Column and Strut

7.1 BASIC CONCEPTS

Column. A bar or a member of a structure inclined at 90° to the horizontal and carrying an axial compressive load is called a column.

Slenderness ratio. The ratio of the equivalent length of the column to the least radius of gyration is called the slenderness ratio.

Buckling load. The minimum axial load at which the column tends to have lateral

displacement & buckle is called the buckling, crippling or critical load.

Equivalent length. It is the length of the column which gives the same bthkhflg otd , as given by a both ends hinged column.

Short Column. A column for which the slenderness ratio is less thani iled a short column.

Medium Column. A column for which the slenderness ratio lies between 32 and 120 is called a medium column.

Long Column. A column for which the slenderness ratiQ is more than 120 called a long column.

Safe load. It is the load under which the column will not buckle.

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7.2 EULER’S THEORY OF COLUMNS

The following assumptions are made in this theory:

1. The column is initially straight and the applied load is truely axial.

2. The material of the column is homogeneous, linear and isotropic.

3. The length of the column is very large as compared to the cross-sectional dimensions of the column.

4. The cross-section of the column is uniform throughout.

5. The shortening of the column due to axial compression is negligible.

6. The self weight of the column is neglected.

7. The ends of the column are frictionless.

There can be three conditions at the end of the column.

2. Pinned end. For such an end, y =0

3. Free end, The column is neither fixed in position nor in direction.

Depending upon the end conditions, there are four types of columns.

• Both ends hinged

• Both ends fixed

• One end fixed and other end hinged

• One end fixed and other end free.

7.2.1 Column hinged at Both Ends

Consider a column having both ends hinged and carrying an axial compressive load P as shown in Fig. 7.1. Taking origin at A, the bending moment at a distance x is:

7.2.2 Column fixed at both ends

Consider the column fixed at both ends as shown in Fig. 7.2 Let MA id M8 be the fixing moments at the ends. At a distance x from A,

7.2.3 Column fixed at one end and hinged at the other

Consider a column hinged at end A and fixed at B as shown in Fig. 7.3. There will be a tbckig m om entM B and a horizontal force RA will have to be applied to A to keep the column in equilibrium. At a distance x from A,

7.2.4 Column fixed at one end and free at the other

Consider a column as shown in Fig. 7.4. Let the horizontal deflection of end A be and fixing moment at end B be MB.

7.2.5 Limitations of Euler’s Theory

1. The Euler’s theory is applicable to columns which are initially exactly straight and the load is truly axial. However, there is always some crookedness in the column and the load may not be exactly axial.

3. This theory does not take into account the axial compressive stress.

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7.3 RANKINE’S FORMULA

This formula takes into account the effect of direct compressive stress. Rankine proposed that crippling load,

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7.4  I.S. FORMULAE

1. Straight line formula

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Example 7.1 A hollow steel column having both ends fixed is 50 mm external diameter and 25 mm internal diameter. If safe compressive stress is 320 MPa and F = 200 GPa, calculate the length.

Solution. For both ends fixed,

                              

                          

                            

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Example 7.2 Using Rankine’s formula, calculate the critical load for a column having both ends hinged when slenderness ratio = 60, E = 200 GPa, a = 320 MPa and A = 100 mm2.

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Example 7.3 Compare the crippling load given by Euler’s and Rankine’s formulae for a tubular strut 2 m long and 30 mm diameter loaded through pin joints at both ends.

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Example 7.4. An I-section beam is shown in Fig. 7.5, simply supported at the ends.

Compute its length, given that when it is subjected to a load of 20 kN per metre length,

it deflects by 10 mm. Also find the safe load when used as a column with both ends

fixed. Assume a factor of safety of 4. Take E = 210 G Pa.

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Torsion of Shaft

When a uniform circular shaft is subjected to a torque it can be shown that every section
of the shaft is subjected to a state of pure shear.

Assumptions
1. The material is homogeneous, i.e. of uniform elastic properties throughout.
2. The material is elastic, following Hook's law with shear stress proportional to
shear strain.
3. The stress does not exceed the elastic limit of proportionality.
4. Circular sections remain circular.
5. Cross sections remain plane.
6. Cross sections rotate as if rigid.

Angle of Twist (θ)

After applying the torque point A moves to point B and made an angleθ .

Modulus of Rigidity (G)
It is a measure of how much deformation the material would undergo subjected to a

shear. G is the property which relates to the stiffness of the material, units N/m2.

Stresses

Composite Shafts

• Series Connection

When two similar or dissimilar shafts of the same or of different materials are connected

together to form one composite shaft, the driving torque being applied at one end and the

resisting torque at the other, the shafts are said to be connected in series.

Consider each component shaft separately, applying the torsion theory to each in turn.

The composite shaft will therefore be as weak as its weakest component.

Parallel Connection

If the driving torque is applied at the junction of the two shafts connected together and
the resisting torque at the other ends of the two shafts, then the shafts are said to be
connected in parallel.


The torque applied in this case is divided between the two shafts but the angle of twist is
the same for each shaft.

Power Transmitted by Shafts
If a shaft carries a torque T N.m and rotates at w rad/s it will do work at the rate:
                                            P = Tw watts

Strain Energy

Strain Energy

Strain Energy of the member is defined as the internal work done in defoming the body by the action of externally applied forces. This energy in elastic bodies is known as elastic strain energy :

Strain Energy in uniaxial Loading

Fig .1

Let as consider an infinitesimal element of dimensions as shown in Fig .1. Let the element be subjected to normal stress s_x.

The forces acting on the face of this element is s_x. dy. dz

where

dydz = Area of the element due to the application of forces, the element deforms to an amount = Î_x dx

  Î_x = strain in the material in x – direction

       

Assuming the element material to be as linearly elastic the stress is directly proportional to strain as shown in Fig . 2.

Fig .2

\ From Fig .2 the force that acts on the element increases linearly from zero until it attains its full value.

Hence average force on the element is equal to ½ s_x . dy. dz.

\ Therefore the workdone by the above force

Force = average force x deformed length

           = ½ s_x. dydz . Î_x . dx

For a perfectly elastic body the above work done is the internal strain energy “du”.

where dv = dxdydz

   = Volume of the element

By rearranging the above equation we can write

The equation (4) represents the strain energy in elastic body per unit volume of the material its strain energy – density ‘u_o' .

From Hook's Law for elastic bodies, it may be recalled that

In the case of a rod of uniform cross – section subjected at its ends an equal and opposite forces of magnitude P as shown in the Fig .3.

Fig .3

ILLUSTRATIVE PROBLEMS

1. Three round bars having the same length ‘L' but different shapes are shown in fig below. The first bar has a diameter ‘d' over its entire length, the second had this diameter over one – fourth of its length, and the third has this diameter over one eighth of its length. All three bars are subjected to the same load P. Compare the amounts of strain energy stored in the bars, assuming the linear elastic behavior.

Solution :

From the above results it may be observed that the strain energy decreases as the volume of the bar increases.

2. Suppose a rod AB must acquire an elastic strain energy of 13.6 N.m using E = 200 GPa. Determine the required yield strength of steel. If the factor of safety w.r.t. permanent deformation is equal to 5.

Solution :

Factor of safety = 5

Therefore, the strain energy of the rod should be u = 5 [13.6] = 68 N.m

Strain Energy density

The volume of the rod is

Yield Strength :

As we know that the modulus of resilience is equal to the strain energy density when maximum stress is equal to s_x .

It is important to note that, since energy loads are not linearly related to the stress they produce, factor of safety associated with energy loads should be applied to the energy loads and not to the stresses.

STATICALLY INDETERMINATE PROBLEMS

 Statically indeterminate problems.
Members for which reaction forces and internal forces can be found out from static
equilibrium equations alone are called statically determinate members or structures.
Problems requiring deformation equations in addition to static equilibrium equations to
solve for unknown forces are called statically indeterminate problems.


















The reaction force at the support for the bar ABC in figure  can be determined
considering equilibrium equation in the vertical direction.




Now, consider the right side bar MNO in figure 1.22 which is rigidly fixed at both the ends.
From static equilibrium, we get only one equation with two unknown reaction forces R1 and
R2.

Hence, this equilibrium equation should be supplemented with a deflection equation which
was discussed in the preceding section to solve for unknowns.
If the bar MNO is separated from its supports and applied the forces , then
these forces cause the bar to undergo a deflection
R1,R2 and P
δMO that must be equal to zero.

δMN and δNO are the deflections of parts MN and NO respectively in the bar MNO.
Individually these deflections are not zero, but their sum must make it to be zero.
Equation 1.19 is called compatibility equation, which insists that the change in length of the
bar must be compatible with the boundary conditions.
Deflection of parts MN and NO due to load P can be obtained by assuming that the
material is within the elastic limit,

 

Substituting these deflections in equation

Combining both the  equations one can get,

 

From these reaction forces, the stresses acting on any section in the bar can be easily
determined.