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13 Solutions 46060 6/11/10 11:56 AM Page 1038 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in wr...

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13 Solutions 46060

6/11/10

11:56 AM

Page 1038

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Determine the critical buckling load for the column. The material can be assumed rigid.

•13–1.

P

L 2 k

Equilibrium: The disturbing force F can be determined by summing moments about point A. a + ©MA = 0;

P(Lu) - Fa

L b = 0 2

A

F = 2Pu Spring Formula: The restoring spring force F1 can be determine using spring formula Fs = kx. Fs = k a

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

2 L b + F1 a Lb - PLu = 01 3 3

L 3

F2 + 2F1 = 3Pu

(1)

k

Spring Force. The restoring spring force A Fsp B 1 and A Fsp B 2 can be determined using the spring formula, 1 2 Lu and x2 = Lu, Fig. b. Thus, 3 3 2 2 = kx1 = ka Lub = kLu 3 3

L 3 A

Fsp = kx, where x1 =

A Fsp B 1

A Fsp B 2 = kx2 = ka Lub = 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), 1 2 kLu + 2 a kLub = 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

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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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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.

•13–5.

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 = 157.74 kip = 158

Ans.

Critical Stress. Euler’s formula is valid only if scr 6 sg. scr =

Pcr 157.74 = = 27.4 ksi 6 sg = 36 ksi A 5.75

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 = 52.29 kip = 52.3 kip

Ans.

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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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.

•13–9.

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

20 ft

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, p2EIy

Pcr =

p2 C 29.0(103) D (26.7)

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 Ix = 385 in4

A = 11.2 in2

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 2

(KxLx)

=

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

= 270.76 kip = 271 kip (Control)

Ans.

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

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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.

•13–13.

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

= 245 kip

Ans.

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.

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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.

•13–17.

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

= 2.924 kip = 2.92 kip

Ans.

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

= 5.968 kip = 5.97 kip

Ans.