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

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647

Problems

10.17 A single compression member of 27-ft effective length is obtained by

connecting two C8 3 11.5 steel channels with lacing bars as shown.

Knowing that the factor of safety is 1.85, determine the allowable

centric load for the member. Use E 5 29 3 10

psi and d 5 4.0 in.

10.18 A column of 22-ft effective length is made by welding two 9 3 0.5-in.

plates to a W8 3 35 as shown. Determine the allowable centric load

if a factor of safety of 2.3 is required. Use E 5 29 3 10

psi.

Fig. P10.17

4.5 in.

Fig. P10.18

10.19 Member AB consists of a single C130 3 10.4 steel channel of

length 2.5 m. Knowing that the pins A and B pass through the

centroid of the cross section of the channel, determine the factor

of safety for the load shown with respect to buckling in the plane

of the figure when u 5 308. Use E 5 200 GPa.

6.8 kN



2.5 m

Fig. P10.19

1.2 m

70

22-mm diamete

18-mm

diameter

Fig. P10.20

4 m

(1) (2) (3) (4)

Fig. P10.21

10.20 Knowing that P 5 5.2 kN, determine the factor of safety for the

structure shown. Use E 5 200 GPa and consider only buckling in

the plane of the structure.

10.21 A rigid block of mass m can be supported in each of the four ways

shown. Each column consists of an aluminum tube that has a 44-

mm outer diameter and a 4-mm wall thickness. Using E 5 70 GPa

and a factor of safety of 2.8, determine the allowable mass for each

support condition.

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648

Columns

10.22 Each of the five struts shown consists of a solid steel rod. (a) Know-

ing that the strut of Fig. (1) is of a 20-mm diameter, determine

the factor of safety with respect to buckling for the loading shown.

(b) Determine the diameter of each of the other struts for which

the factor of safety is the same as the factor of safety obtained in

part a. Use E 5 200 GPa.

Fig. P10.23 and P10.24

900 mm

(1) (2) (3) (4) (5)

 7.5 kN

Fig. P10.22

10.23 A 1-in.-square aluminum strut is maintained in the position shown

by a pin support at A and by sets of rollers at B and C that prevent

rotation of the strut in the plane of the figure. Knowing that L

3 ft, determine (a) the largest values of L

and L

that can be

used if the allowable load P is to be as large as possible, (b) the

magnitude of the corresponding allowable load. Consider only

buckling in the plane of the figure and use E 5 10.4 3 10

psi.

10.24 A 1-in.-square aluminum strut is maintained in the position shown

by a pin support at A and by sets of rollers at B and C that pre-

vent rotation of the strut in the plane of the figure. Knowing that

5 3 ft, L

5 4 ft, and L

5 1 ft, determine the allowable

load P using a factor of safety with respect to buckling of 3.2.

Consider only buckling in the plane of the figure and use E 5

10.4 3 10

psi.

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10.25 Column AB carries a centric load P of magnitude 15 kips. Cables BC

and BD are taut and prevent motion of point B in the xz plane. Using

Euler’s formula and a factor of safety of 2.2, and neglecting the ten-

sion in the cables, determine the maximum allowable length L. Use

E 5 29 3 10

psi.

W10  21

Fig. P10.25

Fig. P10.27 and P10.28

10.26 A W8 3 21 rolled-steel shape is used with the support and cable

arrangement shown in Prob. 10.25. Knowing that L 5 24 ft, deter-

mine the allowable centric load P if a factor of safety of 2.2 is

required. Use E 5 29 3 10

psi.

10.27 Column ABC has a uniform rectangular cross section with b 5

12 mm and d 5 22 mm. The column is braced in the xz plane at

its midpoint C and carries a centric load P of magnitude 3.8 kN.

Knowing that a factor of safety of 3.2 is required, determine the

largest allowable length L. Use E 5 200 GPa.

10.28 Column ABC has a uniform rectangular cross section and is braced

in the xz plane at its midpoint C. (a) Determine the ratio b/d for

which the factor of safety is the same with respect to buckling in

the xz and yz planes. (b) Using the ratio found in part a, design

the cross section of the column so that the factor of safety will be

3.0 when P 5 4.4 kN, L 5 1 m, and E 5 200 GPa.

*10.5 ECCENTRIC LOADING; THE SECANT FORMULA

In this section the problem of column buckling will be approached

in a different way, by observing that the load P applied to a column

is never perfectly centric. Denoting by e the eccentricity of the load,

i.e., the distance between the line of action P and the axis of the

649

10.5 Eccentric Loading; the Secant Formula

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650

Columns

column (Fig. 10.18a ), we replace the given eccentric load by a cen-

tric force P and a couple M

of moment M

5 Pe (Fig. 10.18b). It

is clear that, no matter how small the load P and the eccentricity e,

the couple M

will cause some bending of the column (Fig. 10.19).

As the eccentric load is increased, both the couple M

and the axial

force P increase, and both cause the column to bend further. Viewed

in this way, the problem of buckling is not a question of determining

how long the column can remain straight and stable under an increas-

ing load, but rather how much the column can be permitted to bend

under the increasing load, if the allowable stress is not to be exceeded

and if the deflection y

max

is not to become excessive.

We first write and solve the differential equation of the elastic

curve, proceeding in the same manner as we did earlier in Secs. 10.3

and 10.4. Drawing the free-body diagram of a portion AQ of the

column and choosing the coordinate axes as shown (Fig. 10.20), we

find that the bending moment at Q is

52Py 2

52Py 2 Pe (10.22)

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

y 2

Transposing the term containing y and setting

(10.6)

as done earlier, we write

1 p

y 52p

(10.23)

Since the left-hand member of this equation is the same as that of

Eq. (10.7), which was solved in Sec. 10.3, we write the general solu-

tion of Eq. (10.23) as

y 5 A sin px 1 B cos px 2 e (10.24)

where the last term is a particular solution of Eq. (10.23).

 Pe

(a)(b)

Fig. 10.18 Column with eccentric load.

 Pe

max

 Pe

Fig. 10.19 Deflection of

column with eccentric load.

 Pe

Fig. 10.20

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651

The constants A and B are obtained from the boundary conditions

shown in Fig. 10.21. Making first x 5 0, y 5 0 in Eq. (10.24), we

have

B 5 e

Making next x 5 L, y 5 0, we write

A sin pL 5 e

1 2 cos pL

(10.25)

Recalling that

sin pL 5 2 sin

cos

and

1 2 cos pL 5 2 sin

and substituting into Eq. (10.25), we obtain, after reductions,

A 5 e tan

Substituting for A and B into Eq. (10.24), we write the equation of

the elastic curve:

y 5 e

tan

sin px 1 cos px 2 1

(10.26)

The value of the maximum deflection is obtained by setting

x 5 Ly2 in Eq. (10.26). We have

max

5 e

tan

sin

1 cos

2 1

5 e ±

sin

1 cos

cos

2 1 ≤

max

5 e

sec

2 1

(10.27)

Recalling Eq. (10.6), we write

max

5 e

sec

2 1

(10.28)

We note from the expression obtained that y

max

becomes infinite

when

(10.29)

While the deflection does not actually become infinite, it nevertheless

becomes unacceptably large, and P should not be allowed to reach the

critical value which satisfies Eq. (10.29). Solving (10.29) for P, we have

(10.30)

10.5 Eccentric Loading; the Secant Formula

max

L/2

[ x  0, y  0]

[ x  L, y  0]

Fig. 10.21

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652

Columns

which is the value that we obtained in Sec. 10.3 for a column under

a centric load. Solving (10.30) for EI and substituting into (10.28),

we can express the maximum deflection in the alternative form

max

5 easec

2 1b

(10.31)

The maximum stress s

max

occurs in the section of the column

where the bending moment is maximum, i.e., in the transverse sec-

tion through the midpoint C, and can be obtained by adding the

normal stresses due, respectively, to the axial force and the bending

couple exerted on that section (cf. Sec. 4.12). We have

max

(10.32)

From the free-body diagram of the portion AC of the column

(Fig. 10.22), we find that

max

5 Py

max

1 M

5 P

max

1 e

Substituting this value into (10.32) and recalling that I 5 Ar

, we write

max

1 1

max

1 e

(10.33)

Substituting for y

max

the value obtained in (10.28), we write

max

1 1

sec

(10.34)

An alternative form for s

max

is obtained by substituting for y

max

from

(10.31) into (10.33). We have

max

a1 1

sec

(10.35)

The equation obtained can be used with any end conditions, as long

as the appropriate value is used for the critical load (cf. Sec. 10.4).

We note that, since s

max

does not vary linearly with the load P,

the principle of superposition does not apply to the determination

of the stress due to the simultaneous application of several loads; the

resultant load must first be computed, and then Eq. (10.34) or

Eq. (10.35) can be used to determine the corresponding stress. For

the same reason, any given factor of safety should be applied to the

load, and not to the stress.

Making I 5 Ar

in Eq. (10.34) and solving for the ratio PyA in

front of the bracket, we write

max

1 1

sec

(10.36)

L/2

max

 Pe

Fig. 10.22

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653

where the effective length is used to make the formula applicable to

various end conditions. This formula is referred to as the secant formula;

it defines the force per unit area, PyA, that causes a specified maximum

stress s

max

in a column of given effective slenderness ratio, L

yr, for a

given value of the ratio ecyr

, where e is the eccentricity of the applied

load. We note that, since PyA appears in both members, it is necessary

to solve a transcendental equation by trial and error to obtain the value

of PyA corresponding to a given column and loading condition.

Equation (10.36) was used to draw the curves shown in Fig. 10.23a

and b for a steel column, assuming the values of E and s

shown in

the figure. These curves make it possible to determine the load per

unit area PyA, which causes the column to yield for given values of the

ratios L

yr and ecyr

050

100

150

300

250

200

0.1

0.2

0.4

0.6

0.8

0.1

0.2

0.4

0.6

0.8

100 150

36 ksi

200



P/A (MPa)

P/A (ksi)

0 50 100 150 200





29  10

psi

250 MPa





200 GPa

Euler’s curve



(a)(b)

Fig. 10.23 Load per unit area, PyA, causing yield in column.

We note that, for small values of L

yr, the secant is almost equal

to 1 in Eq. (10.36), and PyA can be assumed equal to

max

1 1

(10.37)

a value that could be obtained by neglecting the effect of the lateral

deflection of the column and using the method of Sec. 4.12. On the

other hand, we note from Fig. 10.23 that, for large values of L

yr,

the curves corresponding to the various values of the ratio ecyr

get

very close to Euler’s curve defined by Eq. (10.139), and thus that the

effect of the eccentricity of the loading on the value of PyA becomes

negligible. The secant formula is chiefly useful for intermediate val-

ues of L

yr. However, to use it effectively, we should know the value

of the eccentricity e of the loading, and this quantity, unfortunately,

is seldom known with any degree of precision.

10.5 Eccentric Loading; the Secant Formula

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654

SAMPLE PROBLEM 10.2

The uniform column AB consists of an 8-ft section of structural tubing hav-

ing the cross section shown. (a) Using Euler’s formula and a factor of safety

of two, determine the allowable centric load for the column and the corre-

sponding normal stress. (b) Assuming that the allowable load, found in part

a, is applied as shown at a point 0.75 in. from the geometric axis of the

column, determine the horizontal deflection of the top of the column and

the maximum normal stress in the column. Use E 5 29 3 10

psi.

SOLUTION

Effective Length. Since the column has one end fixed and one end

free, its effective length is

5 2

8 ft

5 16 ft 5 192 in.

Critical Load. Using Euler’s formula, we write

129 3 10

psi218.00 in

192 in.

5 62.1 kips

a. Allowable Load and Stress. For a factor of safety of 2, we find

all

F.S.

62.1

ips

all

5 31.1 kips >

and

s 5

all

31.1

ips

54 in

s = 8.79 ksi

b. Eccentric Load. We observe that column AB and its loading are

identical to the upper half of the column of Fig. 10.19 which was used in

the derivation of the secant formulas; we conclude that the formulas of Sec.

10.5 apply directly to the case considered here. Recalling that P

all

and using Eq. (10.31), we compute the horizontal deflection of point A:

5 e csec a

b2 1 d5 10.75 in.2csec a

b2 1 d

0.75 in.

2.252 2 1

5 0.939 in. >

The maximum normal stress is obtained from Eq. (10.35):

c1 1

sec a

31.1

ips

3.54 in

c1 1

10.75 in.212 in.

1.50 in.

sec a

8.79 ksi

1 1 0.667

2.252

5 22.0 ksi >

e ⫽ 0.75 in.

8 ft

(a)

(b)

A ⫽ 3.54 in

I ⫽ 8.00 in

r ⫽ 1.50 in.

c ⫽ 2.00 in.

4 in.

e ⫽ 0.75 in.

all

⫽ 31.1 kips

e ⫽ 0.75 in.

⫽ 0.939 in.

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PROBLEMS

655

10.29 An axial load P is applied to the 32-mm-diameter steel rod AB as

shown. For P

5 37 kN and e 5 1.2 mm, determine (a) the deflec-

tion at the midpoint C of the rod, (b) the maximum stress in the

rod. Use E

5 200 GPa.

1.2 m

32-mm

diamete

Fig. P10.29

30 mm

0.6 m

4 mm

Fig. P10.30

10.30 An axial load P 5 15 kN is applied at point D that is 4 mm from

the geometric axis of the square aluminum bar BC. Using E

70 GPa, determine (a) the horizontal deflection of end C, (b) the

maximum stress in the column.

10.31 The line of action of the 75-kip axial load is parallel to the geomet-

ric axis of the column AB and intersects the x axis at x

5 0.6 in.

Using E

5 29 3 10

psi, determine (a) the horizontal deflection

of the midpoint C of the column, (b) the maximum stress in the

column.

20 ft

0.6 in.

75 kips

8  35

Fig. P10.31

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656

Columns

10.32 An axial load P is applied to the 32-mm-square aluminum bar BC

as shown. When P

5 24 kN, the horizontal deflection at end C is

4 mm. Using E

5 70 GPa, determine (a) the eccentricity e of the

load, (b) the maximum stress in the bar.

10.33 An axial load P is applied to the 1.375-in. diameter steel rod AB

as shown. When P

5 21 kips, it is observed that the horizontal

deflection at midpoint C is 0.03 in. Using E

5 29 3 10

psi, deter-

mine (a) the eccentricity e of the load, (b) the maximum stress in

the rod.

32 mm

0.65 m

Fig. P10.32

30 in.

1.375-in.

diamete

Fig. P10.33

10.34 The axial load P is applied at a point located on the x axis at a

distance e from the geometric axis of the rolled-steel column BC.

When P

5 350 kN, the horizontal deflection of the top of the

column is 5 mm. Using E

5 200 GPa, determine (a) the eccentric-

ity e of the load, (b) the maximum stress in the column.

10.35 An axial load P is applied at point D that is 0.25 in. from the geo-

metric axis of the square aluminum bar BC. Using E

5 10.1 3 10

psi, determine (a) the load P for which the horizontal deflection

of end C is 0.50 in., (b) the corresponding maximum stress in the

column.

250  58

3.2 m

Fig. P10.34

1.75 in.

2.5 ft

0 .25 in.

Fig. P10.35

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