E X A M P L E
13.1
A 7.2-m-long A-36 steel tube having the cross section shown in Fig. 13–9
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.
Pcr
Solution
Using Eq. 13–5 to obtain the critical load with Est 200 GPa,
p2EI
Pcr =
L2
22
4 m/1000
p [200(10
kN/m2](–
(75)44 - 14–1–4p12.752
(70)4)(1
mm)4
[29110362) kip>in
] A 14–
p132
B in4
————————————————————————
=
2 2
[24 ft12 in.>ft2]
(7.2 m)
Ans.
228.2 kN
70 mm
75 mm
7.2 m
This force creates an average compressive stress in the column of
scr =
64.5(1000
kip N/kN)
Pcr
228.2 kN
= ———–———————
= 14.3
ksi N/mm2 100 MPa
100.2
2
2
A
(70)22]] mm
[(75) -p12.752
[p132
in2 2
Since cr Y 250 MPa, application of Euler’s equation is appropriate.
Pcr
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.
Solution
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
Iy 15.3 106 mm4. By inspection, buckling will occur about the y–y axis.
Why? Applying Eq. 13–5, we have
2
4 4
2 2
[291106)32kN/m
kip>in
]137.14)in
2 )(1 m/1000 mm)4
](15.3(10
mm
p2EI p2[200(10
——————————–————————————
=
= 512 kip
Pcr =
1887.6 kN
L2
[12 ft12 in.>ft2]
(4 2m)2
When fully loaded, the average compressive stress in the column is
512
kip kN (1000 N/kN)
Pcr
1887.6
= ———–———————
= 56.1 ksi
scr =
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
= 2 ————
250
N/mm
;
2
9.13 in5890mm2
P 1472.5 kN
Ans.
In actual practice, a factor of safety would be placed on this loading.
x
y
y
4m
x
Fig. 13–10
13.3
E X A M P L E
P
x
y
y
x
4m
A W150 24 steel column is 8 m long and is fixed at its ends as shown in
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
midheight. 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
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
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
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.
Applying Eq. 13–11, we have
4m
(a)
Fig. 13–13a
4m
1Pcr2x =
6
in
p
2[29110
[200(1026)ksi]29.1
kN/m2]13.4(10
) m4
———–———————————
=
=
401.7
kip 1653.2 kN (1)
1KL22x
1144 in.2
(42 m)2
1Pcr2y =
6
p2[200(10
[29110326)ksi]9.32
in4
kN/m2]1.83(10
) m4
———–———————————
=
=
262.5
kip 460.8 kN (2)
(2.82 m)2
1KL22y
1100.8 in.2
p2EIx
2
3
4
p2EIy
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
scr =
(b)
262.5
kip 3) N
Pcr
460.8(10
ksi
150.6 N/mm2
= ——————
= 59.3
A
3060
4.43
in2mm2
Since this stress is less than the yield stress, buckling will occur before
the material yields. Thus,
Fig. 13–13b
Pcr 461 kN
2.8 m
Ans.
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
(c)
4 m(1000
mm/m)
144
in.
KL
b = ———–————
= 56.2
60.4
r x
2.56 in.
66.2 mm
2.8
m(1000
mm/m)
KL
100.8
in.
a
b = ————–————
= 69.0
114.3
r y
1.46 in.
24.5 mm
Fig. 13–13
Hence, y–y axis buckling will occur, which is the same conclusion reached
by comparing Eqs. 1 and 2.
a
y – y axis buckling
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
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,
Iy = 23.2110-62 m4.
z
P
y
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
1KL2y =0.715 m2 = 3.5 m.
Applying Eq. 13–11, the critical loads for each case are
p2[7011092 N>m2]161.3110-62 m42
p2EIx
1Pcr2x =
=
1KL22x
110 m22
= 424 kN
p2EIy
p2[7011092 N>m2]123.2110-62 m42
1Pcr2y =
=
1KL22y
13.5 m22
= 1.31 MN
x
5m
(a)
Fig. 13–14a
By comparison, as P is increased the column will buckle about the x–x axis.
The allowable load is therefore
Pcr
424 kN
Pallow =
=
= 141 kN
Ans.
F.S.
3.0
Since
scr =
Pcr
424 kN
=
= 56.5 MPa 6 215 MPa
A
7.5110-32 m2
Euler’s equation can be applied.
Le = 10 m
Le = 3.5 m
x—x axis buckling
y –y axis buckling
(c)
Fig. 13–14c
(b)
Fig. 13–14
E X A M P L E
13.5
The steel column shown in Fig. 13–19 is assumed to be pin-connected at
its top and base. Determine the allowable eccentric load P that can be
applied. Also, what is the maximum deflection of the column due to this
loading? Due to bracing, assume buckling does not occur about the
y axis. Take Est 200(103) MPa, Y 250 MPa.
z
P
25 mm
50 mm
75 mm
y
75 mm
x
4.5 m
Fig. 13–19
Solution
Computing the necessary geometrical properties, we have
Ix =
1
12
in.23 mm)
= 363 in.
(50in.216
mm)(150
414.06 106 mm4
12
A = 12
in.2 =mm)
12 in.
(50in.216
mm)(150
27500 mm4
rx =
36
in.4 106 mm4
14.06
= 1.732
in. 43.30 mm
——–—–——–—
2
B
D12 in.7500
mm2
mm
e = 125in.
KL = 1(4.5
1115 ft2112
in.>ft2= 4500
180 in.
mm)(1000)
mm
180 in.
KL
4500
mm
104
= ———–—= 104
rx
43.30 in.
mm
1.732
Since the curves in Fig. 13–18b have been established for Est 200(103)
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
KL>rx = 104. Using the curve defined by the eccentricity ratio
ec/r2 (25 mm)(75 mm)/(43.30 mm)2 1, we get
P (MPa)
—
A
300
250
ec = 0.1
—
r2
200
0.5
P
83 ksi
MPa
L 12
A
1.0
100
1.5
P (83 MPa)(7500 mm2) 622 500 N 622.5 kN
Ans.
KL
—
—
100
150
200 r
A-36 structural steel
Est = 200(103) MPa, σY = 250 MPa
0
50
We can check this value by showing that it satisfies the secant
formula, Eq. 13–19:
smax
Fig. 13–18
P
P
ec
L
=
c1 + 2 sec a
bd
A
2r A EA
r
144 kip3) N
180
144622.5(10
kip 3) N
4500in.mm
622.5(10
250
36 ⱨ ——–———
c1 +1112(1)
secsec
a ——–—–— ———–————–——–—–—
bd
2 2
3
2
211.732
in.2
A
3
2 2
12
in.
29110
2
ksi112
(2)43.3 mm D 200(10 ) N/mmin.
7500 mm
7500 mm2
[
(
36 ⱨ 83[1
12[1 + sec(1.0586
sec 11.0570rad)]
rad2]
250
250
36 ⱨ 83[1
1211 + sec60.65°]
sec 60.56°2
252.3
250
36 L 36.4
The maximum deflection occurs at the column’s center, where
max 250 MPa.Applying Eq. 13–16, we have
vmax = ecsec a
P L
b - 1d
A EI 2
[ (
) ]
3
4500 mm
) Nin.
180
144 kip 622.5(10
mm sec
———–————–——–—–——–—
b -6 1 d 4 ——–—— 1
= 25
1 in.csec
a
3
24
3
2
2 10 mm
A 29110
D 200(10
) N/mm
2 ksi136
in 2 14.06
25 mm[see 1.0586 rad 1]
25 mm[see 60.65° 1]
26.0 mm
Euler’s formula
Eq. 13–6
ec = 0
—
r2
Ans.
)]
13.6
E X A M P L E
The W200 59 A-36 steel column shown in Fig. 13–20a 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.
z
x
P
200 mm
y
y
x
4m
(a)
Solution
From the support conditions it is seen that about the y–y axis the column
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.
y-y Axis Buckling. From Fig.13–12d the effective length factor is Ky = 0.7,
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
Fig.P13–20a
1Pcr2y =
2.8 m
5136247 N 5136 kN
4m
(b)
y –y axis buckling
x-x Axis Yielding. From Fig. 13–12b, Kx = 2, so (KL)x 2(4 m) 8
m 8000 mm. Again using the table in Appendix B to determine
A 7580 mm2, c 210 mm/2 105 mm, and rx 89.9 mm, and applying
the secant formula, we have
Fig. 13–20b
P
sY =
M= P(200 mm)
or
4m
33
2 in42
p
2 )ksi]
149.1
22[29110
[200(10
N/mm
](20.4)(106) mm4)
————————————————
=
= 1383 kip
1KL22y
1100.8(2800
in.22 mm)2
p2EIy
[
Px
250 —–—
1
7580
1KL2x
Px
Px
ec
c1 + 2 sec a
bd
A
2rx A EA
rx
200 105
8000
Px
———–—
——––——–—
sec ——–—
89.92
2(89.9) D 200(103) · 7580
(
)]
421.2
[1 + sec(1.143
2.979 sec 10.0700
1.895 106 Px[1= Px2.598
103 2Px2]
Solving for Px by trial and error, noting that the argument for sec is in
radians, we get
(c)
x – x axis buckling
Fig. 13–20
Px 419368 N 419.4 kN
Ans.
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.
13.7
E X A M P L E
σ (MPa)
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. 13–22.
If it is used as a pin-supported column, determine the critical load.
270
Solution
The radius of gyration is
σpl = 150
1p>421152
I
=
= 7.5 mm
AA
B p11522
4
r =
and therefore the slenderness ratio is
0.001
11600 mm2
KL
= 80
=
r
7.5 mm
Fig. 13–22
Applying Eq. 13–20 yields
scr =
p2Et
1KL>r22
=
p2Et
18022
= 1.542110 - 32Et
(1)
First we will assume that the critical stress is elastic. From Fig. 13–22,
E =
150 MPa
= 150 GPa
0.001
Thus, Eq. 1 becomes
scr = 1.542110 - 32[15011032] MPa = 231.3 MPa
Since scr 7 spl = 150 MPa, inelastic buckling occurs.
From the second line segment of the s - P diagram, Fig. 13–22, we
have
Et =
270 MPa - 150 MPa
¢s
=
= 120 GPa
¢P
0.002 - 0.001
Applying Eq. 1 yields
scr = 1.542110 - 32[12011032] MPa = 185.1 MPa
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
0.002
Ans.
∋
E X A M P L E
13.8
P
x
y
y
An A-36 steel W250 149 member is used as a pin-supported column,
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.
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,
1116
ft2112 in.>ft2
1(5 m)(1000
mm/m)
KL
= —————————
= 72.45
74.18
r
12.65
(67.4in.2
mm)
P
Fig. 13–27
From Eq. 13–22, we have
a
2p2E
KL
b =
r c B sY
=
3 3
2p
2 ksi]
222[29110
(200)(10
) MPa
——–—–———–—
36
ksiMPa
D
250
B
= 126.1
125.66
Here 0 6 KL>r 6 1KL>r2c, so Eq. 13–23 applies.
c1 sallow =
1KL>r22
21KL>r22c
dsY
515>32 + [13>821KL>r2>1KL>r2c] - [1KL>r23>81KL>rc23]6
2
2 2>21126.12
]36MPa
ksi
172.452
[1 [1
(74.18)
/2(125.66)2]250
= ———————————–—–——–————————
3 3>81126.1233]6
515>32
[13>82172.45>126.12]-[(74.18)
[172.452
/8(125.66) ]}
{(5/3) + [(3/8)(74.18/125.66)]
MPa
= 110.85
16.17 ksi
The allowable load P on the column is therefore
sallow =
P
;
A
PP
16.17 kip>in.
110.85
N/mm2 = —————
2
2
19000
29.4
in.mm
P 2106150 N 2106 kN
Ans.
13.9
E X A M P L E
The steel rod in Fig. 13–28 is to be used to support an axial load of 80 kN.
If Est 210(103) MPa and Y 360 MPa, determine the smallest diameter
of the rod as allowed by the AISC specification. The rod is fixed at both
ends.
d
80 kN
80 kN
5000 mm
Fig. 13–28
Solution
For a circular cross section the radius of gyration becomes
r =
I
d
11>42p1d>224
=
=
2
AA
4
B 11>42pd
Applying Eq. 13–22, we have
a
3 3) MPa]
222[29110
[210(10
KL
2p2E
2p
2 ksi]
107.3
——–—–——–—–
b =
=
= 107.0
D
r c B sY
50
ksi
360
MPa
B
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
sallow =
12p2E
231KL>r22
2 2
3 3
2 2
[29110
2 kip>in
18 3kip
1212p
[210(10
) N/mm
] ]
80(10
)N
———–———————
———–—
=
2
2
2
2
11>42pd
23[0.5115
ft2112
in.>ft21d>42]
23[0.5(5000
mm)/(d/4)]
(1/4)
d
101.86 103
22.92
———–——
= 1.152d
0.0108d2 2 mm
d2 d2
mm
d = 55.42
2.11 in.
Use
d 56 mm
For this design, we must check the slenderness-ratio limits; i.e.,
0.511521122
KL 0.5(5)(1000)
160
= ————–— = 179
r
12.25>42
(56/4)
Since 107.3 179 200, use of Eq. 13–21 is appropriate.
Ans.
E X A M P L E
13.10
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
a 2014-T6 aluminum alloy. Determine the dimensions of its crosssectional area if its width is to be twice its thickness.
60 kN
b
2b
y
x
Solution
750 mm
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:
1(750)
11302
KL
KL
103.9
2598.1
=
=
= ———
ry
3
bb
2Iy>A
211>1222b1b23>[2b1b2]
(1/12)2b(b )/[2b(b)]
(1)
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
N/mm2
= 195
28 ksi
A
60 kN
123kip
60(10
)N
22
——––— = 195
N/mm
28 kip>in
2b1b2
2b(b)
Fig. 13–29
b = 0.463
12.40 in.
mm
Checking the slenderness ratio, we have
2598.1
KL
103.9
= ———= 224.5
209.57 1212
r
0.463
12.40
Try Eq. 13–26, which is valid for KL>r Ú 55:
378125
MPa
54
000 ksi
P
= ———–——
A
1KL>r2
(KL/r)2 2
378125
60(10
123)
54
000
——— = —————
2b1b2
(2598.1/b)2 2
1103.9>b2
2b(b)
b = 27.05
1.05 in.
mm
Ans.
From Eq. 1,
2598.1
103.9
KL
= ———= 99.3
96.07 5555 OK
OK
r
1.05
27.05
Note: It would be satisfactory to choose the cross section with
dimensions 27 mm by 54 mm.
E X A M P L E
13.11
A board having cross-sectional dimensions of 150 mm by 40 mm is used
to support an axial load of 20 kN, Fig. 13–30. If the board is assumed to
be pin supported at its top and bottom, determine its greatest allowable
length L as specified by the NFPA.
20 kN
40 mm
150 mm
y
x
L
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
3718
MPa
540 ksi
P
= –————
A
(KL/d)22
1KL>d2
3
5 kip
20(10
)N
3718
540N/mm
ksi 2
———–———–— = –————–—
15.5mm)(40
in.211.5mm)
in.2
(150
11 L/40
L>1.5mm)
in.222
(1
L = 1336
44.8 in.
mm
Here
1144.8 in.2
(1)1336
mm
KL
= –––——–—= 29.8
33.4
d
1.5
in.
40 mm
Since 26 6 KL>d … 50, the solution is valid.
Ans.
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
2(1600
2180
in.2mm)
KL
=
= 277.1
277.1
3 3
r
2[11>12214
in.212 in.2
in.24
in.] mm]
1
[(1/12)(80 mm)(40
mm)]>[12
]/[(40
mm)80
By inspection, Eq. 13–26 must be used 1277.1 7 552. Thus,
54
000 ksi
ksi MPa
378125
MPa 54 000
378125
= —––——–—
= 0.703
sallow = —––——–—
ksi
4.92 MPa
22
2
1KL>r2
1277.12
(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
1Pe2c
P
+
A
I
P11 in.212
in.2
PP
P(20 mm)(40
mm)
—––————–————
+
= —––——–——
3
40
mm(80
2 in.14
in.2 mm)
(1/12)(40
mm)3
11>12212
in.214mm)(80
in.2
smax =
0.00078125 P
= 0.3125P
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
ends and is subjected to the eccentric load P. Determine the maximum
allowable value of P using the interaction method if the allowable
bending stress is (b)allow 160 MPa. E 200 GPa, Y 250 MPa.
Solution
P
750 mm
y
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
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
4m
1(4 m)(1000
mm/m)
1115
ft112 in.>ft2
KL
104.71
= 120
= —––—————–—
38.2in.mm
r
1.50
Since
a
M = P(750 mm)
P
2p
2 ksi]
2 [29110
(200)(10
) MPa
2p E
KL
——–—–——–——
125.66
=
= 126.1
b =
r c B sY
250ksiMPa
D
C
36
2
22
3 3
Fig. 13–33
then KL>r 6 1KL>r2c and so Eq. 13–23 must be used.
sallow =
[1 - 1KL>r22>21KL>r2c2]sY
[15>32 + [13>821KL>r2>1KL>r2c] - [1KL>r23>81KL>r2c3]]
2 2
2
>21126.12
]36 2ksi
[1
[1 11202
(104.71)
/2(125.66)
] ⴢ 250 MPa
= —––————––—–——————————––––————–—–
3 3
3
3
{(5/3)+ [13>8211202>1126.12]
[(3/8)(104.71)/(125.66)]
[(104.71)
/8(125.66)
]}
[15>32
- [11202
>81126.12
]]
= 85.59
10.28 ksi
MPa
Applying the interaction Eq. 13–31 yields
sb
sa
+
… 1
1sa2allow
1sb2allow
P130 in.213.10
in42 6) mm4
P>5.87
in.22
P/3790
mm
P(750
mm)(157in.2>141.4
mm/2)/17.1(10
—––——–—2 + —––——–————————————
= 1
1
10.28
ksi
22160
ksi N/mm2
85.59
N/mm
P
= 8.43
40.65kip
kN
Ans.
Checking the application of the interaction method for the steel
section, we require
3
8.43 kip>15.87
in.2 mm2)
sa
40.65(10
) N/(3790
= ——–—–——–——–——
= 0.14060.125
0.15 0.15
OK OK
2
2
1sa2allow
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
20 mm
60 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
2160
in.2
2(1200
mm)
KL
= —––——–—
= 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
ksiMPa
3718
MPa 540
3718
sallow = —––——–
= —––—–—
= 0.3375
ksi MPa
2.324
2 2
1KL>d2
1402
(KL/d)22
(40)
Applying Eq. 13–30 with sallow = smax, we have
P
Mc
sallow =
+
A
I
P14
in.213
P(80in.2
mm)(60 mm)
P P
—––——–———————
2.324
N/mm
+
0.3375
ksi2 = —–——–———
3
3 in.16
in.2 mm)
(1/12)(600
mm)(120
mm)3
60
mm(120
11>12213
in.216 in.2
P 3.35 kN
Ans.