Steel connection design explained

1. Analysis of
moment
connections

2. Basic principles
of connection
design
• Provide as direct a load
path as possible
• Avoid complex stress
conditions
• Weld in the shop, bolt
on site

3. Welded
connections

4. Moment connection of an I-Beam
M
• Bending moment is
carried mainly by the
flanges
• Therefore connect
flanges for moment
transfer

5. Moment connection of an I-Beam
M
• Welded connection
• Fillet welds
• Full penetration
welds
• Compression transfer
can also be
accomplished through
direct bearing
Resultant tension force T = M/d
d
C = T

6. Shear connection of an I-Beam
• Shear is carried
mainly by the web
• Therefore connect
the web for shear
transfer V

7. Shear connection of an I-Beam
• Fillet welds in shear
are commonly used
• Connect entire web
and adjust weld size
to suit shear load V

8. Moment connection of a plate
Stress in weld
σ = M (d/2) / I
= M (d/2) / (ad3/12) [kN/m2]
q = σ a
= M (d/2) / (d3/12)
= M (d/2) / I’ [kN/m]
Where
I’ = I/a
Then choose a weld size a that will
carry q
q = σ.a
where a = weld size
M
d

9. Moment connection of a plate
Can also use simplified
approach:
• Break moment into a
force couple
• Choose a suitable weld
size
• Then calculate the
required length of the
weld to carry the tension
force T
M
q = T/l
where l = weld length
d
Resultant tension force T = M/d
C = T

10. Welded shear plate
V
Centroid
of weld
group
e
V
M = V.e

11. Simplified approach
• Break eccentric load
up into a vertical
force along the
vertical weld and a
pair (couple) of
horizontal forces
along the horizontal
welds
• Then choose
lengths of welds to
carry the calculated
forces
V
V.e’/d
V
d
V.e’/d
e’

12. “Stress” calculations
V
M = V.e
V
M = V.e
+

13. “Stress” calculations for vertical force V
V
Divide shear equally
amongst all the weld lines
q = V / (total length of weld)
Choose a weld size that can
carry the “stress” q
Note q is actually a force
per length [kN/m]
qV

14. “Stress” calculations for Moment M = V.e
Treat the weld group as a cross-
section subjected to a torsional
moment
p x y
I’ 2 = I’ 2 + I’ 2
where I’ = I/a
qAx = M yA / I’p
qAy = M xA / I’p
qAM = (q 2 + q
Ax Ay
2)0.5
Similarly for point B
Then select weld size for max. q
qAx
qBx
yA
xB xA
yB
A
qAy qAM
M = V.e
B
qBy
qBM

15. “Stress” calculations for combined V and M
V
M = V.e
qAx
qAy
qAV
qA
A Combine the weld “stress”
components from the vertical
force and the torsional moment
qA = [qAx
2 + (qAV + qAy)2]0.5
Similarly for point B or any other
point that might be critical
Then select weld size for the
maximum value of q
B

16. Example of a complex connection
Column tree for Times Square 4, NYC

17. Bolted connections

18. Moment
splice in a
column

19. Moment splice of an I-Beam
M
• Bolted connection
• Divide tension and
compression resultant
equally between bolts
Resultant tension force T = M/d
d
C = T

20. Shear
connection in
bridge
diaphragm
girder
(Alex Fraser Bridge)

21. Shear connection of an I-Beam
• Bolted connections
to transfer shear are
commonly used
• Connect entire web to
avoid stress
concentrations and
shear lag
V

22. Shear connection via end plate
Coped flanges to fit in
between column
flanges
End plate

23. Moment connection with and end
or base plate

24. Moment connection with fully
welded end plate
M
hi
hmax
Ti
Tmax
Ti = Tmax (hi / hmax)
M = Σ Ti hi
C = Σ Ti

25. Pre-tensioned moment connection

26. Pre-tensioned Moment
Connection
Apply both tension and
compression forces to pre-
tensioned bolts.
Compression force can be
seen as a release of the
tension force.
M
M
TM
Ti
+
=

27. Bolted shear plate
P
Centroid of
bolt group
e
P
M = Pe

28. Vertical load
P
VP
VP
Divide the force by
n, the number of
bolts
VP = P / n

29. Moment
M
FyM
FMi
yi
xi Treat the bolt group as a
cross-section subjected to a
torsional moment
p i
I = Σ A r 2
i
2 2
= Σi A (xi + yi )
and with I’P = IP/A
FxM = M yi / I’p
FyM = M xi / I’p
FMi = (FxM
2 + FyM
2)0.5
Then select a bolt size for the
maximum force FM
ri
bolt area A
bolt i
FxM

30. Combined vertical force and
moment
P
FxM
FyM
VP
Fmax
Fmax = [FxM
2 + (FyM + VP)2]0.5
M = Pe
Then select a bolt size for the
maximum force Fmax

31. Connection Design

32. Bolted Steel Connections
Bolts in tension
6 x 200 =1200 kN
Bolts in shear

33. Failure modes of bolts in shear
Hole bearing
Hole tearout
Bolt shear

34. Bolts in
shear
e d + 2 mm
t
d
Bolt shear
Hole tearout
Hole bearing
Br = φ t n e Fu
r
B = 3 φ t d n Fu
Vr = 0.6 φ n m Ab Fu
A325M bolts: Fu = 830 MPa
A490M bolts: Fu = 1040 MPa
350W steel:
Fu = 450 MPa
φ = 0.67; n = number of bolts; m = number of shear planes

35. Bolts in tension
threads shank
d
A325M 20 mm dia
Tensile strength
Tr = 0.75 φ n Ab Fu
=0.75(0.67)(1)(314)(830) N
= 131 kN

36. Bolts in tension and shear

37. Bolts in combined shear and
tension
2
2
b b u
2
f
f
V 2
 T  0.56 A F 
m2
Where β = an interaction factor derived from test results
= 0.69 for A325 bolts, shear plane through shank
= 0.41 for A325 bolts, shear plane through threads
= 0.56 for A490 bolts, shear plane through shank
= 0.30 for A490 bolts, shear plane through threads

38. Interaction Formula
• A simpler version:
2
2
f
A F 
V 2
 T 2
 0.56
f
m2 1 – –b 2 –b –u 3
 0.75b AbFu 2

f
f
T 2
V m2
Vf/Vr
1.0
r
⏟ r
T 2
T 2 1.0
V 2
r
Tf/Tr
1.0

39. d + 2 mm = 22 mm
Example
e = 45mm
350W steel:
Fu = 450 MPa
t = 12mm
A325M bolt: Fu = 830 MPa
d = 20mm
Hole tear-out
Br = φ t n e Fu Ab= πd2/4=314 mm2
Hole bearing
Br = 3 φ t d n Fu
= (3)(0.67)(12)(20)(3)(450)
= 651 x 103 N = 651 kN
Bolt shear
Vr = 0.6 φ n m Ab Fu
= 0.6(0.67)(3)(2)(314)(830) N
= 630 kN
= (0.67)(12)(3)(45)(450)
= 487 x 103 N = 487 kN
To avoid hole tear-out
e > 3d

40. Slip-critical connections
• When slippage of a connection during normal
service conditions (under unfactored loads) is
undesirable
• Bolts are pretensioned to at least 70% of their
ultimate strength
Vs = c1 ks m n (0.7)(0.75) AbFu
where ks = slip coefficient and c1 = coefficient that relates the specific
initial tension and mean slip to a 5% probability of slip for bolts
installed by the turn-of-the-nut method

41. Values of c1 and ks
Contact surface of
bolted parts
ks
c1
A325 and
A325M
bolts
A490 and
A490M
bolts
Clean mill scale, or blast-
cleaned with class A coatings 0.33 0.82 0.78
Blast-cleaned or blast-cleaned
with class B coatings 0.50 0.89 0.85
Hot-dipped galvanized with
wire brushed surfaces 0.40 0.90 0.85

42. Slip-critical connections in
combined shear and tension
V T
1.0
1.9
Vs nAbFu

43. Welded connections
4 Times Square, NYC

44. Welded connections
Types of welds:

45. Welded truss connections

46. Welded beam connections
Full
penetration
groove
weld
Bolted
connection
for web
Backing
bar
Moment
connection
Simple
connection

47. Strength of welded connection
Tension
Base metal:
Tr = φ Ag Fy
Weld metal:
Tr = φw An Xu
Shear
φw = 0.67
Base metal:
Vr = 0.67 φw Am Fu
Weld metal:
Vr = 0.67 φw Aw Xu

48. Fillet weld
strengths
P
θ = 0o
θ = 90o
P
0.707b
b
Am = b L
Aw = 0.707 b L
Base metal:
Vr = 0.67 φw Am Fu
Weld metal:
Vr = 0.67 φw Aw Xu (1 + 0.5 sin1.5θ)

49. Electrode strengths, Xu
Matching
electrode
strength
Xu [MPa]
Steel grades (G40.21-M)
260 300 350 380 400 480 700
410 X X
480 X X X X
550 X
620 X
820 X

50. Fillet weld example
P Base metal:
Weld strength:
6 mm fillet
150 mm
Vr = 0.67 φw Am Fu
Weld metal:
Vr = 0.67 φw Aw Xu (1 + 0.5 sin1.5θ)
6 mm fillet
P Base metal:
Pr = 0.67 φw Am Fu
= 0.67(0.67)(6)L(450) N
= 1.2 L = 1.2 (550) = 660 kN
Weld metal:
Pr = 0.67 φw Aw Xu (1 + 0.5 sin1.5θ)
= 0.67(0.67)(0.707)(6)L(480)(…..) N
= 0.914 [400 + 1.5(150)]
= 571 kN
0.707b = 4.2 mm
b = 6 mm
Am = b L
Aw = 0.707 b L

51. Example
C
1000 kN
400 kN
1142 kN
800 kN
1000 kN
L = [800 / 0.9] / 2
req
= 444 mm
400 kN
6 mm fillet
Tr = 2 x571
= 1142 kN

52. Example
Mr
A325 20 mm dia
= Tr (0.45) kN
= 756 (0.45)
= 340 kNm
Vr
16 mm
Tr = φ A Fy
= 0.9 (16) (150) (350) N
= 756 kN Vr = 210/2 x 10
= 1050 kN