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F-20

Example F.3a W-Shape Flexural Member with NoncompactFlanges in Strong-Axis Bending

Given:

Select an ASTM A992 W-shape beam with a simple span of 40 feet. The nominal loads are a

uniform dead load of 0.05 kip/ft and two equal 18 kip concentrated live loads acting at the

third points of the beam. The beam is continuously braced. Also calculate the deflection.

Note: A beam with noncompact flanges will be selected to demonstrate that the tabulated

values of the Steel Construction Manualaccount for flange compactness.

Solution:

Material Properties:

ASTM A992 Fy = 50 ksi Fu = 65 ksi

Calculate the required flexural strength at midspan

LRFD ASD

wu = 1.2(0.0500 kip/ft) = 0.0600 kip/ftPu = 1.6(18.0 kips) = 28.8 kips

Mu =

20.0600 kip/ft 40.0 ft

8

+ 40.0 ft

28.8kips3

= 396 kip-ft

wa = 0.0500 kip/ftPa = 18.0 kips

Ma =

20.0500 kip/ft 40.0 ft

8

+ 40.0 ft

18.0kips3

= 250 kip-ft

Select the lightest section with the required strength from the bold entries in Manual Table 3-2

SelectW21u48.

This beam has a noncompact compression flange atFy = 50 ksi as indicated by footnote f in

Manual Table 3-2. This is also footnoted in Manual Table 1-1.

Manual

Table 3-2

ManualTable 1-1

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Check the available strength

LRFD ASD

IbMn = IbMpx = 398 kip-ft > 396 kip-ft o.k. : :pxn

b b

MM= 265 kip-ft > 250 kip-ft o.k.

Manual

Table 3-2

Note: the value Mpx in Table 3-2 include strength reductions due the noncompact nature of theshape

Calculate deflection

Ix = 959 in.4

'max =4

5

384

wl

EI+

3

28

Pl

EI

=

4 3

4

5 0.0500 kip/ft 40.0 ft 12 in./ft

384 29,000 ksi 959 in.+

3 3

4

18.0 kips 40.0 ft 12 in./ft

28 29,000ksi 959in

= 2.66 in.

This deflection can be compared with the appropriate deflection limit for the application.Deflection will often be more critical than strength in beam design.

Manual

Table 1-1

Manual

Table 3-23Diagrams 1

and 9

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F-22

Example F.3b W-Shape Flexural Member with NoncompactFlanges in Strong-Axis Bending

Given:

Verify the results from Example F.3a by calculation using the provisions of the Specification.

Solution:

Material Properties:

ASTM A992 Fy = 50 ksi Fu = 65 ksi

Geometric Properties:

W21u48 Sx = 93.0 in.3 Zx = 107 in.3

Check flange slenderness

O =2

f

f

b

t= 9.47

The limiting width-thickness ratios for the compression flange are:

Opf= 0.38y

E

F=

29,000 ksi0.38

50 ksi= 9.15

Orf= 1.0y

E

F=

29,000 ksi1.0

50 ksi= 24.1

Orf > O > Opf; therefore, the compression flange is noncompact. This could also be determinedfrom the footnote f in Manual Table 1-1.

Calculate the nominal flexural strength, Mn

Since the beam is continuously braced, and therefore not subject to lateral-torsional buckling,the available strength is governed by Section F3

Mp =FyZx = 50 ksi(107 in.3) = 5350 kip-in. or 446 kip-ft.

Mn = pf

0.7pf

p p y x

rf

M M F S O O

O O

Mn = 39.47 9.15

5350kip-in. 5350kip-in 0.7 50ksi 93.0in.24.1 9.15

= 5310 kip-in. or 442 kip-ft.

ManualTable 2-3

ManualTable 1-1

Manual

Table 1-1

Table B4.1

Case 1

Eqn. F3-1

Calculate the available flexural strength

LRFD ASD

Ib = 0.90IbMn = 0.90(442 kip-ft)

= 398 kip-ft > 396 kip-ft o.k.

:b = 1.67Mn:b = 442 kip-ft / 1.67

= 265 kip-ft > 250 kip-ft o.k.

Section F1

Note that these available strength values are identical to the tabulated values in Manual Table3-2, which account for the non-compact flange.