Columns and struts

Definitions:

• Columns and Stanchions: Vertical compression members in buildings.
• Struts: compression members in roof trusses
• Beam: Jib of a crane.
• Beam column: Co beam that is acted on by an axial compressive force in addition to transversely applied loads.

Classification:

• Short Column: A column that fails essentially by direct crushing at ultimate load.

Crushing load Pc = fc.A,

fc= ultimate crushing stress.

• Long columns:

1. Members considerably long in comparison of lateral dimensions.
2. The member essentially fails by buckling or crippling to bending.

Radius of gyration: 

Slenderness Ratio : Effective length/least radius of gyration.

Significance: As slenderness ratio increases, permissible stress or critical stress reduces, consequently, load carrying capacity also reduces.

• Radius of gyration will be least along major axis of cross section.

Eg: for a rectangular column along yy -axis

• For a given area, Tubular section will have maximum radius of gyration.
• H-Section is more efficient than I-Section.
  

Equilibrium of a column: A column is said to have buckled or failed when it reaches “Neutral Equilibrium”.

Euler’s Theory:

• Critical load: The smallest force at which a buckled shape is possible. Prior to this load the column remains straight. The columns buckle in the plane of the major axis of the cross section as shown below:
  
• Assumptions:

1. Column is initially perfectly straight and is axially loaded.
2. Section of column is uniform
3. The material is perfectly elastic, homogeneous, isotropic and obeys hook’s law.
4. Length of column is very large compared to lateral dimension.
5. Direct stress is small compared to bending stress corresponding to buckling condition.
6. Self weight of column is ignorable.
7. The column will fail by buckling alone.

• Euler’s formula for general case: For a general case critical load,
  

Where

l= Effective length

I = Moment of Inertia of section about the axis of least resistance.

E = Young’s Modulus.

• Effective length and critical loads for various boundary conditions compared to a column whose both ends are hinged.

l = Eff. Length

L = actual length

• Limitations of Euler’s formula:

1. Euler’s formula can also be written as
   
   As f and E are constant for a particular material, Euler’s formula is valid for a particular range of slenderness ratio, for e.g. for mild steel whose fc = 3300 Kg/cm2 and E = 2.1 x 106 Kg/cm2 Euler formula is not valid for slenderness ratio less than 80.
2. Euler’s formula are valid only up to proportional limit i.e., in linear inelastic zone

Note:

• The relation between slenderness ratio and corresponding critical stress is hyperbolic.
• According to Euler formulas the critical load does not depend upon strength property of material the only material property involved is the elastic modules ‘E’ which physically represents the stiffness characteristics of the material.

Rankine’s formula:

• It is an empirical formula.
• Takes into account both direct crushing (Pa) load and Euler critical load( PE)
  

Basic Formula :

Rankine’s Co-efficient: is independent of geometry and end conditions, can be modified to incorporate imperfections

Material	fy	Rankine’s Constant
Mild steel	3200	1/7500
Wrough Iron	2500	1/9000
Cast Iron	5500	1/1600

• Rankine’s formula is valid for any type of column.
• No limitations for slenderness ratio.

Straight line formula : (Johnson’s straight line formula)(Empirical)

• It is assumed that allowable stress depends or L/r (slenderness ratio) and varies in a straight line fashion. Applicable to small slenderness ranges

P=A[f-n( λ)]

P = safe load on the column.
A = cross sectional area of column
f= allowable stress in column material
n = constant depends on the material

λ= slenderness ratio

Parabolic formula: (Johnson’s straight line formula) (Empirical)

• Allowable stress is assumed to vary as (L/r)2

P= A[f - Bλ2]

Where:

P = safe load on the column

A= cross sectional area of column

F = allowable stresses in the column material

λ = slenderness ratio

Eccentrically loaded columns:

• Euler’s formula
  

Where

Where σmax = critical stress in the column

P = axial load on the column

e = eccentricity of the column load

I = effective length of column

EI= flexural rigidity

• Rankine’s method:
  

P=Rankine’s load

f= allowable crushing strength of material

e = eccentricity of loading

Yc = distance of compression fibre from centriod

λ = slenderness ratio

r = least radius of gyration with respect to minor axis

• Secant formula:

For standard pinned column

σmax = maximum stress is located at the extreme compression fiber of the middle point (x = L/2)

of the column.

Validity: It applies to column of any length provided the maximum stress does not exceed

the elastic limit.

NOTE:

• In the above equation (‘r’ may not be minimum since it is obtained from the value of ‘I’ associated with the axis around which bending occurs.
• The relation between σmax and ‘P’ is not linear σmax increases faster than ‘p’ . Therefore the solutions for maximum stresses in columns caused by different axial forces cannot be super posed instead, the forces must be superposed first, and then the stresses can be calculated.

Perry’s formula: (approximate formula)

Where

σ = permissible stress in the material(given)

From the above equation ao can be calculated with that safe load is equal to P = σ0A

σ = maximum permissible stress

σo = stress due to direct load

σE = stress due to Euler’s critical load

Yc = distance to extreme compression fibre.

Core of a cross section: The area with in which a direct load to act, so as not to cause tension in any part of cross section.

• Rectangle or square: Middle third rule Core is a rhombus whose diagonals are d/3 and b/3 Are of core = bd/18 i.e., 1/18th of total area.
  
• >> Solid circular section:

1. middle fourth rule
2. core dia is d/4 (i.e., eccentricity limit is d/8 to avoid tension) and core area is 1/16th  of area of circular section.

• >> Hollow circular section

External dia = D

Internal dia = d

Slope and Deflection

Relation between curvature, slope and deflection:

Eg: A simply supported beam of length L is subjected to couples as shown in figure determine maximum slope and deflection at the centre of the beam?

Method of determining slope and deflection:

• Double integration method:(Not suitable for objective type questions.)
• Area moment method:(for cantilevers, slopes and deflections can be determined very quickly.)
• Conjugate beam method:(very much suitable for beams of varying sections, subjected to couples, for cantilevers of S. S . Beams.)
• Macaulay ‘s method:(Also successive integration method.)

NOTE: In double integration or Macaulay’s method two constants of integration C1 and C2 will be obtained. These are determined using end conditions.

Mohr’s Theorem’s: Moment Area Method: 

• Theorem 1: The angle between tangents drawn at any two points on the deflected curve, is equal to the area of M / El diagram between the two points.

i.e.,  θ = area of M / EI diagram.

A = area of B.M.D.

• Theorem 2: The intercept on a vertical line made by two tangents drawn at the two points on the deflected curve, is equal to the moment of M / EI diagram between the two points about the vertical line.

Ax̅ / EI x̅ = distance of C.G. of B.M.D.

E.G: (Suitable for cantilevers) 

Step 1 : To determine slope and deflection at any point.

Step 2 : Draw (BMD) / (EI)

i.e., M / EI

Step 3 : Slope = area of (M / EI) diagram between fixed end point under consideration.

Step 4 : Deflection Ax̅ / EI

A = B.M.D area between fixed end and point under consideration.

x̅ = distance of C.G. of M / EI from point under consideration.

Ex.1

Ex.2

Maxwell’s Law of Reciprocal Deflections:

Consider cantilever beam AB. Let ‘C’ be intermediate point. Then the deflection at due to a point load ‘P’ at B say YCB, is equal deflection at ‘B’ due to a point load ‘P’ a

i.e., YBC

SLOPE & DEFLECTION FOR DIFFERENT LOADING OF BEAMS:

Torsion

>>Torsion: If moment is applied in a plane perpendicular to the longitudinal axis of the beam (or) shaft, it will be subjected to Torsion

Ex: 

• Shaft Transmitting Torque or power.
• L beams
• Portico beams
• Curved beams
• Closed coiled springs.

>>Torsion Formula:

Where

T = Torque applied

θ = Twist of cross section

τs = Maximum shear stress due to torsion

R = Radius of shaft

L = Length of shaft

J = Polar moment of inertia

Assumptions:

• Plane normal sections of shaft remain plane after twisting.
• Torsion is uniform along the shaft
• Material of the shaft is homogeneous, and isotropic.
• Radii remain straight after torsion.
• Stress is proportional to strain i.e., all the stresses are with in elastic limit.

Note: 

• The stress setup at any point in a cross section is one of pure shear or simple shear.
• The longitudinal axis is neutral axis.
• The shear stress will vary linearly from zero at the centre to maximum at the outer surface  (any point on periphery)
  

>>Torsional Section Modulus:

Zp = J/R = Polar Moment of Inertia/ Radius of shaft

As the value of Torsional modulus increases. the Torsional strength increases. For Ex: A hollow circular shaft compared to that of a solid shaft of same area, will have more Torsional strength.

For a solid circular shaft,

For a hollow circular shaft,

D1 = Outer diameter

D2 = Inner diameter.

Torsional Rigidity:

CJ Unit : kg. cm2 or Nmm2

The torsional which produces unit twist per unit length.

Angle of Twist: θ = TL/CJ

>>Power Transmitted by a Shaft: In SI system: Power (P) is measured in watts (W)

>>Design of Shaft: To be safe against maximum permissible shear stress.

Diameter of shaft,

>>Composite Shafts : When two dissimilar shafts are connected together to form one shaft

the shaft is known as composite shaft.

Shafts in Series : If the driving torque applied at one end, and the resisting torque the other end, the shafts are said to have be connected in series.

For such shaft,

• both the parts carry some Torque i.e., T1 = T2
• Total angle of twist at fixed and is sum of separate angles of twist of two shafts.
  

Shaft in Parallel: If the Torque ‘T’ is applied at the junction of two shafts and resisting Torque at their remote ends, the shafts are said to be connected in parallel.

For such a case,

>>Combined bending and Torsion:

• Let a shaft be subjected to a bending moment of ‘M’ and twisting moment ‘T’ at a section.
  
• Equivalent Torque : It is the twisting moment, which acting along produce the maximum shear stress due to combined bending and Torsion.
  
• Equivalent Bending Moment : The bending moment to produce the maximum bending stress equal to greater principle stress
  

>>Comparison of Hollow and Solid Shafts:

• When the areas of solid and hollow sections are equal,
  
• when radius of solid shaft is equal to external radius of hollow shaft,
  
• The ratio of the weight of a hollow shaft and solid shaft of equally strong is
  

>>Strain energy due to torsion: 

Shear Stress Distribution in Beams

The shear stress in any section at a distance ‘y’ from neutral axis is given by

Where

F = Shear force at the given section

A y  = moment of the area of the section above the level under consideration.

b = width of the beam at the level under consideration

I = M.I of the beam section about the N.A

Shear stress Distribution for Beam sections of various shapes:
Rectangular section: τavg

Average shear stress  = (F / bd)

τmax =  (3/2) τave 

For solid circular section:

For triangular section:

A beam of square section is placed horizontally with one diagonal placed horizontally:

For I - Section: