CH 13
CH 13
CH 13
1
A 7.2-m-long A-36 steel tube having the cross section shown in Fig. 13–9 Pcr
is to be used as a pin-ended column. Determine the maximum allowable
axial load the column can support so that it does not buckle.
Solution
Using Eq. 13–5 to obtain the critical load with Est 200 GPa,
70 mm
p2EI
Pcr =
L2
] A 14– B in4
7.2 m 75 mm
22
[29110362) kip>in
p [200(10 kN/m2](– p132 (70)4)(1
(75)44 - 14–1–4p12.7524 m/1000 mm)4
= ———————————————————————— 2 2
[24 ft12 in.>ft2]
(7.2 m)
228.2 kN Ans.
This force creates an average compressive stress in the column of
Pcr 64.5(1000
228.2 kN kip N/kN)
scr = = ———–——————— = 14.3 ksi N/mm2 100 MPa
100.2
A [(75) -p12.752
[p132 2 2
(70)22]] mm
in2 2 Pcr
Since cr Y 250 MPa, application of Euler’s equation is appropriate.
Fig. 13–9
E X A M P L E 13.2
The A-36 steel W200 46 member shown in Fig. 13–10 is to be used as
a pin-connected column. Determine the largest axial load it can support
before it either begins to buckle or the steel yields. x
Solution
y y
From the table in Appendix B, the column’s cross-sectional area and
moments of inertia are A 5890 mm2, Ix 45.5 106 mm4, and 4m
x
Iy 15.3 106 mm4. By inspection, buckling will occur about the y–y axis.
Why? Applying Eq. 13–5, we have
[291106)32kN/m
p2EI p2[200(10
2 2 2
kip>in ]137.14)in
](15.3(10 2 )(1 m/1000 mm)4
4 4
mm
Pcr = = ——————————–———————————— = 512 kip 1887.6 kN
L2 (4 2m)2
[12 ft12 in.>ft2]
When fully loaded, the average compressive stress in the column is
Pcr 512 kip kN (1000 N/kN)
1887.6 Fig. 13–10
scr = = ———–———————= 56.1 ksi 320.5 N/mm2
A 9.13 in2 5890mm2
Since this stress exceeds the yield stress (250 N/mm2), the load P is
determined from simple compression:
P P
36 ksi
250 = 2 ————
N/mm 2
; ; P 1472.5 kN Ans.
9.13 in5890mm2
In actual practice, a factor of safety would be placed on this loading.
E X A M P L E 13.3
P
x
A W150 24 steel column is 8 m long and is fixed at its ends as shown in
y Fig. 13–13a. 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
y midheight. Determine the load it can support so that the column does not
x buckle nor the material exceed the yield stress. Take Est 200 GPa and
4m Y 410 MPa.
Solution
The buckling behavior of the column will be different about the x and y axes
due to the bracing. The buckled shape for each of these cases is shown in
4m Figs. 13–13b and 13–13c. From Fig. 13–13b, the effective length for buckling
about the x–x axis is (KL)x 0.5(8 m) 4 m, and from Fig. 13–13c, for
buckling about the y–y axis, (KL)y 0.7(8 m/2) 2.8 m. The moments of
(a)
inertia for a W150 24 are determined from the table in Appendix B. We
have Ix 13.4 106 mm4, Iy 1.83 106 mm4.
Fig. 13–13a
Applying Eq. 13–11, we have
p2EIx p [200(1026)ksi]29.1 6
2 3 4
2[29110 kN/m2]13.4(10 ) m4
1Pcr2x =
in
= = kip 1653.2 kN (1)
1KL22x 1144 in.2
———–——————————— 401.7
(42 m)2
p2EIy [29110326)ksi]9.32
p2[200(10 in4
kN/m2]1.83(10 6
) m4
1Pcr2y =
4m
= ———–——————————— = kip 460.8 kN (2)
1KL22y 1100.8 in.2
262.5
(2.82 m)2
By comparison, buckling will occur about the y–y axis.
The area of the cross section is 3060 mm2, so the average compressive
stress in the column will be
x – x axis buckling
Pcr 262.5 kip 3) N
460.8(10
(b) scr = = —————— ksi
= 59.3 150.6 N/mm2
A 3060
4.43 in2mm2
Fig. 13–13b Since this stress is less than the yield stress, buckling will occur before
the material yields. Thus,
Pcr 461 kN Ans.
2.8 m Note: From Eq. 13–11 it can be seen that buckling will always occur
about the column axis having the largest slenderness ratio, since a large
slenderness ratio will give a small critical load. Thus, using the data for
the radius of gyration from the table in Appendix B, we have
a b = ———–————
KL 4 m(1000
144 in. mm/m)
= 56.2 60.4
r x 2.56 in.
66.2 mm
a b = ————–————
KL 2.8 m(1000
100.8 in. mm/m)
y – y axis buckling = 69.0 114.3
r y 1.46 in.
24.5 mm
(c)
Hence, y–y axis buckling will occur, which is the same conclusion reached
Fig. 13–13 by comparing Eqs. 1 and 2.
E X A M P L E 13.4
The aluminum column is fixed at its bottom and is braced at its top by cables
so as to prevent movement at the top along the x axis, Fig. 13–14a. If it is z
assumed to be fixed at its base, determine the largest allowable load P that
can be applied. Use a factor of safety for buckling of F.S. = 3.0. Take
Eal = 70 GPa, sY = 215 MPa, A = 7.5110-32 m2, Ix = 61.3110-62 m4,
P
y
Iy = 23.2110-62 m4.
y
Solution
Buckling about the x and y axes is shown in Fig. 13–14b and 13–14c,
respectively. Using Fig. 13–12a, for x–x axis buckling, K = 2, so
1KL2x =215 m2 = 10 m. Also, for y–y axis buckling, K = 0.7, so
x
Le = 3.5 m
P
25 mm
50 mm
y 75 mm
75 mm
x
4.5 m
Fig. 13–19
Solution
Computing the necessary geometrical properties, we have
12
1
Ix = (50in.216 = 363 in.
in.23 mm)
mm)(150 414.06 106 mm4
12
A = 12
(50in.216 in.2 =mm)
mm)(150 27500 mm4
12 in.
36 in.4 106 mm4
14.06
rx = = 1.732
——–—–——–— in. 43.30 mm
B 2
D12 in.7500 mm2
e = 125in.
mm
KL = 1(4.5
1115 ft2112 in.>ft2= 4500
mm)(1000) 180 in.
mm
KL 180 in.
4500 mm
= ———–—= 104
104
rx 43.30 in.
1.732 mm
Since the curves in Fig. 13–18b have been established for Est 200(103) P (MPa)
—
A
MPa, and Y 250 MPa, we can use them to determine the value of
P>A and thus avoid a trial-and-error solution of the secant formula. Here 300
KL>rx = 104. Using the curve defined by the eccentricity ratio 250 ec = 0.1
—
ec/r2 (25 mm)(75 mm)/(43.30 mm)2 1, we get 200
r2
0.5 Euler’s formula
Eq. 13–6
P ec = 0
L 12
83 ksi
MPa 100
1.0 —
r2
A 1.5
Fig. 13–18
c1 + 2 sec a bd
P ec L P
smax =
A r 2r A EA
250 1211
36 ⱨ 83[1 + sec60.65°]
sec 60.56°2
36 L 36.4
250 252.3
vmax = ecsec a b - 1d
P L
A EI 2
= 25 mm sec
1 in.csec[ (
a
144 kip 622.5(10
———–————–——–—–——–—
A 29110 3
D 200(102 ksi136
3
3
180) Nin.
in 2 14.06
) N/mm 24
2 10 mm 2 ) ]
b -6 1 d 4 ——–—— 1
4500 mm
25 mm[see 60.65° 1]
26.0 mm Ans.
E X A M P L E 13.6
z The W200 59 A-36 steel column shown in Fig. 13–20a is fixed at its
x P
base and braced at the top so that it is fixed from displacement, yet free
200 mm
to rotate about the y–y axis. Also, it can sway to the side in the y–z plane.
y
Determine the maximum eccentric load the column can support before
y it either begins to buckle or the steel yields.
Solution
x
From the support conditions it is seen that about the y–y axis the column
4m behaves as if it were pinned at its top and fixed at the bottom and
subjected to an axial load P, Fig. 13–20b. About the x–x axis the column
is free at the top and fixed at the bottom, and it is subjected to both an
axial load P and moment M P(200 mm), Fig. 13–20c.
(a)
y-y Axis Buckling. From Fig.13–12d the effective length factor is Ky = 0.7,
Fig.P13–20a so (KL)y 0.7(4 m) 2.8 m 2800 mm. Using the table in Appendix B to
determine Iy for the W200 59 section and applying Eq. 13–11, we have
p2EIy p
22[29110 2 )ksi]
33
149.1
2 in42
](20.4)(106) mm4)
1Pcr2y =
[200(10 N/mm
= ———————————————— = 1383 kip
1KL22y 1100.8(2800
in.22 mm)2
2.8 m
4m 5136247 N 5136 kN
Solving for Px by trial and error, noting that the argument for sec is in
radians, we get
Fig. 13–20 Since this value is less than (Pcr) 5136 kN, failure will occur about the
x–x axis. Also, 419.4 103 N/7580 mm2 55.3 MPa Y 250 MPa.
E X A M P L E 13.7
A solid rod has a diameter of 30 mm and is 600 mm long. It is made of a σ (MPa)
material that can be modeled by the stress–strain diagram shown in Fig. 13–22.
If it is used as a pin-supported column, determine the critical load.
270
Solution
The radius of gyration is σpl = 150
I 1p>421152 4
r = = = 7.5 mm
AA B p11522
and therefore the slenderness ratio is ∋
0.001 0.002
KL 11600 mm2
= = 80 Fig. 13–22
r 7.5 mm
p2Et p2Et
scr = = = 1.542110 - 32Et
1KL>r22 18022
(1)
First we will assume that the critical stress is elastic. From Fig. 13–22,
150 MPa
E = = 150 GPa
0.001
Since this value falls within the limits of 150 MPa and 270 MPa, it is
indeed the critical stress.
The critical load on the rod is therefore
Pcr = scrA = 185.1 MPa[p10.015 m22] = 131 kN Ans.
E X A M P L E 13.8
P An A-36 steel W250 149 member is used as a pin-supported column,
x y Fig. 13–27. Using the AISC column design formulas, determine the
largest load that it can safely support. Est 200(103) MPa, Y 250 MPa.
y x
Solution
The following data for a W250 149 is taken from the table in Appendix B.
5m
A 19000 mm2 rx 117 mm ry 67.4 mm
Since K = 1 for both x and y axis buckling, the slenderness ratio is largest
if ry is used. Thus,
2p2E
a b =
KL
r c B sY
222[29110
2p 2 ksi]
3 3
(200)(10 ) MPa
= ——–—–———–—
D
B 36 ksiMPa
250
= 126.1
125.66
1KL>r22
c1 - dsY
21KL>r22c
sallow =
515>32 + [13>821KL>r2>1KL>r2c] - [1KL>r23>81KL>rc23]6
-
[1 [1 (74.18)2 2>21126.12
172.452 2
/2(125.66)2]250]36MPa
ksi
= ———————————–—–——–————————
515>32
{(5/3) + [(3/8)(74.18/125.66)]
[13>82172.45>126.12]-[(74.18)
[172.4523 3>81126.1233]6
/8(125.66) ]}
= 110.85 MPa
16.17 ksi
The allowable load P on the column is therefore
P PP
sallow = ; N/mm2 = —————
16.17 kip>in.
110.85 2 2
A 19000
29.4 in.mm
80 kN 80 kN
5000 mm
Fig. 13–28
Solution
For a circular cross section the radius of gyration becomes
I 11>42p1d>224 d
r = = =
AA B 11>42pd 2 4
Applying Eq. 13–22, we have
2p2E 222[29110 2 ksi]
3 3) MPa]
a b =
KL 2p [210(10
= ——–—–——–—– 107.3
= 107.0
r c B sY D
B 50
360ksi
MPa
Since the rod’s radius of gyration is unknown, KL>r is unknown, and
therefore a choice must be made as to whether Eq. 13–21 or Eq. 13–23
applies. We will consider Eq. 13–21. For a fixed-end column K = 0.5, so
12p2E
sallow =
231KL>r22
18 3kip
80(10 )N 1212p
2 2 3 3
[29110
[210(10 2 kip>in
) N/mm 2 2
] ]
———–— = ———–———————
11>42pd
(1/4) d 2 2
23[0.5115 ft2112
23[0.5(5000 2
in.>ft21d>42]
mm)/(d/4)] 2
101.86 103
22.92
———–—— = 1.152d
0.0108d2 2 mm
d2 d2
d = 55.42 mm
2.11 in.
Use
d 56 mm Ans.
For this design, we must check the slenderness-ratio limits; i.e.,
0.511521122
KL 0.5(5)(1000)
= ————–— = 179
12.25>42
160
r (56/4)
Since 107.3 179 200, use of Eq. 13–21 is appropriate.
E X A M P L E 13.10
60 kN A bar having a length of 750 mm is used to support an axial compressive
load of 60 kN, Fig. 13–29. It is pin supported at its ends and made from
b 2b a 2014-T6 aluminum alloy. Determine the dimensions of its cross-
sectional area if its width is to be twice its thickness.
x y
Solution
Since KL 750 mm is the same for both x–x and y–y axis buckling, the
largest slenderness ratio is determined using the smallest radius of
gyration, i.e., using Imin = Iy:
750 mm
KL KL 1(750)
11302 103.9
2598.1
= = = ——— (1)
ry 2Iy>A 211>1222b1b23>[2b1b2]
3
(1/12)2b(b )/[2b(b)] bb
Here we must apply Eq. 13–24, 13–25, or 13–26. Since we do not as yet
know the slenderness ratio, we will begin by using Eq. 13–24.
P
= 195 N/mm2
28 ksi
60 kN A
123kip
60(10 )N 22
Fig. 13–29 ——––— = 195 N/mm
28 kip>in
2b1b2
2b(b)
b = 0.463
12.40 in.
mm
20 kN
40 mm 150 mm
y
x
20 kN
Fig. 13–30
Solution
By inspection, the board will buckle about the y axis. In the NFPA
equations, d = 1.5 in. Assuming that Eq. 13–29 applies, we have
P 3718 MPa
540 ksi
= –————
A 1KL>d2
(KL/d)22
3
5 kip
20(10 )N 3718 ksi 2
540N/mm
———–———–— = –————–—
15.5mm)(40
(150 in.211.5mm)
in.2 11 L/40
(1 in.222
L>1.5mm)
L = 1336
44.8 in.
mm Ans.
Here
KL 1144.8 in.2
(1)1336 mm
= –––——–—= 29.8
33.4
d 1.5 in.
40 mm
Since 26 6 KL>d … 50, the solution is valid.
E X A M P L E 13.12
The column in Fig. 13–32 is made of aluminum alloy 2014-T6 and is used
to support an eccentric load P. Determine the magnitude of P that can
be supported if the column is fixed at its base and free at its top. Use
Eq. 13–30. P
40 mm 20 mm
40 mm
40 mm
1600 mm
Fig. 13–32
Solution
From Fig. 13–12b, K = 2. The largest slenderness ratio for the column
is therefore
KL 2(1600
2180 in.2mm)
= = 277.1
277.1
r 12[11>12214 in.212 in.2
[(1/12)(80 mm)(40
3 3
mm)]>[12 in.24
]/[(40 in.] mm]
mm)80
By inspection, Eq. 13–26 must be used 1277.1 7 552. Thus,
54 000 ksi
378125 MPa 54 000 ksi MPa
378125
sallow = —––——–— = —––——–— = 0.703
1KL>r2 1277.12
22 2 ksi
4.92 MPa
(KL/r) (277.1)2
The actual maximum compressive stress in the column is determined
from the combination of axial load and bending. We have
P 1Pe2c
smax = +
A I
PP P11 in.212 in.2
P(20 mm)(40 mm)
= —––——–—— + —––————–————
40 mm(80
2 in.14 11>12212
in.2 mm) (1/12)(40 3
in.214mm)(80
in.2 mm)3
= 0.3125P
0.00078125 P
Assuming that this stress is uniform over the cross section, instead of just
at the outer boundary, we require
sallow = smax; 4.92 0.00078125P
P 6297.6 N 6.30 kN Ans.
E X A M P L E 13.13
The A-36 steel W150 30 column in Fig. 13–33 is pin connected at its P
ends and is subjected to the eccentric load P. Determine the maximum 750 mm
allowable value of P using the interaction method if the allowable
bending stress is (b)allow 160 MPa. E 200 GPa, Y 250 MPa.
y
Solution x
Here K = 1. The necessary geometric properties for the W150 30 are
taken from the table in Appendix B.
A 3790 mm2 Ix 17.1 106 mm4 ry 38.2 mm d 157 mm
4m
We will consider ry because this will lead to the largest value of the
slenderness ratio. Also, Ix is needed since bending occurs about the x
axis (c 157 mm/2 78.5 mm). To determine the allowable compressive
stress, we have
KL 1(4 m)(1000
1115 mm/m)
ft112 in.>ft2
= —––—————–— 104.71
= 120 M = P(750 mm)
r 38.2in.mm
1.50
Since P
2 2 [29110
22
2 ksi]
3 3
a b =
KL 2p E 2p (200)(10 ) MPa
= ——–—–——–—— 125.66
= 126.1 Fig. 13–33
r c B sY C
D 250ksiMPa
36
= 8.43
P 40.65kip
kN Ans.
Checking the application of the interaction method for the steel
section, we require
sa 3
8.43 kip>15.87
40.65(10 in.2 mm2)
) N/(3790
= ——–—–——–——–—— = 0.14060.125
1sa2allow 2 2
0.15 0.15
OK OK
10.28 kip>in
85.59 N/mm
E X A M P L E 13.14
The timber column in Fig. 13–34 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.
P
60 mm 60 mm
60 mm
20 mm
x
y
1200 mm
Fig. 13–34
Solution
From Fig. 13–12b, K = 2. Here we must calculate KL>d to determine
which equation from Eqs. 13–27 through 13–29 should be used. Since sallow
is determined using the largest slenderness ratio, we choose d 60 mm.
This is done to make this ratio as large as possible, and thereby yields the
lowest possible allowable axial stress. This is done even though bending
due to P is about the x axis. We have
KL 2160
2(1200in.2
mm)
= —––——–— = 40
40
d 360in.mm
The allowable axial stress is determined using Eq. 13–29 since
26 6 KL>d 6 50. Thus,
540 ksi
3718 MPa 540 ksiMPa
3718
sallow = —––——– = —––—–— = 0.3375
2.324
1KL>d2 1402
ksi MPa
(KL/d)22 2 2
(40)
Applying Eq. 13–30 with sallow = smax, we have
P Mc
sallow = +
A I
P P P14 in.213
P(80in.2
mm)(60 mm)
2.324 N/mmksi2 = —–——–——— + —––——–———————
11>12213
0.3375
3 in.16
60 in.2 mm)
mm(120 (1/12)(600 3
mm)(120
in.216 in.2 mm)3
P 3.35 kN Ans.