Beer F.P., Johnston E.R., DeWolf J.T., Mazurek D.F. Mechanics of Materials

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637

and setting I 5 Ar

, where A is the cross-sectional area and r its

radius of gyration, we have

EAr

(10.13)

The quantity Lyr is called the slenderness ratio of the column. It is

clear, in view of the remark of the preceding paragraph, that the

minimum value of the radius of gyration r should be used in comput-

ing the slenderness ratio and the critical stress in a column.

Equation (10.13) shows that the critical stress is proportional

to the modulus of elasticity of the material, and inversely propor-

tional to the square of the slenderness ratio of the column. The plot

of s

versus Lyr is shown in Fig. 10.8 for structural steel, assuming

E 5 200 GPa and s

5 250 MPa. We should keep in mind that no

factor of safety has been used in plotting s

. We also note that, if

the value obtained for s

from Eq. (10.13) or from the curve of

Fig. 10.8 is larger than the yield strength s

, this value is of no inter-

est to us, since the column will yield in compression and cease to be

elastic before it has a chance to buckle.

100

0 100

200

250

300

(MPa)

 250 MPa

E  200 GPa

(L/r)









Fig. 10.8 Plot of critical stress.

Our analysis of the behavior of a column has been based so far

on the assumption of a perfectly aligned centric load. In practice,

this is seldom the case, and in Sec. 10.5 the effect of the eccentricity

of the loading is taken into account. This approach will lead to a

smoother transition from the buckling failure of long, slender col-

umns to the compression failure of short, stubby columns. It will also

provide us with a more realistic view of the relation between the

slenderness ratio of a column and the load that causes it to fail.

10.3 Euler’s Formula for Pin-Ended Columns

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EXAMPLE 10.01

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

wood. Assuming E 5 13 GPa, s

all

5 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.

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

make

5 2.5

100 kN

5 250 kNL 5 2 mE 5 13 GPa

in Euler’s formula (10.11) and solve for I. We have

I 5

1250 3 10

N212 m2

113 3 10

Pa2

5 7.794 3 10

Recalling that, for a square of side a, we have I 5 a

y12, we write

5 7.794 3 10

a 5 98.3 mm < 100 mm

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

s 5

100 kN

10.100 m2

5 10 MPa

Since s is smaller than the allowable stress, a 100 3 100-mm cross section

is acceptable.

(b) For the 200-kN Load. Solving again Eq. (10.11) for I, but

making now P

5 2.5(200) 5 500 kN, we have

15.588

5 15.588 3 10





a 5 116.95 mm

The value of the normal stress is

s 5

200 kN

10.11695 m2

5 14.62 MPa

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

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

resistance to compression. We write

A 5

200 kN

12 MPa

5 16.67 3 10

5 16

67 3 10



5 129

1 mm

A 130 3 130-mm cross section is acceptable.

10.4 EXTENSION OF EULER’S FORMULA TO

COLUMNS WITH OTHER END CONDITIONS

Euler’s formula (10.11) was derived in the preceding section for

a column that was pin-connected at both ends. Now the critical

load P

will be determined for columns with different end

conditions.

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formula (10.11) by using a column length equal to twice the actual

length L of the given column. We say that the effective length L

the column of Fig. 10.9 is equal to 2L and substitute L

5 2L in

Euler’s formula:

(10.119)

The critical stress is found in a similar way from the formula

(10.139)

The quantity L

yr is referred to as the effective slenderness ratio of

the column and, in the case considered here, is equal to 2Lyr.

Consider next a column with two fixed ends A and B supporting

a load P (Fig. 10.10). The symmetry of the supports and of the load-

ing about a horizontal axis through the midpoint C requires that the

shear at C and the horizontal components of the reactions at A and

B be zero (Fig. 10.11). It follows that the restraints imposed upon

the upper half AC of the column by the support at A and by the

10.4 Extension of Euler’s Formula to

Columns with Other End Conditions

 2L

(b)(a)

Fig. 10.9 Column with free end.

Fig. 10.10 Column with

fixed ends.

In the case of a column with one free end A supporting a load

P and one fixed end B (Fig. 10.9a), we observe that the column will

behave as the upper half of a pin-connected column (Fig. 10.9b).

The critical load for the column of Fig. 10.9a is thus the same as for

the pin-ended column of Fig. 10.9b and can be obtained from Euler’s

L/2

Fig. 10.11 Buckled shape

of column with fixed ends.

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Columns

lower half CB are identical (Fig. 10.12). Portion AC must thus be

symmetric about its midpoint D, and this point must be a point of

inflection, where the bending moment is zero. A similar reasoning

shows that the bending moment at the midpoint E of the lower half

of the column must also be zero (Fig. 10.13a). Since the bending

moment at the ends of a pin-ended column is zero, it follows that

the portion DE of the column of Fig. 10.13a must behave as a pin-

ended column (Fig. 10.13b). We thus conclude that the effective

length of a column with two fixed ends is L

5 Ly2.

In the case of a column with one fixed end B and one pin-

connected end A supporting a load P (Fig. 10.14), we must write and

solve the differential equation of the elastic curve to determine the

effective length of the column. From the free-body diagram of the

entire column (Fig. 10.15), we first note that a transverse force V is

exerted at end A, in addition to the axial load P, and that V is statically

indeterminate. Considering now the free-body diagram of a portion AQ

of the column (Fig. 10.16), we find that the bending moment at Q is

M 5 2Py 2 Vx

⫽

(a)(b)

Fig. 10.13

Fig. 10.14 Column

with one end pin-

connected and one

end fixed.

[ x ⫽ 0, y ⫽ 0]

[ x ⫽ L, y ⫽ 0]

[ x ⫽ L, dy/dx ⫽ 0]

Fig. 10.15

Fig. 10.16

L/4

Fig. 10.12

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Substituting this value into Eq. (9.4) of Sec. 9.3, we write

y 2

Transposing the term containing y and setting

(10.6)

as we did in Sec. 10.3, we write

1 p

y 52

(10.14)

This equation is a linear, nonhomogeneous differential equation of

the second order with constant coefficients. Observing that the

left-hand members of Eqs. (10.7) and (10.14) are identical, we

conclude that the general solution of Eq. (10.14) can be obtained

by adding a particular solution of Eq. (10.14) to the solution (10.8)

obtained for Eq. (10.7). Such a particular solution is easily seen

to be

y 52

or, recalling (10.6),

y 52

(10.15)

Adding the solutions (10.8) and (10.15), we write the general solution

of Eq. (10.14) as

5 A sin px 1 B cos px 2

(10.16)

The constants A and B, and the magnitude V of the unknown

transverse force V are obtained from the boundary conditions indi-

cated in Fig. (10.15). Making first x 5 0, y 5 0 in Eq. (10.16), we

find that B 5 0. Making next x 5 L, y 5 0, we obtain

A sin pL 5

(10.17)

Finally, computing

5 Ap cos px 2

and making x 5 L, dyydx 5 0, we have

Ap cos pL 5

(10.18)

10.4 Extension of Euler’s Formula to

Columns with Other End Conditions

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Columns

Dividing (10.17) by (10.18) member by member, we conclude that

a solution of the form (10.16) can exist only if

tan pL 5 pL (10.19)

Solving this equation by trial and error, we find that the smallest

value of pL which satisfies (10.19) is

pL 5 4.4934 (10.20)

Carrying the value of p defined by Eq. (10.20) into Eq. (10.6)

and solving for P, we obtain the critical load for the column of

Fig. 10.14

20.19EI

(10.21)

The effective length of the column is obtained by equating the

right-hand members of Eqs. (10.119) and (10.21):

20.19EI

Solving for L

, we find that the effective length of a column with one

fixed end and one pin-connected end is

5 0.699L < 0.7L.

The effective lengths corresponding to the various end condi-

tions considered in this section are shown in Fig. 10.17.

 0.7L

 0.5L

 2L

 L

B B

one pinned end

(d) Both ends

fixed

(b) Both ends

pinned

(a) One fixed end,

one free end

P P

Fig. 10.17 Effective length of column for various end conditions.

Fig. 10.14

(repeated)

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SAMPLE PROBLEM 10.1

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 5 20 in., E 5 10.1 3 10

psi, P 5 5 kips,

and that a factor of safety of 2.5 is required.

SOLUTION

Buckling in xy Plane. Referring to Fig. 10.17, we note that the effec-

tive length of the column with respect to buckling in this plane is L

5 0.7L.

The radius of gyration r

of the cross section is obtained by writing

A 5 ab

and, since I

5 Ar

r

5 a

112

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

the xy plane is

0.7L

(1)

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

to buckling in this plane is L

5 2L, and the corresponding radius of gyra-

tion is r

5 b

12. 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.

Referring to Eq. (10.139), we note that this will be the case if the two values

obtained above for the effective slenderness ratio are equal. We write

and, solving for the ratio

5 0.35>

b. Design for Given Data. Since F. S. 5 2.5 is required,

F.S.

P 5

2.5

5 kips

5 12.5 kips

Using a 5 0.35b, we have A 5 ab 5 0.35b

and

12,500

0.35b

Making L 5 20 in. in Eq. (2), we have L

5 138.6/b. Substituting for E,

yr, and s

into Eq. (10.139), we write

12,500 l

0.35b

110.1 3 10

psi2

138.6

1.6



0.35b

0.567



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PROBLEMS

644

10.1 Knowing that the spring at A is of constant k and that the bar AB

is rigid, determine the critical load P

Fig. P10.1

Fig. P10.2

10.2 Knowing that the torsional spring at B is of constant K and that

the bar AB is rigid, determine the critical load P

c r

10.3 Two rigid bars AC and BC are connected by a pin at C as shown.

Knowing that the torsional spring at B is of constant K, determine

the critical load P

c r

for the system.

Fig. P10.4

Fig. P10.3

10.4 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 P

for the system.

10.5 The rigid bar AD is attached to two springs of constant k and is in

equilibrium in the position shown. Knowing that the equal and

opposite loads P and P9 remain vertical, determine the magnitude

of the critical load for the system. Each spring can act in either

tension or compression.

P'P'

Fig. P10.5

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Problems

10.6 The rigid rod AB is attached to a hinge at A and to two springs,

each of constant k. If h 5 450 mm, d 5 300 mm, and m 5 200 kg,

determine the range of values of k for which the equilibrium of

rod AB is stable in the position shown. Each spring can act in

either tension or compression.

10.7 The rigid rod AB is attached to a hinge at A and to two springs,

each of constant k 5 2 kips/in., that can act in either tension or

compression. Knowing that h 5 2 ft, determine the critical load.

Fig. P10.6

Fig. P10.7

10.8 A frame consists of four L-shaped members connected by four

torsional springs, each of constant K. Knowing that equal loads P

are applied at points A and D as shown, determine the critical

value P

of the loads applied to the frame.

10.9 Determine the critical load of a round wooden dowel that is

48 in. long and has a diameter of (a) 0.375 in., (b) 0.5 in. Use

E 5 1.6 3 10

psi.

10.10 Determine the critical load of a steel tube that is 5 m long and has

a 100-mm outer diameter and a 16-mm wall thickness. Use E 5

200 GPa.

Fig. P10.8

100 mm

16 mm

Fig. P10.10

10.11 A compression member of 20-in. effective length consists of a solid

1-in.-diameter aluminum 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 5 10.6 3 10

psi.

1.0 in. 1.0 in.

0.5 in.

Fig. P10.11

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Columns

10.12 Two brass rods used as compression members, each of 3-m effec-

tive length, have the cross sections shown. (a) Determine the wall

thickness of the hollow square rod for which the rods have the

same cross-sectional area. (b) Using E 5 105 GPa, determine the

critical load of each rod.

10.13 A column of effective length L can be made by gluing together

identical planks in either of the arrangements shown. Determine

the ratio of the critical load using the arrangement a to the critical

load using the arrangement b.

10.14 Determine the radius of the round strut so that the round and

square struts have the same cross-sectional area and compute the

critical load of each strut. Use E 5 200 GPa.

10.15 A compression member of 7-m effective length is made by welding

together two L152 3 102 3 12.7 angles as shown. Using E 5

200 GPa, determine the allowable centric load for the member if

a factor of safety of 2.2 is required.

60 mm

40 mm

Fig. P10.12

d/3

(a)(b)

Fig. P10.13

25 mm

1 m

Fig. P10.14

102 mm

152 mm

102 mm

Fig. P10.15

10.16 A column of 3-m effective length is to be made by welding together

two C130 3 13 rolled-steel channels. Using E 5 200 GPa, deter-

mine for each arrangement shown the allowable centric load if a

factor of safety of 2.4 is required.

(a)(b)

Fig. P10.16

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