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ch13 column buckling (static)

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FM_TOC 46060
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CONTENTS
To the Instructor
iv
1 Stress
1
2 Strain
73
3 Mechanical Properties of Materials
92
4 Axial Load
122
5 Torsion
214
6 Bending
329
7 Transverse Shear
472
8 Combined Loadings
532
9 Stress Transformation
619
10 Strain Transformation
738
11 Design of Beams and Shafts
830
12 Deflection of Beams and Shafts
883
13 Buckling of Columns
1038
14 Energy Methods
1159
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•13–1.
Determine the critical buckling load for the column.
The material can be assumed rigid.
P
L
2
k
Equilibrium: The disturbing force F can be determined by summing moments
about point A.
a + ©MA = 0;
P(Lu) - F a
L
b = 0
2
A
F = 2Pu
Spring Formula: The restoring spring force F1 can be determine using spring
formula Fs = kx.
Fs = ka
L
kLu
ub =
2
2
Critical Buckling Load: For the mechanism to be on the verge of buckling, the
disturbing force F must be equal to the restoring spring force F1.
2Pcr u =
Pcr =
L
2
kLu
2
kL
4
Ans.
1038
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13–2. Determine the critical load Pcr for the rigid bar and
spring system. Each spring has a stiffness k.
P
Equilibrium: The disturbing forces F1 and F2 can be related to P by writing the
moment equation of equlibrium about point A. Using small angle ananlysis, where
cos u ⬵ 1 and sin u = u,
+ ©MA = 0;
F2 a
L
3
k
L
2
b + F1 a L b - PLu = 01
3
3
L
3
F2 + 2F1 = 3Pu
k
(1)
Spring Force. The restoring spring force A Fsp B 1 and A Fsp B 2 can be determined using
the spring formula,
2
1
Lu and x2 = Lu, Fig. b. Thus,
3
3
2
2
= kx1 = ka Lu b = kLu
3
3
L
3
A
Fsp = kx, where x1 =
A Fsp B 1
A Fsp B 2 = kx2 = ka Lu b =
1
3
Critical Buckling Load. When the mechanism is on the verge of buckling the
disturbing force F must be equal to the restoring force of the spring Fsp. Thus,
F1 = A Fsp B 1 =
2
kLu
3
F2 = A Fsp B 2 =
1
kLu
3
Substituting this result into Eq. (1),
2
1
kLu + 2 a kLu b = 3Pcr u
3
3
Pcr =
5
kL
9
Ans.
1039
1
kLu
3
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13–3. The leg in (a) acts as a column and can be modeled
(b) by the two pin-connected members that are attached to a
torsional spring having a stiffness k (torque兾rad). Determine
the critical buckling load. Assume the bone material is rigid.
a + ©MA = 0;
- P(u)a
P
L
b + 2ku = 0
2
L
—
2
Require:
k
Pcr =
4k
L
Ans.
L
—
2
(a)
*13–4. Rigid bars AB and BC are pin connected at B. If
the spring at D has a stiffness k, determine the critical load
Pcr for the system.
(b)
P
A
Equilibrium. The disturbing force F can be related P by considering the equilibrium
of joint A and then the equilibrium of member BC,
a
B
Joint A (Fig. b)
+ c ©Fy = 0;
FAB cos f - P = 0
FAB =
a
P
cos f
k
D
Member BC (Fig. c)
a
©MC = 0; F(a cos u) -
P
P
cos f (2a sin u) sin f(2a cos u) = 0
cos f
cos f
C
F = 2P(tan u + tan f)
Since u and f are small, tan u ⬵ u and tan f ⬵ f. Thus,
F = 2P(u + f)
(1)
Also, from the geometry shown in Fig. a,
2au = af
f = 2u
Thus Eq. (1) becomes
F = 2P(u + 2u) = 6Pu
Spring Force. The restoring spring force Fsp can be determined using the spring
formula, Fsp = kx, where x = au, Fig. a. Thus,
Fsp = kx = kau
1040
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13–4. Continued
Critical Buckling Load. When the mechanism is on the verge of buckling the
disturbing force F must be equal to the restoring spring force Fsp.
F = Fsp
6Pcru = kau
Pcr =
ka
6
Ans.
1041
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•13–5.
An A-36 steel column has a length of 4 m and is
pinned at both ends. If the cross sectional area has the
dimensions shown, determine the critical load.
25 mm
Section Properties:
A = 0.01(0.06) + 0..05(0.01) = 1.10 A 10 - 3 B m2
Ix = Iy =
10 mm
1
1
(0.01) A 0.063 B +
(0.05) A 0.013 B = 0.184167 A 10 - 6 B m4
12
12
25 mm
Critical Buckling Load: K = 1 for pin supported ends column. Applying Euler’s
formula,
Pcr =
25 mm
25 mm
10 mm
p2EI
(KL)2
p2 (200)(109)(0.184167)(10 - 6)
=
[1(4)]2
= 22720.65 N = 22.7 kN
Ans.
Critical Stress: Euler’s formula is only valid if scr 6 sg.
scr =
Pcr
22720.65
= 20.66 MPa 6 sg = 250 MPa
=
A
1.10(10 - 3)
O.K.
13–6. Solve Prob. 13–5 if the column is fixed at its bottom
and pinned at its top.
25 mm
Section Properties:
A = 0.01(0.06) + 0.05(0.01) = 1.10 A 10 - 3 B m2
10 mm
1
1
Ix = Iy =
(0.01) A 0.063 B +
(0.05) A 0.013 B = 0.184167 A 10 - 6 B m4
12
12
25 mm
Critical Buckling Load: K = 0.7 for one end fixed and the other end pinned
column. Applying Euler’s formula,
Pcr =
p EI
(EL)2
p2 (200)(109)(0.184167)(10 - 6)
=
[0.7(4)]2
= 46368.68 N = 46.4 kN
Ans.
Critical Stress: Euler’s formula is only valid if scr 6 sg.
scr =
25 mm
25 mm
10 mm
2
Pcr
46368.68
= 42.15 MPa 6 sg = 250 MPa
=
A
1.10(10 - 3)
1042
O.K.
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13–7. A column is made of A-36 steel, has a length of 20 ft,
and is pinned at both ends. If the cross-sectional area has
the dimensions shown, determine the critical load.
6 in.
0.25 in.
The cross sectional area and moment of inertia of the square tube is
5.5 in.
A = 6(6) - 5.5(5.5) = 5.75 in2
I =
0.25 in.
1
1
(6)(63) (5.5)(5.53) = 31.74 in4
12
12
0.25 in.
0.25 in.
The column is pinned at both of its end, k = 1. For A36 steel, E = 29.0(103) ksi and
sg = 36 ksi (table in appendix). Applying Euler’s formula,
Pcr =
p2 C 29.0(103) D (31.74)
p2EI
=
(KL)2
C 1(20)(12) D 2
Ans.
= 157.74 kip = 158
Critical Stress. Euler’s formula is valid only if scr 6 sg.
Pcr
157.74
=
= 27.4 ksi 6 sg = 36 ksi
A
5.75
scr =
O.K.
*13–8. A column is made of 2014-T6 aluminum, has a
length of 30 ft, and is fixed at its bottom and pinned at its
top. If the cross-sectional area has the dimensions shown,
determine the critical load.
6 in.
0.25 in.
5.5 in.
The cross-sectional area and moment of inertia of the square tube is
0.25 in.
A = 6(6) - 5.5(5.5) = 5.75 in2
0.25 in.
1
1
I =
(6)(63) (5.5)(5.53) = 31.74 in4
12
12
The column is fixed at one end, K = 0.7. For 2014–76 aluminium, E = 10.6(103) ksi
and sg = 60 ksi (table in appendix). Applying Euler’s formula,
Pcr =
p2 C 10.6(103) D (31.74)
p2EI
=
(KL)2
C 0.7(30)(12) D 2
Ans.
= 52.29 kip = 52.3 kip
Critical Stress. Euler’s formula is valid only if scr 6 sg.
scr =
Pcr
52.3
=
= 9.10 ksi 6 sg = 60 ksi
A
5.75
O.K.
1043
0.25 in.
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•13–9.
The W14 * 38 column is made of A-36 steel and is
fixed supported at its base. If it is subjected to an axial load
of P = 15 kip, determine the factor of safety with respect to
buckling.
P
From the table in appendix, the cross-sectional area and moment of inertia about
weak axis (y-axis) for W14 * 38 are
20 ft
A = 11.2 in2
Iy = 26.7 in4
The column is fixed at its base and free at top, k = 2. Here, the column will buckle
about the weak axis (y axis). For A36 steel, E = 29.0(103) ksi and sy = 36 ksi.
Applying Euler’s formula,
p2 C 29.0(103) D (26.7)
p2EIy
Pcr =
C 2 (20)(12) D 2
=
(KL)2
= 33.17 kip
Thus, the factor of safety with respect to buckling is
F.S =
Pcr
33.17
=
= 2.21
P
15
Ans.
The Euler’s formula is valid only if scr 6 sg.
scr =
Pcr
33.17
=
= 2.96 ksi 6 sg = 36 ksi
A
11.2
O.K.
13–10. The W14 * 38 column is made of A-36 steel.
Determine the critical load if its bottom end is fixed
supported and its top is free to move about the strong axis
and is pinned about the weak axis.
P
From the table in appendix, the cross-sectional area and moment of inertia about
weak axis (y-axis) for W14 * 38 are
A = 11.2 in2
Ix = 385 in4
Iy = 26.7 in4
The column is fixed at its base and free at top about strong axis. Thus, kx = 2. For
A36 steel, E = 29.0(103) ksi and sg = 36 ksi.
Pcr =
p2EIx
(KxLx)
2
=
p2 C 29.0(103) D (385)
C 2 (20)(12) D 2
= 478.28 kip
The column is fixed at its base and pinned at top about weak axis. Thus, ky = 0.7.
Pcr =
p2EIy
2
(KyLy)
=
p2 C 29.0(103) D (26.7)
C 0.7(20)(12) D 2
Ans.
= 270.76 kip = 271 kip (Control)
The Euler’s formula is valid only if scr 6 sg.
scr =
Pcr
270.76
=
= 24.17 ksi 6 sg = 36 ksi
A
11.2
O.K.
1044
20 ft
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13–11. The A-36 steel angle has a cross-sectional area of
A = 2.48 in2 and a radius of gyration about the x axis of
rx = 1.26 in. and about the y axis of ry = 0.879 in. The
smallest radius of gyration occurs about the z axis and is
rz = 0.644 in. If the angle is to be used as a pin-connected
10-ft-long column, determine the largest axial load that can
be applied through its centroid C without causing it to buckle.
y
z
C
x
x
z
y
The least radius of gyration:
r2 = 0.644 in.
scr =
p2E
2
A KL
r B
controls.
K = 1.0
;
p2 (29)(103)
(120) 2
C 1.00.644
D
=
= 8.243 ksi 6 sg
O.K.
Pcr = scr A = 8.243 (2.48) = 20.4 kip
Ans.
*13–12. An A-36 steel column has a length of 15 ft and is
pinned at both ends. If the cross-sectional area has the
dimensions shown, determine the critical load.
8 in.
0.5 in.
0.5 in.
6 in.
0.5 in.
Ix =
1
1
(8)(73) (7.5)(63) = 93.67 in4
12
12
Iy = 2 a
Pcr =
1
1
b (0.5)(83) +
(6)(0.53) = 42.729 in4 (controls)
12
12
p2(29)(103)(42.729)
p2EI
=
2
(EL)
[(1.0)(15)(12)]2
= 377 kip
Ans.
Check:
A = (2)(8)(0.5) + 6(0.5) = 11 in2
scr =
Pcr
377
=
= 34.3 ksi 6 sg
A
11
Therefore, Euler’s formula is valid
1045
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•13–13.
An A-36 steel column has a length of 5 m and is
fixed at both ends. If the cross-sectional area has the
dimensions shown, determine the critical load.
I =
10 mm
10 mm
50 mm
1
1
(0.1)(0.053) (0.08)(0.033) = 0.86167 (10 - 6) m4
12
12
Pcr =
100 mm
p2(200)(109)(0.86167)(10 - 6)
p2EI
=
2
(KL)
[(0.5)(5)]2
= 272 138 N
= 272 kN
scr =
=
Pcr
;
A
Ans.
A = (0.1)(0.05) - (0.08)(0.03) = 2.6(10 - 3) m2
272 138
= 105 MPa 6 sg
2.6 (10 - 3)
Therefore, Euler’s formula is valid.
13–14. The two steel channels are to be laced together
to form a 30-ft-long bridge column assumed to be pin
connected at its ends. Each channel has a cross-sectional
area of A = 3.10 in2 and moments of inertia Ix = 55.4 in4,
Iy = 0.382 in4. The centroid C of its area is located in
the figure. Determine the proper distance d between the
centroids of the channels so that buckling occurs about the
x–x and y¿ – y¿ axes due to the same load. What is the value
of this critical load? Neglect the effect of the lacing.
Est = 2911032 ksi, sY = 50 ksi.
y
0.269 in.
C
d
y
In order for the column to buckle about x - x and y - y at the same time, Iy must
be equal to Ix
Iy = Ix
0.764 + 1.55 d2 = 110.8
d = 8.43 in.
Ans.
Check:
d 7 2(1.231) = 2.462 in.
O.K.
3
p (29)(10 )(110.8)
p2 EI
=
2
(KL)
[1.0(360)]2
Ans.
= 245 kip
Check stress:
scr =
x
C
d 2
Iy = 2(0.382) + 2 (3.10) a b = 0.764 + 1.55 d2
2
Pcr =
1.231 in.
x
Ix = 2(55.4) = 110.8 in.4
2
y¿
Pcr
245
=
= 39.5 ksi 6 sg
A
2(3.10)
Therefore, Euler’s formula is valid.
1046
y¿
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13–15. An A-36-steel W8 * 24 column is fixed at one end
and free at its other end. If it is subjected to an axial load
of 20 kip, determine the maximum allowable length of the
column if F.S. = 2 against buckling is desired.
Section Properties. From the table listed in the appendix, the cross-sectional area
and moment of inertia about the y axis for a W8 * 24 are
A = 7.08 in2
Iy = 18.3 in4
Critical Buckling Load. The critical buckling load is
Pcr = Pallow (F.S) = 20(2) = 40 kip
Applying Euler’s formula,
p2 EIy
Pcr =
40 =
(KL)2
p2 C 29 A 103 B D (18.3)
(2L)2
L = 180.93 in = 15.08 ft = 15.1 ft
Ans.
Critical Stress. Euler’s formula is valid only if scr 6 sY.
scr =
Pcr
40
=
= 5.65 ksi 6 sY = 36 ksi
A
7.08
O.K.
*13–16. An A-36-steel W8 * 24 column is fixed at one
end and pinned at the other end. If it is subjected to an axial
load of 60 kip, determine the maximum allowable length of
the column if F.S. = 2 against buckling is desired.
Section Properties. From the table listed in the appendix, the cross-sectional area
and moment of inertia about the y axis for a W8 * 24 are
A = 7.08 in2
Iy = 18.3 in4
Critical Buckling Load. The critical buckling load is
Pcr = Pallow (F.S.) = 60(2) = 120 kip
Applying Euler’s formula,
Pcr =
120 =
p2EIy
(KL)2
p2 C 24 A 103 B D (18.3)
(0.7L)2
L = 298.46 in = 24.87 ft = 24.9 ft
Ans.
Critical Stress. Euler’s formula is valid only if scr 6 sY.
scr =
Pcr
120
=
= 16.95 ksi 6 sY = 36 ksi
A
7.08
O.K.
1047
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•13–17.
The 10-ft wooden rectangular column has the
dimensions shown. Determine the critical load if the ends
are assumed to be pin connected. Ew = 1.611032 ksi,
sY = 5 ksi.
Section Properties:
10 ft
A = 4(2) = 8.00 in2
4 in.
Ix =
1
(2) A 43 B = 10.667 in4
12
Iy =
1
(4) A 23 B = 2.6667 in4 (Controls !)
12
2 in.
Critical Buckling Load: K = 1 for pin supported ends column. Applying Euler’s
formula,.
Pcr =
p2EI
(KL)2
p2(1.6)(103)(2.6667)
=
[1(10)(12)]2
Ans.
= 2.924 kip = 2.92 kip
Critical Stress: Euler’s formula is only valid if scr 6 sg.
scr =
Pcr
2.924
=
= 0.3655 ksi 6 sg = 5 ksi
A
8.00
O.K.
13–18. The 10-ft column has the dimensions shown.
Determine the critical load if the bottom is fixed and the
top is pinned. Ew = 1.611032 ksi, sY = 5 ksi.
Section Properties:
A = 4(2) = 8.00 in2
10 ft
1
(2) A 43 B = 10.667 in4
Ix =
12
4 in.
2 in.
1
Iy =
(4) A 23 B = 2.6667 in4 (Controls!)
12
Critical Buckling Load: K = 0.7 for column with one end fixed and the other end
pinned. Applying Euler’s formula.
Pcr =
p2EI
(KL)2
p2 (1.6)(103)(2.6667)
=
[0.7(10)(12)]2
Ans.
= 5.968 kip = 5.97 kip
Critical Stress: Euler’s formula is only valid if scr 6 sg.
scr =
Pcr
5.968
=
= 0.7460 ksi 6 sg = 5 ksi
A
8.00
O.K.
1048
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13–19. Determine the maximum force P that can be
applied to the handle so that the A-36 steel control rod BC
does not buckle. The rod has a diameter of 25 mm.
P
350 mm
A
250 mm
45⬚
Support Reactions:
a + ©MA = 0;
P(0.35) - FBC sin 45°(0.25) = 0
FBC = 1.9799P
Section Properties:
A =
p
A 0.0252 B = 0.15625 A 10 - 3 B p m2
4
I =
p
A 0.01254 B = 19.17476 A 10 - 9 B m4
4
Critical Buckling Load: K = 1 for a column with both ends pinned. Appyling
Euler’s formula,
Pcr = FBC =
1.9799P =
p2EI
(KLBC)2
p2(200)(109) C 19.17476(10 - 9) D
[1(0.8)]2
P = 29 870 N = 29.9 kN
Ans.
Critical Stress: Euler’s formula is only valid if scr 6 sg.
scr =
1.9799(29 870)
Pcr
= 120.5 MPa 6 sg = 250 MPa
=
A
0.15625(10 - 3)p
1049
O.K.
C
B
800 mm
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*13–20. The W10 * 45 is made of A-36 steel and is used
as a column that has a length of 15 ft. If its ends are assumed
pin supported, and it is subjected to an axial load of 100 kip,
determine the factor of safety with respect to buckling.
P
Critical Buckling Load: Iy = 53.4 in4 for a W10 * 45 wide flange section and
K = 1 for pin supported ends column. Applying Euler’s formula,
Pcr =
15 ft
p2EI
(KL)2
p2 (29)(103)(53.4)
=
[1(15)(12)]2
P
= 471.73 kip
Critical Stress: Euler’s formula is only valid if scr 6 sg. A = 13.3 in2 for the
W10 * 45 wide-flange section.
scr =
Pcr
471.73
=
= 35.47 ksi 6 sg = 36 ksi
A
13.3
O.K.
Pcr
471.73
=
= 4.72
P
100
Ans.
Factor of Safety:
F.S =
The W10 * 45 is made of A-36 steel and is used
as a column that has a length of 15 ft. If the ends of the
column are fixed supported, can the column support the
critical load without yielding?
•13–21.
P
Critical Buckling Load: Iy = 53.4 in4 for W10 * 45 wide flange section and
K = 0.5 for fixed ends support column. Applying Euler’s formula,
Pcr =
15 ft
p2EI
(KL)2
p2 (29)(103)(53.4)
=
[0.5(15)(12)]2
P
= 1886.92 kip
Critical Stress: Euler’s formula is only valid if scr 6 sg. A = 13.3 in2 for W10 * 45
wide flange section.
scr =
Pcr
1886.92
=
= 141.87 ksi 7 sg = 36 ksi (No!)
A
13.3
Ans.
The column will yield before the axial force achieves the critical load Pcr and so
Euler’s formula is not valid.
1050
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–22. The W12 * 87 structural A-36 steel column has a
length of 12 ft. If its bottom end is fixed supported while
its top is free, and it is subjected to an axial load of
P = 380 kip, determine the factor of safety with respect to
buckling.
W 12 * 87
A = 25.6 in2
Ix = 740 in4
P
Iy = 241 in4 (controls)
12 ft
K = 2.0
Pcr =
p2(29)(103)(241)
p2EI
=
= 831.63 kip
2
(KL)
[(2.0)(12)(12)]2
Pcr
831.63
=
= 2.19
P
380
F.S. =
Ans.
Check:
scr =
=
Pcr
A
831.63
= 32.5 ksi 6 sg
25.6
O.K.
13–23. The W12 * 87 structural A-36 steel column has a
length of 12 ft. If its bottom end is fixed supported while its
top is free, determine the largest axial load it can support.
Use a factor of safety with respect to buckling of 1.75.
W 12 * 87
A = 25.6 in2
Ix = 740 in4
P
Iy = 241 in4
(controls)
K = 2.0
12 ft
Pcr
p2(29)(103)(241)
p2EI
=
=
= 831.63 kip
2
(KL)
(2.0(12)(12))2
P =
Pcr
831.63
=
= 475 ksi
F.S
1.75
Ans.
Check:
scr =
P
831.63
=
= 32.5 ksi 6 sg
A
25.6
O.K.
1051
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*13–24. An L-2 tool steel link in a forging machine is pin
connected to the forks at its ends as shown. Determine the
maximum load P it can carry without buckling. Use a factor
of safety with respect to buckling of F.S. = 1.75. Note from
the figure on the left that the ends are pinned for buckling,
whereas from the figure on the right the ends are fixed.
P
P
1.5 in.
0.5 in.
24 in.
Section Properties:
A = 1.5(0.5) = 0.750 in2
Ix =
1
(0.5) A 1.53 B = 0.140625 in4
12
Iy =
1
(1.5) A 0.53 B = 0.015625 in4
12
P
Critical Buckling Load: With respect to the x - x axis, K = 1 (column with both
ends pinned). Applying Euler’s formula,
Pcr =
p2EI
(KL)2
p2(29.0)(103)(0.140625)
=
[1(24)]2
= 69.88 kip
With respect to the y - y axis, K = 0.5 (column with both ends fixed).
Pcr =
p2EI
(KL)2
p2(29.0)(103)(0.015625)
=
[0.5(24)]2
= 31.06 kip
(Controls!)
Critical Stress: Euler’s formula is only valid if scr 6 sg.
scr =
Pcr
31.06
=
= 41.41 ksi 6 sg = 102 ksi
A
0.75
O.K.
Factor of Safety:
F.S =
1.75 =
Pcr
P
31.06
P
P = 17.7 kip
Ans.
1052
P
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The W14 * 30 is used as a structural A-36 steel
column that can be assumed pinned at both of its ends.
Determine the largest axial force P that can be applied
without causing it to buckle.
•13–25.
P
From the table in appendix, the cross-sectional area and the moment of inertia
about weak axis (y-axis) for W14 * 30 are
A = 8.85 in2
Iy = 19.6 in4
25 ft
Critical Buckling Load: Since the column is pinned at its base and top, K = 1. For
A36 steel, E = 29.0(103) ksi and sg = 36 ksi. Here, the buckling occurs about the
weak axis (y-axis).
P = Pcr =
p2EIy
(KL)2
=
p2 C 29.0(103) D (19.6)
C 1(25)(12) D 2
Ans.
= 62.33 kip = 62.3 kip
Euler’s formula is valid only if scr 6 sg.
scr =
Pcr
62.33
=
= 7.04 ksi 6 sg = 36 ksi
A
8.85
O.K.
13–26. The A-36 steel bar AB has a square cross section.
If it is pin connected at its ends, determine the maximum
allowable load P that can be applied to the frame. Use a
factor of safety with respect to buckling of 2.
a + ©MA = 0;
C
FBC sin 30°(10) - P(10) = 0
FBC = 2 P
+
: ©Fx = 0;
A
1.5 in.
30⬚
B
1.5 in.
FA - 2P cos 30° = 0
1.5 in.
10 ft
FA = 1.732 P
P
Buckling load:
Pcr = FA(F.S.) = 1.732 P(2) = 3.464 P
L = 10(12) = 120 in.
I =
1
(1.5)(1.5)3 = 0.421875 in4
12
Pcr =
p2 EI
(KL)2
3.464 P =
p2 (29)(103)(0.421875)
[(1.0)(120)]2
P = 2.42 kip
Ans.
Pcr = FA(F.S.) = 1.732(2.42)(2) = 8.38 kip
Check:
scr =
Pcr
8.38
=
= 3.72 ksi 6 sg
A
1.5 (1.5)
O.K.
1053
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Page 1054
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13–27. Determine the maximum allowable intensity w of
the distributed load that can be applied to member BC
without causing member AB to buckle. Assume that AB is
made of steel and is pinned at its ends for x–x axis buckling
and fixed at its ends for y–y axis buckling. Use a factor
of safety with respect to buckling of 3. Est = 200 GPa,
sY = 360 MPa.
w
C
1.5 m
B
0.5 m
2m
30 mm
x
Ix =
1
(0.02)(0.033) = 45.0(10 - 9)m4
12
Iy =
1
(0.03)(0.023) = 20(10 - 9) m4
12
x
x-x axis:
Pcr = FAB (F.S.) = 1.333w(3) = 4.0 w
K = 1.0,
Pcr =
L = 2m
p2EI
(KL)2
4.0w =
p2(200)(109)(45.0)(10 - 9)
[(1.0)(2)]2
w = 5552 N>m = 5.55 kN>m
Ans.
(controls)
y -y axis
K = 0.5,
4.0w =
L = 2m
p2 (200)(109)(20)(10 - 9)
[(0.5)(2)]2
w = 9870 N>m = 9.87 kN>m
Check:
scr =
20 mm
y
y
Moment of inertia:
4(5552)
Pcr
=
= 37.0 MPa 6 sg
A
(0.02)(0.03)
O.K.
1054
30 mm
A
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*13–28. Determine if the frame can support a load of
w = 6 kN>m if the factor of safety with respect to buckling
of member AB is 3. Assume that AB is made of steel and is
pinned at its ends for x–x axis buckling and fixed at its ends
for y–y axis buckling. Est = 200 GPa, sY = 360 MPa.
w
C
B
1.5 m
0.5 m
Check x -x axis buckling:
Ix =
1
(0.02)(0.03)3 = 45.0(10 - 9) m4
12
K = 1.0
Pcr
2m
30 mm
x
20 mm
y
y
L = 2m
p2(200)(109)(45.0)(10 - 9)
p2EI
=
=
2
(KL)
((1.0)(2))2
x
A
30 mm
Pcr = 22.2 kN
a + ©MC = 0;
FAB(1.5) - 6(2)(1) = 0
FAB = 8 kN
Preq’d = 8(3) = 24 kN 7 22.2 kN
No, AB will fail.
Ans.
The beam supports the load of P = 6 kip. As a
result, the A-36 steel member BC is subjected to a
compressive load. Due to the forked ends on the member,
consider the supports at B and C to act as pins for x–x axis
buckling and as fixed supports for y–y axis buckling.
Determine the factor of safety with respect to buckling
about each of these axes.
•13–29.
a + ©MA = 0;
P
4 ft
A
3 ft
1
)(1)(3)3
p2(29)(103)(12
p2EI
=
= 178.9 kip
2
(KL)
(1.0(5)(12))2
178.9
= 8.94
20
Ans.
y - y axis buckling:
Pcr =
F.S. =
3 in.
y
x -x axis buckling:
F.S. =
B
C x
3
FBC a b (4) - 6000(8) = 0
5
FBC = 20 kip
Pcr =
4 ft
1
)(3)(1)3
p2 (29)(103)(12
p2EI
=
= 79.51
2
(KL)
(0.5(5)(12))2
79.51
= 3.98
20
Ans.
1055
1 in.
x
y
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13–30. Determine the greatest load P the frame will
support without causing the A-36 steel member BC to
buckle. Due to the forked ends on the member, consider the
supports at B and C to act as pins for x–x axis buckling and
as fixed supports for y–y axis buckling.
P
4 ft
A
3 ft
3
FBC a b (4) - P(8) = 0
5
a + ©MA = 0;
4 ft
B
y
3 in.
C x
1 in.
FBC = 3.33 P
y
x -x axis buckling:
Pcr =
x
1
)(1)(3)3
p2(29)(103)(12
p2EI
=
= 178.9 kip
(KL)2
(1.0(5)(12))2
y - y axis buckling:
Pcr =
1
)(3)(1)3
p2(29)(103)(12
p2EI
=
= 79.51 kip
(KL)2
(0.5(5)(12))2
Thus,
3.33 P = 79.51
P = 23.9 kip
Ans.
13–31. Determine the maximum distributed load that can
be applied to the bar so that the A-36 steel strut AB does
not buckle. The strut has a diameter of 2 in. It is pin
connected at its ends.
w
C
A
2 ft
The compressive force developed in member AB can be determined by writing the
moment equation of equilibrium about C.
a + ©MC = 0;
FAB(2) - w(2)(3) = 0
A = p(12) = p in2
I =
FAB = 3w
4 ft
p 4
p
(1 ) = in4
4
4
Since member AB is pinned at both ends, K = 1. For A36 steel, E = 29.0(103) ksi
and sg = 36 ksi.
Pcr =
p EI
;
(KL)2
p C 29.0(10 ) D (p>4)
2
2
3w =
3
C 1(4)(12) D 2
Ans.
w = 32.52 kip>ft = 32.5 kip>ft
The Euler’s formula is valid only if scr 6 sg.
scr =
3(32.52)
Pcr
=
= 31.06 ksi 6 sg = 36 ksi
p
A
O.K.
1056
B
2 ft
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Page 1057
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*13–32. The members of the truss are assumed to be pin
connected. If member AC is an A-36 steel rod of 2 in.
diameter, determine the maximum load P that can be
supported by the truss without causing the member to buckle.
P
C
B
4 ft
D
A
3 ft
Section the truss through a -a, the FBD of the top cut segment is shown in Fig. a. The
compressive force developed in member AC can be determined directly by writing
the force equation of equilibrium along x axis.
+
: ©Fx = 0;
3
FAC a b - P = 0
5
A = p(12) = p in2
I =
FAC =
5
P (C)
3
p 4
p
(1 ) = in4
4
4
Since both ends of member AC are pinned, K = 1. For A-36 steel, E = 29.0(103) ksi
and sg = 36 ksi. The length of member AC is LAC = 232 + 42 = 5 ft.
Pcr =
p2EI
;
(KL)2
p2 C 29.0(103) D (p>4)
5
P =
3
C 1(5)(12) D 2
P = 37.47 kip = 37.5 kip
Ans.
Euler’s formula is valid only if scr 6 sg.
scr
5
(37.47)
Pcr
3
=
=
= 19.88 ksi 6 sg = 36 ksi
p
A
O.K.
1057
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•13–33.
The steel bar AB of the frame is assumed to be pin
connected at its ends for y–y axis buckling. If w = 3 kN>m,
determine the factor of safety with respect to buckling about
the y–y axis due to the applied loading. Est = 200 GPa,
sY = 360 MPa.
6m
w
B
C
40 mm
40 mm
3m
40 mm
y
x
A
4m
The force with reference to the FBD shown in Fig. a.
a + ©MC = 0;
3
3(6)(3) - FAB a b (6) = 0
5
A = 0.04(0.08) = 3.2(10 - 3) m2
Iy =
FAB = 15 kN
1
(0.08)(0.043) = 0.4267(10 - 6)m4
12
The length of member AB is L = 232 + 42 = 5m. Here, buckling will occur about
the weak axis, (y-axis). Since both ends of the member are pinned, Ky = 1.
Pcr =
p2EIy
(KyLy)2
=
p2 C 200(109) D C 0.4267(10 - 6) D
C 1.0(5) D 2
= 33.69 kN
Euler’s formula is valid only if scr 6 sg.
scr =
33.69(103)
Pcr
= 10.53(106)Pa = 10.53 MPa 6 sg = 360 MPa
=
A
3.2(10 - 3)
O.K.
Thus, the factor of safety against buckling is
F.S =
Pcr
33.69
=
= 2.25
FAB
15
Ans.
1058
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Page 1059
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–34. The members of the truss are assumed to be pin
connected. If member AB is an A-36 steel rod of 40 mm
diameter, determine the maximum force P that can be
supported by the truss without causing the member to buckle.
2m
C
E
D
1.5 m
B
A
2m
P
By inspecting the equilibrium of joint E, FAB = 0. Then, the compressive force
developed in member AB can be determined by analysing the equilibrium of joint
A, Fig. a.
+ c ©Fy = 0;
3
FAC a b - P = 0
5
+
: ©Fx = 0;
5
4
P a b - FAB = 0
3
5
A = p(0.022) = 0.4(10 - 3)p m2
I =
FAC =
5
P (T)
3
FAB =
4
P(c)
3
p
(0.024) = 40(10 - 9) p m4
4
Since both ends of member AB are pinned, K = 1. For A36 steel, E = 200 GPa and
sg = 250 MPa.
Pcr =
p2EI
;
(KL)2
p2 C 200(109) D C 40(10 - 9)p D
4
P =
3
C 1(2) D 2
P = 46.51(103) N = 46.5 kN
Ans.
The Euler’s formula is valid only if scr 6 sg.
scr
4
(46.51)(103)
Pcr
3
= 49.35(106) Pa = 49.35 MPa 6 sg = 250 MPa O.K.
=
=
A
0.4(10 - 3)p
1059
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Page 1060
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13–35. The members of the truss are assumed to be pin
connected. If member CB is an A-36 steel rod of 40 mm
diameter, determine the maximum load P that can be
supported by the truss without causing the member to buckle.
2m
C
E
D
1.5 m
B
A
2m
P
Section the truss through a–a, the FBD of the left cut segment is shown in Fig. a. The
compressive force developed in member CB can be obtained directly by writing the
force equation of equilibrium along y axis.
+ c ©Fy = 0;
FCB - P = 0
A = p(0.022) = 0.4(10 - 3)p m2
FCB = P (C)
I =
p
(0.024) = 40(10 - 9)p m4
4
Since both ends of member CB are pinned, K = 1. For A36 steel, E = 200 GPa and
sg = 250 MPa.
Pcr =
p2EI
;
(KL)2
P =
p2 C 200(109) D C 40(10 - 9)p D
C 1(1.5) D 2
= 110.24(103) N = 110 kN
Ans.
The Euler’s formula is valid only if scr 6 sg.
scr =
110.24(103)
Pcr
= 87.73(106) Pa = 87.73 MPa 6 sg = 250 MPa
=
A
0.4(10 - 3)p
1060
O.K.
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*13–36. If load C has a mass of 500 kg, determine the
required minimum diameter of the solid L2-steel rod AB
to the nearest mm so that it will not buckle. Use F.S. = 2
against buckling.
A
45°
4m
D
Equilibriun. The compressive force developed in rod AB can be determined by
analyzing the equilibrium of joint A, Fig. a.
©Fy¿ = 0; FAB sin 15° - 500(9.81) cos 45° = 0
FAB = 13 400.71 N
Section Properties. The cross-sectional area and moment of inertia of the solid
rod are
A =
p 2
d
4
I =
p d 4
p 4
a b =
d
4 2
64
Critical Buckling Load. Since the rod is pinned at both of its ends, K = 1. Here,
Pcr = FAB (F.S.) = 13400.71(2) = 26801.42 N. Applying Euler’s formula,
Pcr =
p2EIy
(KL)2
26801.42 =
p2 C 200 A 109 B D c
p 4
d d
64
[1(4)]2
d = 0.04587 m = 45.87 mm
Use d = 46 mm
Ans.
Critical Stress. Euler’s formula is valid only if scr 6 sY.
scr =
Pcr
26801.42
=
= 16.13 MPa 6 sY = 703 MPa
p
A
2
A 0.046 B
4
1061
O.K.
60°
B
C
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Page 1062
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
•13–37. If the diameter of the solid L2-steel rod AB is
50 mm, determine the maximum mass C that the rod can
support without buckling. Use F.S. = 2 against buckling.
A
45°
4m
D
Equilibrium. The compressive force developed in rod AB can be determined by
analyzing the equilibrium of joint A, Fig. a.
©Fy¿ = 0; FAB sin 15° - m(9.81) cos 45° = 0
FAB = 26.8014m
B
Section Properties. The cross-sectional area and moment of inertia of the rod are
A =
I =
p
A 0.052 B = 0.625 A 10 - 3 B pm2
4
p
A 0.0254 B = 97.65625 A 10 - 9 B pm4
4
Critical Buckling Load. Since the rod is pinned at both of its ends, K = 1. Here,
Pcr = FAB (F.S.) = 26.8014m(2) = 53.6028m. Applying Euler’s formula,
Pcr =
p2EIy
(KL)2
53.6028m =
p2 c 200 A 109 B d c 97.65625 A 10 - 9 B p d
[1(4)]2
m = 706.11 kg = 7.06 kg
Ans.
Critical Stress. Euler’s formula is valid only if scr 6 sY.
scr =
53.6028(706.11)
Pcr
=
= 19.28 MPa 6 sY = 703 MPa
A
p 0.625 A 10 - 3 B
1062
60°
O.K.
C
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Page 1063
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–38. The members of the truss are assumed to be pin
connected. If member GF is an A-36 steel rod having a
diameter of 2 in., determine the greatest magnitude of load
P that can be supported by the truss without causing this
member to buckle.
H
G
16 ft
P
Support Reactions: As shown on FBD(a).
Member Forces: Use the method of sections [FBD(b)].
FGF = 1.3333P (C)
Section Properties:
A =
I =
p 2
A 2 B = p in2
4
p 4
A 1 B = 0.250p in4
4
Critical Buckling Load: K = 1 for a column with both ends pinned. Applying
Euler’s formula,
Pcr = FGF =
1.3333P =
p2EI
(KLGF)2
p2 (29)(103)(0.250p)
[1(16)(12)]2
P = 4.573 kip = 4.57 kip
Ans.
Critical Stress: Euler’s formula is only valid if scr 6 sg.
scr =
1.3333(4.573)
Pcr
=
= 1.94 ksi 6 sg = 36 ksi
p
A
1063
O.K.
D
C
B
16 ft
FGF (12) - P(16) = 0
E
12 ft
A
+ ©MB = 0;
F
16 ft
P
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Page 1064
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–39. The members of the truss are assumed to be pin
connected. If member AG is an A-36 steel rod having a
diameter of 2 in., determine the greatest magnitude of load
P that can be supported by the truss without causing this
member to buckle.
H
G
16 ft
P
Support Reactions: As shown on FBD(a).
Member Forces: Use the method of joints [FBD(b)].
3
= 0
F
5 AG
FAG = 1.6667P (C)
Section Properties:
LAG = 2162 + 122 = 20.0 ft
A =
I =
p 2
A 2 B = p in2
4
p 4
A 1 B = 0.250p in4
4
Critical Buckling Load: K = 1 for a column with both ends pinned. Applying
Euler’s formula,
Pcr = FGF =
1.6667P =
p2EI
(KLGF)2
p2 (29)(103)(0.250p)
[1(20)(12)]2
P = 2.342 kip = 2.34 kip
Ans.
Critical Stress: Euler’s formula is only valid if scr = sg.
scr =
1.6667(2.342)
Pcr
=
= 1.24 ksi 6 sg = 36 ksi
p
A
1064
O.K.
D
C
B
16 ft
P -
E
12 ft
A
+ c ©Fy = 0;
F
16 ft
P
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Page 1065
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*13–40. The column is supported at B by a support that
does not permit rotation but allows vertical deflection.
Determine the critical load Pcr . EI is constant.
L
B
Pcr
A
Elastic curve:
EI
d2y
= M = -P y
dx2
P
d2y
+
y = 0
EI
dx2
y = C1 sin c
P
P
x d + C2 cos c
xd
A EI
A EI
Boundry conditions:
At x = 0;
0 = 0 + C2;
At x = L;
y = 0
C2 = 0
dv
= 0
dx
P
P
dv
= C1
cos c
L] d = 0;
dx
A EI
A EI
cos c
P
L d = 0;
A EI
For n = 1 ;
Pcr =
C1
P
p
L = na b
A EI
2
P
Z 0
A EI
n = 1, 3, 5
p2
P
=
EI
4L2
p2EI
4L2
Ans.
1065
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w
•13–41.
The ideal column has a weight w (force兾length)
and rests in the horizontal position when it is subjected to the
axial load P. Determine the maximum moment in the column
at midspan. EI is constant. Hint: Establish the differential
equation for deflection, Eq. 13–1, with the origin at the mid
span. The general solution is v = C1 sin kx + C2 cos kx +
1w>12P22x2 - 1wL>12P22x - 1wEI>P22 where k2 = P>EI.
P
L
Moment Functions: FBD(b).
a + ©Mo = 0;
M(x) =
wL
x
wx a b - M(x) - a
bx - Pv = 0
2
2
w 2
A x - Lx B - Pv
2
[1]
Differential Equation of The Elastic Curve:
EI
d2y
= M(x)
dx2
EI
d2y
w 2
=
A x - Lx B - Py
2
dx2
w
d2y
P
y =
+
A x2 - Lx B
EI
2EI
dx2
The solution of the above differential equation is of the form
v = C1 sin a
P
P
w 2
wL
wEI
x b + C2 cos ¢
xb +
x x A EI
A EI
2P
2P
P2
[2]
dv
P
P
P
P
w
wL
= C1
cos ¢
x ≤ - C2
sin ¢
x≤ +
xdx
A EI
A EI
A EI
A EI
P
2P
[3]
and
The integration constants can be determined from the boundary conditions.
Boundary Condition:
At x = 0, y = 0. From Eq. [2],
0 = C2 -
wEI
P2
C2 =
wEI
P2
At x =
L dy
= 0. From Eq.[3],
,
2 dx
0 = C1
P
wEI
P
w L
P L
P L
wL
cos ¢
sin ¢
≤ ≤ + a b A EI
A EI 2
A EI 2
P 2
2P
P2 A EI
C1 =
wEI
P L
tan ¢
≤
A EI 2
P2
1066
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Page 1067
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13–41. Continued
Elastic Curve:
y =
w EI
P L
P
EI
P
x2
L
EI
x≤ +
x≤ +
tan ¢
cos ¢
- x B
≤ sin ¢
R
P P
A EI 2
A EI
P
A EI
2
2
P
However, y = ymax at x =
ymax =
=
L
. Then,
2
P L
P L
P L
EI
w EI
EI
L2
tan ¢
cos ¢
B
≤ sin ¢
≤ +
≤ R
P P
A EI 2
A EI 2
P
A EI 2
8
P
wEI
P L
PL2
sec
- 1R
B
¢
≤
A EI 2
8EI
P2
Maximum Moment: The maximum moment occurs at x =
Mmax =
L
. From, Eq.[1],
2
w L2
L
- L a b R - Pymax
B
2 4
2
= -
wL2
wEI
P L
PL2
- P b 2 B sec ¢
- 1R r
≤ 8
A EI 2
8EI
P
= -
PL
wEI
B sec ¢
≤ - 1R
P
A EI 2
Ans.
13–42. The ideal column is subjected to the force F at its
midpoint and the axial load P. Determine the maximum
moment in the column at midspan. EI is constant. Hint:
Establish the differential equation for deflection, Eq. 13–1.
The general solution is v = C1 sin kx + C2 cos kx - c2x>k2,
where c2 = F>2EI, k2 = P>EI.
F
P
L
2
Moment Functions: FBD(b).
a + ©Mo = 0;
M(x) +
F
x + P(v) = 0
2
M(x) = -
F
x - Pv
2
[1]
Differential Equation of The Elastic Curve:
EI
d2y
= M(x)
dx2
EI
F
d2y
= - x - Py
2
2
dx
d2y
F
P
y = x
+
EI
2EI
dx2
The solution of the above differential equation is of the form,
v = C1 sin a
P
P
F
x b + C2 cos ¢
xb x
A EI
A EI
2P
1067
[2]
L
2
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13–42. Continued
and
dv
P
P
P
P
F
= C1
cos ¢
x ≤ - C2
sin ¢
x≤ dx
A EI
A EI
A EI
A EI
2P
[3]
The integration constants can be determined from the boundary conditions.
Boundary Conditions:
At x = 0, y = 0. From Eq.[2], C2 = 0
At x =
L dy
= 0. From Eq.[3],
,
2 dx
0 = C1
C1 =
P L
P
F
cos ¢
≤ A EI
A EI 2
2P
F
EI
P L
sec ¢
≤
2P A P
A EI 2
Elastic Curve:
y =
F
EI
P L
P
F
sec ¢
x≤ x
≤ sin ¢
2P A P
A EI 2
A EI
2P
=
F
EI
P L
P
sec ¢
x≤ - xR
B
≤ sin ¢
2P A P
A EI 2
A EI
However, y = ymax at x =
ymax =
=
L
. Then,
2
F
EI
P L
P L
L
sec ¢
B
≤ sin ¢
≤ - R
2P A P
A EI 2
A EI 2
2
F
EI
P L
L
tan ¢
B
≤ - R
2P A P
A EI 2
2
Maximum Moment: The maximum moment occurs at x =
Mmax = -
L
. From Eq.[1],
2
F L
a b - Pymax
2 2
= -
FL
F
EI
P L
L
- Pb
tan ¢
B
≤ - Rr
4
2P A P
A EI 2
2
= -
P L
F EI
tan ¢
≤
2 AP
A EI 2
Ans.
1068
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Page 1069
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13–43. The column with constant EI has the end
constraints shown. Determine the critical load for the
column.
P
L
Moment Function. Referring to the free-body diagram of the upper part of the
deflected column, Fig. a,
a + ©MO = 0;
M + Pv = 0
M = - Pv
Differential Equation of the Elastic Curve.
EI
d2v
= M
dx2
EI
d2v
= - Pv
dx2
d2v
P
+
v = 0
2
EI
dx
The solution is in the form of
v = C1 sin a
P
P
x b + C2 cos ¢
xb
A EI
A EI
(1)
dv
P
P
P
P
= C1
cos ¢
x ≤ - C2
sin ¢
x≤
dx
A EI
A EI
A EI
A EI
(2)
Boundary Conditions. At x = 0, v = 0. Then Eq. (1) gives
0 = 0 + C2
At x = L,
C2 = 0
dv
= 0. Then Eq. (2) gives
dx
0 = C1
P
P
cos ¢
L≤
A EI
A EI
C1 = 0 is the trivial solution, where v = 0. This means that the column will remain
straight and buckling will not occur regardless of the load P. Another possible
solution is
cos ¢
P
L≤ = 0
A EI
P
np
L =
A EI
2
n = 1, 3, 5
The smallest critical load occurs when n = 1, then
p
Pcr
L =
A EI
2
Pcr =
p2EI
4L2
Ans.
1069
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Page 1070
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*13–44. Consider an ideal column as in Fig. 13–10c, having
both ends fixed. Show that the critical load on the column
is given by Pcr = 4p2EI>L2. Hint: Due to the vertical
deflection of the top of the column, a constant moment
M¿ will be developed at the supports. Show that
d2v>dx2 + 1P>EI2v = M¿>EI. The solution is of the form
v = C1 sin1 1P>EIx2 + C2 cos1 1P>EIx2 + M¿>P.
Moment Functions:
M(x) = M¿ - Py
Differential Equation of The Elastic Curve:
EI
d2y
= M(x)
dx2
d2y
= M¿ - Py
dx2
EI
M¿
d2y
P
+
y =
2
EI
EI
dx
(Q.E.D.)
The solution of the above differential equation is of the form
v = C1 sin a
P
P
M¿
x b + C2 cos ¢
xb +
A EI
P
A EI
[1]
and
dv
P
P
P
P
= C1
cos ¢
x ≤ - C2
sin ¢
x≤
dx
A EI
A EI
A EI
A EI
[2]
The integration constants can be determined from the boundary conditions.
Boundary Conditions:
At x = 0, y = 0. From Eq.[1], C2 = At x = 0,
M¿
P
dy
= 0. From Eq.[2], C1 = 0
dx
Elastic Curve:
y =
M¿
P
x≤ R
B 1 - cos ¢
P
A EI
and
M¿
P
P
dy
=
sin ¢
x≤
dx
P A EI
A EI
However, due to symmetry
sin B
L
dy
= 0 at x = . Then,
dx
2
P L
a bR = 0
A EI 2
or
P L
a b = np
A EI 2
where n = 1, 2, 3,...
The smallest critical load occurs when n = 1.
Pce =
4p2EI
L2
(Q.E.D.)
1070
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•13–45. Consider an ideal column as in Fig. 13–10d, having
one end fixed and the other pinned. Show that the critical load
on the column is given by Pcr = 20.19EI>L2. Hint: Due to the
vertical deflection at the top of the column,a constant moment
M¿ will be developed at the fixed support and horizontal
reactive forces R¿ will be developed at both supports. Show
that d2v>dx2 + 1P>EI2v = 1R¿>EI21L - x2. The solution
is of the form v = C1 sin 11P>EIx2 + C2 cos 11P>EIx2 +
1R¿>P21L - x2. After application of the boundary conditions
show that tan 1 1P>EIL2 = 1P>EI L. Solve by trial and
error for the smallest nonzero root.
Equilibrium. FBD(a).
Moment Functions: FBD(b).
M(x) = R¿(L - x) - Py
Differential Equation of The Elastic Curve:
EI
d2y
= M(x)
dx2
EI
d2y
= R¿(L - x) - Py
dx2
d2y
P
R¿
+
y =
(L - x)
2
EI
EI
dx
(Q.E.D.)
The solution of the above differential equation is of the form
v = C1 sin a
P
P
R¿
(L - x)
x b + C2 cos ¢
xb +
A EI
P
A EI
[1]
and
dv
P
P
P
P
R¿
= C1
cos ¢
x ≤ - C2
sin ¢
x≤ dx
A EI
A EI
A EI
A EI
P
The integration constants can be determined from the boundary conditions.
Boundary Conditions:
At x = 0, y = 0. From Eq.[1], C2 = -
At x = 0,
R¿L
P
dy
R¿ EI
= 0. From Eq.[2], C1 =
dx
P AP
Elastic Curve:
y =
=
R¿ EI
P
R¿L
P
R¿
sin ¢
x≤ cos ¢
x≤ +
(L - x)
P AP
A EI
P
A EI
P
EI
P
P
R¿
sin ¢
x ≤ - L cos ¢
x ≤ + (Lx) R
B
P AP
A EI
A EI
1071
[2]
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13–45. Continued
However, y = 0 at x = L. Then,
0 =
P
P
EI
sin ¢
L ≤ - L cos ¢
L≤
A EI
A EI
AP
tan ¢
P
P
L≤ =
L
A EI
A EI
(Q.E.D.)
By trial and error and choosing the smallest root, we have
P
L = 4.49341
A EI
Then,
Pcr =
20.19EI
L2
(Q.E.D.)
13–46. Determine the load P required to cause the A-36
steel W8 * 15 column to fail either by buckling or by
yielding. The column is fixed at its base and free at its top.
1 in.
P
Section properties for W8 * 15:
A = 4.44 in2
Ix = 48.0 in4
rx = 3.29 in.
d = 8.11 in.
Iy = 3.41 in4
8 ft
Buckling about y - y axis:
K = 2.0
P = Pcr =
L = 8(12) = 96 in.
p2EIy
(KL)2
Check: scr =
p2(29)(103)(3.41)
=
[(2.0)(96)]2
= 26.5 kip
(controls)
Pcr
26.5
=
= 5.96 ksi 6 sg
A
4.44
Ans.
O.K.
Check yielding about x - x axis:
smax =
P
ec
KL P
c 1 + 2 sec a
bd
A
2r A EA
r
26.5
P
=
= 5.963 ksi
A
4.44
(1) A 8.11
ec
2 B
=
= 0.37463
2
r
(3.29)2
2.0(96)
P
26.5
KL
=
= 0.4184
2r A EA
2(3.29) A 29(103)(4.44)
smax = 5.963[1 + 0.37463 sec (0.4184)] = 8.41 ksi 6 sg = 36 ksi
1072
O.K.
13 Solutions 46060
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Page 1073
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13–47. The hollow red brass C83400 copper alloy shaft is
fixed at one end but free at the other end. Determine the
maximum eccentric force P the shaft can support without
causing it to buckle or yield. Also, find the corresponding
maximum deflection of the shaft.
2m
a
a
P
150 mm
30 mm
Section Properties.
A = p A 0.032 - 0.022 B = 0.5 A 10 - 3 B p m2
I =
20 mm
Section a – a
p
A 0.034 - 0.024 B = 0.1625 A 10 - 6 B p m4
4
0.1625 A 10 B p
I
=
= 0.01803 m
C 0.5 A 10 - 3 B p
AA
-6
r =
e = 0.15 m
c = 0.03 m
For a column that is fixed at one end and free at the other, K = 2. Thus,
KL = 2(2) = 4 m
Yielding. In this case, yielding will occur before buckling. Applying the secant
formula,
smax =
P
ec
KL P
B 1 + 2 sec ¢
≤R
A
2rx A EA
rx
70.0 A 106 B =
70.0 A 106 B =
P
0.5 A 10
-3
Bp
P
0.5 A 10 - 3 B p
D1 +
0.15(0.03)
0.018032
secC
P
4
ST
2(0.01803)A 101 A 109 B C 0.5 A 10 - 3 B p D
a 1 + 13.846 sec 8.8078 A 10 - 3 B 2Pb
Solving by trial and error,
P = 5.8697 kN = 5.87 kN
Ans.
Maximum Deflection.
vmax = e B sec ¢
P KL
≤ - 1R
A EI 2
= 0.15 D sec C
5.8697 A 103 B
4
a b S - 1T
C 101 A 109 B C 0.1625 A 10 - 6 B p D 2
= 0.04210 m = 42.1 mm
Ans.
1073
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Page 1074
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*13–48. The hollow red brass C83400 copper alloy shaft is
fixed at one end but free at the other end. If the eccentric
force P = 5 kN is applied to the shaft as shown, determine
the maximum normal stress and the maximum deflection.
2m
a
a
P
150 mm
30 mm
Section Properties.
A = p A 0.032 - 0.022 B = 0.5 A 10 - 3 B p m2
I =
20 mm
Section a – a
p
A 0.034 - 0.024 B = 0.1625 A 10 - 6 B p m4
4
0.1625 A 10 - 6 B p
I
= 0.01803 m
=
r =
C 0.5 A 10 - 3 B p
AA
e = 0.15 m
c = 0.03 m
For a column that is fixed at one end and free at the other, K = 2. Thus,
KL = 2(2) = 4 m
Yielding. Applying the secant formula,
smax =
P
ec
KL P
B 1 + 2 sec ¢
≤R
A
2r A EA
r
5 A 103 B
5 A 103 B
4
ST
=
secC
D1 +
2(0.01803) C 101 A 109 B C 0.5 A 10 - 3 B p D
0.018032
0.5 A 10 - 3 B p
0.15(0.03)
= 57.44 MPa = 57.4 MPa
Ans.
Since smax 6 sY = 70 MPa, the shaft does not yield.
Maximum Deflection.
vmax = e B sec ¢
P KL
≤ - 1R
A EI 2
= 0.15D sec C
5 A 103 B
4
a b S - 1T
C 101 A 109 B C 0.1625 A 10 - 6 B p D 2
= 0.03467 m = 34.7 mm
Ans.
1074
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Page 1075
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•13–49.
The tube is made of copper and has an outer
diameter of 35 mm and a wall thickness of 7 mm. Using a
factor of safety with respect to buckling and yielding of
F.S. = 2.5, determine the allowable eccentric load P. The
tube is pin supported at its ends. Ecu = 120 GPa, sY =
750 MPa.
2m
P
14 mm
Section Properties:
A =
p
(0.0352 - 0.0212) = 0.61575(10 - 3) m2
4
I =
p
(0.01754 - 0.01054) = 64.1152(10 - 9) m4
4
r =
I
64.1152(10 - 9)
=
= 0.010204 m
AA
A 0.61575(10 - 3)
For a column pinned at both ends, K = 1. Then KL = 1(2) = 2 m.
Buckling: Applying Euler’s formula,
Pmax = Pcr =
p2 (120)(109) C 64.1152(10 - 9) D
p2EI
=
= 18983.7 N = 18.98 kN
(KL)2
22
Critical Stress: Euler’s formula is only valid if scr 6 sg.
scr =
Pcr
18983.7
= 30.83 MPa 6 sg = 750 MPa
=
A
0.61575(10 - 3)
O.K.
Yielding: Applying the secant formula,
smax =
(KL) Pmax
Pmax
ec
B 1 + 2 sec ¢
≤R
A
2r A EA
r
750 A 106 B =
0.61575(10 - 3)
750 A 106 B =
0.61575(10 - 3)
Pmax
Pmax
B1 +
0.014(0.0175)
0.0102042
sec ¢
Pmax
2
≤R
2(0.010204)A 120(109)[0.61575(10 - 3)]
A 1 + 2.35294 sec 0.0114006 2Pmax B
Solving by trial and error,
Pmax = 16 885 N = 16.885 kN (Controls!)
Factor of Safety:
P =
Pmax
16.885
=
= 6.75 kN
F.S.
2.5
Ans.
1075
P
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Page 1076
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13–50. The tube is made of copper and has an outer
diameter of 35 mm and a wall thickness of 7 mm. Using a
factor of safety with respect to buckling and yielding of
F.S. = 2.5, determine the allowable eccentric load P that it
can support without failure. The tube is fixed supported at
its ends. Ecu = 120 GPa, sY = 750 MPa.
2m
P
14 mm
Section Properties:
A =
p
A 0.0352 - 0.0212 B = 0.61575 A 10 - 3 B m2
4
I =
p
A 0.01754 - 0.01054 B = 64.1152 A 10 - 9 B m4
4
r =
I
64.1152(10 - 9)
=
= 0.010204 ms
AA
A 0.61575(10 - 3)
For a column fixed at both ends, K = 0.5. Then KL = 0.5(2) = 1 m.
Buckling: Applying Euler’s formula,
Pmax = Pcr =
p2(120)(109) C 64.1152(10 - 9) D
p2EI
=
= 75 935.0 N = 75.93 kN
(KL)2
12
Critical Stress: Euler’s formula is only valid if scr 6 sg.
scr =
Pcr
75 935.0
= 123.3 MPa 6 sg = 750 MPa
=
A
0.61575(10 - 3)
O. K.
Yielding: Applying the secant formula,
smax =
(KL) Pmax
Pmax
ec
B 1 + 2 sec ¢
≤R
A
2r A EA
r
750 A 106 B =
0.61575(10 - 3)
750 A 106 B =
0.61575(10 - 3)
Pmax
Pmax
B1 +
0.014(0.0175)
0.0102042
sec ¢
2
Pmax
≤R
2(0.010204)A 120(109)[0.61575(10 - 3)]
A 1 + 2.35294 sec 5.70032 A 10 - 3 B 2P B
Solving by trial and error,
Pmax = 50 325 N = 50.325 kN (Controls!)
Factor of Safety:
P =
Pmax
50.325
=
= 20.1 kN
F.S.
2.5
Ans.
1076
P
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Page 1077
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13–51. The wood column is fixed at its base and can be
assumed pin connected at its top. Determine the maximum
eccentric load P that can be applied without causing the
column to buckle or yield. Ew = 1.811032 ksi, sY = 8 ksi.
P
y
4 in.
x
x
P
y 10 in.
10 ft
Section Properties:
A = 10(4) = 40 in2
ry =
Iy =
1
(4)(103) = 333.33 in4
12
Ix =
1
(10)(43) = 53.33 in4
12
Ix
333.33
=
= 2.8868 in.
AA
A 40
Buckling about x -x axis:
P = Pcr =
p2(1.8)(103)(53.33)
p2EI
=
= 134 kip
(KL)2
[(0.7)(10)(12)]2
Check: scr =
Pcr
134
=
= 3.36 ksi 6 sg
A
40
O.K.
Yielding about y- y axis:
smax =
P
ec
KL
P
a 1 + 2 sec a
b
b
A
2r A EA
r
5(5)
ec
=
= 3.0
r2
2.88682
a
0.7(10)(12)
P
P
KL
b
=
= 0.0542212P
2r A EA
2(2.8868) A 1.8(103)(40)
8(40) = P[1 + 3.0 sec (0.0542212P)]
By trial and error:
P = 73.5 kip
Ans.
(controls)
1077
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Page 1078
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*13–52. The wood column is fixed at its base and can
be assumed fixed connected at its top. Determine the
maximum eccentric load P that can be applied without
causing the column to buckle or yield. Ew = 1.811032 ksi,
sY = 8 ksi.
P
y
4 in.
x
x
P
y 10 in.
10 ft
Section Properties:
A = 10(4) = 40 in2
ry =
Iy =
1
(4)(103) = 333.33 in4
12
Ix =
1
(10)(43) = 53.33 in4
12
Iy
333.33
=
= 2.8868 in.
AA
A 40
Buckling about x - x axis:
P = Pcr =
p2(1.8)(103)(53.33)
p2EI
=
= 263 kip
2
(KL)
[(0.5)(10)(12)]2
Check: scr =
Pcr
263
=
= 6.58 ksi 6 sg
A
40
O.K.
Yielding about y -y axis:
smax =
P
ec
KL P
a1 + 2 sec a
bb
A
2r A EA
r
5(5)
ec
=
= 3.0
r2
2.88682
a
0.5(10)(12)
P
P
KL
b
=
= 0.0387292P
2r A EA
2(2.8868) A 1.8(103)(40)
8(40) = P[1 + 3.0 sec (0.0387292P)]
By trial and error:
P = 76.5 kip
Ans.
(controls)
1078
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Page 1079
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
The W200 * 22 A-36-steel column is fixed at its
base. Its top is constrained to rotate about the y–y axis and
free to move along the y–y axis. Also, the column is braced
along the x–x axis at its mid-height. Determine the
allowable eccentric force P that can be applied without
causing the column either to buckle or yield. Use F.S. = 2
against buckling and F.S. = 1.5 against yielding.
•13–53.
100 mm
y
A = 2860 mm2 = 2.86 A 10 - 3 B m2
ry = 22.3 mm = 0.0223 m
Ix = 20.0 A 106 B mm4 = 20.0 A 10 - 6 B m4
c =
bf
2
=
102
= 51 mm = 0.051 m
2
e = 0.1m
Buckling About the Strong Axis. Since the column is fixed at the base and free at
the top, Kx = 2. Applying Euler’s formula,
Pcr =
p2EIx
(KL)x 2
=
p2 c 200 A 109 B d c 20.0 A 10 - 6 B d
[2(10)]2
= 98.70kN
Euler’s formula is valid if scr 6 sY.
scr =
98.70 A 103 B
Pcr
=
= 34.51 MPa 6 sY = 250MPa
A
2.86 A 10 - 3 B
O.K.
Then,
Pallow =
Pcr
98.70
=
= 49.35 kN
F.S.
2
Yielding About Weak Axis. Since the support provided by the bracing can be
considered a pin connection, the upper portion of the column is pinned at both of its
ends. Then Ky = 1 and L = 5 m. Applying the secant formula,
smax =
A KL B y Pmax
Pmax
ec
C 1 + 2 sec B
RS
A
2ry A EA
ry
250 A 106 B =
250 A 106 B =
Pmax
2.86 A 10
-3
Pmax
B
2.86 A 10 - 3 B
D1 +
0.1(0.051)
0.02232
sec C
1(5)
Pmax
ST
2(0.0223)A 200 A 109 B C 2.86 A 10 - 3 B D
c 1 + 10.2556 sec 4.6875 A 10 - 3 B 2Pmax d
Solving by trial and error,
Pmax = 39.376 kN
Then,
Pallow =
Pmax
39.376
=
= 26.3 kN (controls)
1.5
1.5
Ans.
1079
5m
y
x
5m
Section Properties. From the table listed in the appendix, the necessary section
properties for a W200 * 22 are
P
x
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Page 1080
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–54. The W200 * 22 A-36-steel column is fixed at its
base. Its top is constrained to rotate about the y–y axis and
free to move along the y–y axis. Also, the column is braced
along the x–x axis at its mid-height. If P = 25 kN,
determine the maximum normal stress developed in the
column.
100 mm
y
Section Properties. From the table listed in the appendix, necessary section
properties for a W200 * 22 are
A = 2860 mm2 = 2.86 A 10 - 3 B m2
ry = 22.3 mm = 0.0223 m
Ix = 20.0 A 106 B mm4 = 20.0 A 10 - 6 B m4
c =
bf
2
=
102
= 51 mm = 0.051 m
2
e = 0.1m
Buckling About the Strong Axis. Since the column is fixed at the base and free at
the top, Kx = 2. Applying Euler’s formula,
Pcr =
p2EIx
(KL)x 2
=
p2 c 200 A 109 B d c 20.0 A 10 - 6 B d
[2(10)]2
= 98.70kN
Euler’s formula is valid only if scr 6 sY.
scr =
98.70 A 103 B
Pcr
=
= 34.51 MPa 6 sY = 250 MPa
A
2.86 A 10 - 3 B
O.K.
Since P = 25 kN 6 Pcr, the column does not buckle.
Yielding About Weak Axis. Since the support provided by the bracing can be
considered a pin connection, the upper portion of the column is pinned at both of its
ends. Then Ky = 1 and L = 5 m. Applying the secant formula,
smax =
=
(KL) P
P
ec
C 1 + 2 sec B
RS
A
2ry A EA
ry
2.5 A 103 B
2.86 A 10 - 3 B
D1 +
0.1(0.051)
0.02232
secC
25 A 103 B
1(5)
ST
2(0.0223) C 200 A 109 B C 2.86 A 10 - 3 B D
= 130.26 MPa = 130 MPa
Ans.
Since smax 6 sY = 250 MPa, the column does not yield.
1080
y
x
5m
5m
P
x
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Page 1081
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–55. The wood column is fixed at its base, and its top
can be considered pinned. If the eccentric force P = 10 kN
is applied to the column, investigate whether the column
is adequate to support this loading without buckling or
yielding. Take E = 10 GPa and sY = 15 MPa.
x
P
150 mm
25 mm
yx
25 mm
75 mm
5m
Section Properties.
A = 0.05(0.15) = 7.5 A 10 - 3 B m2
Ix =
rx =
1
(0.05) A 0.153 B = 14.0625 A 10 - 6 B m4
12
14.0625 A 10 - 6 B
Ix
= 0.04330 m
=
AA
C 7.5 A 10 - 3 B
1
(0.15) A 0.053 B = 1.5625 A 10 - 6 B m4
12
e = 0.15 m
c = 0.075 m
Iy =
For a column that is fixed at one end and pinned at the other K = 0.7. Then,
(KL)x = (KL)y = 0.7(5) = 3.5 m
Buckling About the Weak Axis. Applying Euler’s formula,
Pcr =
p2EIy
(KL)y 2
=
p2 C 10 A 109 B D C 1.5625 A 10 - 6 B D
3.52
= 12.59 kN
Euler’s formula is valid if scr 6 sY.
scr =
12.59 A 103 B
Pcr
=
= 1.68 MPa 6 sY = 15 MPa
A
7.5 A 10 - 3 B
O.K.
Since Pcr 7 P = 10 kN, the column will not buckle.
Yielding About Strong Axis. Applying the secant formula.
smax =
=
(KL)x P
ec
P
C 1 + 2 sec B
RS
A
2rx A EA
rx
10 A 103 B
7.5 A 10 - 3 B
D1 +
0.15(0.075)
2
0.04330
secC
10 A 103 B
3.5
ST
2(0.04330) C 10 A 109 B C 7.5 A 10 - 3 B D
= 10.29 MPa
Since smax 6 sY = 15 MPa , the column will not yield.
Ans.
1081
75 mm
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Page 1082
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*13–56. The wood column is fixed at its base, and its
top can be considered pinned. Determine the maximum
eccentric force P the column can support without
causing it to either buckle or yield. Take E = 10 GPa
and sY = 15 MPa.
x
P
150 mm
25 mm
yx
25 mm
75 mm
5m
Section Properties.
A = 0.05(0.15) = 7.5 A 10 - 3 B m2
Ix =
1
(0.05) A 0.153 B = 14.0625 A 10 - 6 B m4
12
14.0625 A 10
Ix
=
C 7.5 A 10 - 3 B
AA
-6
rx =
B
= 0.04330 m
1
(0.15) A 0.053 B = 1.5625 A 10 - 6 B m4
12
e = 0.15 m
c = 0.075 m
Iy =
For a column that is fixed at one end and pinned at the other K = 0.7. Then,
(KL)x = (KL)y = 0.7(5) = 3.5 m
Buckling About the Weak Axis. Applying Euler’s formula,
Pcr =
p2EIy
(KL)y
2
=
p2 C 10 A 109 B D C 1.5625 A 10 - 6 B D
3.52
= 12.59 kN = 12.6 kN
Ans.
Euler’s formula is valid if scr 6 sY.
scr
12.59 A 103 B
Pcr
=
=
= 1.68 MPa 6 sY = 15 MPa
A
7.5 A 10 - 3 B
O.K.
Yielding About Strong Axis. Applying the secant formula with P = Pcr = 12.59 kN,
smax =
=
(KL)x P
P
ec
C B 1 + 2 sec B
RS
A
2rx A EA
rx
12.59 A 103 B
7.5 A 10 - 3 B
D1 +
0.15(0.075)
0.043302
secC
12.59 A 103 B B
3.5
ST
2(0.04330) C 10 A 109 B C 7.5 A 10 - 3 B D
= 13.31 MPa 6 sY = 15 MPa
O.K.
1082
75 mm
13 Solutions 46060
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Page 1083
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
The W250 * 28 A-36-steel column is fixed at its
base. Its top is constrained to rotate about the y–y axis and
free to move along the y–y axis. If e = 350 mm, determine
the allowable eccentric force P that can be applied without
causing the column either to buckle or yield. Use F.S. = 2
against buckling and F.S. = 1.5 against yielding.
•13–57.
P e
x
x
y
6m
Section Properties. From the table listed in the appendix, necessary section
properties for a W250 * 28 are
A = 3620 mm2 = 3.62 A 10 - 3 B m2
rx = 105 mm = 0.105 m
Iy = 1.78 A 106 B mm4 = 1.78 A 10 - 6 B m4
c =
260
d
=
= 130 mm = 0.13 m
2
2
e = 0.35 m
Buckling About the Strong Axis. Since the column is fixed at the base and pinned at
the top, Kx = 0.7. Applying Euler’s formula,
Pcr =
p2EIy
(KL)y 2
=
p2 C 200 A 109 B D C 1.78 A 10 - 6 B D
[0.7(6)]2
= 199.18 kN
Euler’s formula is valid only if scr 6 sY.
scr =
199.18 A 103 B
Pcr
=
= 55.02 MPa 6 sY = 250 MPa
A
3.62 A 10 - 3 B
O.K.
Thus,
Pallow =
Pcr
199.18
=
= 99.59 kN
F.S.
2
Yielding About Strong Axis. Since the column is fixed at its base and free at its top,
Kx = 2. Applying the secant formula,
smax =
(KL)x Pmax
Pmax
ec
C 1 + 2 sec B
RS
A
2rx A EA
rx
250 A 106 B =
250 A 106 B =
Pmax
3.62 A 10
-3
Pmax
B
3.62 A 10 - 3 B
D1 +
0.35(0.13)
0.1052
sec C
2(6)
Pmax
ST
2(0.105)A 200 A 109 B C 3.62 A 10 - 3 B D
A 1 + 4.1270 sec (0.0021237)2Pmax B
Solving by trial and error,
Pmax = 133.45 kN
Then,
Pallow =
y
Pmax
133.45
=
= 88.97 kN = 89.0 kN (controls)
1.5
1.5
Ans.
1083
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Page 1084
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–58. The W250 * 28 A-36-steel column is fixed at its
base. Its top is constrained to rotate about the y–y axis and
free to move along the y–y axis. Determine the force P and
its eccentricity e so that the column will yield and buckle
simultaneously.
P e
x
x
y
6m
Section Properties. From the table listed in the appendix, the necessary section
properties for a W250 * 28 are
A = 3620 mm2 = 3.62 A 10 - 3 B m2
rx = 105 mm = 0.105 m
Iy = 1.78 A 106 B mm4 = 1.78 A 10 - 6 B m4
c =
d
260
=
= 130 mm = 0.13 m
2
2
Buckling About the Weak Axis. Since the column is fixed at the base and pinned at
its top, Kx = 0.7. Applying Euler’s formula,
Pcr =
p2EIy
(KL)y 2
=
p2 C 200 A 109 B D C 1.78 A 10 - 6 B D
[0.7(6)]2
= 199.18 kN = 199 kN
Ans.
Euler’s formula is valid only if scr 6 sY.
scr =
199.18 A 103 B
Pcr
=
= 55.02 MPa 6 sY = 250 MPa
A
3.62 A 10 - 3 B
O.K.
Yielding About Strong Axis. Since the column is fixed at its base and free at its top,
Kx = 2. Applying the secant formula with P = Pcr = 199.18 kN,
smax =
(KL)x P
P
ec
C 1 + 2 sec B
RS
A
2rx A EA
rx
250 A 106 B =
199.18 A 103 B
3.62 A 10 - 3 B
D1 +
e(0.13)
0.1052
secC
y
199.18 A 103 B
2(6)
ST
2(0.105) C 200 A 109 B C 3.62 A 10 - 3 B D
e = 0.1753 m = 175 mm
Ans.
1084
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Page 1085
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–59. The steel column supports the two eccentric
loadings. If it is assumed to be pinned at its top, fixed at the
bottom, and fully braced against buckling about the y–y
axis, determine the maximum deflection of the column
and the maximum stress in the column. Est = 200 GPa,
sY = 360 MPa.
130 kN 50 kN
80 mm
120 mm
100 mm
10 mm
6m
A = 0.12(0.1) - (0.1)(0.09) = 3.00 A 10 - 3 B m2
Ix =
1
1
(0.1) A 0.123 B (0.09) A 0.13 B = 6.90 A 10 - 6 B m4
12
12
rx =
Ix
6.90(10 - 6)
=
= 0.047958 m
AA
A 3.00(10 - 3)
For a column fixed at one end and pinned at the other end, K = 0.7.
(KL)x = 0.7(6) = 4.2 m
The eccentricity of the two applied loads is,
e =
130(0.12) - 50(0.08)
= 0.06444 m
180
Yielding About x–x Axis: Applying the secant formula,
(KL)x P
P
ec
B 1 + 2 sec ¢
≤R
A
2rx A EA
rx
180(103)
=
-3
3.00(10 )
y
B1 +
0.06444(0.06)
0.047958
2
sec ¢
180(103)
4.2
≤R
2(0.047958)A 200(109)(3.00)(10 - 3)
= 199 MPa
Ans.
Since smax 6 sg = 360 MPa, the column does not yield.
Maximum Displacement:
ymax = e B sec ¢
P KL
≤ - 1R
A EI 2
= 0.06444 B sec ¢
4.2
180(103)
a
b ≤ - 1R
A 200(109)[6.90(10 - 6)] 2
= 0.02433 m = 24.3 mm
Ans.
1085
10 mm
100 mm
x
Section Properties:
smax =
x
10 mm
y
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Page 1086
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*13–60. The steel column supports the two eccentric
loadings. If it is assumed to be fixed at its top and bottom,
and braced against buckling about the y–y axis, determine
the maximum deflection of the column and the maximum
stress in the column. Est = 200 GPa, sY = 360 MPa.
130 kN 50 kN
80 mm
120 mm
100 mm
10 mm
6m
A = 0.12(0.1) - (0.1)(0.09) = 3.00 A 10 - 3 B m2
Ix =
1
1
(0.1) A 0.123 B (0.09) A 0.013 B = 6.90 A 10 - 6 B m4
12
12
rx =
6.90(10 - 6)
Ix
=
= 0.047958 m
AA
A 3.00(10 - 3)
For a column fixed at both ends, K = 0.5.
(KL)x = 0.5(6) = 3.00 m
The eccentricity of the two applied loads is,
e =
130(0.12) - 50(0.08)
= 0.06444 m
180
Yielding About x–x Axis: Applying the secant formula,
(KL)x P
P
ec
B 1 + 2 sec ¢
≤R
A
2rx A EA
rx
180(103)
=
-3
3.00(10 )
y
B1 +
0.06444(0.06)
0.047958
2
sec ¢
180(103)
3.00
≤R
2(0.047958)A 200(109)(3.00)(10 - 3)
= 178 MPa
Ans.
Since smax 6 sg = 360 MPa, the column does not yield.
Maximum Displacement:
ymax = e B sec ¢
P KL
≤ - 1R
A EI 2
= 0.06444 B sec ¢
3
180(103)
a b ≤ - 1R
9
6
A 200(10 )[6.90(10 )] 2
= 0.01077 m = 10.8 mm
Ans.
1086
10 mm
100 mm
x
Section Properties:
smax =
x
10 mm
y
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Page 1087
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–61. The W250 * 45 A-36-steel column is pinned at its
top and fixed at its base. Also, the column is braced along
its weak axis at mid-height. If P = 250 kN, investigate
whether the column is adequate to support this loading.
Use F.S. = 2 against buckling and F.S. = 1.5 against
yielding.
P
4
250 mm
P
250 mm
4m
Section Properties. From the table listed in the appendix, the necessary section
properties for a W250 * 45 are
A = 5700 mm = 5.70 A 10
2
-3
Bm
2
rx = 112 mm = 0.112 m
Iy = 7.03 A 106 B mm4 = 7.03 A 10 - 6 B m4
c =
266
d
=
= 133 mm = 0.133 m
2
2
The eccentricity of the equivalent force P¿ = 250 +
250
= 312.5 kN is
4
250
(0.25)
4
= 0.15 m
250
250 +
4
250(0.25) e =
Buckling About the Weak Axis. The column is braced along the weak axis at
midheight and the support provided by the bracing can be considered as a pin. The
top portion of the column is critical is since the top is pinned so Ky = 1 and
L = 4 m Applying Euler’s formula,
Pcr =
p2EIy
(KL)y 2
=
p2 C 200 A 109 B D C 7.03 A 10 - 6 B D
[1(4)]2
= 867.29 kN
Euler’s equation is valid only if scr 6 sY.
scr =
867.29 A 103 B
Pcr
=
= 152.16 MPa 6 sY = 250 MPa
A
5.70 A 10 - 3 B
O.K.
Then,
œ
Pallow
=
Pcr
867.29
=
= 433.65 kN
F.S.
2
œ
Since Pallow
7 P¿ , the column does not buckle.
Yielding About Strong Axis. Since the column is fixed at its base and pinned at its
top, Kx = 0.7 and L = 8 m. Applying the secant formula with
œ
= P¿(F.S.) = 312.5(1.5) = 468.75 kN
Pmax
smax =
=
œ
œ
(KL)x Pmax
Pmax
ec
C 1 + 2 sec B
RS
A
2rx A EA
rx
468.75 A 103 B
5.70 A 10 - 3 B
C1 +
0.15(0.133)
0.112
2
sec B
468.75 A 103 B
0.7(8)
RS
2(0.112) C 200 A 109 B C 5.70 A 10 - 3 B D
= 231.84 MPa
Since smax 6 sY = 250 MPa, the column does not yield.
1087
4m
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•13–62. The W250 * 45 A-36-steel column is pinned at its
top and fixed at its base. Also, the column is braced along
its weak axis at mid-height. Determine the allowable force
P that the column can support without causing it either
to buckle or yield. Use F.S. = 2 against buckling and
F.S. = 1.5 against yielding.
P
4
250 mm
P
250 mm
4m
Section Properties. From the table listed in the appendix, the necessary section
properties for a W250 * 45 are
A = 5700 mm2 = 5.70 A 10 - 3 B m2
rx = 112 mm = 0.112 m
Iy = 7.03 A 106 B mm4 = 7.03 A 10 - 6 B m4
c =
The eccentricity of the equivalent force P¿ = P +
d
266
=
= 133 mm = 0.133 m
2
2
P
= 1.25P is
4
P
(0.25)
4
= 0.15 m
P
P +
4
P(0.25) e =
Buckling About the Weak Axis. The column is braced along the weak axis at
midheight and the support provided by the bracing can be considered as a pin. The
top portion of the column is critical is since the top is pinned so Ky = 1 and
L = 4 m. Applying Euler’s formula,
Pcr =
p2EIy
(KL)y
2
=
p2 C 200 A 109 B D C 7.03 A 10 - 6 B D
[1(4)]2
= 867.29 kN
Euler’s equation is valid only if scr 6 sY.
scr
867.29 A 103 B
Pcr
=
=
= 152.16 MPa 6 sY = 250 MPa
A
5.70 A 10 - 3 B
O.K.
Then,
œ
=
Pallow
Pcr
F.S.
867.29
2
= 346.92 kN
1.25Pallow =
Pallow
Yielding About Strong Axis. Since the column is fixed at its base and pinned at its
top, Kx = 0.7 and L = 8 m. Applying the secant formula,
smax =
œ
œ
(KL)x Pmax
Pmax
ec
C 1 + 2 sec B
RS
A
2rx A EA
rx
250 A 106 B =
250 A 106 B =
1.25Pmax
5.70 A 10
-3
1.25Pmax
B
5.70 A 10 - 3 B
C1 +
0.15(0.133)
0.1122
sec B
0.7(8)
1.25Pmax
RS
2(0.112) A 200 A 109 B C 5.70 A 10 - 3 B D
A 1 + 1.5904 sec (0.00082783) 2Pmax B
Solving by trial and error,
Pmax = 401.75 kN
Then,
Pallow =
401.75
= 267.83 kN = 268 kN (controls)
1.5
Ans.
1088
4m
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13–63. The W14 * 26 structural A-36 steel member is
used as a 20-ft-long column that is assumed to be fixed at
its top and fixed at its bottom. If the 15-kip load is applied
at an eccentric distance of 10 in., determine the maximum
stress in the column.
15 kip
Section Properties for W 14 * 26
A = 7.69 in2
d = 13.91 in.
10 in.
20 ft
rx = 5.65 in.
Yielding about x - x axis:
smax =
P
ec
KL
P
b d;
c 1 + 2 sec a
A
2 r AE A
r
15
P
=
= 1.9506 ksi;
A
7.69
K = 0.5
10 A 13.91
ec
2 B
=
= 2.178714
r2
(5.65)2
0.5 (20)(12)
15
KL P
=
= 0.087094
2 r A EA
2(5.65) A 29 (103)(7.69)
smax = 1.9506[1 + 2.178714 sec (0.087094)]
= 6.22 ksi 6 sg = 36 ksi
O.K.
Ans.
*13–64. The W14 * 26 structural A-36 steel member is
used as a column that is assumed to be fixed at its top and
pinned at its bottom. If the 15-kip load is applied at an
eccentric distance of 10 in., determine the maximum stress
in the column.
15 kip
Section Properties for W 14 * 26
A = 7.69 in2
d = 13.91 in.
20 ft
rx = 5.65 in.
Yielding about x - x axis:
smax =
P
ec
KL
P
c 1 + 2 sec a
b d;
A
2 r AE A
r
15
P
=
= 1.9506 ksi ;
A
7.69
K = 0.7
10 A 13.91
ec
2 B
=
= 2.178714
2
r
(5.65)2
0.7 (20)(12)
15
KL P
=
= 0.121931
2 r A EA
2(5.65) A 29 (103)(7.69)
smax = 1.9506[1 + 2.178714 sec (0.121931)]
= 6.24 ksi 6 sg = 36 ksi
10 in.
O.K.
Ans.
1089
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•13–65. Determine the maximum eccentric load P the
2014-T6-aluminum-alloy strut can support without causing
it either to buckle or yield. The ends of the strut are
pin-connected.
P
150 mm
100 mm
a
a
3m
50 mm
100 mm
Section a – a
Section Properties. The necessary section properties are
A = 0.05(0.1) = 5 A 10 - 3 B m2
Iy =
1
(0.1) A 0.053 B = 1.04167 A 10 - 6 B m4
12
4.1667 A 10
Ix
=
C 5 A 10 - 3 B
AA
-6
rx =
B
= 0.02887 m
For a column that is pinned at both of its ends K = 1. Thus,
(KL)x = (KL)y = 1(3) = 3 m
Buckling About the Weak Axis. Applying Euler’s formula,
Pcr =
p2EIy
(KL)y 2
=
p2 C 73.1 A 109 B D C 1.04167 A 10 - 6 B D
32
= 83.50 kN = 83.5 kN
Ans.
Critical Stress: Euler’s formula is valid only if scr 6 sY.
scr =
83.50 A 103 B
Pcr
=
= 16.70 MPa 6 sY = 414 MPa
A
5 A 10 - 3 B
O.K.
Yielding About Strong Axis. Applying the secant formula,
smax =
=
(KL)x P
P
ec
C 1 + 2 sec B
RS
A
2rx A EA
rx
83.50 A 103 B
5 A 10 - 3 B
D1 +
0.15(0.05)
0.028872
83.50 A 10 B
3
ST
2(0.02887) C 73.1 A 109 B C 5 A 10 - 3 B D
3
secC
= 229.27 MPa 6 sY = 414 MPa
O.K.
1090
P
150 mm
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13–66. The W8 * 48 structural A-36 steel column is fixed
at its bottom and free at its top. If it is subjected to the
eccentric load of 75 kip, determine the factor of safety with
respect to either the initiation of buckling or yielding.
75 kip
y
12 ft
Section Properties: For a wide flange section W8 * 48,
A = 14.1 in2
rx = 3.61 in.
Iy = 60.9 in4
d = 8.50 in.
For a column fixed at one end and free at the other and, K = 2.
(KL)y = (KL)x = 2(12)(12) = 288 in.
Buckling About y–y Axis: Applying Euler’s formula,
P = Pcr =
p2EIy
(KL)2y
p2 (29.0)(103)(60.9)
=
2882
= 210.15 kip
Critical Stress: Euler’s formula is only valid if scr 6 sg.
scr =
Pcr
210.15
=
= 14.90 ksi 6 sg = 36 ksi
A
14.1
O. K.
Yielding About x–x Axis: Applying the secant formula,
smax =
36 =
(KL)x Pmax
Pmax
ec
B 1 + 2 sec ¢
≤R
A
2rx A EA
rx
8 A 1.50
Pmax
Pmax
288
2 B
sec ¢
B1 +
≤R
14.1
2(3.61)A 29.0(103)(14.1)
3.612
36(14.1) = Pmax A 1 + 2.608943 sec 0.06238022Pmax B
Solving by trial and error,
Pmax = 117.0 kip (Controls!)
Factor of Safety:
F.S. =
Pmax
117.0
=
= 1.56
P
75
Ans.
1091
8 in.
y
x
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13–67. The W8 * 48 structural A-36 steel column is fixed
at its bottom and pinned at its top. If it is subjected to the
eccentric load of 75 kip, determine if the column fails by
yielding. The column is braced so that it does not buckle
about the y–y axis.
75 kip
y
12 ft
Section Properties: For a wide flange section W8 * 48,
A = 14.1 in2
rx = 3.61 in.
d = 8.50 in.
For a column fixed at one end and pinned at the other end, K = 0.7.
(KL)x = 0.7(12)(12) = 100.8 in.
Yielding About x–x Axis: Applying the secant formula,
smax =
=
(KL)x P
P
ec
B 1 + 2 sec ¢
≤R
A
2rx A EA
rx
8 A 1.50
75
100.8
75
2 B
sec ¢
B1 +
≤R
2
14.1
2(3.61)A
3.61
29.0(103)(14.1)
= 19.45 ksi 6 sg = 36 ksi
O.K.
Hence, the column does not fail by yielding.
Ans.
1092
8 in.
y
x
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*13–68. Determine the load P required to cause the steel
W12 * 50 structural A-36 steel column to fail either by
buckling or by yielding. The column is fixed at its bottom
and the cables at its top act as a pin to hold it.
2 in.
P
25 ft
Section Properties: For a wide flange section W12 * 50,
A = 14.7 in2
rx = 5.18 in.
Iy = 56.3 in4
d = 12.19 in.
For a column fixed at one end and pinned at the other end, K = 0.7.
(KL)y = (KL)x = 0.7(25)(12) = 210 in.
Buckling About y–y Axis: Applying Euler’s formula,
P = Pcr =
p2EIy
(KL)2y
p2 (29.0)(103)(56.3)
=
2102
= 365.40 kip
Critical Stress: Euler’s formula is only valid if scr 6 sg.
scr =
Pcr
365.4
=
= 24.86 ksi 6 sg = 36 ksi
A
14.7
O. K.
Yielding About x–x Axis: Applying the secant formula,
smax =
36 =
(KL)x P
P
ec
B 1 + 2 sec ¢
≤R
A
2rx A EA
rx
2 A 12.19
P
210
P
2 B
sec ¢
B1 +
≤R
14.7
2(5.18)A 29.0(103)(14.7)
5.182
36(14.7) = P A 1 + 0.454302 sec 0.03104572P B
Solving by trial and error,
Pmax = 343.3 kip = 343 kip (Controls!)
Ans.
1093
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•13–69.
Solve Prob. 13–68 if the column is an A-36 steel
W12 * 16 section.
2 in.
P
25 ft
Section Properties: For a wide flange section W12 * 16,
A = 4.71 in2
rx = 4.67 in.
Iy = 2.82 in4
d = 11.99 in.
For a column fixed at one end and pinned at the other end, K = 0.7.
(KL)y = (KL)x = 0.7(25)(12) = 210 in.
Buckling About y–y Axis: Applying Euler’s formula,
P = Pcr =
p 2EIy
(KL)2y
p2 (29.0)(103)(2.82)
=
2102
= 18.30 kip = 18.3 kip‚
Ans.
Critical Stress: Euler’s formula is only valid if scr 6 sg.
scr =
Pcr
18.30
=
= 3.89 ksi 6 sg = 36 ksi
A
4.71
Yielding About x–x Axis: Applying the secant formula,
smax =
=
(KL)x P
P
ec
B 1 + 2 sec ¢
≤R
A
2rx A EA
rx
2 A 11.99
18.30
210
18.30
2 B
seca
bR
B1 +
4.71
2(4.67)A 29.0(103)(4.71)
4.672
= 6.10 ksi 6 sg = 36 ksi
O.K.
1094
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13–70. A column of intermediate length buckles when the
compressive stress is 40 ksi. If the slenderness ratio is 60,
determine the tangent modulus.
p2 Et
A KrL B 2
scr =
40 =
;
a
KL
b = 60
r
p2 Et
(60)2
Et = 14590 ksi = 14.6 (103) ksi ‚
Ans
s(ksi)
13–71. The 6-ft-long column has the cross section shown
and is made of material which has a stress-strain diagram
that can be approximated as shown. If the column is
pinned at both ends, determine the critical load Pcr for the
column.
0.5 in.
55
0.5 in.
5 in.
25
0.5 in.
3 in.
Section Properties: The neccessary section properties are
A = 2[0.5(3)] + 5(0.5) = 5.5 in2
P (in./in.)
0.001
I = 2B
ry =
1
1
(0.5) A 33 B R +
(5) A 0.53 B = 2.3021 in4
12
12
2.3021
Iy
=
= 0.6470 in.
AA
A 5.5
For the column pinned at both of its ends, K = 1. Thus,
1(6)(12)
KL
= 111.29
=
ry
0.6470
Critical Stress. Applying Engesser’s equation,
scr =
p2Et
a
KL
r
b
=
p2Et
111.292
= 0.7969 A 10 - 3 B Et
(1)
From the stress - strain diagram, the tangent moduli are
(Et)1 =
25 ksi
= 25 A 103 B ksi
0.001
(Et)2 =
(55 - 25) ksi
= 10 A 103 B (ksi)
0.004 - 0.001
0 … s 6 25 ksi
25ksi 6 s … 40 ksi
Substituting (Et)1 = 25 A 103 B into Eq. (1),
scr = 0.7969 A 10 - 3 B c 25 A 103 B d = 19.92 ksi
Since scr 6 sY = 25 ksi, elastic buckling occurs. Thus,
Pcr = scr A = 19.92(5.5) = 109.57 kip = 110 kip‚
Ans.
1095
0.004
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s(ksi)
*13–72. The 6-ft-long column has the cross section shown
and is made of material which has a stress-strain diagram
that can be approximated as shown. If the column is fixed
at both ends, determine the critical load Pcr for the column.
0.5 in.
55
0.5 in.
5 in.
25
0.5 in.
3 in.
Section Properties. The neccessary section properties are
A = 2[0.5(3)] + 5(0.5) = 5.5 in2
P (in./in.)
0.001
I = 2B
ry =
1
1
(0.5) A 33 B R +
(5) A 0.53 B = 2.3021 in4
12
12
Iy
2.3021
=
= 0.6470 in.
AA
A 5.5
For the column fixed at its ends, K = 0.5. Thus,
0.5(6)(12)
KL
= 55.64
=
ry
0.6470
Critical Stress. Applying Engesser’s equation,
From the stress - strain diagram, the tangent moduli are
(Et)1 =
25 ksi
= 25 A 103 B ksi
0.001
(Et)2 =
(55 - 25)ksi
= 10 A 103 B ksi
0.004 - 0.001
0 … s 6 25 ksi
25 ksi 6 s … 40 ksi
Substituting (Et)1 = 25 A 103 B ksi into Eq. (1),
scr = 3.1875 A 10 - 3 B c 25 A 103 B d = 79.69 ksi
Since scr 7 sY = 25 ksi, the inelastic buckling occurs. Substituting (Et)2 into Eq. (1),
scr = 3.1875 A 10 - 3 B c 10 A 103 B d = 31.88 ksi
Since 25 ksi 6 scr 6 55 ksi, this result can be used to calculate the critical load.
Pcr = scr A = 31.88(5.5) = 175.31 kip = 175 kip
Ans.
1096
0.004
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s (MPa)
•13–73.
The stress-strain diagram of the material of a
column can be approximated as shown. Plot P兾A vs. KL兾r
for the column.
350
200
Tangent Moduli. From the stress - strain diagram,
(Et)1 =
(Et)2 =
200 A 106 B
0.001
0 … s 6 200MPa
= 200 GPa
(350 - 200) A 106 B
0.004 - 0.001
= 50 GPa
0
P (in./in.)
0.001
0.004
200MPa 6 s … 350 MPa
Critical Stress. Applying Engesser’s equation,
scr
P
=
A
p2Et
(1)
KL 2
a
b
r
If Et = (Et)1 = 200 GPa, Eq. (1) becomes
p2 C 200 A 109 B D
1.974 A 106 B
P
=
=
MPa
A
KL 2
KL 2
a
b
a
b
r
r
when scr =
P
= sY = 200 MPa, this equation becomes
A
200 A 106 B =
p2 C 200 A 109 B D
a
KL 2
b
r
KL
= 99.346 = 99.3
r
If Et = (Et)2 = 50 GPa, Eq. (1) becomes
P
=
A
p2 c 50 A 109 B d
0.4935 A 106 B
MPa
KL 2
KL 2
¢
≤
¢
≤
r
r
P
= sY = 200 MPa, this
when scr =
A
equation gives
200 A 106 B =
=
p2 C 50 A 109 B D
a
KL 2
b
r
KL
= 49.67 = 49.7
r
Using these results, the graphs of
KL
P
vs.
is shown in Fig. a can be plotted.
r
A
1097
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s (MPa)
13–74. Construct the buckling curve, P兾A versus L兾r,
for a column that has a bilinear stress–strain curve in
compression as shown. The column is pinned at its ends.
260
140
0.001
Tangent modulus: From the stress–strain diagram,
(Et)1 =
140(106)
= 140 GPa
0.001
(Et)2 =
(260 - 140)(106)
= 40 GPa
0.004 - 0.001
Critical Stress: Applying Engesser’s equation,
scr
p2Et
P
=
A
L 2
a b
r
[1]
Substituting (Et)1 = 140 GPa into Eq. [1], we have
p2 C 140(109) D
P
=
A
ALB2
r
P
=
A
When
1.38(106)
A Lr B 2
MPa
L
P
= 140 MPa, = 99.3
r
A
Substitute (Et)2 = 40 GPa into Eq. [1], we have
p2 C 40(109) D
P
=
A
ALB2
r
P
=
A
When
0.395(106)
A Lr B 2
MPa
L
P
= 140 MPa, = 53.1
r
A
1098
0.004
P (mm/mm)
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13–75. The stress-strain diagram for a material can be
approximated by the two line segments shown. If a bar
having a diameter of 80 mm and a length of 1.5 m is made
from this material, determine the critical load provided the
ends are pinned. Assume that the load acts through the axis
of the bar. Use Engesser’s equation.
s (MPa)
1100
200
0.001
E1 =
200 (106)
= 200 GPa
0.001
E2 =
1100 (106) - 200 (106)
= 150 GPa
0.007 - 0.001
Section properties:
I =
p 4
c;
4
r =
p 4
I
0.04
c
4 c
=
= =
= 0.02 m
AA
C p c2 2
2
A = pc2
Engesser’s equation:
1.0(1.5)
KL
=
= 75
r
0.02
scr =
p2 Et
A
B
KL 2
r
=
p2 Et
(75)2
= 1.7546(10 - 3) Et
Assume Et = E1 = 200 GPa
scr = 1.7546 (10 - 3)(200)(109) = 351 MPa 7 200 MPa
Therefore, inelastic buckling occurs:
Assume Et = E2 = 150 GPa
scr = 1.7546 (10 - 3)(150)(109) = 263.2 MPa
200 MPa 6 scr 6 1100 MPa
O.K.
Critical load:
Pcr = scr A = 263.2 (106)(p)(0.042) = 1323 kN‚
Ans
1099
0.007
P (mm/mm)
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*13–76. The stress-strain diagram for a material can be
approximated by the two line segments shown. If a bar
having a diameter of 80 mm and a length of 1.5 m is made
from this material, determine the critical load provided the
ends are fixed. Assume that the load acts through the axis of
the bar. Use Engesser’s equation.
s (MPa)
1100
200
0.001
E1 =
200 (106)
= 200 GPa
0.001
E2 =
1100 (106) - 200 (106)
= 150 GPa
0.007 - 0.001
Section properties:
I =
p 4
c;
4
r =
p 4
I
0.04
c
4 c
=
= =
= 0.02 m
AA
C p c2 2
2
A = pc2
Engesser’s equation:
0.5 (1.5)
KL
=
= 37.5
r
0.02
scr =
p2 Et
A
B
KL 2
r
=
p2 Et
(37.5)2
= 7.018385(10 - 3) Et
Assume Et = E1 = 200 GPa
scr = 7.018385 (10 - 3)(200)(109) = 1403.7 MPa 7 200 MPa
NG
Assume Et = E2 = 150 GPa
scr = 7.018385 (10 - 3)(150)(109) = 1052.8 MPa
200 MPa 6 scr 6 1100 MPa
O.K.
Critical load:
Pcr = scr A = 1052.8 (106)(p)(0.042) = 5292 kN
Ans.
1100
0.007
P (mm/mm)
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•13–77.
The stress-strain diagram for a material can be
approximated by the two line segments shown. If a bar
having a diameter of 80 mm and length of 1.5 m is made
from this material, determine the critical load provided one
end is pinned and the other is fixed. Assume that the load
acts through the axis of the bar. Use Engesser’s equation.
s (MPa)
1100
200
0.001
E1 =
200 (106)
= 200 GPa
0.001
E2 =
1100 (106) - 200 (106)
= 150 GPa
0.007 - 0.001
Section properties:
I =
p 4
c ;
4
r =
p 4
I
0.04
c
4 c
=
= =
= 0.02 m
AA
C pc2
2
2
A = pc2
Engesser’s equation:
0.7 (1.5)
KL
=
= 52.5
r
0.02
scr =
p2 Et
A
B
KL 2
r
=
p2Et
(52.5)2
= 3.58081 (10 - 3) Et
Assume Et = E1 = 200 GPa
scr = 3.58081 (10 - 3)(200)(109) = 716.2 MPa 7 200 MPa
NG
Assume Et = E2 = 150 GPa
scr = 3.58081 (10 - 3)(150)(109) = 537.1 MPa
200 MPa 6 scr 6 1100 MPa
O.K.
Critical load:
Pcr = scr A = 537.1 (106)(p)(0.042) = 2700 kN
Ans.
1101
0.007
P (mm/mm)
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13–78. Determine the largest length of a structural A-36
steel rod if it is fixed supported and subjected to an axial
load of 100 kN. The rod has a diameter of 50 mm. Use the
AISC equations.
Section Properties:
A = p A 0.0252 B = 0.625 A 10 - 3 B p m2
I =
p
A 0.0254 B = 97.65625 A 10 - 9 B p m4
4
r =
97.65625(10 - 9)p
I
=
= 0.0125 m
AA
A 0.625(10 - 3)p
Slenderness Ratio: For a column fixed at both ends, K = 0.5. Thus,
0.5L
KL
=
= 40.0L
r
0.0125
AISC Column Formula: Assume a long column.
sallow =
100(103)
0.625(10 - 3)p
=
12p2E
2
23 A KL
r B
12p2 C 200(109) D
23(40.0L)3
L = 3.555 m
KL
KL
2p2E
= 40.0(3.555) = 142.2 and for A–36 steel, a
b =
r
r e A sg
KL
KL
2p2[200(109)]
=
= 125.7. Since a
b …
… 200, the assumption is correct.
r e
r
A 250(106)
Here,
Thus,
L = 3.56 m
Ans.
1102
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13–79. Determine the largest length of a W10 * 45
structural steel column if it is pin supported and subjected
to an axial load of 290 kip. Est = 29(103) ksi, sY = 50 ksi.
Use the AISC equations.
Section Properties: For a W10 * 45 wide flange section,
A = 13.3 in2
ry = 2.01 in
Slenderness Ratio: For a column pinned at both ends, K = 1. Thus,
a
1(L)
KL
b =
= 0.49751L
r y
2.01
AISC Column Formula: Assume a long column,
sallow =
12p2E
2
23 A KL
r B
12p2 C 29(103) D
290
=
13.3
23(0.49751L)2
L = 166.3 in.
KL
KL
2p2E
= 0.49751 (166.3) = 82.76 and for grade 50 steel, a
b =
r
r c A sg
KL
KL
2p2[29(103)]
6 a
b , the assumption is not correct.
=
= 107.0. Since
r
r c
50
A
Thus, the column is an intermediate column.
Here,
Applying Eq. 13–23,
B1 sallow =
(KL>r)2
2(KL>r)2c
R sg
3(KL>r)
(KL>r)3
5
+
3
8(KL>r)c
8(KL>r)3c
B1 -
(0.49751L)2
R (50)
2(107.02)
290
=
13.3
3(0.49751L)
(0.49751L)3
5
+
3
8(107.0)
8(107.03)
0 = 12.565658 A 10 - 9 B L3 - 24.788132 A 10 - 6 B L2 - 1.743638 A 10 - 3 B L + 0.626437
Solving by trial and error,
L = 131.12 in. = 10.9 ft
Ans.
1103
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*13–80. Determine the largest length of a W10 * 12
structural A-36 steel section if it is pin supported and is
subjected to an axial load of 28 kip. Use the AISC equations.
For a W 10 * 12,
A = 3.54 in2
ry = 0.785 in.
s =
28
P
=
= 7.91 ksi
A
3.54
Assume a long column:
sallow =
a
a
12p2E
23(KL>r)2
KL
12p2(29)(103)
12p2E
=
= 137.4
b =
r
A 23(7.91)
A 23sallow
KL
2p2(29)(103)
2p2E
=
= 126.1,
b =
r c A sg
A
36
KL
KL
7 a
b
r
r c
Long column.
KL
= 137.4
r
L = 137.4 ¢
r
0.785
≤ = 137.4 ¢
≤ = 107.86 in.
K
1
= 8.99 ft
Ans.
•13–81.
Using the AISC equations, select from Appendix B
the lightest-weight structural A-36 steel column that is 14 ft
long and supports an axial load of 40 kip.The ends are pinned.
Take sY = 50 ksi.
Try, W6 * 15 (A = 4.43 in2
ry = 1.46 in.)
a
KL
2p2(29)(103)
2p2E
b =
=
= 107
r c A sY
50
A
a
(1.0)(14)(12)
KL
b =
= 115.1,
ry
1.46
a
KL
KL
b 7 a
b
ry
r c
Long column
sallow =
12p2(29)(103)
12 p2E
=
= 11.28 ksi
2
23(KL>r)
23(115.1)2
Pallow = sallowA
= 11.28(4.43) = 50.0 kip 7 40 kip
O.K.
Use W6 * 15
Ans.
1104
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13–82. Using the AISC equations, select from Appendix B
the lightest-weight structural A-36 steel column that is 12 ft
long and supports an axial load of 40 kip. The ends are fixed.
Take sY = 50 ksi.
A = 2.68 in2
Try W6 * 9
a
ry = 0.905 in.
KL
2p2(29)(103)
2p2E
b =
=
= 107
r c A sY
A
50
0.5(12)(12)
KL
=
= 79.56
ry
0.905
KL
KL
6 a
b
ry
r c
Intermediate column
sallow =
KL>r 2
C 1 - 12 A (KL>r)c
B D sg
KL>r
KL>r 3
C 53 + 38 A (KL>r)c
B - 18 A (KL>r)c
B D
=
2
C 1 - 12 A 79.56
126.1 B D 36 ksi
1 79.56 3
C 53 + 38 A 79.56
126.1 B - 8 A 12.61 B D
= 15.40 ksi
Pallow = sallowA
= 15.40(2.68)
= 41.3 kip 7 40 kip
O.K.
Use W6 * 9
Ans.
1105
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13–83. Using the AISC equations, select from Appendix B
the lightest-weight structural A-36 steel column that is 24 ft
long and supports an axial load of 100 kip.The ends are fixed.
Section Properties: Try a W8 * 24 wide flange section,
A = 7.08 in2
ry = 1.61 in
Slenderness Ratio: For a column fixed at both ends, K = 0.5. Thus,
a
AISC
Column
0.5(24)(12)
KL
b =
= 89.44
r y
1.61
Formula:
For
A–36
steel,
a
KL
2p2E
b =
r c A sg
KL
KL
2p2[29(103)]
6 a
b , the column is an intermediate
= 126.1. Since
r
r c
A
36
column. Applying Eq. 13–23,
=
(KL>r)2
B1 sallow =
R sg
3(KL>r)
(KL>r)3
5
+
3
8(KL>r)c
8(KL>r)3c
B1 =
2(KL>r)2c
(89.442)
2(126.12)
R (36)
3(89.44)
(89.443)
5
+
3
8(126.1)
8(126.13)
= 14.271 ksi
The allowable load is
Pallow = sallowA
= 14.271(7.08)
= 101 kip 7 P = 100 kip
O.K.
W8 * 24
Ans.
Thus, Use
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*13–84. Using the AISC equations, select from Appendix B
the lightest-weight structural A-36 steel column that is 30 ft
long and supports an axial load of 200 kip.The ends are fixed.
a
A = 14.1 in2
ry = 2.08 in.
Try W8 * 48
KL
2 p2 (29)(103)
2 p2E
b =
=
= 126.1
r c A sg
A
36
0.5 (30)(12)
KL
=
= 86.54
ry
2.08
a
KL
KL
b 6 a
b intermediate column.
ry
r c
b 1 - 12 B
sallow =
b 53 + 38 B
KL
r
A KL
r Bc
e1 =
e 53
+
3
8
C
1
2
KL
r
A KL
r Bc
2
R r sg
R - 18 B
KL
r
A KL
r Bc
2
C 86.54
126.1 D f36
86.54
126.1
D - C
D
1 86.54 3
f
8 126.1
3
R r
= 14.611 ksi
Pallow = sallow A = 14.611 (14.1) = 206 kip 7 P = 200 kip
Use
O.K.
W 8 * 48
Ans.
1107
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A W8 * 24 A-36-steel column of 30-ft length is
pinned at both ends and braced against its weak axis at midheight. Determine the allowable axial force P that can be
safely supported by the column. Use the AISC column
design formulas.
•13–85.
Section Properties. From the table listed in the appendix, the necessary section
properties for a W8 * 24 are
A = 7.08 in2
rx = 3.42 in.
ry = 1.61 in.
Slenderness Ratio. For a column pinned at both of its ends, K = 1. Here,
Lx = 30(12) = 360 in. and Ly = 15(12) = 180 in. Thus,
¢
1(360)
KL
= 105.26
≤ =
r x
3.42
¢
1(180)
KL
= 111.80 (controls)
≤ =
r y
1.61
AISC
=
C
Column
2p2 C 29 A 103 B D
36
Formulas.
For
= 126.10. Since
¢
A-36
steel
¢
KL
KL
≤ 6 ¢
≤ , the
r y
r c
KL
2p2E
≤ =
r c A sY
column
is
an
intermediate column.
(KL>r)2
B1 sallow =
R sY
3(KL>r)
(KL>r)3
5
+
3
8(KL>r)c
8(KL>r)c 3
C1 -
=
2(KL>r)c 2
111.802
2 A 126.102 B
S(36)
3(111.80)
5
111.803
+
3
8(126.10)
8 A 126.103 B
= 11.428 ksi
Thus, the allowable force is
Pallow = sallowA = 11.428(7.08) = 80.91 kip = 80.9 kip
1108
Ans.
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13–86. Check if a W10 * 39 column can safely support an
axial force of P = 250 kip. The column is 20 ft long and is
pinned at both ends and braced against its weak axis at
mid-height. It is made of steel having E = 29(103) ksi and
sY = 50 ksi. Use the AISC column design formulas.
Section Properties. From the table listed in the appendix, the necessary section
properties for a W10 * 39 are
A = 11.5 in2
rx = 4.27 in.
ry = 1.98 in.
Slenderness Ratio. For a column pinned at both of its ends, K = 1. Here,
Lx = 20(12) = 240 in. and Ly = 10(12) = 120 in. Thus,
¢
1(240)
KL
= 56.21
≤ =
r x
4.27
¢
1(120)
KL
= 60.606 (controls)
≤ =
r y
1.98
AISC
=
Column
2p2 c 29 A 103 B d
S
50
Formulas.
For
A-36
= 107.00 . Since ¢
steel
¢
KL
2p2E
≤ =
r c A sY
KL
KL
≤ 6 ¢
≤ , the column is an
r y
r c
intermediate column.
B1 sallow =
2(KL>r)c 2
R sY
3(KL>r)
(KL>r)3
5
+
3
8(KL>r)c
8(KL>r)c 3
C1 -
=
(KL>r)2
60.6062
2 A 107.002 B
S(50)
3(60.606)
5
60.6063
+
3
8(107.00)
8 A 107.003 B
= 22.614 ksi
Thus, the allowable force is
Pallow = sallowA = 22.614(11.5) = 260.06 kip 7 P = 250 kip
Thus, a W10 * 39 column is adequate.
1109
O.K.
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13–87. A 5-ft-long rod is used in a machine to transmit
an axial compressive load of 3 kip. Determine its smallest
diameter if it is pin connected at its ends and is made of a
2014-T6 aluminum alloy.
Section properties:
A =
p 2
d ;
4
I =
p d 4
pd4
a b =
4 2
64
pd4
r =
I
d
64
=
=
AA
C p4 d2
4
sallow =
P
=
A
p
4
3
3.820
=
d2
d2
Assume long column:
1.0 (5)(12)
240
KL
=
=
d
r
d
4
sallow =
54 000
A
B
KL 2
r
;
3.820
54000
=
d2
C 240 D 2
d
d = 1.42 in.
Ans.
KL
240
=
= 169 7 55
r
1.42
O.K.
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*13–88. Check if a W10 * 45 column can safely support
an axial force of P = 200 kip. The column is 15 ft long and
is pinned at both of its ends. It is made of steel having
E = 29(103) ksi and sY = 50 ksi. Use the AISC column
design formulas.
Section Properties. Try W10 * 45. From the table listed in the appendix, the
necessary section properties are
A = 13.3 in2
ry = 2.01 in.
Slenderness Ratio. For a column pinned at both of its ends, K = 1. Thus,
a
1(15)(12)
KL
b =
= 89.552
r y
2.01
KL
2p2E
AISC Column Formulas. Here, a
b =
=
r c A sY
S
KL
KL
Since a
b 6 a
b , the
r y
r c
2p2 c29 A 103 B d
50
= 107.00.
column is an intermediate column.
(KL>r)2
B1 sallow =
R sY
3(KL>r)
(KL>r)3
5
+
3
8(KL>r)c
8(KL>r)c 3
C1 -
=
2(KL>r)c 2
89.5522
2 A 107.002 B
S(50)
3(89.552)
5
89.5523
+
3
8(107.00)
8 A 107.003 B
= 17.034 ksi
Thus, the allowable force is
Pallow = sallowA = 17.034(13.3) = 226.55 kip 7 P = 200 kip
O.K.
Thus,
A W10 * 45 can be used
Ans.
1111
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•13–89.
Using the AISC equations, check if a column
having the cross section shown can support an axial force of
1500 kN. The column has a length of 4 m, is made from A-36
steel, and its ends are pinned.
20 mm
350 mm
300 mm
10 mm
Section Properties:
A = 0.3(0.35) - 0.29(0.31) = 0.0151 m2
Iy =
1
1
(0.04) A 0.33 B +
(0.31) A 0.013 B = 90.025833 A 10 - 6 B m4
12
12
ry =
Iy
90.02583(10 - 6)
=
= 0.077214 m
AA
A
0.0151
Slenderness Ratio: For a column pinned at both ends, K = 1. Thus,
a
AISC
=
Column
2p2[200(109)]
1(4)
KL
b =
= 51.80
r y
0.077214
Formula:
= 125.7. Since
A 250(10 )
column. Applying Eq. 13–23,
6
For
a
KL
2p2E
b =
r c A sg
(KL>r)2
2(KL>r)2c
R sg
3(KL>r)
(KL>r)3
5
+
3
8(KL>r)c
8(KL>r)3c
B1 =
steel,
KL
KL
6 a
b , the column is an intermediate
r
r c
B1 sallow =
A–36
(51.802)
2(125.72)
20 mm
R (250)(106)
3(51.80)
(51.803)
5
+
3
8(125.7)
8(125.73)
= 126.2 MPa
The allowable load is
Pallow = sallowA
= 126.2 A 106 B (0.0151)
= 1906 kN 7 P = 1500 kN
O.K.
Thus, the column is adequate.
Ans.
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13–90. The A-36-steel tube is pinned at both ends. If it
is subjected to an axial force of 150 kN, determine the
maximum length that the tube can safely support using the
AISC column design formulas.
100 mm
80 mm
Section Properties.
A = p A 0.052 - 0.042 B = 0.9 A 10 - 3 B p m2
I =
r =
p
A 0.054 - 0.044 B = 0.9225 A 10 - 6 B p m4
4
0.9225 A 10 - 6 B p
I
=
= 0.03202 m
AA
C 0.9 A 10 - 3 B p
Slenderness Ratio. For a column pinned at both of its ends, K = 1. Thus,
1(L)
KL
=
= 31.23L
r
0.03202
AISC Column Formulas.
sallow =
12p2E
23(KL>r)2
150 A 103 B
.9 A 10 - 3 B p
=
12p2 C 200 A 109 B D
23(31.23L)2
L = 4.4607 m = 4.46 m
Here,
KL
= 31.23(4.4607) = 139.33.
r
2p2 C 200 A 109 B D
= 125.66. Since a
250 A 10 B
long column is correct.
=
C
6
Ans.
For
A-36
steel
a
KL
2p2E
b =
r c A sY
KL
KL
b 6
6 200, the assumption of a
r c
r
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13–91. The bar is made of a 2014-T6 aluminum alloy.
Determine its smallest thickness b if its width is 5b. Assume
that it is pin connected at its ends.
600 lb
b
5b
8 ft
Section Properties:
A = b(5b) = 5b2
Iy =
1
5 4
(5b) A b3 B =
b
12
12
ry =
5 4
Iy
23
12 b
=
=
b
AA
C 5b2
6
600 lb
Slenderness Ratio: For a column pinned at both ends, K = 1. Thus,
a
1(8)(12)
332.55
KL
=
b =
23
b
r y
6 b
Aluminum (2014 - T6 alloy) Column Formulas: Assume a long column and apply
Eq. 13–26.
sallow =
54 000
(KL>r)2
0.600
54 000
=
2
5b
A 332.55 B 2
b
b = 0.7041 in.
Here,
KL
KL
332.55
=
7 55, the assumption is correct. Thus,
= 472.3. Since
r
r
0.7041
b = 0.704 in.
Ans.
1114
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*13–92. The bar is made of a 2014-T6 aluminum alloy.
Determine its smallest thickness b if its width is 5b. Assume
that it is fixed connected at its ends.
600 lb
b
5b
8 ft
Section Properties:
A = b(5b) = 5b2
Iy =
1
5 4
(5b) A b3 B =
b
12
12
ry =
Iy
23
12 b
=
=
b
AA
C 5b2
6
5
600 lb
4
Slenderness Ratio: For a column fixed at both ends, K = 0.5. Thus,
a
0.5(8)(12)
166.28
KL
=
b =
r y
b
23
6 b
Aluminum (2014 - T6 alloy) Column Formulas: Assume a long column and apply
Eq. 13–26.
sallow =
54 000
(KL>r)2
0.600
54 000
=
5b2
A 166.28 B 2
b
b = 0.4979 in.
Here,
KL
KL
166.28
= 334.0. Since
=
7 55, the assumption is correct.
r
r
0.4979
Thus,
b = 0.498 in.
Ans.
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•13–93.
The 2014-T6 aluminum column of 3-m length has
the cross section shown. If the column is pinned at both
ends and braced against the weak axis at its mid-height,
determine the allowable axial force P that can be safely
supported by the column.
15 mm
170 mm
15 mm
15 mm
100 mm
Section Properties.
A = 0.1(0.2) - 0.085(0.17) = 5.55 A 10 - 3 B m2
Ix =
1
1
(0.1) A 0.23 B (0.085) A 0.173 B = 31.86625 A 10 - 6 B m4
12
12
Iy = 2 c
rx =
ry =
1
1
(0.015) A 0.13 B d +
(0.17) A 0.0153 B = 2.5478 A 10 - 6 B m4
12
12
31.86625 A 10 - 6 B
Ix
=
= 0.07577
AA
C 5.55 A 10 - 3 B
2.5478 A 10 - 6 B
Iy
=
= 0.02143 m
AA
C 5.55 A 10 - 3 B
Slenderness Ratio. For a column pinned at both of its ends, K = 1. Here, Lx = 3 m
and Ly = 1.5 m. Thus,
a
(1)(3)
KL
b =
= 39.592
r x
0.07577
a
(1)(1.5)
KL
b =
= 70.009 (controls)
r y
0.02143
2014-T6 Alumimum Alloy Column Formulas. Since a
KL
b 7 55, the column can
r y
be classified a long column,
sallow = D
373 A 103 B
T Mpa
= C
373 A 103 B
S MPa
a
KL 2
b
r
70.0092
= 76.103 MPa
Thus, the allowed force is
Pallow = sallowA = 76.103 A 106 B C 5.55 A 10 - 3 B D = 422.37 kN = 422 kN
1116
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13–94. The 2014-T6 aluminum column has the cross
section shown. If the column is pinned at both ends and
subjected to an axial force P = 100 kN, determine the
maximum length the column can have to safely support the
loading.
15 mm
170 mm
15 mm
15 mm
100 mm
Section Properties.
A = 0.1(0.2) - 0.085(0.17) = 5.55 A 10 - 3 B m2
Iy = 2 c
ry =
1
1
(0.015) A 0.13 B d +
(0.17) A 0.0153 B = 2.5478 A 10 - 6 B m4
12
12
2.5478 A 10 - 6 B
Iy
=
= 0.02143 m
AA
C 5.55 A 10 - 3 B
Slenderness Ratio. For a column pinned at both of its ends, K = 1. Then,
a
1(L)
KL
b =
= 46.6727L
r y
0.02143
2014-T6 Alumimum Alloy Column Formulas. Assuming a long column,
sallow = D
100 A 103 B
373 A 103 B
5.55 A 10 - 3 B
a
KL 2
b
r
= C
T MPa
373 A 103 B
(46.672L)2
S A 106 B Pa
L = 3.083 m = 3.08 m
Since a
Ans.
KL
b = 46.6727(3.083) = 143.88 7 55, the assumption is correct.
r y
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13–95. The 2014-T6 aluminum hollow section has the
cross section shown. If the column is 10 ft long and is fixed
at both ends, determine the allowable axial force P that can
be safely supported by the column.
4 in.
3 in.
Section Properties.
A = p A 22 - 1.52 B = 1.75p in2
r =
I =
p 4
A 2 - 1.54 B = 2.734375p in4
4
I
2.734375p
=
= 1.25 in.
AA
A 1.75p
Slenderness Ratio. For a column fixed at both of its ends, K = 0.5. Thus,
0.5(10)(12)
KL
=
= 48
r
1.25
2014-T6 Aluminum Alloy Column Formulas. Since 12 6
KL
6 55, the column can
r
be classified as an intermediate column.
sallow = c 30.7 - 0.23a
KL
b d ksi
r
= [30.7 - 0.23(48)] ksi
= 19.66 ksi
Thus, the allowable load is
Pallow = sallowA = 19.66 A 106 B (1.75p) = 108.09 kip = 108 kip
1118
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*13–96. The 2014-T6 aluminum hollow section has the
cross section shown. If the column is fixed at its base and
pinned at its top, and is subjected to the axial force
P = 100 kip, determine the maximum length of the column
for it to safely support the load.
4 in.
3 in.
Section Properties.
A = p A 22 - 1.52 B = 1.75p in2
r =
I =
p 4
A 2 - 1.54 B = 2.734375p in4
4
I
2.734375p
=
= 1.25 in.
AA
A 1.75p
Slenderness Ratio. For a column fixed at its base and pinned at its top, K = 0.7.
Thus,
0.7(L)
KL
=
= 0.56L
r
1.25
2014-T6 Aluminum Alloy Column Formulas. Assuming an intermediate column,
sallow = c 30.7 - 0.23a
KL
b d ksi
r
100
= 30.7 - 0.23(0.56L)
1.75p
L = 97.13 in. = 8.09 ft
Ans.
KL
= 0.56(97.13) = 54.39 6 55, the assumption of an intermediate column
r
is correct.
Since
1119
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•13–97.
The tube is 0.25 in. thick, is made of a 2014-T6
aluminum alloy, and is fixed at its bottom and pinned at its
top. Determine the largest axial load that it can support.
P
x
y
6 in.
x
6 in.
y
10 ft
Section Properties:
P
A = 6(6) - 5.5(5.5) = 5.75 in2
I =
1
1
(6) A 63 B (5.5) A 5.53 B = 31.7448 in4
12
12
r =
I
31.7448
=
= 2.3496 in.
A 5.75
AA
Slenderness Ratio: For a column fixed at one end and pinned at the other end,
K = 0.7. Thus,
0.7(10)(12)
KL
=
= 35.75
r
2.3496
Aluminium (2014 –∑ T6 alloy) Column Formulas: Since 12 6
KL
6 55, the
r
column is classified as an intermediate column. Applying Eq. 13–25,
sallow = c 30.7 - 0.23a
KL
b d ksi
r
= [30.7 - 0.23(33.75)]
= 24.48 ksi
The allowable load is
Pallow = sallowA = 22.48(5.75) = 129 kip
Ans.
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13–98. The tube is 0.25 in. thick, is made of a 2014-T6
aluminum alloy, and is fixed connected at its ends.
Determine the largest axial load that it can support.
P
x
y
6 in.
x
6 in.
y
10 ft
Section Properties:
P
A = 6(6) - 5.5(5.5) = 5.75 in2
I =
1
1
(6) A 63 B (5.5) A 5.53 B = 31.7448 in4
12
12
r =
I
31.7448
=
= 2.3496 in.
AA
A 5.75
Slenderness Ratio: For column fixed at both ends, K = 0.5. Thus,
0.5(10)(12)
KL
=
= 25.54
r
2.3496
Aluminium (2014 – T6 alloy) Column Formulas: Since 12 6
KL
6 55, the
r
column is classified as an intermediate column. Applying Eq. 13–25,
sallow = c 30.7 - 0.23 a
KL
b d ksi
r
= [30.7 - 0.23(25.54)]
= 24.83 ksi
The allowable load is
Pallow = sallowA = 24.83(5.75) = 143 kip
Ans.
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13–99. The tube is 0.25 in. thick, is made of 2014-T6
aluminum alloy and is pin connected at its ends. Determine
the largest axial load it can support.
P
x
y
6 in.
x
6 in.
y
Section Properties:
A = 6(6) - 5.5(5.5) = 5.75 in2
I =
1
1
(6) A 63 B (5.5) A 5.53 B = 31.7448 in4
12
12
r =
I
31.7448
=
= 2.3496 in.
AA
A 5.75
10 ft
P
Slenderness Ratio: For a column pinned as both ends, K = 1. Thus,
1(10)(12)
KL
=
= 51.07
r
2.3496
Aluminum (2014 – T6 alloy) Column Formulas: Since 12 6
KL
6 55, the
r
column is classified as an intermediate column. Applying Eq. 13–25,
sallow = c 30.7 - 0.23 a
KL
b d ksi
r
= [30.7 - 0.23(51.07)]
= 18.95 ksi
The allowable load is
Pallow = sallowA = 18.95(5.75) = 109 kip
Ans.
*13–100. A rectangular wooden column has the cross
section shown. If the column is 6 ft long and subjected to an
axial force of P = 15 kip, determine the required minimum
1
dimension a of its cross-sectional area to the nearest 16
in.
so that the column can safely support the loading. The
column is pinned at both ends.
a
2a
Slenderness Ratio. For a column pinned at both of its ends, K = 1. Then,
(1)(6)(12)
KL
72
=
=
a
a
d
NFPA Timber Column Formula. Assuming an intermediate column,
sallow = 1.20 c 1 -
1 KL>d 2
a
b d ksi
3 26.0
15
1 72>a 2
= 1.20 c 1 - a
b d
2a(a)
3 26.0
a = 2.968 in.
Use a = 3 in.
Ans.
KL
72
KL
=
= 24. Since 11 6
6 26, the assumption is correct.
d
3
d
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•13–101.
A rectangular wooden column has the cross
section shown. If a = 3 in. and the column is 12 ft long,
determine the allowable axial force P that can be safely
supported by the column if it is pinned at its top and fixed at
its base.
a
2a
Slenderness Ratio. For a column fixed at its base and pinned at its top K = 0.7.
Then,
0.7(12)(12)
KL
=
= 33.6
d
3
NFPA Timer Column Formula. Since 26 6
KL
6 50, the column can be classified
d
as a long column.
sallow =
540 ksi
540
=
= 0.4783 ksi
2
(KL>d)
33.62
The allowable force is
Pallow = sallowA = 0.4783(3)(6) = 8.61 kip
Ans.
13–102. A rectangular wooden column has the cross
section shown. If a = 3 in. and the column is subjected to
an axial force of P = 15 kip, determine the maximum
length the column can have to safely support the load. The
column is pinned at its top and fixed at its base.
a
2a
Slenderness Ratio. For a column fixed at its base and pinned at its top, K = 0.7.
Then,
KL
0.7L
=
= 0.2333L
d
3
NFPA Timber Column Formula. Assuming an intermediate column,
sallow = 1.20 c 1 -
1 KL>d 2
a
b d ksi
3 26.0
15
1 0.2333L 2
= 1.20 c 1 - a
b d
3(6)
3
26.0
L = 106.68 in. = 8.89 ft
Ans.
KL
KL
= 0.2333(106.68) = 24.89. Since 11 6
6 26, the assumption is
d
d
correct.
Here,
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13–103. The timber column has a square cross section and
is assumed to be pin connected at its top and bottom. If it
supports an axial load of 50 kip, determine its smallest side
dimension a to the nearest 12 in. Use the NFPA formulas.
14 ft
a
Section properties:
A = a2
sallow = s =
50
P
= 2
A
a
Assume long column:
sallow =
50
=
a2
C
540
2
A KL
d B
540
(1.0)(14)(12)
a
D
2
a = 7.15 in.
(1.0)(14)(12)
KL
KL
=
= 23.5,
6 26
d
7.15
d
Assumption NG
Assume intermediate column:
sallow = 1.20 B 1 -
1 KL>d 2
a
b R
3 26.0
2
50
1
a
b R
= 1.20 B 1 - a
2
a
26.0
3
1.0(14)(12)
a = 7.46 in.
1.0(14)(12)
KL
KL
=
= 22.53, 11 6
6 26
d
7.46
d
Assumption O.K.
1
Use a = 7 in.
2
Ans.
1124
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*13–104. The wooden column shown is formed by gluing
together the 6 in. * 0.5 in. boards. If the column is pinned
at both ends and is subjected to an axial load P = 20 kip,
determine the required number of boards needed to form
the column in order to safely support the loading.
P
6 in.
0.5 in.
9 ft
Slenderness Ratio. For a column pinned at both of its ends, K = 1. If the number of
the boards required is n and assuming that n(0.5) 6 6 in. Then, d = n(0.5). Thus,
(1)(9)(12)
KL
216
=
=
n
d
n(0.5)
P
NFPA Timber Column Formula. Assuming an intermediate column,
sallow = 1.20 B 1 -
1 KL>d 2
a
b R ksi
3 26.0
1 216>n 2
20
= 1.20 B 1 - a
b R
[n(0.5)](6)
3 26.0
n2 - 5.5556n - 23.01 = 0
Solving for the positive root,
n = 8.32
Use n = 9
Ans.
KL
KL
216
=
= 24. Since n(0.5) = 9(0.5) = 4.5 in. 6 6 in. and 11 6
6 26,
d
9
d
the assumptions made are correct.
Here,
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•13–105.
The column is made of wood. It is fixed at its
bottom and free at its top. Use the NFPA formulas to
determine its greatest allowable length if it supports an
axial load of P = 2 kip.
P
2 in.
y
y
x
x
4 in.
Slenderness Ratio: For a column fixed at one end and free at the other end, K = 2.
Thus,
L
2(L)
KL
=
= 1.00L
d
2
NFPA Timber Column Formulas: Assume a long column. Apply Eq. 13–29,
sallow =
540
ksi
(KL>d)2
2
540
=
2(4)
(1.00L)2
L = 46.48 in
Here,
KL
KL
= 1.00(46.48) = 46.48. Since 26 6
6 50, the assumption is correct.
d
d
Thus,
L = 46.48 in. = 3.87 ft
Ans.
13–106. The column is made of wood. It is fixed at its
bottom and free at its top. Use the NFPA formulas to
determine the largest allowable axial load P that it can
support if it has a length L = 4 ft.
P
2 in.
y
x
y
x
4 in.
Slenderness Ratio: For a column fixed at one end and free at the other end, K = 2.
Thus,
L
2(4)(12)
KL
=
= 48.0
d
2
NFPA Timber Column Formulas: Since 26 6
KL
6 50, it is a long column. Apply
d
Eq. 13–29,
sallow =
=
540
ksi
(KL>d)2
540
48.02
= 0.234375 ksi
The allowable axial force is
Pallow = sallowA = 0.234375[2(4)] = 1.875 kip
Ans.
1126
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13–107. The W14 * 53 structural A-36 steel column
supports an axial load of 80 kip in addition to an eccentric
load P. Determine the maximum allowable value of P based
on the AISC equations of Sec. 13.6 and Eq. 13–30. Assume
the column is fixed at its base, and at its top it is free to sway
in the x–z plane while it is pinned in the y–z plane.
z
80 kip
x
y
12 ft
Section Properties: For a W14 * 53 wide flange section.
A = 15.6 in2
Ix = 541 in4
d = 13.92 in.
rx = 5.89 in.
ry = 1.92 in.
Slenderness Ratio: By observation, the largest slenderness ratio is about y -y axis.
For a column fixed at one end and free at the other end, K = 2. Thus,
a
2(12)(12)
KL
b =
= 150
r y
1.92
Allowable Stress: The allowable stress can be determined using AISC Column
2p2[29(103)]
2p2E
KL
b =
=
= 126.1. Since
Formulas. For A–36 steel, a
r c
B sY
B
36
KL
KL
b …
… 200, the column is a long column. Applying Eq. 13–21,
a
r c
r
sallow =
12p2E
23(KL>r)2
12p2(29.0)(103)
=
23(1502)
= 6.637 ksi
Maximum Stress: Bending is about x - x axis. Applying we have
smax = sallow =
6.637 =
Mc
P
+
A
I
P(10) A 13.92
P + 80
2 B
+
15.6
541
P = 7.83 kip
Ans.
1127
P
y
x 10 in.
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*13–108. The W12 * 45 structural A-36 steel column
supports an axial load of 80 kip in addition to an eccentric
load of P = 60 kip. Determine if the column fails based on
the AISC equations of Sec. 13.6 and Eq. 13–30. Assume that
the column is fixed at its base, and at its top it is free to sway
in the x–z plane while it is pinned in the y–z plane.
z
80 kip
x
y
12 ft
Section Properties: For a W12 * 45 wide flange section,
A = 13.2 in2
d2 = 12.06 in.
Ix = 350 in4
rx = 5.15 in.
ry = 1.94 in.
Slenderness Ratio: By observation, the largest slenderness ratio is about y- y axis.
For a column fixed at one end and free at the other end, K = 2. Thus,
a
2(12)(12)
KL
b =
= 148.45
r y
1.94
Allowable Stress: The allowable stress can be determined using AISC Column
2p2[29(103)]
2p2E
KL
b =
=
= 126.1. Since
Formulas. For A–36 steel, a
r c
B sY
B
36
KL
KL
b …
… 200, the column is a long column. Applying Eq. 13–21,
a
r c
r
sallow =
12p2E
23(KL>r)2
12p2(29.0)(103)
=
23(148.452)
= 6.776 ksi
Maximum Stress: Bending is about x - x axis. Applying Eq. 1 we have
smax =
=
Mc
P
+
A
I
60(10) A 12.06
140
2 B
+
13.2
350
= 20.94 ksi
Since smax 7 sallow, the column is not adequate.
1128
P
y
x 10 in.
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•13–109.
The W14 * 22 structural A-36 steel column is
fixed at its top and bottom. If a horizontal load (not shown)
causes it to support end moments of M = 10 kip # ft,
determine the maximum allowable axial force P that can be
applied. Bending is about the x–x axis. Use the AISC
equations of Sec. 13.6 and Eq. 13–30.
P
x
M
y
y
x
12 ft
Section properties for W14 * 22:
A = 6.49 in2
d = 13.74 in2
Ix = 199 in4
ry = 1.04 in.
M
Allowable stress method:
P
0.5(12)(12)
KL
=
= 69.231
ry
1.04
a
KL
KL
2p2E
KL
2p2(29)(103)
b =
=
= 126.1,
6 a
b
r c
r
r c
B sY
B
36
y
Hence,
B1 (sa)allow =
B 53
+
3
8
smax = (sa)allow =
16.510 =
1
2
¢
KL
r
2
A KL
r B
2
A KL
r Bc
A KL
r Bc
-
c1 -
≤ R sY
1
8
3
A KL
r B
3
A KL
r Bc
=
R
c 53 +
3
8
1
2
2
A 69.231
126.1 B d36
1 69.231 3
A 69.231
126.1 B - 8 A 126.1 B d
= 16.510 ksi
My c
P
+
A
Ix
10(12)(13.74
P
2 )
+
6.49
199
P = 80.3 kip
Ans.
1129
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13–110. The W14 * 22 column is fixed at its top and
bottom. If a horizontal load (not shown) causes it to support
end moments of M = 15 kip # ft, determine the maximum
allowable axial force P that can be applied. Bending is
about the x–x axis. Use the interaction formula with
1sb2allow = 24 ksi.
P
x
M
y
y
x
12 ft
Section Properties for W 14 * 22:
A = 6.49 in2
d = 13.74 in2
Ix = 199 in4
ry = 1.04 in.
M
P
Interaction method:
0.5(12)(12)
KL
=
= 69.231
ry
1.04
a
KL
2p2E
KL
KL
2p2(29)(103)
b =
=
= 126.1,
6 a
b
r c
ry
r c
B sY
B
36
Hence,
B1 (sa)allow =
B 53
sa =
+
3
8
1
2
¢
KL
r
2
A KL
r B
2
A KL
r Bc
A KL
r Bc
-
c1 -
≤ R sY
1
8
P
P
=
= 0.15408 P
A
6.49
3
A KL
r B
3
A KL
r Bc
=
R
c 53 +
3
8
A
1
2
2
A 69.231
126.1 B d36
69.231
126.1
B
-
1
8
A
B
69.231 3
d
126.1
= 16.510 ksi
15(12) A 13.74
Mxc
2 B
=
= 6.214 ksi
sb =
Ix
199
sb
sa
+
= 1.0
(sa)allow
(sb)allow
6.2141
0.15408 P
+
= 1.0
16.510
24
P = 79.4 kip
Ans.
1130
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13–111. The W14 * 43 structural A-36 steel column
is fixed at its bottom and free at its top. Determine the
greatest eccentric load P that can be applied using
Eq. 13–30 and the AISC equations of Sec. 13.6.
40 kip
16 in.
10 ft
Section properties for W14 * 43:
A = 12.6 in2
d = 13.66 in.
Iy = 45.2 in4
ry = 1.89 in.
b = 7.995
Allowable stress method:
2(10)(12)
KL
=
= 126.98
ry
1.89
a
2p2E
KL
KL
KL
2p2 (29)(103)
b =
=
= 126.1, 200 7
7 a
b
r c
ry
r c
B sY
B
36
(sa)allow =
12p2(29)(103)
12p2E
=
= 9.26 ksi
2
23(KL>r)
23(126.98)2
smax = (sa)allow =
P
My c
P
+
A
Iy
P(16) A 7.995
P + 40
A B
9.26 =
+
12.6
45.2
P = 4.07 kip
Ans.
1131
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*13–112. The W10 * 45 structural A-36 steel column is
fixed at its bottom and free at its top. If it is subjected to a
load of P = 2 kip, determine if it is safe based on the AISC
equations of Sec. 13.6 and Eq. 13–30.
40 kip
16 in.
10 ft
Section Properties for W10 * 45:
A = 13.3 in2
d = 10.10 in.
Iy = 53.4 in4
ry = 2.01 in.
b = 8.020 in.
Allowable stress method:
2.0(10)(12)
KL
=
= 119.4
ry
2.01
a
KL
2p2E
2p2(29)(103)
b =
=
= 126.1
r c
B sY
B
36
KL
KL
6 a
b
r
r c
(sa)allow Ú
(sa)allow =
10.37 Ú
c1 5
3
+
A
1 (KL>r)
2 (KL>r)3 d sY
3 KL>r
8 KL>rc
My c
P
+
A
Iy
2
C 1 - 12 A 119.4
126.1 B D 36
2
B -
c
3
1 (KL>r)
8 (KL>rc)3
=
5
3
+
3
8
1 119.4 3
A 119.4
136.1 B - 8 A 126.1 B
P
= 10.37 ksi
2(16) A 8.020
42
2 B
+
13.3
53.4
10.37 Ú 5.56
O.K.
Column is safe.
Yes.
Ans.
1132
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The A-36-steel W10 * 45 column is fixed at its
base. Its top is constrained to move along the x–x axis but
free to rotate about and move along the y–y axis. Determine
the maximum eccentric force P that can be safely supported
by the column using the allowable stress method.
•13–113.
12 in.
y
Section Properties. From the table listed in the appendix, the section properties for
a W10 * 45 are
bf = 8.02 in.
rx = 4.32 in.
Iy = 53.4 in4
ry = 2.01 in.
Slenderness Ratio. Here, Lx = 24(12) = 288 in. and for a column fixed at its base
and free at its top, Kx = 2. Thus,
a
2(288)
KL
b =
= 133.33 (controls)
r x
4.32
Since the column is fixed at its base and pinned at its top, Ky = 0.7 and
Ly = 24(12) = 288 in. Then,
a
0.7(288)
KL
b =
= 100.30
r y
2.01
Allowable Stress. The allowable stress will be determined using the AISC column
formulas. For A-36 steel,
2p2 C 29 A 103 B D
KL
2p2E
KL
KL
b =
=
= 126.10. Since a
b 6 a
b 6 200,
r c
r c
r x
B sY
C
36
the column is classified as a long column.
a
sallow =
=
12p2E
23(KL>r)2
12p2 C 29 A 103 B D
23(133.332)
= 8.400 ksi
Maximum Stress. Bending is about the weak axis. Since M = P(12) and
bf
8.02
=
= 4.01 in,
c =
2
2
sallow =
Mc
P
+
A
I
8.400 =
[P(12)](4.01)
P
+
13.3
53.4
y
x
24 ft
A = 13.3 in2
P
P = 8.604 kip = 8.60 kip
Ans.
1133
x
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Page 1134
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13–114. The A-36-steel W10 * 45 column is fixed at its
base. Its top is constrained to move along the x–x axis but
free to rotate about and move along the y–y axis. Determine
the maximum eccentric force P that can be safely supported
by the column using an interaction formula. The allowable
bending stress is (sb)allow = 15 ksi.
12 in.
bf = 8.02 in.
rx = 4.32 in.
y
24 ft
Iy = 53.4 in4
ry = 2.01 in.
Slenderness Ratio. Here, Lx = 24(12) = 288 in and for a column fixed at its base
and free at its top, Kx = 2. Thus,
a
2(288)
KL
b =
= 133.33 (controls)
r x
4.32
Since the column is fixed at its base and pinned at its top, Ky = 0.7 and
Ly = 24(12) = 288 in. Then,
a
0.7(288)
KL
b =
= 100.30
r y
2.01
Allowable Stress. The allowable stress will be determined using the AISC column
2p2 C 29 A 103 B D
2p2E
KL
= 126.10. Since
formulas. For A-36 steel, a
b =
=
C
r c
B sY
36
a
KL
KL
b 6 a
b 6 200, the column is classified as a long column.
r c
r x
sallow =
=
12p2E
23(KL>r)2
12p2 C 29 A 103 B D
23 A 133.332 B
= 8.400 ksi
Interaction Formula. Bending is about the weak axis. Here, M = P(12) and
bf
8.02
=
= 4.01 in.
c =
2
2
P>A
Mc>Ar2
+
= 1
(sa)allow
(sb)allow
P>13.3
+
8.400
P(12)(4.01) n C 13.3 A 2.012 B D
15
y
x
Section Properties. From the table listed in the appendix, the section properties for
a W10 * 45 are
A = 13.3 in2
P
= 1
P = 14.57 kip = 14.6 kip
Ans.
14.57>13.3
sa
=
= 0.1304 6 0.15
(sa)allow
8.400
O.K.
1134
x
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13–115. The A-36-steel W12 * 50 column is fixed at its
base. Its top is constrained to move along the x–x axis but
free to rotate about and move along the y–y axis. If the
eccentric force P = 15 kip is applied to the column,
investigate if the column is adequate to support the loading.
Use the allowable stress method.
12 in.
bf = 8.08 in.
rx = 5.18 in.
y
24 ft
Iy = 56.3 in4
ry = 1.96 in.
Slenderness Ratio. Here, Lx = 24(12) = 288 in. and for a column fixed at its base
and pinned at its top, Kx = 2. Thus,
a
2(288)
KL
b =
= 111.20 (controls)
r x
5.18
Since the column is fixed at its base and pinned at its top, Ky = 0.7 and
Ly = 24(12) = 288 in. Thus,
a
0.7(288)
KL
b =
= 102.86
r y
1.96
Allowable Stress. The allowable stress will be determined using the AISC column
formulas. For A-36 steel, a
a
2p2 C 29 A 103 B D
KL
2p2E
= 126.10. Since
b =
=
C
r c A sY
36
KL
KL
b 6 a
b , the column can be classified as an intermediate column.
r x
r c
B1 sallow =
2(KL>r)C 2
R sY
3(KL>r)
(KL>r)3
5
+
3
8(KL>r)C
8(KL>r)C 3
C1 -
=
(KL>r)2
111.202
2 A 126.102 B
S(36)
3(111.20)
5
111.203
+
3
8(126.10)
8 A 126.103 B
= 11.51 ksi
Maximum Stress. Bending is about the weak axis. Since, M = 15(12) = 180 kip # in.
bf
8.08
=
= 4.04 in.,
and c =
2
2
smax =
y
x
Section Properties. From the table listed in the appendix, the section properties for
a W12 * 50 are
A = 14.7 in2
P
180(4.04)
P
Mc
15
+
=
+
= 13.94 ksi
A
I
14.7
56.3
Since smax 7 sallow, the W12 * 50 column is inadequate according to the allowable
stress method.
1135
x
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*13–116. The A-36-steel W12 * 50 column is fixed at its
base. Its top is constrained to move along the x–x axis but
free to rotate about and move along the y–y axis. If the
eccentric force P = 15 kip is applied to the column,
investigate if the column is adequate to support the loading.
Use the interaction formula. The allowable bending stress is
(sb)allow = 15 ksi.
12 in.
bf = 8.08 in.
rx = 5.18 in.
y
24 ft
ry = 1.96 in.
Slenderness Ratio. Here, Lx = 24(12) = 288 in and for a column fixed at its base
and pinned at its top, Kx = 2. Thus,
a
2(288)
KL
b =
= 111.20 (controls)
r x
5.18
Since the column is fixed at its base and pinned at its top, Ky = 0.7 and
Ly = 24(12) = 288 in. Then,
a
Allowable
=
C
Axial
2p2 C 29 A 103 B D
36
0.7(288)
KL
b =
= 102.86
r y
1.96
Stress.
For
= 126.10. Since a
A-36
steel,
a
KL
2p2E
b =
r c A sY
KL
KL
b 6 a
b , the column can be
r x
r c
classified as an intermediate column.
C1 sallow =
2(KL>r)c 2
SsY
3(KL>r)
(KL>r)3
5
+
3
8(KL>r)c
8(KL>r)c 3
C1 -
=
(KL>r)2
111.202
2 A 126.102 B
S(36)
3(111.20)
5
111.203
+
3
8(126.10)
8 A 126.103 B
= 11.51 ksi
Interaction Formula. Bending is about the weak axis. Here, M = 15(12)
bf
8.08
= 180 kip # in. and c =
=
= 4.04 in.
2
2
P>A
Mc>Ar2
15>14.7
+
+
=
(sa)allow
(sb)allow
11.51
180(4.04) n C 14.7 A 1.962 B D
15
y
x
Section Properties. From the table listed in the appendix, the section properties for
a W12 * 50 are
A = 14.7 in2
P
= 0.9471 6 1
P = 14.57 kip = 14.6 kip
Ans.
15>14.7
sa
=
= 0.089 6 0.15
(sa)allow
11.51
O.K.
Thus, a W12 * 50 column is adequate according to the interaction formula.
1136
x
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•13–117.
A 16-ft-long column is made of aluminum alloy
2014-T6. If it is fixed at its top and bottom, and a
compressive load P is applied at point A, determine the
maximum allowable magnitude of P using the equations of
Sec. 13.6 and Eq. 13–30.
P
A
4.25 in.
x
0.5 in.
y
8 in.
Section properties:
A = 2(0.5)(8) + 8(0.5) = 12 in2
Ix =
1
1
(8)(93) (7.5)(83) = 166 in4
12
12
Iy = 2 a
ry =
1
1
b (0.5)(83) +
(8)(0.53) = 42.75 in4
12
12
Iy
42.75
=
= 1.8875 in.
AA
A 12
Allowable stress method:
0.5(16)(12)
KL
KL
= 50.86, 12 6
=
6 55
ry
ry
1.8875
sallow = c 30.7 - 0.23a
KL
bd
r
= [30.7 - 0.23(50.86)] = 19.00 ksi
smax = sallow =
19.00 =
Mx c
P
+
A
Ix
P(4.25)(4.5)
P
+
12
166
P = 95.7 kip
Ans.
1137
y
x 8 in.
0.5 in.
0.5 in.
13 Solutions 46060
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Page 1138
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13–118. A 16-ft-long column is made of aluminum alloy
2014-T6. If it is fixed at its top and bottom, and a
compressive load P is applied at point A, determine the
maximum allowable magnitude of P using the equations of
Sec. 13.6 and the interaction formula with 1sb2allow = 20 ksi.
P
A
4.25 in.
x
0.5 in.
y
8 in.
Section Properties:
A = 2(0.5)(8) + 8(0.5) = 12 in2
Ix =
1
1
(8)(93) (7.5)(83) = 166 in4
12
12
Iy = 2 a
ry =
1
1
b (0.5)(83) +
(8)(0.53) = 42.75 in4
12
12
Iy
42.75
=
= 1.8875 in.
AA
A 12
Interaction method:
0.5(16)(12)
KL
KL
= 50.86, 12 6
=
6 55
ry
ry
1.8875
sallow = c 30.7 - 0.23a
KL
bd
r
= [30.7 - 0.23(50.86)]
= 19.00 ksi
sa =
P
P
=
= 0.08333P
A
12
sb =
P(4.25)(4.50)
Mc
=
= 0.1152P
Ix
166
sb
sa
+
= 1.0
(sa)allow
(sb)allow
0.08333P
0.1152P
+
= 1
19.00
20
P = 98.6 kip
Ans.
1138
y
x 8 in.
0.5 in.
0.5 in.
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Page 1139
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13–119. The 2014-T6 hollow column is fixed at its base
and free at its top. Determine the maximum eccentric
force P that can be safely supported by the column. Use the
allowable stress method. The thickness of the wall for the
section is t = 0.5 in.
6 in.
P
3 in.
6 in.
8 ft
Section Properties.
A = 6(3) - 5(2) = 8 in2
Ix =
1
1
(3) A 63 B (2) A 53 B = 33.1667 in4
12
12
rx =
33.1667
Ix
=
= 2.036 in.
AA
A
8
Iy =
1
1
(6) A 33 B (5) A 23 B = 10.1667 in4
12
12
ry =
Iy
10.1667
=
= 1.127 in.
AA
A
8
Slenderness Ratio. For a column fixed at its base and free at its top, K = 2. Thus,
a
2(8)(12)
KL
b =
= 170.32
r y
1.127
Allowable Stress. Since a
KL
b 7 55, the column can be classified as a long
r y
column.
sallow =
54 000 ksi
54 000 ksi
=
= 1.862 ksi
(KL>r)2
170.312
Maximum Stress. Bending occurs about the strong axis so that M = P(6) and
6
c = = 3 in.
2
sallow =
1.862 =
Mc
P
+
A
I
C P(6) D (3)
P
+
8
33.1667
P = 2.788 kip = 2.79 kip
Ans.
1139
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Page 1140
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*13–120. The 2014-T6 hollow column is fixed at its base
and free at its top. Determine the maximum eccentric force
P that can be safely supported by the column. Use the
interaction formula. The allowable bending stress is
(sb)allow = 30 ksi. The thickness of the wall for the section is
t = 0.5 in.
6 in.
P
3 in.
6 in.
8 ft
Section Properties.
A = 6(3) - 5(2) = 8 in2
Ix =
1
1
(3) A 63 B (2) A 53 B = 33.1667 in4
12
12
rx =
33.1667
Ix
=
= 2.036 in.
AA
A
8
Iy =
1
1
(6) A 33 B (5) A 23 B = 10.1667 in4
12
12
ry =
Iy
10.1667
=
= 1.127 in.
AA
A
8
Slenderness Ratio. For a column fixed at its base and pinned at its top, K = 2. Thus,
a
2(8)(12)
KL
b =
= 170.32
r y
1.127
KL
b 7 55, the column can be classified as the column is
r y
classified as a long column.
Allowable Stress. Since a
sallow =
54000 ksi
54000 ksi
=
= 1.862 ksi
(KL>r)2
170.312
Interaction Formula. Bending is about the strong axis. Since M = P(6) and
6
= 3 in,
c =
2
P>A
Mc>Ar2
+
= 1
(sa)allow
(sb)allow
P>8
+
1.862
[P(6)](3) n C 8 A 2.0362 B D
30
= 1
P = 11.73 kip = 11.7 kip
Ans.
1140
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
•13–121.
The 10-ft-long bar is made of aluminum alloy
2014-T6. If it is fixed at its bottom and pinned at the top,
determine the maximum allowable eccentric load P that can
be applied using the formulas in Sec. 13.6 and Eq. 13–30.
P
x
1.5 in.
1.5 in.
x
Section Properties:
A = 6(4) = 24.0 in2
Ix =
1
(4) A 63 B = 72.0 in4
12
Iy =
1
(6) A 43 B = 32.0 in4
12
ry =
Iy
32.0
=
= 1.155 in.
AA
A 24
Slenderness Ratio: The largest slenderness ratio is about y- y axis. For a column
pinned at one end fixed at the other end, K = 0.7. Thus,
a
0.7(10)(12)
KL
b =
= 72.75
r y
1.155
Allowable Stress: The allowable stress can be determined using aluminum
KL
(2014 –T6 alloy) column formulas. Since
7 55, the column is classified as a long
r
column. Applying Eq. 13–26,
sallow = c
=
54 000
d ksi
(KL>r)2
54 000
72.752
= 10.204 ksi
Maximum Stress: Bending is about x - x axis. Applying Eq. 13–30, we have
smax = sallow =
10.204 =
P
Mc
+
A
I
P(1.5)(3)
P
+
24.0
72.0
P = 98.0 kip
Ans.
1141
y
3 in.
2 in.
y
2 in.
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Page 1142
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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
13–122. The 10-ft-long bar is made of aluminum alloy
2014-T6. If it is fixed at its bottom and pinned at the top,
determine the maximum allowable eccentric load P that
can be applied using the equations of Sec. 13.6 and the
interaction formula with 1sb2allow = 18 ksi.
P
x
1.5 in.
1.5 in.
x
Section Properties:
A = 6(4) = 24.0 in2
Ix =
1
(4) A 63 B = 72.0 in4
12
Iy =
1
(6) A 43 B = 32.0 in4
12
rx =
Ix
72.0
=
= 1.732 in.
AA
A 24.0
ry =
Iy
32.0
=
= 1.155 in.
AA
A 24.0
Slenderness Ratio: The largest slenderness radio is about y -y axis. For a column
pinned at one end and fixed at the other end, K = 0.7. Thus
a
0.7(10)(12)
KL
b =
= 72.75
r y
1.155
Allowable Stress: The allowable stress can be determined using aluminum
KL
7 55, the column is classified as a long
(2014 –T6 alloy) column formulas. Since
r
column. Applying Eq. 13–26,
(sa)allow = c
=
54 000
d ksi
(KL>r)2
54 000
72.752
= 10.204 ksi
Interaction Formula: Bending is about x - x axis. Applying Eq. 13–31, we have
Mc>Ar2
P>A
+
= 1
(sa)allow
(sb)allow
P(1.5)(3)>24.0(1.7322)
P>24.0
+
= 1
10.204
18
P = 132 kip
Ans.
1142
y
3 in.
2 in.
y
2 in.
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13–123. The rectangular wooden column can be
considered fixed at its base and pinned at its top. Also, the
column is braced at its mid-height against the weak axis.
Determine the maximum eccentric force P that can be safely
supported by the column using the allowable stress method.
6 in.
6 in.
Section Properties.
Ix =
dx = 6 in.
5 ft
1
(3) A 63 B = 54 in4
12
dy = 3 in.
Slenderness Ratio. Here, Lx = 10(12) = 120 in. and for a column fixed at its base
and pinned at its top, K = 0.7. Thus,
a
0.7(120)
KL
b =
= 14
d x
6
Since the bracing provides support equivalent to a pin, Ky = 1 and
Ly = 5(12) = 60 in. Then
a
1(60)
KL
b =
= 20 (controls)
d y
3
KL
6 26, the column can be classified as the column
d
is classified as an intermediate column.
Allowable Stress. Since 11 6
sallow = 1.20c 1 = 1.20 c 1 -
1 KL>d 2
a
b d ksi
3 26.0
1 20 2
a
b d ksi = 0.9633 ksi
3 26.0
Maximum Stress. Bending occurs about the strong axis. Here, M = P(6) and
6
c = = 3 in.
2
sallow =
0.9633 =
6 in.
3 in.
5 ft
A = 6(3) = 18 in2
P
P
Mc
+
A
I
[P(6)](3)
P
+
18
54
P = 2.477 kip = 2.48 kip
Ans.
1143
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*13–124. The rectangular wooden column can be
considered fixed at its base and pinned at its top. Also, the
column is braced at its mid-height against the weak axis.
Determine the maximum eccentric force P that can be
safely supported by the column using the interaction
formula. The allowable bending stress is (sb)allow = 1.5 ksi.
6 in.
6 in.
5 ft
Section Properties.
rx =
Ix =
1
(3) A 63 B = 54 in4
12
Ix
54
=
= 1.732 in.
AA
A 18
dx = 6 in.
dy = 3 in.
Slenderness Ratio. Here, Lx = 10(12) = 120 in. and for a column fixed at its base
pinned at its top, K = 0.7. Thus,
a
0.7(120)
KL
b =
= 14
d x
6
Since the bracing provides support equivalent to a pin, Ky = 1 and
Ly = 5(12) = 60 in. Then
a
1(60)
KL
b =
= 20 (controls)
d y
3
KL
6 26, the column can be classified as the
d
column is classified as an intermediate column.
Allowable Axial Stress. Since 11 6
sallow = 1.20c 1 = 1.20 c 1 -
1 KL>d 2
a
b d ksi
3 26.0
1 20 2
a
b d ksi = 0.9633 ksi
3 26.0
Interaction Formula. Bending occurs about the strong axis. Since M = P(6) and
6
c = = 3 in.
2
P>A
Mc>Ar2
+
= 1
(sa)allow
(sb)allow
P>18
+
0.9633
[P(6)](3) n C 18 A 1.7322 B D
1.5
6 in.
3 in.
5 ft
A = 6(3) = 18 in2
P
= 1
P = 3.573 kip = 3.57 kip
Ans.
1144
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•13–125.
The 10-in.-diameter utility pole supports the
transformer that has a weight of 600 lb and center of gravity
at G. If the pole is fixed to the ground and free at its
top, determine if it is adequate according to the NFPA
equations of Sec. 13.6 and Eq. 13–30.
G
15 in.
18 ft
2(18)(12)
KL
=
= 43.2 in.
d
10
26 6 43.2 … 50
Use Eq. 13–29,
sallow =
540
540
=
= 0.2894 ksi
(KL>d)
(43.2)2
smax =
Mc
P
+
A
I
smax =
(600)(15)(5)
600
+
2
p (5)
A p4 B (5)4
smax = 99.31 psi 6 0.289 ksi
O.K.
Yes.
Ans.
1145
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13–126. Using the NFPA equations of Sec. 13.6 and
Eq. 13–30, determine the maximum allowable eccentric
load P that can be applied to the wood column. Assume that
the column is pinned at both its top and bottom.
P
0.75 in.
6 in.
3 in.
12 ft
Section Properties:
A = 6(3) = 18.0 in2
Iy =
1
(6) A 33 B = 13.5 in4
12
Slenderness Ratio: For a column pinned at both ends, K = 1.0. Thus,
a
1.0(12)(12)
KL
b =
= 48.0
d y
3
Allowable Stress: The allowable stress can be determined using NFPA timber
KL
column formulas. Since 26 6
6 50, it is a long column. Applying Eq. 13–29,
d
sallow =
=
540
ksi
(KL>d)2
540
= 0.234375 ksi
48.02
Maximum Stress: Bending is about y -y axis. Applying Eq. 13–30, we have
smax = sallow =
0.234375 =
P
Mc
+
A
I
P(0.75)(1.5)
P
+
18.0
13.5
P = 1.69 kip
Ans.
1146
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13–127. Using the NFPA equations of Sec. 13.6 and
Eq. 13–30, determine the maximum allowable eccentric
load P that can be applied to the wood column. Assume that
the column is pinned at the top and fixed at the bottom.
P
0.75 in.
6 in.
3 in.
12 ft
Section Properties:
A = 6(3) = 18.0 in2
Iy =
1
(6) A 33 B = 13.5 in4
12
Slenderness Ratio: For a column pinned at one end and fixed at the other end,
K = 0.7. Thus,
a
0.7(12)(12)
KL
b =
= 33.6
d y
3
Allowable Stress: The allowable stress can be determined using NFPA timber
KL
column formulas. Since 26 6
6 50, it is a long column. Applying Eq. 13–29,
d
sallow =
=
540
ksi
(KL>d)2
540
= 0.4783 ksi
33.62
Maximum Stress: Bending is about y -y axis. Applying Eq. 13–30, we have
smax = sallow =
0.4783 =
Mc
P
+
A
I
P(0.75)(1.5)
P
+
18.0
13.5
P = 3.44 kip
Ans.
1147
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*13–128. The wood column is 4 m long and is required to
support the axial load of 25 kN. If the cross section is square,
determine the dimension a of each of its sides using a factor
of safety against buckling of F.S. = 2.5. The column is
assumed to be pinned at its top and bottom. Use the Euler
equation. Ew = 11 GPa, and sY = 10 MPa.
25 kN
a
4m
a
1
a4
(a) A a3 B =
, P = (2.5)25 = 62.5 kN and K = 1
12
12 cr
for pin supported ends column. Applying Euler’s formula,
Critical Buckling Load: I =
Pcr =
62.5 A 10
3
B =
p2EI
(KL)2
a
p2(11)(109) A 12
B
4
[1(4)]2
a = 0.1025 m = 103 mm
Ans.
Critical Stress: Euler’s formula is only valid if scr 6 sY.
scr =
62.5(103)
Pcr
=
= 5.94 MPa 6 s Y = 10 MPa
A
0.1025(0.1025)
1148
O.K.
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•13–129. If the torsional springs attached to ends A and
C of the rigid members AB and BC have a stiffness k,
determine the critical load Pcr.
P
k
A
L
2
B
Equilibrium. When the system is given a slight lateral disturbance, the configuration
shown in Fig. a is formed. The couple moment M can be related to P by considering
the equilibrium of members AB and BC.
Member AB
+ c ©Fy = 0;
a + ©MA = 0;
By - P = 0
(1)
By a
(2)
L
L
sin u b + Bx a cos u b - M = 0
2
2
Member BC
a + ©MC = 0; - By a
L
L
sin u b + Bx a cos u b + M = 0
2
2
(3)
Solving Eqs. (1), (2), and (3), we obtain
Bx = 0
By =
2M
L sin u
M =
PL
sin u
2
Since u is very small, the small angle analysis gives sin u ⬵ u. Thus,
M =
PL
u
2
(4)
Torslonal Spring Moment. The restoring couple moment Msp can be determined
using the torsional spring formula, M = ku. Thus,
Msp = ku
Critical Buckling Load. When the mechanism is on the verge of bucklling M must
equal Msp.
M = Msp
Pcr L
u = ku
2
Pcr =
2k
L
Ans.
1149
L
2
k
C
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13–129. Continued
1150
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13–130. Determine the maximum intensity w of the
uniform distributed load that can be applied on the beam
without causing the compressive members of the supporting
truss to buckle. The members of the truss are made from
A-36-steel rods having a 60-mm diameter. Use F.S. = 2
against buckling.
w
B
A
C
2m
Equilibrium. The force developed in member BC can be determined by considering
the equilibrium of the free-body diagram of the beam AB, Fig. a.
3
w(5.6)(2.8) - FBC a b(5.6) = 0 FBC = 4.6667w
5
a + ©MA = 0;
The Force developed in member CD can be obtained by analyzing the equilibrium
of joint C, Fig. b,
+ c ©Fy = 0;
FAC a
+
: ©Fx = 0;
4
12
4.6667w a b + 7.28a b w -FCD = 0
5
13
5
3
b - 4.6667w a b = 0
13
5
FAC = 7.28w (T)
FCD = 10.4533w (C)
Section Properties. The cross-sectional area and moment of inertia of the solid
circular rod CD are
A = p A 0.032 B = 0.9 A 10 - 3 B p m2
I =
p
A 0.034 B = 0.2025 A 10 - 6 B p m4
4
Critical Buckling Load. Since both ends of member CD are pinned, K = 1. The
critical buckling load is
Pcr = FCD (F.S.) = 10.4533w(2) = 20.9067w
Applying Euler’s formula,
Pcr =
p2EI
(KL)2
20.9067w =
p2 C 200 A 109 B D C 0.2025 A 10 - 6 B p D
[1(3.6)]2
w = 4634.63 N>m = 4.63 kN>m
Ans.
Critical Stress: Euler’s formula is valid only if scr 6 sY.
scr =
20.907(4634.63)
Pcr
=
= 34.27 MPa 6 sY = 250 MPa
A
0.9 A 10 - 3 B p
1151
O.K.
3.6 m
D
1.5 m
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13–131. The W10 * 45 steel column supports an axial load
of 60 kip in addition to an eccentric load P. Determine the
maximum allowable value of P based on the AISC equations
of Sec. 13.6 and Eq. 13–30. Assume that in the x–z plane
Kx = 1.0 and in the y–z plane Ky = 2.0. Est = 2911032 ksi,
sY = 50 ksi.
z
P
60 kip
x
y
10 ft
Section properties for W 10 * 45:
A = 13.3 in2
d = 10.10 in.
rx = 4.32 in.
ry = 2.01 in.
Ix = 248 in4
Allowable stress method:
a
1.0(10)(12)
KL
b =
= 27.8
r x
4.32
a
2.0(10)(12)
KL
b =
= 119.4
r y
2.01
a
2p2(29)(103)
KL
2p2E
b =
=
= 107
r c
B sg
B
50
(controls)
KL
KL
7 a
b
r
r c
(sa)allow =
12p2(29)(103)
12p2E
=
= 10.47 ksi
2
23(KL>r)
23(119.4)4
smax = (sa)allow =
Mc
P
+
A
I
P(8) A 10.10
P + 60
2 B
+
10.47 =
13.3
248
P = 25.0 kip
Ans.
1152
y
x 8 in.
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*13–132. The A-36-steel column can be considered pinned
at its top and fixed at its base. Also, it is braced at its
mid-height along the weak axis. Investigate whether a
W250 * 45 section can safely support the loading shown.
Use the allowable stress method.
600 mm
10 kN
4.5 m
Section Properties. From the table listed in the appendix, the necessary section
properties for a W250 * 45 are
A = 5700 mm2 = 5.70 A 10 - 3 B m2
d = 266 mm = 0.266 m
Ix = 71.1 A 106 B mm4 = 71.1 A 10 - 6 B m4
rx = 112 mm = 0.112 m
ry = 35.1 mm = 0.0351 mm
Slenderness Ratio. Here, Lx = 9 m and for a column fixed at its base and pinned at
its top, Kx = 0.7. Thus,
¢
0.7(9)
KL
= 56.25
≤ =
r x
0.112
Since the bracing provides support equivalent to a pin, Ky = 1 and Ly = 4.5 m.
Then,
¢
1(4.5)
KL
= 128.21 (controls)
≤ =
r y
0.0351
Allowable Stress. For A-36 steel, a
Since a
2p2 C 200 A 109 B D
KL
2p2E
b =
=
= 125.66.
r c
C 250 A 106 B
B sY
KL
KL
b 6 a
b 6 200, the column can be classified as a long column.
r c
r y
sallow =
12p2 C 200 A 109 B D
12p2E
=
= 62.657 MPa
2
23(KL>r)
23(128.21)2
Maximum Stress. Bending occurs about the strong axis. Here, P = 10 + 40
0.266
d
= 50 kN, M = 40(0.6) = 24 kN # m and c =
=
= 0.133 m,
2
2
smax =
50 A 103 B
24 A 103 B (0.133)
P
Mc
+
=
+
= 53.67 MPa
A
I
5.70 A 10 - 3 B
71.1 A 10 - 6 B
Since smax 6 sallow, the column is adequate according to the allowable stress
method.
1153
4.5 m
40 kN
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•13–133.
The A-36-steel column can be considered pinned
at its top and fixed at its base. Also, it is braced at
its mid-height along the weak axis. Investigate whether a
W250 * 45 section can safely support the loading shown.
Use the interaction formula. The allowable bending stress is
(sb)allow = 100 MPa.
600 mm
10 kN
4.5 m
Section Properties. From the table listed in the appendix, the necessary section
properties for a W250 * 45 are
A = 5700 mm2 = 5.70 A 10 - 3 B m2
Ix = 71.1 A 10 B mm = 71.1 A 10
6
4
-6
d = 266 mm = 0.266 m
4.5 m
Bm
4
rx = 112 mm = 0.112 m
ry = 35.1 mm = 0.0351 mm
Slenderness Ratio. Here, Lx = 9 m and for a column fixed at its base and pinned at
its top, Kx = 0.7. Thus,
a
0.7(9)
KL
b =
= 56.25
r x
0.112
Since the bracing provides support equivalent to a pin, Ky = 1 and Ly = 4..5 m.
Then,
a
Allowable
=
C
1(4.5)
KL
b =
= 128.21 (controls)
r y
0.0351
Axial
2p2 C 200 A 109 B D
250 A 10
6
B
Stress.
For
= 125.66. Since a
A–36
steel,
a
KL
2p2E
b =
r c
B sY
KL
KL
b 6 a
b 6 200, the column can be
r c
r y
classified as a long column.
sallow =
12p2 C 200 A 109 B D
12p2E
=
= 62.657 MPa
23(KL>r)2
23(128.21)2
Interaction Formula. Bending is about the strong axis. Here, P = 10 + 40 = 50 kN,
0.266
d
M = 40(0.6) = 24 kN # m and c =
=
= 0.133 m,
2
2
Mc>Ar2
P>A
+
=
(sa)allow
(sb)allow
50 A 103 B n 5.70 A 10 - 3 B
62.657 A 106 B
+
24 A 103 B (0.133) n C 5.70 A 10 - 3 B A 0.1122 B D
100 A 106 B
= 0.5864 6 1
sa
=
(sa)allow
50 A 103 B n 5.7 A 10 - 3 B
62.657 A 106 B
O.K.
= 0.140 6 0.15
O.K.
Thus, a W250 * 45 column is adequate according to the interaction formula.
1154
40 kN
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13–134. The member has a symmetric cross section. If it is
pin connected at its ends, determine the largest force it can
support. It is made of 2014-T6 aluminum alloy.
0.5 in.
2 in.
P
5 ft
Section properties:
A = 4.5(0.5) + 4(0.5) = 4.25 in2
P
1
1
(0.5)(4.53) +
(4)(0.5)3 = 3.839 in4
I =
12
12
r =
I
3.839
=
= 0.9504 in.
AA
A 4.25
Allowable stress:
1.0(5)(12)
KL
=
= 63.13
r
0.9504
KL
7 55
r
Long column
sallow =
54000
54000
=
= 13.55 ksi
(KL>r)2
63.132
Pallow = sallowA
= 13.55(4.25) = 57.6 kip
Ans.
1155
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13–135. The W200 * 46 A-36-steel column can be
considered pinned at its top and fixed at its base. Also, the
column is braced at its mid-height against the weak axis.
Determine the maximum axial load the column can support
without causing it to buckle.
6m
Section Properties. From the table listed in the appendix, the section properties for
a W200 * 46 are
A = 5890 mm2 = 5.89 A 10 - 3 B m2
Iy = 15.3 A 106 B mm4 = 15.3 A 10 - 6 B m4
Ix = 45.5 A 106 B mm4 = 45.5 A 10 - 6 B m4
Critical Buckling Load. For buckling about the strong axis, Kx = 0.7 and Lx = 12 m.
Since the column is fixed at its base and pinned at its top,
Pcr =
p2EIx
(KL)x 2
=
p2 c 200 A 109 B d c 45.5 A 10 - 6 B d
[0.7(12)]2
= 1.273 A 106 B N = 1.27 MN
For buckling about the weak axis, Ky = 1 and Ly = 6 m since the bracing provides
a support equivalent to a pin. Applying Euler’s formula,
Pcr =
p2EIy
(KL)y 2
=
p2 c 200 A 109 B d c 15.3 A 10 - 6 B d
[1(6)]2
= 838.92 kN = 839 kN (controls)Ans.
Critical Stress. Euler’s formula is valid only if scr 6 sY.
scr =
838.92 A 103 B
Pcr
=
= 142.43 MPa 6 sY = 250 MPa
A
5.89 A 10 - 3 B
1156
O.K.
6m
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*13–136. The structural A-36 steel column has the cross
section shown. If it is fixed at the bottom and free at the top,
determine the maximum force P that can be applied at A
without causing it to buckle or yield. Use a factor of safety
of 3 with respect to buckling and yielding.
P
20 mm
A
10 mm
Section properties:
100 mm
-3
2
4m
©A = 0.2(0.01) + 0.15 (0.01) + 0.1(0.01) = 4.5(10 ) m
150 mm
A
100 mm
10 mm
0.005 (0.2)(0.01) + 0.085 (0.15)(0.01) + 0.165 (0.1)(0.01)
©xA
x =
=
©A
4.5(10 - 3)
= 0.06722 m
1
(0.2)(0.013) + 0.2 (0.01)(0.06722 - 0.005)2
12
Iy =
+
1
(0.01)(0.153) + 0.01 (0.15)(0.085 - 0.06722)2
12
+
1
(0.1)(0.013) + 0.1 (0.01)(0.165 - 0.06722)2
12
= 20.615278 (10 - 6) m4
1
1
1
(0.01)(0.23) +
(0.15)(0.013) +
(0.01)(0.13)
12
12
12
Ix =
= 7.5125 (10 - 6) m4
ry =
Iy
20.615278(10 - 6)
=
= 0.0676844
AA
A
4.5 (10 - 3)
Buckling about x - x axis:
Pcr =
p2(200)(109)(7.5125)(10 - 6)
p2 EI
=
2
(KL)
[2.0(4)]2
= 231.70 kN
scr =
(controls)
231.7 (103)
Pcr
= 51.5 MPa 6 sg = 250 MPa
=
A
4.5 (10 - 3)
Yielding about y- y axis:
smax =
P
ec
KL P
c 1 + 2 sec a
b d;
A
2r A EA
r
e = 0.06722 - 0.02 = 0.04722 m
0.04722 (0.06722)
ec
=
= 0.692919
0.0676844
r2
2.0 (4)
P
P
KL
=
= 1.96992 P (10 - 3) 2P
2r A EA
2(0.0676844) A 200 (109)(4.5)(10 - 3)
250(106)(4.5)(10 - 3) = P[1 + 0.692919 sec (1.96992P (10 - 3) 2P)]
By trial and error:
P = 378.45 kN
Hence,
Pallow =
231.70
= 77.2 kN
3
Ans.
1157
10 mm
100 mm
13 Solutions 46060
6/11/10
11:56 AM
Page 1158
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•13–137.
The structural A-36 steel column has the cross
section shown. If it is fixed at the bottom and free at the top,
determine if the column will buckle or yield when the load
P = 10 kN. Use a factor of safety of 3 with respect to
buckling and yielding.
P
20 mm
A
10 mm
100 mm
Section properties:
4m
©A = 0.2 (0.01) + 0.15 (0.01) + 0.1 (0.01) = 4.5 (10 - 3) m2
0.005 (0.2)(0.01) + 0.085 (0.15)(0.01) + 0.165 (0.1)(0.01)
©xA
= 0.06722 m
=
©A
4.5 (10 - 3)
Iy =
1
(0.2)(0.013) + 0.2 (0.01)(0.06722 - 0.005)2
12
ry =
+
1
(0.01)(0.153) + 0.01 (0.15)(0.085 - 0.06722)2
12
+
1
(0.1)(0.013) + 0.1 (0.01)(0.165 - 0.06722)2 = 20.615278 (10 - 6) m4
12
1
1
1
(0.01)(0.23) +
(0.15)(0.013) +
(0.01)(0.13) = 7.5125 (10 - 6) m4
12
12
12
Iy
BA
20.615278 (10 - 6)
=
B
4.5 (10 - 3)
= 0.067843648 m
Buckling about x -x axis:
Pcr =
p2(200)(109)(7.5125)(10 - 6)
p2 EI
=
= 231.70 kN
2
(KL)
[2.0(4)]2
scr =
231.7 (103)
Pcr
= 51.5 MPa 6 sg = 250 MPa
=
A
4.5 (10 - 3)
Pallow =
O.K.
Pcr
231.7
=
= 77.2 kN 7 P = 10 kN
FS
3
Hence the column does not buckle.
Yielding about y- y axis:
smax =
P =
P
KL
ec
P
bd
c 1 + 2 sec a
A
2r A EA
r
A
100 mm
10 mm
x =
Ix =
150 mm
e = 0.06722 - 0.02 = 0.04722 m
10
= 3.333 kN
3
3.333 (103)
P
= 0.7407 MPa
=
A
4.5 (10 - 3)
0.04722 (0.06722)
ec
= 0.689815
=
(0.067844)
r2
2.0 (4)
P
KL
3.333 (103)
=
= 0.1134788
2 r AE A
2(0.06783648) A 200 (109)(4.5)(10 - 3)
smax = 0.7407 [1 + 0.692919 sec (0.1134788)] = 1.25 MPa 6 sg = 250 MPa
Hence the column does not yield!
No.
Ans.
1158
10 mm
100 mm
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