Crippling.ppt

Crippling

In column buckling (Fig. 1), all points in the cross-section undergo the same lateral deflection, i.e., the cross-section remains unchanged all along the column. In other words, the column buckles as a whole. This happens in long columns.

In short columns with cross-sections composed of flange plate elements like formed or extruded sections,, local buckling takes place (at a lower load) before column buckling happens. Here one or more flanges buckle as panels independent of the others. For example, in Fig. 2, the flange BCDE has buckled whereas the flange containing the leg AB has not. The compressive stress at which this happens is same as the compressive buckling stress of BCDE treated as a panel simply supported along the loaded edges BC and DE and the corner edge BE and free on the fourth edge CD. In the figure, the face BCGDEF assumes the deflected position of BCG'DEF. That is, in local buckling, column cross-section changes and varies along its length.

Crippling is a failure phenomenon which happens after local buckling in which corner sections progressively take greater and greater share of load as the load is further increased until failure takes place

Fig.1 Column Buckling

Source: Bruhn

A

B

Buckled shape

Fig. 2 Local Buckling Fig. 3 Post-local Buckling

As the load is increased, the flange containing the leg AB also buckles as a panel. Then the column would have undergone local buckling totally. At this point, the compressive stress would be constant throughout the cross-section. Even after both the legs AB and BC have buckled, the column continues to take more load without failing. As the load is increased, the corner regions which are relatively stiffer start taking more and more share of load as in Fig. 3. Later at some load, the column fails. That is called crippling and the corresponding load is called crippling load

Crippling

Fig.1 Column Buckling Fig. 2 Local Buckling Fig. 3 Post-local Buckling

A

B

Buckled shape

Crippling

A few things to note about crippling:

• Crippling is relevant only for short columns

• The crippling load depends only on the cross-section of the column and not on its length

• The crippling load cannot be determined by theoretically. It is determined by empirical formulae substantiated by experiments

• Crippling load is always greater than the local buckling stress. For a very short column, it can, in the extreme case, be equal to the local buckling stress

Formulae for crippling

The following are the two most commonly used methods of determining the crippling load:

1. Gerard Method - applicable to sections in which all the segments are of the same thickness as in formed sections

2. Method of segments – applicable to sections in which all the segments need not be necessarily of the same thickness as in extruded sections – Also applicable to formed sections where the cross-section is not one of the standard ones in the list specified in Gerard method

Gerard Method

For sections with distorted unloaded edges as angles, tubes, V groove plates, multi-corner sections and stiffened panels,

(1)

0.851/2

CY

2

CY

CS

F

E

A

gt0.56

F

F

Crippling

(2)

For sections with straight unloaded edges such as plates, tee, cruciform and H sections,

0.401/2

CY

2

CY

CS

F

E

A

gt0.67

F

F

For 2 corner sections, Z, J, and channel sections,

0.751/3

CY

2

CY

CS

F

E

A

t3.2

F

F

FCS = Crippling stress for section (psi)

FCY = Compressive yield stress (psi)

t = Element thickness (inches)

A = Section area (in.2)

E = Young’s modulus of elasticity

g = Number of flanges which compose the composite section, plus the number of cuts necessary to divide the section into a series of flanges as defined in fig. 4

(3)

Crippling

A flange is defined as a plate element with one end being a corner and the other end free. An intersection with another plate element is considered to be a corner. If both ends are corners, then a cut is introduced to convert it into two flanges.

In addition to the above criterion in counting flanges and cuts, one must also take into consideration whether the lips and bulbs provide supporting condition to the adjacent elements as in the next slide.

Equations (1) – (3) are also represented in the form of charts in Figs, C7.7, C7.8 and C7.9 of Bruhn

Fig. 4 Definition of flanges and cuts

Angle

Basic sectiong = 2

Plate

Cut

1 Cut2 Flanges3 = g

Cuts

Tube

4 Cuts8 Flanges12 = g

Basic sectiong = 3

T-Section

0 Cuts4 Flanges4 = g

Cruciform

Cut

H-Section

1 Cut6 Flanges7 = g

Crippling

C7.9 Restraint produced by Lips and Bulbs.

Quite often in formed sections, the flange element which has a free edge is rather small in width as illustrated in Fig. 5. Also for extruded sections, a bulb is often used as illustrated in Fig 6. The question then arises, is the lip or bulb sufficiently large enough to provide a simple support to the adjacent plate element? If it is not sufficiently large as to provide the simply supported condition, counting the number of flanges and cuts must be done in a different way as explained in slide 11.

We first proceed to see what are the criteria or the test/tests to be passed by the lip or bulb to provide the simply supported condition to the adjacent element.

Fig 5Fig 6

Crippling

The lower limit for the length of the lip is governed from the condition that it should be long enough to have sufficient bending stiffness to provide the simply supported condition. This limit is given by

where the various dimensions are defined in Fig. 7. Note that t f = tL = t in equation (4), because lip is used

generally in formed sections for which the flange and lip are of the same thickness.

(4)

Fig. 7. Lip and bulb dimensions

t

b5

t

b

t

b0.910 fL

3

L

Crippling

The upper and lower limits for the lip length are shown in Fig. 8 where the permissible length should lie in the hatched portion

But the lip cannot be too long! If its length is more, its buckling stress will be less. Another condition for the lip to be able to provide the simply supported condition is that its buckling stress should be not less than that of the adjacent plate element. This condition provides an upper limit to the lip length. This upper limit is given by

(5)

Fig. 8 Upper and lower limits for lip length

f

f

L

L

t

b0.328

t

b

LIP PROVIDES AT LEAST SIMPLE SUPPORT FOR FLANGE (ACCEPTABLE ZONE)

As an alternative to Fig. 8, one could also use the following figure from Niu to check whether the lip qualifies to provide support to the adjacent segment. The abcissa is same in both the figures, but the only difference is that the ordinate in the figure below is bL/bF rather than bL/t. It makes no difference because both figures are applicable only to formed sections where the thickness is constant. Both figures give the same result

Crippling

(of Niu)

Crippling

Fig. 7(b)

In case of a bulb, unlike in a lip, increase in diameter increases both the bending stiffness and buckling stress. So, there is only a lower bound for the bulb size which is given by the following equation:

The minimum bulb diameter as given by eq. (6) is represented graphically in Fig. 9

(6)

Fig. 9 Minimum bulb diameter to provide simply supported condition

In extruded section, a circular bulb is often used to stiffen a free edge as illustrated in fig. 7(b)

t

b7.44

t

D0.374

t

D1.6

t

D f

234

Crippling

A

B C

A

B

D

Method of counting flanges and cuts if the lip or bulb does not qualify to provide support to the adjacent plate element

Consider the following figure showing three channel sections with unequal flanges AB and CD. Let us consider the role of CD as a lip to the plate element BC. In Fig. (10a), the lip CD satisfies both the criteria defined by equations (4) and (5). So, CD qualifies to provide support to BC and C is a corner. Now, we have to introduce a cut in the middle of BC to make it into 2 flanges BE and EC (A flange should have one corner and one free end). So, flanges + cuts = 4 + 1 =5

(a)

E

In Fig. 10(b), CD is a little short and fails the criterion of eq. (4). So, BC is to be considered free at C and must be treated as one flange. Then, flanges + cuts = 3 + 0 = 3

In Fig. (10c), CD is a little too long and fails the test of eq. (5). So, again, CD fails the lip test. That is, BC is to be considered free at C and must be treated as one flange. Then, flanges + cuts = 3 + 0 = 3

Note that in both cases Figs. 10(b) and 10(c), CD qualifies as a flange because BC provides support to CD at C by automatically passing both the criteria of eqs. (4) and (5). Note that here the roles of BC and CD as flange and lip are interchanged. The implication is that the area of CD must be included in the sectional area A in equations (1) – (3)

A

B C

A

D

(a)

DA

B C

(c)

C

D

(b)

B

AA

B C

D

(a)

DA

B CC

D

(b)

B

AA

B C

DD

A

B CC

D

B

AA

B C

D

A

B C

D

Fig. 10 Qualification of lip to provide support

C

D

B

A

(b)(b)

Crippling

Crippling loads calculated from equations (1) to (3) should be limited to the values as given in the following Table:

Type of section Max. FCS

Angles .7 FCY

V Groove plates FCY

Multi-corner sections, including tubes .8 FCY

Stiffened panels FCY

Tee, Cruciform and H Sections .8 FCY

2 corner sections. Zee, J, Channels .9 FCY

C7.7 Correction for Cladding.

Since many formed sections are made from alclad sheets, the clad covering acts to reduce the value of crippling stress and thus a correction factor η must be used to take care of this reduction in strength. This correction is

Where, σcl = Cladding yield stress

σcr = Crippling stress

f = Ratio of total cladding thickness to total thickness. f = 0.10 for alclad 2024-T3 and .08 for alclad 7075-T3.

3f1

fσσ

31

ηcr

cl

Table 1

Crippling

Problem 1: Determine the crippling stress for the formed section shown in Fig if the material is Aluminum alloy 2024-T3. FCY=40000, EC=10700000

Also calculate the crippling stresses if each of the lips were (i) 0.225” long and (ii) 0.35” long

Solution:

Since the elements of the cross-section are of constant thickness, Gerard method is applicable. Also, since the section is a multi-cornered one, equation (1) applies. For a formed section of this type, the cross-sectional area will be generally given. Since it is not given here, let us calculate it assuming that corners are all right-angled and sharp. The section is symmetric about the vertical centerline.

A = 2 × 0.04 × [(0.325 – 0.02) + (1.0 - 0.04) + (1.5 – 0.04) + (0.625 - 0.02)] = 0.266 sq. in.

Let us first check the lips. bL = 0.305, bf = 0.96 and t = 0.04

Apply eq. (4)t

b5

t

b

t

b0.910 fL

3

L

.325

A B C D

8.395.04

.305

.04

.3050.910 LHS

3

125.38

.04

.3055RHS

Since LHS>RHS the eq. (4) is satisfied

Crippling

Now check eq. (5) which is equivalent to bL/bf<0.328

bL/bf=0.305/0.96=0.318<0.328, so eq. (5) is also satisfied

That means, the 0.325 in. lips satisfy both the conditions (4) and (5) and lie in the acceptable area shown by hatched portion in Fig.8. They therefore provide simply supported condition to the two 1.0 in. horizontal plate elements AB and CD at the bottom and we can introduce a cut in each of these elements.

g = number of flanges + number of cuts = 12 + 5 =17

Eq. (1) gives,

867.040,000

1010.7

0.266

04.0170.56

F

E

A

gt0.56

F

F0.851/262

0.851/2

CY

2

CY

CS

This value of FCS/FCY could have been straightaway read off from the chart in Fig C7.7 of Bruhn.

FCS=0.867×40,000=34681 psi

From table C7.1, it is recommended for multi-corner sections that FCS maximum be limited to 0.8 FCY unless tests can prove higher values.

FcsMAX=0.8×40,000=32,000 psi. Since this is less than the above calculated values it should be used.

Crippling

Now let us calculate the crippling stresses if each of the lips were (i) 0.225“ long and (ii) 0.35 “ long

Case (i): bL = 0.225”

Solution:

The section is symmetric about the vertical centerline.

A = 2 X 0.04 X [(0.225 – 0.02) + (1.0 - .04) + (1.5 – 0.04) + (0.625 - 0.02)] = 0.258 sq. in.

Let us first check the lips. bL = 0.205, bf = 0.96 and t = 0.04

Apply eq. (4)

LHS < RHS. So, eq. (4) is not satisfied and the design point lies below the lower curve of Fig. 8. This is a case of the lip being too short. Hence, the lips cannot provide simply supported condition to the adjacent elements AB and CD and we cannot have cuts in these elements.

g = number of flanges + number of cuts = 12 + 3 =15

Eq. (1) gives,

= 0.56 X [ (15 X 0.042 / 0.258) ( 10.7 X 106 / 40,000)½]0.85 = 0.8

t

b5

t

b

t

b0.910 fL

3

L

37.1170.04

0.205

0.04

0.2050.910 LHS

3

120

0.04

0.2055RHS

8.040,000

1010.7

0.258

04.0510.56

F

E

A

gt0.56

F

F0.851/262

0.851/2

CY

2

CY

CS

Crippling

This value of Fcs/Fcy could have been straightaway read off from the chart in Fig. C7.7 of Bruhn.

Fcs = 0.8 X 40,000 = 32,000 psi.

The above calculated value of crippling stress is correct and should be accepted because it is just equal to the limiting value of 0.8 Fcy as given in Table 1 for multi-cornered sections.

Case (ii): bL = 0.35”

Solution:

The section is symmetric about the vertical centerline.

A = 2 X 0.04 X [(0.35 – 0.02) + (1.0 - .04) + (1.5 – 0.04) + (0.625 - 0.02)] = 0.268 sq. in.

Let us first check the lips. bL = 0.33, bf = 0.96 and t = 0.04

Apply eq. (4)

LHS > RHS. So, eq. (4) is satisfied and the design point lies above the lower curve of Fig. 8. Now, we have to try the next test of eq. (5) and check whether the design point lies below the upper curve as it should. The condition represented by eq. (5) is equivalent to bL/bf ≤ 0.328.

bL/bf = 0.35/0.96 = 0.364 > 0.328. Hence, the lip fails the test of upper limit for the lip length. The design point lies in the unacceptable region above the upper curve of Fig.8. The implication is that the lip is a little too long to qualify for supporting the adjacent element and we cannot have cuts in the adjacent elements AB and CD.

t

b5

t

b

t

b0.910 fL

3

L

7.5020.04

0.33

0.04

0.330.910 LHS

3

120

0.04

0.2055RHS

Crippling

This value of Fcs/Fcy could have been straightaway read off from the chart in Fig. C7.7 of Bruhn.

Fcs = 0.775 X 40,000 = 31,000 psi

The above calculated value of crippling stress is correct and should be accepted because it is less than the limiting value of 0.8 Fcy as given in Table 1 for multi-cornered sections.

g = number of flanges + number of cuts = 12 + 3 =15

Eq. (1) gives,

775.040,000

1010.7

0.268

04.0510.56

F

E

A

gt0.56

F

F0.851/262

0.851/2

CY

2

CY

CS

Method of segments (Source: Michael Niu)

• This method is applicable to sections in which all the segments need not be necessarily of the same thickness as in extruded sections

• Also applicable to formed sections where the cross-section is not one of the standard ones in the list specified in Gerard method

• The section is broken down into individual segments, as shown in Fig.11. Each segment has a width b and a thickness t and will have either one or no edge free

Fig. 11 Method of segments

• The allowable crippling stress Fcc for each segment is found from the applicable material test curve. Such curves are available for for different materials and different b/t ratios. Niu gives plots for typical aircraft materials, 2024 and 7075 (Figs. 10.7.6 and Fig. 10.7.7 of Niu) .

Alternately, Fcc for each segment can also be found from Gerard formula, eq. (1), with the use of appropriate value of g (= 1 or 3, for one edge free and no edge free, respectively)

As a third alternative, Fig. 10.7.10 of Niu can be used in place of Gerard formula

(The relevant figures from Niu are reproduced in the subsequent slides for easy reference)

nn

ccnnncc

2211

2cc221cc11cc

tb

FtbF or

........)tbtb(

........)FtbFtb(F

where: b1, b2, … - Lengths of individual segments

t1 , t2 , … - Individual segment thicknesses

Fcc1 , Fcc2 , … - Allowable crippling stresses of individual segments

• Eqns. (4) and (5) defining the lower and upper limits for the lip length (to provide simply supported condition to the adjacent segment ) hold good in the previous step

• The allowable crippling stress for the entire section is computed by taking a weighted average of the allowables for each segment

(7)

(of Niu)

Dotted (one edge free)

1. 2024-T3 Clad2. 7075-T6

(of Niu)

(of Niu)

Note: The cut-off value of Fcc/Fcy is wrongly marked corresponding to the ultimate stress Ftu. It should be marked at unity corresponding to the yield stress Fcy

cy

ccn

F

FccnF

ccnnn Fbtnntb

n

n

t

b

nn

ccnnncc tb

FtbF

nb nt

The value of Fccn/Fcy in the above table could also have been read-off from Fig. 10.7.7 of Niu since the material is known to be 7075-T6 and it is an extruded section. Alternatively, it could also have been calculated from the Gerard formula, eq. (1).

Also, note that the cut-off stress is taken as Fcy and not Ftu confirming that the marking of cut-off stress in Fig. 10.7.10 is not correct

Since this is a formed section with constant thickness, one could simply use the Gerard method using the relevant formula from equations (1) to (3) whichever is applicable. However, it can also be solved by the segment method as illustrated by Niu and shown next.

First, let us carry out lip test on segment 1 to see whether it provides at least simply supported condition to segment 2.

bL = 0.25, bF = 0.863, tL = tF = t = 0.0255, bL/t = 9.8, bF/t = 33.8

0.910(bL/t)3 - bL/t = 846.7, 5 bF/t = 169, So, 0.910(bL/t)3 - bL/t > 5 bF/t

That is, the condition for minimum lip length, eq. (4) is satisfied

The upper limit for the lip length as defined by eqn. (5) is

which for the present section of constant thickness reduces to bL/bf ≤ 0.328. The actual value of bL/bf

= 0.25/0.863 = 0.29 ≤ 0.328. Hence, the test for upper limit of the lip length is also satisfied. Hence, segment 1 passes the lip test and qualifies to provide at least simply supported condition to segment 2 and be counted as a separate segment by itself.

Whether segment 1 passes the lip test or not can also be checked from the figure given by Niu (See Fig. 8 of this PPT). Locating the point bL/t = 9.8 and bf/t =33.8 in this figure, one can see that the point lies within the permissible range (Of course, some extrapolation will be required to locate the point on the plot).

So, we now proceed to the method of segments by (i) counting segment 1 as a separate segment with one end free and the other end simply supported and (ii) considering segment 2 as simply supported at both ends.

f

f

L

L

t

b0.328

t

b

The value of Fccn/Fcy in the above table could also have been calculated from the Gerard formula, eq. (1).

ccnnn FbtccnFnb nt nntb

n

n

t

b

nn

ccnnncc tb

FtbF

Since the thickness is constant, let us also solve the problem by the Gerard method. Since it is a Z-section, eq. (3) should be used.

or, psi

By the method of segments, we got the answer as 18,500 psi. The difference could be because it is strictly not a Z-section, because of the extra lip. In such a case, it is recommended to use the method of segments.

231.070000

10x5.10

1151.0

0255.02.3

F

E

A

t3.2

F

F75.03/162

0.751/3

CY

2

CY

CS

170,16000,70x231.0FCS