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

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657

Problems

10.36 An axial load P is applied at a point located on the x axis at a distance

5 12 mm from the geometric axis of the W310 3 60 rolled-steel

column BC. Assuming that L

5 3.5 m and using E 5 200 GPa,

determine (a) the load P for which the horizontal deflection at end

C is 15 mm, (b) the corresponding maximum stress in the column.

10.37 Solve Prob. 10.36, assuming that L is 4.5 m.

10.38 The line of action of an axial load P is parallel to the geometric

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

5 0.8 in. Using

5 29 3 10

psi, determine (a) the load P for which the hori-

zontal deflection of the midpoint C of the column is 0.5 in.,

(b) the corresponding maximum stress in the column.

310  60

Fig. P10.36

22 ft

0.8 in.

8  40

Fig. P10.38

10.39 A brass pipe having the cross section shown has an axial load P

applied 5 mm from its geometric axis. Using E

5 120 GPa, deter-

mine (a) the load P for which the horizontal deflection at the mid-

point C is 5 mm, (b) the corresponding maximum stress in the

column.

10.40 Solve Prob. 10.39, assuming that the axial load P is applied 10 mm

from the geometric axis of the column.

10.41 The steel bar AB has a

-in. square cross section and is held

by pins that are a fixed distance apart and are located at a distance

5 0.03 in. from the geometric axis of the bar. Knowing that at

temperature T

the pins are in contact with the bar and that the

force in the bar is zero, determine the increase in temperature for

which the bar will just make contact with point C if d

5 0.01 in.

Use E

5 29 3 10

psi and a coefficient of thermal expansion

5 6.5 3 10

/8F.

10.42 For the bar of Prob. 10.41, determine the required distance d for

which the bar will just make contact with point C when the tem-

perature increases by 120

8F.

120 mm

t  6 mm

2.8 m

Fig. P10.39

e  0.03 in.

in.

4 in.

Fig. P10.41

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Columns

10.43 An axial load P is applied to the W10 3 30 rolled-steel column BC

that is free at its top C and fixed at its base B. Knowing that the

eccentricity of the load is e

5 0.5 in. and that for the grade of steel

used s

5 36 ksi and E 5 29 3 10

psi, determine (a) the mag-

nitude of P of the allowable load when a factor of safety of 2.4 with

respect to permanent deformation is required, (b) the ratio of the

load found in part a to the magnitude of the allowable centric load

for the column. (Hint: Since the factor of safety must be applied

to the load P, not to the stress, use Fig. 10.23 to determine P

10.44 Solve Prob. 10.43, assuming that the length of the column is

reduced to 5 ft.

10.45 A 3.5-m-long steel tube having the cross section and properties shown

is used as a column. For the grade of steel used s

5 250 MPa and

5 200 GPa. Knowing that a factor of safety of 2.6 with respect

to permanent deformation is required, determine the allowable

load P when the eccentricity e is (a) 15 mm, (b) 7.5 mm. (See hint

of Prob. 10.43).

10 ⫻ 30

L ⫽ 7.5 ft

Fig. P10.43

127 mm

A ⫽ 3400 mm

I ⫽ 7.93 ⫻ 10

–6

r ⫽ 48.3 mm

3.5 m

P⬘

Fig. P10.45 and P10.46

10.46 Solve Prob. 10.45, assuming that the length of the tube is increased

to 5 m.

10.47 A 250-kN axial load P is applied to a W200 3 35.9 rolled-steel

column BC that is free at its top C and fixed at its base B. Knowing

that the eccentricity of the load is e

5 6 mm, determine the largest

permissible length L if the allowable stress in the column is

80 MPa. Use E

5 200 GPa.

10.48 A 100-kN axial load P is applied to the W150 3 18 rolled-steel

column BC that is free at its top C and fixed at its base B. Knowing

that the eccentricity of the load is e

5 6 mm, determine the largest

permissible length L if the allowable stress in the column is 80 MPa.

Use E

5 200 GPa.

Fig. P10.47 and P10.48

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659

Problems

10.49 Axial loads of magnitude P 5 20 kips are applied parallel to the

geometric axis of the W8 3 15 rolled-steel column AB and inter-

sect the x axis at a distance e from the geometric axis. Knowing

that s

all

5 12 ksi and E 5 29 3 10

psi, determine the largest

permissible length L when (a) e

5 0.25 in., (b) e 5 0.5 in.

P⬘

Fig. P10.49 and P10.50

4.0 ft

12 kips

Fig. P10.51

10.50 Axial loads of magnitude P 5 135 kips are applied parallel to the

geometric axis of the W10 3 54 rolled-steel column AB and inter-

sect the x axis at a distance e from the geometric axis. Knowing

that s

all

5 12 ksi and E 5 29 3 10

psi, determine the largest

permissible length L when (a) e

5 0.25 in., (b) e 5 0.5 in.

10.51 A 12-kip axial load is applied with an eccentricity e

5 0.375 in. to

the circular steel rod BC that is free at its top C and fixed at its base

B. Knowing that the stock of rods available for use have diameters

in increments of

in. from 1.5 in. to 3.0 in., determine the lightest

rod that can be used if s

all

5 15 ksi. Use E 5 29 3 10

psi.

10.52 Solve Prob. 10.51, assuming that the 12-kip axial load will be

applied to the rod with an eccentricity e 5

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10.53 An axial load of magnitude P 5 220 kN is applied at a point located

on the x axis at a distance e

5 6 mm from the geometric axis of

the wide-flange column BC. Knowing that E

5 200 GPa, choose

the lightest W200 shape that can be used if s

all

5 120 MPa.

10.54 Solve Prob. 10.53, assuming that the magnitude of the axial load

is P

5 345 kN.

10.55 Axial loads of magnitude P

5 175 kN are applied parallel to the

geometric axis of a W250 3 44.8 rolled-steel column AB and inter-

sect the axis at a distance e

5 12 mm from its geometric axis.

Knowing that s

5 250 MPa and E 5 200 GPa, determine the

factor of safety with respect to yield. (Hint: Since the factor of safety

must be applied to the load P, not to the stresses, use Fig. 10.23 to

determine P

1.8 m

Fig. P10.53

3.8 m

P⬘

Fig. P10.55

10.6 DESIGN OF COLUMNS UNDER A CENTRIC LOAD

In the preceding sections, we have determined the critical load of a

column by using Euler’s formula, and we have investigated the defor-

mations and stresses in eccentrically loaded columns by using the

secant formula. In each case we assumed that all stresses remained

below the proportional limit and that the column was initially a

straight homogeneous prism. Real columns fall short of such an ide-

alization, and in practice the design of columns is based on empirical

formulas that reflect the results of numerous laboratory tests.

Over the last century, many steel columns have been tested

by applying to them a centric axial load and increasing the load

until failure occurred. The results of such tests are represented in

660

Columns

10.56 Solve Prob. 10.55, assuming that e 5 0.16 mm and P 5 155 kN.

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Fig. 10.24 where, for each of many tests, a point has been plotted with

its ordinate equal to the normal stress s

at failure, and its abscissa

equal to the corresponding value of the effective slenderness ratio,

yr. Although there is considerable scatter in the test results, regions

corresponding to three types of failure can be observed. For long col-

umns, where L

yr is large, failure is closely predicted by Euler’s for-

mula, and the value of s

is observed to depend on the modulus of

elasticity E of the steel used, but not on its yield strength s

. For very

short columns and compression blocks, failure occurs essentially as a

result of yield, and we have s

< s

. Columns of intermediate length

comprise those cases where failure is dependent on both s

and E.

In this range, column failure is an extremely complex phenomenon,

and test data have been used extensively to guide the development of

specifications and design formulas.

Empirical formulas that express an allowable stress or critical

stress in terms of the effective slenderness ratio were first introduced

over a century ago, and since then have undergone a continuous

process of refinement and improvement. Typical empirical formulas

previously used to approximate test data are shown in Fig. 10.25. It

is not always feasible to use a single formula for all values of L

yr.

Most design specifications use different formulas, each with a defi-

nite range of applicability. In each case we must check that the for-

mula we propose to use is applicable for the value of L

yr for the

10.6 Design of Columns under a Centric Load

Short

columns

Intermediate columns Long columns

Euler’s critical stress

/r)









Fig. 10.24 Plot of test data for steel columns.

Gordon-Rankine formula:

1

Parabola:

Straight line:











( )













( )

Fig. 10.25 Plots of empirical formulas for column critical stress.

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662

Columns

column involved. Furthermore, we must determine whether the for-

mula provides the value of the critical stress for the column, in which

case we must apply the appropriate factor of safety, or whether it

provides directly an allowable stress.

Specific formulas for the design of steel, aluminum and wood

columns under centric loading will now be considered. Photo 10.2

shows examples of columns that would be designed using these for-

mulas. The design for the three different materials using Allowable

Stress Design is first presented. This is followed with the formulas

needed for the design of steel columns based on Load and Resistance

Factor Design.†

Photo 10.2 The water tank in (a) is supported by steel columns and the

building under construction in (b) is framed with wood columns.

(a)

(b)

Structural Steel—Allowable Stress Design. The formulas

most widely used for the allowable stress design of steel columns

under a centric load are found in the Specification for Structural

Steel Buildings of the American Institute of Steel Construction.‡

As we shall see, an exponential expression is used to predict s

all

for columns of short and intermediate lengths, and an Euler-based

relation is used for long columns. The design relations are devel-

oped in two steps:

1. First a curve representing the variation of s

with Lyr is

obtained (Fig. 10.26). It is important to note that this curve does not

incorporate any factor of safety.§ The portion AB of this curve is

defined by the equation

5 30.658

(10.38)

†In specific design formulas, the letter L will always refer to the effective length of the

column.

‡Manual of Steel Construction, 13th ed., American Institute of Steel Construction,

Chicago, 2005.

§In the Specification for Structural Steel for Buildings, the symbol F is used for

stresses.

0.39

4.71



Fig. 10.26 Steel column design.

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663

where

(10.39)

The portion BC is defined by the equation

5 0.877s

(10.40)

We note that when Lyr 5 0, s

5 s

in Eq. (10.38). At point B,

Eq. (10.38) joins Eq. (10.40). The value of slenderness Lyr at the

junction between the two equations is

5 4.71

(10.41)

If Lyr is smaller than the value in Eq. (10.41), s

is determined from

Eq. (10.38), and if Lyr is greater, s

is determined from Eq. (10.40).

At the value of the slenderness Lyr specified in Eq. (10.41), the stress

5 0.44 s

. Using Eq. (10.40), s

5 0.877 (0.44 s

) 5 0.39 s

2. A factor of safety must be introduced to obtain the final AISC

design formulas. The factor of safety specified by the specification is

1.67. Thus

all

(10.42)

The formulas obtained can be used with SI or U.S. customary

units.

We observe that, by using Eqs. (10.38), (10.40), (10.41), and

(10.42), we can determine the allowable axial stress for a given grade

of steel and any given value of Lyr. The procedure is to first compute

the value of Lyr at the intersection between the two equations from

Eq. (10.41). For given values of Lyr smaller than that in Eq. (10.41),

we use Eqs. (10.38) and (10.42) to calculate s

all

, and for values

greater than that in Eq. (10.41), we use Eqs. (10.40) and (10.42) to

calculate s

all

. Figure 10.27 provides a general illustration of how s

varies as a function of Lyr for different grades of structural steel.

0 50 100 150 200

L/r

all



Fig. 10.27 Steel column design for

different grades of steel.

10.6 Design of Columns under a Centric Load

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EXAMPLE 10.02

Determine the longest unsupported length L for which the S100 3 11.5

rolled-steel compression member AB can safely carry the centric load

shown (Fig. 10.28). Assume s

5 250 MPa and E 5 200 GPa.

From Appendix C we find that, for an S100 3 11.5 shape,

A 5 1460 mm

r

5 41.7 mmr

5 14.6 mm

If the 60-kN load is to be safely supported, we must have

all

60 3 10

1460

2 6

5 41.1 3 10

We must compute the critical stress s

. Assuming Lyr is larger than the

slenderness specified by Eq. (10.41), we use Eq. (10.40) with (10.39) and

write

5 0.877 s

5 0.877

1200 3 10

Pa2

1.731 3 10

Using this expression in Eq. (10.42) for s

all

, we write

all

1.67

1.037 3 10

Equating this expression to the required value of s

all

, we write

1.037 3 10

5 1.41 3 10

PaL

r 5 158.8

The slenderness ratio from Eq. (10.41) is

5 4.71

200 3 10

250 3 10

5 133.2

Our assumption that Lyr is greater than this slenderness ratio was correct.

Choosing the smaller of the two radii of gyration, we have

14.6 3 10

5 158.8L 5 2.32 m

P  60 k

Fig. 10.28

Aluminum. Many aluminum alloys are available for use in struc-

tural and machine construction. For most columns the specifications

of the Aluminum Association† provide two formulas for the allow-

able stress in columns under centric loading. The variation of s

all

with Lyr defined by these formulas is shown in Fig. 10.29. We note

that for short columns a linear relation between s

all

with Lyr is used

and for long columns an Euler-type formula is used. Specific formu-

las for use in the design of buildings and similar structures are given

below in both SI and U.S. customary units for two commonly used

alloys.

†Specifications for Aluminum Structures, Aluminum Association, Inc., Washington, D.C.,

2010.

L/r

all

 C

 C



all



(L/r)

all





Fig. 10.29 Aluminum column

design.

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665

Alloy 6061-T6:

r , 66: s

all

20.3 2 0.127

ksi (10.43)

140 2 0.874

MPa (10.439)

r $ 66:

all

51,400 ksi

354 3 10

MPa

(10.44)

Alloy 2014-T6:

r , 55: s

all

30.9 2 0.229

ksi (10.45)

213 2 1.577

MPa (10.459)

r $ 55:

all

55,400 ksi

382 3 10

MPa

(10.46)

Wood. For the design of wood columns the specifications of the

American Forest & Paper Association† provides a single equation

that can be used to obtain the allowable stress for short, intermedi-

ate, and long columns under centric loading. For a column with a

rectangular cross section of sides b and d, where d , b, the variation

of s

all

with Lyd is shown in Fig. 10.30.

10.6 Design of Columns under a Centric Load

†National Design Specification for Wood Construction, American Forest & Paper Associa-

tion, American Wood Council, Washington, D.C., 2005.

‡In the National Design Specification for Wood Construction, the symbol F is used for

stresses.

L/d

500

all



Fig. 10.30 Wood column design.

For solid columns made from a single piece of wood or made

by gluing laminations together, the allowable stress s

all

5 s

(10.47)

where s

is the adjusted allowable stress for compression parallel to

the grain.‡ Adjustments used to obtain s

are included in the speci-

fications to account for different variations, such as in the load dura-

tion. The column stability factor C

accounts for the column length

and is defined by the following equation:

1 1 1s

(10.48)

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Columns

The parameter c accounts for the type of column, and it is equal to

0.8 for sawn lumber columns and 0.90 for glued laminated wood

columns. The value of s

is defined as

0.822E

(10.49)

Where E is an adjusted modulus of elasticity for column buckling.

Columns in which Lyd exceeds 50 are not permitted by the National

Design Specification for Wood Construction.

EXAMPLE 10.03

Knowing that column AB (Fig. 10.31) has an effective length of 14 ft, and

that it must safely carry a 32-kip load, design the column using a square

glued laminated cross section. The adjusted modulus of elasticity for the

wood is E 5 800 3 10

psi, and the adjusted allowable stress for compres-

sion parallel to the grain is s

5 1060 psi.

1 1 1s

1 1 21.98 3 10

210.902

1 1 21.98 3 10

210.902

21.98 3 10

0.90

14 ft

P  32 kips

Fig. 10.31

We note that c 5 0.90 for glued laminated wood columns. We must

compute the value of s

. Using Eq. (10.49) we write

0.822E

0.8221800 3 10

psi2

168 in./d

5 23.299d

psi

We then use Eq. (10.48) to express the column stability factor in terms

of d, with (s

) 5 (23.299d

y1.060 3 10

) 5 21.98 3 10

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