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CE311 FALL 2018 Lesson 8: Column Buckling I

"Always, there is football." – Cristal Choi, former exchange student who lived in Prof. Kurtz’s house, explaining “all American holidays” to a new exchange student.

LESSON 8: COLUMN BUCKLING IWednesday, September 12, 2018

LESSON OBJECTIVES 1. Determine the controlling slenderness KL/r for a pinned-braced column.2. Calculate the Euler Elastic Buckling Stress Fe for a column, considering both axes of buckling.3. Describe the conditions for which the Euler expression accurately predicts column buckling.4. Calculate the nominal buckling load Pcr for a column per AISC equations, considering the controlling axis of

buckling.

TODAY’S SCOPE – THE SIMPLE BRACED FRAMEToday’s scope is a column in a simple-braced frame. A simple braced frame uses pinned connections and is stabilized by the truss-action of diagonal bracing. Bracing prevents the relative translation of the top and the bottom of the column. In contrast, unbraced frames are characterized by the relative translation of the top and the bottom of the column and will be covered in a subsequent lesson.

SOME NOMENCLATURE (for more, see the INDEX tab in the steel manual)In structural steel design, the letter F indicates a stress, P indicates a compressive force, T indicates a tensile force.Fcr – critical buckling stress (ksi). When columns are very long (long enough to remain elastic when buckling occurs), then

Fcr is based on Fe, the Euler buckling stressFe – Euler buckling stress (ksi). The Euler equation predicts the buckling stress, assuming that the column is elastic

(generally true for very long columns).Fy – yield strength (ksi). This is the maximum elastic stress that steel can sustain, in tension. Ix – moment of inertia about the member’s x-axis (in4)Iy – moment of inertia about the member’s y-axis (in4)KLx – effective column length (inches) for buckling about the x-axis – this is the distance over which 0 to of a sine curve is

completed by the column’s buckled shape.KLy – effective column length (inches) for buckling about the y-axis – this is the distance over which 0 to of a sine curve is

completed by the column’s buckled shape.rx – radius of gyration about the member’s x-axis (inches)ry – radius of gyration about the member’s y-axis (inches)Pcr – critical buckling load (kips). This equals the Euler buckling load if the column is very slender and remains elastic during

buckling.

Pe – Euler buckling load,

Pe=π2 EI(KL )2

=Agπ2 E

(KLr )

2

Pn – nominal column capacity (kips). This is the same as Pcr because all columns in compression fail by buckling.

READING The online version of this handout contains 8 pages of reading Textbook (Segui), pp. 109 to 121.

REFERENCE AISC Steel Manual Chapter E in the specifications (16.1-31 to 16.1-33) AISC Steel Manual Part One (Table 1-1) AISC Table 4-22

24 ft.Brace

12 ft.

12 ft.

24 ft.Brace

12 ft.

12 ft.

A B

A B

100k 100k


HOMEWORK (Due Friday. Standard assignment. Presentation counts)Terms: Pn is the nominal buckling strength (units = kips). This is the actual predicted buckling (failure) load of the column. Pn/ is the (ASD) Allowable load, where is the safety factor ( =1.67); i.e., an engineer would not want the applied (actual) load to exceed the allowable load, as this would be unsafe. Pn is the (LRFD) Design Strength (it is not an allowable load!!! If your column were to actually have this load on it, then you should get the Hell out of the building fast), whereis the strength reduction factor (=0.9 for column buckling); i.e., the engineer would not want the factored load to exceed the design strength, as this would be unsafe.

0. (Do this but do not hand in). Go to the steel manual, Part 2 and flip through the pages; this is where you will find the most general information about steel design (LRFD, ASD, fabrication tolerances, materials, etc.). Find page 2-10. This is where the load combinations are found. Go to AISC Tables 2-4 and 2-5 to answer the following:

a. What is the “preferred ASTM material specification” for W shapes? What is the minimum yield strength for this? What is the minimum tensile strength for this?

b. What is the “preferred ASTM material specification” for L shapes? What is the minimum yield strength for this? What is the minimum tensile strength for this?

Note: from now on, unless stated otherwise, you are to assume the “preferred ASTM material specification” for all shapes.

0.’ (Do this but do not hand in). For a 24-ft long W12x96 column, two possible bracing ideas are shown, A and B. Circle the bracing idea that is “smart.”

0.’’ (Do this but do not hand in). Prove to yourself that “Braces are blind to gravity loads”; i.e., the brace does not have a force in it, due to the vertical load.

a. Do equilibrium on Point A. From this, determine the force in AB (remember zero-force members?)b. Knowing the force in AB, do equilibrium on Joint B. From this determine the force in the brace.

0.’’’ (Do this but do not hand in). Let’s suppose that we know that the actual applied load on a column is 100 kips. Let’s also suppose that we know that the Pn of the column is 105 kips. Is the column safe?0.’’’’ (Do this but do not hand in). Let’s suppose that we know that the factored load on a column is 100 kips. Let’s also suppose that we know that the Pn of the column is 105 kips. Is the column safe?

18’

30’30’


1. Compute the allowable buckling stress Fcr/ for a column that has a slenderness of 100 if Fy=50ksi. Then, go to Table 4-22 in the AISC manual and find the allowable buckling stress Fcr/ for a column that has a slenderness of 100 (note: from now on, Table 4-22 is a good table to use to save yourself time. Put a tab there. Avoid hand-calculating Fcr unless Fy is an odd specification, not found in the table.)

2. The 6” diameter Extra-Strong Pipe section (Pipe 6 x-strong) is subjected to axial forces PDead = 55 kips and PLive = 55 kips. Determine if the section is adequate per ASD. Then determine if the section is adequate per LRFD. Use the “preferred material specification”.

3. Determine if an HSS4x3x3/8 column is adequate per ASD. Then determine if the section is adequate per

LRFD. The single-story 1-bay building frame has a Roof Dead Load pressure of 25psf. This includes the

proper allowance for all roof weights: fill beams, girders, the roof deck, the roofing material, etc.(it’s all included in the 25 psf number)

Snow Load pressure: 50psf (note: that’s a lot. This building must be in Canada) The story height is 18’ but pay close attention to the bracing and note the way it braces the column’s

weak-axis. The bay lengths are 30’, each.Note: the nominal dimensions of the HSS (“hollow structural shape”) are 4x3x3/8, but use the real dimensions1 and properties given in the AISC Part One Tables. As described in the previous problems (refer to problem 0, above), use the “preferred material specification” for rectangular HSS.

1 Always use the decimal dimensions of cross-sections, for calculations. The fractional dimensions are for detailers and architects, only. Example: a W16x26 has a depth of 15.7” or 15 ¾”. Engineers use the 15.7”. Detailers and architects use the 15 ¾”.

HSS4x3x3/8 Column

Braces

Pn

Very Slender ColumnLong Length, KL

Low Cross-Section ILow Pn

Buckled Shape

Pn

Less Slender ColumnShorter Length, KL

Higher Cross-Section IHigher Pn

Buckled ShapeKL

KL


SUPPLEMENTARY READING:

Objective: Calculate the Euler Elastic Buckling Load Pe for a column and State the limitations of the Euler Expression.

If an increasing axial force is applied to a column that is very short and stocky, then it will eventually fail because of material failure such as concrete crushing (concrete columns) or steel yielding (steel columns). However, if the column is somewhat slender, then increasing axial force will result in failure due to buckling at the buckling stress Fcr, rather than by material failure.

The buckling strength of a column depends upon how slender the column is. Columns that are very slender buckle at stresses that are lower than columns that are shorter and stockier. Column slenderness is quantified by the slenderness

ratio KLr

, where KL is the effective length of the column and r is the radius of gyration of its cross-section.

THREE RANGES OF SLENDERNESS, KLr

For most engineering materials, three general ranges of column slenderness can be identified: Short Columns . Columns that are very short and stocky do not buckle. The maximum stress that a short

column can reach in axial compression is the stress corresponding to the material’s ultimate strength Fu. Slender or Elastic Columns . Columns that are very slender will buckle in an elastic manner when subjected to

their maximum load; that is, the buckling stress is sufficiently low so that the material does not exhibit any yielding or damage. Upon unloading, these columns will elastically rebound to their unloaded position, showing no damage. Elastic column buckling stress is reasonably predicted by the Euler expression:

F e=π2 E

( KLr )

2

Where:E = Elastic Modulus (psi or ksi)KL = Effective column length (inches or feet)r = Radius of Gyration (inches or feet)KLr = Slenderness Ratio (dimensionless)

Figure 1


Note that the strength of the material (e.g., material yield stress Fy or material ultimate strength Fu) is not present in the Euler expression. For elastic buckling, the strength of the material is irrelevant because buckling occurs at a stress that is well-below the material strength.

Intermediate or Inelastic Columns . These columns have slenderness that is in-between short columns and elastic columns. For these columns, buckling is influenced by the strength of the material and this buckling occurs as part of the column cross-section undergoes yielding or damage in some way.

Typical Column Buckling Strength Curve

Column Slenderness (KL/r)

Buck

ling

Stre

ss F

cr (

ksi)

Short Columns

Intermediate or Inelastic

Columns

Slender or ElasticColumns

Figure 2

The effective length KL will be covered in greater depth with subsequent objectives and you will see that the effective length depends on the support conditions of the column (pinned, fixed, free, etc.). However, columns that are pinned on both ends are the simple case that is handled with this objective because the effective length KL is simply the physical length of the column L.

Pn

Buckled Shape

Pin

Pin

KL = L

Pinned-Pinned ConditionK = 1

KL = L

Pe = ?

5 ¼”


The radius of gyration r=√ IA

where I is the moment of inertia of the cross-section and A is the area of the cross-section.

Note that: Radius of gyration r is NOT the same as the radius r (be careful with this distinction, when analyzing circular

cross-sections such as pipes). Because Ix is not generally equal to Iy, rx is not generally the same as ry.

To: Calculate the Euler Elastic Buckling Load Pe for a column and State the limitations of the Euler Expression Determine the effective length(s), KL (with this objective’s examples, the effective lengths KLx and KLy are equal,

but with subsequent objectives, you will see that KLx is not generally equal to KLy) Determine the moment(s) of inertia I of the cross-section (with this objective’s examples, Ix is equal to Iy, but with

subsequent objectives, you will see that Ix is not generally equal to Iy). Based on the I and A of the cross-section,

compute the radius of gyration using r=√ IA

Compute the Euler stress Fe=

π2 E

(KLr )

2 , using consistent units.

Axial force = (axial stress) x (area). Therefore, the Euler buckling load Pe = FeA. Know that the Euler stress is:

o A good predictor of buckling stress when the column is very slender so that the buckling stress is substantially less than the ultimate or yield stress of the material. A rule of thumb is that, if Fe is less than one-third of the material’s ultimate or yield stress, than the Euler stress is an accurate predictor of buckling.

o A poor predictor of buckling stress when the column is very stocky. Note the fact that the Fe goes to infinity when KL/r→0. Clearly, the Euler stress is a very poor predictor when KL/r is small.

Example 1Problem: Determine the Euler buckling load Fe for the timber column and state whether or not the Euler buckling load is a reasonable and accurate predictor of the actual buckling load.Given: Timber column cross-section has nominal dimensions of 6”x6”, with actual dimensions of 5¼”x 5¼”. The modulus of the elasticity E = 1500 ksi, while the ultimate strength of the timber is 6000 psi. The column is 8-ft long, pin-supported on each end.

Solution: Being careful with consistent units!


KL = 8’ = 96”

I= b h3

12=

(5.25 )4

12=63.31¿4

A=(5.25 )2=27.56 i n2

r=√ IA

=√ 63.3127.56

=1.516∈.

KLr

=96 } over {1.516=63.32

F e=π2 E

( KLr )

2 =π 2(1500)(63.32 )2

=3.692 ksi

Pe=F e A=(3.692ksi ) (27.56 i n2 )=102 kips Soln.

However, the Euler buckling stress Fe=3.692ksi is more than 60% of the material’s ultimate stress of 6 ksi. Therefore, the Euler buckling stress and load are not accurate/reasonable predictors of the actual buckling load.

Pe = ?

25-ftA A

6”

Elevation View (Side View)

Pin-Ended 25-ft-long Column

Cross-Section View A-A6x½” HSS

0.465”


Example 2Problem: Determine the Euler buckling load Fe for the round steel HSS 6”x0.5”(“hollow structural shape”) column and state whether or not the Euler buckling load is a reasonable and accurate predictor of the actual buckling load.Given: The HSS 6”x0.5”column has a 6” outer diameter and a nominal wall thickness of 0.5”, but an actual wall thickness of 0.465” (refer to AISC Table 1-13 for other properties). The modulus of the elasticity of steel is 29000 ksi. The yield strength of the steel is 42ksi. The column is 25-ft long, pin-supported on each end.

Solution: Being careful with consistent units!KL=25'=300

For a circle: I=πr 4

4

I=π (ro

4−ri4 )

4 =π ((3)4−(2.535)4 )

4 =31.18 in4

A=π (ro2−r i

2)=π ((3 )2−(2.535)2 )=8.086i n4

r=√ IA

=√ 31.188.086

=1.964∈.

KLr

=300 } over {1.964=152.8

F e=π2 E

( KLr )

2 =π 2(29000)

(152.8 )2=12.26 ksi

Pe=F e A=(12.26 ksi ) ( 8.086 in2 )=99.2 kips Soln.

The Euler buckling stress Fe=12.26ksi is about 30% of the material’s yield stress of 42 ksi. Therefore, the Euler buckling stress and load are probably accurate/reasonable predictors of the actual buckling stress and load.

Objective: Determine the controlling slenderness KL/r for a column.Ordinary column buckling is also known as flexural buckling due to the fact that buckling is resisted by the flexural stiffness of the column, which will be greater if the EI of the column is greater. For a fixed column area and column material, the EI of the section is maximized if the cross-sectional area is spread out as much as possible, as is the case when a hollow pipe section is used as a column. Whereas the pipe section shown in Figure 3a below has the same area as the solid circular section shown in Figure 3b, the hollow pipe section is a vastly superior column because its moment of inertia is so much larger. The result is a column with far greater axial load-carrying capability.

Hollow Pipe SectionEfficient Column Choice

Illustration:Outside Diameter = 5”

Wall Thickness = 0.209”Area, A = 3.14 in2

Moment of Inertia, I = 9.02 in4Radius of Gyration, r = 1.69 in.

Solid Circular SectionInefficient Column Choice

Illustration:Diameter = 2”

Area, A = 3.14 in2Moment of Inertia, I = 1.57 in4

Radius of Gyration, r = 0.707 in.

Both sections have the same area. They use the same amount of material. But the hollow section has an I that is 5.7 times larger and a

radius of gyration that is 2.4 times larger.

(A) (B)

x

y y

x

LA A

Cross-Section View A-AWide-Flange Section

LA A

Two Views of the Same Column

View 1Buckling causes bending about the

strong axis

View 2Buckling causes bending about the

weak axis

View 22

1

View 1

Conclusion: The column is weaker with respect to the y-y axis, so this axis will control

P


AXES – WEAK AXIS AND STRONG AXISThe previous figure illustrates that a column shape in which material is spread out as much as possible will be the most efficient column shape, because it maximizes the flexural stiffness, while using the same amount of material. In the previous figure, the moments of inertia for the x and y axes were equal: Ix = Iy. In general, this is not the case for all columns. For the majority of columns, one axis of bending has a higher moment of inertia than the other.

The I-shaped section below illustrates this. The I-shape has Ix that is 10 times larger Iy. Though the column could buckle such that it causes bending about either axis, the weak axis will control.

If the shape previously shown were to be subjected to an increasing axial force until buckling occurs, the shape would fail with bending about the y-y axis because its bending resistance is much less about this axis. For this reason, the y-y axis is commonly called the weak axis and the x-x axis is called the strong axis. One would also say that the weak axis controls.

PINNED SUPPORTS = BRACES

Figure 3

Figure 4

Colum

n

Beam

Brace

A

B

C

D E

F

G

H I

J

K

L M

N

O

P Q

R

S

T

KL

KL

KL

Truss action in this bayPrevents horizontal movement of all joints All joints are “braced”. No joints can move horizontally

KL

KL

KL

A Simple-Braced Frame

Column Weak Axes KLy=6

ft

Column Strong Axes KLx=12 ft

Two views of the same structure. Note that,

for weak-axis bending the braces are closer

6 ft.12 ft.

Figure 6


KL is the effective length of a column. For a pin-ended column, the effective length IS the length of the column L. Though we may represent these as lines with pin-symbols on both ends,

What does this symbol really represent?

The symbol of a pin-ended column really represents a column that is braced by pin-ended truss-action, which prevents the translation of the column at the braced point. Because the truss-action is pin-ended, however, the bracing does not prevent the rotation of the column at the braced point. In order for the rotation of the column to also be prevented, the columns would have to be rigidly connected (as opposed to pin-connected) to large, stiff elements such as a deep girder. When a frame is braced and consists exclusively of pin-connected beams, columns, and braces, it is known as a simple-braced frame. As shown in Figure 4, all of the columns in a simple-braced frame have effective length KL equal to the length of the column between the braced points.

AXES – KLX AND KLY

The previous figure showed the bracing for one planar view of a simple-braced frame and illustrated that the effective length KL is the distance between the braces. However, the bracing is not necessarily the same for other planes in the structure. For example, Figure 6 illustrates that the designer may elect to brace the structure so that KL for one axis of buckling is different from KL of the other axis of buckling. This is usually an intentional strategy employed by the designer, as the use of additional bracing is used to strengthen the weak axis of the column.

Figure 4


CONTROLLING SLENDERNESS - ( KLr )

xvs .( KL

r )y

You have seen that KLx is not generally equal to KLy and that rx is not generally equal to ry. You also know that the buckling

strength of a column depends on the slenderness ratio ( KLr ). While a column can buckle about either its x-axis or its y-

axis, it will be controlled by whichever axis has the greater slenderness ratio ( KLr ).

To Determine the controlling slenderness KL/r for a column:1. Determine KLx and KLy for the column.2. Determine rx and ry for the column cross-section.

3. Compute the slendernesses ( KLr )

x∧( KL

r )y. The greater ( KL

r )❑

, either ( KLr )

x∨(KL

r )y, will control buckling.

Example: If ( KLr )

x=100∧( KL

r )y=80 , then the column will fail by buckling about the x-axis and the buckling

stress will be determined based on ( KLr )=100.

Column Weak Axes KLy=6

ft

Column Strong Axes KLx=12 ft

7 ft.14 ft.

Base Col.Base Col.

Two views of the same structure. Note that,

for weak-axis bending the braces are closer


ExampleDetermine the controlling axis of buckling for one of the base columns. Two planar views are shown for the same structure.Given: The base column is a W12x72. Refer to AISC Table 1-1 for required properties.

Solution: KLx = 14 ft = 168” KLy = 7 ft = 84”

From AISC Table 1-1 rx=5.31 inches ry = 3.04 inches

Compute slendernesses:

( KLr )

x=( 168

5.31 )=31.64

( KLr )

y=( 84

3.04 )=27.63

The controlling slenderness is 31.64. Buckling will occur about the x-axis. The buckling load can be computed

using ( KLr )

❑=31.64

AISC BUCKLING EQUATIONSAs illustrated in Figure 2, the Euler Expression is valid for long slender columns that buckle elastically; i.e., no yielding of the steel takes place so that the column would elastically rebound once the load is removed. For somewhat shorter, stockier columns, buckling is not accurately predicted by the Euler Expression. Instead, the column behavior is influenced by the yielding of the material. Hence, less slender columns buckle such that they exhibit partial yielding and permanent deformation.

The AISC column buckling expressions are shown below. Note that they incorporate the Euler Expression, Fe.

If F y

F e≥ 2.25 ELASTIC BUCKLING. Fcr=0.877 Fe

If F y

F e≤ 2.25 INELASTIC BUCKLING. F cr=[0.658

F y

Fe ] F y


Based on these, the nominal column strength Pn is simply FcrA, where A is obviously the cross-sectional area of the column.

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