CCHHAAPPTTEERR 1100

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.

PROBLEM 10.1

Knowing that the spring at A is of constant k and that the bar AB is rigid, determine the critical load cr .P

SOLUTION

Let θ be the angle change of bar AB.

sinF kx kL θ= =

2

0: cos 0

sin cos sin 0

BM FL Px

kL PL

θ

θ θ θ

Σ = − =

− =

Using

2sin and cos 1, 0kL PLθ θ θ θ θ≈ ≈ − =

2( ) 0kL PL θ− = crP 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 cr .P

SOLUTION

Let θ be the angle change of bar AB.

, sinM K x L Lθ θ θ= = ≈

0: 0 0BM M Px K PLθ θ= − = − =

( ) 0K PL θ− = cr /P 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 crP for the system.

SOLUTION

Let θ be the angle change of each bar.

BM Kθ=

0: 0B A

A

M K F L

KF

L

θθ

= − =

=

Bar AC. cr1 1

0: 02 2C AM P L LFθΣ = − =

crAF

Pθ

= crK

PL

=

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,

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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 crP for the system.

SOLUTION

Let δ be the deflection of point C.

Using free body AC and

1 3

0 : 03C A A

PM LR P R

L

δδ= − + = =

Using free body BC and

2

0: 03C BM LR PδΣ = − =

3

2BP

RL

δ=

Using both free bodies together,

0: 0

3 30

29

02

x A BF R R k

P Pk

L LP

kL

δδ δ δ

δ

Σ = + − =

+ − =

− =

cr2

9

kLP =

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,

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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 2 kip/in.,k = that can act in either tension or compression. Knowing that 2 ft,h = determine the critical load.

SOLUTION

Let θ be the small rotation angle.

3

4

D

C

B

x h

x h

x h

θθθ

≈≈≈

3C C

D D

F kx kh

F kx kh

θθ

= ≈= ≈

0: 3 0A D C BM hF hF PxΣ = + − =

2 2 59 4 0,

2kh kh hP P khθ θ+ − = =

Data: 2.0 kip/in. 2ft 24 in.k h= = =

5

(2.0)(24)2

P = 120.0 kipsP =

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

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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 610.6 10 psiE = × .

SOLUTION

Solid: 4

2 4

4 4 2 64 oo

S o sd

A d I dπ π π = = =

Hollow: ( )2 2 23 3

4 4 4 4H o i S oA d d A dπ π= − = =

2 21 10.5 in.

4 2i o i od d d d= = =

Solid rod: 4 4(1.0) 0.049087 in64SIπ= =

2 2 6

3cr 2 2

(10.6 10 )(0.049087)12.839 10 lb

(20)SEI

PL

π π ×= = = ×

Hollow rod: ( )4

4 4 4 41(1) 0.046019 in

64 64 2

π π = − = − =

H o iI d d

2 2 6

3cr 2 2

(10.6 10 )(0.046019)12.036 10 lb 12.04 kips

(20)HEI

PL

π π ×= = = × =

(a) 3 3

3

12.839 10 12.036 100.0625

12.839 10S H

S

P P

P

− × − ×= =×

Percent reduction = 6.25%

(b) For the hollow rod, cr 12.04 kipsP =

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

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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).

41

12aI d=

2 2 4

cr, 2 212a

e e

EI EdP

L L

π π= =

Arrangement (b). 3

3min

3 4

1 1( ) ( )

12 3 12 3

1 19( )

12 3 324

yd d

I I d d

dd d

= = +

+ =

2 2 4

cr, 2 2

19

324 b

e e

EI EdP

L L

π π= =

cr,

cr,

1 324 27

12 19 19a

b

P

P= ⋅ = cr,

cr,

1.421a

b

P

P=

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

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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

200 GPa,E = determine the allowable centric load for the member if a factor of safety of 2.2 is required.

SOLUTION

Angle L152 × 102 × 12.7: 2

6 4 6 4

3060 mm

7.20 10 mm 2.59 10 mm

50.3 mm 24.9 mm

x y

A

I I

y x

=

= × = ×

= =

Two angles: 6 6 4(2)(7.20 10 ) 14.40 10 mmxI = × = ×

6 2 6 42[(2.59 10 ) (3060)(24.9) ] 8.975 10 mmyI = × + = ×

6 4 6 4min 8.975 10 mm 8.975 10 myI I −= = × = ×

2 2 9 6

3cr 2 2

(200 10 )(8.975 10 )361.5 10 N 361.5 kN

(7.0)e

EIP

L

π π −× ×= = = × =

crall

361.5

. . 2.2

PP

F S= = all 164.0 kNP =

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,

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PROBLEM 10.20

Knowing that 5.2 kN,P = determine the factor of safety for the structure shown. Use 200E = GPa and consider only buckling in the plane of the structure.

SOLUTION

Joint B: From force triangle,

5.2

sin 25 sin 20 sin 135

3.1079 kN (Comp)

2.5152 kN (Comp)

BCAB

AB

BC

FF

F

F

= =° ° °

==

Member AB: 4 4

3 4

9 4

185.153 10 mm

4 2 4 2

5.153 10 m

ABd

Iπ π

−

= = = ×

= ×

2 2 9 9

,cr 2 2

3

,cr

(200 10 )(5.153 10 )

(1.2)

7.0636 10 N 7.0636 kN

7.0636. . 2.27

3.1079

ABAB

AB

AB

AB

EIF

L

FF S

F

π π −× ×= =

= × =

= = =

Member BC: 4 4

22

4 2 4 2BCd

Iπ π = =

3 4 9 4

2 2 2 2

2 2 9 9

,cr 2

3

,cr

11.499 10 mm 11.499 10 m

1.2 1.2 2.88 m

(200 10 )(11.499 10 )

2.88

7.8813 10 N 7.8813 kN

7.8813. . 3.13

2.5152

BC

BCBC

BC

BC

BC

L

EIF

L

FF S

F

π π

−

−

= × = ×

= + =

× ×= =

= × =

= = =

Smallest F.S. governs. . . 2.27F S =

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

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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 629 10E = × psi.

SOLUTION

4

4

W10 22: 118 in

11.4 in

x

y

I

I

× =

=

3

3 3cr

15 10 lb

( . .) (2.2)(15 10 ) 33 10 lb

P

P F S P

= ×

= = × = ×

Buckling in xz-plane. 0.7eL L=

2

cr 2cr0.7(0.7 )

y yEI EIP L

PL

π π= =

6

3

(29 10 )(11.4)449.2 in.

0.7 33 10L

π ×= =×

Buckling in yz-plane. 2eL L=

2

cr 2

6

3cr

(2 )

(29 10 )(118)505.8 in.

2 2 33 10

x

x

EIP

L

EIL

P

π

π π

=

×= = =×

Smaller value for L governs. 449.2 in.L = 37.4 ftL =

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,

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PROBLEM 10.27

Column ABC has a uniform rectangular cross section with 12b = mm and 22d = 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 200E = GPa.

SOLUTION

3 3cr

2

cr 2cr

( . .) (3.2)(3.8 10 ) 12.16 10 N

ee

P F S P

EI EIP L

PL

π π

= = × = ×

= =

Buckling in xz-plane. cr

eEI

L LP

π= =

3 3 3 4

9 4

1 1(22)(12) 3.168 10 mm

12 12

3.168 10 m

I db

−

= = = ×

= ×

9 9

3

(200 10 )(3.168 10 )0.717 m

12.16 10L π

−× ×= =×

Buckling in yz-plane. cr

22 2e

e

L EIL L L

P

π= = =

3 3 3 4

9 4

9 9

3

1 1(12)(22) 10.648 10 mm

12 12

10.648 10 m

(200 10 )(10.648 10 )0.657 m

2 12.16 10

I bd

Lπ

−

−

= = = ×

= ×

× ×= =×

The smaller length governs. 0.657 mL = 657 mmL =