Columns

ColumnsWhat you should know

• Before you start with this unit, you should be able to do the following:

• Solve elementary differential equations.• Calculate tensile and bending stresses.• Determine the second moment of area about

the horizontal and vertical axes passing through the centroid of built-up sections.

• Define the boundary conditions.

Expected Outcomes

• Upon Completion of this unit, you should be able to do the following:

• Derive and apply the Euler formula. • Determine the slenderness ratio and effective

slenderness ratio of a slender strut.• Determine the validity limit for the Euler formula.• Derive and apply the Rankine-Gordon (Rankine)formula

for slender struts.• Apply the Johnson equation for intermediate struts.• Apply the Perry-Robertson equation for intermediate

struts.

Definition of a strut

• Struts are defined as long compression members which fail by buckling before the induced compressive stress reaches yield point.

• The buckling load is defined as the axial load which will keep the strut in its bent form.• The calculation of the cross sectional area of a compression member is usually based on the buckling load.

P

P

FF

Strut

Example of a buckled column

• The diagram shows the buckling of a slender column subjected to axial loading.

• The phenomenon of buckling can be extended to other structures other than columns.• If you step on top an empty aluminum can, the thin cylinder willBuckle and the can will collapse.

Analysis of Strut

• The first two theories to be considered in the analysis of strut are the Euler and Rankine-Gordon theories.

• The formulas derived in these theories are based on the following limitations:

• 1) The deflection is very small.• 2) The column is perfectly straight before the load

is applied.• 3) The column material obeys Hooke’s law.• 4) The load is applied axially.

Euler formula

• Euler’s theory applies to very long struts where the effects of the direct compressive stress may be ignored.

• This was first published by Leonhard Euler in the eighteen century.

• The experiment was conducted on an ideal column with no imperfection with both ends pinned.

Euler formula

• Consider the pin-jointed strut MN, which is kept bent by the compressive force P.

• Let the deflection at a distance x from M be y.

• From equation 6.3;

Euler formula

Euler formula

• PE is the Euler buckling load for a pin-jointed strut.

• 15.1

End Fixity(Other boundary conditions)

End Fixity

• a) Both ends pinned.• b) One end fixed one end free.• c) One end fixed one end pinned .• d) Both ends fixed.

Effective length

• This is considered to be the distance between the points of contra fixture which develop as buckling occurs, the position of these points depending on the end fixing conditions.

• Effective length(le)• n is a constant depending on the end fixing

conditions.• The crippling load for the pin-jointed strut (fig. (a))

was calculated in equation (15.1)• effective length le for this case is l for n=1.

n

Critical buckling length

• This is the minimum length of strut which will buckle under a certain specified load.

• For pinned ends, when n = 1:• • (15.2)

C

2

2

e

EIPE

P

EIC

Critical length ratio (CLR)

• CLR

• For pinned ends:

• CLR:

endspinnedforlengthbucklingCritical

lengthbucklingCritical

1

P

EIP

EI

End fixing conditions

• One end fixed one end free:

• CLR:

2

2

2

2

25.0

2

EIPE

EIPE

P

EIC 5.0

5.05.0

P

EIP

EI


• One end fixed one end pinned:

• CLR:

2

2

2

2

2

707.0

EIPE

EIPE

P

EIC 41.1

41.141.1

P

EIP

EI


• Both ends fixed:

• • CLR:

2

2

2

2

4

2

EIPE

EIPE

P

EIC 2

22

P

EIP

EI

Example 15.1

• Calculate the buckling load for a bar 1 m long and 20 mm in diameter using the Euler formula and assuming

• (a) both ends pin-jointed; • (b) both ends fixed;• (c) one end fixed and the other end pinned;• (d) one end fixed and the other end free.• (e) For case (a), calculate the maximum central

deflection before the yield strength of 300 MPa is reached (E = 200 GPa).

Solution

Example 15.2

• A column consists of a hollow tube of length 2,5 m and with an outside diameter equal to I,5 times that of the inside diameter. The column must carry an axial load of 40 kN without buckling. Using a factor of safety of 5, calculate the dimensions of the tube if one end is fixed and the other end is free.

• Use the Euler formula and take E = 200 GPa.

Solution

Example 15.3

• A strut is 9 m long with a cross-section as used in example 9.2. Use Euler’s formula and calculate the buckling load if;

• (a)the ends are ball-jointed;• (b)the ends are pinned, restricting rotation

about the YY axis (E = 200 GPa).

Solution

Slenderness ratio

• The slenderness ratio of a specific column determines which buckling zone the beam is in (simple compression, inflexible buckling, flexible buckling) and also check method used for determination of the safety coefficient.

• The slenderness ratio is the ratio of the length of the strut l and k the radius of gyration about the axis which buckling will occur.

• This is written as:k

l

Slenderness ratio

• The smaller radius of gyration if the end fixing conditions about the XX and YY axes are the same is used for the slenderness ratio.

• The effective slenderness ratio:

• This is used for the general expression, so that struts of the same material but with different end fixings may be compared with one another.

k

le

Slenderness ratio

• If the slenderness ratio lies between 0 and 20, use pure crushing load equation:

• If the slenderness ratio lies between 20 and 120, use the Rankine formula.

• If the slenderness ratio is above 120, use the Euler formula.

Area

LoadStress

Validity limit for the Euler formula

• The Euler formula is dependent on E, any results obtained from this formula where the limit of

• proportionality is exceeded, will not be valid.• The yield stress is normally used as an

approximation of the proportional limit stress.• Let Sy be the proportional limit stress.• From experimental results it was found that the

point where the Euler curve becomes invalid is in the region:


• The effective slenderness ratio at this point is termed the validity limit for the Euler formula,

• 15.3

Steel struts of intermediate length

• The Euler formula can only be used for long struts with the effective slenderness ratio larger than the validity limit of the Euler formula.

• Many attempts have been establish a formula for steel struts of intermediate length.

• Since it is impossible to manufacture a perfect strut, some other approaches are used to

analyze steel struts of intermediate length.

Rankine-Gordon (Rankine) formula

• If PE is the buckling load given by the Euler formula.

• PC the load that will cause the proportional

• limit stress Sy in a short strut.

• And PR the load that will cause failure (not necessarily buckling).

• According to the Rankine formula,


• If the length of the strut approaches zero, then 1/PE will approach zero and PR will approach PC.

• If the length of the strut is increased, 1/PE will become larger 1/PC than and PR will approach PE.


• Therefore,

• (15.4)

• = This is the Rankine constant for strut with pinned ends.

The Johnson (parabolic) formula

• If the slenderness ratio is less than the critical slenderness ratio, the column is classified as a short column.

• In short columns, failure occurs by compression without appreciable buckling and at stresses exceeding the proportional limit.

• For this condition, Johnson's formula is applicable• This method is valid in the zone of inelastic

buckling.

The Johnson (parabolic) formula• If a parabola is fitted to the Euler curve at the

validity limit of the Euler formula.• The tangents to the two curves at the point of

intersection are parallel.• The slope of the tangent at the point of

intersection is called tangent modulus Et

• The critical load for intermediate struts according to the Johnson formula is:

Construction of a strut failure lines

Line

Line

Tangent point

Short strut line

Empirical zone

Safe Euler region

Safe Johnson region


• A is the cross-sectional area of the strut • C = 0,25 for one end fixed and other end free.• C = 1 for both ends pinned.• C = 2 for one end fixed and other end pinned. • C = 4 for both ends fixed. • A factor of safety should be used to account

for imperfection in strut manufacture


Advantages

• The advantage of the Johnson formula is that it provides results that closely resemble experimental

values.• For a strut with a

slenderness ratio larger than indicated in the figure, the Euler formula must be used.

Disadvantages

• The disadvantage is that this formula is only valid for struts with a slenderness ratio less than indicated in the figure.

Typical Johnson and Euler curves

Perry-Robertson formula

• This formula is based on the assumption that no strut will be perfectly straight.

• This formula is used in the Steel Construction Handbooks of most countries.

• For steel:

• (15.6)

• where A is the cross-sectional area of the strut.

Perry-Robertson formula

• And the Euler bucking stress is:

• Pc/A is the critical stress without any factor of safety built into the formula.

• Use a factor of safety of I,7 when using this formula.

• The advantage of this formula is that it can be used for intermediate and long struts.

Typical Eulerand Perry-Robertson curves

Comparison of the Euler and Rankine-Gordon formulae

• The Euler buckling load cannot be used if the effective slenderness ratio is less than the validity limit.

• The figure indicates that the stress at which buckling occurs for struts with a large slenderness ratio becomes very small.

Comparison of the Euler and Rankine-Gordon formulae

• Hence a slender strut will buckle at a small compressive stress.

• This condition can be improved by increasing the second moment of area or using a material with a larger modulus of elasticity.

Example 15.4

• A steel strut of rectangular cross-section is 1,5 m long. The width of the section is 2,5 times the

• thickness and it must carry a load of 80 kN with a factor safety of 3.

• (a) Assuming the ends to be built in, calculate the cross-sectional dimensions of the strut using the

Rankine formula.• Take the yield stress as 300 MPa and the Rankine

constant I for pinned ends as 1/7 500 (E = 200 GPa).

Example 15.4

• (b) Use MATHCAD to determine the cross-sectional dimensions using the Rankine formula.

• (c) Calculate the safe load using the Euler formula. Is this result valid?

• (d) Calculate the required dimensions for the strut using the other valid formulae discussed in this unit.

• (e) Plot the curves for the different valid formulae discussed in this unit.

Solution

• (a) W = 2,5 T, where W is the width and T the thickness of the strut. The smallest radius of gyration k occurs about the XX axis (fig 15.7).

• (1)

Solution

Solution

• The MATHCAD Solution can be referred to the textbook.

Solution

• e)

Example 15.5

A hollow cast iron column with fixed ends is subjected to a load of I MN with a safety of 2,5.

• The column is 4,5 m long and has an external diameter of 250 mm.

• (a) Calculate the thickness of the metal required using the Rankine formula.

• Take the Rankine constant for fixed ends and the proof stress for cast iron as 175 MPa.

• (b) Calculate the safe load of this column according to the Euler formula (E = 85 GPa).

• (c) Which of these two answers should be used?

Solution

• (a) Since the Rankine constant is given for fixed ends, the effective length le must be replaced by l, the length of the strut, in equation 15.4.

Solution

Solution

• The effective slenderness ratio is less than the validity limit for the Euler formula and hence the safe load of I MN should be used.

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