Metal Structures

Lecture XIV

Joints - bolted and welded(part II)

Contents

Interactions → #t / 3

Truss nodes → #t / 7

Splice joints of hollow sections → #t / 47

Stiffeners → #t / 57

L section; additional rules → #t / 83

I-beam, welded joints; additional rules → #t / 84

Pins; additional rules → #t / 86

Column head → #t / 88

Examination issues → #t / 95

Joints

Small parts of members, where are contact between two or

more members. There are many specific phenomenons on

there short part of beams, columns, etc. Calculation

according to level of cross-section and level of element.

Example from Ist design project: bearing resistance and

punching resistance (effects of contact between bolts and

plates and memebers), slip resistance and block tearing

(effect of contact between plates and members).

Photo: Author

Interactions

→ #3 / 69

Joints - more examples:

• vertical stiffeners;

• support on masonry structures;

• contact with concrete base;

• rigid connection beam-column;

• and many many others;Photo: Author

→ #3 / 70

• Trusses nodes are welded joints. There are contact between part of nodes. Way of

calculations is presented on lecture #14;

• Stiffeners can be used for different reasons (support of slender web, joint with

secondary beam, reinforcement of column-beam joint, reinforcement of beam in

place of support on wall or on column). General rules for calculations are presented

on lecture #14;

• Column head is welded joint and part could be bolted joint. Way of calculations

for both parts is presented on lecture #14;

Intercaction / contact element-element, bolted, pinned and welded joints

(continuation)

Photo: EN 1993-1-8 fig 7.3, 7.4 Photo: Author

→ #11 / 36

• Additional rules for welded joint between L section and plate are presented on

lecture #14;

• Additional rules for welded joint between two I-beam are presented on lecture

#14;

• Additional rules for pin ended members and contact bearing stresses for pins are

presented on lecture #14;

• Bolted joint for R&CHS are presented on lecture #14.

Intercaction / contact element-element, bolted, pinned and welded joints

(continuation)

→ #11 / 37

Truss nodes

For steel memebers used as a part of truss, requirements for resistance and stability

must be satisfied (→ #13). But, additionally, many requirements for local behaviour of

joints must be satisfied. For joints, different types of local instability and local

concentration of stresses are very important. These local phenomenons make, that

many types of cross-sections are banned for truss.

PN B 03200 EN 1993

Elements Each types of cross-sections are accepted

Joints No additional requirements Additional requirements → many

types of cross-sections are not

accepted

Cross-sections of truss members

→ #13 / 11

Modern types of cross-sections

(EN)

Old types of cross-sections (PN-B)

Chords

Web

members

Photo: Author

→ #13 / 12

Additional requirements for truss nodes

(EN 1993-1-8 7.1):

Chords → ✓ I ;

Web members → ✓ ;

Deformation ends of element are not accepted;

fy (✓) ≤ 460 MPa;

fy (✓) > 355 MPa → fy, design = 0,9 fy;

t (✓) ≥ 2,5 mm;

tchord (✓) ≤ 25 mm;

Compressed chords and web memebers → Ist or IInd class of cross-section;

bi ≥ 30o;

Distances between web members (eccentricitire) must be respected (→ #t / 11 - 14);

Shape of joints must be respected (EN 1993-1-8 fig. 7.1), (→ #t / 15 - 16);

(Length of member) / (depth of member) > 6 (EN 1993-1-8 5.1.5.(3));

Photo: tatasteelconstruction.com

Photo: Author

D

Permissible:

g ≥ t1 + t2

Permissible:

q / p ≥ 0,25

Not permissible:

g < t1 + t2

or

q / p < 0,25

t1t2

g

p

q

EN 1993-1-8 7.1Photo: Author

Additionally

d1 ≤ d2

and

fy,1 t1 ≤ fy,2 t2

EN 1993-1-8 7.1

Photo: Author

Results:

e

e

There is possible, that we must trace other

axis of member to satisfy requiremen for

g ≥ t1 + t2 or q / p ≥ 0,25

It makes eccentricities. Eccentricities make

non-zero values of bending moment.

Photo: Author

Photo: EN 1993-1-8 fig. 5.3

Acceptable limits for

eccentricities:

-0,55 a0 ≤ e ≤ +0,25 a0

a0 = h0 or d0

EN 1993-1-8 (5.1a), (5.1b)

Types of permissible joints

Photo: EN 1993-1-8 fig. 7.1

For each type of joint, many additional requirements must be fulfilled. These requirements are

presented in few tables in EN 1993-1-8; symbols are explained in EN 1993-1-8 1.5.(4), (5), (6).

Joint Table Comments

Chord Web members

CHS CHS 7.1 -

RHS CHS, RHS 7.8, 7.9 -

I-beam CHS, RHS 7.20 -

C-section CHS, RHS 7.21 C-section for chord is acceptet, but in this situation local

bending moments must be taken into consideration (this

means, this structure is not ideal truss).

Geneally, requirements presentes in tables, are as follow:

min ≤ (depth of HS) / (thickness of its wall) ≤ max

There are three groups of requirements for nodes of truss (EN 1993-1-8 5.1.5):

Acceptable values of eccentricities (→ #t / 14);

Loads applied according to ideal truss (→ #13 / 3);

General and additional requirements, permissible shapes of joints (→ #t / 10, 15, 16).

Consequences of satisfied or not satisfied (partial satisfied) are various, in dependence of group

of requirements. It must be taken into consideration in calculation model of truss. There are

three basis models of calculation:

1. Ideal truss 2. Continous chords 3. Frame

(hinge joints) (hinge/rigid joints) (rigid joints)

Photo: Author

Additionally, there are three possibilities for ideal truss in case of local bending moments:

1a. No bending moments 1b. Bending moments act on

chord only

1c. Bending moments act on

joint and chord

Ideal truss: axial forces only

for elements and joints.

Axial forces only for most

part of elements; axial forces

and bending moments for part

of chords; axial forces for

joints.

Axial forces only for most

part of elements; axial forces

and bending moments for part

of chords; axial forces and

bending moments for joints.

Model 2: axial forces only for web members; axial forces and bending moments for chords;

axial forces and bending moments for joints.

Model 3: axial forces and bending moments for each members and joints.

Photo: Author

Models of calculations for various values of eccentricities:

Member Eccentricities

0 Inside

acceptable

limits (#t / 14)

Outside

acceptable limits

(#t / 14)

Compression

chord

1a 1b 1c

Tension chord 1a 1b = 1a 1c = 1a

Brace

member

1a 1b = 1a 1c = 1a

Joint 1a 1b = 1a 1c

No difference between

1b or 1c and 1a (ideal

truss) for this part of

structure

In case of continous loads (for example: truss purlin), or applied out of nodes – model 2.

In case of not satisfied general, requirements or permissible shapes of joints – model 3.

Algorithm

Photo: Author

→ #3 / 84

There are many ways of local destrucion for truss nodes. According to EN 1993-1-8,

there are 6 modes of failure, presented on fig. 7.3, 7.4.

Photo: eqclearinghouse.org

Photo: scielo.br

Photo: offshoremechanics.asmedigitalcollection.asme.org

Sections 1. Chord face failure

F M

CHS - CHS

RHS - RHS

C/R HS - I / H - -

Photo: EN 1993-1-8 fig 7.3, 7.4

Sections 2. Chord wall / web failure

F M

CHS - CHS

RHS - RHS

C/R HS - I / H

Photo: EN 1993-1-8 fig 7.3, 7.4

Sections 3. Chord shear failure

F M

CHS - CHS

RHS - RHS

C/R HS - I / H

Photo: EN 1993-1-8 fig 7.3, 7.4

Sections 4. Punching shear

F M

CHS - CHS

RHS - RHS

C/R HS - I / H - -

Photo: EN 1993-1-8 fig 7.3, 7.4

Sections 5. Brace failure

F M

CHS - CHS

RHS - RHS

C/R HS - I / H

Photo: EN 1993-1-8 fig 7.3, 7.4

Sections 6. Local buckling

F M

CHS - CHS

RHS - RHS

C/R HS - I / H

Photo: EN 1993-1-8 fig 7.3, 7.4

For various shapes of nodes and various cross-sections of members (CHS, RHS, I-beam),

various modes of failure are the most dangerous. Resistance of node is defined in

dependence of the most dangerous modes.

Nodes are loaded by axial forces (Ni, Ed) from web members, first of all. In many cases, local

bending moments must be taken into consideration. For flat truss we analysed only bending

moments in plane (ip) of truss (Mip, i, Ed). Additionally, for multi-chords truss, bending

moments out of plane (op) of truss (Mop, i, Ed) must be taken into consideration.

There are different formulas for resistance of truss nodes (Ni, Rd, Mip, i, Rd, Mop, i, Rd) for

different shapes of nodes and different modes of failue. These formulas are presented on EN

1993-1-8, tab. 7.2-7.7, 7.10-7.19, 7.21, 7.22, 7.24. Way of check of resistance depends on

type of cross-sections of members (CHS, RHS, I-beam).

Web memebers

- chord

Chapter Tables General formula Comment

CHS - CHS 7.4 7.2, 7.3, 7.4,

7.5, 7.6, 7.7

Ni, Ed / Ni, Rd +

(Mip, i, Ed / Mip, i, Rd)2 +

Mop, i, Ed / Mop, i, Rd ≤ 1,0

Gusset plates are

possible

Tab. 7.4 - I/H is not a

chord

C/R HS - RHS 7.5 7.10, 7.11,

7.12, 7.13,

7.14, 7.15,

7.16, 7.17,

7.18, 7.19,

Ni, Ed / Ni, Rd +

Mip, i, Ed / Mip, i, Rd +

Mop, i, Ed / Mop, i, Rd ≤ 1,0

Gusset plates are

possible

C/R HS - I / H 7.6 7.21, 7.22 Ni, Ed / Ni, Rd +

Mip, i, Ed / Mip, i, Rd ≤ 1,0

I / H chord is not

recommended for

multi-chords trusses

C/R HS - C 7.7 7.24 Different formulas Secondary moments

should be taken into

account

Modern truss: for joints must be calculated:

Stiffness (lecture # 20, #21)

Resistance of connecting elements (welds / bolts; lecture #9, #11)

Resistance of joint (lecture #t)

Old type truss (designed „under rule” of old Polish Standard) - only connecting

elements were calculated; gusset plates were drawed; stifeness of nodes were not

calculated.

Old type:

Axes must intersect in one point;

Ends of members must be as close as possible each other (10-15 mm place for welds);

Photo: Author

There must be marked length of connecting elements (length of weld / space for bolts) along

members;

Outline of gusset plate = ends of connecting elements;

Photo: Author

Reflex angles are not accepted.

Photo: Author

Old type and modern trusses cam be designed with gusset plates or without gusset plates

Photo: Konstrukcje stalowe, K. Rykaluk,

Dolnośląskie Wydawnictwo Edukacyjne

Wrocław 2001

Photo: Konstrukcje stalowe, K. Rykaluk,

Dolnośląskie Wydawnictwo Edukacyjne

Wrocław 2001

There is need additional flat surface ("mounting table") for support of purlins when

we use CHS

Photo: Konstrukcje stalowe, K. Rykaluk,

Dolnośląskie Wydawnictwo Edukacyjne

Wrocław 2001

Deformations: elongation (tensile force) and abridgement (compressive force):

Transmission of forces form chords to supports and zero force members:

Photo: Author

→ #13 / 87

Members with big value of forces

should goes to nodes as close as

posible.

Photo: Author

Example: web members CHS,

chords HEB

Photo: Author

Example: web members CHS,

chords CHS

Support of truss

There are two most important types of supports:

Steel columns;

(concrete columns – rare case)

Steel columns;

Concrete columns;

Masonry walls

Photo: Author

Joints truss - steel column

Photo: Author Photo: vijaylaxmiengineering.in

Photo: aleo.com

Way of calculation → #t / 46

Joints truss - steel column

Photo: Author

Photo: wasatchsteel.blogspot.com

Photo: fireengineering.com

Joints truss - concrete / masonry structure

Photo: Author

Photo: tboake.com

Photo: structuremag.org Way of calculation → #12 / 78, #12 / 95

There is big horizontal reaction for symmetrical support. Because of this reaction, truss will

behave as for unsymmetrical supports after short time of exploatation (deformations of

columns, local destruction of masonry or concrete structure around anchor bolts).

Calculation as for symmetrical supports means, that forces taken under consideration in

design project are completely different than real forces in structure: for bottom chord real are

much more bigger than theoretical. This means big probability destruction of bottom chord

and collapse total structure.

Better way is to model the truss with unsymmetrical supports - this is closer its real

behaviour.

Photo: Author→ #13 / 85

There are special design way for diversity pinned and roller supports for massive

structures or structures with big loads (bridges).

Photo: web.mit.edu

Photo: web.mit.edu

Photo: .tatasteelconstruction.com

Photo: .texasescapes.com

Photo: . fbcdn-photos-g-a.akamaihd.net

Photo: wikipedia

Because of susceptibility of column /

wall, and / or anchor bolts, there are

spring supports for horizontal direction

and rotation in real truss.

One roller and one pinned support is

good approximation and idealisation

for this situation.

Changing the static scheme due to the

susceptibility of structural elements is common

in the analysis of steel structures. For example,

difference between pinned and rigid column

base is only number and position of anchor

bolts.

Photo: Author

Photo: j-p.com.ua

Photo: 1.bp.blogspot.com

Splice joints of hollow sections

Photo: Author

Because of transport loading gauge, long structre should be divided into transport

members. There is good idea, that max length of membes (L 1 or L2) should not be greated

than 12,00 m. Members are connected each other by splice joints on construction site.

For I-beam chords, there can be adopted shear joint (→ Des #1).

Photo: gsi-eng.eu

Photo: encrypted-tbn0.gstatic.com

Photo: zs4-sanok.pl

For hollow sections, there is used tension joint.

The problem is, that in EN 1993-1-8 tension joint is presented for I-beams only. The

same in old Polish Standard PN B 03200.

Generalisations of both methods are presented on literature.

Generalisation of Literature Comments Page

PN B 03200 J. Bródka, M. Broniewicz,

Konstrukcje stalowe z rur,

Arkady 2001

Separated procedures for CHS and

RHS;

No clear influence of longitudinal

stiffeners for resistance;

Similar methods, only few small

differeces;

#t / 50 –

52

EN 1993-1-8 Access Steel SN044a-EU

Design models for splices

in structural hollow,

Internet edition

#t / 53 -

56

PN B 03200, CHS

tptpri

re

rp

r0

e1 e2

p2

re = ri + t

r0 = re + e2

rp = r0 + e1

rp = 2 r0 - ri

tp = max (t1 ; t2)

t1 = √[(2 NEd / (fyp k)]

k = [k1 + √(k22 - k1

2)] / (2 k1)

k1 = ln (r0 / ri)

k2 = k1 + 2

NEdNEd

t2 = 1,2 √[(c SRt) / (fyp beff)]

beff = min (2pm ; 4m + 1,25 e)

m = r0 - ri

e = min (rp - r0 ; 1,25m)

c = m - 0,5d

SRt = As min (0,65fub ; 0,85fyb)

Photo: Author

n = max (n1 ; n2)

n1 = NEd k3 / SRt

n2 = NEd / SRt

k3 = 1 - 1 / k + 1 / (k k4)

k4 = ln (rp / r0)

Additionally, welds should be calculated according to Lec #9 example 4.

Requirements satisfy → no additionaly condition for NEd / NRd

PN B 03200, RHS

Photo: Author

NEd NEd

tp tp

A B C

p2 p2e1 e1

e2 e2 e2

e3 e3e3

For cases A and B:

• M16, M20 or M24 are recommended;

• tp ≥ d (case A, n = 4)

• tp ≥ d + 3 mm (case B, n = 8)

• NRd = 0,8 n SRt

• SRt → #t / 50

Additionally, welds should be calculated according

to Lec #9 example 4.

NEd / NRd ≤ 1,0

√{K NEd / [n (1+d)]} ≤ tp ≤ √(K NEd / n)

d = 1 - d / p2

K = 4000 bred / (fyp p2)

bred = e2 + tRHS - d / 2

NRd = tp2 (1 + d a) n / K

a1 = [(K SRt / tp2) - 1] (e3 + 0,5 d) / [d (e3 + e2 + tRHS)]

e1 = 0,5 p2

e3 ≤ 1,25 e2

PN B 03200, RHS

Photo: Author

NEd NEd

tp tp

A B C

p2 p2e1 e1

e2 e2 e2

e3 e3 e3

For case C:

Neff = NEd {1 + bred d a2 / [a3 (1 + d a2 )]} / n

a2 = [K NEd / (n tp2) -1 ] / d

a3 = e3 + d / 2

SRt → #t / 50NEd / NRd ≤ 1,0

Neff / SRt ≤ 1,0

Additionally, welds should be calculated according

to Lec #9 example 4.

EN 1993-1-8, CHS

tptpri

re

rp

r0

e1 e2

p2

2,2 d0 ≤ p2 ≤ min (14 tp ; 200 mm)

d0 = d + 2 mm (d ≤ 24 mm)

d0 = d + 3 mm (d > 24 mm)

1,2 d0 ≤ e2 ≤ 1,5 - 2,0 d

1,2 d0 ≤ e1

NEdNEd

Photo: Author

Additionally, welds should be calculated according to Lec #9 example 4.

NRd = min (NRd1 ; NRd2)

NRd1 = tp2 fyp p k / (2 gM0)

k → #t / 50

NRd2 = n Ft, Rd / k3

k3 = 1 - 1 / k + 1 / (k k5)

k5 = ln (reff / r0)

reff = re + e2 + eeff

eeff = min (e2 ; 1,25 e1)NEd / NRd ≤ 1,0

EN 1993-1-8, RHS

Photo: Author

NEd NEd

tp tp

A B C

p2 p2e1 e1

e2 e2 e2

e3 e3e3

Cases A and B are not recommended. SHS are accepted in case C.

12 mm ≤ tp ≤ 26 mm

4 ≤ n ≤ 2 + 2 hRHS / p2

d0 = d + 2 mm (d ≤ 24 mm)

d0 = d + 3 mm (d > 24 mm)

1,2 d0 ≤ e2 ≤ 1,5 - 2,0 d

1,2 d0 ≤ e1

2,2 d0 ≤ p2 ≤ min (5,0 d ; 14 tp ; 200 mm)

EN 1993-1-8, RHS

Photo: Author

NEd NEd

tp tp

A B C

p2 p2e1 e1

e2 e2 e2

e3 e3 e3

For case C:

EN 1993-1-8, RHS

Photo: Author

NEd NEd

tp tp

A B C

p2 p2e1 e1

e2 e2 e2

e3 e3 e3

For case C, continuation:

√{K NEd / [n (1+d)]} ≤ tp ≤ √(K NEd / n)

K → #t / 52

NRd = min (n Ft, Rd ; n Bp,Rd ; N1, Rd)

Ft, Rd → #10

Bp,Rd → #11

Additionally, welds should be calculated

according to Lec #9 example 4.

N1, Rd = tp2 (1 + d a1) n / (K gM2)

a1 → #t / 52

NEd / NRd ≤ 1,0

Stiffeners

• Support for transverse beams

(connection between primary and

secondary beams);

• Support for slender web

(prevention of web local

instability);

• Support for slender flange

(prevention of flange local

instability);

• Increasing of web resistance

under shear forces.

Photo: Author

Support for transverse beams (connection between primary and secondary beams)

Photo: Author

Support for slender web (prevention of web local instability);

Support for slender flange (prevention of flange local instability)

Probability of instability for compressed flange and / or compressed part of web is much lower

for little sub-panels than for one big panel.

Photo: Author

Increasing of web resistance under shear forces.

Photo: Author

Photo: Author

1. No end post

2. Non-rigid end post

3. Rigid end post

4. Transverse stiffener

5. Intermediate support

stiffener

6. Longitudinal stiffener

7. Column transverse

stiffener

8. Diagonal stiffener

Location of stiffeners

Photo: Author

Vertical (2, 3, 4, 5): over

supports, in connections

between primary and

secondary beam and

under big transversal

loads;

Transversal (7): on axes

of flanges;

Longitudinal (6): h / hc=

1/3 - 1/2;

Diagonal: on joints beam-

columns.

h

h

hc

hc

compression

compressiontension

tension

Geometry of stiffeners:

a

hw

bs

ts

tw

Photo: Author

Conditions:

Independent of the load Load-dependent

Conditions: Stiffeners: Stiffeners: Conditions:

Thickness of adjacent

elements

(#t / 65)

2, 3, 4, 5, 7 2, 4, 5, 7 Contact stress

(#t / 70)

Class of the cross-section

(#t / 66)

2, 3, 4, 5, 6,

7, 8

2, 4, 5, 7 Axial compression

(#t / 71 - 77)

Prevent torsional buckling of

stiffener (#t / 67)

2, 4, 5, 7 6 Cross-sectional resistance

(#t / 78)

Rigid supports for web panel

(#t / 68)

2, 4, 5, 7 8 Diagonal stiffener

(#t / 79 - 81)

Rigid end post

(#t / 69)

3 2, 3, 4, 5, 6,

7, 8

Welds

(#t / 82)

Thickness of adjacent elements:

ts ≥ thickness of secondary beam web

or

ts ≥ thickness of beam flange

Photo: Author

Photo: Author

Class of the cross-section:

There is no special requirements in EN 1993, but each formula for stiffeners is

as for cross-section of no-VIth class of cross-section. Because of this stiffeners

should have Ist, IInd or IIIrd class of cross-section.

bs / ts ≤ 14 e

Prevent torsional buckling of stiffener

EN 1993-1-5 (9.3)

JT / Jp ≥ 5,3 fy / E

JT = bs ts3 / 3

Jp = bs3 ts / 3 + bs ts

3 / 12

Rigid supports for web panel

EN 1993-1-5 9.3.3 (9.6)

a / hw ≥ √2 → Jst ≥ 1,50 hw3 tw

3 / a2

a / hw < √2 → Jst ≥ 0,75 hw tw3

Jst = 2 [ bs3 ts / 12 + bs ts (bs + tw)2 / 4 ]

Rigid end post

EN 1993-1-5 9.3.1 (3)

EN 1993-1-5 fig. 9.6

Two couples of stiffeners: 2 As ; Ws, x

e ≥ 0,1 hw

As ≥ 4 hw tw2 / e (two couples of plates)

Ws, x ≥ 4 hw tw2 (hot rolled I-beam as end post)

Contact stresses

EN 1993-1-5 9.4 (2)

Fs, Ed / (2 cs ts fy) ≤ 1,0

cs

bs

Photo: Author

Stiffeners are treated as bar, compressed by axial force Ns, Ed;

We analyse flexural buckling about x-x axis;

We take into consideration cross-section of stiffeners and cooperating

part of web (┼ cross-section);

Axial compression

EN 1993-1-5 9.2.1

EN 1993-1-5 9.4 (2)

Axial force Ns, Ed contains imperfections

of stiffeners;

Additionally we must analyse

imperfections of web - represented by

additional load q;

We must analyse interaction between

axial force Ns, Ed, buckling about x-x and

bending moment Ms, Ed (q).Photo: Author

Photo: lmsteelfab.com

Ns, Ed = max (Fs, Ed + DNst ; V*Ed + DNst ; DNst )

Fs, Ed - transverse force, acts on stiffeners (from secondary beam, from supports...);

there is possible that Fs, Ed = 0 (for stiffeners used as support for slender web or flange only

→ #t / 73 );

DNst = sm b2 / p2 - from imperfections of stiffeners (b = hw for analysis of stiffeners);

V*Ed = max [VEd - fyw hw tw / (lw gM1 √3) ; 0] - part of shear force over web

resistance of shearing;

VEd - shear force at the distance 0,5 hw from the edge of the panel with the largest

shear force;

gM1 = 1,0;

lw → #t / 74;

sm → #t /75;

_

_

Ns, Ed = DNst (for stiffeners used as support for

slender web or flange only)

Ns, Ed > DNst

Requirements from #t / 71 may be assumed to

be satisfied provided when:

Jst ≥ (1 + 300 w0 u / b) (sm b4) / (E p4)

Jst = 2 [ bs3 ts / 12 + bs ts (bs + tw)2 / 4 ]

Total procedure of calculations is necessary (#t

/ 71 - #t / 77):

q = p sm (w0 + wel) / 4

sm → #t /75;

w0, u, wel → #t / 76;

b = hw

EN 1993-1-5 A.3

a = a / hw

a < 1,0

kt kzts + 4,00 + 5,34 / a2 kzts + 5,35 + 4,00 / a2

a ≥ 1,0

kzts = max { [2,1 3√ (Jst / hw)] / tw ; [9 hw2 4√ (Jst / (hw t3

w))] / a2 }

Jst - about axis z for longitudinal stiffeners;

If no longitudinal stiffeners, kzts = 0

Slenderness factor for web resistance:

Photo: Author

lw = hw / (86,4 tw e) lw = hw / (37,4 tw e √kt )

_ _

sm = (scr, c / scr, p) (1 / a1 + 1 / a2) Neq- axial / b

scr, c = p2 E tw2 / [ 12 (1 - n2) a2 ] ≈ 190 000 (tw / a)2 [MPa]

scr, p = ks (28,4 e)2 fy (tw / b)2 ≈ 190 000 ks (tw / b)2 [MPa]

a = (a1 + a2) / 2 (usually a = a1 = a2)

b = hw

ks - for web,

according to EN

1993-1-5 tab. 4.1

Neq- axial → #t / 76;

w0 = s / 300

s = min (a1 ; a2 ; b)

wel = b / 300

u = max [ 1,0 ; p2 E emax gM1 / (fy 300 b) ]

emax = bs / 2

b = hw

Neq- axial = max (NEd, 1 ; smax A / 2 )

smax is analised when MEd or MEd + NEd,1 is

applied to beam.

Photo: Author

Ms, Ed = q hw2 / 8

cx = cx (c, ┼, lcr) (according to lecture #5)

lcr = 0,75 hw

NRd = A┼ fy / gM0

MRd = W┼, x, el fy / gM0

Ns, Rd / (cx NRd) + Ms, Ed / MRd ≤ 1,0 - D0, x

Class of cross-section ┼ 1 or 2 3 or 4

D0, x 0,1 + 0,2 [ (W┼, x, pl / W┼, x, el) - 1] 0,1

tw

ts 15 e tw15 e tw

EN 1993-1-1 NA.20

Photo: Author

Cross-sectional resistance (longitudinal stifeners)

EN 1993-1-5 9.3.4

There are two possibilities of calculations:

1.

Longitudinal stiffeners are

treated as a part of cross-section;

their geometry is added to

geometry of beam;

2.

NEd, eq = scomp, max Acomp / 2

NRd = 2 bh-s th-s fy

NEd,eq / NRd ≤ 1,0

Photo: Author

Diagonal stiffeners

Photo: fgg.uni-lj.si

Photo: Author

Possibility of increasing resistance and stiffness of column web in shear;

Rarely use, first of all in exterior columns of frames;

No information in Eurocode;

No clear guidance in the literature;

Photo: microstran.com.au

MEd

MEd

hIc

hIb

a

F1F1

F1

F2

F2

F2

F3

F1 = MEd / hIc

F2 = MEd / hIb

F3 = √ (F12 + F2

2)

Diagonal stiffener

under tension

Photo: Author

MEd

hIb

hIc

a

F1F1

F1

F3

F2

F2F2

Diagonal stiffener

under compression

(more often used)

According to supposition in literature, if diagonal stiffness and its welds have

enough resistant to bear tension F3:

resistance of column web in shear Vwp, Rd → ∞ (→ #12);

stiffness of column web in shear k2 → ∞ (→ #21, #22);

Photo: microstran.com.au

Photo: skcthailand.com

Photo: chodor-projekt.net

Welds

Vertical and transversal stiffeners:

Lecture #9 example 1, FV = max (Fs, Ed ; V*Ed ; 0) → #t / 72;

Longitudinal stiffeners:

Lecture #9 example 6a, s1, t1 according to max values on analysed member;

Diagolnal stiffeners:

Lecture #9 example 2, Fy = F3 → #t / 80;

Resistance of L section, connected by one leg:

NRd = Aeff fy / γM1

L - section; additional rules

Photo: Author

Recommendation:

b1 ≈ b – 30 mm

b2 ≈ b + 30 mm

t1 = t2 b2 / b1

For distribution of cross-section forces:

t = (t1 + t2) / 2

b = (b1 + b2) / 2

Afp = t b

Jfp = 2 [b t3 / 12 + b t (h / 2)2]

For this project is possible to take the same geometry of flange plates and web plates as for bolted connection.

Photo: Author

I-beam, welded joints; additional rules

→ Des #1 / 62

min 10 mm

min 30o

min 10 mm

min 30o

h ≥ 25 mm; h ≥ 3 t r ≥ 25 mm; r ≥ 3 t

t t

h r r

t

Photo: Author

→ #8 / 17

Geometrical requirements for pin ended members

EN 1993-1-8 tab 3.9

Pins

FEdd0

a

c

When thickness t is given:

a ≥ [FEd gM0 / (2 t fy)] + 2 d0 / 3

c ≥ [FEd gM0 / (2 t fy)] + d0 / 3

FEd

d02,5 d0

0,75 d0

1,6 d0

0,3 d0

1,3 d0

When geometry is given:

t ≥ 0,7 √ [FEd gM0 / fy]

d0 ≤ 2,5 t

Photo: Author

sh, Ed / fh, Rd ≤ 1,0

sh, Ed = 0,591 √ [E FEd, ser (d0 - d) / (d2 t)]

fh, Rd = 2,5 fy / gM6, ser

Contact bearing stress for pins

EN 1993-1-8 3.13.2

Jb / Jc ≥ 20 20 > Jb / Jc ≥ 10 10 > Jb / Jc

No rocker Flat rocker Round rocker

Cap plate bending

Cap plate - column welds

Rocker - beam compression

Rocker - cap plate welds

Rocker - cap plate

compression

Cap plate - bending

Cap plate - column welds

Cap plate - column

compression

Rocker - beam compression

Rocker - cap plate welds

Rocker - cap plate

compression

Cap plate bending

Cap plate - column welds

Cap plate - column

compression

Column head

Hinge support for bean on columnPhoto: Author

Rocker-beam compression

EN 1337-6

Structural bearings; rocker bearings

Beam (round rocker):

(NEd / bf) /[23 r fu2 / (E γ

M)] ≤ 1,0 γ

M= 1,0

or (flat rocker):

(NEd / bf) / [fy (2 tf + bf) / γM] ≤ 1,0 γ

M= 1,1

Rocker: (round and flat)

(NEd / bf ) / [fy (2 tr + bf) / γM ] ≤ 1,0 γ

M= 1,1

tf

tr

r

b

L bf

L ≈ bf ≈ hchc

Photo: Author

Alternatively, according to PN B 03200:

Contact flat - flat element:

NEd / Acontact ≤ 1,25 fy

Contact round - flat element:

0,42 √ [ E NEd / (bf r) ] ≤ 3,6 fy

Rocker-cap plate compression

NEd / ( L b fy ) ≤ 1,0

Rocker - narrow thick member; stress between

rocker and cap plate has nearly constant value

s s

Photo: Author

Welds between cap plate

and column

Lecture #9, example #3

Welds between roocker and

cap plate

Photo: Author

Cap plate bending

With rocker:

tcp ≥ min { tf ; √[ (3 NEd l12) / (a b fy) ] - tr }

Without rocker:

tcp ≥ min {tf ; √[ (3 NEd l2) / (a b1 fy) ]}

tcp

tcptcp

Calculated as short

cantilever

Photo: Author

Cap plate-column compression

Cap plate - wide thin member; stress between cap plate and column has non-linear

character; we assume big constant value in central part and zero at ends of flange. Constant

value of stress exists in distance from end of column.

NEd / Aeff ≤ fy

2 tcp + b

2 tcp + b

b

tcp

s

s

s

sPhoto: Author

Calculation of resistance for truss nodes

Modes of failure for truss nodes

The role and placement of horizontal and vertical stiffeners

Calculation of vertical stiffeners

Calculation of columns heads

Examination issues

Whether-in-plane - obciążenie przęsłowe w prętach kratownicy

Out-of-plane - obciążenie prostopadłe do płaszczyzny kratownicy

Closely spaced build-up members - pręt wielogałęziowy

Chord size failure - zniszczenie przystykowe pasa

Chord size wall / web failure - zniszczenie boków / środnika pasa

Chord shear failure - ścięcie pasa

Punching shear - przebicie

Brace failure - zniszczenie elementu skratowania

Local buckling - wyboczenie miejscowe

Gussed plate - blacha węzłowa

Splice joint - styk montażowy

Head - głowica

Stiffener - żebro

Contact strength - docisk

Rocker - płytka centrująca

Cap plate - płyta głowicy

Thank you for attention

© 2017 Tomasz Michałowski, PhD

tmichal@usk.pk.edu.pl