LRFD – Floor beamLRFD – Floor beam

Unbraced top flangeUnbraced top flange

Lateral Torsion BucklingLateral Torsion Buckling

We have to check if there is plastic failure (yielding) or lateral-torsion buckling.

This depends on the length between the lateral braces, related to the limiting lengths.

Lp is the limiting length for plastic failure

Lr is the limit length for torsional buckling.

If Lb < Lp it is plastic failure

If Lp < Lb < Lr we have a different failure criteria

If Lb > Lr we use the lateral buckling stress criteria

Plastic FailurePlastic Failure

If Lb < Lp

Mn = Mp = y Zx

Zx is the plastic section modulus about the x axis

LLpp < L < Lbb < L < Lrr

€

Mn = Cb M p − M p − 0.7σ ySx( )Lb − LpLr − Lp

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥≤ M p

LLbb > L > Lrr

Mn = crSx ≤ Mp

The following definitions The following definitions applyapply

€

cr =Cbπ

2E

Lbrts

⎛ ⎝ ⎜ ⎞

⎠ ⎟2 1 + 0.078

Jc

Sxho

Lbrts

⎛

⎝ ⎜

⎞

⎠ ⎟

2

€

Lp =1.76 ry E

σ y

€

Lr =1.95 rts E

0.7σ y

J c

Sxh0

1+ 1+ 6.760.7σ ySxh0

EJc

⎛

⎝ ⎜

⎞

⎠ ⎟

2

cc

For a doubly symmetric I-shape c=1

For a channel,

Where h0 = distance between flange centroids

€

c =h0

2

IyCw

Conservative simplificationsConservative simplifications

€

cr =Cbπ

2ELbrts

⎛ ⎝ ⎜ ⎞

⎠ ⎟

€

Lr = πrtsE

0.7σ y

€

rts2 =

IyCwSx

A beam of A992 steel with a span of 20 feet supports a stub pipe column with a factored load combination of 55 kips

A992 Steel: structural steel, used in US for I-beams. Density = 7.85 g/cm3. Yield strength = 50 ksi.

No flooring – no lateral bracing on top No flooring – no lateral bracing on top flangeflange

Find max moment.

Assume beam weighs 50 lbs/ft

From distributed load, Mmax = w L2/8

From point load, Mmax = P L / 4

Mmax = 55,000 * (20/4) + 50 * (20^2)/8 = 277.5 kip-ft

Use trial methodUse trial method

• Find a beam that has a Mp of at least 277.5 kip-ft

• Need to check if it will fail in plastic mode (Mp) or from flange rotation (Mr)

• Tables will show limiting unbraced lengths.

• Lp is full plastic capacity

• Lr is inelastic torsional buckling.

• If our length is less than Lp, use Mp. If greater than Lr, use Mr

Selected W Shape Properties Selected W Shape Properties – Grade 50– Grade 50

Prop W18x35 W18x40 W21x50 W21x62

Mp (kip-ft) 249 294 416 540

Lp (ft) 4.31 4.49 4.59 6.25

Lr (ft) 11.5 12.0 12.5 16.7

Mr (kip-ft) 173 205 285 381

Sx (in3) 57.6 68.4 94.5 127

Iy (in4) 15.3 19.1 24.9 57.5

ho (in) 17.28 17.38 20.28 20.39

ry (in) 1.22 1.27 1.30 1.77

J (in4) 0.506 0.81 1.14 57.5

Cw 1140 1440 2560 5970

W18 x 40 looks promisingW18 x 40 looks promising

294 > 277.5

But, Lp = 4.49. Our span is 20 feet.

And, Lr = 12.0 again, less than 20’

Mr = 205, which is too small.

W21x50 has Lr = 12.5, and Mr = 285.

That could work!

Nominal flexural design Nominal flexural design stressstress

Mn = cr Sx

The buckling stress, cr , is given as

€

cr =Cbπ

2E

Lbrts

⎛ ⎝ ⎜ ⎞

⎠ ⎟2 1 + 0.078

Jc

Sxho

Lbrts

⎛

⎝ ⎜

⎞

⎠ ⎟

2

Terms in the equationTerms in the equation

rts = effective radius of gyration

h0 = distance between flange centroids

J = torsional constant (torsional moment of inertia)

Cw = warping constant

c = 1.0 for doubly symmetric I-shape

Effective radius of gyrationEffective radius of gyration

€

rts2 =

IyCwSx

€

rts = 24.9 • 2560

94.5 = 1.635 in

So the critical stress isSo the critical stress is

€

cr =1.0 π 2 29,000

20 x 12( )1.635

⎛

⎝ ⎜

⎞

⎠ ⎟2 1 + 0.078

1.14 • 1.0

94.5 • 20.28

20 • 12

1.635

⎛

⎝ ⎜

⎞

⎠ ⎟2

= 18.77 ksi

Then the nominal moment isThen the nominal moment is

Mn = cr Sx

= 18.77 • 94.5 = 1,774 kip-in = 147.9 kip-ft

We need 277.5!! If we had the AISC design manual, they show

unbraced moment capabilities of beams. We would have selected W21x62, which turns

out to handle 315.2 kip-ft unbraced.