0 •..

INTERACTION CURVES FOR CIR~AR COl,UMN SECTIONS

by

Robert L~ Ketter

tor

~ethlehem Steel CompanyBethlehem, Pennsylvania

JUly 9, 1951

Fritz Engineering LaboratoryDepartment of Civll Engineering and Mechanics

Lehigh UniversityBethlehem, Pennsylvania

Fo L. Report No o CT -SO

."

-.•

I

- 1 ..,

INTRODUCTION

This report is divided into three parts. The first

presents the derivation of equations for pr~dicting failure

of a column, fixed at one end with moment and axial thrust

applied at the other, ~ere initial yield is used as a

criterion .of failure. Also included is a derivation of the

collapse solution using simple plastic theory where L/r~Oo

The second part presents sample calculations necessary

for plotting the interaction curves for a 1/2 inch rod,I

L/r =' 120.

The third part contains a complete set of curves for

bar sizes ranging from 3/a " to 7/a tl in diameter, varying

by sixteenths of an inch for slenderness ratios of 80, 90,

100, 110" and 120.

The modUlUS of elasticity has been assumed equal to

29,500,000 psi with an elastic limit of 36,000 psi. All

calculations are on file at the Fritz Engineering Laboratory,

Lehigh University, Bethlehem, Pennsylvania.

JO.

... 2 -

I ... DERIVATION

INITI~. YIELD INTERACTION CURVE:*. -Consider the column with axial load and moment as shown

below:

j-XP MA~.

~~BP

~: ' -I· L -I

The generaJ. equation of the deflection curve** for such

loading is:

'1= - ~ [Sin lot _ ~l + !A l-Sin k(L""'X) - <L'"'!X>]P Sin kL LJ P Sin kL L

------(1)

.-

If this expression is differentiated with respect to x an

equation 1s obtained for the slope ~t any point along the

deflected beam. By letting x =L and setting the resulting

slope at point B = 0, we can solve for MB in terms of

MA·

&f3 = y' = M~ tp ~~.-m + 6EI

u =~2

"...,

where;

(ValuesElastic

tjJ = ~ [s~ 2u - ~J~ =~ [~-T~2UJ·

of \\l and Ware tabulated in Timoshenko' sStability', pages 499-605).

**

Based on' Progress Rerrort L, WELDED CONTINUOUS FRAMESAN"D THEIR COMPONENTS, 'Interaction Curves for Columns II ,

by Robert L 0 Ketter and Lynn S. Beedle. . .

Timoshenko, S., "Theory of Elastic Stabilityll, McGrawHil1 D New York, 1936, page 12.

"1=

"1=

...

- 3 -

Sinee 9B = 0 ,MEL M~

~~ = 6'EI 0Therefore.

~MAMB =SUbstituting this value in the general equation for deflection~

(Eq. 1), an equation for a fixed - end condition is obtaIned.

_~ ..t [Sin kx -:'J + ~ [Sin k(L~) _(i.-.x:)lp 2~ Sin kL L p. Sin kL L J~ rSin k(L-x) _(L-x) _L(Sin lot _ ~)\ I -------------(2)P l Sin kL, L 24' Sin kL LLJ . . .

Differentiating Equation (2) twice with respect to x, the

curvature of the mamber is obtained as a function of distance

along the member.

.'

"I' =

"I" =

~ r -k (Cosk(L..ox)\ + 1 _Uk Cos kx .1~1p ISin ttL '\ 'J L 241'\ Sin lit -L')J-lilA ~ +~ . rJ -ks Sin kx,~)r LSin KL (-Sin k(L-::::) -~ Sin kt ""J

MA tt2 [ ."In = P Sin kL - Sin k(L-x)

To find the point of maximum moment in order that initial

yield can be investigated for this section, the expression for

moment, Equation (3), is differentiated w.ith respect to x and

-.•)

Since the curvature = ~ M!EI,

M = -El.!A. 2. . 1 [- Sin k(L-.x)P EI Sin KL

M = S~ HI. ~1n k(L-x) - ~ Sin kx] ---~~-~~--------~--(3)

set equal to zero o The solution of the resulting equation will

.. 4 -

determine the distance from the point' of applied moment to

the section of maximum mam~nt in terms of the axial load on

the column.

M' =-s~ ~ ~os k(L-x) + !..cos kxJ = 02~

Cos k(L-x)

Therefore.

or

x* =

= - -1. Cos lex2l\'

Tan -1 [-:atr- Cos kLlSin !CL J

k..--....- ..--...--------------..-( 4)

.'

' .•1

To find the axial load,' P, that will be required to

~intain a moment gradient of zero at the end of the column.

(the maximum load that can be applied and still have the

maximum mOOlent occur at the end of the column), set x = 0

in equation (4). Then,

o2\\' = - Cos }d,

Solution occurs when 2 u = 2'.33, where 2 u = kL ::: LMTherefore,

SUbstituting the value of LB as determined fram Euler buckling

load for a condition of one end fixed p the other pinned, PE,;

Included in this report is a curve for determining x..See page 'l.?. ell

... 5 -

, .

p -(2033)B E I

1f 2 E I(0.'1)'& ~,

..

...Therefor. as long as P is less tha.D or equal to (0.27 ~E') the

maximum mO+llent mll occur at the end of the column.

To plot the interaction curve# consider the basic' equation

for yielding at a section sUbjected to both bending momant and

axial load.

~y c "~ + ~ax ------~----------------------(5)

Where Mmax' for our particular case, woUld be determined from

Equation (3) substi tuting in value.s of :x determined from"

Equation ( 4) for sel.ected values of P.

Applying the above equation to m~bers of circular cross·

,seotion.

or

'Pcry = - +1Td8,.

M' . ::max

32

.\I

Where Mmax is determined by equation (3). Therefore.

WI _[a-XTrdIJ- 4Pd] Sin kL

,,! - 32[81n k(L-x) 4/SlY 'Sin ~

Using the peconnnended value of (J y = 36,000 psi,

- Pd1 Sin kLk(L-x) - 2 Sin lot

- 6 -

SUbstituting value of d for the various columns und~r

consideration in Equation (6) resl..:uts in the following

initial yield interaction curve equations

...d Interaction Curve Equation-

3/8 11 MA =O.0469(3.9757-P)(Q)

7/16 11 MA =O.0547(5.4ll9-P)(Q)

1/2 11 MA =O.0625{7.0686-P){Q)

9/16" MA =O.0703(8.9463-P)(Q)

5/8" MA =O.0781(1l.045-P)(Q)

.11/16" MA =O.0859(13.364-P)(Q)

3/4 11 MA - 0.0938(150905-P)(Q)

13/16" MA =0.1016(18.666-P)(Q) ,

7/8 11 MA =O.1094(2l.648-P)(Q)

-.. .

Q = Sin k L

Sin k(L-x) - : lV Sin kx

Note: Terms are defined on page 2.

... 7 -

"SIMILE PLASTIC TFIEORy"COLLAPSE Interaction Curves*for L/r =0.

Consider a column with axial load and end moment applied

...as shom b~low: p

L/r 0

For the l·imiting case where the axial load, P, is equal

to zero, (a beamproplem), the ultimate moment that can be

applied to the column, using the simple plastic theory, is

where Z is defined as the statical moment of the section.

.''For the other limiting case, that where the applied end

moment, Mo , is equal to zero, the maximum. axial load the~

column can carry i~

For the case where axial load and end moment both exist p

(investigating only the case of c~nplete plasticity), a stress

I .. I

I '! /1

i ---.Jr-:~ Iy

That due... to Mo

occur:

-- -+ --~.~ -.

That dueto P

'~.. J<:r. . i'j

. Total Stress.Distribution

distribution similar to. that shown below Would

ko-Y""I

I . I_ ...--!{_. ~/,/(!~

I !!'I~'If!).'. /~;/I

Based on Progress Report L, tiV'ELDED CONTnnJOUS cFRoAl~t'~Srr '"AND THEIE COMPONENTS, "Interaction Curves foruuu ...by Robert L o Ketter and L~ln S. Beedle.

..

- 8 -

The cross-sectional area, Al' supporting the axial load,

P, is shoWn as the shaded portion in the above figure. The

remaining area determines the moment, Moo

!l.:L '" 2~o¥(d/2). - (Yo)·· (d/2)·Sin-1 (d~\1

Bince P = er-yAl

P =20"Y ~~(dI'<O.-<1ol· + (a,.g).s1n-l(&~1--------(7)The 8upport1n \ or08s ....ect1onal moment 18

Mo = 4 (J1 ( '1 i<d/2) s - y2 dy \

. }~oOr,

Mo '" + 4/3 <fy td/ 2 ). - (10)·J3/2 --------------(8)

By selecting various values of 10 and·~ubstituting these

in equations (7) and (8), ex.pressic)]:~.s for P and Mo are

determined in terms of the di~-neter, d, of the column. The

fol1o~ng were usedl

'Yo P Mo

d/16 40488d2 5.866da

.{ile 8.905dB 5 o448drS,

dl4 17.219d2 · 3.893da

3d/e 24019'7ds l.'738dlS

7d/l6 26.804dft a.682d'"

.'

...

- 9 -

II. SAMPLE SOLUTION

Using' a 1/2rr rod. L/r = 120, for illustration.

SECTIONAL PROPERTIES:

d = 0.50 in. L = 15.00 in.

A - 0.1963 in. P-y = 7.0686 kips-,Ix = 0~003068 in04 My = 0.4417 in. kips

r.x = 0.1250 in. Mp = 0.7499 in. kips

S = 0.01227 in.8 PE , = 8.100 kips

Z = 0.02083 in.80.27PE' = 20187 kips

INITIAL YIELD FAILURE: Using the Equation shown on page 6.

Sin kLMo =. 0.0625 (7 0 0686 - P) S'in k(L -x) - i72\\i< Sin k.x}

r------..- r-'---''T___._._ .......,_

(1) (2) ( 3) (4) (5) (6) ( 7) ( 8)p PIEI {(2) =k krt=2u .xlt x L-x k(L-~.-

2.3782.5 0.02762 001662 2.493 0.0460 0.690 14.3103.5 0.03867 0.1966 2.949 0.1465 2.198 12.802 2.517 I4.5 0.04972 0.2230 3.345 0.2130 3.195 11.805 2.633 I

5.5 0.06077 0.2465 30698 0.2605 3.908 11.092 2.7346.5 0.07182 0.2680 4.020 0.2985 4.478 10'.522 2.820---( g') (10) (11) , (12) - (13) (14.)

!Or. Sin k{L-x) Sin kx 0 '4J ll~

0.1147 0.69036 0.11445 3.0190 2.0704 4.1408·0.4321 0.58347 0.41877 9.9422 5.5628 11.12560.7125 0.48556 ' 0.65372 -9.9835 -4.3639 -8.'12780.9633 0.39493 0.82107 -3.5136 -1.0863 -2 ..1726102001 0.31457 0.93204 -203147 -004353 -o.G706..

(15) (16) . (1;;"')'··-r-=.J].ss-· '~ 19 ) ~20)--

0/2ty :0/2~Sln kx) (10)-(16) Sin kL (18~'(17) 7.686-P

0.7291 0.0834 0.6070 , 0.60279 0.9931 ' 4.56860.8936 0.3742 0.2093 0.18984 i 0.9070 3.56861.1439 . 0.7478 -0.2622 -0.20357 0.7764 2.5686

. 1.6172 1.3278 -0.9329 -0.5294.9 0.5676 1.56862.6587 2.4:780 -2.1634 -0.770'74 '0.4563 0.5686

- 10 -

(21 )--,--" (22)'-'·~··

0.0625x(20) (2l)x{19)= M~

0.2855 0.28350.2230 0.20230.1605 0.12460.0980 0.05560,0355 0.0162

Plotting these points give the following curve.

p

p

I . a.SM~

- 11 -

SIMPLE PLASTIC THEORY COLLAPSE (L/r=O):

..,

. (Using equations shown on page

- ._...__...........

Yo P M

d/16 1.122 0.733,

d/8 2.226 0.681

d/4 4.305 0.487

B.d/s 6.049 0.217

7d/16 6.701 I 0.085

8 .)

Plotting these p.oints glve the following curve,.

s.o

p

~o

a,s

... Figo 1

Fig. 2

Fig. 3

Fig. 4

Flg. 5

Fig. 6

Fig. 7

Figc 8

Fig. 9

Fig. 10 -

-12-

o

III FIG'UJ{ES

Interaction curve f'or.3/a ll rodL/r= 80, 90, 100, 110 and 120

Interaction curve for 7/16" rodL Ir = 80, 90, 100, 110 an·d 120

"Interaction curve for liz" rodL/r = eo, 90, 100, 110 and 120

Interaction curve for 9/1Su rodLIT:' ::: 80, 90, 100, 110 and 120

Interaction curve for 5/8" rodL'Ir ::: 80, 90, 100, 110 and 120

Interaction curve .for 11/16u rodL/r = 80, 90, 100, 110 and 120

Interaotion curve for 3/4" rodL/r ::: eo, 90, 100, 110 and 120

-Interaction curve for 13/16" rodL/r ::: 80, 90, 100, 110 m d 12'0

Interaction curve .for 7/eu rodL/r = 80, 90, 100, 110 and 120

Curve for determining distance toseotion of maximum moment

page

14

15

16

17

19

20

21

22

- I"~-.1\\5

R,L.K.

...

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16 i?L. ~.

..

[)

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Ia.~ I

~ ~

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o 0(\1 i'J

Dl

FIG. '2

-------- ---~

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..

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