steel and timber structures (ceng4123)...the lost parts) in accordance with the procedures specified...

STEEL AND TIMBER STRUCTURES (CENG4123)

PART TWO: DESIGN OF STRUCTURAL STEEL MEMBERS

November 2, 2017

Addis Ababa institute of Technology 1

Presentation

Outline

November 2, 2017Addis Ababa institute of Technology

2

Topic 1: Tension Members

Topic 2: Compression Members

Topic 3: Flexural Members

Topic 4: Beam-Column Members



3

Introduction: What are Compression members


4

Compression members are the most common of structural elements carrying either axial load with or without bending moment. This

topic deals with for the cases where members are carrying only axial load with eccentricity considered to be less than 1/1000 of the

member length.Compression members in buildings:

It’s virtually impossible to avoid

some eccentricity in compression

members.

Introduction: Compression in Steel Members


5

Compression members in supporting structures:


6


Compression members in bridges:

Compression member

Tension Member


7


Compression members in trusses:


8

Introduction: Tension in Steel Members

Compression members in frame bracings.

Depending on the direction of the load being applied either of the

members may be a compression member where as the other acts

as tension member.


9


Usually members in compression are made with rolled profiles or built up sections .


10


The resistance of a steel member subject to axial compression depends on;

► the cross section resistance or

► the occurrence of instability phenomena, such as flexural buckling, torsional buckling or flexural-torsional

buckling.

In general, the design for

compression is governed

by the second condition

(instability phenomena) as

steel members are

usually of medium to

high slenderness.


11

Introduction: Buckling of compressed steel members

The resistance of a compression member decreases as its length increases, in contrast to the axially loaded tension

member whose resistance is independent of its length.

Thus, the compressive resistance of a very slender member

may be much less than its tensile resistance, as shown


12


This decrease in resistance is caused by the action of the applied compressive load N which causes bending in a

member with initial curvature.

The curvature and the lateral deflection of a

compression member increase with the load

For the hypothetical limiting case of a perfectly

straight elastic member, there is no bending until

the applied load reaches the elastic buckling value

Ncr (EC3 refers to Ncr as the elastic critical force).

At this load, the compression member begins to

deflect laterally, as shown in the Figure, and

these deflections grow until failure occurs at the

beginning of compressive yielding.


13


The instability or buckling phenomenon can take place flexurally, torsionally or with a combination of a flexural and a torsional behavior:

Compression members having typical I- or

H-shaped cross-section with two axes of

symmetry are generally interested by

flexural buckling

Cruciform sections, T-sections, angles and, in

general, all cross-sectional shapes in which all

the elements converge into a single point, are

generally sensitive to torsional buckling

phenomena.

Cross-sections with one axis of

symmetry are prone to flexural-

torsional buckling in many cases

instead of the torsional one, owing to

the fact that both cross-sectional

centroid and shear center lie on the

axis of symmetry but are often not

coincident.

Design According to EC3: Approach


14

► Compression members subject to

▪ Axial compression only

▪ No bending

► However in practically real columns are

subject to

▪ Eccentricities of axial loads

▪ Transverse forces

► The treatment distinguishes between

▪ Stocky columns, and

▪ Slender columns

Design According to EC3: Stocky Columns


15

► The characteristics of stocky columns are

▪ Very low slenderness

▪ Unaffected by overall buckling

► The compressive strength of stocky columns is

▪ Dictated by the cross-section

▪ a function of the section classification

Design According to EC3: Resistance


16

NEd ≤ Nc,Rd

,

0

,

0

y

c Rd

M

eff y

c Rd

M

A fN

A fN

► where A and Aeff represent the gross area and the effective cross sectional area, respectively, and

► fy is the yield strength, with

► γM0 =1.0 representing the material partial safety factor.

cross-sections of class 1, 2 or 3:,

cross-sections of class 4

► Local instability phenomena only penalize the axial force-carrying capacity for cross-sections belonging to class 4

because failure occurs at a stress level considerably smaller than the yielding stress.

► When the cross-section of the compression member is characterized by a single axis of symmetry, an additional

flexural action ΔM Ed may arise, due to the eccentricity between the gross cross-section centroid (on which the axial

force is nominally applied) and the centroid of the resisting cross-section. (See Next Slides)

Strength design for a compression member subjected to a centric axial force N Ed at a given cross section is performed

by comparing the demand to the axial resistance capacity N c,Rd , that is::

,

0

,

0

y

c Rd

M

eff y

c Rd

M

A fN

A fN

Design According to EC3: Effective area Ac,eff


17

For class 4 cross-sections it is assumed that parts of the area under compression due to local

instability phenomena do not have any resistance (lost area):

Typically, the compressed portions of the cross-sections, which have to be neglected for the

resistance checks, are the parts close to the free end of an outstand flange or the central part of an

internal compressed element.

Gross and effective cross-sections in the

case of axial load

ΔM Ed=N x NEd



18

From the design point of view, it is necessary to evaluate the effective cross-section (i.e. gross section minus all

the lost parts) in accordance with the procedures specified in EN 1993-1-5 (Design of steel structures – Part 1–

5: Plated structural elements).

In case of a class 4 circular hollow cross-section, reference has to be made to EN 1993-1-6 (Design of steel

structures – Part 1–6: Strength and Stability of Shell Structures).

The effective area of a compressed plate A eff can be obtained from the gross area, A g , as:

,c eff cA A

Reduction factor ρ is defined as:

► Internal compression elements: ► Outstand compression elements:



19

Rules for the evaluation

of the effective width of

internal compression

elements.



20

Rules for the evaluation of the effective width

of outstand compression elements.

Design According to EC3: Summary


21

Cross-section Classification Summary

i. Find f y from the product standards.

ii. Get ε from Table 5.2.

iii. Substitute the value of ε into the class limits in Table 5.2 to work out the class of the flange and web .

iv. Take the least favorable class from the flange outstand, web in bending and web in compression

results to get the overall section class

Cross-section Resistance Check Summary

i. Determine the design compression force (NEd).

ii. Choose a section and determine the section classification .

iii. Determine N c,Rd , using the following equation for Class 1,2 and 3 sections, and Class 4 sections

respectively.

iv. Carry out the cross-sectional resistance check.

,

0

,

0

y

c Rd

M

eff y

c Rd

M

A fN

A fN

Worked Example: Example on cross-section resistance in compression


22

S 535 for t≤40mm

Material Properties:

► fy = 355 MPa

► fu = 510 MPa

► E = 210 GPa

► h = 254.1mm

► b = 254.6mm

► tw= 8.6mm

► tf = 14.2mm

► r = 12.7mm

254 x 254 x 73 UKC

► A = 9310mm2

Solution [I]. Section classification

Step1.2: Evaluate the slenderness ratio (c/t or d/t)

Step1.3: Evaluate the parameter ε.

(2 ) 110.37.77

2 14.2

w

f

b t rc

t t

235 2350.81

355yf

Step1.4: Determine class of the outstand element

Step1.1: Identify the element type.7.77 10 8.14 Class 2

c

t

Flange is outstand and the web is Internal element

The whole cross-section according to the least

favorable classification is CLASS 2

outstand

2 (2 ) 200.323.39

8.6

f

w

h t rc

t t

Internal

Step1.5: Determine class of the internal element

23.29 33 26.85 Class 1c

t

Step1.6: Determine class of the cross section

Example 3.1. A 254 x 254 x 73 UKC is to be used as a short ( λ≤0.2) compression member. Calculate the resistance of

the cross-section in compression, assuming grade S355 steel.For a nominal material thickness (tf = 14.2 mm and tw = 8.6 mm) of less than or equal to 16 mm, the

nominal value of yield strength fy for grade S355 steel is found from EN 10025- 2 to be 355 N/mm2.

Solution [II]. Cross-section compression resistance

Step2.1: Determine N c,Rd .

3

,

0

9310 35510 3305

1.00

y y

c Rd

M

A fN kN

Thank you for your kind attention!

End of Class Seven! Questions?

November 2, 2017

23

Addis Ababa institute of Technology

STEEL AND TIMBER STRUCTURES (CENG4123)

PART TWO: DESIGN OF STRUCTURAL STEEL MEMBERS

November 5, 2017

Addis Ababa institute of Technology 1

Presentation

Outline


2

Topic 1: Tension Members


Topic 3: Flexural Members

Topic 4: Beam-Column Members



3

Buckling : Elastic compression members


4

Buckling of straight members:

If its stable here then,

will return to

original position

when Unloaded,

If its unstable here then, will collapse,

If its neither stable nor

unstable here then,

neutral equilibrium

The load Ncr at which a straight compression member buckles

laterally can be determined by finding a deflected position which is

one of equilibrium.

Euler has done this assuming half sinusoidal buckling curve and

came up with the expression for Ncr;

The elastic buckling load Ncr and the elastic buckling stress cr.

→ In which i=√(I/A) is the radius of gyration

The buckling load varies inversely as the

square of the slenderness ratio L/i, as

shown in Figure, in which the dimensionless

buckling load Ncr/Ny is plotted against the

generalized slenderness ratio

Buckling : Thin-walled section properties


5

Buckling : Elastic compression members


6

Bending of members with initial curvature:

Real structural members are not perfectly straight, but have small initial curvatures, which causes it to bend from

the commencement of application of the axial load, and this increases the maximum stress in the member.

As the deflections v increase with the load N, so also do the bending moments and the stresses.

The limiting axial load NL at which the compression member first yields (due to a combination of axial plus bending

stresses) is given by;

in which

For stocky members, the limiting load NL approaches the squash load Ny,

while for slender members the limiting load approaches the elastic buckling

load Ncr.

Buckling : Inelastic compression members


7

Tangent modulus theory of buckling:

The analysis of a perfectly straight elastic compression member discussed in the pervious slides applies only to a material

whose stress–strain relationship remains linear.

However, the buckling of an elastic member of a non-linear material, such as that whose stress–strain relationship is shown

in the figure, can be analyzed by a simple modification of the linear elastic treatment.

It is only necessary to note that the small bending

stresses and strains, which occur during buckling, are

related by the tangent modulus of elasticity Etcorresponding to the average compressive stress N/A

instead of the initial modulus E.

Ncr,t is obtained by substituting Et for E, whence

It can be seen that the deviation of Ncr,t from Ncrincreases as the slenderness ratio L/i decreases.

Buckling : Inelastic compression members


8

Buckling of members with residual stresses:

The presence of residual stresses in an intermediate length steel compression member may cause a significant reduction in its

buckling resistance.

Residual stresses are established during the cooling of a hot-rolled or welded steel member (and during plastic deformation such as

cold-rolling).

Residual stress could;

► Cause yielding with

axial stresses,

► Reduce effective area

Buckling : Real Compression Members


9

The conditions under which real members act differ in many ways from the idealized conditions assumed in the pervious

slides for the analysis of the elastic buckling of a perfect member.

Real members;

► are not perfectly straight,

► their loads are applied eccentrically, and

► accidental transverse loads may act,

These imperfections can be represented by an increased equivalent

initial curvature which has a similar effect on the behavior of the member

as the combined effect of all of these imperfections.(Curve A)

Real members;

► A real member also has residual stresses, and

► its elastic modulus E and yield stress fy may vary throughout the

member.

These can be represented by an equivalent set of the residual

stresses.(Curve B)

Thus the real member behaves as a member with equivalent initial curvature (curve A)until the elastic limit is reached. It then

follows a path which is similar to and approaches that of a member with equivalent residual stresses (curve B).

Design for Instability According to EC3: Approach


10

► Compression members subject to

▪ Axial compression only

▪ No bending

► However in practically real columns are

subject to

▪ Eccentricities of axial loads

▪ Transverse forces

► The treatment distinguishes between

▪ Stocky columns, and

▪ Slender columns

Design for Instability According to EC3: Slender Columns


11

►The characteristics of slender columns are

▪ Columns of medium slenderness are very sensitive to the effects of imperfections

▪ Inelastic buckling occurs before the Euler buckling load due to various imperfections

▪ Initial out-of-straightness

▪ Residual stresses

▪ Eccentricity of axial applied loads

▪ Strain-hardening



12

NEd ≤ Nb,Rd

,

1

,

1

y

b Rd

M

eff y

b Rd

M

A fN

A fN

► where A and Aeff represent the gross area and the effective cross sectional area, respectively, and

► fy is the yield strength, with

► γM1 =1.0 representing the material partial safety factor.

► X is the reduction factor for the relevant buckling mode. It is obtained from the following expression:

cross-sections of class 1, 2 or 3:,


In compression members it must also be verified that:

where Nb,Rd is the design buckling

resistance of the compression member

and this generally controls design.

22

1, but 1.0

2

0.5 1 0.2

where is the non-dimensional slenderness coefficient,

given by.

1

1

1/

//

cry cr

effcreff y cr

LAf N

i

A ALA f N

i

cross-sections of class 1,

2 or 3:,




13

► is the imperfection factor;

► Ncr is the elastic critical load (Euler’s critical load) for the relevant buckling mode;

► Lcr is length of the corresponding buckling mode;

► i is the radius of gyration of the cross section;

► And

22

1, but 1.0

2

0.5 1 0.2

where is the non-dimensional slenderness coefficient,

given by.

1

1

1/

//

cry cr

effcreff y cr

LAf N

i

A ALA f N

i

cross-sections of class 1,

2 or 3:,


Where;

12

/ 93.9 ;

235 / with in /

y

y y

E f

f f N mm

Design According to EC3: Elastic buckling load & buckling Length


14

The elastic buckling load Ncr varies with;

► the member geometry,

► loading, and

► restraints,

but there is no guidance given in EC3 for determining either Ncr or Lcr.

2

cr

cr

EIN

L

But the following equation is generally

applicable for compression members;

It should be noted that it is often

necessary to consider the member

behavior in each principal plane, since

the effective lengths Lcr,y and Lcr,z may

also differ, as well as the radii of gyration

iy and iz.

The buckling lengths for isolated members, for several support conditions.

Design According to EC3: European design buckling curves


15

The effect of imperfections is included by the imperfection factor , which assumes values of 0.13, 0.21, 0.34,

0.49 and 0.76 for curves a0 , a, b, c and d (European design buckling curves), respectively

The imperfection factor and the

associated buckling curve to be adopted in

design of a given member depends;

► on the geometry of the cross sections,

► on the steel grade,

► on the fabrication process and

► on the relevant buckling plane,

as described in Table 6.2 of EC1993-1-1

(see next slide);



16



17

Design According to EC3: Slenderness for torsional and torsional-flexural buckling


18

In compression members with open cross sections, account should be taken of the possibility that resistance to

torsional or flexural-torsional buckling could be less than the resistance to flexural buckling.

The design process for these members is very similar to that for flexural buckling, the non-dimensional

slenderness coefficient being replaced by the non-dimensional slenderness coefficient for Torsion, evaluated

by the following expressions;

/ ;

/

T y cr

T eff y cr

Af N

A f N

cross-sections of class 1, 2 or 3,


where Ncr is the lower of the values Ncr,T and Ncr,TF. Ncr,Tis the elastic critical load for torsional buckling and Ncr,TF is

the elastic critical load for flexural-torsional buckling.

For both phenomena, the imperfection coefficient can be taken as corresponding

to flexural buckling about the z axis, obtained from Table 6.2 of EC1993-1-1

2

, 2 2

2

, , , , , , ,

1

14

2

Wcr T T

c ET

cr TF cr y cr T cr y cr T cr y cr T

EIN GI

i L

N N N N N N N

For members with symmetric cross section with

respect to the y axis, the torsional and flexural-

torsional buckling critical load is given by;

Design According to EC3: Slenderness for torsional and torsional-flexural buckling


19

2

, 2 2

2

, , , , , , ,

1

14

2

Wcr T T

c ET

cr TF cr y cr T cr y cr T cr y cr T

EIN GI

i L

N N N N N N N

For members with symmetric cross section with

respect to the y axis, the torsional and flexural-

torsional buckling critical load is given by;

► ic is the radius of polar gyration given by ; ic2=yc

2+(Iy+Iz)/A

► GIT is the stiffness of the section in uniform torsion;

► IT is the torsion constant;

► EIW is the warping stiffness;

► IW is the warping constant;

► LET is an equivalent length that depends on the restrictions to torsion and warping at the end sections;

► N cr,y is the critical load for flexural buckling about the y axis

► is a factor given by =1-(yc/ic)2, where yC is the distance along the y axis between the shear center

and the centroid of the section.

Where;

Design According to EC3: Summary


20

Worked Example: Example on resistance in compression against buckling


21

S 335 for t≤40mm Material Properties:

► fy = 355 MPa

► fu = 510 MPa

► E = 210 GPa

Example 3.2. Design the column BD of the steel structure represented in the figure below, using a HEB cross

section in S 355 steel, according to EC1993-1-1. The column is fixed at the base and hinged at section B (with respect

to the two principal axis of the cross section). Cross section B is fixed in both horizontal directions, in the plane

of the structure (due to the beam itself) and in the perpendicular plane (because of secondary bracing members).

Loading s already factored for ULS.

Solution

Step1: Compute the design applied compressive

axial force N Ed.



22

► h = 254.1mm

► b = 254.6mm

► tw= 10mm

► tf = 17mm

► r = 10mm

HEB 240

► A = 106cm2

► Iy = 11260cm4

► Iz = 3923cm4

► iy = 10.31cm

► iz = 6.08cm

Step2: Select a preliminary cross section.

Assuming class 1,2 or 3 cross sections, and considering

minimum cross sectional resistance.

As it is expected that buckling resistance will govern the

member design, a HEB 240 in S 355 steel is proposed (class 1

in pure compression), with the following properties

(geometrical and mechanical):

Step3: Check for instability.

Step3.1: Identify the Buckling length in both direction.

According to the support conditions, the buckling lengths

are equal in both planes, given by:

Because the buckling lengths are equal in both planes, the

orientation of the cross section is arbitrary. For constructional

reasons, the section is placed as shown in the Figure, with the

strong axis (y axis) in the perpendicular direction to the

plane of the structure.

a-a

a a



23

Step3.3: Determine the slenderness coefficients.

Since the selected section is section of class 1:

1

2

1

5.6 10.71

10.31 10 76.4

cry

y

y

y

L

i

1

6 3

1

1

/

210 10 / 355 10

76.4

yE f

1

2

1

5.6 11.21

6.08. 10 76.4

crzz

z

z

L

i

Ab

ou

t a

xis

y-y

Ab

ou

t a

xis

z-z

Step3.4: Calculation of the reduction factor Xmin

22

2 2

1

1

1.48 1.48 1.21

0.43

z

zz z

z

z

2

2

0.5 1 0.2

0.5 1 0.49 1.21 0.2 1.21

1.48

z z z

z

z

Step4: Section Verification.



24

Alternatively and conservatively the

reduction factor, X ,for each buckling

axis can be calculated from the buckling

curve provided in EC1993-1-1Xz=0.43

Xy=0.78


25

Worked Example: Example on checking a UB compression member

Example 3.3. The 457 × 191 UB 82 compression member of S275 steel of is simply supported about both principal

axes at each end (Lcr,y = 12.0 m), and has a central brace which prevents lateral deflections in the minor principal

plane (Lcr,z = 6.0 m). Check the adequacy of the member for a factored axial compressive load corresponding to a

nominal dead load of 160 kN and a nominal imposed load of 230kN.

S 275 for t≤40mm Material Properties:

► fy = 275 MPa

► fu = 430 MPa

► E = 210 GPa

Solution


axial force N Ed.

Step2: Classify the cross-section.

Flange = external or outstand element.

Web = internal or stiffened element.

(2 ) 191.3 9.9 (2 10.4)5.026

2 2 16.0

w

f

b t rc

t t

5.026 9 Class 1c

t

Flange = external or outstand element.

41.118 42 Class 4c

t

Web = internal or stiffened element.

(2 ) (2 ) 460 (2 16.0) (2 10.2)41.118

9.9

f

w

h t rc

t t

and so the cross section is Class 4 (slender).


26


Step3: Compute the Effective area. the cross-section Aeff .

Step4: Compute the Cross-section compression resistance Nc,Rd.

,

0

,

10067 2752768 561

1.0

eff y

c Rd

M

c Rd Ed

A fN

N kN kN N


Step5: Compute the Buckling resistance of the Member Nb,Rd.

Since the selected section is section of class 4: Slenderness coefficient


27


Select the buckling curve and corresponding “” value

Xz=0.305

Xy=0.83

Buckling will occur about the minor (z) axis. For a rolled UB section

(with h/b > 1.2 and tf ≤ 40mm), buckling about the z-axis, use

buckling curve (b) with α = 0.34.

and so the member is satisfactory!


28

Worked Example: Example on buckling resistance of a CHS compression member

Example 3.4. A hot finished circular hollow section (CHS) member is to be used as an internal column in a multi-storey

building. The column has pinned boundary conditions at each end, and the inter-storey height is 4m, as shown. The

critical combination of actions results in a design axial force of 2110kN. Assess the suitability of a hot-rolled 244.5 x 10

CHS in grade S355 steel for this application.

SolutionS 355 for t≤40mm


► fy = 355 MPa

► fu = 510 MPa

► E = 210 GPa

Step1: Classify the cross-section.

Tubular sections (Table 5.2, sheet 3):

Step2: Compute the Cross-section compression resistance Nc,Rd.

Step3: Compute the Buckling resistance of the Member Nb,Rd.

X=0.74


29

Worked Example: Example on buckling resistance of a CHS compression member

Elastic critical force and non-dimensional slenderness for flexural

buckling

For a hot-rolled CHS, use buckling curve a (Table 6.5 (Table 6.2

of EN 1993-1-1)). For buckling curve a, = 0.21


30

Worked Example: Example on Designing an RHS compression member

Example 3.5. Design a suitable hot-finished RHS of S355 steel to resist the loading of example 3.3.S 355 for t≤40mm Material Properties:

► fy = 355 MPa

► fu = 510 MPa

► E = 210 GPa

Solution


axial force N Ed.

Step2: Select a cross section based on an assumed reduction

value.

All plate members = internal or stiffened element.

33 28.25 38 Class 2c

t

(2 ) (2 ) 25.0 (2 8.0) (2 4.0)28.25

8.0

c h t r

t t

and so the cross section is Class 2.

Try a 250 × 150 × 8 RHS, withA = 60.8cm2, iy = 9.17cm,

iz = 6.15cm, t = 8.0mm, r=4.0mm.

Step3: Classify the cross-section.Hence, verification is only required for buckling resistance


31

Worked Example: Example on Designing an RHS compression member

Step4: Verify the Buckling resistance of the Member Nb,Rd Vs Nb,Ed.

Since the selected section is section of class 2: Slenderness

coefficient

Select the buckling curve and corresponding “” value

Buckling will occur about the major (y) axis. For a hot-finished

RHS, use buckling curve (a) with α = 0.21Xy=0.296

Xz=0.484

and so the member is satisfactory!


32

Worked Example: Example on buckling of double angles

Example 3.6. Two steel 125 × 75 × 10 UA are connected together at 1.5 m intervals to form the long compression

member whose properties are given. The minimum second moment of area of each angle is 49.9 cm4. The member is

simply supported about its major axis at 4.5 m intervals and about its minor axis at 1.5 m intervals. Determine the

elastic buckling load of the member Ncr.


► E = 210 GPa

Solution

Step1: Compute the elastic buckling load about

the major and minor axis, N cr,y and Ncr,Z for the

whole section.

Member buckling about the major axis:

Member buckling about the minor axis:

Step2: Compute the elastic buckling

load, N cr for a single angle about its

own minor axis.

To calculate the elastic buckling load

following equation is generally

applicable for compression members;

2

cr

cr

EIN

L

and so for both angles

2Ncr,min=2 × 459.7= 919kN < 1520kN.

The lowest buckling load of 919 kN

corresponds to the case where each

unequal angle buckles about its own

minimum axis.

Thank you for your kind attention!

End of Class Eight! Questions?

November 5, 2017

33

Addis Ababa institute of Technology

Class 7-Part-2 Design of Structural Steel Members - Compression Memebers - 30102017Class 8-Part-2 Design of Structural Steel Members - Compression Members - 07112017

steel and timber structures (ceng4123)...the lost parts) in accordance with the procedures specified...

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