aect460 lecture 3

7/28/2019 AECT460 Lecture 3

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Lecture 3 Page 1 of 6

Lecture 3 ASD and LRFD Princip les

Load and Resistance Factor Design (LRFD)see AISC p. 2-6 and 16.1-213 thru 217

The LRFD was developed in the 1980s as an alternative design method to the tried-and-true Allowable Stress Design (ASD) method. It is based on a limit state

philosophy. A limit state is a term used to describe a condition in which thestructure ceases to perform as intended. A strength limit state defines the safetyagainst failure due to loading and a serviceability limit state is a functionalrequirement such as deflection, drift or vibration.

In general, the LRFD method uses a statistical approach in determining factoredloads that are compared against ultimate member strengths. In other words:

iQi


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LRFD Load Factors: see AISC p. 2-8

The following 6 load factors are used to obtain the most severe factored loads:

1) 1.4D2) 1.2D + 1.6L + 0.5(Lr or S or R)3) 1.2D + 1.6(Lr or S or R) + (0.5L or 0.8W)

4) 1.2D + 1.6W + 0.5L + 0.5(Lr or S or R)5) 1.2D + 1.0E + 0.5L + 0.2S6) 0.9D + (1.6W or 1.0E)

where: D = service dead loadsL = service floor live loadLr = service roof live loadS = snow loadR = rainwater loadW = wind load

E = earthquake load

Example 1GIVEN: A flat roof is framed with 24-0 long W18x40 beams spaced 8-0 o.c. Theservice applied roof dead load is 25 PSF and the applied service roof live load = 20PSF (per IBC ch. 16). The service wind load on the flat roof is -8 PSF (uplift).

REQUIRED:1) Determine the maximumLRFD factored uniform load on the beam, wu.

2) Determine the maximum LRFD factored moment on the beam, Mu.

24-0

wu

W

18x40

W

18x40

Typ.

W

18x40

24-0

8-0

Roof Framing Plan

Beam Loading F.B.D


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Step 1 Determine D, L r and W in terms of PLF:

D = DL(Trib. Width) + Beam wt.= 25 PSF(8 ft) + 40 PLF= 240 PLF

Lr = LL(Trib. Width)

= 20 PSF(8 ft)= 160 PLF

W = -8 PSF(8 ft)= -64 PLF

Step 2 Determine maximum FACTORED uniform load, wu:

1) 1.4D1.4(240 PLF) = 336 PLF

2) 1.2D + 1.6L + 0.5(Lr or S or R)1.2(240 PLF) + 0.5(160 PLF) = 368 PLF

3) 1.2D + 1.6(Lr or S or R) + (0.5L or 0.8W)

1.2(240 PLF) + 1.6(160 PLF) =544 PLF USE

4) 1.2D + 1.6W + 0.5L + 0.5(Lr or S or R)1.2(240 PLF) + 0.5(160 PLF) = 368 PLF

5) 1.2D + 1.0E + 0.5L + 0.2S1.2(240 PLF) = 288 PLF

6) 0.9D + (1.6W or 1.0E)0.9(240 PLF) + 1.6(-64) = 114 PLF OR 318 PLF

Step 3 Determine maximum FACTORED moment on beam, Mu:

Mu =8

2Lwu

=8

)"0'24)(544( 2PLF

= 39,168 ft-lb

Mu = 39.2 kip-ft

Do not use neg. wind if itreduces pos. gravity loads

Do not use neg. wind if itreduces pos. gravity loads


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Al lowable Stress Design (ASD) See AISC p. 2-7 and 16.1-216 thru 217

The Allowable Stress Design (ASD) method is based on the concept that the stresslevels in a component do not exceed established specific allowable stresses underservice loads. For any single component, there may be several different allowablestress limits that must be checked.

The basic design equation for ASD is as follows:

n

i

RQ

where: Qi = working or service load (see IBC ch. 16)Rn = nominal resistance strength of member

= (omega) safety factor, see AISC p. 2-10= 1.67 for limit-states involving yielding= 2.00 for limit-states involving rupture

=

5.1

ASD Load Factors: see AISC p. 2-9

The following 7 load factors are used to obtain the most severe loads:

1) D2) D+L3) D+(Lr or S or R)

4) D+0.75L+0.75(Lr or S or R)5) D+(W or 0.7E)6) D+0.75(W or 0.7E)+0.75L+0.75(Lr or S or R)7) 0.6D+(W or 0.7E)

where: D = service dead loadsL = service floor live loadLr = service roof live loadS = snow loadR = rainwater load

W = wind loadE = earthquake load


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Example 2GIVEN: Similar to Example 1, a flat roof is framed with 24-0 long W18x40 beamsspaced 8-0 o.c. The service applied roof dead load is 25 PSF and the appliedservice roof live load = 20 PSF (per IBC ch. 16). The service wind load on the flatroof is -8 PSF (uplift).

REQUIRED:1) Determine the maximumASD service uniform load on the beam, w.2) Determine the maximumASD service moment on the beam, Mmax.

Step 1 Determine D, L r and W in terms of PLF:

D = DL(Trib. Width) + Beam wt.= 25 PSF(8 ft) + 40 PLF= 240 PLF

Lr = LL(Trib. Width)

= 20 PSF(8 ft)= 160 PLF

W = -8 PSF(8 ft)= -64 PLF

Step 2 Determine maximum SERVICE uniform load, w:

1) D240 PLF

2) D+L240 PLF + 0 = 240 PLF

3) D+(Lr or S or R)

240 PLF + 160 PLF =400 PLF USE

4) D+0.75L+0.75(Lr or S or R)240 PLF + 0 + 0.75(160 PLF) = 360 PLF

W18x40

W18x40

Typ.

W18x40

24-0

8-0Roof Framing Plan

24-0

w

Beam Loading F.B.D


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5) D+(W or 0.7E)240 PLF + (-64 PLF) = 176 PLF240 PLF (-64 PLF) = 304 PLF

6) D+0.75(W or 0.7E)+0.75L+0.75(Lr or S or R)240 PLF + 0 + 0.75(160 PLF) = 360 PLF

7) 0.6D+(W or 0.7E)0.6(240 PLF) + (-64 PLF) = 80 PLF0.6(240 PLF) (-64 PLF) = 208 PLF

Step 3 Determine maximum SERVICE moment on beam, Mmax:

Mmax =8

2wL

=8

)"0'24)(400( 2PLF

= 28,800 ft-lb

Mmax = 28.8 kip-ft

Do not use neg. wind if it

reduces pos. gravity loads

aect460 lecture 3

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