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The column is essentially a compression member, but the manner in which it tends to fail and the amount of load which causes failure depend on:
The first point is obvious - a steel column can carry a greater load than timber column of similar size.
Columns having a large cross-section area compared to the height are likely to fail by crushing. These "short columns" have been dealt with earlier.
Buckling of Slender Columns
If a long, thin, flexible rod is loaded axially in compression, it will deflect a noticeable amount. This phenomenon is called buckling and occurs when the stresses in the rod are still well below those required to cause a compression/ shearing-type failure. Buckling is dangerous in that it is sudden and once started is progressive.
Although the buckling of a column can be compared with the bending of a beam, there is an important difference in that the designer can choose the axis about which a beam bends, but normally the column will take the line of least resistance and buckle in the direction where the column has the least lateral unsupported dimension.
Since the loads on columns are never perfectly axial and the columns are not perfectly straight, there will always be small bending moments induced in the column when it is compressed.
There may be parts of the cross section area where the sum of the compressive stresses caused by the load on the column could reach values larger than the allowable or even the ultimate strength of the material.
Therefore the allowable compressive strength d cw is reduced by a factor kl , which depends on the slenderness ratio and the material used.
PbW = kl x d cw x A Where:
Pbw = allowable load with respect to buckling
kl = reduction factor, which depends on the slenderness ratio
d cw = allowable compressive stress
A = cross-section area of the column
When the load on a column is not axial but eccentric, a bending stress is induced in the column as well as a direct compressive stress. This bending stress will have to be considered when designing the column with respect to buckling.
As stated earlier, the relationship between the length of the column, its lateral dimensions and the end fixity conditions will strongly affect the resistance of the column to buckling. An expression called slenderness ratio has been developed to describe this relationship:
A = KL / r = l / r where:
A = slenderness ratio
K = effective length factor, whose value depends on how the ends of the column are fixed
L = length of the column
r = radius of gyration (r = I / A)
I = effective length of the column (K x L)
There are four types of end conditions for a column or strut:
1 Total freedom of rotation and side movement - like the top of a flagpole. This is the weakest end condition.
The consideration of the two end conditions together results in the following theoretical values for the effective length factor. (Kp = factor usually used in practice).
Types of end conditions
Columns and struts with both ends fixed in position and effectively restrained in direction would theoretically have an effective length of half the actual length. However, in practice this type of end condition is almost never perfect and therefore somewhat higher values for K are used and can be found in building codes. In fact, in order to avoid unpleasant surprises, the ends are often considered pinned (Kp = 1.0) even if the ends in reality are restrained or partially restrained in direction.
Types of end conditions II
The effective length can be different with respect to the different cross-sectional axes:
1. A timber strut which is restrained at the centre has only half. The effective length when buckling about the y-y axis, as when buckling about the x-x axis. Such a strut can therefore have a thickness less than its width.
2. In the structural framework, the braces will reduce the effective length to 1 when the column A-B is buckling sideways, but since there is no bracing restricting buckling forwards and backwards, the effective length for buckling in these directions is 31. Similarly, the bracing struts have effective lengths of l/2 d and d respectively.
3. The leg of a frame, which is pinned to the foundation has the effective length 1 = 2 L, but if the top is effectively secured for sideways movement, the effective length is reduced to 1 = L.
4. In a system of post and lintel where the bottom of the post is effectively held in position and secured in direction by being cast in concrete, the effective length / = 2 L.
Axially Loaded Timber Columns
Timber columns are designed with the following formulae:
A =KL / r and Pbw = kl x d cw x A
Note that in some building codes a value of slenderness ratio in the case of sawn timber is taken as the ratio between the effective length and the least lateral width of the column l / b
Design a timber column which is 3 metres long, supported as shown in the figure and loaded with a compressive load of 15kN. Allowable compressive stress (d cw) for the timber is 5.2N/mm²
Table 4.5 Reduction Factor, (k), for Stresses with Respect to the Slenderness Ratio for Wood Columns
|Slend-erness Ratio||l / b||2.9||5.8||8.7||11.5||14.4||17.3||20.2||23.0||26.0||28.8||34.6||40.6||46.2||52.0|
|l / r||10||20||30||40||50||60||70||80||90||100||120||140||160||180|
b = least dimension of cross section, r = radius of gyration
1 In this case the end conditions for buckling about the x-x axis are not the same as about the y-y axis. Therefore both directions must be designed for buckling. (Where the end conditions are the same, it is sufficient to check for buckling in the direction which has the least radius of gyration).
Find the effective length for buckling about both axis. Buckling about the x-x axis, both ends pinned:
lx = 1.0 x 3000 = 3000mm
Buckling about the y-y axis, both ends fixed:
Iy = 0.65 x 3000 = 1950mm
2 Choose a trial cross section, which should have its largest lateral dimension resisting the buckling about the axis with the largest effective length. Try 50 x 125mm. The section properties are:
A = b x d = 50 x 125 = 6250mm²
3 Find the allowable load with regard to buckling on the column for buckling in both directions.
l x = lx / rx = 3000 / 36.1 = 83 :. kl x = 0.41 (from table 4.5)
l y = ly / ry = 1950 / 14.4 = 135 :. kl x = 0.16 (from table 4.5)
Pw=kl x s c x A
Pwx = 0.41 x 5.2 x 6250mm² = 13325N
Pwy = 0.16 x 5.2 x 6250mm² = 5200N
4 The allowable load with respect to buckling is smaller than the actual load. Therefore a bigger cross section has to be chosen. Try 75 x 125 mm and repeat steps 2 and 3.
A = 75 x 125 = 9375mm²
Find the allowable buckling load for the new cross section:
l x =lx/rx =3000/36.1 = 83 gives kl x = 0.41
l y =ly/ry =1950 / 21.7 = 90 gives kl y = 0.35
Pwx = 0.41 x 5.2 x 9375 = 19988N say 20kN
Pwy = 0.35 x 5.2 x 9375 = 17063N say 17kN
The allowable load with respect to buckling on the column with cross section 75 x 125mm is therefore 17kN. This is bigger than the actual load, but further iterations to find exactly the section to carry the 15kN are not necessary.
The compressive stress in the chosen cross section will be:
s c = F / A = 9375 = 1.6N/mm²
This is much less than the allowable compressive stress which made no allowance for slenderness.
Axially Loaded Steel Columns
The allowable loads for steel columns with respect to buckling can be calculated in the same manner as for timber. However, the relation between the slenderness ratio and the reduction factor (k') is slightly different as seen in Table 4.6.
Table 4.6 Reduction factor (kl ) for Stresses with Respect to the Slenderness Ratio for Steel Columns
Calculate the safe load on a hollow square steel stanchion, whose external dimensions are 120 x 120mm. The walls of the column are 6mm thick and the allowable compressive stress ace = 150N/mm². The column is 4 metres high and both ends are held effectively in position, but one is also restrained in direction.
The effective length of the column l = 0.85L = 0.85 x 4000 = 3400mm.
l =l / r =3400 / 46.6 = 73 gives Kl = 0.72
Pw = kl x s cw x A = 0.72 x 150 (1202 - 1082) = 295kN.
Axially Loaded Concrete Columns
Most building codes permit the use of plain concrete only in short columns, i.e., columns where the ratio of the effective length to least lateral dimension does not exceed 15, i.e. l/r C 15. If the slenderness ratio is between 10 and 15, the allowable compressive strength must be reduced. The tables of figures relating to l/ b in place of a true slenderness ratio are only approximate, since radii of gyration depend on both b and d values in the cross section, and must be used with caution. In the case of a circular column:
b = D / 4 x (12)1/2 0.87D, where
D = the diameter of the column.
Table 4.7 Permissible Compressive Stress (Pcc ) in concrete for Columns (N/mm²)
|Concrete Mix||Slenderness ratio, 1/b|
Higher values of stress may be permitted, depending on levels of supervision of work.
A concrete column, with an effective length of 4 metres has a cross section of 300 x 400mm. Calculate the allowable axial load, if a nominal concrete mix 1:2:4 is to be used.
Slenderness ratio l/b = 400 / 300 = 13.3
Hence table gives Pcc = 3.47 N/mm² by interpolation.
Pw = Pcc x A = 3.47 x 300 x 400 = 416.4 kN.
Eccentrically Loaded Timber and Steel Columns
Where a column is eccentrically loaded, two load effects have to be considered:
The axial compressive stress caused by the load. The bending stresses caused by the eccentricity of the load.
Obviously, by the law of superposition, the added stresses of the two load affects must be below the allowable stress.
Therefore s /Pcw + f / fw £ 1 i.e.
(axial comp. stress) / (allowable comp. stress) + (bending stress) / (allowable bending stress) £ 1
s c / (kl x s cw) + f / fw £ 1 which can be transferred to
P1 / (Kl x A) + (s cw / fw) x M / Z £ s cw
Example 2.1 Determine within 25mm the required diameter of a timber post loaded as shown in the figure. The bottom of the post is fixed in both position and direction by being cast in a concrete foundation. Allowable stresses for the timber used are s cw = 9 N/mm² and fw = 10N/mm².
The load of 5kN on the cantilever causes a bending moment of M = F x e = 5kN x 0.5m = 2.5kNm in the post below the cantilever.
The effective length of the post = L x K = 3000 x 2.1 = 6300mm. Try with a post having the diameter 200mm.
The cross sectional properties are:
A = (p D2) / 4 = (p x 2002) / 4 = 31400mm²
A = (p D3) / 32 = (p x 2003) / 32 = 785400mm²
r = D / 4 = 200 / 4 = 50mm
The slenderness ratio =l / r= 6300 / 50 = 126
interpolation in Table 4.5 gives kl = 0.18
P / (Kl x A) + (s cw / fw) x M / Z £ s cw
30000 / (0.18 x 31400) +(9 / 10) x 2.5 x 106 / 167480 = 8.17N/mm2 £ 9N/mm2
If the post has a diameter of 200mm, it will be able to carry the loads, but the task was to determine the diameter within 25mm. Therefore a diameter of 175mm must also be tried.
l = 43 75 = 144 kl = 0.13
30000 / (0.18 x 24050) +(9 / 10) x 2.5 x 106 / 167480 = 23 N/mm2 £ 9N/mm2
This diameter is too small, so a diameter of 200mm should be chosen. It will be appreciated that the choice of effective length based on end fixity has a great effect on the solution.
Plain and Centrally Reinforced Concrete Walls
Basically walls are designed in the same manner as columns, but there are a few differences. A wall is distinguished from a column by having a length which is more than five times the thickness.
Plain concrete walls should have a minimum thickness of 100mm.
Where the load on the wall is eccentric, the wall must have centrally placed reinforcement of at least 0.2% of the cross section area if the eccentricity ratio e/ b exceeds 0.20. This reinforcement may not be included in the load carrying capacity of the wall.
Many agricultural buildings have walls built of blocks or bricks. The same design approach as that shown for plain concrete with axial loading can be used. The maximum allowable compressive stresses must be ascertained, but the reduction ratios can be used as before.
Example 22 Determine the maximum allowable load per metre of a 120mm thick wall, with an effective height of 2.8 metres and made from concrete grade C 15, a) when the load is central, b) when the load is eccentric by 20mm.
Slenderness ratio l / b = 2800 /120 = 23.3
Pcw = 2.8 - 3.3/5 (2.8 - 2.0) = 2.27N/mm² = 2.27MN/m²
Allowable load Pw = A x Pcw = 1.0 x 0.12 x (1.06 x 106) / 1000 = 272.4kN/m wall
A double interpolation gives:
Pcw = 1 .06N/mm² = 1 .06MN/ m²
Allowable load Pw= 1.0 x 0.12 x (1.06 x 106) / 1000 = 127.2kN/m wall
Central reinforcement is not required since e / b < 0.20
Table 4.8 Allowable Compressive Stress, Pcw for Concrete Used in Walls (N/mm²)
|Concrete Grade or Mix||Slenderness Ratio l/b||Ratio of Eccentricity of the load e/ b|
|Plain Concrete Walls||Centrally Reinforced Concrete Walls|
Higher values of stress may be permitted, depending on levels of supervision work.
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