STEEL AND TIMBER STRUCTURES (CENG4123) · 2018. 8. 10. · DESIGN OF STRUCTURAL STEEL MEMBERS November 5, 2017 Addis Ababa institute of Technology 1. ... Tension members are the most

STEEL AND TIMBER STRUCTURES (CENG4123)

PART TWO: DESIGN OF STRUCTURAL STEEL MEMBERS

November 5, 2017

Addis Ababa institute of Technology 1

Presentation

Outline

November 5, 2017Addis Ababa institute of Technology

2

Topic 1: Tension Members

Topic 2: Compression Members

Topic 3: Flexural Members

Topic 4: Beam-Column Members

Topic 1: Tension Members


3

Introduction: What are Tension members


4

Tension members are the most efficient and economical of all structural elements.

Tension members are all around us.

Introduction: What are Tension members


5

The spider’s web is a good example of a tension structure.

Tension members are the most efficient and economical of all structural elements.


6

Introduction: Tension in Steel Members

Man made tension elements: Cable stayed and Suspension Bridges.


7


Man made tension elements: Truss members in tension.


8


Man made tension elements: frame bracings members in tension.


9


Usually members in tension are made with hot-rolled profiles, typically angles or channels: in other cases, cold-formed

profiles can be conveniently used.

The cross sectional arrangement of axially stressed tension members

is structurally unimportant so long as the net cross sectional are is

sufficient to carry the design loads and the shape can be conveniently

connected to other members in the structure


10


Load carrying capacity of tension members is essentially governed by:

► distribution of the residual stresses due to the manufacturing

process;

► connection details of the element ends.

The load carrying capacity at the connection location depends on the effective area.

The effective cross-sectional area has to be evaluated

according to standards provisions.

The design of members under tensile force can be

based on the selection of a member with a cross-section

greater than the minimum area A min , which can be

evaluated on the basis of tensile design load N, such as

min

d

NA

f

where f d is the design tension limit strength.

In tension members problem of buckling

does not occur but due consideration

should be taken so that sagging

problem is avoided.


11


Failure is assumed to occur when:

► Insufficient gross-sectional area of member away from joint.

► Insufficient net cross-sectional area of the joint

Holes for rivet or bolts in tension affect the member in two ways

► Reduce the area of the cross-section

► Result in non-uniform strain in the cross-section in the neighborhood of the hole

Design According to EC3: Approach


12

Members in tension subjected to the design axial force NEd must satisfy the following condition at every section,

in accordance with European provisions:

Ned ≤ Nt,Rd

,

0

,

2

0.9

y

pl Rd

M

net uu Rd

M

A fN

A fN

min

► where A and Anet represent the gross area and the net area in correspondence of the holes, respectively, and

► fy and fu are the yield and ultimate strength, respectively, with

► γM0 =1.0 and γM2 =1.25 representing the material partial safety factors.

is associated with ductile failure due to the attainment of the yield

strength,

is related to a brittle failure in the connection section (governed by

the attainment of the ultimate strength).

Design tension resistance, Nt,Rd of the cross-section has to be assumed to be the minimum between

i. the plastic resistance of the gross cross-section, Npl,Rd , and

ii. the ultimate resistance of the net cross-section in correspondence of the connection, Nu,Rd.

Design According to EC3: Effective area


13

Anet for Non staggered fasteners for plates

𝐴net = 𝐴g −𝑑𝑜𝑡

Ag – gross sectional area

do - hole diameter

t – thickness

Rivet

diameter (d)

mm

Hole

diameter (do)

mm

≤ 14 d + 1

14< d ≤22 d + 2

=24 d + 2

≥ 27 d + 3



14

Anet for staggered fasteners for plates

Rivet

diameter (d)

mm

Hole

diameter (do)

mm

≤ 14 d + 1

14< d ≤22 d + 2

=24 d + 2

≥ 27 d + 3

If holes are staggered, failure may occur

► Along line A-B-E

► Along zigzag line A-B-C-D

Net area is calculated by simple empirical formula,

𝐴net = 𝐴g −𝑑𝑜𝑡 +𝑆2𝑡

4𝑃

Where: Ag – gross sectional area

do - hole diameter

t – thickness

S – pitch

P – spacing between center of hole or gauge distance



15

Anet Angle connected with its two legs

Rivet

diameter (d)

mm

Hole

diameter (do)

mm

≤ 14 d + 1

14< d ≤22 d + 2

=24 d + 2

≥ 27 d + 3

The method of solution assumes one leg of the angle to be rotated and brought in

the plane of the second angle and we get

► Gross width of the angle in its position would be the sum of the length of the two

legs less the angle thickness

► Gauge distance for rivet or bolts holes in the two legs is the sum of the gauge

distance in each leg less the angle thickness



16

Anet Channel and T-Sections

Rivet

diameter (d)

mm

Hole

diameter (do)

mm

≤ 14 d + 1

14< d ≤22 d + 2

=24 d + 2

≥ 27 d + 3

► For single channel connected through the web or singe tee connected through

the flange, the effective area should be calculated by

𝐴net = 𝑎1 +3𝑎1

3𝑎1 + 𝑎2𝑎2

Where a1 – is the net sectional area of the connected leg

a2 – is the net sectional area of the unconnected leg a2

a1



17

Anet Double Angel

Rivet

diameter (d)

mm

Hole

diameter (do)

mm

≤ 14 d + 1

14< d ≤22 d + 2

=24 d + 2

≥ 27 d + 3

a) For double angle connected back to back or space between

𝐴net = 𝑎1 +5𝑎1

5𝑎1 + 𝑎2𝑎2

b) For double angle or tee placed back to back and connected to each side of a

gusset or to each side of a rolled section, the effective area should be

calculated as for plate using previous equations

𝐴net = 𝐴g −𝑑𝑜𝑡 +𝑆2𝑡

4ℎ

Design According to EC3: Ultimate resistance of the net cross-

section for special cases


18

For Non staggered fasteners for angels

A single angle in tension connected by a single row of bolts in one leg may be treated as concentrically loaded

over an effective net section for which the design ultimate resistance, Nu,Rd , has to be determined as:

• with one bolt

2 0

,

2

2.0 0.5u Rd

M

e dN

• with two bolts

2,

2

net uu Rd

M

A fN

• with three bolts

3,

2

net uu Rd

M

A fN

Pitch p1 ≤2.5d0 5.0d0

Two

bolts

2=0.4 2=0.7

Three

bolts or

more

3=0.5 3=0.7

For intermediate value of p1,

linear interpolation may be used.

Design According to EC3: : Ultimate resistance of the net cross-

section for special cases


19

For staggered fasteners for angles

The total area to be deducted should be taken as the greater of:

► The maximum sum of the sectional area of the holes (A f ) in any

cross-section perpendicular to the member axis

► The sum of the sectional areas of all holes in any diagonal or

multi-linear line extending progressively across the member or

part of the member less s2t/(4p) for each gauge space in the

chain of holes, which can be expressed as

2

04

s tt n d

p

where, n is the number of holes along the considered line

Design According to EC3: Summary


20

Tension Member Design Steps Summary

i. Determine the design axial load N Ed

ii. Choose a section.

iii. Find f y and f u from the product standards.

iv. Get the gross area A and the net area Anet.

v. Substitute the values into the equations to work out N pl,Rd and N u,Rd.

vi. The design tensile Resistance is the lesser of the values of N pl,Rd and N u,Rd.

vii. Carry out the tension verification:

,

0

,

2

0.9

y

pl Rd

M

net uu Rd

M

A fN

A fN

For angles connected by a single

row of bolts, use the required

equation to work out N u,Rd

Which will depend on the number

of bolts.

2 0

,

2

2.0 0.5u Rd

M

e dN

2,

2

net uu Rd

M

A fN

3

,

2

net uu Rd

M

A fN

For 1 bolt: For 2 bolts: For 3 or more bolts:

,

,

1.0t Ed

t Rd

N

N

Worked Example: Example on Angle in Tension


21

Example 2.1. Verify, according to the EC3 Code, the strength of a single equal leg

angle L 120 × 10 mm in tension connected on one side via one line of two M16 bolts

in standard holes (Figure, dimensions in mm). Bolts connect only one side of the

angle to a gusset plate. The angle is subjected to a design axial load N Ed of 350 kN.

► Ag= 2318mm2

Solution:

Step2: Calculate the available tensile rupture strength N u,Rd.

Step3: Compute the design tension resistance, Nt,Rd of the

cross-section .

Step4: Section verification.

Step1: Calculate the available tensile yield strength N pl,Rd.

S 235 Material Properties:

► fy = 235 MPa

► fu= 355 MPa

► E= 210 GPa

L 120 x10 mm

Geometric Properties:

► Bolt diameter: d= 16mm

► Standard hole: d0= 17mm

► Hole distance: p1= 70mm

3

,

0

2318 23510 544.7

1.0

y

pl Rd

M

A fN kN

32

,

2

0.594 2318 10 17 35510 367.5

1.25

net uu Rd

M

A fN kN

N t,Rd = min(N pl,Rd, N u,Rd)=367.5kN

,

,

3501.0... !

367.5

t Ed

t Rd

N kNOK

N kN

Since P1 = 70 = 4.118d0 with linear

interpolation 2 = 0.594

Worked Example: Example on calculating net area A net


22

Example 2.2. Calculate the net area A net of the bolted section of the plate represented in the Figure. Assume a plate

with thickness t and the remaining dimensions (in mm), as indicated in the Figure.

Solution:


► Standard hole: d0=15mm

(4)

Failure line 1 225 2 15 195netA t t

Failure line 2

260225 4 15 2 205

4 45netA t t

Failure line 3

260225 5 15 4 230

4 45netA t t

Failure line 4 225 3 15 180netA t t

The net area of the plate is given by the minimum value,

A net = 180t

Worked Example: Example on a Joint of a Tension Chord of a Trussed

Beam


23

Example 2.3. Verify, in accordance with EC3, the splice connection in Figure (dimensions in mm), which connects the

end of two members of the chord of a trussed beam and transfers a design axial tension load N Ed of 2250 kN.


► fy = 235 MPa

► fu= 355 MPa

► E= 210 GPa

The flanges of the beam are

composed by 340 × 16 mm

plates and a plate 260 × 12 mm

forms the beam web. Single

cover plates 340 × 16 mm are

bolted to the beam flange in

normal holes (d 0 = 26).




Beam


24

Solution:

Step1: Calculate gross area of the chord and cover plates:22 2 340 16 260 12 14000gb f f w wA b t b t mm

22 340 16 10880gpA mm

Step2: Calculate the available tensile yield strength N pl,Rd.

3

,

0

10880 23510 2556.8 2250

1.0

gp y

pl Rd Ed

M

A fN kN N kN

Step3: Calculate the available tensile rupture strength N u,Rd.

For the verification of brittle collapse, and due to the presence of

staggered holes, the area to deduce from the gross resisting area

has to be considered as the minimum b/n the possible failure lines

(see figure):2

0

2 22

2

2 2 16 26 832 , OR

80 16(16 4 26) 2 1024

4 4 80

10880 2 1024 8832net

t b mm

s tt n b mm

p

A mm

3

,

2

8832 3550.9 0.9 10 2257.5

1.25

net uu Rd

M

A fN kN

Step4: Section verification.

,

,

22501.0... !

2257.5

t Ed

t Rd

N kNOK

N kN


Beam


25

Example 2.4. A flat bar, 200 mm wide and 25 mm thick, is to be used as a tie. Erection conditions require that the bar

be constructed from two lengths connected together with a lap splice using six M20 bolts, as shown in Figure Calculate

the tensile strength of the bar, assuming grade S275 steel.


► fy = 275 MPa

► fu= 430 MPa

► E= 210 GPa



Solution:

Step1: Calculate gross area and net area of the plate:

2200 25 5000gA mm

the total area to deduce from the gross resisting area has

to be considered as the minimum between the possible

failure lines (see figure):2

0

2 22

2

25 22 550 , OR

90 25(25 2 22) 594

4 4 100

5000 594 4406net

t b mm

s tt n b mm

p

A mm

Step2-4: The tensile resistance, N t,Rd ,of the plate:

,

0

, 1325

g y

pl Rd

M

pl Rd

A fN

N kN

, 1325t RdN kN,

2

,

0.9

1364

net uu Rd

M

u Rd

A fN

N kN

Worked Example: Example on design of a Tension Chord of a Trussed

Member


26

Example 2.5. Consider the chord AB of the steel truss, indicated in Figure, assuming it is submitted to a

design tensile axial force of N Ed = 220 kN. The cross section consists of two angles of equal legs, in steel grade

S235. Design chord AB assuming bolted connection:


► fy = 235 MPa

► fu= 360 MPa

► E= 210 GPa



Solution:

Step1: The chord, made up by angles of equal legs,

is connected by 2 bolts only in one leg. The following

design conditions must be ensured: :

2, , , ,

0 2

, with min ;g y net u

Ed t Rd t Rd pl Rd u Rd

M M

A f A fN N N N N

Step2: Calculate minimum gross area and the

corresponding net area required:3

2

0

235 10220 9.36

1.0

g y g

g

M

A f AkN A cm

Since P1 = 100 = 5d0 , 2 = 0.70

22

2

2

0

0.7 360220 10.91

1.25

2 10.91 3.6 , with t in cm

net u netnet

M

net g g

A f AkN A cm

A A td A t cm

Hence, the later governs the angle section selection

Worked Example: Example on design of a Tension Chord of a Trussed

Member


27

Example 2.5. Consider the chord AB of the steel truss, indicated in Figure, assuming it is submitted to a

design tensile axial force of N Ed = 220 kN. The cross section consists of two angles of equal legs, in steel grade

S235. Design chord AB assuming bolted connection:


► fy = 235 MPa

► fu= 360 MPa

► E= 210 GPa



Step3: Select an angle section :

Adopting a cross section with enhanced resistance,

for example, two angles (60x60x6 mm (Ag = 13.82

cm2 and A net =11.66 cm2)

Step4: Section verification:

,

0

, 324.8

g y

pl Rd

M

pl Rd

A fN

N kN

, 235.1 220 ...t RdN kN kN OK

,

2

,

0.9

235.1

net uu Rd

M

u Rd

A fN

N kN

Thank you for your kind attention!

End of Class Six! Questions?

November 5, 2017

28

Addis Ababa institute of Technology

Documents

STEEL AND TIMBER STRUCTURES (CENG4123) · 2018. 8. 10. · DESIGN OF STRUCTURAL STEEL MEMBERS November 5, 2017 Addis Ababa institute of Technology 1. ... Tension members are the most