Yu W., LaBoube R.A. Cold-Formed Steel Design

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BENDING STRENGTH AND DEFLECTION 155

Figure 4.49 Effective design widths of compression and tension ﬂanges for shear lag.

The nominal moment about the x axis is

= 1.469(50) = 73.45 in.-kips

2. Nominal Moment Based on Shear Lag Consideration.

According to Figs. 4.44 and 4.47,

8 − 2

(

0.06

)

= 3.94 in.

5 × 12

3.94

= 15.23 < 30

Because the L/w

ratio is less than 30 and the member

carries a concentrated load, additional consideration

for shear lag is needed. Using Table 4.6,

Effective design width

Actual width

= 0.845

Therefore the effective design widths of compression

and tension ﬂanges between webs are (Fig. 4.49)

0.845[8 − 2(0.06)] = 6.6586 in.

Using the full areas of webs, the moment of inertia

about the x axis is

= 4[3.22356(0.06)(2.5 − 0.03)

+ 0.01166(2.5 − 0.073)

]

+ 2

(0.06)(4.6925)

= 6.046 in.

and the section modulus is

2.5

= 2.418 in.

The nominal moment is

= 2.418(50) = 120.9in.-kips

3. Nominal Moment for Design. From the above calcula-

tion, the nominal moment for design is 73.45 in.-kips.

Shear lag does not govern the design.

Figure 4.50 Flange curling of I-beam subject to bending.

4.42

4.2.7.2 Flange Curling When a beam with unusually

wide and thin ﬂanges is subject to bending, the portion of

the ﬂange most remote from the web tends to deﬂect toward

the neutral axis. This is due to the effect of longitudinal

curvature of the beam and bending stresses in both ﬂanges.

This subject was studied by Winter in 1940.

4.42

Let us consider an I-beam which is subject to pure

bending as shown in Fig. 4.50. The transverse compo-

nent q of the ﬂange force f

t per unit width can be

determined by

q =

tdφ

EI/M

(4.124)

where f

= average bending stress in ﬂanges

t = ﬂange thickness

dφ,dl,r

= as shown in Fig. 4.50

E = modulus of elasticity

I = moment of inertia of beam

d = depth of beam

If the value of q is considered to be a uniformly

distributed load applied on the ﬂange, the deﬂection or

curling at the outer edge of the ﬂange can be computed

by the conventional method for a cantilever plate,

156 4 FLEXURAL MEMBERS

namely,

= 3









(1 − μ

) (4.125)

where c

= deﬂection at outer edge

= projection of ﬂange beyond web

D = ﬂexural rigidity of plate, =Et

/12(1 − μ

)

By substituting E = 29.5 × 10

ksi and μ = 0.3 in Eq.

(4.125), the following formula for the maximum width of

an unusually wide stiffened or unstiffened ﬂange in tension

and compression can be obtained:



1800td



100c



0.061tdE/f



100c

/d (4.126)

where c

= permissible amount of curling, in.

= average stress in full unreduced ﬂange width,

ksi (w

, t,andd were deﬁned previously)

When members are designed by the effective design

width procedure, the average stress equals the maximum

stress times the ratio of the effective design width to

the actual width. Equation (4.126) is included in Section

B1.1(b) of the North American Speciﬁcation to limit the

width of unusually wide ﬂanges.

The above formula for determining w

is derived on

the basis of a constant transverse component q. As soon

as ﬂange curling develops, the distance from the ﬂange

to the neutral axis becomes smaller at the outer edge

of the ﬂange. This results in the reduction of bending

stresses. Therefore the values of q vary along the ﬂange

as shown in Fig. 4.50. Since the amount of c

is usually

limited to a small percentage of the depth, the error in the

determination of w

by using Eq. (4.126) is negligible and

on the conservative side.

The above approximate treatment for I-beams can also

be applied to the design of box and U-type beams, except

that for the latter the ﬂanges of the closed box beams may

be regarded as simple plates freely supported at webs and

that w

is to be measured as half of the distance between

webs. Using the same analogy, one can determine the ﬂange

curling c

for closed box sections as follows:

384

(

)

= 5











1 − μ



(4.127)

A comparison of Eqs. (4.125) and (4.127) indicates that

the use of Eq. (4.126), which is derived on the basis of

I-beams, to determine w

for box beams, may result in

a possible error of 13%. This is because

√

5/3 = 1.13.

However, this discrepancy can be reduced if the restraint

of webs and the variable values of the transverse component

q are taken into consideration.

No speciﬁc values are given by the North American

Speciﬁcation for the maximum permissible amount of

curling. However, it is stated in the AISI Commentary that

the amount of curling that can be tolerated will vary with

different kinds of sections and must be established by the

designer. An amount of curling on the order of 5% of the

depth of the section is usually not considered excessive.

Assuming c

/d = 0.05, Eq. (4.126) can be simpliﬁed as

= 0.37



tdE

In general, the problem of ﬂange curling is not a critical

factor to limit the ﬂange width. However, when the appear-

ance of the section is important, the out-of-plane distortion

should be c losely controlled in practice.

Example 4.16 Determine the amount of curling for the

compression ﬂange of the hat section used in Example 4.3

when the section is subjected to the allowable moment. Use

the ASD method.

SOLUTION. The curling of the compression ﬂange of

the hat section can be computed by Eq. (4.126). In the

calculation,

(

15.0 − 2 ×0.105

)

= 7.395 in.

t = 0.105 in.

d = 10.0in.

40.69

1.67

4.934

14.415

= 8.34 ksi

Using Eq. (4.126),

7.395 =



0.061 × 0.105 × 10 ×29,500

8.34



100c

= 26.77

√

= 0.0058 in.

4.3 DESIGN OF BEAM WEBS

4.3.1 Introduction

Not only should thin-walled cold-formed steel ﬂexural

members be designed for bending strength and deﬂection

as discussed in Section 4.2 but also the webs of beams

should be designed for shear, bending, combined bending

and shear, web crippling, and c ombined bending and web

DESIGN OF BEAM WEBS 157

crippling. In addition, the depth of the web should not

exceed the maximum value permitted by Section B1.2 of

the North American Speciﬁcation.

The maximum allowable depth-to-thickness ratio h/t for

unreinforced webs is limited to 200, in which h is the

depth of the ﬂat portion of the web measured along the

plane of the web and t is the thickness of the web. When

bearing stiffeners are provided only at supports and under

concentrated loads, the maximum depth-to-thickness ratio

may be increased to 260. When bearing stiffeners and

intermediate shear stiffeners are used simultaneously, the

maximum h/t ratio is 300. These limitations for h/t ratios

are established on the basis of the studies reported in Refs.

3.60 and 4.56–4.60. If a web consists of two or more sheets,

the h/t ratios of the individual sheets shall not exceed the

maximum allowable ratios mentioned above.

The following discussions deal with the minimum

requirements for bearing and shear stiffeners, the design

strength for shear and bending in beam webs, the load or

reaction to prevent web crippling, and combinations of

various types of strengths.

4.3.2 Stiffener Requirements

Section C3.7 of the 2007 edition of the North American

Speciﬁcation, provides the following design requirements

for attached bearing stiffeners and shear stiffeners. When

the bearing stiffeners do not meet these requirements, the

load-carrying capacity for the design of such members can

be determined by tests.

(a) Bearing Stiffeners. For beams having large h/t ratios,

bearing stiffeners attached to beam webs at supports or

under concentrated loads can be designed as compression

members. The nominal strength, P

, for bearing stiffeners

is the smaller of the values determined by 1 and 2 as

follows:

= F

(4.128)

2. P

= Nominal axial load evaluated according to Section

5.8 with A

replaced by A



= 2.00 (ASD)



0.85 (LRFD)

0.80 (LSD)

where

⎧

⎪

⎨

⎪

⎩

18t

+ A

for bearing stiffeners at interior

support or under concentrated load (4.129)

10t

+ A

for bearing stiffeners

at end support (4.130)

= cross-sectional area of bearing stiffeners

= lower value of F

for beam web,

or F

for stiffener section

⎧

⎪

⎨

⎪

⎩

t +A

for bearing stiffeners at

interior support or under

concentrated load (4.131)

t +A

for bearing stiffeners

at end support (4.132)

= 25t



0.0024





+ 0.72



≤ 25t (4.133)

= 12t



0.0044





+ 0.83



≤ 12t (4.134)

= length of bearing stiffener

t = base steel thickness of beam web

In addition, the speciﬁcation stipulates that w/t

ratios

for the stiffened and unstiffened elements of cold-formed

steel bearing stiffeners should not exceed 1.28



E/F

and

0.42



E/F

, respectively. In the above expressions, F

the yield stress and t

is the thickness of the stiffener steel.

It should be noted that Eq. (4.128) is used to prevent end

crushing of bearing stiffeners, while the second P

is used

to prevent column buckling of the combined web stiffener

section. The equations for computing the effective areas

and A

and the effective widths b

and b

are adopted

from Nguyen and Yu.

4.59

Figures 4.51 and 4.52 show the

effective areas A

and A

of the bearing stiffeners.

(b) Bearing Stiffeners in C-Section Flexural Members. For

two-ﬂange loading (Figs. 4.66c and 4.66d )ofC-section

ﬂexural members with bearing stiffeners that do not meet

the above requirements of Section 4.3.2a, the nominal

strength P

should be determined as follows:

= 0.7(P

+ A

) ≥ P

(4.135)



= 1.70 (ASD)



0.90 (LRFD)

0.80 (LSD)

where P

is the nominal web crippling strength for C-

section ﬂexural members calculated in accordance with Eq.

(4.170) for single web members, at end or interior locations;

is the effective area of the bearing stiffener subjected

to uniform compressive stress, calculated at yield stress;

and F

is the yield stress of the bearing stiffener steel.

Equation (4.135) is based on the research conducted by Fox

and Schuster at the University of Waterloo.

1.299,4.229–4.231

The program investigated the behavior of 263 stud- and

truck-type bearing stiffeners in cold-formed steel C-section

158 4 FLEXURAL MEMBERS

Figure 4.51 Effective area A

of bearing stiffener: (a) at end support; (b) at interior support and

under concentrated load.

Figure 4.52 Effective area A

of bearing stiffener: (a) at end support; (b) at interior support and

under concentrated load.

ﬂexural members subjected to two-ﬂange loading at both

interior and end locations. This equation is applicable

within the following limits:

1. Full bearing of the stiffener is required. If the bearing

width is narrower than the stiffener such that one of the

stiffener ﬂanges is unsupported, P

shall be reduced by

50%.

2. Stiffeners are C-section stud or track members with a

minimum web depth of 3

in. (89 mm) and minimum

base steel thickness of 0.0329 in. (0.84 mm).

3. The stiffener is attached to the ﬂexural member web with

at least three fasteners (screws or bolts).

4. The distance from the ﬂexural member ﬂanges to the ﬁrst

fastener(s) is not less than d /8, where d is the overall

depth of the ﬂexural member.

5. The length of the stiffener is not less than the depth of

the ﬂexural member minus

in. (9 mm).

6. The bearing width is not less than 1

in. (38 mm).

to satisfy the following requirements for spacing, moment

of inertia, and gross area:

1. Spacing a between Stiffeners. When shear stiffeners are

required, the spacing shall be based on the nominal

shear strength V

permitted by Section 4.3.3.2e and the

following limits:

a ≤



260

h/t



h (4.136)

a ≤ 3h (4.137)

2. Moment of Inertia I

of Shear Stiffeners. With reference

to an axis in the plane of the web, the moment of inertia

of a pair of attached stiffeners or of a single stiffener

shall satisfy the following requirements:

≥ 5ht



−

0.7a



(4.138)

≥





(4.139)

3. Gross area A

of Shear Stiffeners. The area A

shall

satisfy the requirement of Eq. (4.140):

≥

1 − C



−

(

a/h

)

(

a/h

)



1 +

(

a/h

)



YDht

(4.140)

where

⎧

⎪

⎨

⎪

⎩

1.53Ek

(h/t)

when C

≤ 0.8 (4.141)

1.11

h/t



when C

> 0.8 (4.142)

⎧

⎪

⎨

⎪

⎩

4.00 +

5.34

(a/ h)

when a/h ≤ 1.0 (4.143)

5.34 +

4.00

(a/ h)

when a/h > 1.0 (4.144)

and a = distance between shear stiffeners

DESIGN OF BEAM WEBS 159

Y = yield stress of web steel/yield stress of

stiffener steel

D = 1.0 for stiffeners furnished in pairs

= 1.8 for single-angle stiffeners

= 2.4 for single-plate stiffeners

Most of the above requirements for shear stiffeners

are based on the AISC Speciﬁcation

1.148

and the study

reported in Ref. 4.59.

(d) Nonconforming Stiffeners. According to Section C3.7.4

of the North American speciﬁcation, the available strength

of members with stiffeners that do not meet the require-

ments of Section 4.3.2a, 4.3.2b, or 4.3.2c, such as stamped

or rolled-in stiffeners, shall be determined by tests in accor-

dance with Chapter F of the speciﬁcation or rational engi-

neering analysis in accordance with Section A1.2(b) of the

Speciﬁcation.

4.3.3 Shear

4.3.3.1 Shear Stress In the design of beams, the actual

shear stress developed in the cross section of the beam

can be calculated by using the following well-known

equation

4.61

(4.145)

where f

= actual shear stress

V = total external shear force at a section

Q = static moment of area between the extreme

ﬁber and the particular location at which the

shear stress is desired, taken about neutral axis

I = moment of inertia of entire cross-sectional

area about neutral axis

t = width of section where shear stress is

desired

Even though Eq. (4.145) gives the exact value at any

location, it has been a general practice to use the average

value in the gross area of the web as the shear stress for

design purposes. This a verage shear stress can be computed

by using the following equation:

(4.146)

where h = depth of the ﬂat portion of the web measured

along the plane of the web

= thickness of the web

The use of Eqs. (4.145) and (4.146) is illustrated in Example

4.17.

Example 4.17 Determine the shear stress distribution at

the end supports of the uniformly loaded channel shown in

Fig. 4.53. Assume that the load is applied through the shear

center of the cross section so that torsion is not involved.

∗

See Appendix B for a discussion of the s hear center.

SOLUTION

1. Exact Shear Stress Distribution Using Eq. (4.145 ). For

simplicity, use square corners and midline dimensions

as shown in Fig. 4.54 for computing the exact shear

stresses at various locations of the section.

(a) Shear stress at points 1 and 4:

V = R

= 1.5kips

= Q

= 0

(

)

1,4

= 0

(b) Shear stress at points 2 and 3:

= Q

= 1.4325

(

0.135

)



× 6.865



= 0.664 in.

(

)

2,3

1.5

(

0.664

)

7.84

(

0.135

)

= 0.941 ksi

= Q

+ 0.135



× 6.865



× 6.865



= 1.46 in.

(

)

1.5

(

1.46

)

7.84

(

0.135

)

= 2.07 ksi

2. Average Shear Stress on Beam Web by Using Eq.

(4.146)

1.5

6.73

(

0.135

)

= 1.65 ksi

From the above calculation it can be seen that for the

channel section used in this example the average shear

stress of 1.65 ksi is 25% below the maximum value of

2.07 ksi.

4.3.3.2 Shear Strength of Beam Webs without Holes

(a) Shear Yielding. When a beam web with a relatively

small h/t ratio is subject to shear stress, the shear capacity

of the beam is probably governed by shear yielding. The

maximum shear stress at the neutral axis can be computed

by Eq. (4.147):

√

(4.147)

∗

When the load does not pass through the shear center, see Appendix B.

160 4 FLEXURAL MEMBERS

Figure 4.53 Example 4.17.

Figure 4.54 Shear stress distribution.

in which τ

is the yield stress in shear and F

is the yield

stress in tension.

The nominal shear strength for yielding can be deter-

mined by the shear stress given in Eq. (4.147) and the web

area, ht, as follows:



√



(

)

∼

0.60F

ht (4.148)

where V

is the nominal shear strength, h is the depth of

the ﬂat portion of the web, and t is the thickness of the

web.

(b) Elastic Shear Buckling. For webs with large h/t ratios,

the shear capacity of the web is governed by shear buck-

ling. Figure 4.55 shows a typical pattern of shear failure.

4.56

Based on studies by Southwell and Skan on shear buck-

ling of an inﬁnitely long plate, the plate develops a series

of inclined waves, as shown in Fig. 4.56.

4.62,4.63

The

elastic critical shear buckling stress can be computed by

Eq. (4.149)

∗



1 − μ



(

h/t

)

(4.149)

where k

= shear buckling coefﬁcient

E = modulus of elasticity of steel

μ = Poisson’s ratio

h = depth of plate

t = thickness of plate

In Eq. (4.149), the value of k

varies with the supporting

conditions and the aspect ratio a/h (Fig. 4.57), in which a

is the length of the plate. For a long plate the value of k

∗

The problem of shear buckling of plane plates has also been studied by

Timoshenko and other investigators. For additional information see Refs.

3.1 and 4.63.

DESIGN OF BEAM WEBS 161

Figure 4.55 Typi cal shear failure pat tern (h/t = 125).

4.56

Figure 4.56 Shear buckling of inﬁnitely long plate

4.63

:(a) simply supported edges; (b)ﬁxed

edges.

was found to be 5.34 for simple supports and 8.98 for ﬁxed

supports, as listed in Table 3.2.

Substituting μ = 0.3 in Eq. (4.149),

0.904k

(h/t)

ksi (4.150)

Thus if the computed theoretical value of τ

is less than

the proportional limit in shear, the nominal shear strength

for elastic buckling can be obtained from Eq. (4.151):

0.904k

(h/t)

(h/t) =

0.904k

(4.151)

moderate h/t ratios, the computed theoretical value of τ

162 4 FLEXURAL MEMBERS

Figure 4.57 Shear buckling stress coefﬁcient of plates versus

aspect ratio a/h.

3.1

may exceed the proportional limit in shear. The theo-

retical value of the critical shear buckling stress should

be reduced according to the change in the modulus of

elasticity. Considering the inﬂuence of strain hardening

observed in the investigation of the strength of plate girders

in shear,

4.64

Basler indicated that Eq. (4.152) can be used

as the reduction formula:

√

cri

(4.152)

where τ

= 0.8τ

= 0.8(F

√

3) is the proportional limit

in shear and the initial critical shear buckling stress is

given as

cri



1 − μ



(

h/t

)

(4.153)

By substituting the values of τ

and τ

cri

into Eq. (4.152),

one can obtain Eq. (4.154) for the shear buckling stress in

the inelastic range, that is,

0.64



h/t

(4.154)

Consequently, the nominal shear strength in the inelastic

range can be obtained from Eq. (4.155):

0.64



h/t

(ht) = 0.64t



E (4.155)

(d) Safety Factors. Prior to 1996, the AISI ASD speciﬁca-

tion employed three different safety factors (i.e., 1.44 for

yielding, 1.67 for inelastic buckling, and 1.71 for elastic

buckling) for determining the allowable shear stresses in

order to use the same allowable values for the AISI

and AISC Speciﬁcations. To simplify the design of shear

elements for using the allowable stress design method, the

safety factor for both elastic and inelastic shear buckling

was taken as 1.67 in the 1996 edition of the AISI Speciﬁ-

cation. The safety factor of 1.50 was used for shear yielding

to provide the allowable shear stress of 0.40F

, which has

been used in steel design for a long time. The use of such

a smaller safety factor of 1.50 for shear yielding was justi-

ﬁed by long-standing use and by the minor consequences

of incipient yielding in shear as compared with those asso-

ciated with yielding in tension and compression.

In the 2001 and 2007 editions of the North American

Speciﬁcation, the constant used in Eq. (4.154) for deter-

mining the inelastic shear buckling stress was reduced

slightly from 0.64 to 0.60 on the basis of Craig’s calibra-

tion of the test data of Laboube and Yu.

1.346,4.56,4.232,4.233

For the purpose of simplicity, a single safety factor of

1.60 was used in Section C 3.2.1 of the Speciﬁcation for

shear yielding, elastic and inelastic shear buckling for the

ASD method with a corresponding resistance factor of

0.95 for LRFD and 0.80 for LSD. With these minor revi-

sions of safety factor and design equations, for ASD, the

North American Speciﬁcation reduces 6% of the a llow-

able inelastic shear buckling strength as compared with

the AISI 1996 ASD Speciﬁcation. For beam webs with

large h/t ratios, the North American Speciﬁcation allows

4% increase of the allowable s trength for the elastic shear

buckling.

(e) North American Design Criteria for Shear Strength of

Webs without Holes. Based on the foregoing discussion of

the shear strength of beam webs, the 2007 edition of the

North American Speciﬁcation includes the following design

provisions in Section C3.2.1 for the ASD, LRFD, and LSD

methods

1.345

C3.2.1 Shear Strength [Resistance] of Webs

without Holes

The nominal shear strength [resistance], V

, shall be calculated

in accordance with Eq. (4.156). The safety factor and resistance

factors given in Section C3.2.1 of the speciﬁcation shall be

used to determine the allowable shear strength or design shear

strength [factored shear resistance] in accordance with the

applicable design method in Section 3.3.1, 3.3.2 or 3.3.3:

= A

(4.156)



= 1.60 (ASD)

= 0.95 (LRFD)

= 0.80 (LSD)

DESIGN OF BEAM WEBS 163

(a) For h/t ≤



= 0.60F

(4.157a)

(b) For



<h/t≤ 1.51



0.60



(h/t)

(4.157b)



12(1 − μ

)(h/t)

0.904Ek

(h/t)

(4.157c)

where V

is the nominal shear strength [resistance] and

= area of web element = ht (4.158)

where h = depth of the ﬂat portion of the web measured

along the plane of the web

t = web thickness

= nominal shear stress

E = modulus of elasticity of steel

= shear buckling coefﬁcient calculated in

accordance with (1) or (2) as follows:

1. For unreinforced webs, k

= 5.34.

2. For beam webs with bearing stiffeners satisfying the

requirements of Section 4.3.2

⎧

⎪

⎨

⎪

⎩

4.00 +

5.34

(a/ h)

when

≤ 1.0 (4.159a)

5.34 +

4.00

(a/ h)

when

> 1.0 (4.159b)

where a = shear panel length for unreinforced web

element

= clear distance between transverse sti ffeners

for reinforced web element

= design yield stress as determined in

accordance with Section 2.2

μ = Poisson’s ratio, =0.3

For a web consisting of two or more sheets, each sheet shall

be considered as a separate element carrying its share of the

shear force.

For the ASD method, the allowable shear stresses in webs

are shown in Fig. 4.58 by using E = 29,500 ksi (203 GPa,

or 2.07×10

kg/cm

). Table 4.7 gives the allowable shear

stresses for F

= 33 and 50 ksi (228 and 345 MPa, or 2320

and 3515×10

kg/cm

). Examples 4.18–4.20 illustrate the

applications of Eqs. (4.156)–(4.159).

Example 4.18 Use the ASD and LRFD methods to deter-

mine the available shear strength for the I-section used in

Example 4.1. Use F

= 50 ksi.

Figure 4.58 Allowable shear stress in webs for the ASD method.

SOLUTION

A. ASD Method.

The depth-to-thickness ratio of each individual web

element is

8 − 2(0.135 + 0.1875)

0.135

7.355

0.135

= 54.48

Based on the North American design criteria, the value of

for unreinforced webs is 5.34. Therefore, according to

Section C3.2.1 of the North American Speciﬁcation,





(29,500)(5.34)

= 56.13

1.51



= 1.51



(29,500)(5.34)

= 84.76

Since h/t <



, use Eqs. (4.156) and (4.157a) to

compute the nominal shear strength of each web element

as follows:

= A

= (ht)(0.6F

)

= (7.355 × 0.135)(0.60 × 50) = 29.79 kips

164 4 FLEXURAL MEMBERS

Table 4.7 Allowable Shear Stresses for ASD Method, ksi

= 33 ksi F

= 50 ksi

a/h a/h

h/t 0.5 1.0 2.0 3.0 >3 0.5 1.0 2.0 3.0 >3

50 12.4 12.4 12.4 12.4 12.4 18.8 18.8 18.8 18.8 18.8

60 12.4 12.4 12.4 12.4

12.4 18.8 18.8 18.8 18.3 17.5

70 12.4 12.4

12.4 12.4 12.2 18.8 18.8 16.4 15.6 15.0

80 12.4 12.4 11.6 11.1 10.7 18.8 17.4 14.3

13.7 13.2

90 12.4

12.4 10.4 9.9 9.5 18.8 15.5 12.7 11.9 11.0

100 12.4 11.3 9.3

8.9 8.6 18.8 13.9 10.6 9.6 8.9

110 12.4 10.3

8.5 8.0 7.4 18.8 12.7 8.7 8.0 7.4

120 12.4 9.4 7.3 6.7 6.2

18.8 10.8 7.3 6.7 6.2

130 12.4

8.7 6.3 5.7 5.3 17.6 9.2 6.3 5.7 5.3

140 12.4 7.9 5.4 4.9 4.5 16.4 7.9 5.4 4.9 4.5

150

12.4 6.9 4.7 4.3 4.0 15.3 6.9 4.7 4.3 4.0

160 11.6 6.1 4.1 3.8 3.5 14.3 6.1 4.1 3.8 3.5

170 11.0 5.4 3.7 3.3 3.1 13.5 5.4 3.7 3.3 3.1

180 10.4 4.8 3.3 3.0 2.7

12.7 4.8 3.3 3.0 2.7

190 9.8 4.3 2.9 2.7 2.5 11.7 4.3 2.9 2.7 2.5

200 9.3 3.9 2.6 2.4 2.2 10.6 3.9 2.6 2.4 2.2

220

8.5 3.2 2.2 2.0 1.8 8.7 3.2 2.2 2.0 1.8

240 7.3 2.7 1.8 1.7 1.5 7.3 2.7 1.8 1.7 1.5

260 6.3 2.3 1.6 1.4 1.3 6.3 2.3 1.6 1.4 1.3

280 5.4 2.0 1.3 1.2 1.1 5.4 2.0 1.3 1.2 1.1

300 4.7 1.7 1.2 1.1 1.0 4.7 1.7 1.2 1.1 1.0

Notes:

1. Values above the single underlines are based on Eq. (4.157a); values between the single underlines and double underlines are based

on Eq. (4.157b); and values below the double underlines are based on Eq. (4.157c). For the case of a/h > 3, k

= 5.34.

2. 1 ksi = 6.9 MPa or 70.3 kg/cm

The allowable shear strength for the I-section having two

webs is



(

29.79

)

1.60

= 37.24 kips

B. LRFD Method.

Using the same nominal shear strength computed in item

A, the design shear strength for the I-section having double

webs is

2φ

= 2(0.95)(29.79) = 56.60 kips

Example 4.19 Use the ASD and LRFD methods to deter-

mine the available s hear strength for the channel section

used in Example 4.2. Use F

= 50 ksi.

SOLUTION

A. ASD Method.

For the given channel section, the depth-to-thickness ratio

of the web is

10 − 2

(

0.075 + 0.09375

)

0.075

9.6625

0.075

= 128.83

Based on k

= 5.34 and F

= 50 ksi,

1.51



= 1.51



(29,500)(5.34)

= 84.76

Since h/t > 1.51



, use Eq. (4.157c) to compute

the nominal shear strength V

as follows:

= A

= (ht)



0.904Ek

(h/t)



= (9.6625 × 0.075)



0.904 × 29500 × 5.34

(128.83)



= 6.22 kips

The allowable shear strength using the ASD method for the

design of the given channel section is



6.22

1.60

= 3.89 kips