document36

Problem Set [Professor Video-1][Professor Video-2][Professor Note]

[Faculty Video][Faculty Note]

Two rigid bars AC and BC are connected as shown to a spring of constant k.

Knowing that the spring can act in either tension or compression, determine the

critical load Pcr for the system.

1.

Determine (a) the critical load for the steel strut, (b) the dimension d for which the

aluminum strut will have the same critical load. (c) Express the weight of the

aluminum strut as a percent of the weight of the steel strut.

2.

Strength of Materials/ Unit 10/ Module 1 Column- I

Subject/Unit Name/Module Name http://10.20.3.1:8080/engineering/I_YEAR/EM/M36...

1 of 33 Friday 19 August 2011 10:05 AM

A compression member of 1.5-m effective length consists of a solid 30-mm-diameter

brass rod. In order to reduce the weight of the member by 25%, the solid rod is

replaced by a hollow rod of the cross section shown. Determine (a) the percent

reduction in the critical load, (b) the value of the critical load for the hollow rod. Use

E = 105 GPa.

3.

Additional Problem Set

The rigid AB is attached to a hinge at A and to two springs, each of constant k = 2

kips/in. that act in either tension or compression. Knowing that h = 2ft, determine the

1.



critical load.

A column of 6-m effective length is to be made from three plates as shown. Using E

= 200 GPa, determine the factor of safety with respect to buckling fro a centric load

of 16kN.

2.

Examples

Two rigid bars, each of length L, forming a straight vertical member as shown in

Fig (a), have torsional springs of stiffness k at ideal pinned joints B and C.

Determine the critical vertical force Pcr and the shape of the buckled member.

1.



Solution:

In order to determine the critical buckling force Pcr, the system must be given a

displacement compatible with the boundary conditions. Such a displacement

with positive sense is shown as A'B'C in Fig. (a). Bar BC rotates through an

angle δθ1 and bar AB independently rotates through an angle δθ

2.

Therefore, this system has two degrees of freedom. Free-body diagrams for

members AB and BC in deflected positions are drawn in Fgs. (b) and (c).

Then, assuming that the member rotations are infinitesimal, equations of

equilibrium are written for each member. In writing these equations, it should

be noted that M1 = k(δθ

2-δθ

1), where the terms in parentheses constitute the

infinitesimal rotation angle between the two bars. On this basis,

and

Rearranging,



These two homogeneous linear equations possess a trivial solution, δθ1=δθ2= 0, as well as a nonzero solution if the determinate of the coefficients is

zero, i.e.,

On expanding this determinant, one obtains the characteristic equation

The roots of such an equation are called eigenvalues, the smallest of which is

the critical buckling load. In this case, there are two roots:

and Pcr = P

1.

Substituting the roots into either one of the simultaneous equations

determines the ratios between the rotations of the bars. Thus, for P1,

δθ2/δθ1= 1.62, and for P2, δθ2/δθ1= -0.62. The corresponding deflected

modes are shown in Figs. (d) and (e). The one in Fig (d) corresponds to Pcr.

These mode shapes are called eigenvectors and are often written in matrix

form as

where δθ1 is an arbitrary constant.

A 2-m-long pin-ended column of square cross section is to be made of wood.

Assuming E = 13 GPa, σall= 12 MPa, and using a factor of safety of 2.5 in

computing Euler's critical load for buckling, determine the size of the cross

section if the column is to safely support (a) a 100-kN load, (b) a 200-kN load.

2.

Solution:



(a) For the 100-kN Load. Using the given factor of safety, we make

Pcr = 2.5(100 kN) = 250 kN L =2m E = 13 GPa

in Euler's formula (Eq. 10) and solve for I. We have

Recalling that, for a square of side a, we have I = a4/12, we write

a4/12 = 7.794 X 10

-6m

4 a = 98.3 mm ≈ 100 mm

We check the value of the normal stress in the column:

Since σ is smaller than the allowable stress, a 100X100-mm cross section is

acceptable.

(b) For the 200-kN Load. Solving again Eq. (10) for I, but making now Pcr =

2.5(200) = 500 kN, we have

I = 15.588 X 10-6 m

4

a4/12 = 15.588 X 10

-6 a = 116.95 mm

The value of the normal stress is

Since this value is larger than the allowable stress, the dimension obtained is

not acceptable, and we select the cross section on the basis of its resistance

to compression. We write



a2 = 16.67 X 10

-3 m

2 a = 129.1 mm

A 130 X 130-mm cross section is acceptable.

An aluminum column of length L and rectangular cross section has a fixed end

B and supports a centric load at A. Two smooth and rounded fixed plates

restrain end A from moving in one of the vertical planes of symmetry of the

column, but allow it to move in the other plane. (a) Determine the ratio a/b of the

two sides of the cross section corresponding to the most efficient design

against buckling. (b) Design the most efficient cross section for the column,

knowing that L = 20 in., E = 10.1 X 106 psi, P = 5 kips, and that a factor of safety

of 2.5 is required.

3.

Solution:

Buckling in xy Plane. Referring to Fig. 14 we note that the effective length of

the column with respect to buckling in this plane is Le = 0.7L. The radius of

gyration rz of the cross section is obtained by writing

Ix = (1/12) ba

3 A = ab

and, since



The effective slenderness ratio of the column with respect to buckling in the xy

plane is

(1)

Buckling in xz Plane. The effective length of the column with respect to

buckling in this plane is Le = 2L and the corresponding radius of gyration is

Thus,

(2)

a. Most Efficient Design. The most efficient design is that for which

the critical stresses corresponding to the two possible modes of buckling are

equal. We note that will be the case if the two values obtained above for the

effective slenderness ratio are equal. We write

and, solving for the ration a/b,

b. Design for Given Data. Since F.S = 2.5 is required,

Pcr

= (F.S)P = (2.5)(5 kips) = 12.5 kips

Using a = 0.35b, we have A = ab = 0.35 b2 and



Making L= 20 in. in Eq(2), we have Le/ry = 138.6/b. Substituting for E, L

e/r, and

σcr into Eq.(10) we write

b = 1.620 in. a = 0.35b = 0.567 in.

Faculty Notes

1 Introduction

The selection of structural and machine elements is based on three characteristics:

strength, stiffness, and stability. The procedures of stress and deformation analysis in

state of stable equilibrium were discussed in some detail in the preceding chapters. But

not all structural systems are necessarily stable. For example, consider a square-ended

metal rod of say 10 min in diameter. If such a rod were made 20 mm long to act as an

axially compressed member, no question of instability would enter, and a considerable

force could be applied. On the other hand, if another rod of the same material were made

1000 mm long to act in compression, then, at a much smaller load than the short price

could carry, the long rod would buckle laterally and could collapse. A slender measuring

stick, if subjected to an axial compression, could fail in the same manner. The

consideration of material strength alone is not sufficient to predict the behaviour of such

members. Stability considerations are primary in some structural systems.

The Phenomenon of structural instability occurs in numerous situations where

compressive stresses are present. Thin sheets, although fully capable of sustaining

tensile loadings, are very poor in transmitting compression. Narrow beams, unbraced

laterally, can turn sidewise and collapse under an applied load. Vacuum tanks, as well as

submarine hulls, unless properly designed, can severely distort under external pressure

and can assume shapes that differ drastically from their original geometry. A thin walled

tube can winkle like tissue paper when subjected either to axial compression or a torque;

see Fig. 1. During some stages of



firing, the thin casings of rockets are critically loaded in compression. These are crucially

important problems for engineering design. Moreover, often the buckling or wrinkling

phenomena observed in loaded members occur rather suddenly. For this reason, many

structural instability failures are spectacular and very dangerous.

A vast number of the structural instability problems suggested by the preceding

listing of problems are beyond the scope of this text. Essentially, only the column problem

will be considered here.

For convenience, this chapter is divided into two parts. Part A is devoted to the theory

of column buckling, and part B deals with design applications First, however, examples of

possible instabilities that may occur in straight prismatic members with different cross

sections will be discussed. This will be followed by establishing the stability criteria for



static equilibrium. The purpose of the next two introductory sections is to clarify for the

reader the aspects of column instability considered in the remainder of the chapter.

2 Examples of Instability

Analysis of the general instability problem of even straight prismatic columns discussed

in this chapter is rather complex, and it is important to be aware, at least in a qualitative

way, of the complexities involved to understand the limitations of the subsequently

derived equations. Buckling of straight columns is strongly influenced by the type of cross

section and some considerations of this problem follow.

In numerous engineering applications, compression members have tubular cross

sections. If the wall thickness is thin, the plate-like elements of such members can buckle

locally. An example of this behavior is illustrated in Fig. 2(a) for a square thin-walled tube.

At a sufficiently large axial load, the side walls tend to subdivide into a sequence of

alternating inward and outward buckles. As a consequence, the plates carry a smaller

axial stress in the regions of large amount of buckling displacement away from corners;

see Fig 2(b). For such cases, it is customary to approximate the complex stress

distribution by a constant allowable stress acting over an effective width w next to the

corners or stiffeners. In this text, except for the design of aluminum-alloy columns, it will

be assumed that the thicknesses of a column plate element are sufficiently large to

exclude the need for considering this local buckling phenomenon.



Some aspects requiring attention in a general column instability problem are

illustrated in Fig. 3. Here the emphasis is placed on the kind of buckling that is possible in

prismatic members. A plank of limited flexural but adequate torsional stiffness subjected

to an axial compressive force is shown to buckle in a bending mode; see Fig 3(a). If the

same plank is subjected to end moments, Fig. 3(b), in addition to a flexural buckling

mode, the cross sections also have a tendency to twist. This is a torsion-bending mode of

bucking, and the same kind of buckling may occur for the eccentric force P, as shown in

Fig. 3(c). Lastly, a pure torsional buckling mode is illustrated in Fig . 3(d). This occurs

when the torsional stuffiness of a member is small. As we know that thin-walled open

sections are generally poor in torsional stiffness. In contrast, thin-walled tubular members

are excellent for resisting torques and are torsionally stiff. Therefore, a tubular member,

such as shown in Fig. 2, generally, will not exhibit torsional buckling. A number of the

open thin-walled sections Fig. 4 are next examined for their susceptibility to torsional

buckling.



Two sections having biaxial symmetry, where centroids C and shear centers S

coincide, are shown in Fig 4(a). Compression members having such cross sections

buckle either in pure flexure, Fig 3 (a) or twist around S, Fig. 3 (d). For thin-walled

members, when the torsional stiffness is smaller than the flexural stiffness, a column may

twist before exhibiting flexural buckling. Generally, this is more likely to occur in columns

with cruciform cross sections than in I-shaped sections. However, the torsional mode of

buckling generally does not control the design, since the usual rolled or extruded metal

cross sections are relatively thick.



The cross sections shown in Figs. 4(b) and (c) have centroids C and shear centers S

in different locations. Flexural buckling would occur for the sections in Fig. 4(b) if the

smallest flexural stiffness around the major principal axis is less than the torsional

stiffness. Otherwise, simultaneous flexural and torsional buckling would develop, with the

member twisting around S. For the sections in Fig. 4(c), buckling always occurs in the

latter mode. In the subsequent derivations, it will be assumed that the wall thickness of

members are sufficiently large to exclude the possibility of torsional or torsional-flexural

buckling. Compression members having cross sections of the type shown in Fig. 4 (c) are

not considered.

The following interesting cases of possible buckling of straight members are also

excluded from consideration in this text. One of these is shown in Fig. 5, where two bars

with pinned joints at the ends from a very small angle with the horizontal. In this case, it is

possible that applied force P can reach a magnitude such that the deformed compressed

bars become horizontal. Then, on a slightly further increase in P, the bars snap-through

to a new equilibrium position. This kind of instability is of great importance in shallow

thin-walled shells and curved plates. Another possible buckling problem is shown in Fig.

6, where a slender circular bar is subjected to torque T. When applied torque T reaches a

critical value, the bar snaps into a helical spatial curve. This problem is of importance in

the design of long slender transmission shafts.

3 Criteria for Stability of Equilibrium



In order to clarify the stability criteria for static equilibrium, consider a rigid vertical bar

with a torsional spring of stiffness k at the base, as shown in Fig. 7(a). The behaviour of

such a bar subjected to vertical force P and horizontal force F is shown in Fig. 7(b) for a

large and a small F. The question then arises: How will this system behave if F = 0?

To answer this question analytically, the system must be deliberately displaced a

small (infinitesimal) amount consistent with the boundary conditions. Then, if restoring

forces are greater than the forces tending to upset the system, the system is stable, and

vice versa.

The rigid bar shown in Fig. 7(a) can only rotate. Therefore, it has only one degree of

freedom. For an assumed small rotation angle θ, the restoring moment is k θ, and, with F= 0, the upstting moment is PL sinθ ≈ PLθ. Therefore, if

kθ > PLθ the system is stable (1)

and if kθ < PLθ the system is unstable (2)

Right at the transistion point, kθ = PLθ, and the equilibrium is neither stable norunstable, but is nutral. The force asssociated with this condition is the critical, or buckling,load, which will be designated P

cr. Fot the bar system considered,



Pcr = k/L (3)

In the presence of horizontal force F, the P - θ curves are as shown by the dashedlines in Fig. 7(b) becoming asymptotic to the horizontal line at P

cr. Similar curves would

result by placing the vertical force P eccentrically with respect to the axis of the bar. Ineither case, even for unstable systems, θ cannot become infinitely large, as there isalways a point of equilibrium at θ somewhat less than π. The apparent discrepancy in thegraph is caused by assuming in Eqs. 1 and 2 that θ is small and that sin θ ≈ θ, and cos θ≈ 1. The condition found for neutral equilibrium when F = 0 can be further eleboratedupon by making reference to Fig. 8.

It is convenient to relate the process for determining the kind of stability to a ballresting on differently shaped frictionless surfaces; see Fig. 8. In this figure, in all threecases, the balls in position 1 are in equilibrium. In order to determine the kind ofequilibrium, it is necessary to displce the balls an infinitesimal distance δθ to either side.



In the first case, Fig. 8(a), the ball would roll back to its initial position, and the equilibriumis stable. In the second case, Fig. 8(b), the ball once displced will not return to its initialposition, and the equilibrium is unstable. In the last case, Fig. 8(c), the ball can remain inits displaced position, where it is again in equilibrium. Such an equilibrium is neutral.Therefore, by analogy, a structural system is in a state of neutral equilibrium when it hasat least two neighboring equilibrium positions an infinitedimal distance apart. Thiscriterion for neutral equilibrium is applicable only for infinitesimal displacements, as atlarge displacements, different conditions may prevail (Fig. 5).

Based on the previous reasoning, the horizontal line for F = 0 shown in Fig. 7(b) ispurely schematic for defining P

cr. Theoretically it has meaning only within an infinitesimal

distance from the vertical axis.

To demonstate this again, consider the rigid vertical bar shown in Fig. 7(a) and set F= 0. Then, in order to determine neutral equilibrium, displace the bar in either directionthrough an angle δθ (not through the angle θ shown in the figure) and formulate theequation of equilibrium:

PL δθ - k δθ = 0 or (PL - k) δθ = 0 (4)

This equation has two distinct solutions: first, when δθ = 0 and P is arbitrary, FIg. 7(b),and, second, when the expression in parentheses vanishes. This second solution yieldsPcr = k/L. For this value of the axial force, δθ is arbitrary. Therefore, there are two

equilibrium positions at Pcr. One of these is for a straight bar, and the other for a bar

inclined at an angle δθ. Since at Pcr, there are these two branches of the solution, such a

point is called the bifurcation (branch) point.

In the previous illustration, the rigid bar has only one degree of freedom, since for anarbitrary infinitesimal displacement, the system is completely described by angle δθ. Aproblem with two degrees of freedom is analyzed in the following example.

Before proceeding with the derivation for critical column loads based on the conceptof neutral equilibrium, it is significant to examine the meaning of such analyses. Criticalloads do not describe the postbuckling process. However, by using the exact (nonlinear)differential equations for curvature, it can be shown that for elastic columns, one can findequilibrium positions above P

cr. The results of such an analysis are illustrated in Fig 9.

Note, especially, that increasing Pcr by mere 1.5 percent causes a maximum sideways

deflection of 22 percent of the column length. For practical reasons, such enormousdeflections can seldom be tolerated. Moreover, the material usually cannot resist theinduced bending stresses. Therefore, failure of real columns would be inelastic.Generallythere is little additional post -buckling strength for real coulmns, and the use of P

cr for

column capacity is acceptable. This contrasts with the behavior of plates and shell wheresignificant post-buckling strength may develop.

Another illustration of the meaning of Pcr

in relation to the behavior of elastic and



elastic-plastic column based on nonlinear analyses is shown in Fig 10. In these plots,columns that are initially bowed into sinusoidal shapes with a maximum center deflection

of are Δo are considered. The paths of equilibrium for these cases vary, depenling on the

extent of the intial curvature. However, regardless of the magnitude of Δo, critical load P

cr

serves as an asymptote for columns with a small amount of curvature, which areaommonly encountered in engineering problems; see Fig 10(b). It is to be noted that aperfectly elastic initially straight long column with pinned ends, upon buckling intoapproximately a complete "circle," attains the intolerable deflection of 0.4 of the columnlength. Behavior of elastic-plastic column can reach P

cr and thereafter drop precipitously

in its carrying capacity. Column imperfections such as crookedness drastically theessential parameter for determining column capacity. With appropriate safeguards,design procedures can be devised employing this key parameter.

BUCKLING THEORY FOR COLUMNS

4. Euler Load for Columns with Pinned Ends

At the critical load, a column that is circular or tubular in its cross-sectional area maybuckle sideways in any direction. In the more general case, a compression member doesnot possess equal flexural rigidity in all directions. The moment of inertia is a maximumaround tone centroidal axis and of the cross-sectional area a minimum around theother;see Fig. 11. The significant flexural rigidity EI of a column depends on the minimumI, and at the critical load a column buckles either to one side or the other in the plane ofthe major axis. The use of a minimum I in the derivation that follows is understood.



Consider the ideal perfectly straight column with pinned supports at both ends; seeFig. 12(a). The least force at which a buckled mode is possible is the critical or Eulerbuckling load.

In order to determine the critical load for this column, the compressed column isdisplaced as shown in FIg. 12(b). In this position, the bending moment according to thebeam sign convention is -Pv. By substituting this value of moment into the differentialequation for the elastic curve for the initially straight column becomes

(5)

by letting λ2 = P/EIand transposing, gives



(6)

This is an equation of the same from as the one for simple harmonic motion, and its

solution is

v = A sin λx +B cos λx (7)

where A and B are arbitrary constants that must be determined from the boundary

conditions. These conditions are

v(0) = 0 and v(L) =0

Hence, v(0) = 0 = A sin 0 + B cos 0 or B = 0

and v(L) = 0 = A sin λL (8)

This equation can be satisfied by taking A = 0. However, with A and B each equal to

zero, as can be seen from Eq. 7, this is solution for a straight column, and is usually

referred to as a trivial solution. An alternative solution is obtained by requiring the sine

term in Eq. 8 to vanish. This occurs when λn equals nπ, where n is an integer. Therefore,since λ was defined as , the , the nth critical force P

n that makes the deflected

shape of the column possible follows from solving L = nπ. Hence,

(9)

These Pn's are the eigenvalues for this problem. However, since in stability problems

only the least value of Pn is importance, n must be taken as unity, and the critical or Euler

load Pcr for an initially perfectly straight elastic column with pinned ends becomes

(10)

where E is the elastic modulus of the material, I is the least moment of inertia of theconstant cross-sectional area of a column,and L is its length. This case of a columnpinned at both ends is often referred to as the fundamental case.



According to Eq. 7, at the critical load, since B = 0, the equation of the buckled elasticcurve is

v= A sin λx (11)

This is the characterstic, or eigrnfunction, of this problem, and since λ=nπ /L, n canassume any integer value. There is an infinite number of such functions. In this linearizedsolution, amplitude A of the buckling mode remains indeterminate. FOr the fundamentalcase n = 1, the elastic curve is a half-wave sine curve. This shape and the modescorresponding to n = 2 and 3 are shown in FIg. 13. The higher modes have no physicalsignificance in buckling problems, since the least critical buckling load occurs at n = 1.

5. Euler Loads for COlumns with Different End Restraints

The same procedure as that discussed before can be used to determine the critical axialloads for columns with different boundary conditions. The solutions of these problems arevery sensitive to the ends restraints. Consider, for example, a column with one end fixedand the other pinned, as shown in FIg. 14, where the buckled column is drawn in adeflected position. Here the effect of unknown end moment M

o and the reactions must be

considered in setting up the differential equation for the elastic curve at the critical load:



(12)

Letting λ2 = P/EI as before, and tranposing, gives

(13)

The homogeneous solution of this differential equation, i.e., when the right side zero, is

the same as that given by Eq. 7. The particular solution, due to the nonzero right side, is

given by dividing the term on that side by λ2. The complete solution then becomes

v = A sin λx + B cos λx + (Mo/P)(1-x/L) (14)

where A and B are arbitrary constants, and Mo is the unknown moment at the fixed end.

The three kinematic boundary conditions are



v(0) = 0 v(L) = 0 and v' (0) =0

Hence, v(0) = 0 = B+ Mo/P

v(L) = 0 = A sin λL + B cos λL

and v'(0) = 0 = Aλ - Mo/PL

Solving these equations simultaneously, one obtains the following transcendental

equation

λL = tan λL (15)

which must be satisfied for a nontrival equilibrium shape of the column at the critical load.

the smallest root of Eq. 15 is

λL = 4.493

from which the corresponding least eigenvalue or critical load for a column fixed at one

end pinned at the other is

(16)

It can be shown that in the case of a column fixed at both ends, Fig. 15(d), the critical load

is

(17)

The last two equations show that by restraining the ends the critical loads are

substantially larger than those in the fundamental case, Eq. 10. On the other hand, the

critical load for a free-standing column, Fig 15(b), with a load at the top is

(18)



In this extreme case, the critical load is only one-fourth of that for the fundamental case.

All the previous formulas can be made to resemble the fundamental case, provided

that the effective column lengths are used instead of the actual column length. This length

turns out to be the distance between the inflection points on the elastic curves. The

effective column length Le for the fundamental case is L, but for the cases discussed it is

0.7L, 0.5L, and 2L, respectively. For a general case, Le = KL, where K is the effective

length factor, which depends on the restraints. Hence, a more general form of the Euler

formula, incorporating the concept of the effective column length Le

(19)

In constrast to the classical cases shown Fig. 15 actual compression members are

seldom truly pinned or completely fixed against rotation at the ends. Because of the

uncertainty regarding the fixity of the ends, columns are often assumed to be pin-ended.

With the exception of the case shown in Fig. 115(b), where it cannot be used, ths

procedure is conservative.



Procedure Summary

Column buckling loads in this and the preceding section are found using the same

curvature-moment relation that was derived for the deflection of beams. However, the

bending moments are written for axially loaded columns in slightly deflected positions.

Mathematically this results in entirely different kind of second order differential equation

than that for beam flexure. The solution of this equation shows that, for the same load, two

neighboring equilibrium configurations are possible for a column. One of these

configurations corresponds to a straight coumn, the other to a slightly bent column. The

axial force associated simultaneously with the bent and straight shape of the column is

the critical buckling load. This occurs at the bifurcation (branching) point of the solution.

In the developed formulation, the columns are assumed to be linearly elastic, and to

have the same cross section throughout the column length. Only the flexural

deformations of a column are considered.

For the second order differential equations considered in this treatment the same

kinematic boundary conditions are applicable as for beams in flexure.

Elastic buckling load formulas are truly remarkable. Although they do not depen on

the strength of a material, they determine the carrying capacity of columns. The only

material property invoved is the elastic modulus E, which physically represents the

stiffness characteristic of a material.

The previous equations do not apply if the axial column stress exceeds the

proportional limit of the material.

Professor Note

Elastic stability: Euler's Buckling load Axial loading shay to a

column is vertical compared to a beam which is horizontal. A

column is long and slender.

A column can be subjected to constrains at ends A&B.

The constrains on support A & B are hinged supports. They

restrict lateral movements in Y direction at the support.

They alloe bending of beam an axial force p acts at the end A

with a reaction force at end B

Intially the column is in straight line AB



When load p is applied also it is in a point C. When this force is applied the column

deflects from straight line to adeflected position

When load Q IN Y direction acts at point C. When this force is applied the column deflects

from straight line to adeflected position.

When load Q is removed the deffromed column reverts back to its original position of

straight line AB

If we keep increasing the axial load P and test the column elastic load PQ we find the the

column is no elastic any more. The column remains in the deformed shape when Q is

removed and does not reverback.

The load PQ which causes column to be instable under axial load is called Euler's critical

load or the buckling load

We come across compressive loads on members as in buildings, long towers, reentering

space vehicle etc. The different names the are known by are

Columns and stanchions are vertical members of building frames.

A post is a general term used for a compression member

A Strut is a compressive member of truss.

A boom is the principal compression member of a crane Minar Qutub,

charminar: In middle east building

Axially loaded compression members

The stress in the column is

If at a load the column crushes it is called the crushing load. If fc be the

ultimate crushing strength then compressive crushing load pc = f

cA

A short column crushes in compression

Permissible stress =

Example 1



Wrought iron buckling stress of and strong timber has .

Its factor of safety required is 2. Find the cross sectional area of the column if the load is

20 KN. To avoid trussing.

Permissible crushing stress

Load = 20KN;

compressing stress

Cross sectional area

Timber needs more area for

Timber Assume a column buckled under load per area is called buckling load or critical

load or crippling load A is the area of cross section or column of length, l, which has

deflected by e, at the middle of the column.

Axial stress on column =

Bending stress on column =

Total Stress

When Z = sectional modules of section about axis of bend



When Pmax

reaches the crushing stress fc, the column member will fail. The reason for

failure is not compression stress Pc, alone but the combined compression P

c and

bending Pb.

Euler's Theory of long columns:

Look from centre like concave +ve

Ends A and B can have kind of B.C's

Both ends pinned1.

One end fixed others end free2.

When both ends are fixed3.

When one end is fixed the other is pinned4.

Case 1:

When both ends of the column are pinned or fixed bending moment equation at point x

on the beam



Solution to above equation is

Deflection

Find c1 & c

2 from B.C's

At x = 0, B deflection is zero y = 0

cos (0) = 1, sin (0) = 0

y = c1 + 0 = 0 or c

1 = 0

At x = l, y = 0

Trival solution

( Case called harmonics

possible solutions)

The least practical value =

P is the critical Euler load beyond which the given column will buckle.

Assumptions mode in Euler's theory for crippling load

column is intially axially loaded and has axial compressive load1.

column has uniform section2.

column material is perfectly elastic, homogeneous, isotropic and obeys hooke's3.

The length of the column is very large compared to lateral dimensions4.



Self weight I ignorable5.

column will fair by buckling alone6.

Example 2:

A mild steel tube 4ms long, 30 mm internal diameter and 4 mm thick is used as a strut

with both ends hinged. Find collapsing load take

For both ends hinged

For hollow pipe

Example 3:

Find the shortest length for pin ended steel column having a

cross section of 60mm X 10mm for which Euler's formula

Applies E = 2 X 105 N/mm

2 critical proportional limit is 250

N/mm2

We need to take the minimum of moment of inertia

As Ixx is the least of the two it will give lower values of P

a to be used in design



Euler's = Critical stress =

Critical proportionality limit is when

Critical proportional limit is the stress when critical Euler stress is reached

'l' is 15.39 time largest lateral dimension of 100 mm = 15.39 for the load PG = 250 X 60 X

100 = 1500 KN column will crush.

Effective Length of a column

For a given end condition the effective length of an column of the same material and

section with the hinged ends having the value of the crippling load equal to that of the

given column l = actual column length; L = effective length



Equivalent length L for different boundary conditions

Fixed-free boundary conditions makes A flexible column can take critical load 1/4 times

that the stiff Fixed- Fixed column



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