MORGAN STATE UNIVERSITY

SCHOOL OF ARCHITECTURE AND PLANNING

LECTURE IV

Dr. Jason E. Charalambides

Elastic and InelasticColumn Buckling cont.

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Let's address an example ofa column that is braced! Bracing essentially

shortens what we addressas the “unbraced length”" What if the column is braced in

one direction and not braced inanother?

" It should better be braced on theweak axis otherwise the wholeconcept is fallacious, i.e. theweak axis is the governing axistherefore the bracing does nothelp.

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Let's see it in an exercise

! Given a W10x54 column ofA992 steel with pinconnection on top andbottom, height of 15' andbracing at mid-height onweak axis, determine thecompressive load capacityof the column.

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Let's see it in an exercise

Braced Wide Flange Shapesubjected to axial loadingProblem Statement:Determine the capacity in axial loading of the given Wshape. The element is pinned at top and bottom with nointermediate bracing, therefore having an unbraced lengthof 15ft in both directions.Use A992 steel

Area Ag 15.8in2:=

Unbraced length on x axis: Lux 15ft:=

Unbraced length on y axis: Luy 7.5ft:=

radius of gyration y ry 2.56in:=

radius of gyration x rx 4.37in:=

K factor K 1:=

Young's Modulus of Elasticity E 29000ksi:=

Bolt diameter db 0.875in:=

Yield Stress: Fy 50ksi:=

Ultimate Strength: Fu 65ksi:=

Factor of Safety phi ϕ 0.9:=

Solution:1) Determining the governing slenderness ratio

λxK Lux⋅

rx:=

15ft 12⋅inft

4.37in41.19= λx 41.189931=

λyK Luy⋅

ry:=

7.5ft 12⋅inft

2.56in35.156= λy 35.156=

r min rx ry, ( ):= r 2.56 in⋅= governing radius of gyrationis not determinant

λ max λx λy, ( ):= λ 41.19= governing unbraced lengthis determinant

The above was already obvious but it was carried on just to "academicallly" justify the numbers

2) Calculating Euler's Buckling Stress

FEπ2 E⋅

λ( )2:=

3.142 29000⋅ ksi

41.19( )2168.529 ksi⋅= FE 168.7 ksi⋅=

3) Determining if the buckling will be elastic or inelastic.

Buckling if λ( ) 4.71EFy

≤ "Inelastic", "Elastic",

:= Buckling "Inelastic"=

Alternatively we can also follow the process below:

FyFE

0.296= Buckling ifFyFE

2.25≤ "Inelastic", "Elastic",

:= Buckling "Inelastic"=

4) Calculating the Buckling Stress (Fcr) and the load capacity of the section:

Fcr 0.658

Fy

FE

Fy⋅:= 0.658

50ksi

168.7ksi

50⋅ ksi 44.167 ksi= Fcr 44.167 ksi⋅=

ΦPn ϕ Ag⋅ Fcr⋅:= 0.9 15.8⋅ in2 44.167⋅ ksi 628.055 kip= ΦPn 628.051 kip⋅=

Solution Method 2: Using Table 4-22:1) Determining the governing slenderness ratio

λxK Lux⋅

rx:=

15ft 12⋅inft

4.37in41.19= λx 41.19=

2) Using table we locate the KL/r valuecorresponding to the Fy used for factrized criticalstress:

ΦFcr 39.7ksi:=

Note: From our previous calculations:

Fcr 44.167 ksi⋅=

Therefore:

ϕ Fcr⋅ 39.75 ksi⋅=

3) Calculating the capacity of the element:

ΦPn ϕ Ag⋅ Fcr⋅:=

ΦPn 628.051 kip⋅=

Solution Method 3:Using Table 4-1 for W shapes pp 4-12 to4-23:Oh you can surely do this too! Yes itis KLx/rx but all you need to do ismultiply that by ry

λx ry⋅ 8.787 ft=

Using the above length for the W10x54we can interpolate the value of 628 kip

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Efficiency in Axial vsFlexural member design! Here we see four sections of about the

same cross sectional area. From left toright we see better forms for axial design tobetter forms for flexural design.

! See moment of inertia for each.

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Efficiency in Axial vsFlexural member design! An efficient flexural member will have:

! An efficient axial member will have:

d≫b f rx≫r y

d≈b frxr y

≈1.6 tο3

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Taking Two Shapes asExamples for Axial Loading ! For the HSS shape @ Fy=42ksi, Lu=15':

For the W 14x30 @ Fy=50ksi, Lu=15':

I xI y

=1

rx=r y=2.95

ΦPn=236 kip

rxr y

≈5.5

I xI y

=14.85

ΦPn=not even listed iηTable 4−1,using Table 4−22≈133kip

r y=1.49

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In Class Exercise

! Given KLx=30', KLy=15' (column isbraced ad mid height), and Pu=1700Kips, find the lightest W shape fromcolumn tables.

! We can start by referring to whatwould be the most efficient Wshapes:" W8, W10, W12, & W14

! Note: Since KLx≠KLy a designer hasto start with just an assumption." Let's assume that the weak axis

controls:

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In Class Exercise

! Try W14" For KLy=15' → W14x159, From Table 4-1, → 1810k" Transform to KLx →

Strong axis controls" Checking for 18+ ft → Φpn<1700 kip NO GO

! Try next heavier section, W14x176." Transform to KLx → " Strong axis controls" Checking for 18.75 ft → Φpn=1853 OK

! Lightest W14 is the W14x176.

rxr y

=1.6→ 30 ft

1.6=18.75 ft>15 ft

rxr y

=1.6→ 30 ft

1.6=18.75 ft>15 ft

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In Class Exercise

! Try W12" For KLy=15' → W12x170, From Table 4-1, → 1790kip

" Transform to KLx →

Strong axis controls again" Checking for 16.85ft → Φpn<1688kip NO GO

! Try next heavier section, W12x190." Needless to try because already the W14x176 is lighter!

! Try W10." For KLy=15ft → No section with Φpn≥1700kip

! Lightest W section is the W14x176.

rxr y

=1.78→ 30 ft

1.78=16.85 ft>15 ft