = # (T) 0 Calculation of Member Forces AXIAL FORCES: Joint # # 808.9# 808.9# 96.49# 96.49# 35.50# 35.50# + ΣF V = = ΣF H = = # 50.3# 0.45# Level II: Section 04 Version.0 Copyright WTCA 7

8 Calculation of Member Forces # 5# 793.8# 96.49# # 50.3# 50.3# 9 8 Calculation of Moments TOP CHORD The top chord panel point moment is calculated as follows: bending & moment length factor top chord uniform load panel length w = 7 pli Q = 0.90 L = 86. in. M pp = w(ql) 8 = 7[0.90(86.)] = 554. in.-lbs. 8 TPI -00 Table D 3 Calculation of Moments TOP CHORD The top chord mid-panel moment is calculated as follows: M mp = w(ql) 8 7(0.709 x 88.5) = = in.-lbs. 8 Q = α (cot θ) β = 0.58 (cot.6) 0.3 = TPI -00 Table D c = 0.5 TPI -00 Table D footnote 4. L = L T + L T + c(s a ) = (4.3) = 88.5 in. 4 Level II: Section 04 Version.0 Copyright WTCA 8

9 Calculation of Moments BOTTOM CHORD The bottom chord moment is calculated at mid-panel only. M bc = w(ql) 8 w = pli Q =.0 L = 88.5 in. = (.0 x 88.5) = in.-lbs. 8 TPI -00 Table D4 5 Member Design Given: The calculated axial forces Find: The required lumber to resist those forces Tension Web Compression Web Top Chord Bottom Chord 6 Tension Web Design f t = P t A = 5# (.5in.)(3.5in.) f # t = P t = A (.5in.)(3.5in.) = 0 psi = 53.4 psi F 9 = 5# f t F ' t F 38 = # Level II: Section 04 Version.0 Copyright WTCA 9

10 Tension Web Design Use the same lumber in the tension webs Design for the largest stress f t = 53.4 psi f t F ' t load duration.5 factor wet.0 service factor F ' t = F t (C D ) (C t ) (C M ) = F t (.5) temperature factor.0 base design value 8 Tension Web Design 53.4 psi F t (.5) x4 #3 Southern Pine: (.5) F t = 475 psi 3 x4 #3 SP x4 #3 SP 9 98 Compression Web Design f t c F ' tc TPI -00 (E7.3-) x4 #3 Southern Pine: F c = 975 psi f c = P c A = 96.49# = psi (.5in.)(3.5in.) 3 F 8 = 96.49# Level II: Section 04 Version.0 Copyright WTCA

11 Compression Web Design ( ) ( + F / F c ) ( F /F c ) + F / F c F' c = F c -.6 design value for compression parallel to grain adjusted by modification factors F c = F c (C D ) (C t ) (C M )= F c (.5).0 = 975 (.5) =.5 psi 3 Compression Web Design F' c = F.5 psi (0.3) E' F = (L ' / d) c.5 psi.5 psi ( + F / F c ) ( + F / F c ) ( F /F c ).6 - critical buckling design value for compression members L ' = 0.8 (L w ) = 74.6 in. actual centerline of the 93.8 web member in. effective buckling length of the compression member L w 93.8 in in. 3 Compression Web Design F' c = F c ( + F / F c ) ( + F / F c ) ( F /F c ) allowable x 6 psi modulus of elasticity (0.3) E' F = = (0.3)(.4 x 6 psi) = 69.7 psi (L ' / d) (74.6 in./.5 in.) in cross-sectional.5 in. dimension of compression member 33 Level II: Section 04 Version.0 Copyright WTCA

12 Compression Web Design 69.7 psi.5 psi + F / F F' c = F c.6.5 psi F' c =.5.5 psi ( c ) ( ) ( c ) - + F / F c F /F psi (( / 69.7/.5) ) ( / ) ( /.5) (69.7/.5) 65.08/.5) F ' c = psi 34 Compression Web Design stress in the member psi f c F ' c adjusted psi lumber design value 3 x4 #3 # SP Compression Web Design F c = F c (C D ) (C t ) (C M ) = F c (.5) = 850 (.5) = 7.5 psi x4 #3 SP: F c = 975 psi x4 # SP: F c = 850 psi x4 #3 SP: F c =.5 psi x4 # SP: F c = 7.5 psi F = (0.3) E' (L / d) = (0.3)(.7 x 6 psi) = psi (74.6 in./.5 in.) 36 Level II: Section 04 Version.0 Copyright WTCA

13 Compression Web Design F' c = F F' c = c ( + F / F c ) ( + F / F c ) ( F /F c ) ( ( / /.5 7.5) ) ( /.5) ( / 7.5) ( 65.08/.5 (06.08/7.5) ) F ' c = 0.85 psi 37 Compression Web Design stress in the member psi f c F ' c adjusted 0.85 lumber psi design value L ' /d in..5 in. 38 Top Chord Design Requires that two checks be made: At panel point conditions Negative moment At mid-panel conditions Positive moment Continuous 3 / 8 in. or thicker plywood sheathing attached to the narrow face of the chord 39 Level II: Section 04 Version.0 Copyright WTCA 3

14 Top Chord Panel Point F b = 850 psi F t = 600 psi F c = 0 psi E =.8 x 6 psi F = # (C) M pp = 554. in.-lbs. x4 SS SP Top Chord Panel Point Experiences both axial compression & bending Combined Stress Analysis actual compression stress parallel to grain 56.5 (P/A) psi f ( c ) F [ + ' c actual bending stress f bx f c F ' b( - allowable design value for compression parallel to grain F CEx )] f c = P A =709.77# 5.5 in = 56.5 psi f by + f F ' by[ c - - F CEy zero f ( bx F be ) ].00 allowable design value for bending NDS 0/05 Equation Top Chord Panel Point F ' c = F c = 0 psi (C D ) = 0 psi (.5) = 45 psi f bx = M pp S 554. in.-lbs. = = psi in. 3 F b = ( + F be / F F b ) b[.9 - ( + F be / F b ) (F be / F b ) 0.95 ] S = bh 6 4 Level II: Section 04 Version.0 Copyright WTCA 4

15 Top Chord Panel Point F ' b = ( + F be / F F b ) b[.9 - ( + F be / F b ) (F be / F b ) 0.95 ] design value for bending adjusted by modification factors F b = F b (C D ) (C M ) (C t ) (C r ) = 850 psi (.5) = psi Top Chord Panel Point F ' b = F b F ' b = psi 44 Top Chord Panel Point f c f ( ) bx F ' + c ( ).00 F ' b() ( ) psi 56.5 psi 45 psi ( ) = psi 45 Level II: Section 04 Version.0 Copyright WTCA 5

16 Top Chord Mid-Panel F = # (C) M mp = in.-lbs Top Chord Mid-Panel Combined Stress actual compression stress parallel to 56.5 grain (P/A) psi f ( c ) F [ + ' c actual bending 8.36 stress (vertical) psi F ' b( - allowable design value for 45 compression psi parallel to grain f bx = M mp S = f bx f c F CEx )] actual bending stress (horizontal) 0 psi f by + f F ' by[ c - - F CEy in.-lbs in. 3 = 8.36 psi f ( bx F be )].00 NDS 0/05 Equation Top Chord Mid-Panel 56.5 psi 8.36 psi f f bx f by ( c ) F [ + f ' c F ' c b( - + f F ' c by[ psi f F CEx )] bx F CEy ( F be ) ] psi critical buckling design value for compression members allowable modulus of elasticity F CEx = (0.3)E' ( L' d ) buckling stiffness factor E ' = E (C T ) (C M ).0 effective buckling length angle of the chord with respect to the C T = + 300L' L ' = QL (secθ h ) horizontal ke L ' = 0.90 (87.75 in.) [sec(.6 )] = 85.0 in (85.0 in.) C T = (.8 x 6 psi) =.84 E ' = E (C T ) (C M ) = (.7 x 6 ) (.84) (.0)=.03 x 6 psi 48 Level II: Section 04 Version.0 Copyright WTCA 6

17 Top Chord Mid-Panel 56.5 psi 8.36 psi f f bx f 0 psi by ( c ) F [ + f ' c F ' c b( - F CEx )] + f F ' c f by[ bx - - F CEy F be 45 psi critical buckling design value for compression members.03 x 6 psi F CEx = (0.3)E' ( L' d ) = (0.3) (.03 x 6 psi) 85.0 in. ( ( )] in. 3.5 in. ) = 3.9 psi critical dimension 3.5 in. 49 Top Chord Mid-Panel 56.5 psi 8.36 psi 0 psi f f bx f by ( c ) F [ + f ' c F ' c b( - F CEx )] + f F ' c f by[ bx - - F CEy ( F be )] psi allowable design value for bending 3.9 psi F ' b = F b = psi 50 Top Chord Mid-Panel f ( c ) F [ ' + c F ' b( - f bx f c F CEx )] psi 8.36 psi ( 45 psi ) [ psi 56.5 psi 3.9 psi ( - ) ] x4 SS SP = Level II: Section 04 Version.0 Copyright WTCA 7

18 Bottom Chord Design Subjected to combined axial load and flexure Generally carries a tensile load May also carry a bending load actual tension stress parallel to grain allowable design value for tension f t F ' t NDS 0/05 Equation 3.9- & 3.9- actual bending stress + f b.00 F b design value for bending adjusted by modification f b -f t.00 factors except C L F b design value for bending adjusted by modification factors except C V 5 Bottom Chord Design F 8-9 = F 9- = 50.3 # (T) M bc = in.-lbs. 3 Only analyzed at mid-panel F b = 850 psi F t = 50 psi F c = 850 psi E =.7 x 6 psi 50.3# x4 # SP 50.3# Bottom Chord Design actual tension stress parallel to grain allowable design value for tension f t F ' t + f b F b f b -f t F b f t = P A = 50.3# 5.5 in = psi F ' t = F t (C D ) (C t ) (C M ) = 50 psi (C D ) = 50 psi (.5) = 7.5 psi 54 Level II: Section 04 Version.0 Copyright WTCA 8

19 Bottom Chord Design f b = M S f t F ' t + f b F b f b -f t F b actual bending stress = in.-lbs in 3 = psi.5 design value for bending adjusted by modification factors except C L F b = F b (C D ) (C M ) (C t ) (C F ) (C V ) (C fu ) (C i ) (C r ) (C c ) (C f ) F b = 850 psi (.5) = 7.5 psi 55 Bottom Chord Design f t F ' t + f b F b.00 f b -f t F b.00 design value for bending adjusted by modification factors except C V F b = F b (C D ) (C M ) (C t ) (C F ) (C L ) (C fu ) (C i ) (C r ) (C c ) (C f ) F b = 850 psi (.5) = 7.5 psi 56 Bottom Chord Design f t F ' t + f b F b psi psi + = psi 7.5 psi f b -f t F b psi psi.00 = psi x4 # SP 57 Level II: Section 04 Version.0 Copyright WTCA 9

20 Quiz 58 Plate Calculations Chapter 8 Metal Connector Plate Analysis # # 9 8 Lateral Resistance Shear Tension 60 Level II: Section 04 Version.0 Copyright WTCA 0

21 Lateral Resistance Heel Reduction Factor (H R ) Developed to allow for moments Multiplied by lateral resistance values H R = ( tanθ -.0) ANSI/TPI -00 (E8.4-) 0.65 H R 0.85 H R = [ tan(.6 ) -.0] = Lateral Resistance minimum required metal connector plate contact area for each member A p = P 0.8(V ' LR) force in the wood member ANSI/TPI -00 E8.4- allowable lateral resistance per metal connector plate unit adjusted per all applicable factors V ' LR = V LR (C D ) (C M ) (C q ) (H R ) load duration factor wet service factor quality control factor heel reduction factor 6 Lateral Resistance Metal connector plates are proprietary products Design values developed by test Listed in evaluation reports Assume design values for this analysis V LRAA = 00 psi V LRAE = 40 psi V LREA = 60 psi V LREE = 50 psi 63 Level II: Section 04 Version.0 Copyright WTCA

22 Lateral Resistance Minimum Axial Design Force 375 pounds ANSI/TPI F - = # 375# F 9- = # 375# 64 Lateral Resistance A p = # P 0.8(V ' LR) = 9.93 in. 0 in. 4" x 6" pending a shear check V ' LR = V LR (C D ) (C M ) (C q ) (H R ) = 6 psi 00 psi quality control factor 65 Shear V s = [(R s )(F vs )(t )] ANSI/TPI -00 E8.6- force parallel to the joint across the shear plane calculated shear length of a metal connector plate average shear resistance ANSI/TPI effectiveness proprietary Section value ratio P s l = [(R s )(F vs )(t )] ASTM allowable 33,000 A653/A 653 shear psim-98 stress of the metal mean thickness of the ANSI/TPI -00 Section metal connector plate Minimum ANSI/TPI steel -00 thickness Section of in. the metal connector plate t = t - t c 0.95 coating ANSI/TPI 0.00 thickness -00 in. Section AISI indicates that the steel may be 95% 66 Level II: Section 04 Version.0 Copyright WTCA

23 Shear t = t - t c 0.95 = " - (0.00") 0.95 = in. 67 Shear shear length of the plate P s l = [(R s )(F vs )(t )] P s # l = = [(Rs )(F vs )(t )] [(0.5)(33,000 psi)( in.)] =.4 in. 4x6 plate: l = 7.0 in..4 in. 4" x 6" lateral resistance and shear checks 68 Tension ANSI/TPI -00 Section 8.5 Plates are designed based on the orientation of the plates relative to the load direction V t = [(R t )(F st )(t )] ANSI/TPI -00 E8.5- P t W p = [(R t )(F st )(t )] 69 Level II: Section 04 Version.0 Copyright WTCA 3

24 Tension axial tensile force in the wood member gross width of the plate average shear resistance ANSI/TPI effectiveness proprietary Section value ratio P t = [(R t )(F st )(t )] W p allowable 33,000 tensile psi stress of the metal mean thickness of the in. metal connector plate 70 Tension gross width of the plate P t W p = [(R t )(F st )(t )] P t W p = [(Rt )(F st )(t )] # = [(0.5)(33,000 psi)( in.)] =.97 in..97" or wider plate would be specified for each side Plate length would be determined based on lateral resistance checks 7 Quiz 7 Level II: Section 04 Version.0 Copyright WTCA 4

25 Feedback Level II: Section 04 Version.0 Copyright WTCA 5