�� 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