ebt 252-lecture 5

EBT 252/4: STRENGTH OF MATERIALS - J. B. JOHNSON FORMULA - AISC COLUMN FORMULAS

By

DR. SRI RAJ RAJESWARI MUNUSAMY

PPK BAHAN, UNIMAP

E-mail: rajeswari@unimap.edu.my

2/12/2013 1

J.B.JOHNSON FORMULA

The Euler formula does not apply for the

intermediate columns.

Hence, many semi-empirical formulas

have been developed.

J.B.Johnson formula is used extensively in

steel structure design and machine

design.

2/12/2013 2

2/12/2013 3

- J. B. Johnson formula is the equation of parabola with its vertex at the

point on the vertical axis with ordinate equal to y.

-The parabola is tangent to the Euler curve at the transition slenderness

ratio kL/r = Cc, which equals to ½ of the yield stress, y of the steel.

kL/r

The value of transition slenderness ratio, Cc can be

determined as follows :-

Thus,

The J. B. Johnson formula is:

2/12/2013 4

𝐂𝐜 =𝟐𝛑𝟐𝐄

𝛔𝐲 𝐄𝐪. 𝟗

𝛔𝐜𝐫 =𝐏𝐜𝐫

𝐀= 𝟏 −

𝐤𝐋𝐫

𝟐

𝟐𝐂𝐜𝟐 𝛔𝐲 𝐄𝐪. 𝟏𝟎

The Euler formula applies when kL/r is

greater than Cc and the J.B.Johnson formula

applies when kL/r is less than Cc.

For kL/r = Cc, both formulas give the same

result.

The Euler formula applies to all materials,

whereas the J.B.Johnson formula applies

mainly to ductile steel.

2/12/2013 5

Example 1

Determine the allowable compressive

load of a 4-in., standard weight steel

pipe that is 25ft long. The column is

made of A36 steel with y = 36 ksi and

is welded to fixed supports at both

ends. Use F.S. = 2 and E=29 x 103 ksi.

2/12/2013 6

Solution:

From Table A-5(a) in Appendix 1, for a 4-in., standard

weight steel pipe,

A = 3.17 in.2

r = 1.51 in

The slenderness ratio is:

2/12/2013 7

From Eq.9, the value of the transition slenderness

ratio Cc is :

Since kL/r < Cc, the J.B.Johnson formula applies.

From Eq.10, we find,

2/12/2013 8

Thus,

2/12/2013 9

THE AISC COLUMN FORMULAS

The American Institute of Steel

Construction (AISC) manual gives formulas

for calculating the allowable compressive

stresses to be used in steel column design.

The AISC column formulas are essentially

the critical buckling stresses from the Euler

and J.B.Johnson formulas divided by the

factor of safety.

2/12/2013 10

The AISC formulas are :-

1. For long columns :

2. For intermediate and short columns :

2/12/2013 11

𝛔𝐚𝐥𝐥𝐨𝐰 =𝛑𝟐𝐄/(𝐤𝐋/𝐫)𝟐

𝐅. 𝐒=

𝛑𝟐𝐄/(𝐤𝐋/𝐫)𝟐

𝟏. 𝟗𝟐 𝐄𝐪. 𝟏𝟏

𝛔𝐚𝐥𝐥𝐨𝐰 =

𝟏 −𝐤𝐋/𝐫 𝟐

𝟐𝐂𝐜𝟐 𝛔𝐲

𝐅. 𝐒 𝐄𝐪. 𝟏𝟐

Where the factor of safety, F.S. is computed from:

2/12/2013 12

* F.S varies from 5/3 (or 1.67) when kL/r = 0 to 23/12(or 1.92)

when kL/r = Cc

𝐅. 𝐒 =𝟓

𝟑+

𝟑𝐤𝐋𝐫

𝟖𝐂𝐜−

𝐤𝐋𝐫

𝟑

𝟖𝐂𝐜𝟑 𝐄𝐪. 𝟏𝟑

Table 1 shows the value of the AISC recommended

effective length factor k for steel column design

when the end-supporting conditions are

approximated.

2/12/2013 13

End Conditions Pinned

Ends

Fixed Ends Fixed,

Pinned Ends

Fixed, Free

Ends

Theoretical k

value

1.0 0.5 0.7 2.0

AISC

recommended

k value

1.0 0.65 0.8 2.10

Table 1: AISC Recommended k Values

Values of the allowable compressive

stress computed from the AISC

formulas corresponding to y = 36 ksi

and y= 50 ksi are tabulated for kL/r

values from 1 to 200 in Tables 2 and 3.

2/12/2013 14

2/12/2013 15

Table 2: AISC Allowable Compressive Stress for Steel Columns for

y = 36 ksi (250 MPa)

Table 3: AISC Allowable Compressive Stress for Steel Columns for

y = 50 ksi (345 MPa)

2/12/2013 16

Example 1 Determine the allowable axial compressive load

for a 10-ft long standard L6 X 4 X ½ steel angle of

A36 steel if the supporting conditions are (a)

pinned at both ends or (b) fixed at both ends.

Use the AISC formulas and the recommended k

values.

Solution:

From the Appendix 2, Table A-4(a), for an L6 X 4 X

½ steel angle, A= 4.75in2 and the least radius of

gyration is rz=0.870in.

2/12/2013 17

2/12/2013 18

Since kL/r >Cc, Eq.11 applies. Thus,

Or from Table 19-2, for y=36ksi and kL/r =138(rounded to the

nearest whole number for use in the table, interpolation is not necessary), the allowable compressive stress is allow =

7.84ksi, the same as calculated above. Thus,

𝐏𝐚𝐥𝐥𝐨𝐰 = 𝛔𝐚𝐥𝐥𝐨𝐰𝐀 = (𝟕. 𝟖𝟒𝐤𝐢𝐩𝐬/𝐢𝐧𝟐)(𝟒. 𝟕𝟓𝐢𝐧𝟐)= 𝟑𝟕. 𝟑𝐤𝐢𝐩𝐬

2/12/2013 19

Table 2

2/12/2013 20

2/12/2013 21

Table 2,

Example 2 A 3-m column having an L127 X 127 X12.7

angle section (Refer to Appendix 3) is

made of A242 steel with E=200GPa and

y=345MPa. The column is fixed at both

ends. Calculate the allowable axial

compressive load using the AISC formulas

and the recommended k values. Use the

allowable stress listed in Table 19-3 to verify

the computations.

2/12/2013 22

𝐀 = 𝟑. 𝟎𝟔 × 𝟏𝟎−𝟑𝐦𝟐; 𝐫𝐦𝐢𝐧 = 𝐫𝐳 = 𝟎. 𝟎𝟐𝟓𝐦

𝐅𝐨𝐫 𝐋𝟏𝟐𝟕 × 𝟏𝟐𝟕 × 𝟏𝟐. 𝟕.

Table 3,

Solution :

𝐅𝐨𝐫 𝐟𝐢𝐱𝐞𝐝 𝐞𝐧𝐝𝐬, 𝐤 = 𝟎. 𝟔𝟓

𝐤𝐋

𝐫=

(𝟎. 𝟔𝟓)(𝟑𝐦)

𝟎. 𝟎𝟐𝟓𝐦= 𝟕𝟖

𝐂𝐜 =𝟐𝛑𝟐𝐄

𝛔𝐲=

𝟐𝛑𝟐(𝟐𝟎𝟎 × 𝟏𝟎𝟗)

𝟑𝟒𝟓 × 𝟏𝟎𝟔= 𝟏𝟎𝟔. 𝟗 = 𝟏𝟎𝟕

𝐤𝐋

𝐫< 𝐂𝐜 ; 𝐓𝐡𝐞𝐫𝐞𝐟𝐨𝐫𝐞 𝐉. 𝐁. 𝐉𝐨𝐡𝐧𝐬𝐨𝐧 𝐟𝐨𝐫𝐦𝐮𝐥𝐚 𝐚𝐩𝐩𝐥𝐢𝐞𝐬

𝐅. 𝐒 =𝟓

𝟑+

𝟑𝐤𝐋𝐫

𝟖𝐂𝐜−

𝐤𝐋𝐫

𝟑

𝟖𝐂𝐜𝟑 =

𝟓

𝟑+

𝟑 𝟕𝟖

𝟖 𝟏𝟎𝟕−

𝟕𝟖 𝟑

𝟖 𝟏𝟎𝟕 𝟑

2/12/2013 23

2/12/2013 24

Table 3,

2/12/2013 25

Appendix 1 2/12/2013 26

Appendix 2 2/12/2013 27

Appendix 3 2/12/2013 28

Appendix 4 2/12/2013 29