mechanical design an integrated approach Part 5 ppt

Determining the Deflection of a Beam by the Rayleigh-Ritz Method ‘A simply supported beam is under a uniform load of intensity w, as shown in Figure 5.15.. Employing Castigliano’s theor

This results in'a,, = —32PL°/miz* ET, The beam deflection is found by inserting the value of ay,

ẹ : obtained into Eq (5.60):

: ‘Comments: At the free end (x = L), retaining only the first three terms of the solution, we have

- the value of the maximum deflection Ving = PL5/3.001ET The exact solution owing to bending is PIA BEL

*5.10 THE RAYLEIGH-RITZ METHOD The Rayleigh-Ritz method is a convenient approach for determining solutions by the prin- ciple of minimum potential energy This technique was suggested by Lord Rayleigh in

1877 and extended by W Ritz in 1909 Here we discuss application of the method to beam

and plate bending problems The procedure also is used in determining beam and plate buckling loads in the next chapter In a beam bending problem, the Rayleigh-Ritz method may be described as follows

First choose an expression, such as polynomials or trigonometric series, for the de- flection curve containing undetermined coefficients a), (m = 1, 2, ) This function must satisfy the geometric (deflection v and slope @) boundary conditions Static (moment Mand shear V) boundary conditions need not be fulfilled Obviously, a proper choice of the de- flection expression is important to ensure good accuracy for the final solution Next, using

the selected displacement expression, obtain the potential energy function I in terms of

dm Note that the values of a, govern the variation of the potential energy

Inasmuch as the potential energy must be minimum at equilibrium, the Rayleigh-Ritz method is stated in the following form:

On =Ũ 31 0 (5.62)

: Odin The preceding represents a set of algebraic equations solved to give the coefficients of am

Finally, carrying these values into the assumed function for deflection, we find the solution for a given problem The advantages of the Rayleigh-Ritz technique lie in the relative ease

with which mixed edge conditions can be treated This method is among the simplest for solving plate and shel! deflections by hand calculation

_ EXAMPLE 5.13 Determining the Deflection of a Beam by the Rayleigh-Ritz Method

‘A simply supported beam is under a uniform load of intensity w, as shown in Figure 5.15 Develop

an expréssion for the deflection curve, using the Fourier series

Solution::: Assume a solution for deflection v of the form given by Eq (5.56):

3,5, + The maximum deflection occurs at the midspan of the beam and is found by inserting this value of a,, into the first equation:

_ 404 1 i

The minus sign means a downward deflection

Comment: Retaining only the first term, ¥ingx = ~wL'/76.5E7 The exact solution due to bend-

ing is —wL*/76.8ET (case 8 of Table A.9)

Determination of the Deflection of a Simply Supported Rectangular Plate Using th Rayleigh-Ritz Method ~ ng gay

Ä simply supported rectangular plate is under a uniform load of intensity p, (Figure 5.16) Derive an

expression for the deflection surface

In this expression, G,,, até: coefficients, to be determined: The integers m and # represent half-

o sinusoidal curves in thé x and y directions,-respectively Clearly, the boundary conditions

‘The quantity A represents the area of the plate surface co

nối Tý can be shown: that [5] introducing Eq: (5.64) into (5.27) and (5.66), considering the orthogo-

oS nality relations (5.57), and integrating, we > obtain

Substituting Eq: (5.69) into Eq (5.64) leads to the expression for the deflection In so doing and car-

tying the résulting equation for the displacement into Eqs (4.48), we obtain the components for the

moments: :

" REFERENCES

Langhaar, H L Energy Methods in Applied Mechanics, Melbourne, FL: Krieger, 1989

2 Oden, J T., and E A Ripperger Mechanics of Elastic Structures, Ind ed New York: McGraw-

Hill, 1981

3 Sokolnikoff, I S Mathematical Theory of Elasticity, 2nd ed Melbourne, FL: Krieger, 1986

4 Ugural, A C., and S K Fenster Advanced Strength and Applied Elasticity, 4th ed Upper Sad-

dle River, NJ: Prentice Hail, 2003

Ugural, A C Stresses in Plates and Shells, 2nd ed New York: McGraw-Hill, 1999

Faupel, J H., and F E Fisher Engineering Design, 2nd ed New York: Wiley, 1981

7 Boresi, A P., and R J Schmidt Advanced Mechanics of Materials, 6th ed New York: Wiley,

9, Ugural, A C Mechanics of Materials New York: McGraw-Hill, 1991

10 Fitigge, W., ed Handbook of Engineering Mechanics New York: McGraw-Hill, 1961

lL Budynas, R G Advanced Strength and Applied Stress Analysis, 2nd ed New York: McGraw-

Hill, 1999

12 Cook, R D., and W C Young Advanced Mechanics of Materials, 2nd ed Upper Saddle River,

NJ: Prentice Hall, 1999

13 Chou, P C., and N J, Pagano Elasticity New York: Dover, 1992

14 Timoshenko, S P., and S Woinowsky-Krieger Theory of Plates and Shells New York:

17 West, H H Fundamentals of Structural Analysis, 2nd ed New York: Wiley, 2002

18 Sack, R L Structural Analysis New York: McGraw-Hill, 1984

19 Peery, D J., and J J Azar Aircraft Structures, 2nd ed New York: McGraw-Hill, 1982

Figures P5.1 and P5.2, Determine the strain energy of the beam caused by the shear deformation

5.3 An overhanging beam ABC of modulus of elasticity E is loaded as shown in Figure P5S.3

Taking into account only the effect of normal stresses, determine (a) The total strain energy

(b) The maximum strain energy density

Assumption: The beam has a rectangular cross section of width b and depth A

5.4 Asimply supported rectangular beam of depth h, width b, and length L is under a uniform load

w (Figure P5.2) Show that the maximum strain energy density due to bending is given by mày = 45U /8V, The quantities U and V represent the strain energy and volume of the beam, respectively

A cantilevered spring of constant flexural rigidity EZ is loaded as depicted in Figure P5.7

Applying Castigliano’s theorem, determine the vertical deflection at point B

Assumption: The strain energy is attributable to bending alone

P

Figure P5.7 Figure P5.8 shows a compound beam with a hinge at C It is composed of two portions: a beam

BC, simply supported at B, and a cantilever AC, fixed at A Employing Castigliano’s theorem,

determine the deflection vp at the point of application of the load P

Hinge

< aS

A steel I-beam is fixed at B and supported at C by an aluminum alloy tie rod CD of cross- sectional atea A (Figure P5.10) Using Castigliano’s theorem, determine the tension P in the rod caused by the distributed load depicted, in terms of w, L, A, E,, E,, and 1, as needed

and 5.12 A bent frame is supported and loaded as shown in Figures P5.11 and PS.12

Employing Castigliano’s theorem, determine the horizontal deflection 5, for point A

Assumption: The effect of bending moment is considered only

A semicircular arch is supported and loaded as shown in Figure P5.13 Using Castigliano’s

theorem, determine the horizontal displacement of the end B

Assumption: The effect of bending moment is taken into account alone

Figure P5.13

A frame is fixed at one end and loaded at the other end as depicted in Figure P5.14 Apply

Castigliano’s theorem to determine

(a) The horizontal deflection 64 at point A

(b) The slope 6, at point A

Assumption: The effects of axial force as well as shear are omitted

(b) The angle of twist at point B

Assumption: The effect of bending moment is considered only

A simple trass carries a concentrated force P as indicated in Figure P5 16 Applying the work-

energy method, determine the horizontal displacement of the joint D

Given: P =.60 kN, A= 250 mm’, E = 210 GPa,

Figure P5.16

‘The basic truss shown in Figure PS.17 carries a vertical load 2P and a horizontal load P at joint

B Apply Castigliano’s theorem, to obtain horizontal displacement 5¢ of point C

5.18

5.19

$.21 5.22

Figure P5.17

A pin-connected truss supports the loads, as shown in Figure P5.18 Using Castigliano’s

theorem, find the vertical and horizontal displacements of the joint C

Figure P5.18 The truss ABC shown in Figure P5.19 is subjected to a vertical load P Employing

Castigliano’s theorem, determine the horizontal and vertical displacements of joint C

1 AALS

L- 3.2m eze=4m "

Figure P5.19

A planar truss supports a horizontal load P, as shown in Figure P5.20, Apply Castigliano’s

theorem to obtain the vertical displacements of joint D

A three-member truss carries load P, as shown in Figure P5.21 Applying Castigliano’s theorem, determine the force in each member

A curved frame of a structure is fixed at one end and simply supported at another, where a

horizontal load P applies (Figure P5.22) Determine the roller reaction F at the end B, using Castigliano’s theorem

Assumption: The effect of bending moment is considered only

5.25 Acantilevered beam with a rectangular cross section carries concentrated loads P at free end and at the center as shown in Figure P5.25 Determine, using Castigliano’s theorem,

(a) The deflection of the free end, considering the effects of both the bending and shear

(b) The error, if the effect of shear is neglected, for the case in which L = 5h and the beam

is made of ASTM-A36 structural steel

5.23 Two-hinged frame ACB carries a concentrated load P at C, as shown in Figure P5.23

load P, as shown in Figure P5.26 What are the horizontal H and vertical F reactions? Use

(b) The horizontal reaction R at B, if the support B is a fixed pin Assumption: The strain energy is attributable to bending onty

Assumption: The strain energy is attributable to bending alone

Figure P5.26

ị 5.24 Figure P5.24 shows a structure that consists of a cantilever AB, fixed at A, and bars BC and

ay CD, pin-connected at both ends Find the vertical deflection of joint C, considering the effects

5.27 A pin-connected structure of three bars supports a load W at joint D (Figure P5.27) Apply

Castigliano’s theorem to determine the force in each bar

5.29 Figure P5.29 shows a fixed-ended beam subjected to a concentrated load P at its midlength

Using the Rayleigh-Ritz method determine the expression for the deflection curve v

Assumption: The deflection curve is of the form

Figure P5.30 Using the principle of virtual work, determine the expression for the deflection

curve u

Assumption: The deflection curve has the form = a sinax/L, where a is to be found

Figure P5.30 5.31 Applying Castigliano’s first theorem, find the load W required to produce a vertical displace-

ment of } in at joint D in the pin-connected structure depicted in Figure P5.27

Given: h = 10 ft, a = 30°, E = 30 x 10° psi, As | in’, Assumption: Each member has the same cross-sectional area A

CHAPTERS © ENERGY METHODS IN DESIGN

A cantilevered beam is subjected to a concentrated load P at its free end (Figure 5.14) Apply the principle of virtual work to determine

(a) An expression for the deflection curve v

(b) The maximum deflection and the maximum slope

Assumption: Deflection curve of the beam has the form v = ax*(3L — x)/2L°, where ais a

constant

Resolve Problem 5.32, employing the Rayleigh-Ritz method

Asimply supported beam is loaded as shown in Figure 5.13 Using the Rayleigh-Ritz method,

determine the deflection at point A

Assumption: The deflection curve of the beam is of the form = ax(L — x), in which ais a

constant

A simply supported rectangular plate carries a concentrated load P at its center (Figure P5.35)

Employing the Rayleigh-Ritz method, derive the equation of the deflection surface w

Assumption: The deflection surface is of the form given by Eq (5.64)

Figure P5.35

231

out experiencing a sudden change in configuration A buckling response leads to instability and collapse of the member Some designs may thus be governed by the possible instability

of a system that commonly arises in buckling of components Here, we are concerned primarily with the column buckling, which presents but one case of structural stability [1-15] Critical stresses in rectangular plates are discussed briefly in Section 6.10 The problem of buckling in springs is examined in Section 14.6 Buckling of thin-walled cylin- ders under axial loading and pressure’vessels are taken up in the last section of Chapter 16, after discussing the bending of shells

Both equilibrium and energy methods are applied in determining the critical load The choice depends on the particulars of the problem considered Although the equilibrium

approach gives exact solutions, the results obtained by the energy approach (sometimes

approximate) usually is preferred due to the physical insight that may be more readily

gained A vast number of other situations involve structural stability, such as the buckling

of pressure vessels under combined loading; twist-bend buckling of shafts in torsion;

lateral buckling of deep, narrow beams; buckling of thin plates in the form of an angle or channel in compression Analysis of such problems is mathematically complex and beyond the scope of this text

may take place for a load that is 1% of the compressive load alone that would cause failure

based on a strength criterion That is, consideration of material strength (stress level) alone

is insufficient to predict the behavior of such a member Railroad rails, if subjected to an

axial compression because of temperature rise, could fail similarly

PIN-ENDED COLUMNS

Consider a slender pin-ended column centrically loaded by compressive forces P at each end (Figure 6.1a) In Figure 6.1b, load P has been increased sufficiently to cause a small lateral deflection This is a condition between stability and instability or neutral equilib- rium The bending moment at any section is MW = — Py So, Bq (4.14) becomes

pinned ends

For simplification, the following notation is used:

P

The solution of Eq (6.2) is

v= Asinkx + Bcoskx (a)

The constants A and B are obtained from the end conditions

When » = 1, solution of Eq (b) results in the value of the smallest critical load, the Euler buckling load:

EI

This is also called the Euler column formula The quantities /, L, and E are moment of in- ertia of the cross-sectional area, original length of the column, and modulus of elasticity, respectively Note that the strength is not a factor in the buckling load Introducing the fore- going results back in Eq (a), we obtain the buckled shape of the column as

CHAPTER6 © BUCKLING DesiGn OF MEMBERS

The value of the maximum deflection, Umax = A, is undefined Therefore, the critical load sustains only a small lateral deflection [1]

It is clear that £/ represents the flexural rigidity for bending in the plane of buckling

Jf the column is free to deflect in any direction, it tends to bend about the axis having the smallest principal moment of inertia 7 By definition, { = Ar?, where A is the cross- sectional.area and r is the radius of gyration about the axis of bending, We may consider the r of an area to be the distance from the axes at that entire area could be concentrated and yet have the same value for / Substitution of the preceding relationship into Eq (6.4) gives

We seek the minimum value of P,; hence, the smallest radius of gyration should be used

in this equation The quotient L/r, called the slenderness ratio, is an important parameter

in the classification of columns

COLUMNS WITH OTHER END CONDITIONS

For columns with various combinations of fixed, free, and pinned supports, the Euler for- mula can be written in the form

ET

E2

in which L, is called the effective length, As shown in Figure 6.2, it develops that the ef-

fective length is the distance between the inflection points on the elastic curves In a like

manner, Eq (6.5) can be expressed as

Figure 6.2 Effective lengths of columns for various end conditions: (a) fixed-free;

() pinned-pinned; (c) fixed-pinned, (d) fixed-fxed; (e) fixed-nonrotating

235

or pins, to achieve the required ideal end condition

Minimum AISI recommended actual effective lengths for steel columns [5] are as follows:

Le =21L (fixed-free) Le=L (pinned-pinned)

Lạ = 0.80L (fixed-pinned) (3

Lạ =0.65L (fixed-fixed)

Lạ =1.2L (fixed-nonrotating) Note that only a steel column with pinned-pinned ends has the same actual length and the theoretical value noted in Figure 6.2b Also observe that steel columns with one or two fixed ends always have actual lengths longer than the theoretical values The foregoing

apply to end construction, where ideal conditions are approximated The distinction

between the theoretical analyses and empirical approaches necessary in design is discussed

in Sections 6.6 and 6.7

6.3 CRITICAL STRESS IN 4 COLUMN

As previously pointed out, a column failure is always sudden, total, and unexpected There

is no advance warning The behavior of an ideal column is often represented on a plot of average critical stress P.,/A versus the slenderness ratio L,/r (Figure 6.3) Such a repre- sentation offers a clear rationale for the classification of compression bars The range of L./r is a function of the material under consideration

[4 columns bats columns

Hence, the Euler’s load of Eq (6.7) is appropriate in this case, and the critical stress is

(6.8)

The corresponding portion CD of the curve (Figure 6.3) is labeled as Euler’s curve The smallest value of the slenderness ratio for which Euler’s formula applies is found by equat- ing og; to the proportional limit or yield strength of the specific material:

Le

( `) =n |= ride Sy (6.9) For instance, in the case of a structural steel with E = 210 GPa and S, = 250 MPa, this equation gives (L./r), = 91

We see from Figure 6.3 that very slender columns buckle at low levels of stress; they are much less stable than short columns Equation (6.9) shows that the critical stress is in- creased by using a material of higher modulus of elasticity E or by increasing the radius of gyration r A tubular column, for example, has a much larger value of r than a solid column

of the same cross-sectional area However, there is a limit beyond which the buckling

strength cannot be increased The wali thickness eventually becomes so thin as to cause the

member to crumble due to a change in the shape of a cross section

SHORT COLUMNS OR STRUTS

Compression members having low slenderness ratios (for instance, steel rods with L./r < 30) show essentially no instability and are called short columns For these bars, failure occurs by yielding or crushing, without buckling, at stresses above the proportional limit of the material Therefore, the maximum stress

+

represents the strength limit of such a column, shown by horizontal line AB in Figure 6.3

This is equal to the yield strength or ultimate strength in compression

INTERMEDIATE COLUMNS

Most structural columns lie in a region between the short and long classifications, repre- sented by part BC in Figure 6.3 Such intermediate columns fail by inelastic buckling at stress levels above the proportional limit Substitution of the tangent modulus £;, slope of the stress-strain curve beyond the proportional or yield point, for the elastic modulus E

is the only modification necessary to make Eq (6.7) applicable in the inelastic range

Hence, the critical stress may be expressed by the generalized Euler buckling formula, the

237

Over the years, many other formulas have been proposed and employed for interme- diate columns A number of these formulas are based on the use of linear or parabolic

relationships between the slenderness ratio and the critical stress The parabolic J B

Johnson formula has the form:

where A represents the cross-sectional area of the column The relation (6.12) seems to be

the preferred one among designers in the machine, aircraft, and structural steel construction fields Despite much scatter in the test results, the Johnson formula has been found to agree

reasonably well with experimental data However, the dimensionless form of tangent mod- ulus curves have very distinct advantage when structures of new materials are analyzed [3]

Note that Eqs (6.8), (6.9), and (6.11) or (6.12) determine the ultimate stresses, not the working stresses It is therefore necessary to divide the right side of each formula by an ap- propriate factor of safety, often 2 to 3, depending on the material, to determine the allow- able values Some typical relationships for allowable stress are introduced in Section 6.6

EXAMPLE 6.1 The Most Efficient Design of a Rectangular Column

Asteel column of length L and an a x b rectangular cross section is fixed at the base and supported

at the top, as shown in Figure 6.4 The column must resist a load P with a factor of safety n with

respect to buckling

(a): What is the ratio of a/b for the most efficient design against buckling?

(b): Design the most efficient cross section for the column, using L = 400 mm, E = 200 GPa, P= ISKN, andn= 2

Assumption: Support restrains end A from moving in the yz plane but allows it to move in the xz

plane

CHAPTER6 @ BUCKLING DESIGN OF MEMBERS

Solution: The radius of gyration r of the cross section is

Similarly, we obtain r, = b//12 The effective lengths of the column shown in Figure 6.4 with

respect to buckling in the xy and xz planes, from Figures 6.2c and 6.2a, are L, = 0.7L and L, = 2L, respectively Thus

Le 0.7L

Ty a/V/12

%» b/VI2

(a) For the most effective design, the critical stresses corresponding to the two possible modes

of buckling are to be identical Referring to Eq (6.8), it is concluded therefore that

Development of Specific Johnson's Formula

Derive specific Johnson formula for the intermediate sizes of columns having

(a) Round cross sections

(b) Rectangular cross sections

EXAMPLE 6.2

(b).- In: the case -of: a-rectangular-section of height A and width b with h <6: A= bh,

1= bh3/12; hence,r? h°/12: Introducing these into Eq (6.12a),

Given: ' L = 15 in., a=tin, b= lin,

Analysis of the Load-Carrying Capacity of a Pin-Ended Braced Column

A steel column braced at midpoint C as shown in Figure 6.5 Determine the allowable load P.) on the

basis of a factor of safety 1

Sy = 36 ksi, E = 30 x 10° psi

Assumptions: Bracing acts as simple support in the xy plane Use n = 3

Solution::° Reférring té Example 6.1, the cross sectional area properties are

CHAPTER6 @ BUCKLING DESIGN OF MEMBERS

- Buckling in the’ xz- plane (unrestrained by the brace) The slenderness ratio is L/ry = 15/0.289 =.51.9:<.91' ảnd the Johnson equation is valid per Eq (6.9) Relation (6.12b) results in

Pai h fs 2 6818 2.273 kips nt 3

241

Case Study 6-1 | BUCKLING STRESS ANALYSIS IN A Basic TRuss

A plane truss, consisting of members AB, BC, and AC, is subjected to vertical and horizontal loads P, and P) at

point B and supported at A and C, as shown in Figure 6.6a

Although the members are actually joined together by riv-

eted or welded connections, it is usual to consider them pinned together (Figure 6.6b) Further information

on trusses may be found in Section 5.6 Determine the

stresses in each member

Given: The geometry of the assembly is known Each

member is 6061-T6 aluminum alloy pipe of outer diame- ter D and inner diameter d

Data

ĐQ=50KN, Ðœ=25kN,

E=70GPa, Š, = 260 MPa, (from Table B.1}

Figure 6.6 (a) Basic truss

{b) Schematic representation of basic

truss

Assumptions:

1 The weight of members is small compared to the

applied loading and is neglected

Case Study (CONCLUDED)

2 The loading is static

3 Each member is treated as a two-force member

4 Friction in the pin joints is ignored

Solution: See Figure 6.6b

The axial forces in members, applying the method of

joints, are found as

Fag = 10 KN (C), Feo = 55 kN (C), Fac = 33 KN (T)

For a tube cross section, we have the following

The critical stresses from Eq (6.8), for members AB and

BC, respectively, are

x? (70 x 10°) q9 = 18.36 MPa

(Øe)An =

z?(70 x 10)

1255 = 32.63 MPa

Cac =

Comments: Interestingly, 18.36 and 32.63 MPa, com-

pared with the yield strength of 260 MPa, demonstrate the significance of buckling analysis in predicting a safe working load A somewhat detailed FEA analysis of the member forces, displacements, and design of a basic truss are discussed in Case Study 17-1

| 6.4 INITIALLY CURVED COLUMNS

‘In an actual structure, it is not always possible for a column to be perfectly straight As might

be expected, the load-carrying capacity and deflection under load of a column are signifi- cantly affected by even a small initial curvature To determine the extent of this influence, consider a pin-ended column with the unloaded form described by a half sine wave:

This is shown by the dashed lines in Figure 6.7, where a, represents the maximum initial displacement

vi = €; Sinkx + co coskx + B sin TT

The constants c¡ and c¿ are evaluated, from the end condidons ø4(0) = (1) = Ú, as

€¡ = ca = 0 The column deflection is then

v= a, sin = Sẻ + Bsi Ee sin —— = in——

° L L 1—b sin + (6-18)

This equation indicates that the axial force P causes the initial deflection of the column to

increase by the factor {/(1 ~ 6) Since b < 1, this factor is always greater than unity

Clearly, if b = 1, deflection becomes infinitely large Note that an initially curved column deflects with any applied load P in contrast to a perfectly straight column that does not bend until P,, is reached

The quantity S represents the section modulus 7 /c, in which c is the distance measured in the y direction from the centroid of the cross section to the outermost fibers

pinned ends

PARTE ® FUNDAMENTALS

By imposing the yield strength in compression (and tension) S, a8 Omax, we write

Eq, (6.19) in the form

solving a quadratic or by trial and error for P The allowable load P,y can then be obtained

by diviđing Py by an appropriate factor of safety n

6.5 ECCENTRIC LOADS AND THE SECANT FORMULA

In the preceding sections, we deal with the buckling of columns for which the load acts

at the centroid of a cross section We here treat columns under an eccentric load This situation is obviously of great practical importance because, frequently, problems occur in which load eccentricities are unavoidable

Let us consider a pinned-end column under compressive forces applied with a small eccentricity e from the column axis (Figure 6.8a) We assume the member is initially straight and that the material is linearly elastic By increasing the load, the column deflects

as depicted by the dashed lines in the figure The bending moment at distance x from the midspan equals M = —P(v + e) Then, the differential equation for the elastic curve ap-

pears in the form ,

2 est +P@+ø)=0 dx?

(b) Graph of the secant formula

CHAPTER6 ® BUCKLING DESIGN OF MEMBERS

‘The boundary conditions are v(L./2) = v(~L/2) = 0 This equation is solved by follow- ing a procedure in a manner similar to that of Sections 6.2 and 6.3 In so doing, expressed

in the terms of the critical load P., = x? EI/L’, the midspan (x = 0) deflection is found as

(6.21)

We observe from the foregoing expression that, as P approaches P.,, the maximum deflec- tion goes to infinity It is therefore concluded that P should not be allowed to reach the crit- ical value found in Section 6.2 for a column under a centric load

The maximum compressive stress Oya, takes place at x = 0 on the concave side of the column Hence,

Omen = P + Moyax

A I

The quantity r is the radius of gyration and ¢ is the distance from the centroid of the cross section to the outermost fibers, both in the direction of eccentricity Carrying Mmax =

—P(Umax + @) and Eq (6.21) into the foregoing expression,

Ø max = af 1+ sec mye „2 2V P„ (6.22a)

ST Tiệc LY EY AL pe OrV AE) 6.22b (6.22b)

This expression is referred to as the secant formula The term ec/r? is called the eccen- tricity ratio

We now impose the yield strength S, as omax and hence P = P, Then, Eq (6.22b) can

(ay The maximum deflection

(b) The factor of safety n against yielding

Given: The geometry of the column and applied loading are known

E = 29 x 10° psi,

= 80 Kips] 6 in + = 20 kips = 100 kips

Solution: See Figure 6.9 and Table A.7

The loading may be replaced by a statically equivalent load P = 100 kips acting with an

eccentricity e== 1.2 in (Figure 6.9) Using the properties of an S8 x 23 section given in Table A.7,

CHAPTER6 © BUCKLING Desicn or MEMBERS

(by Ina like manner, through the use of Eq (6.22a), we have

Comments: Since thé maximum stress and the slenderness ratio are within the elastic limit of 36:kại and the slenderness ratio of about 200, the secant formula is applicable

6.6 DESIGN OF COLUMNS UNDER

A CENTRIC LOAD

In Sections 6.2 and 6.3, we obtain the critical load in a column by applying Euler’s formula

This is followed by the investigation of the deformations and stresses in initially curved columns and eccentrically loaded columns by using the combined axial load and bending for- mula and the secant formula, respectively In each case, we assume that all stresses remain

below the proportional point or yield limit and that the column is a homogeneous prism

The foregoing idealizations are important in understanding column behavior How- ever, the design of actual columns must be based on empirical formulas that consider the data obtained by laboratory tests Care must be used in applying such “special purpose for- mulas.” Specialized references should be consulted prior to design of a column for a par- ticular application Typical design formulas for centricaily loaded columns made of three different materials follow These represent specifications recommended by the American Institute of Stee] Construction (AISC), the Aluminum Association, and the National Forest Products Association (NFPA) Various computer programs are readily available for the

analysis and design of columns with any cross section (including variable) and any bound-

Column formulas for aluminum 6061-T6 alloy [6]

Solution: See Table A.6

Le ‘A suitable size for a prescribed shape may conveniently be obtained using tables in the AISC

Gai = 19 ksi = 130 MPa (= s 9.5) (6.26a) manual Howevet, we use a trial-and-error procedure here Substituting the given data, Eq (6.25a)

= (Le/ry? MPa (= > 66) (6.26¢) : Using Table A.6, we select a W150 x 24 section having an area A = 3060 mm? (greater than

2720 mm?) and with a minimum r of 24.6 mm The value of L,/r = 4/0.0246 = 163 is greater than

C = 126, which was obtained in the foregoing Therefore, applying Eq (6.24b),

z?Q@00 x 10?) Column formulas for timber of a rectangular cross section [7]

, Ou = ETE (F < 50) (6.27) a T9263)

in which d is the smallest side dimension of the member The allowable stress is not to ex- 408 KN, and a column with a larger A, a larger r, or both must be selected

ceed the value of stress for compression parallel to grain of the timber used

Note that, for the structural steel columns, in Eqs (6.24) and (6.25), C defines the Second Try: Consider a W150 x 37 section (see Table A.6) with A = 4740 mm? and minimum

r= 38,6 mm For this case, L./r = 4/0.0386 = 104 is less than 126 From Eq (6.24a), we have

fo limiting value of the slenderness ratio between intermediate and long bars This is taken to

an correspond to one-half the yield strength S, of the steel By Eq (6.8) we therefore have 5 3/104 1/1043

==®+tglxczl—TzlIx>xzz] =t

L [ee ng (5) § (5) 3

Clearly, by applying a variable factor of safety, Eq (6.25b) renders a consistent formula for ,

intermediate and short columns Also observe in Eq (6.26) that for short and intermediate aluminum columns, oq is constant and linearly related to L./r For long columns, a Euler-

type formula is applied in both steel and aluminum columns [8] Equation (6.27) for timber Comment: A W150 x 37 steel section is acceptable

columns is also a Euler formula, adjusted by a suitable factor of safety

The permissible load for this section, 86.3 x 4740 = 409 KN, is slightly larger than the design load

EXAMPLE 6.5 Design of a, Wide-Flange Steel Column AN ECCENTRIC LOAD

Select the lightest wide-flange steel section to support an axial load of P on an effective length of Lạ Recall from Section 6.5 that the secant formula is a rational equation and applies for all

es column ths ther hand, this for is quite di i si

Given: $,—250MPa, £=200GPa, (from Table B.1), | mn lengths On the ot 2 1: rmula is quite difficult to employ in design, even

using computers to facilitate the computation Among various approaches employed in P= 408 KN; T¿ = 4m designing eccentrically loaded columns, the so-called interaction method seems the most

simple In this technique, the maximum stress in a member owing to an eccentric com-