project in steel design

8/13/2019 Project in Steel Design

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Steel Truss ProjectDesign

Design of Howe Truss as to its Bolted Connections

Submi t ted by:

Mercy Dawn T. Chavez

CE-5 student

Subm it ted to:

Engr. Romeo Santos

Instructor


2/19


3/19

Structural nalysis

Description of the truss:

Bay width, x = 9m @ 1.5 interval

Bay length, truss to truss spacing = 4m

Height of the truss = 3m

Roof slope, = = 33.69Length of top chord= S = = 5.408m

S1 =

= 1.803m

S2 = = 0.902m

Given Data:

Roof corr. GI sheet no. 26 = 0.043 Kn/m2

Purlins 4 deep channel @ 0.4 on center both ways = 10.79 kg/m

Ceiling marine plywood attached to 2x2

Studs for ceiling frame spaced @ 0.4 on center both ways = 5.7 kn/m3

Roof life load = 1.9 KPa


4/19

Calculation:

Loadings to b e carried by the truss

Live loads

LL = 1.9(4)(1.5) = 11.4 kn

LL1 = 11.4kn

LL2 = 5.4 kn

Dead loads

Weight of GI sheet

DL1 = 1.803 (4) (0.043) = 0.258 KnDL2 =

= 0.129 kn

Weight of Purlins = (10.79kg/m) (9.81m/s2) (4m) = 423.40 kn

No. of purlins =

= 13.5 say 13

DL1 =

(1.8) (

= 1.52 kn

DL2 = 0.76 kn

Weight of marine plywood


5/19

=0.00635 m

DL =

= 0.415kn/m (1.5) = 0.218 kn

Weight of studs

2x2= 4 in2= 0.00258 m2

(

) = 60

(

) = 9.426 Kn/m3

DL = (

(4m) +

(9m)) (0.00288m2) (9.426 Kn/m3) = 4.377 kn

DL1 = 14.591 Kn/ 30m = 0.486 Kn/m (1.5m) = 0.729 kn

DL2 = 0.365 kn

Weight of truss

W =aL (1 +

) a, width = 4m = 13.123 ft

L, length = 9 m = 29.53 ft

W =

(13.123 ft) (29.53 ft) (1 +

) = 521.07 kg

= 521.07 kgs (9.81 m/s2) = 5,111.7 N

DL = 5.11 Kn/ 9m = 0.568 Kn/m (1.5) = 0.852 kn

Wind load

Based from NSCP 1992, sec.2.3

P= CeCqqsI

w/r: Cq = 0.9 for outward ; 0.3 for inward

Ce = 0.7

qs = 2000 Pa

I = 1.00

Inward: P= CeCqqsI = 0.7(0.3) (20000) (1) = 0.42 Kpa

Outward: P= CeCqqsI = 0.7(0.9) (2000) (1) = 1.26 Kpa

Joints concentrated loads:

For inward:

CL1 = 0.42 (1.803) (4) = 3.03 kn


6/19

CL2 = 3.03/2 = 1.515 kn

For outward:

CL1 = 1.26 (1.803) (4) = 9.09 kn

CL1 = 9.09/2 = 4.54 kn

Total loads in top chord

F1 = live load + weigh of purlins + weight of G.I. sheet

F1 = [(7.6 kn/m) + (1.019 kn/m)] (1.019 kn/m) + 0.258

F1 = (8.619 kn/m) (1.803 kn/m) + (0.258 kn) = 15.798 KN

F1/2 = 7.899 KN

Total loads in bottom chord

W1 = weight of marine plywood + weight of the studs + live load

W1 = [(0.012 kn/m) + (0.486 kn/m) + (7.6 kn/m)] (3 m) = 24.294 KN

Getting the values for reaction:

= 0 + clockwise9Bv = 15.798(7.5) + (15.798) (3) + (15.798) (4.5) + 15.798(6) + (15.798) (7.5) + (7.899)

(9) + 24.294(1.5) + (24.294) (1.5) + (24.294) (3) + 24.294(4.5) + (24.294) (6) + (24.294)

(7.5)

Av = Bv = 108.129 KN right and left support reactions


7/19

Design of Howe Truss MembersSteel Properties referred to ASEP 2004

Design of Top chord

Pmax = 13.76 KN (compression)

Try L 2x 2 x

(double angle)

Fy = 248Mpa

Properties of steel section:

I = 0.42 x 10

6

mm

4

A = 1045 mm2

L = 1803 mm

Solution:

R = = = 20.048 mm =

= 89.934 mm

C = =

C = 126.169

< C

Fs =+

-

Fs =+

-

Fs = 1.889MPa

= [1-

] (

)

= [1- ] ( ) 97.934 MPa

Allowable = 97.934 MPa=

Actual = 13.167 MPa97.934 MPa> 13.167 MPa

Therefore the assumed design is safe!

Design of Bottom Chord

See solution of top chord for allowablestress in bottom chord

Pmax = 62.33 KN (tension)

Try L 2x 2 x

(double angle)


Actual = 59.646 MPa97.934 MPa>59.646MPa


8/19

Therefore the assumed design is safe


9/19

Design of web members

[Vertical and diagonal braces]

Pmax = 31.73KN (tension)

Try 65 x 65 x 10

Fy = 248Mpa

L = 3000 mm (the longest length above all web members)

Properties of steel section:

A = 448 mm2

Rx= 19.76 mm

Solution:

C = =

C = 126.169

= = 155.360 mm

> C

= = = 42.668 MPa


Actual = 26.245 MPa

42.668 MPa> 26.245 MPa


Pmax = 23.26 KN (compression)see solution above

Allowable = 42.668 MPa

= Actual = 19.239 MPa42.668 MPa> 19.239 MPa



10/19

Design of Bolts Connection

Components uses:

Gusset Plate = to prevent from tearing

Bolts = to prevent from shearing

The reason why the minimum number of bolt/rivet is 2, for in case of maintenance in

changing oxidized bolts, still theres a bolt that could support the members.

Data:

Fp= 1.5 Fu

= 1.5 (414)

= 621MPa

Fv= 207 MPa

Minimum edge distance; Le is not < 1.5d = 1.5 (20) = 30 mm

Maximum distance (Le) of bolts from center to center; 12t is > 150 mm = 12 (5) = 60mm

Therefore 60mm is not < 30 mm o.k.

Note: Shear in bolts for double shearAv = 2Area of bolts x number of bolts

Detailed of joints: - see A3 for the figure design

At Joint 1

Member 1-2 = 62.33 Kn

For shearing

Ps = AFv (2) = (

KNNo. of Bolts =

= 0.4792 pcs

For bearing

Pb = plate thickness x bolt dia. x Fp (2)

Pb = (5) (20) (621) (2) = 124.2 Kn

No. of Bolts = = 0.5022 pcs


11/19

Therefore use 2-20 mm boltsMember 1-8= 13.76 Kn

For shearing

Ps = KNNo. of Bolts =

= 0.106 2 pcs

For bearing

Pb = 124.2 Kn


Therefore use 2-20 mm bolts

At Joint 2

Member 2-1; use 2-20 mm boltsMember 2-3; use 2-20 mm boltsMember 2-8= 1.79 Kn

For shearing

Ps = 130.062Kn

No. of Bolts =

= 0.014 2 pcs For bearing

Pb = 124.2 Kn

No. of Bolts == 0.0142 pcs

Therefore use 2-20 mm bolt

At Joint 3

Member 3-2; use 3-20 mm boltsMember 3-4= 35.34 Kn

For shearing

Ps = 130.062 Kn

No. of Bolts =

= 0.2722 pcs


12/19

For bearing

Pb = 124.2 Kn

No. of Bolts =

= 0.2852 pcs

Therefore use 2-20 mm boltsMember 3-9= 11.10 Kn

For shearing

Ps = 130.062 KN

No. of Bolts=

= 0.0852 pcs For bearing

Pb = 124.2 Kn


Therefore use 2-20 mm boltsMember 3-8 = 16.78 Kn

For shearing

Ps = 130.062 Kn

No. of Bolts =

= 0.1292 pcs For bearing

Pb = 124.2 Kn


Therefore use 2-20 mm bolt

At Joint 4

Member 4-3; use 2-20 mm boltsMember 4-5; use 2-20 mm bolts

Member 4-9 = 23.26 Kn

For shearing

Ps = 130.062 Kn

No. of Bolts =

= 0.1792 pcs


13/19

For bearing

Pb = 124.2 Kn


Therefore use 2-20 mm boltsMember 4-10 = 31.73 Kn

For shearing

Ps = 130.062 KN

No. of Bolts =

= 0.244

2 pcs

For bearing

Pb = 124.2 Kn


Therefore use 2-20 mm boltsMember 4-11; use 2-20 mm bolts

At Joint 8

Member 8-1; use 2-20 mm boltsMember 8-2; use 2-20 mm boltMember 8-3; use 2-20 mm boltMember 8-9 = 13.76 Kn

For shearing

Ps = KNNo. of Bolts =

= 0.106 2 pcs

For bearing member

Pb = 124.2 Kn


Therefore use 2-20 mm

bolts

At Joint 9

Member 9-8; use 2-20 mm boltsMember 9-3; use 2-20 mm boltsMember 9-4; use 2-20 mm bolts

Member 9-10; use 2-20 mm boltsAt Join t 10

Member 10-9; use 2-20 mm boltsMember 10-11; use 2-20 mm bolts

Member 10-4; use 2-20 mm bolts


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Design spacing of bolts in the gusset plate

Note: Max. Le = 12t = 12(5) = 60 mm

At Joint 1

Member 1-8 Le for gusset plate

Min Le (edge distance) = is not < 1.5 (21)

=

= 13.29 is < 31.5: say 40mm

Spacing (S):

S + ; S 13.29 + ; S 23.79 say 80 mm



= = 60.22 is > 31.5: say 60mm

Spacing (S):

S + ; S 60.22 + ; S 70.72 say 100 mm

At Joint 2



=

= 1.729 is < 31.5: say 60mm

Spacing (S):

S + ; S 60.22 + ; S 70.72 say 100 mm



=

= 47.188 is > 31.5: say 60mm


15/19

Spacing (S):

S + ; S 47.188 + ; S 57.688 say 80 mm

At Joint 3



=

= 34.145 is > 31.5: say 40mm

Spacing (S):

S + ; S 34.145 + ; S 44.645 say 80 mm



=

= 10.725 is not < 31.5: say 40mm

Spacing (S):

S + ; S 10.725 + ; S 57.688 say 60 mm



=

= 16.216 is < 31.5: say 40mm

Spacing (S):

S + ; S 16.216 + ; S 26.716 say 60 mm

At Joint 4

Member 4-5 see solution member 3-4



=

= 18.077 is < 31.5: say 40mm

Spacing (S):


16/19

S + ; S 18.077 + ; S 44.645 say 60 mm



=

= 30.567 is < 31.5: say 40mm

Spacing (S):

S +

; S 30.567 +

; S 41.157 say 60 mm



=

= 21.507 is < 31.5: say 40mm

Spacing (S):

S + ; S 21.507 + ; S 41.157 say 60 mm

At Joint 5



Min Le (edge distance) =

is not < 1.5 (21)

=

= 21.507 is < 31.5: say 40mm

Spacing (S):

S + ; S 21.507 + ; S 41.157 say 60 mm



=

= 8.957 is < 31.5: say 40mm

Spacing (S):


17/19

S + ; S 8.957 + ; S 19.457 say 60 mm

At Joint 6



At Joint 7



=

= 10.570 is < 31.5: say 40mm

Spacing (S):

S + ; S 10.570 + ; S 21.070 say 60 mmAt Joint 8


At Joint 9



=

= 7.507 is < 31.5: say 40mm

Spacing (S):

S + ; S 7.507 + ; S 17.727 say 60 mmAt Join t 10



=

= 10.754 is < 31.5: say 40mm

Spacing (S):

S + ; S 10.754 + ; S 21.524 say 60 mmMember 11-12 see solution member 7-12


18/19


19/19

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