EXAMPLE 13.2
Copyright © 2011 Pearson Education South Asia Pte Ltd
A W150×24 steel column is 8 m long and is fixed at its ends as shown in Fig. 13–11a. Its load-carrying capacity is increased by bracing it about the y–y (weak) axis using struts that are assumed to be pin connected to its mid-height. Determine the load it can support so that the column does not buckle nor the material exceed the yield stress. Take Est = 200 GPa and σY = 410 MPa.
EXAMPLE 13.2 (cont)
Copyright © 2011 Pearson Education South Asia Pte Ltd
• Effective length for buckling about the x–x and y–y axis is
• From the table in Appendix B,
• Applying Eq. 13–11,
Solutions
mm 2800m 8.22/87.0
mm 4000m 485.0
y
x
KLKL
46
46
mm 1083.1
mm 104.13
y
x
I
I
2 62
2 2
2 62
2 2
200 13.4 101653.2 kN (1)
4000
200 1.83 10460.8 kN (2)
2800
cr xx
cr yy
EIPKL
EIPKL
EXAMPLE 13.2 (cont)
Copyright © 2011 Pearson Education South Asia Pte Ltd
• By comparison, buckling will occur about the y–y axis.
• The average compressive stress in the column is
• Since this stress is less than the yield stress, buckling will occur before the material yields.
• Thus,
• From Eq. 13–12 it can be seen that buckling will always occur about the column axis having the largest slenderness ratio, since it will give a small critical stress.
Solutions
32
460.8 10150.6 N/mm 150.6 MPa
3060cr
crPA
461 kN (Ans)crP
13.4 SECANT FORMULA
Copyright © 2011 Pearson Education South Asia Pte Ltd
• For design of a column subjected to eccentric load, consider the moment-curvature equation
2 2 2
''
or '' where
EIV M P e vPv v eEI
13.4 SECANT FORMULA
Copyright © 2011 Pearson Education South Asia Pte Ltd
• The solution is
• Since v = 0 at x = 0, so C2 = e
• Since v = 0 at x = L,
So,
2 2 2
''
or '' where
EIV M P e vPv v eEI
exCxCv cossin 21
21 cos 2sin and sin 2sin cos2 2 2L L LL L
2tan1
LeC
SECANT FORMULA (cont)
Copyright © 2011 Pearson Education South Asia Pte Ltd
• Hence,
• And
• To find we require
1cossin
2tan xxLev
max 2 sec 2 1x Lv e L
crP
2sec L
EIPcr
22
L
EIPcr
2
2 crEIP
L
SECANT FORMULA (cont)
Copyright © 2011 Pearson Education South Asia Pte Ltd
• Hence,
• And
• Note: A nonlinear relationship occurs between the load P and the defection v. As a result, the principle of superposition does not apply here.
1cossin
2tan xxLev
12sec2max Lev Lx
SECANT FORMULA (cont)
Copyright © 2011 Pearson Education South Asia Pte Ltd
• Maximum moment occurs at the column’s midpoint, i.e.
• Hence the maximum stress is
max
max 2
sec2
Or, 1 sec2
P Mc P Pe c LA I A I
P ec L PA r r EA
2secor max
LPeMvePM
SECANT FORMULA (cont)
Copyright © 2011 Pearson Education South Asia Pte Ltd
Design curves:• The above equation for maximum stress σmax is
transcendental, and cannot be solved explicitly. • Graphs to aid designer are available.
EXAMPLE 13.4
Copyright © 2011 Pearson Education South Asia Pte Ltd
The W200 x 59 A-36 steel column shown in Fig. 13–17a is fixed at its base and braced at the top so that it is fixed from displacement, yet free to rotate about the y–y axis. Also, it can sway to the side in the y–z plane. Determine the maximum eccentric load the column can support before it either begins to buckle or the steel yields.
EXAMPLE 13.4 (cont)
Copyright © 2011 Pearson Education South Asia Pte Ltd
• For y–y axis buckling, it is subjected to an axial load P.
• For x–x axis yielding, it is subjected to an axial load P and moment M.
Solutions
(Ans) kN 4.419N 419368
10143.1sec598.2110895.1
2sec1
36
2
x
xx
x
x
x
x
xY
PPP
EAP
rKL
rec
AP
kN 51362800
104.20102002
632
2
2
y
yycr KL
EIP
13.5 INELASTIC BUCKLING
Copyright © 2011 Pearson Education South Asia Pte Ltd
• In engineering practice, columns are generally classified according to the type of stresses developed within the column at the time of failure.
• Long slender columns will become unstable when the compressive stress remain elastic. It is referred as elastic instability.
• Intermediate Columns fail due to inelastic instability, meaning that compressive stress at failure is greater than the material’s proportional limit (σpl).
• Short Columns do not become unstable; rather the material simply yields or fractures.
INELASTIC BUCKLING (cont)
Copyright © 2011 Pearson Education South Asia Pte Ltd
• Consider a column having a slenderness ratio KL/r. The formula for critical stress depends on the value of the slenderness ratio.– When KL/r ≥ (KL/r)pl, apply the Euler’s formula.– When KL/r ≤ (KL/r)pl, apply the Engesser formula.
EXAMPLE 13.5
Copyright © 2011 Pearson Education South Asia Pte Ltd
A solid rod has a diameter of 30 mm and is 600 mm long. It is made of a material that can be modeled by the stress–strain diagram shown in Fig. 13–19. If it is used as a pin-supported column, determine the critical load.
EXAMPLE 13.5 (cont)
Copyright © 2011 Pearson Education South Asia Pte Ltd
• The radius of gyration is
• The slenderness ratio is
• Engesser equation states that
• For elastic critical stress, thus .
• Inelastic buckling occurs since
Solutions
805.7
6001
rKL
mm 5.715
154/2
4
AIr
t
tcr E
rKLE 3
2
2
10542.1
GPa 150001.0
150E MPa 3.231cr
MPa 150 plcr
EXAMPLE 13.5 (cont)
Copyright © 2011 Pearson Education South Asia Pte Ltd
• From the second line segment of the graph,
• Applying the value, we have
• Since this value falls within the limits of 150 MPa and 270 MPa, it is critical stress.
• The critical load on the rod is therefore
Solutions
MPa 1.1851012010542.1 33 cr
GPa 120001.0002.0
150270
tE
(Ans) kN 131015.01.185 2 AP crcr
13.6 ALLOWABLE STRESS FOR CONCENTRIC LOADING
Copyright © 2011 Pearson Education South Asia Pte Ltd
• Practically, columns are not perfectly straight as assumed in theory. By performing experimental tests on a large number of axially loaded columns, the results may be plotted and a design formula developed by curve-fitting the mean of the data.
• Note: The experimental curve is similar to that determined from the secant formula. It accounts for the influence of an “accidental” eccentricity ratio on the column’s strength.
ALLOWABLE STRESS FOR CONCENTRIC LOADING (cont)
Copyright © 2011 Pearson Education South Asia Pte Ltd
• For steel columns:
33
2
2
2
2
2
8//83352
1
2
21
cc
Yc
allow
Yc
cY
rKLrKLrKLrKLrKLrKL
Er
KL
rKLE
2
212 for 200
23allow
c
E KL KLr rKL r
for c
KL KLr r
ALLOWABLE STRESS FOR CONCENTRIC LOADING (cont)
Copyright © 2011 Pearson Education South Asia Pte Ltd
• For aluminum columns:
rKL
rKL
rKL
rKLr
KL
allow
allow
allow
55 378125
5512 MPa 628.15.214
120 MPa 195
2
ALLOWABLE STRESS FOR CONCENTRIC LOADING (cont)
Copyright © 2011 Pearson Education South Asia Pte Ltd
• For timber columns:
5026 MPa 3718
2611 MPa 0.26/
31125.8
110 MPa 25.8
2
2
dKL
dKL
dKLdKL
dKL
allow
allow
allow
EXAMPLE 13.6
Copyright © 2011 Pearson Education South Asia Pte Ltd
An A-36 steel W250 x 149 member is used as a pin-supported column, Fig. 13–24. Using the AISC column design formulas, determine the largest load that it can safely support.
EXAMPLE 13.6 (cont)
Copyright © 2011 Pearson Education South Asia Pte Ltd
• From Appendix B,
• Since k = 1 for both x and y axis buckling,
• From Eq. 13–22, we have
• Here , so
• The allowable load P on the column is therefore
Solutions
18.744.67
50001
rKL
mm 4.67 mm 117 mm 19000 2 yx rrA
crKLrKL //0
66.125250
1020022 322
Yc
Er
KL
2
2
3 3
12
110.85 MPa5 3 3 8 / / 8
Y
callow
c c
KL rKL r
KL r KL r KL r KL r
(Ans) kN 210619000
85110 ; PP.AP
allow
13.7 DESIGN OF COLUMNS FOR ECCENTRIC LOADING
Copyright © 2011 Pearson Education South Asia Pte Ltd
• Method 1: Use available column formula
• Method 2: Use Interaction Formula
IMc
APmax where M = Pe such that σmax ≤ σallow
1a b
a ballow allow
where σa denotes the stress due to axial load, σb denotes the stress due to bending.
EXAMPLE 13.10
Copyright © 2011 Pearson Education South Asia Pte Ltd
The column in Fig. 13–29 is made of aluminum alloy 2014-T6 and is used to support an eccentric load P. Determine the maximum magnitude of P that can be supported if the column is fixed at its base and free at its top. Use Eq. 13–30.
EXAMPLE 13.10 (cont)
Copyright © 2011 Pearson Education South Asia Pte Ltd
• From Fig. 13–10b, K = 2.The largest slenderness ratio for the column is therefore
• By inspection, Eq. 13–26 must be used (277.1 > 55). Thus,
• The maximum compressive stress in the column is determined from the combination of axial load and bending.
• Assuming that this stress is uniform over the cross section, we require
Solutions
1.277
8040/408012/1
160023
r
KL
MPa 92.4
1.277378125
/378125
22 rKLallow
PI
cPeAP 00078125.0max
(Ans) kN 30.600078125.0.924 ;max PPallow
EXAMPLE 13.12
Copyright © 2011 Pearson Education South Asia Pte Ltd
The timber column in Fig. 13–31 is made from two boards nailed together so that the cross section has the dimensions shown. If the column is fixed at its base and free at its top, use Eq. 13–30 to determine the eccentric load P that can be supported.
EXAMPLE 13.12 (cont)
Copyright © 2011 Pearson Education South Asia Pte Ltd
• Since K = 2,
• Since 26 < KL/d < 50, the allowable axial stress is
• With σallow = σmax,
Solutions
MPa 324.2
403718
/3718
22 dKLallow
4060
12002
dKL
(Ans) kN 35.3
12060121)60)(80(
12060324.2 3
P
PPI
McAP
allow