Splices and connections Spli ces ces and conn ecti ons
1- Design of connections of vertical and diagonal: ECP 130
a- At the connection away from X.G.:
vertical and diagonal -1 Number Number per one side = n =
Fmax Ps
/2
distributed on 2 flanges
Fmax = Maximum allowable stress * total area Maximum Maximum allowable stress stres s = Fc = 1.6 – ( 8.5 * 10
-5
) for Compression member member
F sr = ……… T min min
= Ft =
1−
≤
0.58Fy for tension member member
T max flange -2
Arrange bolts in the 2 flanges. ( )
⇐
( 7 rows ) .
b Fl
≥
2*3φ + t w
⇒
case of 2 columns
b Fl
≥
4*3φ + t w
⇒
case of 4 columns
b Fl
≥
6*3φ + t w
⇒
case of 6 columns
1.5d 1.5d 1.5d
3d 1.5d
1/37
Splices and connections
s s s e n c k c i h T g B i g
w o T e k h t c . T h i c t e b n t d d l e e e f f e r e W t d i f f t u B
Y 1 : 5
s s e n c k c i h T a l l m S
Butt Weld
X R=10 X
/ o 5 . / 1 o / < 3 / <
Y
20cm
Gusset plate with big thickness Diagonal member
o
/ <15
Vertical member
2/37
Splices and connections : -3 1- Centerline
2- Upper or lower chord
3- Vertical
4- Diagonal
edge and pitch ( diagonal )
-4
gusset
Min e,p
Arranging bolts on min edge and pitch pitch
3/37
Splices and connections
Formation of gusset plate according to arrangment of bolts of diagonal (mostly critical)
Very important note: The gusset plate has only 3 shapes no more. If there is no diagonal, the angle of gusset plate will be 15
Min e,p
0
Min e,p
Min e,p
3 differnt shapes of Gusset plate
vertical edge = 2 d
4/37
Splices and connections b- Design of connection at position of X .G.( with vertical member i- If the connection is not subjected to moment :
i.e. All connections of deck bridge & all connections of through bridge except at position of end portal frame bracing. This means that every face is designed alone.
vertical and diagonal -1 Number per one side = n =
Fmax Ps
/2
distributed on 2 flanges
flange -2 Arrange bolts in the 2 flanges.
: -3 1- Centerline
2- Upper or lower chord
3- Vertical
4- Diagonal
edge and pitch ( diagonal )
-4
gusset
Min e,p
Arranging bolts on min edge and pitch
5/37
Splices and connections
Formation of gusset plate according to arrangment of bolts of diagonal (mostly critical) 5- Draw head plate and XG
vertical
(Height of XG is designed from loads on XG, while height of head plate = hxg + 4cm (2cm from each side)
a-
If hhead plate < hgusset plate:
6/37
Splices and connections
Gusset plate XG – 1
vertical XG 16 edge & head plate – 2 head n1 XG pitch Gusset plate plate C – Check n2 of XG from drawing. From drawing in this page n 2 Connecting head plate with Gusset plate) = 16 bolt R XG
∴
n2
≤ P s :
Gusset plate vertical n1 XG n1 of vertical in this drawing = 20 bolt
7/37
Splices and connections
b-
If hhead plate > hgusset plate
Gusset plate XG – 1 Gusset 20 n1 – 2 head gusset ( ) plate
vertical head plate plate gusset plate head plate Gusset plate packing – 3 XG head plate Gusset plate
8/37
Splices and connections
Min e,p
Draw the connection between the vertical & upper or lower chord first, then draw X.G. vertical member
n
Stringer
n
2
n
1
2
n
n
1
1
cross girder
vertical member
cross girder lower chord
Cr oss-section of thr ough bridge
diagonal of X-frame
Cross-section of Deck br idge
9/37
Splices and connections ii) If the connection is subjected to moment :
i.e. All connections of Pony Bridge and connections of through bridge at position of portal frame bracing. The inner face of bolts is designed the maximum strength of the vertical
While the outer face of bolts designed on M & Q where Q = R XG M=
M=
C 100
h+ w
h2 2
1 1 Ru * * h 2 3
for pony bridge
where w = 0.1*0.5*S
for through bridge with closed vertical frame
(Loaded case because of R XG Ru
M
Ru
CL of upper beam
s s u r t f o l V
s s u r t f o l V
2h/3 Ru /2
M
CL of upper beam
Ru /2 Ru /2
Ru /2
2h/3
s s u r t f o l V
CL of XG
h/3 CL of XG
h/3
Pony bridge
CL of XG Through bridge
1
M = Ru * l 2
for through bridge with inclined open frame
Every face is designed alone. Inner face for bolts between vertical and gusset plate. And outer face between head plate and gusset
vertical and diagonal -1 Number per one side = n =
Fmax Ps
/2
distributed on 2 flanges
flange -2 Arrange bolts in the 2 flanges.
: -3 1- Centerline
2- Upper or lower chord
3- Vertical
4- Diagonal
10/37
s s u r t f o l V
h
Splices and connections
edge and pitch ( diagonal )
-4
gusset
Formation of gusset plate according to arrangment of bolts of diagonal (mostly critical) vertical
5- Draw head plate and XG
(Height of XG is designed from loads on XG, while height of head plate = hxg + 4cm (2cm from each side)
XG haunch
11/37
Splices and connections
gusset plate n1
n 1 is number of bolts in one side (one flange) of the I-beam = 20 bolt
n 2 : check :
R XG n2
≤ P s
n2 = 8*4 = 32bolts
T ext , b, M ≤ 0.8 T where T ext , b, M =
1 f 1 + f 2 * * B * X ≤ 0.8 T n 2
n is the no. of bolts in one row M Where f 1 = y1 , f 2
=
M y 2 , I
y 2
where I =
=
H 2
BH 3 12
,
y1
=
H 2
-x
The diagonal is designed as before
12/37
Splices and connections upper chord
backing
n
1
n
1
H
ss girder f 2
X rail
vertical member B
f 1
bracket
n
2
n
cross girder
1
n
1
lower chord
Cross section of pony bridge
13/37
Splices and connections Design of CH ORD Spli ce
The field splice for upper and lower chord will be used each 16m (max). There is no field splice for diagonal and vertical
•
The splice for vertical and diagonal is used only when number of bolts in one row is more than 7.
The splice may be ei ther si ngl e shear or doubl e shear
a- Splice with plate from one side (single shear):
Key drawing 14/37
Splices and connections (1) Dimensioning:
= t f Take b sp = b f t 1
For flange:
= H web − 2cm Take H sp * t 2 = { H web * t web ) H sp
For web:
(2) Number of bolts:
For flange:
n1 =
For web:
n2
=
(bf * t f ) * (Fc or Ft ) Ps (single shear) (H web * t w ) * (Fc or Ft ) Ps (single shear)
FC for upper chord and 0.58Fy for lower chord. If the number of bolts in one row is greater than 7, so we can use 2 splice plates.
b- Splice with plates from both sides (double shear): (1) Dimensioning of splice plates:
For flange: 1 2
1 2
b f t f = b f t 1
b f t f = 2 * 0.4 b f t 2
For web:
∴
t1=
1 2
t f
get t 2
h w t w = 2 * h s.pl * t 3
t min = 8 mm (2) Number of bolts:
For flange:
n1
=
(b f * t f ) * (Fc or Ft ) 2Ps (double shear)
For web:
n2
=
(H web * t w ) * (Fc or Ft ) 2Ps (double shear)
FC for upper chord and 0.58Fy for lower chord.
15/37
Splices and connections
n2
t3 t2 n1
t1
c- Splice If there is stiffener in the upper or lower chord:
We must make splice for the stiffener
b1
h
n 3
b2
16/37
Splices and connections n 3 (no. of bolts of stiffener)
n3
=
b st * t st * ( F c orF t ) P s
Look at the drawing of the stiffener For pony & through bridges:
X.G upper chord stiffener X.G lower chord For deck bridge: vise versa
Splice for vertical and diagonal: vertical 7 diagonal XG 7 XG XG . haunch XG
XG vertical diagonal – 1 gusset 7 7 - : plate 1- Centerline
2- Upper or lower chord
3- Vertical
4- Diagonal
% 20 7 – 2 1.2 *[n required – 7 * no of bolts in one row] diagonal splice plate – 3
:
Assume after design number of bolts for D1 = 40, For D2 = 18 bolts and for v = 17 bolt and 4 bolts in each row
4 40 – 1 7 < 10 gusset plate diagonal 7 – 2 ( ) XG 17/37
Splices and connections
V D
D 1
2
7 XG diagonal vertical – 3 3 gusset vertical 20 D2 20 D1 haunch gusset
V D
D 1
2
18/37
Splices and connections
– 4 ndiff = (40 – 4 * 7 ) * 1.2 = 14.4
Taken 16 bolts
4 rows
4 gusset
diagonal 4 Splice plate – 5 packing gusset
In case of using splice plate, packing will be used at the position of connection as shown.
V D
D 1
2
7 vertical
19/37
Splices and connections Ex ample 1:
Design & draw to Scale 1:10 the given joint using H.S.B M24 (Ps = 6.29 t). Ldiagonal = 3.5m
Double Track
480
320
376
10
45
388
10
St 44
-985t
-945t
16
22
Pony bridge
0
+ 1 - 7 4 4 6 t t
16
22 Vertical
Diagonal
Solution:
Since the vertical and diagonal are subjected to both tension and compression, so we will use the bigger of F c and Ft . For diagonal: I x I y
=
1* 38.83 12
= 2 * 1.6 *
+ 2 * 32 *1.6 (
323 12
38.8 2
+
1.6 2 4 ) = 46651 cm 2
4
= 8738 cm
A = 32 * 1.6 * 2 + 38.8 * 1 = 141.2 cm r x
=
λin =
46651 141.2 7.9
r y
= 18.2 cm
0.7 * 350
=
8738 141.2
λout =
= 31 -5
2
2
= 7.9 cm
1.2 * 350 18.2
= 23.1
2
FC = 1.6 – 8.5 * 10 * 31 = 1.52 t / cm
20/37
Splices and connections Diagonal, class II, n = 200000 ∴ F sr = 2.8 t / cm 2.8 2 2 = 1.83 t / cm > 1.6 t / cm − 76
=
F max
2
1−
144
∴ F design = 141.2 * 1.6 = 225.92 t n=
225.92 6.29
= 35.9 bolts per 2 flanges
i.e 18 bolts per flange
b flange = 32 cm , (2 bolts) 6 d + t w = 6 * 2.4 + 1 = 15.4 cm < 32 (4 bolts) 12 d + t w = 12 * 2.4 + 1 = 29.8 cm < 32 So we have 4 bolts each row
No. of rows =
18 4
= 4.5
Take 5 rows (40 bolts in 2 flanges) For vertical: I x
=
1* 37.6 3 12
= 2 * 2.2 *
I y
+ 2 * 48 * 2.2 ( 483 12
37.6 2
+
2.2 2 4 ) = 88067 cm 2
4
=40550 cm
A = 48 * 2.2 * 2 + 37.6 * 1 = 248.8 cm r x
88067
=
248.8
2
r y
= 18.8 cm
=
40550 248.8
= 12.8 cm
0
L vertical = 3.5 sin 45 = 2.47 m
λin =
0.7 * 247 12.89
λout =
= 13.5 -5
2
FC = 1.6 – 8.5 * 10 * (15.8) = 1.58 t / cm
1.2 * 247 18.8
= 15.8
2
∴ F sr as before So ∴ F design = 248.8* 1.6 = 398.1 t n=
398.1 6.29
= 63.3 bolts per 2 flanges
i.e 31.6 bolts per flange
b flange = 48 cm , (4 bolts) = 12 d + t w = 12 * 2.4 + 1 = 29.8 cm < 48
21/37
Splices and connections (6 bolts) = 6*3 d + t w = 18 * 2.4 + 1 = 44.2 cm < 48 6 bolts per one row
No. of rows =
31.6 6
= 6 rows < 7
USE 6 rows * 6 bolts * 2 flanges = 72 bolts
L
g
70 140 140
4 0 7 5 7 5 7 5 7 5 4 0
140 140
Upper chord
140 70 480
320
22/37
Splices and connections Ex ample 2:
Design the given diagonal joint using H.S.B M24 (Ps = 6.29 t). Ldiagonal = 3.5m. Double Track. Pony bridge. Assume vertical as previous example.
360
14
-985t
-945t
3.6
45
388
0
t e n s i o n
16 Diagonal Solution: 2
A = 38.8 * 1.4 + 2 * 36 * 3.6 = 313.52 cm 2
F t = 1.6 t / cm < F max of fatigue
∴F design = 313.52 * 1.6 = 501.6 t n=
501.6 6.29
= 80 bolt
b flange = 36 cm
(40 bolt each flange)
, (4 bolts) = 12 d + t w = 12 * 2.4 + 1 = 29.8 cm < 36 cm (6 bolts) = 18 d + t w = 18 * 2.4 + 1 = 44.2 cm >36 cm
∴Use 4 bolts per row Assume n 1 = 7 rows
No. of rows =
40 4
= 10 rows > 7
use Splice
n1 = 7*4 = 28 bolts
n2 = (40-28)*1.2 = 14.4 bolts
No. of rows = 14.4 / 4 = 4 rows
23/37
Splices and connections Ex ample 3:
Pony Bridge, steel 52, M 24, Maximum compression force in upper
chord is 900 t, Force is the shown vertical = 110t. hxg = 100cm It is required to design & draw the connection between the vertical, lower chord & X.G.
St 52
360 3.6
388
14
110t
m 5
16
A
5m
Diagonal
Solution:
A = 38.8 * 1 + 2 * 30 * 1.6 = 134.1 cm
2
Maximum & minimum forces are not given F max = 134.1 * 0.58 * 3.6 = 280 t For M 24, steel 52, case I, P S = 6.94 t h = 5 – 0.5 = 4.5m n1
=
280 6.94
900 4.5 2 * 4.5 + 0.1* 5* 0.5* M = = 43 m t 100 2
/ 2 = 20.2 bolts each side
30 cm > 12 * 2.4 + 1 = 29.8 taken 4 bolts
Packing
each row no. of rows =
20.2 = 6 rows 4
End of G.Pl. m m 0 0 1 * 8
For n 2 : take haunch 20 cm So n 2 = 11 * 4 = 44 bolts B = 30 cm (b f of X.G)
50 100 50 100
f 2 f 1
0 7 1
100
24/37
H
Splices and connections Important note: The maximum force in the vertical is the maximum reaction of the X.G (vertical carry only X.G) Qext , b
=
110 44
T ext , b, M =
= 2.5 < 6.94 t
f + f 2 *( 1 ) * B * X 4 2
1
B = 30 cm
X = 2 + 5 + 10 =17 cm
H = 100 + 2 + 20 + 2 = 124 cm
I =
f 1 =
4300 124 2 * = 0.055 t / cm 4766560 2
f 2 =
4300 124 *( − 17) = 0.04 t / cm2 4766560 2
T ext , b, M =
1 4
*(
0.055 + 0.04 2
30 *124 3 12
4
= 4766560 cm
) * 30 * 17 = 6.06 t < 0.8 * 22.23 = 17.8 t
O.K safe we can use no haunch.
25/37
Splices and connections
upper chord
rail vertical member bracket
n = 6 row 1 n2= 11 row cross girder
lower chord
26/37
Splices and connections Ex ample 4:
Design & draw to Scale 1:10 the given joint using H.S.B M24 (Ps = 6.29 0
t). Ldiagonal = 3.5m inclined at 45 , Single track, deck bridge, st 52, the section of X.G is IPE 600. The connection is at the position of X-frame. R XG = 150 t 480
320 22
376
10
388
10
16
22 Vertical
A
16
Dia onal
Solution:
For diagonal: as given in example 1 4
4
Ix= 46651 cm r x= 18.2 cm
2
Iy= 8738 cm
A = 141.2 cm
λin
r y = 7.9 cm -5
2
λout = 23.1
= 31
2
FC = 1.6 – 8.5 * 10 * 31 = 1.52 t / cm
Diagonal, class II, n = 200000 ∴ F sr = 2.8 t / cm F max
=
2
2.8 2 2 = 1.83 t / cm > 1.6 t / cm − 76
1−
144
∴ F design = 141.2 * 1.6 = 225.92 t n=
225.92 6.29
= 35.9 bolts per 2 flanges
So we have 4 bolts each row
i.e 18 bolts per flange
No. of rows =
18 4
= 4.5
Take 5 rows (40 bolts in 2 flanges)
27/37
Splices and connections For vertical: n1 is the same as for example 1. 4
4
Ix = 88067 cm
Iy =40550 cm
r x= 18.8 cm
r y = 12.8 cm
λin
λout = 15.8
= 13.5
-5
2
A = 248.8 cm
0
L vertical = 3.5 sin 45 = 2.47 m
2
FC = 1.6 – 8.5 * 10 * (15.8) = 1.58 t / cm
So ∴ F design = 248.8* 1.6 = 398.1 t
∴ F sr as before n1
=
398.1 6.29
2
= 63.3 bolts per 2 flanges
i.e 31.6 bolts per flange
b flange = 48 cm ,(6 bolts) = 6*3 d + t w = 18 * 2.4 + 1 = 44.2 cm < 48 6 bolts per one row
No. of rows =
31.6 6
= 6 rows < 7
USE 6 rows * 6 bolts * 2 flanges = 72 bolts Design of n2: Assume n 2 = 6 * 4 = 24 bolts For n2 :
150 6.94
= 22 bolts < 24
L
g
70 140 Head PL.
140 140 170
4 0 7 5 7 5 7 5 7 5 4 0
Upper chord
100 70 480
Stiffener PL.
320
28/37
Splices and connections
Ex ample 5:
Single track through bridge, R XG = 170 t, M 24, st 52, b = 40cm. L=80m. For X.G. hxg = 100cm, tw = 1cm, bf = 30cm, tf = 2.5cm Design & draw the marked connections in different views
1
4 1 4 4 t
m 8
3
7 5 t
1 1 0 t
m 6
2
8m
8m 8m Upper plan
Solution: F or connection 1:
200
A =
Estimation of diagonal:
144 0.58 F y
A f = (68.6 – 38 * 1) / 2 = 14.3 cm A = 20 * 2 + 38 = 78 cm n
=
163.8 6.94
2
2
10
2
= 68.6 cm
b f = 20 cm (min)
10
380
F max = 78 * 2.1 = 163.8 t
/ 2 = 11.8 bolts each flange
10 Diagonal
29/37
Splices and connections Use 2 bolts each row à no. of rows =
11.8 2
= 6 rows
For upper chord: There is no difference, so it will be designed on
maximum strength. Force = 75 cos 45 = 53 t
60
Assume the section is as shown 40 t
30
=
t = 2.6 cm
3.6
40
t=2.6cm 2
A = (60 + 2 * 40) * 2.6 = 364 cm
60 * 2.6 * 1.3 + 2 * 40 * 206 * (2.6 + 20)
& &= y
Ix
364
=
2
60*2.6*(13.5-1.3)
+2
40
= 13.5
403 2 + 40 * 2.6(22.6 − 13.5) 2.6 * 12
=
4
68177cm Iy = 2.6 *
6030 12
2
4
+ 2*40*2.6*20 = 130000 cm
Lin = lout = 0.85*8 = 6.8 m R x =
68177 364
λin
= 13.7m -5
2
2
Fc = 2.1 – 13.5 *10 * 49.7 = 1.77t/cm lg = 114.5cm 643 4 * S *114.5
=
680 13.7
= 49.7
Fmax = 1.77* 364= 643t
from drawing
= 0.2 * 5.2
Smin = 1.4cm = 14mm
For vertical V1: No X.G but there is horizontal beam subjected to moment.
F = 144 cos45 = 102 t (comp.)
30/37
Splices and connections For the design of vertical and the connection of the vertical (n1) we will use maximum force in the member, due to D + L + I, so use loaded case of wind, For design of horizontal beam and its connection (n2), we will use unloaded case of wind, because the connection is subjected to wind only. 5 1 . 3
4
4
Loaded case
4
Unloaded case /
For loaded case (for vertical): h = 8 – 1.0 – 0.35 – 3.5 = 3.15 m
W w = 0.1 [0.5 * 4 + 0.5 * 3.15] = 0.36 t / m R u = 0.36 * 80 / 2 = 14.4 t M=
h=8-
/
1 2
= 7.5
2 14.4 2 * 7.5 * * h = = 36 m t 2 3 2 3
Ru
Ru 2h 2 3
Closed frame
Ru 1h 2 3 Estimation of vertical member: 40 t w
=
64
400 20
t w = 1.2 cm
3.6
Assume F C = 1.8 t / cm
F = 102 t comp.
2
12
360
20 Vertical 31/37
Splices and connections 102 36 + 2 0.4 = 78 cm2 ∴ A flange = 1.8 Take b = 20 t,
take t = 2 cm,
b = 40 cm
A = 2 * 40 * 2 + 36 * 1.2 = 203.2 cm I x
=
1.2 * 363 12
I y
= 2*2*
r x
=
40 3 12
62426 203.2
λin =
+ 2 * 40 * 2 * (
2
+
2 2 4 ) = 62426 cm , 2
4
= 21333 cm
r y
= 17.5 cm
0.7 * 800 10.2
36
2
λout =
= 54.9 < 90 -5
2
21333
=
203.2
= 10.2 cm
0.85 * 800 17.5
= 38.9
2
F C = 2.1 – 13.5 * 10 * 54.9 = 1.69 t / cm n1 using maximum strength:
=
1.69 * 203.2 6.94
/ 2 = 25 bolts each side
∴ n1 (between Vl. and gusset) = 25 bolts each side Use 4 bolts each row * 7 bolts each row.
For n2 (between horizontal beam and gusset) There is no shear (neglect O W). For moment, unloaded case is critical /
W w = 0.2* 0.5 * 4 *2 = 0.8 t / m M=
R u = 0.8 * 80 / 2 = 32 t
2 32 2 * 7.5 * * h = = 80 m t 2 3 2 3
Ru
Horizontal beam: Sx
=
M x 8000 3 = = 4570 cm use H E B 550 1.8 1.8
If we make beam without haunch so the arrangement of bolts will be as shown
32/37
Splices and connections upper chord
114.5
100 60
H = 55 + 2 * 2 = 59 cm I =
30 * 593 12
f 2 =
HEB550 Without haunch
3 6 5 * 7 2 = 3 6 0 3 6
X = 2 +2.9+3 + 5 = 12.9 cm 4
= 513447.5 cm
f 1 =
8000 59 2 * = 0.46 t / cm 513447.5 2
8000 59 * ( − 12.9) = 0.26 t /cm2 513447.5 2
1 0.26 + 0.46 T ext, b,M = ( ) * 30 * 12.9 = 35 t > 0.8 * 22.23 t = 17.8 t 4 2 So we have to use haunch. Take 7 rows (as vertical) Take h haunch = 20cm and t f of stiff = 2 cm
114.5
100 60
upper chord
3 6 5 * 7 2 = 3 6 0 3 6
HEB550
80 65
Packing
H = 2 + 55 + 20 + 2 + 2 = 81 cm
X = 2 + 2 + 6.5 +4 = 14.5cm
33/37
Splices and connections I =
30 * 813 12
f 2 =
f 1 =
4
= 1328603 cm
8000 81 2 = 0.24 t / cm * 1328603 2
8000 81 * ( − 14.5) = 0.16 t / cm2 1328603 2
1 0.16 + 0.24 T ext, b,M = ( ) * 30 * 14.5 = 21.8 t > 17.8 t 4 2
unsafe
Increase one or 2 rows & recheck
F or connection 2:
Since force in diagonal is smaller and we designed the previous diagonal using min. section, (use same section) For vertical, we have to estimate new section because the force decreases and there is no moment. Diagonal as before (same section), 2 bolts * 6 rows Vertical: F = 75 cos45 = 53t
40 t w
64
=
t w = 1.2 cm
3.6
Assume F C = 1.6 t / cm
∴ A flange =
2
53 1.6
200
2
= 33 cm
10
A f = 33 – 40 * 1.2 = -ve
A = 38 * 1.2 + 2 * 20 * 1 = 85.6 cm l in = 0.7 * 8 = 5.6 m I x
I y
= =
1.2 * 383 12 1* 20 3
λin =
12
+ 2 * 20 * (
2
2
+
=
20697
r y
=
1333.3
4
3.95
= 142 > 90
Vertical
1 2 4 ) = 20697 cm , r x 2
* 2 = 1333.3 cm
0.7 * 800
10
l out = 0.85 * 8 = 6.8 m 38
380
12
Take minimum flange 20 * 1 cm
85.6
85.6
= 15.5 cm
= 3.95 cm
unsafe
34/37
Splices and connections Use flange 30 * 2 cm
c
note
t f
=
A = 36 * 1.2 + 30* 2 * 2 = 163.2 cm I x I y r x
=
1.2 * 363 12 2 * 30
=
+
2
21 3.6
= 11
300 20
2 2 4 ) = 47986 cm 2
360
4
* 2 = 9000 cm
20
47986 163.2
λin =
2
2
= 6.7 <
12
3
12
=
+ 2 * (30 * 2) * (
36
(30 − 1.2 − 2 *1) / 2
0.7 * 800 7.4
r y
= 17.4 cm
9000
=
163.2
λout =
= 75.7 < 90 -5
2
= 7.4 cm 0.85 * 800 18
Vertical
= 37.8
2
F C = 2.1 – 13.5 * 10 * 75.7 = 1.32 t / cm F max = 1.32 * 163.2 = 215 t n1
=
215 6.94
/ 2 = 16 bolts each side
b f1 = 30 cm, 12 * 2.4 + 1 = 29.8 (take 4 bolts each row) No. of rows =
16 4
= 4 rows.
n2 between X.G. and gusset plate: Only shear is affecting n2 =
R xg P s
=
170 6.94
= 24.5
taken
No. of rows =
24.5 4
= 7 rows.
Estimation of lower chord: 40
Lower chord is a zero member N – truss [Although this joint is a loaded joint, but the
40
t=2.6
X.G has no effect in the horizontal direction] For connection 2: L 3 – L 2 = 110 cos 45 = 78 t 78 4 *1 * l g
= 0.2 * 5.2
25
l g = 19 cm (min) < 100 cm ok
35/37
Splices and connections
cross girder
lower chord
100cm
F or connection 3:
The connection has X.G & subjected to moment because this connection is in the vertical frame supporting the upper bracing. The lower connection is subjected to moment & shear. The shear is the maximum reaction of X.G = 170t Diagonal: as before 6rows*2 bolts each row For vertical n1 as in connection 1: (same member) Use 4 bolts each row * 7 bolts each row.
45 95 50 100 6 3
6 3
2 7 * 5
114.5
n 2 (for X.G): Assume n = 4 bolts per row in 7 rows (as n 1) and add 3*4 bolts in the packing. n2 = 4 bolts per row in 10 rows
n 2 = 40 bolt
36/37
