Application of Notional Loads on Structures | EHF

Introduction

Notional loads are horizontal forces added to a structure to account for the effects of geometric imperfections and inaccuracies introduced during its construction. It is the minimum lateral force that must be resisted by a structure in order to ensure stability in the event of geometric imperfection in a structural element. The application of notional loads is meant to serve as an alternative to modelling the actual ‘out of plumbness’ and ‘lack of straightness’ of the structure. Thus, instead of changing the geometry of the structure, an equivalent de-stabilizing load is added to the structure. Perhaps, one of the benefits of handling this issue with loads rather than geometry is that, it is easier and quicker to adjust the loading on structures than modify their stiffness matrix.

Principles

As stated in the last section, notional loads are horizontal forces that must be added to a structure to account for the effect of inaccuracies and imperfections. These imperfections could occur due to either, out of plumpness of columns or out of straightness of elements within the structure. Whatever the case, this effects must be accounted for within the structural analysis and design of the structure. This is done using equivalent horizontal forces. This equivalent horizontal forces are normally considered as a proportion of the vertical forces.

Since notional loads represent forces that are due to imperfections, some materials will be more susceptible to the phenomena than others. For this reason, notional loads are linked to the material the structure is being constructed from.

The Eurocodes for steel and concrete structures have sections within them that are dedicated to deriving notional horizontal loads within structures. The next section will explain how notional loads are applied to steel and concrete structures in the Eurocode.

Application of Notional Loads to Steel Frames

Steel frames are very sensitive to notional loads. Rightly so, most steel frames are composed of prefabricated elements, inaccuracies and eccentricities within their connections are inevitable and very impactful. Consequently, their effect must be considered when designing steel frames.

Clause 5.3.2.(3) in BS EN 1993-1-1 is the section that accounts for imperfections in steel structures. It uses a coefficient (Ø) which is multiplied by the vertical forces to obtain the equivalent horizontal forces.

This coefficient is defined as:

\phi ={ \phi  }_{ o }{ \alpha  }_{ h }{ \alpha  }_{ m }

o) is defined as the sway angle at which a structure rotates due to the application of a notional load. It has a constant value of 0.005

αh is a factor related to the storey height of the vertical elements in the structure, defined as:

{ \alpha  }_{ h }=\frac { 2 }{ \sqrt { h }  } \quad but\quad 0.66\le { \alpha  }_{ h }\le 1

‘h’ is the height of the structure

αm takes into account the numbers of vertical elements in a row and is defined as

{ \alpha  }_{ m }=\sqrt { 0.5\left( 1+\frac { 1 }{ m }  \right)  }

‘m’ is the number of vertical elements in a row

The equivalent horizontal load is applied at each floor level and is defined as:

{ F }_{ h }=\phi { V }_{ ED }

Where: ‘Fh‘ is the equivalent horizontal load and ‘VEd‘ is the vertical load on a floor. Ø is as defined earlier

Where the overall applied lateral load is more than 15% of the vertical load in a member then the notional horizontal load can be ignored. This is expressed as Hd ≥ 0.15 Vd as defined in the above referenced clause.

Clause 5.3.2(4)B in BS EN1993:1-1

Application of Notional Loads to Concrete Frames

The approach to imperfections in concrete frames is almost the same as with steel frames. The only key difference is that there is some provision for horizontally aligned elements as well as those that are vertical.

Clause 5.2.(5) of BS EN1992-1-1 deals with imperfections in concrete frames using a coefficient (Ѳ) in similar fashion to steel frames, the equivalent load is also function of the vertical force on the structure. However, there is a slight difference in the calculation of this coefficient for concrete frames which takes into account the length of elements as well as their height.

Here, Ѳ is defined as:

\theta ={ \theta  }_{ o }{ \alpha  }_{ h }{ \alpha  }_{ m }

o) is defined as the sway angle the same way for steel structures Ѳo=0.005

αh is a factor related to the length or height of the vertical elements in the structure, defined as:

{ \alpha  }_{ h }=\frac { 2 }{ \sqrt {l}  } \quad but\quad 0.66\le { \alpha  }_{ h }\le 1

‘l’ is the length or height of the elements within the structure

αm is as defined in the case of steel frames. ‘m’ is the number of vertical elements contributing to the horizontal force that is being exerted on the structure, defined as:

{ \alpha  }_{ m }=\sqrt { 0.5\left( 1+\frac { 1 }{ m }  \right)  }

The equivalent horizontal load, applied at each floor levels is defined as:

{ F }_{ h }=\theta { V }_{ ED }

Notional Loads on Isolated Elements within a Frame

The notional loading due to imperfection discussed in the preceding sections are those occurring in the structure due to irregularities in the frame, known as global imperfections. It is possible for geometric imperfections to also exist within individual elements, this must also be checked. However, these imperfections are usually considered in the design of the elements themselves and are not necessarily assessed as equivalent horizontal loads.

Worked Example

A 6-storey office building is to be constructed using a steel frame structure having a 7.5m x 8m grids layout. The design wind load on this building is 3.5kN/m2 and the design value of the vertical actions on the floors is 14.5kN/m2 while the roof loading is 8.5kN/m2. The structure is being braced using steel bracing and has a movement joint at 30m. Figure 1 shows the overall dimensions of the structure. Determine whether a notional load should be applied to this structure and if so, what is their magnitude.

We are going to first determine the need to include notional loading in the analysis of this frame. This is done by comparing the total applied horizontal load due to wind load in both directions against 15% of the vertical load’s at each level.

The wind load on the structure is defined as:

{ W }_{ k }={ q }_{ w }{ A }_{ ref }

Where Wk is the wind load; qw is the design wind pressure; Aref is the reference area

Wind load acting on longer side

This is the wind load acting on the bracing in the shorter direction of the structure.

{ W }_{ k }=3.0\times (22.5\times 30)=2025kN
@ Roof=\frac { 3.75\times 0.5 }{ 22.5 } \times 2025=168.7kN
@Floors =\frac { 3.75 }{ 22.5 } \times 2025=337.5kN
Wind load acting on shorter side

This is the wind load acting on the bracing in the longer portion of the structure.

{ W }_{ k }=3.0\times (22.5\times 16)=1080kN
@Roof =\frac { 3.75\times 0.5 }{ 22.5 } \times 1080=90kN
@Floor =\frac { 3.75 }{ 22.5 } \times 1080=180kN

{ V }_{ Ed }=qA

Where: VEd is the vertical load per level; q is the level design load; A is the area of the structure

Roof Level
{ V }_{ Ed }=8.5\times (30\times 16)=4080kN
0.15{ V }_{ Ed } =0.15\times 4080=612kN
Floor Levels
{ V }_{ Ed }=14.5\times (30\times 16)=6960kN
0.15{ V }_{ Ed }=0.15\times 6960=1044kN

Verification

@Roof:612kN>168.7kN
\& \quad 612>90kN
@Floor :1044kN>337.5kN
\& \quad 1044>180kN\\

Therefore notional horizontal load must be considered in both directions.

Equivalent Horizontal Forces

{ F }_{ h }=\phi { V }_{ ED }
\phi ={ \phi  }_{ o }{ \alpha  }_{ h }{ \alpha  }_{ m }
{ \phi  }_{ o }=0.005
{ \alpha  }_{ h }=\frac { 2 }{ \sqrt { h }  } =\frac { 2 }{ \sqrt { 22.5 }  }
=0.42\ge 0.66=0.66
{ \alpha  }_{ m }=\sqrt { 0.5\left( 1+\frac { 1 }{ m }  \right)  }
=\sqrt { 0.5\left( 1+\frac { 1 }{ 15 }  \right)  } =0.73
\phi ={ \phi  }_{ o }{ \alpha  }_{ h }{ \alpha  }_{ m }
=0.005\times 0.66\times 0.73=0.0024

Therefore the equivalent horizontal force to be applied at each level is 0.24% of the vertical loads applied at the level.

{ F }_{ h,roof }=\phi _{ Ed }{ V }_{ Ed }
=0.0024\times 4080=9.8kN
{ F }_{ h,floors }=\phi { V }_{ Ed }
=0.0024\times 6960=16.70kN

Finally, this load must be added to the wind loads in both directions when assessing the structure for lateral loads. Consequently, the load is shared by the lateral stability system according to their stiffness’s.

Thank You!!!

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