ARCH 631 Note Set 21.3 F2014abn

1

Examples:

Steel

Example 1 (AISC Design Examples vV13.0)

Select a W18x50 from Table 3-2. Determined the required flexural strength with the self weight.

LRFD ASD

wu-s.w. = 1.2(0.040 kip/ft) = 0.048 kip/ft

ftkiip

2kip/ft

ftkip 273.48

ft) (35.00.048266

adjusteduM

wu-s.w. = 0.040 kip/ft)

ftkiip

2kip/ft

ftkip .11908

ft) (35.00.040841

adjusteduM

273.4 kip-ft o.k. 190.1 kip-ft o.k.

ARCH 631 Note Set 21.3 F2014abn

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Example 1 (continued)

I required is in this grouping, with the W21x44 (bold) the most economical. But this section must be 18 inches maximum, and the W18 x 46 does not have enough (even though it has enough

moment capacity of 340 k-ft (bMpx)) Look to the next section above for a W18

with I > 748 in4.

273.4 kip-ft 190.1 kip-ft

ARCH 631 Note Set 21.3 F2014abn

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Example 1 (continued)

For Example 1 (F1-1a), the unbraced length was zero. There is no zero on the chart, so the far left is used starting at the moment required of 273.4 k-ft. When a W18 is not encountered with a greater moment capacity going up on the page, going to the right will intersect with a W18 line.

For this example (F1-2a), the unbraced length is 11.7 ft. The same procedure applies, starting at a moment required of 273.4 k-ft. It shows that the W18x50 is not the most economical for the capacity with the dashed line. The most economical would have been the W21x48, but it is too deep. The W18x55 looks “economical” with the solid line, but it weighs 5 more pounds per foot, and this design is governed by the deflection.

F1-1a

F1-2a

273.4

273.4 kip-ft o.k. 190.1 kip-ft o.k.

ARCH 631 Note Set 21.3 F2014abn

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Top values are total factored distributed load from strength and deflection criteria.

Values below in gray are for live load deflection limit (unfactored).

LRFD STANDARD LOAD TABLE FOR OPEN WEB

STEEL JOISTS, K-SERIES Based on a 50 ksi Maximum Yield Strength – Loads shown in Pounds per Linear Foot (plf)

256 288 + 72 600 + 960

+ 360 = 13,744 lbs.

256 288 + 72 600

960 660

6773

6971

6971 465

6971 360+600+960 288+72

14.03

6971(14.03) – 360(11.03) -

960(7.03) – 600(5.03) - (288+72)(14.03)

2

48,634

48,634 = 432.3

465

Select 18K7 for total load (502) and live load (180) and call it: 18K9SP

Example 2 (LRFD)

ARCH 631 Note Set 21.3 F2014abn

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Example 3 (AISC Design Examples vV13.0)

ARCH 631 Note Set 21.3 F2014abn

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Example 3 (continued)

ARCH 631 Note Set 21.3 F2014abn

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Example 4 (LRFD) Investigate the accepatbility of a W16 x 67 used as a beam-column under the unfactored

loading shown in the figure. It is A992 steel (Fy = 50 ksi). Assume 25% of the load is dead

load with 75% live load.

SOLUTION:

DESIGN LOADS (shown on figure):

Axial load = 1.2(0.25)(350k)+1.6(0.75)(350k)=525k

Moment at joint = 1.2(0.25)(60 k-ft) + 1.6(0.75)(60 k-ft) = 90 k-ft

Determine column capacity and fraction to choose the appropriate interaction equation:

925966

1215.

in.

)(ft

r

kL ftin

x

and 73462

1215

in.

)(ft

r

kL ftin

y

(governs)

k.in.)ksi.(AFPP gcrcncc 85600719530 2

2087085600

525..

k.

k

P

P

c

r so use 019

8.

M

M

M

M

P

P

nyb

uy

nxb

ux

nc

u

There is no bending about the y axis, so that term will not have any values.

Determine the bending moment capacity in the x direction:

The unbraced length to use the full plastic moment (Lp) is listed as 8.69 ft, and we are over that so of we don’t want to determine it from formula, we can find the beam in the

Available Moment vs. Unbraced Length tables. The value of Mn at Lb =15 ft is 422 k-ft.

Determine the magnification factor when M1 = 0, M2 = 90 k-ft:

016090

0604060

2

1 ...M

M..C

ftk

ftk

m

k.,

.

in.)ksix(

rKl

EAPe 46958

925

71910302

232

2

2

1

01640486955251

60

1 1

1 ..)k.k(

.

)PP(

CB

eu

m

USE 1.0 Mu = (1)90 k-ft

Finally, determine the interaction value:

01061422

90

9

8870

9

8...

M

M

M

M

P

Pftk

ftk

nyb

uy

nxb

ux

nc

u

525 k

This is NOT OK. (and outside error tolerance).

The section should be larger.

525 k

90 k-ft 90 k-ft

525 k

525 k