Timber and Steel Design
Lecture 11
Bolted Connections
Riveted Connections
Types of Joints
Failure of Joints
Bearing & Friction connections
Truss Joints
Shear and Tension on Bolt
Mongkol JIRAVACHARADET
SURANAREE
INSTITUTE OF ENGINEERING
UNIVERSITY OF TECHNOLOGY
SCHOOL OF CIVIL ENGINEERING
Riveted Connections
rivet = round ductile steel bar (shank) with a head at one end
d
Snap
Pan
Flat
countersunk
d = Shank diameter
= Nominal diameter
Round
countersunk
Riveting Procedure
Rivets are heated before driving
φ hot rivet = φ hole
φ
Head is formed by
- Hand hammer
- Hydraulic pressure
- Pneumatic pressure
On cooling, the rivet shrinks both length
and diameter
Connected parts become tighter:
- Tension in rivet
- Compression in plates
Friction between plates called
“clamping action”
Forms of riveted joints
P
Lap Joint
P
P
Rivet in single shear
fv = P / A
P
Eccentricity in lap joint
P
P
e
P
P
Undesired bending causes tension in rivets
Double riveted lap joint
Butt Joint
P
P/2
P
P/2
P/2
Rivet in double shear
P
fv = P/2A
Butt joint v.s. Lap joint:
- กก
ก
- !ก"#!!
$
- %"!&
!ก'(
!ก
Bolted Connections
P/2
กก
กก
P/2
W section
P
P/2
(a) ก
(c)
(e)
(b)
ก
!
(d)
(f)
ก
Failure of Joints
Case 1: Shear failure of rivets
P/2
P
P
P
P/2
Single shear in lap joint
Double-shear in butt joint
Shear stress in rivet exceed the limit
Case 2: Shear failure of plates
Insufficient edge distance
Shearing out
Case 3: Tension or Tearing failure of plates
P
P
Tensile stress at net cross section exceeds tensile strength
Case 4: Bearing failure of plate
P
P
Plate may be crushed when bearing stress in plate exceeds the limit
Bolted Connections
P/2
P
P/2
Friction from clamping pressure
using high strength bolt
Beam splice moment connection
Bearing-type Connections
d
2
Shearing Strength: (π / 4) d Fv
Ft (ก.ก./.2)
กก
A307
A325 (ก"#$%
!)
A325 (ก&"#$%
A490 (ก"#$%
!)
!)
A490 (ก&"#$%
!)
!
Fv (ก.ก./.2)
Friction-type Bearing-type
1,400
-
700
3,100
1,200
1,480
3,100
1,200
2,100
3,800
1,480
1,970
3,800
1,480
2,800
Threads Excluded from Shear Plane or not ?
Diameter = d
Shear
planes
Bearing
stress
P
2P
P
Bearing Strength
Bearing area = d x t
t
Excessive bearing results in shear tear-out failure mode.
Failure surface
Allowable bearing strength Ra :
d
Ra = bearing area × Fp = d t Fp
Failure surface
Le
For standard or short-slotted holes → Fp = 1.2 Fu
Ra/2
Fp =
Ra
Ra/2
Le Fu
≤ 1.2Fu
2d
for Le < 1.5d
Ra = 0.5 Le t Fu ≤ 1.2 d t Fu
Or
Ra = 0.6 Lc t Fu ≤ 1.2 d t Fu
For long-slotted holes perpendicular to the load → Fp = 1.0 Fu
Spacing and Edge-Distances Requirements
Marking for Bolt
P
g
s
P
s
A307
Pitch: center-to-center of bolts parallel to axis of member
Gage: center-to-center of bolts lines perpendicular to axis of member
Edge distance: shortest distance from center of bolt to edge
A325
min Le = 1.5-2 diameter
Distance between bolts: shortest distance between bolts
min distance = 3 diameter
A490
Spacing and Edge-Distances Requirements
Lc2
Lc1
2
Lc = clear distance, in the direction
parallel to the applied load.
Le = edge-distance to center of the hole
1
s = center-to-center spacing of holes
Le
h = d + 2 mm = hole diameter
s
For edge bolts(2), use Lc = Le – h/2
For other bolts(1), use Lc = s – h
s ≥ 2 23 d
(preferably 3d )
Le ≥ value from AISC Table J3.4
(1.5d - 2d )
11-1 ก
P ก !"
#ก
A36, ก"ก
( A325 22 .., ",-, "ก
(ก,ก
"./, ,,กก( 1.5d, ,,,,(
ก"ก
(กก( 3d
1:
22 mm. bolts
P
P
1.0 cm
1.0 cm
P
30 cm
Ag = 1.0 (30) = 30
3.2
An = 30 – 2 (2.5) (1.0) = 25 3.2 = Ae
P = 0.60 (2.5) (30.0) = 45 P
P = 0.50 (4.0) (25.0) = 50 ก"ก
(4"./
,ก"(:
"./:
P = 0.25π (2.2) 2 (2.1) (4) = 31.93 ก:
P = (1.0) (2.2) (1.2) (4.0) (4) = 42.24 กก !!" P
= 31.93
11-2 ,5
ก"ก
( A325 19 .. ",-31"ก
(
ก,ก"./(ก(ก ? 5"
#ก A36 ,
,,
,,,,(
ก"ก
(" 89 ก
1.0 cm
P/2 = 40 ton
P/2 = 40 ton
19 mm. bolts (A = 2.84 cm2)
2.0 cm
P = 80 ton
1.0 cm
ก"ก
(4"./
,ก::
ก
"./
ก"ก
(1( = 2(2.84)(2.1) = 11.93 ก
ก
ก"ก
(1 ( = (2.0)(1.9)(1.2)(4.0) = 18.24 (
ก"ก
(ก = 80/11.93 = 7+
ก
ก 8 $" 9 ('()*ก!ก+,
)
(!/0)
11-3 1: ,ก!"
#ก "./ V ก, 80 ,,ก
ก"ก
( A325 22 .. (A=3.8 3.2) " 8
ก ,,
,,,
ก"ก
(D" 89 ก
,
"ก
(ก,ก"./ 5"
#ก A36 , Eกกกก
PL2x40
45.2 cm
W500x128
Ix = 71000
tf = 1.8
48.8 cm
2 cm
s
I = 71,000 + 2(2×40)(25.4)2
= 174,226 3.4
q = VQ/I
= (80)(2×40)(25.4)/174,226
2 cm
PL2x40
= 0.93 /3.
"./
ก"ก
(( = (2)(3.8)(2.100) = 15.96 (!/0)
ก
ก"ก
(( = (2)(2.2)(1.8)(1.2)(4.0) = 38.02 s = 15.96/0.93 = 17 3.
( 15 20. c-c)
05 AISC ก smax ก"ก
(:G: ,ก"ก:
478 / Fy
/ 24 ":(
/ 30 3. H5:ก(
0ก/,' S 0$:
24 ":( = 24(1.8) = 43.2 3. > 15 3.
478 t f
Fy
=
478(1.8)
= 17.2 3. > 15 3.
2,500
Smax = 30 3. > 15 3.
OK
OK
OK
Slip-critical(Friction-type) Connections
5".,ก
ก"ก
(ก
:/ A325 , A490
P
N
F
P
P
N
F
P
1ก
ก"ก
(ก ()
',ก
ก (0.0.)
A325
A490
19
22
25
32
12.7
17.7
23.2
32.3
15.9
22.3
29.1
46.4
AISC360-05 Specification for Structural Steel Buildings
5",
d + 2 ..
",- ("!IGก
)
d + 8 ..
",J"IK ("!IGก
)
D (ก( × ()
(d + 2 ..) × (d + 10 ..)
(d + 2 ..) × 2.5 d
( (ก( × ()
d + 2 mm
d + 10 mm
(a) Standard
d + 2 mm
2.5 d
d + 8 mm
(c) Short slot
d + 2 mm
(b) Oversized
(d) Long slot
+/,ก
ก@Aก
- :G:,กH9ก"/DIG
- /Dก9",5 Gusset plate 5(
Gusset plate
Truss Joint
Double member with a single connector piece, called a gusset plate.
Members must have sufficient "bite" into the plate for bolts or weld contact.
Converging centerlines enable equilibrium without member moments.
'
$ ก B #)ก
*+ #'(กก A325
!ก
19 !.!. *3!4
ก*
ก3
5
6ก+3ก
12 !.!.
2Ls 75x75x10
A
1L
E
7 ton
50
x
x
50
8
2Ls 90x90x10
B
28 ton
C
1L 50x50x8
20 ton
D
8 ton
)*$ ก7กก A325 ก*
ก3
ก7
:
Fv = 2,100 กก./=!.2
ก7ก
: Fp = 1.2(4,000) = 4,800 กก./=!.2
+,$+$ AB: P = 20 '
'(
( 2Ls 75x75x10 &
กก
*
ก7
*กก A325 19 !.!. 1 P = 2×
ก7ก
π
4
× d 2 × Fv = 2 ×
π
4
× 1.92 × 2.1 = 11.91 ton Control
P = d × 2t × Fp = 1.9 × 2(1.0) × 4.8 = 18.24 ton
7
กก(ก = 20/11.91 = 1.68 2 '
+,$+$ BC: P = 28 '
'(
( 2Ls 90x90x10 กก
*
π
π
ก7
*
P = 2×
ก7ก
P = d × 2t × Fp = 1.9 × 2(1.0) × 4.8 = 18.24 ton
4
× d 2 × Fv = 2 ×
4
× 1.92 × 2.1 = 11.91 ton Control
7
กก(ก = 28/11.91 = 2.53 3 '
+,$+$ BD: P = 8 '
π
'(
( 1Ls 50x50x8 กก
π
ก7
*
P=
ก7ก
P = d × t × Fp = 1.9 × 0.8 × 4.8 = 7.30 ton
4
× d 2 × Fv =
4
× 1.92 × 2.1 = 5.95 ton
Control
7
กก(ก = 8/5.95 = 1.35 2 '
+,$+$ BE: P = 7 '
π
'(
( 1Ls 50x50x8 กก
π
ก7
*
P=
ก7ก
P = d × t × Fp = 1.9 × 0.8 × 4.8 = 7.30 ton
4
× d 2 × Fv =
4
× 1.92 × 2.1 = 5.95 ton
7
กก(ก = 7/5.95 = 1.18 2 '
Control
Detail Drawing
A
2Ls 75x75x10
B
2Ls 90x90x10
C
E
Gusset plate 20 mm
1L 50x50x8
1L
50
x
50
x8
Bolt A325×19 mm
D
More ckeck for:
- Effective area of tension members (0.5Fu Ae)
- Block shear failure of tension members
Timber and Steel Design
Lecture 11
Bolted Connections
Riveted Connections
Types of Joints
Failure of Joints
Bearing & Friction connections
Truss Joints
Shear and Tension on Bolt
Mongkol JIRAVACHARADET
SURANAREE
UNIVERSITY OF TECHNOLOGY
INSTITUTE OF ENGINEERING
SCHOOL OF CIVIL ENGINEERING
Riveted Connections
rivet = round ductile steel bar (shank) with a head at one end
d
Snap
Pan
Flat
countersunk
d = Shank diameter
= Nominal diameter
Round
countersunk
Riveting Procedure
Rivets are heated before driving
φ hot rivet = φ hole
φ
Head is formed by
- Hand hammer
- Hydraulic pressure
- Pneumatic pressure
On cooling, the rivet shrinks both length
and diameter
Connected parts become tighter:
- Tension in rivet
- Compression in plates
Friction between plates called
“clamping action”
Forms of riveted joints
P
Lap Joint
P
P
Rivet in single shear
fv = P / A
P
Eccentricity in lap joint
P
P
e
P
P
Double riveted lap joint
Undesired bending causes tension in rivets
Butt Joint
P
P/2
P
P/2
P/2
Rivet in double shear
P
fv = P/2A
Butt joint v.s. Lap joint:
- กก
ก
- !ก"#!!
$
- %"!&
!ก'(
!ก
P/2
Bolted Connections
กก
กก
P/2
W section
P
P/2
(a) ก
(c)
ก
(b)
(d)
ก
(e)
!
(f)
Failure of Joints
Case 1: Shear failure of rivets
P/2
P
P
Single shear in lap joint
P
P/2
Double-shear in butt joint
Shear stress in rivet exceed the limit
Case 2: Shear failure of plates
Insufficient edge distance
Shearing out
Case 3: Tension or Tearing failure of plates
P
P
Tensile stress at net cross section exceeds tensile strength
Case 4: Bearing failure of plate
P
P
Plate may be crushed when bearing stress in plate exceeds the limit
Bolted Connections
P/2
P
P/2
Friction from clamping pressure
using high strength bolt
Beam splice moment connection
Bearing-type Connections
d
2
Shearing Strength: (π / 4) d Fv
Ft (ก.ก./.2)
กก
A307
A325 (ก"#$%
!)
A325 (ก&"#$%
A490 (ก"#$%
!)
!)
A490 (ก&"#$%
!)
!
Fv (ก.ก./.2)
Friction-type Bearing-type
1,400
-
700
3,100
1,200
1,480
3,100
1,200
2,100
3,800
1,480
1,970
3,800
1,480
2,800
Threads Excluded from Shear Plane or not ?
Diameter = d
Shear
planes
Bearing
stress
P
2P
P
Bearing Strength
Bearing area = d x t
t
Excessive bearing results in shear tear-out failure mode.
Failure surface
Allowable bearing strength Ra :
d
Ra = bearing area × Fp = d t Fp
Failure surface
For standard or short-slotted holes → Fp = 1.2 Fu
Le
Ra/2
Fp =
Ra
Ra/2
Le Fu
≤ 1.2Fu
2d
for Le < 1.5d
Ra = 0.5 Le t Fu ≤ 1.2 d t Fu
Or
Ra = 0.6 Lc t Fu ≤ 1.2 d t Fu
For long-slotted holes perpendicular to the load → Fp = 1.0 Fu
Spacing and Edge-Distances Requirements
Marking for Bolt
P
g
s
P
s
A307
Pitch: center-to-center of bolts parallel to axis of member
Gage: center-to-center of bolts lines perpendicular to axis of member
Edge distance: shortest distance from center of bolt to edge
A325
min Le = 1.5-2 diameter
Distance between bolts: shortest distance between bolts
min distance = 3 diameter
A490
Spacing and Edge-Distances Requirements
Lc2
Lc1
2
Lc = clear distance, in the direction
parallel to the applied load.
Le = edge-distance to center of the hole
1
s = center-to-center spacing of holes
Le
h = d + 2 mm = hole diameter
s
For edge bolts(2), use Lc = Le – h/2
For other bolts(1), use Lc = s – h
s ≥ 2 23 d
(preferably 3d )
Le ≥ value from AISC Table J3.4
(1.5d - 2d )
11-1 ก
P ก !"
#ก
A36, ก"ก
( A325 22 .., ",-, "ก
(ก,ก
"./, ,,กก( 1.5d, ,,,,(
ก"ก
(กก( 3d
1:
22 mm. bolts
P
1.0 cm
P
P
1.0 cm
30 cm
Ag = 1.0 (30) = 30
3.2
An = 30 – 2 (2.5) (1.0) = 25 3.2 = Ae
P = 0.60 (2.5) (30.0) = 45 P
P = 0.50 (4.0) (25.0) = 50 ก"ก
(4"./
,ก"(:
"./:
P = 0.25π (2.2) 2 (2.1) (4) = 31.93 ก:
P = (1.0) (2.2) (1.2) (4.0) (4) = 42.24 กก !!" P
= 31.93
11-2 ,5
ก"ก
( A325 19 .. ",-31"ก
(
ก,ก"./(ก(ก ? 5"
#ก A36 ,
,,
,,,,(
ก"ก
(" 89 ก
1.0 cm
P/2 = 40 ton
P/2 = 40 ton
19 mm. bolts (A = 2.84 cm2)
2.0 cm
P = 80 ton
1.0 cm
ก"ก
(4"./
,ก::
ก
"./
ก"ก
(1( = 2(2.84)(2.1) = 11.93 (!/0)
ก
ก
ก"ก
(1 ( = (2.0)(1.9)(1.2)(4.0) = 18.24 (
ก"ก
(ก = 80/11.93 = 7+
ก
ก 8 $" 9 ('()*ก!ก+,
)
11-3 1: ,ก!"
#ก "./ V ก, 80 ,,ก
ก"ก
( A325 22 .. (A=3.8 3.2) " 8
ก ,,
,,,
ก"ก
(D" 89 ก
,
"ก
(ก,ก"./ 5"
#ก A36 , Eกกกก
PL2x40
48.8 cm
W500x128
Ix = 71000
tf = 1.8
2 cm
45.2 cm
PL2x40
s
I = 71,000 + 2(2×40)(25.4)2
= 174,226 3.4
q = VQ/I
= (80)(2×40)(25.4)/174,226
2 cm
= 0.93 /3.
"./
ก"ก
(( = (2)(3.8)(2.100) = 15.96 (!/0)
ก
ก"ก
(( = (2)(2.2)(1.8)(1.2)(4.0) = 38.02 s = 15.96/0.93 = 17 3.
( 15 20. c-c)
05 AISC ก smax ก"ก
(:G: ,ก"ก:
478 / Fy
/ 24 ":(
/ 30 3. H5:ก(
0ก/,' S 0$:
24 ":( = 24(1.8) = 43.2 3. > 15 3.
478 t f
Fy
=
OK
478(1.8)
= 17.2 3. > 15 3.
2,500
OK
Smax = 30 3. > 15 3.
OK
Slip-critical(Friction-type) Connections
5".,ก
ก"ก
(ก
:/ A325 , A490
P
N
F
P
P
N
F
P
1ก
ก"ก
(ก ()
',ก
ก (0.0.)
A325
A490
19
22
25
32
12.7
17.7
23.2
32.3
15.9
22.3
29.1
46.4
AISC360-05 Specification for Structural Steel Buildings
5",
d + 2 mm
d + 2 ..
",- ("!IGก
)
d + 8 ..
",J"IK ("!IGก
)
D (ก( × ()
(d + 2 ..) × (d + 10 ..)
(d + 2 ..) × 2.5 d
( (ก( × ()
d + 10 mm
(a) Standard
d + 2 mm
2.5 d
d + 8 mm
(c) Short slot
d + 2 mm
(b) Oversized
(d) Long slot
+/,ก
ก@Aก
- :G:,กH9ก"/DIG
- /Dก9",5 Gusset plate 5(
Gusset plate
Truss Joint
Double member with a single connector piece, called a gusset plate.
Members must have sufficient "bite" into the plate for bolts or weld contact.
Converging centerlines enable equilibrium without member moments.
'
$ ก B #)ก
*+ #'(กก A325
!ก
19 !.!. *3!4
ก*
ก3
5
6ก+3ก
12 !.!.
2Ls 75x75x10
A
1L
E
7 ton
50
x
x
50
8
2Ls 90x90x10
B
28 ton
C
1L 50x50x8
20 ton
D
8 ton
)*$ ก7กก A325 ก*
ก3
ก7
:
Fv = 2,100 กก./=!.2
ก7ก
: Fp = 1.2(4,000) = 4,800 กก./=!.2
+,$+$ AB: P = 20 '
'(
( 2Ls 75x75x10 &
กก
*
ก7
*กก A325 19 !.!. 1 P = 2×
ก7ก
π
4
× d 2 × Fv = 2 ×
π
4
× 1.92 × 2.1 = 11.91 ton Control
P = d × 2t × Fp = 1.9 × 2(1.0) × 4.8 = 18.24 ton
7
กก(ก = 20/11.91 = 1.68 2 '
+,$+$ BC: P = 28 '
'(
( 2Ls 90x90x10 กก
*
π
π
ก7
*
P = 2×
ก7ก
P = d × 2t × Fp = 1.9 × 2(1.0) × 4.8 = 18.24 ton
4
× d 2 × Fv = 2 ×
4
× 1.92 × 2.1 = 11.91 ton Control
7
กก(ก = 28/11.91 = 2.53 3 '
+,$+$ BD: P = 8 '
π
'(
( 1Ls 50x50x8 กก
π
ก7
*
P=
ก7ก
P = d × t × Fp = 1.9 × 0.8 × 4.8 = 7.30 ton
× d 2 × Fv =
4
4
× 1.92 × 2.1 = 5.95 ton
Control
7
กก(ก = 8/5.95 = 1.35 2 '
+,$+$ BE: P = 7 '
π
'(
( 1Ls 50x50x8 กก
π
ก7
*
P=
ก7ก
P = d × t × Fp = 1.9 × 0.8 × 4.8 = 7.30 ton
4
× d 2 × Fv =
4
× 1.92 × 2.1 = 5.95 ton
Control
7
กก(ก = 7/5.95 = 1.18 2 '
Detail Drawing
A
2Ls 75x75x10
B
2Ls 90x90x10
C
E
1L 50x50x8
1L
50
x
50
x8
Bolt A325×19 mm
Gusset plate 20 mm
D
More ckeck for:
- Effective area of tension members (0.5Fu Ae)
- Block shear failure of tension members