Design of Steel Elements in Tension to Eurocode

Steel elements are often required to sustain tensile forces, these elements are referred to as tension members or tie. Example of tension members includes steel bracings in buildings, roof truss members, cables etc. The design of steel elements in tension is straightforward. This is because the behaviour of steel in tension is closely related to the stress-strain behaviour of a material subjected to uniaxial tensile forces.

In this post, the design of steel elements in tension using Eurocode 3 will be presented.

Designing for Tension to EC-3

Steel elements exclusively subjected to tensile forces are under a uniaxial stress state. This is defined in Clause 6.2.3(1) that a steel element subjected to tension must satisfy the following expression:

\frac { { N }_{ Ed } }{ { N }_{ t,Rd } } \le 1.0

Where NEd= the design tensile force ; Nt,Rd=the design tension resistance.

Tension Members with Welded Connections

For sections without holes (welded connections), the design tension resistance Nt,Rd should be taken as the design plastic resistance of the cross-section Npl,Rd.

{ N }_{ pl,Rd }=\frac { A{ f }_{ y } }{ { \gamma }_{ M0 } }

Where: A is the gross cross-section area, fy is the yield strength of steel and γM0 is the partial factor of safety defined as 1.0 in the National Annexe.

Tension members Welded on one leg

Members that comprise angles connected by welding only in one leg can be treated as being concentrically loaded. Resistance can be determined using the expression given above but based on an effective cross-section area. According to Clause 4.13 of BS EN 1993-1-1, this is determined as:

  • for angles of equal legs or unequal legs that are connected by the larger leg, the area of the effective section may be considered as equal to the gross area.
  • for angles of unequal legs, connected by the smaller leg, the area of the effective section should be taken as equal to the gross area of an equivalent angle, with legs that are equal to the smaller of the legs
Tension Members with Bolted Connections

For section with holes (bolted connections), the design tension resistance is taken as the lesser of the following.

  • The plastic resistance of the gross section as defined above; and
  • The design ultimate resistance of the net cross-section at holes for fasteners which is defined as
{ N }_{ u,Rd }=0.9\frac { { A }_{ net }{ f }_{ u } }{ { \gamma }_{ M2 } }

Where: Anet = is the net cross section area; fu is the ultimate strength of steel and γM2 is the partial factor of saftey defined as 1.25 in the National Annaex.

Tension Members Bolted on one leg

However, there are further requirements. Unsymmetrically connected members in tension such as angle iron connected by one leg, T or channel sections. The eccentricity in joints and the effects of the spacing and edge distances of the bolts must be taken into account in determining the design resistance.

Angle sections bolted on one legs from which the reduction factors can be determined. tension member
Figure 1: Angles bolted on one leg

Clause 3.10.3(2) of BS-EN 1993-1-1 recommends that a single angle in tension connected by a single row of bolts see Figure 1, may be treated as concentrically loaded over an effective net section for which the design ultimate resistance should be determined as follows

{ N }_{ u,Rd }=\frac { 2.0\left( { e }_{ 2 }-0.5{ d }_{ o } \right) t{ f }_{ u } }{ { \gamma }_{ M2 } } \quad for\quad 1\quad bolt
{ N }_{ u,Rd }=\frac { { \beta }_{ 2 }{ { A }_{ net }f }_{ u } }{ { \gamma }_{ M2 } } \quad for\quad 2\quad bolts
{ N }_{ u,Rd }=\frac { { \beta }_{ 3 }{ { A }_{ net }f }_{ u } }{ { \gamma }_{ M2 } } \quad for\quad 3\quad bolts\quad or\quad more
Where
  • do is the hole diameter;
  • e2 is the distance of the centre of the fastener holes to the adjacent edge of the angle, perpendicular to the direction of load transfer (See figure 1).
  • γM2 is a partial safety factor as defined before.
  • t is the thickness of the leg of an angle.
  • fu is the ultimate strength of steel.

The net area Anet is obtained from (clause 6.2.2) of BS-EN 1993-1-1. In angles of unequal legs, connected by the smaller leg, Anet should be taken as equal to the net section area of an equivalent equal-leg angle of leg size equal to that of the smaller leg.

Parameters β1and β2 are reduction factors which are defined depending on the distance between holes (pitch p1), according to Table 1; for values of 2.5d0 < p1 < 5d, these parameters can be determined by linear interpolation.

Reductionn factors for tension member bolted to one leg of angle bar
Table 1: Reduction factors

Finally, it should be noted that no matter the value given by the ultimate resistance of steel, it is always limited by the plastic resistance of the section.

Procedure for designing a Tension Member

  • At the ultimate limit state, determine the design axial tensile force NEd using appropriate partial factors.
  • Select the steel grade
  • Select a section such that A>NEd/fy
  • Determine the design tensile resistance (Nt,Rd) for sections without holes (Nt,Rd=Npl,Rd). For section with holes Nt,Rd= min of (Npl,Rd, Nu,Rd).
  • Verify that NEd/Nt,Rd<1.0.

Worked Example

The maximum tensile force in the diagonals of a steel truss NEd = 350kN. The cross-section consists of two angles of equal legs, in steel grade S275. Design the diagonals assuming two distinct possibilities for the connections. (a) Welded connection on one leg. (b) Bolted connection on one leg.

Figure 2: Worked Example of Tension memeber
Welded connection and bolted connection of a tension member
Figure 3: Welded Connection & Bolted Connection
Welded Connection

The diagonals are made up by two angles of equal legs, but the connection is made only in one leg of the angle. Thus, the effective area can be considered equal to the gross area. Therefore, the following conditions must be satisfied.

{ N }_{ t,Rd }={ N }_{ pl,Rd }\le { N }_{ Ed }=\frac { A{ f }_{ y } }{ { \gamma }_{ M0 } }
{ A }\ge \frac { { N }_{ Ed }{ \gamma }_{ M0 } }{ { f }_{ y } } =\frac { 350\times { 10 }^{ 3 } }{ 275 } =12.72{ cm }^{ 2 }
Try\quad 2\times \left( 60\times 60\times 6 \right) \quad Equal\quad Angle
A=2\times 6.91=13.82{ cm }^{ 2 }
{ N }_{ t,Rd }={ N }_{ pl,Rd }=\frac { A{ f }_{ y } }{ { \gamma }_{ M0 } } =\frac { 13.82\times { 10 }^{ 2 }\times 275 }{ 1.0 } =380.05kN
\frac { { N }_{ Ed } }{ { N }_{ t,Rd } } =\frac { 350 }{ 380.05 } =0.92<1\quad (o.k)
Adopt\quad 2\times \left( 60\times 60\times 6 \right) \quad EA-Section
Bolted Connection

In this case, the diagonals are made up of angles of equal legs connected by 2 bolts only in one leg. According to clause 3.10.3 of EC3-1-8, the following design conditions must be ensured.

{ { N }_{ Ed }\le N }_{ t,Rd }=min\left[ { N }_{ pl,Rd }=\frac { A{ f }_{ y } }{ { \gamma }_{ M0 } } ;{ \quad N }_{ u,Rd }=\frac { { \beta }_{ 2 }{ A }_{ net }{ f }_{ u } }{ { \gamma }_{ M2 } } \right]
Try\quad 2\times \left( 60\times 60\times 6 \right) \quad Equal\quad Angle
A=2\times 6.91=13.82{ cm }^{ 2 }
{ d }_{ o }=22mm;\quad 2.5{ d }_{ o }=55mm;\quad { 5d }_{ o }=110m
as\quad { p }_{ 1 }=100mm<110mm\quad \therefore \quad { \beta }_{ 2 }=0.6\quad (by\quad interpolation)

The net area of the bolted section made up of two angles is given by

{ A }_{ net }=A-2{ d }_{ o }t=18.06\times { 10 }^{ 2 }-\left( 2\times 8\times 22 \right) =11.18{ cm }^{ 2 }
{ N }_{ u,Rd }=\frac { { \beta }_{ 2 }{ A }_{ net }{ f }_{ u } }{ { \gamma }_{ M2 } }
=\frac { 0.6\times 11.18\times { 10 }^{ 2 }\times 360 }{ 1.25 } =193.19kN<({ N }_{ Ed }=350kN)

Since, NEd =350kN > Nu,Rd =193.19kN ; therefore, the chosen cross section is not appropriate. By adopting a cross section with enhanced resistance, for example, two angles 80x80x8 mm (A =24.6 cm2 and Anet =21.08 cm2), then

{ N }_{ pl,Rd }=\frac { A{ f }_{ y } }{ { \gamma }_{ M0 } } =\frac { 24.6\times { 10 }^{ 2 }\times 275 }{ 1.0 } =676.5kN
{ N }_{ u,Rd }=\frac { { { \beta }_{ 2 }A }_{ net }{ f }_{ u } }{ { \gamma }_{ M2 } } =\frac { 0.6\times 21.08\times 360 }{ 1.25 } =364.3kN
{ N }_{ t,Rd }=min\left( 676.5kN;\quad 364.3kN \right) =364.3kN
\frac { { N }_{ Ed } }{ { N }_{ t,Rd } } =\frac { 350 }{ 364.3 } =0.96<1.0\quad o.k
Adopt\quad 2\times \left( 80\times 80\times 8 \right) \quad EA-Section

Also see: Designing a Laterally Restrained Steel Beam

Thank you!

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