CHAPTER 10
PROBLEM 10.1 Knowing that the spring at A is of constant k and that the bar AB is rigid, determine the critical load Pcr .
SOLUTION Let θ be the angle change of bar AB.
F = kx = kL sin θ
ΣM B = 0: FL cos θ − Px = 0 kL2 sin θ cos θ − PL sin θ = 0
Using sin θ ≈ θ
and cosθ ≈ 1, kL2θ − PLθ = 0 (kL2 − PL)θ = 0
Pcr = kL
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PROBLEM 10.2 Knowing that the torsional spring at B is of constant K and that the bar AB is rigid, determine the critical load Pcr .
SOLUTION Let θ be the angle change of bar AB. M = Kθ , x = L sin θ ≈ Lθ M B = 0: M − Px = 0 Kθ − PLθ = 0 ( K − PL) θ = 0
Pcr = K/L
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PROBLEM 10.3 Two rigid bars AC and BC are connected by a pin at C as shown. Knowing that the torsional spring at B is of constant K, determine the critical load Pcr for the system.
SOLUTION
Let θ be the angle change of each bar.
M B = Kθ
M B = 0: Kθ − FA L = 0 FA =
Bar AC.
Kθ L
ΣM C = 0: Pcr Pcr =
FA
θ
1 1 Lθ − LFA = 0 2 2 Pcr =
K L
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PROBLEM 10.4 Two rigid bars AC and BC are connected as shown to a spring of constant k. Knowing that the spring can act in either tension or compression, determine the critical load Pcr for the system.
SOLUTION Let δ be the deflection of point C. Using free body AC and 1 3
M C = 0 : − LRA + Pδ = 0
RA =
3Pδ L
Using free body BC and ΣM C = 0:
2 LRB − Pδ = 0 3
RB =
3Pδ 2L
Using both free bodies together, ΣFx = 0: RA + RB − kδ = 0 3 Pδ 3 Pδ + − kδ = 0 L 2L 9 P 2 L − k δ = 0 Pcr =
2kL 9
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PROBLEM 10.7 The rigid rod AB is attached to a hinge at A and to two springs, each of constant k = 2 kip/in., that can act in either tension or compression. Knowing that h = 2 ft, determine the critical load.
SOLUTION Let θ be the small rotation angle. xD ≈ hθ xC ≈ 3hθ xB ≈ 4hθ FC = kxC ≈ 3khθ FD = kxD ≈ khθ ΣM A = 0: hFD + 3hFC − PxB = 0 kh 2θ + 9kh 2θ − 4hP = 0, P =
Data:
5 kh 2
k = 2.0 kip/in. h = 2 ft = 24 in. P=
5 (2.0)(24) 2
P = 120.0 kips
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PROBLEM 10.11 A compression member of 20-in. effective length consists of a solid 1-in.diameter aluminum rod. In order to reduce the weight of the member by 25%, the solid rod is replaced by a hollow rod of the cross section shown. Determine (a) the percent reduction in the critical load, (b) the value of the critical load for the hollow rod. Use E = 10.6 × 106 psi .
SOLUTION Solid:
AS =
Hollow:
AH = di2 =
Solid rod:
IS = Pcr =
Hollow rod:
IH =
Pcr =
π 4
do2
π
(d 4
Is = 2 o
1 2 do 4
π 64
π do
π
d o4 = 4 2 64
)
− di2 = di =
4
3 3π 2 AS = do 4 44
1 do = 0.5 in. 2
(1.0) 4 = 0.049087 in 4
π 2 EI S 2
L
π
(d 64
4 o
π 2 EI H 2
L
=
π 2 (10.6 × 106 )(0.049087) (20)
)
− di4 =
=
2
π
1 (1) 4 − 64 2
4
= 0.046019 in 4
π 2 (10.6 × 106 )(0.046019) (20)
= 12.839 × 103 lb
2
(a)
PS − PH 12.839 × 103 − 12.036 × 103 = = 0.0625 PS 12.839 × 103
(b)
For the hollow rod,
= 12.036 × 103 lb = 12.04 kips
Percent reduction = 6.25% Pcr = 12.04 kips
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PROBLEM 10.13 A column of effective length L can be made by gluing together identical planks in either of the arrangements shown. Determine the ratio of the critical load using the arrangement a to the critical load using the arrangement b.
SOLUTION Arrangement (a). Ia = Pcr, a =
Arrangement (b).
1 4 d 12
π 2 EI L2e
I min = I y =
=
π 2 Ed 4 12 L2e
1 d 3 1 d (d ) + ( d ) 12 3 12 3 +
Pcr, b = Pcr, a Pcr,b
=
π 2 EI L2e
=
3
1 d 3 19 4 d (d ) = 12 3 324
19π 2 Ed 4 324 L2e
1 324 27 ⋅ = 12 19 19
Pcr, a Pcr,b
= 1.421
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PROBLEM 10.15 A compression member of 7-m effective length is made by welding together two L152 × 102 × 12.7 angles as shown. Using E = 200 GPa, determine the allowable centric load for the member if a factor of safety of 2.2 is required.
SOLUTION Angle L152 × 102 × 12.7:
A = 3060 mm 2 I x = 7.20 × 106 mm 4
I y = 2.59 × 106 mm 4
y = 50.3 mm x = 24.9 mm
Two angles:
I x = (2)(7.20 × 106 ) = 14.40 × 106 mm 4
I y = 2[(2.59 × 106 ) + (3060)(24.9) 2 ] = 8.975 × 106 mm 4 I min = I y = 8.975 × 106 mm 4 = 8.975 × 10−6 m 4 Pcr = Pall =
π 2 EI L2e
=
π 2 (200 × 109 )(8.975 × 10−6 )
Pcr 361.5 = F .S . 2.2
(7.0) 2
= 361.5 × 103 N = 361.5 kN Pall = 164.0 kN
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PROBLEM 10.20 Knowing that P = 5.2 kN, determine the factor of safety for the structure shown. Use E = 200 GPa and consider only buckling in the plane of the structure.
SOLUTION Joint B:
From force triangle, FBC FAB 5.2 = = sin 25° sin 20° sin 135° FAB = 3.1079 kN (Comp) FBC = 2.5152 kN (Comp)
Member AB:
I AB =
π d
4
4
π 18 3 4 = = 5.153 × 10 mm 42 4 2 = 5.153 × 10−9 m 4
FAB ,cr =
π 2 EI AB L2AB
=
π 2 (200 × 109 )(5.153 × 10−9 ) (1.2) 2
= 7.0636 × 103 N = 7.0636 kN F .S . =
Member BC:
I BC =
FAB ,cr FAB
π d
=
7.0636 = 2.27 3.1079
4
π 22 = 42 4 2
4
= 11.499 × 103 mm 4 = 11.499 × 10−9 m 4 L2BC = 1.22 + 1.22 = 2.88 m 2 FBC ,cr =
π 2 EI BC L2BC
=
π 2 (200 × 109 )(11.499 × 10−9 ) 2.88 3
= 7.8813 × 10 N = 7.8813 kN FBC ,cr 7.8813 = = 3.13 F .S . = 2.5152 FBC
Smallest F.S. governs.
F .S . = 2.27
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PROBLEM 10.25 Column AB carries a centric load P of magnitude 15 kips. Cables BC and BD are taut and prevent motion of point B in the xz plane. Using Euler’s formula and a factor of safety of 2.2, and neglecting the tension in the cables, determine the maximum allowable length L. Use E = 29 × 106 psi.
SOLUTION W10 × 22: I x = 118 in 4 I y = 11.4 in 4
P = 15 × 103 lb Pcr = ( F .S .) P = (2.2)(15 × 103 ) = 33 × 103 lb
Buckling in xz-plane.
Le = 0.7 L Pcr =
L=
Buckling in yz-plane.
(0.7 L)
2
L=
π
EI y
0.7
Pcr
(29 × 106 )(11.4) = 449.2 in. 0.7 33 × 103
π
Le = 2 L Pcr = L=
Smaller value for L governs.
π 2 EI y
π 2 EI x (2 L) 2
π 2
EI x π = 2 Pcr
L = 449.2 in.
(29 × 106 )(118) = 505.8 in. 33 × 103 L = 37.4 ft
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PROBLEM 10.27 Column ABC has a uniform rectangular cross section with b = 12 mm and d = 22 mm. The column is braced in the xz plane at its midpoint C and carries a centric load P of magnitude 3.8 kN. Knowing that a factor of safety of 3.2 is required, determine the largest allowable length L. Use E = 200 GPa.
SOLUTION Pcr = ( F .S .) P = (3.2)(3.8 × 103 ) = 12.16 × 103 N Pcr =
Buckling in xz-plane.
π 2 EI
Le = π
L2e
L = Le = π
EI Pcr
EI Pcr
1 3 1 db = (22)(12)3 = 3.168 × 103 mm 4 12 12 = 3.168 × 10−9 m 4
I=
L =π
Buckling in yz-plane.
Le = 2 L L =
I=
L=
The smaller length governs.
(200 × 109 )(3.168 × 10−9 ) = 0.717 m 12.16 × 103 Le π = 2 2
EI Pcr
1 3 1 bd = (12)(22)3 = 10.648 × 103 mm 4 12 12 = 10.648 × 10−9 m 4
π 2
(200 × 109 )(10.648 × 10−9 ) = 0.657 m 12.16 × 103
L = 0.657 m
L = 657 mm
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.
