truss work

Truss (1) layout

Usually the height of truss is recommended to be in the range (1/12-1/8), we chose truss height 1.9m which is about (1/10.8). And slope = 18.4% which is good for drainage.

Loads;

1. Snow loads: snow load was taken to be = 1.4 kn/m2.2. Live loads: for roof truss= .75 kn/m23. Dead load: weight of truss is calculated by computer program,

Weight of metal deck = 0.1 kn/m2 according to sheets manuals, Weight of finishing (fall ceiling, mechanical and electrical) = 0.25 kn/m2 Purlins self weight = 0.1 kn/m2 Truss self weight is calculated by SAP program.

Dead load = 0.45 kn/m2

4. Wind loads: wind load was calculated according to British code.(details in )

Point loads on joints:

For internal joint →Tributary area = 1.93 x 8.2 = 15.99 m2

For external joint →Tributary area = 8 m2

D.L Int.j =.45x15.99 Ext.j=.45x8

7.19 kn3.6 kn

L.L Int.j =.75x15.99Ext.j=.75x8

11.99 kn6 kn

S.L Int.j =1.4x15.99Ext.j=1.4x8

22.4 kn 11.2 kn

W.L Wy: Int.j = -.482x15.99xcos11.2 Ext.j = -.482x8xcos11.2Wx: =± .482x15.99xsin11.2

-7.56 kn-3.77 kn±1.5 kn

Results from SAP:

Deflection:

Maximum deflection should not exceed L/360 = 4.22 cm

From SAP2000:

Maximum deflection = 2.25 cm < max O.K

Table: Steel Design 1 - Summary Data - AISC-LRFD99, Part 1 of 2

Frame DesignSect DesignType Status Ratio Pu

KN

1 TS4X4X5/16 Column No Messages 0.195103 -32.226

2 TS2.5X2.5X3/16 Column No Messages 0.375563 51.349

3 TS2.5X2.5X3/16 Column No Messages 0.071025 -9.941







10 TS4X4X3/16 Beam No Messages 0.877974 382.755








18 TS4X4X5/16 Brace No Messages 0.185005 -17.521









Frame DesignSect DesignType Status Ratio Pu

KN


27 TS2.5X2.5X3/16 Brace No Messages 0.658291 -101.658

28 TS2.5X2.5X3/16 Brace No Messages 0.043506 16.315




32 TS2.5X2.5X3/16 Brace No Messages 0.658291 -101.658


Check sections manually for truss 1:

Take two elements;

Element 20 (TS 4x4x5/16)

Compression force= 474Kn=106 kips, fy=50 ksi, fu=65 ksi, L=1.93m=6.33 ft

Slenderness ratio = KLr

¿1× 6.33× 12

1.48<¿ 200

=51.3 ¿ 200 OK

λc= KLrπ

√( fyE

) =1× 6.33× 12

1.48× π√ ( 50

29000) = 0.678

λc ¿1.5

Then; Fcr= .658λc 2fy = 41.24 ksi

ΦcPn= .85xAgxFcr= 0.85x5.36x41.24 = 187.9 kips

187.9 >106 kips OK

Element 30 (TS 2.5x2.5x3/16)

Tensile force= 78.4 Kn=17.3 kips, fy=50 ksi, fu=65 ksi, L=2.69m=8.83 ft, Ag= 1.64 in2

Yield criterion:

ΦTn=0.9xfyxAg = 0.9 x 50 x 1.64 = 73.8 kips >17.2 kips OK

Fracture criterion:

Ae = 0.85 Ag =1.39 in2

ΦTn=0.75xfuxAe = 0.75 x 65 x 1.39 = 67.9kips >17.3 kips OK

Slenderness ratio = Lr=8.83 ×12

0.93 ¿ 300

= 113.9 ¿ 300 OK

Element 17 (TS 4x4x3/16)

Tensile force= 473 Kn=104 kips, fy=50 ksi, fu=65 ksi, L=1.9m=6.23ft, Ag= 2.77 in2

Yield criterion:

ΦTn=0.9xfyxAg = 0.9 x 50 x 2.77 = 124.6 kips >104 kips OK

Fracture criterion:

Ae = 0.85 Ag =2.35 in2

ΦTn=0.75xfuxAe = 0.75 x 65 x 2.35 = 114.5kips >104 kips OK

Slenderness ratio = Lr=6.12 ×12

1.54 ¿ 300

= 47.7 ¿ 300 OK

Design of Purlins and Sag rods:

Purlines

WD= WD- Self weight of truss= .45 kn/m2

W(S or L) => S¿ L so snow load control: WS= 1.4kn/m2

WU= 1.2x.45 +1.6x1.4= 2.78 kn/m2

Spaces between Purlines = 1.93m so W total/m =1.93x2.78 =5.36 kn/m

Fx/m= 5.36 x sin11.2= 1.04 kn/m

Fy/m= 5.36 x cos11.2= 5.26 kn/m

Mu= Wl2/8 = 5.26x8.22/8 = 44.2 kn.m = 31.96 kip.ft

Mn= 31.96/.9= 35.5 kip.ft

Zx=Mnx12/fy=8.52 in3

Choose W8x10

Sag Rods;

Spacing between sag rods =13

(bay)=13

x 8.25=2.75m

Tu horizontal on purlins = 1.04 kn/m

Sag rod will be design on top rod so tributary area for load will be from top to peak of the truss.

Wu=2.78 kn/m

Tu= 2.78 x (2.75x7.73) x sin11.2=11.4 kn= 2.51 kips

Require Ag= 2.51

.75 x .75 x65=0.069 in2

Use for all sag rods Φ1/2’’ (Ag=0.196 in2) at spacing 2.75 m.

Welded connection:

Pu = 78.5 KN = 17.72 kips

Pu/ in weld = 17.27/(2..5*4) = 1.73 kips/in

Use E70 fillet weld and try 3/16” thickness (minimum from table J2.4 LRFD)

ΦRn = 0.75 (0.707 × 3/16 × [0.6×70 (1+0.5sin351.5)] = 5.11 kips / in > 1.73 O.K

Check strength of base metal

ΦRn = ΦFbm = 0.9×0.6×fy×t = 0.9×0.6×50×3/16=5.06 kip/in > 1.73 O.K

Cut off connection:

Bolted connection with two gusset plate above and under the tube section:

Use A325 bolts and the threads are included in shear plan => fv = 48 kis

Try Φ ¾” diameter bolts

Ru = 405 KN = 89.1 kips (tension)

Φ Rn = 0.75 × 48 × (л ×0.752 /4) × N ≥ 89.1 kips

N = number of bolts = 5.6 => so use 6 Φ 3/4” bolts (3 in each side)

To find plate thickness (t) Check bearing, yielding and fracture and choose max t:

1. Bearing: Edge bolts : ΦRn= 0.75( 1.2×2×t×65) ≤ 1.75(2.4×0.75×t×65)

= 117t > 87.75t (control)

Interior bolt: ΦRn= 0.75( 1.2×2.5×t×65) ≤ 1.75(2.4×0.75×t×65)

= 146t > 87.75t (control)

So total ΦRn = 6 × 87.75t = 526 t > 89.1 kips

t > 0.17 in => t= 1/4”

2. Yielding:

ΦRn= 2(0.9×50×4in×t) > 89.1 kips => t> 0.25 => t=1/4”

3. Fracture:

ΦRn= 2(0.75×Ae×65) > 89.1 kips

Ae = [4-(3/4+1/8)]t = 25t/8

ΦRn= 2(0.75×25t/8×65) > 89.1 kips => t > 0.3” = 5/16”

Take t = 5/16”

Check fracture for tube section:

Ru=89.1kips , Ag=2.77in2

An= 2.77-2×(3/4+1/8)×3/16=2.442 in2 , Ae= UAn = 0.75×2.44=1.83 in2

ΦRn= 0.75×65×1.83=89.3kips >89.1 kips O.K

Base plate:

Fy= 50ksi , fc’=28Mpa=4Ksi , Pu=177.6KN=39.1Kips

Dimension of base plate:

Column dimension = 40×40 cm2

Bearing on less than full area of concrete => use 6” ×7.5” base plate

Pp=0.85×4×(6×7.5) ×√ 15.752

(7.5 ×6) ≤1.7×4×(6×7.5)

= 359 kips > 306 kips (control)

Φc Pp = 0.6×306 = 183.6 kips > 177.6 KN O.K

Thickness of base plate (t):

t ≥ L(√ 2 Pu0.9 BNfy

), L = max of(m,n,λn’)

n= (B-0.8×4)/2 =(6-0.8×4)/2 = 1.4

m= (N-0.95×4)/2 =(7.5-0.95×4)/2 = 1.85

λn’= 1×1/4×√ (4 × 4) = 1

t = 1.85 (√ 2× 39.50.9× 6 ×7.5 ×50

) = 0.365”

Check bearing:

Tension in tube = 338.5 KN = 75 kips

Use Φ7/8” anchorage bolts => N= 75/[0.75 × 48 × (л ×(7/8)2 /4)]=3.46

so use 4Φ7/8” bolts

ΦRn = Φ×2.4×d×t×fu => t= 75/(0.75×2.4×7/8×65) =0.73”

So use 1” thickness of base plate

Weld size:

75 kips < 0.75×(0.707×w) ×[(2×3.75×0.6×70)+(4×0.6×70×1.5) => w > 0.249in , Use ¼ in

Truss (2) layout

Point loads on joints:

For internal joint →Tributary area = 2.14 x (6.9+5.15)/2 = 12.9 m2

For external joint →Tributary area = 6.45 m2

D.L Int.j =.45x12.9 Ext.j=.45x6.45

5.8 kn2.9 kn

L.L Int.j =.75x12.9Ext.j=.75x6.45

9.67 kn4.84 kn

S.L Int.j =1.4x12.9Ext.j=1.4x6.45

18.1 kn 9.03 kn

W.L Wy: Int.j = -.482x12.9 xcos11.2 Ext.j = -.482x6.45 xcos11.2Wx: =± .482x12.9 xsin11.2

-6.1 kn-3.05 kn±1.21 kn


Frame DesignSect DesignType

Status Ratio Pu(KN)

4 TS4X4X5/16 Column No Messages 1.043684 Pu

5 TS2.5X2.5X3/16 Column No Messages 0.813513 KN



11 TS2X2X3/16 Beam No Messages 0.461858 -13.105



17 TS2X2X3/16 Brace No Messages 0.612059 10.965

18 TS2X2X3/16 Brace No Messages 0.765675 26.583





Design of Purlins and Sag rods:

Purlines

WD= WD- Self weight of truss= .45 kn/m2

W(S or L) => S¿ L so snow load control: WS= 1.4kn/m2

WU= 1.2x.45 +1.6x1.4= 2.78 kn/m2

Spaces between Purlines = 2.14m so W total/m =2.14x2.78 =5.95 kn/m

Fx/m= 5.95 x sin11.2= 1.16 kn/m

Fy/m= 5.95 x cos11.2= 5.84 kn/m

Mu= Wl2/8 = 5.84x6.92/8 = 34.7 kn.m = 25.1 kip.ft

Mn= 25.1/.9= 27.9 kip.ft

Zx=Mn x 12/fy=6.69 in3

Choose W6x12

Sag Rods;

Spacing between sag rods =13

(bay)=13

x 6.9=2.3m

Tu horizontal on purlins = 1.16 kn/m

Sag rod will be design on top rod so tributary area for load will be from top to peak of the truss.

Wu=2.78 kn/m

Tu= 2.78 x (2.3x6.41) x sin11.2=7.96 kn =1.75 kips

Require Ag= 1.75

.75 x .75 x65=0.048 in2

Use for all sag rods Φ 1/4’’ (Ag=0.049 in2) at spacing 2.3 m.

truss work

Documents