Derivation of Actions on Concrete Columns-Worked Example

Introduction

Before a column can be designed, the actions to which it must support must be determined. This includes the axial compressive forces and bending moments applied. This post concerns the derivation of these actions, and a practical example of a 4 storey building will be used to illustrate the process.

Worked Example

The structure shown in figure 1 is a four storey having a typical floor plan. The suspended floors are made up of solid-one way slabs and beams. Preliminary sizing of the frame elements have been carried out and is indicated on the figure. The storey heights =3.5m

To illustrate the process the actions on the central column C3 will be obtained

Fig: 1 Worked Example

This was the worked example in the post How to Analyze Elements in Frames, you might want to review this post before attempting to digest the content of this post, because some of the principles presented here have been defined already.

As stated earlier the actions on columns are of two forms, axial forces and bending moments. The axial forces can be obtained either using the tributary area method or the vertical reaction from a sub-frame analysis. While the bending moments are obtained by analyzing a sub-frame.

Axial Actions

In this post, we are going to utilize the tributary area method to obtain the axial forces. The concept has been laid-down before, you might also like to review the post before progressing with this post. Tributary Area of Structural Elements

Roof
  • Permanent Actions gk= 1.5kN/m2
  • Variable Actions qk= 1.5kN/m2
  • Beam self weight = 5.15kN/m
Floors
  • Permanent Actions gk= 6.25 kN/m2
  • Variable Actions qk= 4.5 kN/m2
  • Beam Self weight =5.15kN/m

This values are taken from the previous post

Tributary Areas

The tributary area of column C3 is given as

Tributary\quad Area=\left( \frac { 7.5+7.5 }{ 2 } \right) \times \left( \frac { 4.8+4.8 }{ 2 } \right) =36{ m }^{ 2 }
Length\quad of\quad beam\quad supported\quad =\left( \frac { 7.5+7.5 }{ 2 } \right) =7.5{ m }
Axial Actions from Roof

Permanent actions on column

Roof\quad load\quad =\quad { g }_{ k,roof }\times Tributary\quad area\quad 1.5\times 36=54kN
Beam\quad self\quad weight\quad =\quad 5.15\times 7.5=38.63kN
Column\quad self\quad weight=\quad 0.35^{ 2 }\times (3.5-0.55)\times 25=9.03kN
{ G }_{ k,roof }= 54+ 38.63+9.03=101.66kN

Variable actions on column

{ Q }_{ k,roof }=0.75\times 36=27kN
Axial Actions from Floors

Permanent actions on column

Foor\quad load\quad =\quad { g }_{ k,floor }\times Tributary\quad area\quad =6.25\times 36=225kN
Beam\quad self\quad weight\quad =\quad 5.15\times 7.5=38.63kN
Column\quad self\quad weight=\quad 0.35^{ 2 }\times (3.5-0.55)\times 25=9.03kN
{ G }_{ k,floor }= 225+ 38.63+9.03=272.66kN

Variable actions on column

{ Q }_{ k,roof }=4.5\times 36=162kN
Imposed Load Reduction Factors

BS-EN 1990-1-1 allows a reduction in the variable actions on structures in category A-D based on the liklihood of all floors being fully occupied at the same time. The reduction factor to be applied is given by the lesser of the following

{ \alpha }_{ A }=1-\frac { A }{ 1000 } \ge 0.75\quad \quad ;\quad { \alpha }_{ n }=1-\frac { n }{ 10 } for\quad 1\le n\le 5
  • A is the area of the structure that is supported
  • n is the number of stories supported

To obtain the reduced variable actions, we simply multiply the cumulative imposed load at each floor by the reduction factor. Table 1 shows the load takedown for column C3

Design Value of Axial Actions

By careful observation of the values from the load-take down table, we can see, that the permanent actions are less than 4.5times the variable actions. Therefore equation 6.10b of BS EN 1990-1-1 is the critical load combination to use.

{ N }_{ Ed }=1.35\xi { G }_{ K }+1.5{ Q }_{ K }
3-R\quad { N }_{ Ed }=(1.35\times 0.925\times 101.6)+(1.5\times 20.25)=157.32kN
2-3\quad { N }_{ Ed }=(1.35\times 0.925\times 374.32)+(1.5\times 141.75)=680.06kN
1-2\quad { N }_{ Ed }=(1.35\times 0.925\times 646.98)+(1.5\times 263.25)=1202.8kN
G-1\quad { N }_{ Ed }=(1.35\times 0.925\times 919.64)+(1.5\times 359.1)=1687.06kN

Moments on Column

The moment on this column needs to be determined at two locations, first at the roof and secondly at the floors. Since the floors are typical, the moment needs to be determined only for one floor. Any of the sub-frames described in the preceding post can be used to determine the column moments, however, for this example subframe 3 particularly suited for obtaining column moments will be used. Analysis of this type of sub-frame, gives the column moments as

\\ { M }_{ col }=\frac { { k }_{ col } }{ \sum { { k }_{ cols } } +0.5\sum { { k }_{ beams } } } \cdot \left( \frac { { w }_{ max }{ l_{ 1 } }^{ 2 } }{ 12 } -\frac { { w }_{ min }{ { { l }_{ 2 } }^{ 2 } } }{ 12 } \right) \quad where\quad { l }_{ 1 }>{ l }_{ 2 }

Recall that the maximum and minimum actions on the frame C-C were obtained before as

{ w }_{ max }=82kN/m\quad ;\quad { w }_{ min }=46.3kN/m

Similarly we can obtain the maximum and minimum actions on the frame at the roof using the same process as

{ w }_{ max }=21kN/m\quad ;\quad { w }_{ min }=15kN/m
Moment at the roof level

The subframe that will be used to calculate the moment on the columns at the roof is shown in figure 2

Fig: 2 Column Subframe at Roof Level

To obtain the column moments, we must now, calculate the stiffnesses of the interconnected elements at joint C.

{ k }_{ col }=\frac { { I }_{ col } }{ l } =\frac { (350\times { 350 }^{ 3 })/12 }{ 3500 } =0.357\times { 10 }^{ 6 }m{ m }^{ 3 }
{ k }_{ R,BM }=\frac { { I }_{ R,BM } }{ l } =\frac { (300\times { 550 }^{ 3 })/12 }{ 7500 } =0.555\times { 10 }^{ 6 }m{ m }^{ 3 }
{ M }_{ col,yy }=\frac { 0.357 }{ 0.357+0.5(0.555+0.555) } \cdot \left( \frac { 21\times { 7.5 }^{ 2 } }{ 12 } -\frac { 15\times { 7.5 }^{ 2 } }{ 12 } \right) =11kN.m
Moments at the Floor levels

The sub-frame used to determine the moment at the floor level is shown as figure 3. The column moments is obtained using the same proceess.

Fig 3: Subframe at Floor Levels
{ k }_{ col }=\frac { { I }_{ col } }{ l } =\frac { (350\times { 350 }^{ 3 })/12 }{ 3500 } =0.357\times { 10 }^{ 6 }m{ m }^{ 3 }
{ k }_{ Fl,BM }=\frac { { I }_{ Fl,BN } }{ l } =\frac { (300\times { 550 }^{ 3 })/12 }{ 7500 } =0.555\times { 10 }^{ 6 }m{ m }^{ 3 }

Note, at the floor levels, the frame has an upper and a lower column, the column moments are to be shared according to the stiffnesses of this columns, in this case the columns are of equal stiffnesses, therefore the value of the moments at the top is equal to the value at the bottom

{ M }_{ col,yy }=\frac { 0.357 }{ 2(0.357)+0.5(0.555+0.555) } \cdot \left( \frac { 82\times { 7.5 }^{ 2 } }{ 12 } -\frac { 46.3\times { 7.5 }^{ 2 } }{ 12 } \right) \\\\=46.86kN.m

With this, the design actions necessary to design this column have been obtained. Table 2 presents a summary of these actions.

The moments have only been determined in the y-y axis, this is because there are no beams framing into this column about the z-z axis, which makes the column uniaxially loaded. For biaxially loaded column the equivalent frame in the z-z axis must be considered to determine the column moments in that direction, in any case, the process is the same.

THANKYOU!!!

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