Beam Design for Moment Asd Lrfd (Example)

Lecture 6 - Page 1 of 13

Lecture 6 – Beam Design for Moment Beams, or sometimes referred to as “flexure” members, are designed on the basis of moment. In ASD and LRFD, the design strength of a beam in flexure is called the “nominal flexural moment”, Mn. This Mn is multiplied by φ (LRFD) or divided by Ω (ASD) to obtain the available moment capacity. This available moment capacity must be greater than the maximum applied factored (LRFD) or service (ASD) moment. Beams are designed on the basis of the following LRFD references:

• AISC Part 3 • AISC Spec Chapter F p. 16.1-44

1. Beam Design Considering Yielding:

Assuming a beam is adequately laterally braced, it will fail by yielding on the compression flange. Most beams are laterally braced by the metal decking that is attached to the compression flange as shown below:

1) LRFD Beam Design:

Available Design Flexural Strength = φbMn Where: φb = 0.90 Mn = nominal flexural moment = Mp = Plastic moment = FyZx

Zx = plastic section modulus, in3 = from properties

Concrete slab over metal decking

Metal decking puddle-welded to top flange of beam


2) ASD Beam Design:

Available Design Flexural Strength = b

nMΩ

Where: Ωb = 1.67 Mn = nominal flexural moment = Mp = Plastic moment = FyZx

Zx = plastic section modulus, in3 = from properties Example 1 (LRFD) GIVEN: A W16x26 steel beam using A992 steel is continuously laterally braced, and experiences a FACTORED moment = 104 KIP-FT. REQUIRED: 1) Determine the design flexural moment, φbMn for the beam. 2) Determine if the beam is adequate.

Step 1 – Determine φbMn for the beam:

φbMn = 0.90(FyZx) since it is continuously laterally braced = 0.90(50 KSI)(44.2 in3) = 1989 KIP-IN φbMn = 165.8 KIP-FT

Step 2 – Determine if the beam is adequate:

Since φbMn = 165.8 KIP-FT > 104 KIP-FT → beam is adequate

From properties, AISC p. 1-21


Example 2 (LRFD) GIVEN: The W16x26 beam from Example 1. REQUIRED: Determine the design flexural moment, φbMn for the beam using the LRFD “Zx” Table 3-2 (see AISC p. 3-11 thru 3-19) Step 1 – Refer to AISC p. 3-18 for W16x26:

Look in the LRFD column φbMpx = 166 KIP-FT

Beam in BOLD is lightest weight for that grouping


Example 3 (ASD) GIVEN: An A992 steel beam “A” is continuously laterally braced and carries a superimposed SERVICE (i.e., not factored) floor live load = 100 PSF and a superimposed SERVICE dead load = 85 PSF. Assume initially the beam weighs 30 PLF and check your results. REQUIRED: 1) Design the lightest weight steel beam using the “Zx” Table 3-2. 2) Design the lightest weight W14 steel beam using the “Maximum Total

Uniform Load Table 3-6” (see AISC p. 3-33 thru 3-95).

Step 1 – Determine SERVICE uniform load, “w” on beam:

By inspecting the ASD load factors AISC p. 2-9, the maximum applied uniform load w = D + L Where: D = uniform dead load, PLF = 7’(85 PSF) + 30 PLF = 625 PLF L = uniform live load, PLF = 7’(100 PSF) = 700 PLF w = (625 PLF) + (700 PLF) = 1325 PLF = 1.325 KLF

Gird

er “A

”

Beam “A”

32’-0”

3@7’

-0” =

21’

-0”

Assumed beam wt.


Step 2 – Determine maximum SERVICE moment, Ma:

Ma = 8

2wL

= 8

)'32(325.1 2KLF

Ma = 169.6 KIP-FT

Step 3 – Select lightest weight beam from Table 3-2:

From AISC p. 3-17 look under ASD column b

pxMΩ

to find a

BOLD moment that is equal or larger than the calculated Ma:

Use W16x40 → b

pxMΩ

= 182 KIP-FT > 169.6 KIP-FT

→ (NOTE: If the assumed beam weight of 30 PLF were increased to 40 PLF, the revised Ma = 170.9 KIP-FT which is still less than 182 KIP-FT)

R2 = 21.2 KIPS R1 = 21.2 KIPS

32’-0”

w = 1.325 KLF


Step 4 – Determine total SERVICE uniform load on beam:

W = Total service load on beam, KIPS = w x span = 1.325 KLF(32’-0”) = 42.4 KIPS

Step 5 – Select lightest W14 beam using Table 3-6 “Maximum Total Uniform Load” tables:

From AISC p. 3-67, look find span = 32’ then read across to find the lightest weight beam having maximum ASD total uniform load > 42.4 KIPS. Use W14x43 → max. total unif. load = 43.4 KIPS > 42.4 KIPS → (NOTE: Using this method, the lightest possible beam is W16x40 → max. total unif. load = 45.5 KIPS > 42.4 KIPS)


2. Beam Design Considering “Lateral-Torsional Buckling”

The compression flange of a beam behaves like a column – it is susceptible to buckling if not adequately laterally braced. This phenomenon of flange buckling of a beam is referred to as “lateral-torsional buckling.” Consider a skinny yard stick under loading. If it is not properly braced, it will twist and fail under a much smaller load than if it were adequately braced. Therefore, wider-flanged beams are better at resisting lateral-torsional buckling than narrow-flanged beams.

It has been determined that the relationship between distance between lateral bracing and moment capacity looks like the following:

Distance between lateral supports

Mom

ent c

apac

ity


The AISC has developed graphs of LRFD and ASD beam design

moments, φbMn, and b

nMΩ

respectively, for beam shapes that has been

mathematically altered from the graph above (for simplicity), and looks like the following: (See AISC p. 5-7 and AISC p. 16.1-33)

Where: φb = 0.90 (LRFD) Ωb = 1.67 (ASD)

Lp = y

y FEr76.1 AISC p. 16.1-48

Lr = y

ts FEr7.0

π AISC p. 16.1-48

E = 29000 KSI Fy = yield stress, KSI

Mp = (FyZx) AISC p. 16.1-47 Mr = 0.7(FySx) AISC p. 16.1-269 Sx = Section modulus about “x” axis from properties, in3

φbMp and b

pMΩ

φbMr and b

rMΩ

Mom

ent c

apac

ity

Lp Lr

Distance between lateral supports

Real curve

AISC curve


Example 4 (LRFD) GIVEN: A W14x43 steel girder using A992 steel. It is laterally braced at Lb = 8’-0” increments by beams framing into the side. It experiences a maximum FACTORED moment, Mu = 250 KIP-FT. REQUIRED: 1) Draw the graph of design moments vs. unbraced length. 2) Determine the design moment, φbMn for an actual unbraced length Lb =

10’-0” using formula F2-2 from AISC p. 16.1-47. 3) Determine the design moment using the “Beam Available Moments vs.

Unbraced Length” graphs on AISC p. 3-125.

Step 1 – Draw graph of design moments vs. unbraced length:

From AISC p. 3-17, pick values of Lp, Lr, φbMp, and φbMr

Step 2 - Determine the design moment, φbMn for an unbraced length Lb = 8’-0” using formula F2-2 from AISC p. 16.1-32

Mn = ⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛

−

−−−

pr

pbxyppb LL

LLSFMMC )7.0(

Where: Cb = 1.0 (conservative) Mp = FyZx = (50KSI)(69.6 in3) = 3480 Kip-In

φbMn = ????

φbMrx = 164 KIP-FT

φbMpx = 261 KIP-FT

Lp = 6.68 ft. Lr = 20.0 ft.

Real curve

Lb = 10’-0”

AISC curve


Mn = ⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛

−−

−−− )'68.6'20'68.6'10)6.62)(50(7.03480((34800.1 3inKSIInKip

= 3159 Kip-In = 263.3 Kip-Ft Therefore, since φ = 0.9: φMn = 0.9(263.3 Kip-Ft) φMn = 237 Kip-Ft

Since φMn = 237 Kip-Ft < Mu = 250 Kip-Ft, beam is UNACCEPTABLE


Step 3 – Determine FACTORED moment strength using Table 3-10:

From AISC p. 3-125: For a W14x43 with Fy = 50 KSI Unbraced length Lb = 10’-0”

Lb = 10’-0”

φMn ≅ 237 Kip-Ft

LRFD


Example 5 (ASD) and (LRFD) GIVEN: A steel girder is laterally braced at Lb = 10’-0, and experiences the service loads as shown below. REQUIRED: Design the lightest weight A992 wide-flange beam using the “Available Moment vs. Unbraced Length” graphs.

Step 1 – Determine ASD maximum SERVICE moment:

w = D + L = (900 PLF) + (1700 PLF) = 2600 PLF = 2.6 KLF Ma = Maximum applied SERVICE moment

= 8

2wL

= 8

)"0'30)(6.2( 2−KLF

Ma = 292.5 Kip-Ft.

30’-0”

Service DL = 900 PLF (incl. beam wt.) Service LL = 1700 PLF


Step 2 – Determine LRFD maximum FACTORED moment:

wu = 1.2D +1.6 L = 1.2(900 PLF) + 1.6(1700 PLF) = 3800 PLF = 3.8 KLF Mu = Maximum applied FACTORED moment

= 8

2Lwu

= 8

)"0'30)(8.3( 2−KLF

Mu = 427.5 Kip-Ft.

Step 3 – Design lightest beam using “Available Moment vs. Unbraced Length” graph:

From AISC p. 3-121 and 3-119, read up from the bottom at Lb = 10’-0” and look up until you hit a solid line for the lightest weight beam exceeding 292.5 Kip-Ft (ASD) and 427.5 Kip-Ft (LRFD).

ASD:

Use W21x62 → ftKipM n −≈Ω

316

LRFD: Use W21x62 → ftKipM n −≈ 474φ

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Beam Design for Moment Asd Lrfd (Example)