Buckling of Columns

Columns are the most important structural elements used in buildings. The material given here helps to understand the basics

1. Pont du Gard (France) -Roman times:
Thebridgeisconstructedfromlimestoneblocks
fittedtogetherwithoutmortarandsecured
withironclamps.Thethreetieredstructure
avoidstheneedforlongcompressivemembers.
Royal Border Bridge (England):
Openedin1850, thebridgecontinuesbeing
usedtoday.Theincreasedslendernessofthe
columnscomparedtothePontduGardreflect
technologicalimprovementsovermany
centuries.

2. The advance from masonry to the slender metal compressive members which make
up each column requires substantial bracing to prevent buckling . Opened in 1857;
Closed in 1964.
Crymlyn Viaduct (U.K.)

3. Thesteeringrod(behindthesteering
wheel)isdesignedtofail.Thatis,therod
bucklesduringacarcrashtoprevent
impalingthedriver.
Thecolumnsofabuildingare
designedsothattheydonotbuckle
undertheweightofabuilding.

4. Columns
•At the end of this topic, you will be able to:
•Classify columns.
•Explain the phenomena of buckling of columns.
•Determine the axial load that a column can withstand just before
buckling.

5. •Long slender members subjected to axial
compressive force are called columns.
•The lateral deflection that occurs is
called buckling. Occurs due to …..
•Geometry or shape, materials, boundary conditions,
and imperfections if any, all affect the stability of
columns.
•The maximum axial load a column can
support when it is on the verge of
buckling is called the critical load, P
cr.
Columns

6. Columns
•Columns are generally subdivided into the following three types according
to how they fail:
•Short columnsfail by crushing (e.g., yielding). Even if loaded eccentrically,
a short column undergoes negligible lateral deflection, so that it can be
analyzed as a member subjected to combined axial loading and bending.
•Long columnsfail by buckling. If the axial load is increased to a critical
value, the initially straight shape of a slender column becomes unstable,
causing the column to deflect laterally and eventually collapse.
•Intermediate columnsfail by a combination of crushing and buckling.

7. Determining the Critical Load
•The formula for the critical load of a column was derived in 1757 by
Leonhard Euler, the great Swiss mathematician. Euler’s analysis was based
on the differential equation of the elastic curve
•With ΣM
A= 0,
•Solving for P, we get

8. Determining the Critical Load
•Where,the effective length L
eof the column is determined by the types of
end supports.
Effective
length, L
e=
L 2L 0.7L 0.5L
P
cr=

9. 9
Effective Length and Actual length of Column
Effective length: The distance between the zero- moment points.

10. P P
L
One End Pinned
& One End Fixed
P P
L
One End Free
& One End Fixed
P P
L
Fixed Ends
End Constraints:
L/2
2L
()
2
2
L
EI
P
z
Cr
π
=
()
2
2
4
L
EI
P
z
Cr
π
=
()
2
2
4L
EI
P
z
Cr
π
=
()
2
2
2
L
EI
P
z
Cr
π
=
L2≈
P P
L
Free or Pinned Ends

11. •The stress in the column just before it buckles may be found by substituting
I = A r
2
•Where, A is the cross-sectional area and ris the least radius of gyration of the
cross section. Thus,
•Where, σ
cris called the critical stress and the ratio Le/r is known as the
slenderness ratio of the column.
•Thus, P
crshould be interpreted as the maximum sustainable load only if σ
cr< σ
pl;
where σ
plis the proportional limit of the material.
•A column always tends to buckle in the direction that offers the least resistance to
bending. For this reason, buckling occurs about the axis that yields the largest
slenderness ratio L
e/r, which is usually the axis of least moment of inertia of the
cross section.
Determining the Critical Load

12. PROBLEM:
Calculatethesafecompressiveloadonahollowcastironcolumn(oneend
rigidlyfixed,otherendhinged)of15cmexternaldiameter,10cminternal
diameter,and10minlength.UseEuler’sformulawithafactorofsafetyas5
andE=95kN/mm
2
.(Ans:74.79kN)
PROBLEM:
Asolidroundbar4mlongand5cmindiameterwasfoundtoextendby
4.6mmunderatensileloadof50kN.Ifthisbarisusedasastrutwithboth
endshinged,determinethebucklingandsafeloadforthebar.Takefactorof
safetyas4.(Ans:4190N;1047.5N)
PROBLEM:
Acolumnhasasectionof15cmx20cmand6minlength. Bothitsends
arefixed. DeterminethesafeloadbyEuler’sformulaforafactorof
safetyof3.TakeE=17.5kN/mm
2
.(Ans:359.8kN)
Determining the Critical Load

13. •Limitations in Euler’s Formula:
•In practice, the ideal conditions are never [ i.e. the strut is initially straight and the end load being applied axially
through centroid] reached. There is always some eccentricity and initial curvature present.
•The column will suffer a deflection which increases with load and consequently a bending moment
is introduced which causes failure before the Euler's load is reached.
•Rankine’s (or Rankin Gordon ) formula:
•Itisanempiricalformulausedforthecalculationofultimateloadbothforshortandlong
columns.Itgivestheultimateloadthatcolumncanbearbeforefailure.Ifcolumnis
short,calculatedloadwillbeknownascrushingload.And,loadwillbebucklingor
critical(orcrippling)load,incaseoflongcolumn.
Crippling or Critical load,????????????=
σ????????????∗????????????
????????????+????????????
????????????????????????
????????????
????????????
where,
σ
c=critical(crippling)stress;
α=Rankine’sconstant=
σ????????????
????????????
????????????
????????????
,
A=c/sareaofthecolumn,
L
e/r=slendernessratio;L
e=effectivelengthofthecolumn,r=radiusofgyration.
Determining the Critical Load –Rankine’s formula

14. Rankine’s Constant for various materials

15. Ahollowcastironcolumn4.5mlongwithbothendsfixed,istocarryanaxialloadof250kNunderworking
conditions.Theinternaldiameteris0.8timestheouterdiameterofthecolumn.UsingRankine–Gordon’s
formula,determinethediametersofthecolumnadoptingafactorofsafetyof4.Assumeσ
c,thecompressive
strengthtobe550N/mm2andRankine’sconstantα=1/1600.
Solution:

16. PROBLEM:
FindtheEuler’scripplingloadforahollowcylindricalsteelcolumnof38mm
externaldiameter&2.5mmthick.Thelengthofthecolumnis2.3manditis
hingedatitsbothends.TakeE=205GPa.Alsodeterminethecripplingloadby
Rankine’sformulausingσ
c=335N/mm
2
andα=1/7500. (Ans:16882.3 N;
17116.3 N)
PROBLEM:
AhollowcolumnofcastironwhoseO.Dis200mmhasathicknessof20mm.It
is4.5mlongandisfixedatbothitsends.CalculatethesafeloadbyRankine’s
formulawithafactorofsafetyof4.Calculatealsotheslendernessratioandthe
ratioofEuler’sandRankine’scriticalloads.Takeσ
c=550N/mm
2
,α=1/1600,
andE=80GPa.(Ans:877356.4 N;35.15; 2.0607)
Determining the Critical Load –Rankine’s formula

17. PROBLEM:
Acolumnofabuildinglooksnotsafe.CEOofacompanyhiredcivilengineertocheck
whetherthecolumnissafeornot.Columnisofmildsteelwhoselengthis3meters
andbothendsarefixed.Loadcomingonthatcolumnis400N.Criticalstresscoming
onthatcolumnis320×10^2Newton/metersquare.Cross-sectionofcolumniscircular
with24cmODand20cmID.Now,checkwhethercolumnissafeornot?
SOLUTION:
External diameter of a column = D = 24 cm; Internal diameter of a column = d = 20 cm
Length of column = L = 3m = 300 cm; Critical Stress = σ
c= 320 x 10^6 Newton / sq.meter
As the column is of mild steel, value of Rankin’s constant, α= 1/7500; Pcr= ?
Effective Length, Le = L/2 = 300 / 2 = 150 cm
Cross-section Area, A = π/4(D
2
−d
2
)= 138.23 cm
2
Radius of Gyration, r =
????????????
????????????
=
8432.035
138.23
= 7.81 cm
Determining the Critical Load –Rankine’s formula

18. Crushing Load:
Using Rankine’s formula, ????????????=
σ
????????????∗????????????
????????????+????????????
????????????????????????
????????????
????????????
????????????=
????????????????????????∗????????????????????????????????????.????????????????????????
????????????+????????????/????????????????????????????????????????????????
????????????????????????????????????
????????????.????????????????????????
????????????
Crushing Load = P = 4216 N
Using a FoSof 5, we can get the safe load, Ps = 4216 / 5 = 843.5 N.
The column is safe.
Determining the Critical Load –Rankine’s formula