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�� Eccentric ShearEccentric Shear
�� Axial TensionAxial Tension
�� Shear and TensionShear and Tension
�� End Plate ConnectionsEnd Plate Connections
Timber and Steel DesignTimber and Steel Design
Lecture Lecture 1122 Bolted Connections IIBolted Connections II
Mongkol JIRAVACHARADET
S U R A N A R E E INSTITUTE OF ENGINEERING
UNIVERSITY OF TECHNOLOGY SCHOOL OF CIVIL ENGINEERING
Bolts Subjected to Eccentric ShearBolts Subjected to Eccentric Shear
Elastic AnalysisElastic Analysis
c.g.
e P
c.g.
e P
P
P c.g. c.g.
e P
P
P
P
V
M
When the load P does not pass through
the center of gravity of a bolt group.
e P
d4d1
d3d2
r4
r1
r3
r2
44332211c.g. drdrdrdrPeM +++==
4
4
3
3
2
2
1
1
d
r
d
r
d
r
d
r===
∑∑∑∑====
2
442
332
222
11 ,,,
d
Mdr
d
Mdr
d
Mdr
d
Mdr
Moment Resistant of Bolt GroupMoment Resistant of Bolt Group
Forces on bolts proportions to C.G. distance
1 1
1 1
sin
sin
r r H
d d v
θθ
= =
1 1
2 2
1 1
r v Md v MvH
d dd d
= = = ∑ ∑
2
MhV
d=∑
Vertical and Horizontal ComponentsVertical and Horizontal Components
h
v
V
H
r1
c.g.
d1
θ
θ
��������� 12-1 ������������� �ก�ก ��������ก�������ก ����������������� ������� �ก�����!��"#$%%�� ��&�ก
15 cm P=12 ton
10 cm
10 cm
10 cm
10 cm
P/8
P/8 P/8
P/8P/8
P/8
P/8
V
V
V
V
V
V
V
H2
H1H1
H1H1
H2H2
P/8
V
H2
C.G.
15 cm
5 cm
������� ���ก�ก����������� ����ก ������������ �����ก��������ก������
�� �� !"�����#$����������ก�ก������%��&��'�����!��������'�����#���� ���ก����'�
������ก��
2 2 2d h v= +∑ ∑ ∑2 2 2 28(5) 4(5 15 ) 1,200d = + + =∑
2
2
240(15)3.0 ton
1200
240(5)1.0 ton
1200
/ 8 12 / 8 1.5 ton
MvH
d
MhV
d
P
= = =
= = =
= =
∑
∑
���(�������ก�ก������%�������'�(����ก�&��%��&��'�)������ก�&
2 23.0 2.5 3.9 tonr = + =
e = 15 + 5 = 20 cm
M = Pe = (12)(20) = 240 t-cm
C.G.
P/8
V
H
15 cm
5 cm
��������� 12-2 ����������� P ��ก����������'�#�����������������(�����%)�� ��� "�'�#������*������# +ก A36 ��� �ก�ก ��� A325 1��� 22 �.�. ������"��&�3�� �ก ���������"��%�45�� ������� �ก�����!��"#$%%�� ��&�ก
5 @ 8 cm
e = 40 cm P
16 cm
Critical bolts
�����
1. �������: M = Pe = 40P &��-;�.
2. ���������ก����������ก� !"#���$���2 2 212(8) 768 cmhΣ = =
2 2 2 2 24(4) 4(12) 4(20) 2,240 cmvΣ = + + =
2 2 2 2768 2,240 3,008 cmd h vΣ = Σ + Σ = + =
3. ��&�'��ก�()������ก�ก�����ก*� �������&�� ������ "%�����������1��
2
40 (20)0.266
3,008
Mv PH P
d= = =Σ 2
40 (8)0.106
3,008
Mh PV P
d= = =Σ
P/12 = 0.083P &��
4. )����,��-����ก�ก�����ก*�
2 2 2 2( /12) 0.266 (0.106 0.083) 0.326r H V P P P= + + = + + =
��ก&���� �.1 � �ก�ก ��� A325 1��� 22 �.�. �����"��&�3�� �ก ���������"��%�45�� ��%ก���45��������)��&�� " 5.63 &�� �����?�
0.326P = 5.63 &��
�?*�#��ก��ก������ P = 17.3 ��� �
Design of Single Line Fasteners under MomentDesign of Single Line Fasteners under Moment
How many bolts required (n=?)
Assuming R is the force in the
outermost fastener
T
C
Typ. p
npnp3
2
M
2d
dMR
Σ=
Average load/length = R/p @ outermost fastener
Load/length @ extreme fiber
−
=1n
n
p
R
R/p
(R/p)×n/(n-1)
−
=122
1
n
n
p
RnpT
)1(63
2 3
−=
=n
pnRnpTM
Solving equation for n :pR
M
n
n
Rp
Mn
616≅
−=
From Steel Structures Design and Behavior 4th Ed. (1996) by Charles G. Salmon pp.150-156
Example 4.12.6 : Determine the required number of 22-mm-diam A325 bolts for one
vertical line of fasteners shown in figures below. Assume it to be a bearing type
connection with threads included in the shear planes (A325-N)
15 cm20 ton
12 mm
plate(n –1) @ 10 cm
Angle 6 mm
thick
Solution:
Design strength of A325-N 22 mm
Double shear:
R = 2(π/4)(2.2)2(2.1) = 15.97 ton
Bearing:
R = 1.2(2.2)(1.2)(4.0) = 12.67 ton
(control)
Estimate the number of bolts required, 77.31067.12
152066=
×××
==pR
Mn
The R value has not been adjusted for the direct shear; try 4 fasteners
Next Step : Check !
Check the adequacy using an elastic analysis15 cm
20 ton
12 mm
plate
4 -A325-N
3 @
10 cm
Moment: M = 20×15 = 300 t-cm
Σd2 = Σv2 = 2×52 + 2×152 = 500 cm2
Moment component:
←=×
=Σ
= ton 0.9500
153002d
MvRx
Direct shear component: Rs = 20/4 = 5.0 ton ↓
Then, the resultant: ] ton 12.67 [ ton 3.1059 22 <=+=R OK
Example 4.12.7 : Determine the required number of 19-mm-diam A325 bolts in
standard holes for the bracket plate, assuming 4 vertical rows. Assume it to be a
bearing type connection with threads included in the shear planes (A325-N)
24 ton40 cm
88 14
8
8
8
8
Solution:
Design strength of A325-N 19 mm
Single shear:
R = (π/4)(1.9)2(2.1) = 5.95 ton
Bearing:
R = 1.2(1.9)(1.2)(4.0) = 10.94 ton
Half load carried by each plate: P = 24/2 = 12 ton
12 mm plate
on each
flange
Load per line of fasteners: P/4 = 12/4 = 3 ton per line
Estimate number of fasteners: 89.3895.5
40366=
×××
==pR
Mn
Try 4 bolts per row
(control)
Check the adequacy using an elastic analysis
Moment: M = 12×40 = 480 t-cm
Σh2 = 8×72 + 8×152 = 2,192 cm2
Moment component:
←=×
=Σ
= ton 66.1472,3
124802d
MvRx
Direct shear component: Rs = 12/16 = 0.75 ton ↓
Then, the resultant: ] ton 5.95 [ ton 27.3)75.007.2(66.1 22 <=++=R OK
12 ton40 cm
88 14
8
8
8
Σv2 = 8×42 + 8×122 = 1,280 cm2
Σd2 = Σh2 + Σv2 = 2,192 + 1,280 = 3,472 cm2
↓=×
=Σ
= ton 07.2472,3
154802d
MhRy
Fasteners Acting in Axial TensionFasteners Acting in Axial Tension
T
Tensile strength: T = Ft Ab
where Ab = fastener gross cross-sectional area
Ft = allowable tensile stress
= 1,400 kg/cm2 for A307
= 3,100 kg/cm2 for A325
= 3,800 kg/cm2 for A490
Diameter = d
Ab = (π/4)d2
Example 4.13.2 : Determine the required number of 19-mm.-diam A490 bolts for the
connection.
60 ton60 ton
Solution:
Tension strength per bolts, T = (π/4) × 1.92 × 3.8 = 10.77 ton
Number n of bolts required, n = 60/10.77 = 5.57, Say 6
USE 6 – 19 mm.-diam. A490 bolts ��
Combined Shear and TensionCombined Shear and Tension
2 angles
2 angles join the beam web
to the column flange
large moment transmitted
through the flanges of beam
Structural tee
2 angles
Allowable Shear and Tensile StressesAllowable Shear and Tensile Stresses
ft
Fv
Ft
fv
elliptical interaction curve
22Safe vv
v
tt fF
F
Ff −=
AISC approximates
vt fCCf 21 Safe −=
400,18.1820,1 ≤− vf
( ) 2239.4080,3 vf− ( ) 22
15.2080,3 vf−
( ) 2275.3780,3 vf− ( ) 22
82.1780,3 vf−
Allowable Tensile Stress Ft for Bolts
Subject to Combined Shear and Tension
Type of Bolt
A307
A325
A490
ก���� ������������ � ก������� ������������ �
Example 7.9 : A WT250x44.8 is used as a bracket to transmit a 25 ton load to a
W350x137 column as shown. Four 22-mm. diameter A325 bolts with thread in shear
are used. Both the column and bracket are of A36 steel. Determine the adequacy of
the connection.
From Steel Design 4th Ed. (2007) by William T. Segui pp. 401
WT250x44.8
4
3
25 tonW350x137
Solution:
For shearing stress,
Total shear force = (3/5)25 = 15 tons
Ab = (π/4)(2.2)2 = 3.80 cm2
For bearing stress,
ksc 98780.34
000,115=
××
=vf
Fv = 0.4(2,500) = 1,000 ksc
> 987 ksc OK
A = (2.2)(1.6) = 3.52 cm2 (controlled by flange of tee)
ksc 065,152.34
000,115=
××
=pf
Fp = 1.2(2,500) = 3,000 ksc > 1,065 ksc OK
For tensile stress,
Total tensile force = (4/5)25 = 20 tons
Ab = (π/4)(2.2)2 = 3.80 cm2
ksc 316,180.34
000,120=
××
=tf
WT250x44.8
4
3
25 tonW350x137
ksc 282,298739.4080,339.4080,3 2222 =×−=−= vt fF
A325 bolts with thread in shear plane:
> [ft = 1,316 ksc] OK
The connection is adequate as a bearing connection.
���กก��������������� ������� !���" �#���$���กก��������������� ������� !���" �#���$
Pe
N.A.
Eccentric Bolts under Shear + TensileEccentric Bolts under Shear + Tensile
Shear stress in each bolt:Shear stress in each bolt:
An
Pfv =
Tensile stress from moment:Tensile stress from moment:
I
ceP
I
cMft ==
M = Pe
��������� 12-3 &�����%���&��#���������*���ก�# +ก A36 (�ก&�������� �ก�ก ��� A325-N 1��� 22 �.�. &��ก�%�Aก��� #�������%�?*�#��ก%����ก 12 &�� #�����ก�Aก��� 30 ;�. �"�"#�����������"#����� �ก�ก ��� 8 ;�. "�"�"#��������% 10 ;�.
P
10 cm
8 cm
8 cm
8 cm
����� 1. ����������������0�1�����ก�ก���
I = 3.8 [4(4)2 + 4(12)2] = 2,432 ;�.4
#�������B�������1�� A325-N = 3,100 กก./;�.2
2. 2����)��!3���ก�!134�-����ก�ก������(�� !
(12,000)(30)(12)1,776 ksc < 3,100 ksc
2,432t
Pecf
I= = = OK
3. 2����)���56������-27 A325-N: Fv
= 1,480 กก./;�.2 #�������45���� �ก�ก ���!5�
12,000395 ksc < 1,480 ksc
(8)(3.8)vf = = OK
&�����%#�������B�������#��������ก���D������� $ (��ก&������� 11.5):
Type of Bolt ก���� ������������ � ก������� ������������ �
A307
A325
A490
14008.11820 ≤− vf
( ) 2239.43080 vf− ( ) 22
15.23080 vf−
( ) 2275.33780 vf− ( ) 22
82.13780 vf−
2 23,080 4.39t vF f= −
2 23,080 4.39 395 2,967 ksc > 1,776 ksc = − × = OK
Example 4.15.1 : Determine the service load capacity P for the connection in the
figure below, if the fasteners are 19-mm.-diam. A325-X bolts subject to shear and
tension in a bearing-type connection with no threads in the shear plane.
From Steel Structures: Design & Behavior 4th Ed. (1996) by Charles G. Salmon pp. 174
2–L125×90×13
14 cm
12
3
4
9 cm 9 cm
P
7.5 cm
3@8=22.5
4
4
12.5 cm
Solution:
Bolts moment of inertia:
I = 2.84(4×42+4×122) = 1,818 cm4
Tensile stress on top bolts:
PP
I
Pecft 0495.0
818,1
125.7=
××==
Shear stress on each bolt: PP
A
Pfv 0440.0
84.28=
×=
Σ=
PffF tvt 0495.015.208.3 : X-A325 max,
22 ==−=
Solve equation for P = 37.9 ton
Therefore, the service load capacity P is 37.9 ton ����
Example 4.15.2 : For the connection of bracket in figure below, determine the
number of 22-mm.-diam. A325-N. Use 7.5-cm vertical pitch.
7.5 cm pitch
P = 28 ton20 cm
p(n
–1)
np
Solution:
Shear strength of one bolt (single shear):
22 cm 8.3)2.2(4
==π
bA
R = 3.8(1.48) = 5.63 ton
Tensile strength of one bolt:
R = 3.8(3.1) = 11.8 ton
Load per vertical line of bolt:
P = 28/2 = 14 ton/line
M = (28x20)/2 = 280 ton-cm/line
Approximate number of bolt per line:
alone for required 4.4
5.78.11
28066
M
Rp
Mn
=
×
×==
alone shear for required 5.2
63.5
14
=
==R
Pn
Try 10 bolts (5 per line)
A325-N: ∅∅∅∅ 22 MM. 10 bolts (5 per line)
7.5 cm
7.5 cm
7.5 cm
7.5 cm
Moment of inertia (one line):
I = 3.8[2(7.5)2 + 2(15)2] = 2,137.5 cm4
Tensile stress:
ksc 3,100ksc 965,15.137,2
)15(10280 3
<=×
==I
Mcft OK
Shear stress:
ksc 1,480ksc 737)8.3(5
1014 3
<=×
=Σ
=b
vA
Pf OK
Check Shear + Tension Interaction:
2222 )737(39.4080,339.4080,3 −=−= vt fF
= 2,665 ksc > [ ft = 1,965 ksc ] OK
End Plate ConnectionsEnd Plate Connections
is a popular beam-to-beam and beam-to-column connection
that has been in used since the mid-1950’s.
Four-bolt unstiffened
(moment resisting)
Simple – shear only
From Segui & AISC1989
EndEnd--plate Shear Connectionsplate Shear Connections
Example 7-10 Design the end-plate shear connection for a
W450x76 beam framing into the flange of a W200x56.2
column. All structural steel is A36. The end reaction is 18 ton.
Use 19-mm-diameter A325N bolts. Use 10-mm-thick plate and assume that the
welding is adequate.
Solution:
Capacity of 19-mm A325N in single shear = (π/4)(1.9)2(1.48) = 4.20 ton/bolt
Bearing capacity of 19-mm A325N = 1.2(4.0)(1.9)(1.0) = 9.12 ton/bolt
Use edge distance = 3 cm, 3/1.9 = 1.58d > 1.5d OK
Use bolt spacing = 6 cm, 6/1.9 = 3.16d > 3d OK
Therefore, shear controls, and the required number of bolts is
n = 18/4.2 = 4.28 bolts (use 6 bolts)
The required plate length (vertically) is
2(6) + 2(3) = 18 cm
6 cm
3 cm 3 cm
6 cm
6 cm
3 cm
3 cm
Design of momentDesign of moment--resisting endresisting end--plate connectionsplate connections
2. Determine tension force in the beam flange.
tf
tpd
M
Tf
Cf
f
ftd
MT
−=
3. Select bolts to resist Tf . Additional bolts
are placed on compression side for beam
reaction, with min. of two.
4. Consider split-tee = portion of beam flange and
adjacent plate subjected to tensile load.
TfPf
Pfw
1 1s Pes = distance from load line to point of inflection
= Pe/2 = (Pf – 0.25db – 0.707w)/2
1. Determine whether connection moment is at least 60% of beam moment strength.
5. Maximum moment in the split-tee at load line:
Me = F1s
where
F1 = shear force = Tf /2
s = Pe/2 = (Pf – 0.25db – 0.707w)/2
6. Required plate thickness:
1 1s
Point of inflection
F1
Me
tp
yb
pp
e FFtb
M75.0
6/2==
Max. bending stress = allowable stress
yp
ep
Fb
Mt
8=
Check shear in the plate:
[ ]yv
pp
v FFtb
Ff 40.01 =≤=
bp
Pf
Pf
Example 7.14 : Design an end-plate for a W400x56.6 beam. This connection must
be capable of transferring the full capacity. Use A36 steel, E70XX electrodes, and
A325 bolts.
Since a W400x56.6 is compact, the allowable bending stress is
Fb = 0.66Fy = 0.66(2,500) = 1,650 ksc
and the moment capacity is
M = Fb Sx = 1.65(1,010) = 1,667 Ton-cm
The flange force is
ton 3.431.16.39
667,1=
−=
−=
f
ftd
MT
Bolts: Try two rows of two bolts each:
Allowable tensile stress A325: Ft = 3,100 ksc
Required bolt area = 43.3/(4×3.1) = 3.49 cm2
Use 22 mm diameter A325 bolts (A = 3.80 cm2)
End plate thickness,
cm 37.35.225
97.8888=
××
==yp
pFb
Mt
End plate width bp ⇒ bf + 5 = 25 cm
1 1s
Point of inflection
F1
Me
tp
Shear force F1 = Tf /2 = 43.3/2 = 21.7 ton
s = Pe /2 = (Pf – 0.25db – 0.707w)/2
s = (5 – 0.25×2.2 – 0.707×0.5)/2 = 4.1 cm
End plate moment, Me = F1 s = 21.7×4.1 = 88.97 t-cm
Use 1 3/8 inch plate (tp = 3.5 cm)
Check shear in the plate:
[ ]ksc 000,140.0
ksc 2485.325
107.21 3
1
=≤
=××
=
y
pp
F
tb
F
OKOK