COLUMNS & STRUTS CLASS NOTES
FOR MECHANICAL ENGINEERING
Column is a vertical member under
axial compressive load. Type
of column is decided by a
slenderness ratio. For various
types of columns, value of
slenderness ratio depends on
the end conditions and material
of the column. Horizontal and
inclined long columns are struts.
There are three types of columns.
Fig. Types of columns, short, medium & long (depends on
slenderness ratio and material)
Firstly short column
Secondly Medium column
Thirdly long column
Struts are long compression members. These are horizontal
or inclined position. These are not vertical. Their analysis
is carried by Euler formula.
DIFFERENT END FIXING OF COLUMNS AND STRUTS
(i) Hinged at both ends
(ii) Hinged at one end & free at the other end
(iii) Fixed at one end while free at the other end
(iv) Fixed at both ends
MINIMUM RADIUS OF GYRATION, kmin
kmin = (Imin/A)0.5
Rectangular section column
k min = (db3/12)
(d is larger side and b is smaller side of the rectangle)
Square Section Column
k min = (a/(12)0.5) where a is the side of the square cross section
Solid Circular Section column
k min = (I/A)0.5 = d/4
Hollow Circular Hollow Section Column
k min = (IH/A) = (d02 + di2)0.5/4
Equivalent length of a column
Fig. Equivalent length of long columns with different end fixings
It depends on the end fixing condition of the column.
The symbol of equivalent symbol is Le.
(a) Column with both ends hinged
Its entire length buckles to form a bow. Therefore
equivalent length of a column with both ends hinged is ‘L’.
Equivalent length takes care of the end fixing conditions.
Different equivalent length for different end fixing conditions.
Equivalent lengths forming a bow are:
(i) Both ends hinged Le= L,
(ii) One end fixed and other free Le= 2L
(iii) One end fixed other hinged Le= L / √2
(iv) Both ends fixed Le= L/2
Minimum Radius of Gyration
kmin is least of ( Ixx /A)0.5 and ( Iyy/A)0.5
Columns and struts use it. It is the ratio of equivalent
length Le and the least radius of gyration k min.
Slenderness ratio = Le/k min. It has no units.
Table: Values of slenderness ratio
Type of column
> 30 & ≤ 120
> 10 & ≤ 30
> 30 & ≤ 50
For STEEL COLUMNS
Firstly Le / kmin ≤ 30, short column
Secondly Le/kmin >30 but<120, medium column
Thirdly Le / kmin > 120, long column
Buckling in columns
It is similar to bending in appearance. But it occurs under an AXIAL load ( load parallel to length). Buckling is bending due to the axial load. Bending occurs due to a load perpendicular to length in a loaded beam. Medium and long columns have buckling. A long column fail by buckling alone. A medium column first contracts then buckles. A short column only contracts.
Fail by contraction in case of ductile column. These fail by crushing in case of brittle materials.
Formula used is P = σyp A ductile materials
P = σ ultimate A brittle materials
Short column have the maximum load carrying capacity.
These columns fail by lateral displacement at a much lower stress than the yield stress. This failure due to lateral displacement is called buckling. Buckling is also referred as instability. Long columns have the least load carrying capacity. If possible, long column may be avoided. Euler equation governs long columns. Buckling is instability.
EULER’S FORMULA FOR LONG COLUMNS
PEULER=P critical=P crippling=P buckling= π2EA/(Le/k min)2
PEULER=P cr = P crippling=P buckling , N (NEWTONS)
E = Young’s modulus, N/mm2
Take E steel= 200 x 1000 N/mm2
A Area of cross section in mm2
Le is effective length. It is the length in which BOW is completed as shown.
These columns fail by a combination of little contraction followed by buckling. Rankine-Gordon formula is used for medium columns.
RANKING GORDON FORMULA IS AN EMPIRICAL FORMULA FOR MEDIUM COLUMN
P Rankine=P cr = P crippling=P buckling= σypA/(1+ α(Le/k min)2)
α= σyp/ π2E
Where P Rankine=P cr = P crippling=P buckling = (N) NEWTONS
σ yp,= YIELD STRESS IN COMPRESSION, N/mm2 , Rankin’s constant
α is RANKINE’s CONSTANT, No units
Le/k min = Slenderness ratio, no units
JOHNSON’s EMPIRICAL FORMULA FOR MEDIUM COLUMNS
It is for medium columns. It is applicable for columns with slenderness ratio equal or less than critical slenderness ratio.
Critical slenderness ratio =(Le/k min)critical = (2π2E/σyp)0.5
Critical slenderness ratio will be different for different materials.
(σcr)JOHN = σyp[1— (σyp (Le/ k min)2)/ 4π2E]
(P cr)JOHN = σyp A[1— (σyp (Le / k min)2)/ 4π2E]
P safe = P allow=P cr /FOS
σsafe = σallow = σcr/FOS
Column and struts are structural
members. Slenderness ratio and
end fixing play crucial role. The
strength decreases with the
increase of slenderness ratio.
Fixed end increases strength.
Eccentric loading decreases
load carrying capacity of the
columns and struts. Equilibrium
of a column is stable, unstable and neutral.
The load at which buckling starts is
buckling /crippling/critical/Euler load
Eccentric Loaded Columns
Fig. Axial & eccentrically loaded columns
It was assumed that the column is under true axial load. But in actual practice, load is not truly axial. It is always under an eccentric load. The loading is not acting along the axis. The eccentricity produces bending stresses in addition to the axial compressive stress. The resultant stress increases. Thus loading carrying capacity of eccentrically loaded columns decreases. Maximum stress under eccentrically loaded columns is
σmax = (P/A) [1± (e(d/2)/k2min) Sec[(Le/2)(P/EI)0.5]
e = eccentricity
P = Axial load
Le = Equivalent length
E = Young’s modulus
I = Moment of inertia of the column
Difference between a column and a strut
Column is a vertical compression member. It is a short or medium or long column on the basis of slenderness ratio. Whereas STRUT is only a long compression member and which is horizontal or inclined to horizontal.
Assumptions used in deriving Euler formula for long columns
σ (buckling) = π2 E/ (le/k min)2
(i) Stresses and strains are within elastic limit.
(ii) Load applied is truly axial.
(iii) A transverse plane remains a transverse plane before and after the load application.
(iv) Material is homogeneous, isotropic and continuous.
(v) Initially column is straight
(vi) Negligible shortening due to axial compression
Effect of lateral load on the bucking of columns
It increases the chances of buckling. Thus decreases the bucking strength of the material.
Limitations of Euler formula
(i) It assumes initial straight column.
There is always small curvature in the column.
(ii) it assumes truly axial load. There is always some eccentricity in load application.
Therefore results may not agree with the experimental data.
https://www.mesubjects.net/wp-admin/post.php?post=13106&action=edit MCQ Columns & Struts