Perry Formula Strut

Consider the B-B axis:

>(1.0x3.0) = 3.0 m >(0.85 x 5.0) = 4.25 m The effective buckling length .■. Lb-b = 4.25 m

Slenderness

212.7

Since 1.4Xc is the largest value this should be used to determine the value of pc using the

Perry strut formula.

Examples in structural analysis 510 limfllqg sLcndamcss h = O.Sf^J:/^)65 ^ |0,2 x {i? x 205000/275)91] -17.15 r)»fj(jJ - ^yiOQO - 5.5(63.02 - |7.1SytO0O = 0.257

lh m

Pv.Py

44E.7

(jgjjx 275)

m.4 Wmni1

Critical value ejPjiir — 1 W.4 M/mmL Compression resistance K = (to* AJ = {19SA X 21* X ilfytf = Ml kN

A similar approach is taken when designing battened struts, the corresponding Clause in BS 5950-1:2000 Structural Use of Steelwork in Building is as follows:

"The slenderness X of a main component (based on its minimum radius of gyration) between end welds or end bolts of adjacent battens should not exceed 50. The slenderness Xb of the battened strut about the axis perpendicular to the plane of the battens should be calculated from:

where Xm is the ratio LE/r of the whole member about that axis. If Xb is less than 1.4Xc the design should be based on Xb=1.4Xc. "

The application of this is illustrated in the solution to Problem 7.5.

6.9 Example 6.4 Compound Section

A column ABCE of a structure is shown in Figure 6.22. The column is 15.0 m long and supports a roof beam DEF at E. The beam carries a load of w kN/m length along its full length DEF. The column is fabricated from a 152x152x23 UC with plates welded continuously to the flanges as shown. Using the data given determine:

(i) the compression resistance of the column, and

(ii) the maximum value of w which can be supported.

 152x152x23 UB Universal Beam Cross-sectional Area (A) 29.2 cm2 Radius of Gyration (ryy) 3.70 cm Radius of Gyration (rxx) 6.54 cm 2nd Moment of Area (Ixx) 1250 cm4 2nd Moment of Area (Iyy) 400 cm4 J.Oni 15.0 fh

Yield Stress py=275 N/mm2

Young's Modulus E=205 kN/mm2

Robertson Constants: Figure 6.22

Solution: Perry strut formula: Note: Since the same curve is used for both the A-A and the B-B axes in this case (i.e. a=5.5), the compression resistance will correspond to the axis with the highest slenderness value, i.e. the one which produces the lowest pc value. Consider the A-A axis: Consider the B-B axis:

Buckling length

The effective buckling S tenderness =

Lb-B^O-O x 6,0) - 6.0 m > (0.S5 x 9.0) = 7,65 m Lb k = 7.65 til length 7650 61.1

Since XA-a is the largest value this should be used to determine the value of pc using the

Perry strut formula.

 IV*! _ (a* xMSKIO* ' m [ 135.77; ,

109.5 N/ilim ij = t4,jL- E 5.5(135.77 - 17,15)/1000 = 0.652

109.5 N/ilim

Limiting slendemess k> = * ° (W * X ¿05000/275)°f] = 17.15

PlPv

= 30.28 N/ram f+ty-PtPrV 228L2 +(228.2*-109.&x275j"

Critical value of pc=80.28 N/mm2 Compression resistance Pc=(pcX^g)=(80.28x6.92x103)/103=555.5 kN (ii) Examples in structural analysis 514 The maximum value of the vertical reaction at E=555.5 kN

6.10 Problems: Buckling Instability

A selection of column cross-sections is indicated in Problems 6.1 to 6.7 in addition to the position of the restraints about the x-x and y-y axes. Using the data given and the equation for the European Column Curves, (the Perry strut formula) determine the value of the compressive strength pc and hence the compression resistance, for each section. Data:

 Problem No. py (N/mm2) E (kN/mm2) Robertson Constant x-x y-y 6.1 275 205 3.5 2.0 6.2 255 205 3.5 3.5 6.3 275 205 5.5 5.5 6.4 255 205 3.5 3.5 6.5 275 205 5.5 5.5 6.6 275 205 5.5 5.5 6.7 255 205 5.5 5.5

Buckling instability 515 Table 6.3- Section Pro perty Data

 Section Property Section 533x210x82 UB 457x152x52 UB 200x90x30 Channel 150x100x10 Hollow Section Overall Depth (D) 528.3 mm 449.8 mm 200.0 mm 100.0 mm Overall Breadth (B) 208.8 mm 152.4 mm 90.0 mm 50.0 mm Cross-sectional Area (A) 105 cm2 66.6 cm2 37.9 cm2 Gyration (ryy) - - 2.88 cm 3.01 cm 2nd Moment of Area (Ixx) 47500 cm4 21400 cm4 2520 cm4 1160.0 cm4 2nd Moment of Area (Iyy) 2010 cm4 645 cm4 314 cm4 614.0 cm4 Problem 6.1 Problem 6.2

taEMH.

9,fi mm

Rtslnhl

;itnml ihf

A-A inis

533 K 2t0v 82 UfJ Problem 6.4

KiiLmnL atom die .VA Jivis hittrresl 1.25 m cchIke

2 i 100 x DO k JO channel sections

Kbinill

hittrresl 1.25 m cchIke

Kbinill

2 i 100 x DO k JO channel sections

Problem 6.5 Problem 6.6 2/JWx SO* JD channel j(M *vil]JOJ V.! "IV Due* Problem 6.7

6.11 Solutions: Buckling Instability Solution

Topic: Uiicklin° 1 us [li bi IJ l\ I'rnlilviu ¡M(im(«rf

3-0)= 1,7 m The effective buckling length ,"- ™ 2jO m 2000

Limiting slendemtss At - 0,2<,tEffK}"' [0,2 * ( ¿r >: 205000075?") -17,15

Con KiiJcr the i-\ aiisT [n = i.5) EJneklinj? length x 4.0) = <f.Qin

£(0.85 x4J0) = 3.4 m Ttie effective buckling lenyli ,". " 4.0 ni

TilA