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

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*Structural Steel—Load and Resistance Factor Design. As

we saw in Sec. 1.13, an alternative method of design is based on the

determination of the load at which the structure ceases to be useful.

Design is based on the inequality given by Eq. (1.26):

1 g

# fP

(1.26)

The approach used for the design of steel columns under a centric

load using Load and Resistance Factor Design with the AISC Speci-

fication is similar to that for Allowable Stress Design. Using the criti-

cal stress s

, the ultimate load P

is defined as

5 s

A (10.50)

The determination of the critical stress s

follows the same approach

used for Allowable Stress Design. This requires using Eq. (10.41) to

determine the slenderness at the junction between Eqs. (10.38) and

Eq. (10.40). If the specified slenderness Lyr is smaller than the value

from Eq. (10.41), Eq. (10.38) governs, and if it is larger, Eq. (10.40)

governs. The equations can be used with SI or U.S. customary

units.

We observe that, by using Eq. (10.50) with Eq. (1.26), we can

determine if the design is acceptable. The procedure is to first

determine the slenderness ratio from Eq. (10.41). For values of Lyr

smaller than this slenderness, the ultimate load P

for use with

Eq. (1.26) is obtained from Eq. (10.50), using s

determined from

Eq. (10.38). For values of Lyr larger than this slenderness, the ulti-

mate load P

is obtained by using Eq. (10.50) with Eq. (10.40). The

Load and Resistance Factor Design Specification of the American

Institute of Steel Construction specifies that the resistance factor f

is 0.90.

Note: The design formulas presented throughout Sec. 10.6 are

intended to provide examples of different design approaches.

These formulas do not provide all the requirements that are

needed for many designs, and the student should refer to the

appropriate design specifications before attempting actual

designs.

Since the column must carry 32 kips, which is equal to s

, we use

Eq. (10.47) to write

all

ips

5 s

5 1.060C

Solving this equation for C

and substituting the value obtained into the

previous equation, we write

30.1

1 1 21.98 3 10

210.902

1 1 21.98 3 10

210.902

21.98 3 10

0.90

Solving for d by trial and error yields d 5 6.45 in.

667

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SAMPLE PROBLEM 10.3

Column AB consists of a W10 3 39 rolled-steel shape made of a grade of

steel for which s

5 36 ksi and E 5 29 3 10

psi. Determine the allowable

centric load P (a) if the effective length of the column is 24 ft in all direc-

tions, (b) if bracing is provided to prevent the movement of the midpoint

C in the xz plane. (Assume that the movement of point C in the yz plane

is not affected by the bracing.)

SOLUTION

We first compute the value of the slenderness ratio from Eq. 10.41 corre-

sponding to the given yield strength s

5 36 ksi.

5 4.71

29 3 10

36 3 10

5 133.7

a. Effective Length 5 24 ft. Since r

, r

, buckling will take place in

the xz plane. For L 5 24 ft and r 5 r

5 1.98 in., the slenderness ratio is

124 3 122 in

1.98 in.

288 in

1.98 in.

5 145.5

Since Lyr . 133.7, we use Eq. (10.39) in Eq. (10.40) to determine s

5 0.877 s

5 0.877

129 3 10

ksi2

1145.52

5 11.86 ksi

The allowable stress, determined using Eq. (10.42), and P

all

are

all

11.86 ksi

5 7.10 ksi

all

5 s

all

A 5 17.10 ksi2111.5 in

25 81.7 kips

b. Bracing at Midpoint C. Since bracing prevents movement of point

C in the xz plane but not in the yz plane, we must compute the slenderness

ratio correspoinding to buckling in each plane and determine which is larger.

xz Plane: Effective length 5 12 ft 5 144 in., r 5 r

5 1.98 in.

Lyr 5 (144 in.)y(1.98 in.) 5 72.7

yz Plane: Effective length 5 24 ft 5 288 in., r 5 r

5 4.27 in.

Lyr 5 (288 in.)y(4.27 in.) 5 67.4

Since the larger slenderness ratio corresponds to a smaller allowable load,

we choose Lyr 5 72.7. Since this is smaller than Lyr 5 133.7, we use Eqs.

(10.39) and (10.38) to determine s

129 3 10

ksi2

172.72

5 54.1 ksi

0.658

136 ksi

54.1 ksi2

36 ksi 5 27.3 ksi

We now calculate the allowable stress using Eq. (10.42) and the allowable

load.

all

27.3 ksi

5 16.32 ksi

all

5 s

all

A 5 116.32 ksi2111.5 in

2 P

all

5 187.7 ksi >

W10  39

A  11.5 in

 4.27 in.

 1.98 in.

24 ft

12 ft

Buckling in xz plane Buckling in yz plane

24 ft

(a)

12 ft

(b)

668

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669

SAMPLE PROBLEM 10.4

Using the aluminum alloy 2014-T6, determine the smallest diameter rod

that can be used to support the centric load P 5 60 kN if (a) L 5 750 mm,

(b) L 5 300 mm.

SOLUTION

For the cross section of a solid circular rod, we have

I 5



A 5 pc



r 5

a. Length of 750 mm. Since the diameter of the rod is not known, a

value of Lyr must be assumed; we assume that Lyr . 55 and use Eq. (10.46).

For the centric load P, we have s 5 P/A and write

5 s

all

382 3 10

MPa

60 3 10

382 3 10

0.750

5 112

5 3 10



5 18

31 mm

For c 5 18.44 mm, the slenderness ratio is

750 mm

18.31 mm

5 81.9 . 55

Our assumption is correct, and for L 5 750 mm, the required diameter is

d 5 2c 5 2

18.31 mm

5 36

6 mm>

b. Length of 300 mm. We again assume that Lyr . 55. Using

Eq. (10.46), and following the procedure used in part a, we find that c 5

11.58 mm and Lyr 5 51.8. Since Lyr is less than 55, our assumption is wrong; we

now assume that Lyr , 55 and use Eq. (10.459) for the design of this rod.

5 s

all

213 2 1.577

MPa

60 3 10

5 c213 2 1.577 a

0.3 m

bd 10

5 11

95 mm

For c 5 11.95 mm, the slenderness ratio is

300

11.95 mm

5 50.2

Our second assumption that Lyr , 55 is correct. For L 5 300 mm, the

required diameter is

d 5 2c 5 2

11.95 mm

5 23

9 mm>

P ⫽ 60 kN

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PROBLEMS

670

10.57 Using allowable stress design, determine the allowable centric load

for a column of 6-m effective length that is made from the follow-

ing rolled-steel shape: (a) W200 3 35.9, (b) W200 3 86. Use

5 250 MPa and E 5 200 GPa.

10.58 A W8 3 31 rolled-steel shape is used for a column of 21-ft effec-

tive length. Using allowable stress design, determine the allowable

centric load if the yield strength of the grade of steel used is

(a) s

5 36 ksi, (b) s

5 50 ksi. Use E 5 29 3 10

psi.

10.59 A steel pipe having the cross section shown is used as a column.

Using the allowable stress design determine the allowable centric

load if the effective length of the column is (a) 18 ft,

(b) 26 ft. Use s

5 36 ksi and E 5 29 3 10

psi.

10.6 0 A column is made from half of a W360 3 216 rolled-steel shape,

with the geometric properties as shown. Using allowable stress

design, determine the allowable centric load if the effective length

of the column is (a) 4.0 m, (b) 6.5 m. Use s

5 345 MPa and

5 200 GPa.

⫽ 0.28 in.

6.0 in.

Fig. P10.59

A ⫽ 13.75 ⫻ 10

⫽ 26.0 ⫻ 10

⫽ 141.0 ⫻ 10

Fig. P10.60

10.61 A compression member has the cross section shown and an effec-

tive length of 5 ft. Knowing that the aluminum alloy used is 2014-

T6, determine the allowable centric load.

4.0 in.

⫽ 0.375 in.

Fig. P10.61

50 mm

20 mm

Fig. P10.62

10.62 Using the aluminum alloy 2014-T6, determine the largest allowable

length of the aluminum bar AB for a centric load P of magnitude

(a) 150 kN, (b) 90 kN, (c) 25 kN.

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671

Problems

10.63 A sawn lumber column with a 7.5 3 5.5-in. cross section has an

18-ft effective length. Knowing that for the grade of wood used the

adjusted allowable stress for compression parallel to the grain is

5 1200 psi and that the adjusted modulus E 5 470 3 10

psi,

determine the maximum allowable centric load for the column.

10. 64 A column having a 3.5-m effective length is made of sawn lumber

with a 114 3 140-mm cross section. Knowing that for the grade of

wood used the adjusted allowable stress for compression parallel to

the grain is s

5 7.6 MPa and the adjusted modulus E 5 2.8 GPa,

determine the maximum allowable centric load for the column.

10.65 A compression member of 8.2-ft effective length is obtained by

bolting together two L5 3 3 3

-in. steel angles as shown. Using

allowable stress design, determine the allowable centric load for

the column. Use s

5 36 ksi and E 5 29 3 10

psi.

10.66 and 10.67 A compression member of 9-m effective length is

obtained by welding two 10-mm-thick steel plates to a W250 3 80

rolled-steel shape as shown. Knowing that s

5 345 MPa and

5 200 GPa and using allowable stress design, determine the

allowable centric load for the compression member.

Fig. P10.65

Fig. P10.67Fig. P10.66

10.68 A column of 18-ft effective length is obtained by connecting four

L3 3 3 3

-in. steel angles with lacing bars as shown. Using allow-

able stress design, determine the allowable centric load for the

column. Use s

5 36 ksi and E 5 29 3 10

psi.

10.69 An aluminum structural tube is reinforced by bolting two plates to

it as shown for use as a column of 1.7-m effective length. Knowing

that all material is aluminum alloy 2014-T6, determine the maxi-

mum allowable centric load.

8 in.

Fig. P10.68

54 mm

8 mm

6 mm 6 mm

34 mm

8 mm 8 mm

8 mm

Fig. P10.69

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672

Columns

10.70 A rectangular column with a 4.4-m effective length is made of

glued laminated wood. Knowing that for the grade of wood used

the adjusted allowable stress for compression parallel to the grain

is s

5 8.3 MPa and the adjusted modulus E 5 4.6 GPa, deter-

mine the maximum allowable centric load for the column.

10.71 For a rod made of the aluminum alloy 2014-T6, select the smallest

square cross section that may be used if the rod is to carry a 55-kip

centric load.

216 mm

140 mm

Fig. P10.70

10.72 An aluminum tube of 90-mm outer diameter is to carry a centric

load of 120 kN. Knowing that the stock of tubes available for use

are made of alloy 2014-T6 and with wall thicknesses in increments

of 3 mm from 6 mm to 15 mm, determine the lightest tube that

can be used.

20 in.

P 55 kips

Fig. P10.71

2.25 m

90-mm outside

diameter

120 kN

Fig. P10.72

10.73 A 72-kN centric load must be supported by an aluminum column

as shown. Using the aluminum alloy 6061-T6, determine the mini-

mum dimension b that can be used.

0.45 m

Fig. P10.73

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673

Problems

10.74 The glued laminated column shown is free at its top A and fixed

at its base B. Using wood that has an adjusted allowable stress for

compression parallel to the grain s

5 9.2 MPa and an adjusted

modulus of elasticity E

5 5.7 GPa, determine the smallest cross

section that can support a centric load of 62 kN.

10.75 An 18-kip centric load is applied to a rectangular sawn lumber col-

umn of 22-ft effective length. Using sawn lumber for which the

adjusted allowable stress for compression parallel to the grain is s

1050 psi and the adjusted modulus is E

5 440 3 10

psi, determine

the smallest cross section that can be used. Use b

5 2d.

2 m

Fig. P10.74

Fig. P10.75

10.76 A glue laminated column of 3-m effective length is to be made

from boards of 24 3 100-mm cross section. Knowing that for the

grade of wood used, E

5 11 GPa and the adjusted allowable stress

for compression parallel to the grain is s

5 9 MPa, determine

the number of boards that must be used to support the centric

load shown when (a) P

5 34 kN, (b) P 5 17 kN.

10.77 A column of 4.5-m effective length must carry a centric load of

900 kN. Knowing that s

5 345 MPa and E 5 200 GPa, use

allowable stress design to select the wide-flange shape of 250-mm

nominal depth that should be used.

10.78 A column of 4.6-m effective length must carry a centric load of

525 kN. Knowing that s

5 345 MPa and E 5 200 GPa, use

allowable stress design to select the wide-flange shape of 200-mm

nominal depth that should be used.

10.79 A column of 22.5-ft effective length must carry a centric load of

288 kips. Using allowable stress design, select the wide-flange

shape of 14-in. nominal depth that should be used. Use s

5 50 ksi

and E

5 29 3 10

psi.

10.80 A square steel tube having the cross section shown is used as a

column of 26-ft effective length to carry a centric load of 65 kips.

Knowing that the tubes available for use are made with wall thick-

nesses ranging from

in. to

in. in increments of

in., use allow-

able stress design to determine the lightest tube that can be used.

Use s

5 36 ksi and E 5 29 3 10

psi.

10.81 Solve Prob. 10.80, assuming that the effective length of the column

is decreased to 20 ft.

100 mm

24 mm

Fig. P10.76

6 in.

Fig. P10.80

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674

Columns

10.82 A centric load P must be supported by the steel bar AB. Using allow-

able stress design, determine the smallest dimension d of the cross

section that can be used when (a) P

5 108 kN, (b) P 5 166 kN. Use

5 250 MPa and E 5 200 GPa.

10.83 Two 3

3 2

-in. angles are bolted together as shown for use as a

column of 6-ft effective length to carry a centric load of 54 kips.

Knowing that the angles available have thicknesses of

, and

in.,

use allowable stress design to determine the lightest angles that can

be used. Use s

5 36 ksi and E 5 29 3 10

psi.

3 d

1.4 m

Fig. P10.82

in.

Fig. P10.83

10.84 Two 89 3 64-mm angles are bolted together as shown for use as a

column of 2.4-m effective length to carry a centric load of 180 kN.

Knowing that the angles available have thicknesses of 6.4 mm, 9.5 mm,

and 12.7 mm, use allowable stress design to determine the lightest

angles that can be used. Use s

5 250 MPa and E 5 200 GPa.

*10.85 A column with a 5.8-m effective length supports a centric load,

with ratio of dead to live load equal to 1.35. The dead load factor

is g

5 1.2, the live load factor g

5 1.6, and the resistance factor

5 0.90. Use load and resistance factor design to determine the

allowable centric dead and live loads if the column is made of the

following rolled-steel shape: (a) W250 3 67, (b) W360 3 101. Use

5 345 MPa and E 5 200 GPa.

*10.86 A rectangular steel tube having the cross section shown is used as

a column of 14.5-ft effective length. Knowing that s

5 36 ksi and

5 29 3 10

psi., use load and resistance factor design to deter-

mine the largest centric live load that can be applied if the centric

dead load is 54 kips. Use a dead load factor g

5 1.2, a live load

factor g

5 1.6 and the resistance factor f 5 0.90.

*10.87 A steel column of 5.5-m effective length must carry a centric dead

load of 310 kN and a centric live load of 375 kN. Knowing that

5 250 MPa and E 5 200 GPa, use load and resistance factor

design to select the wide-flange shape of 310-mm nominal depth

that should be used. The dead load factor g

5 1.2, the live load

factor g

5 1.6, and the resistance factor f 5 0.90.

*10.88 The steel tube having the cross section shown is used as a column of

15-ft effective length to carry a centric dead load of 51 kips and a

centric live load of 58 kips. Knowing that the tubes available for use

are made with wall thicknesses in increments of

in. from

in. to

in., use load and resistance factor design to determine the lightest

tube that can be used. Use s

5 36 ksi and E 5 29 3 10

psi. The

dead load factor g

5 1.2, the live load factor g

5 1.6, and the

resistance factor f

5 0.90.

89 mm 89 mm

64 mm

Fig. P10.84

7 in.

5 in.

in.



Fig. P10.86

6 in.

Fig. P10.88

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675

10.7 DESIGN OF COLUMNS UNDER AN

ECCENTRIC LOAD

In this section, the design of columns subjected to an eccentric load

will be considered. You will see how the empirical formulas devel-

oped in the preceding section for columns under a centric load can

be modified and used when the load P applied to the column has

an eccentricity e which is known.

We first recall from Sec. 4.12 that an eccentric axial load P

applied in a plane of symmetry of the column can be replaced by an

equivalent system consisting of a centric load P and a couple M of

moment M 5 Pe, where e is the distance from the line of action of

the load to the longitudinal axis of the column (Fig. 10.32). The

normal stresses exerted on a transverse section of the column can

then be obtained by superposing the stresses due, respectively, to

the centric load P and to the couple M (Fig. 10.33), provided that

10.7 Design of Columns under an

Eccentric Load

M  Pe

Fig. 10.32 Column with eccentric load.

centric





bending



Fig. 10.33 Stresses on column

transverse section.

the section considered is not too close to either end of the column,

and as long as the stresses involved do not exceed the proportional

limit of the material. The normal stresses due to the eccentric load

P can thus be expressed as

s 5 s

centric

bendin

(10.51)

Recalling the results obtained in Sec. 4.12, we find that the maxi-

mum compressive stress in the column is

max

(10.52)

In a properly designed column, the maximum stress defined by

Eq. (10.52) should not exceed the allowable stress for the column.

Two alternative approaches can be used to satisfy this requirement,

namely, the allowable-stress method and the interaction method.

a. Allowable-Stress Method. This method is based on the

assumption that the allowable stress for an eccentrically loaded col-

umn is the same as if the column were centrically loaded. We must

have, therefore, s

max

# s

all

, where s

all

is the allowable stress under

a centric load, or substituting for s

max

from Eq. (10.52)

# s

all

(10.53)

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676

Columns

The allowable stress is obtained from the formulas of Sec. 10.6

which, for a given material, express s

all

as a function of the slender-

ness ratio of the column. The major engineering codes require that

the largest value of the slenderness ratio of the column be used to

determine the allowable stress, whether or not this value corresponds

to the actual plane of bending. This requirement sometimes results

in an overly conservative design.

EXAMPLE 10.04

A column with a 2-in.-square cross section and 28-in. effective length is

made of the aluminum alloy 2014-T6. Using the allowable-stress method,

determine the maximum load P that can be safely supported with an

eccentricity of 0.8 in.

We first compute the radius of gyration r using the given data

2 in.

5 4 in

I 5

2 in.

5 1.333 in

r 5

1.333 in

4 in

5 0.5774 in.

We next compute Lyr 5 (28 in.)y(0.5774 in.) 5 48.50.

Since Lyr , 55, we use Eq. (10.48) to determine the allowable

stress for the aluminum column subjected to a centric load. We have

all

30.9 2 0.229

48.50

5 19.79 ksi

We now use Eq. (10.53) with M 5 Pe and c 5

2 in.

5 1 in. to

determine the allowable load:

4 in

0.8 in.

1 in.

333 in

# 19.79 ksi

P # 23.3

ips

The maximum load that can be safely applied is P 5 23.3 kips.

b. Interaction Method. We recall that the allowable stress for a

column subjected to a centric load (Fig. 10.34a) is generally smaller

than the allowable stress for a column in pure bending (Fig. 10.34b),

since the former takes into account the possibility of buckling.

Therefore, when we use the allowable-stress method to design an

eccentrically loaded column and write that the sum of the stresses

due to the centric load P and the bending couple M (Fig. 10.34c)

must not exceed the allowable stress for a centrically loaded column,

the resulting design is generally overly conservative. An improved

method of design can be developed by rewriting Eq. 10.53 in the

form

# 1

(10.54)

and substituting for s

all

in the first and second terms the values

of the allowable stress which correspond, respectively, to the

(a)(c)(b)

P'P'

M M

Fig. 10.34 Column load possibilities.

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