C
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 Pcr.
SOLUTION
Let θ be the angle change of bar AB. sin
F=kx kL= θ 2
0: cos 0 sin cos sin 0
B M FL Px kL PL θ θ θ θ Σ = − = − = Using 2
sinθ θ≈ and cosθ≈1, kLθ−PLθ = 0 2
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.
, sin M =Kθ x L= θ ≈Lθ 0: 0 0 B M = M−Px= Kθ−PLθ= (K PL− ) θ = 0 Pcr =K L/
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 10.3
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.
B M =Kθ 0: 0 B A A M K F L K F L θ θ = − = = Bar AC. 0: cr 1 1 0 2 2 C A M P Lθ LF Σ = − = cr A F P θ = cr K P L =
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 0 : 0 3 C A A P M LR P R L δ δ = − + = =
Using free body BC and 2 0: 0 3 C B M LR Pδ Σ = − = 3 2 B P R L δ = Using both free bodies together,
0: 0 3 3 0 2 9 0 2 x A B F R R k P P k L L P k L δ δ δ δ δ Σ = + − = + − = − = cr 2 9 kL 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 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 10.7
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.
3 4 D C B x h x h x h θ θ θ ≈ ≈ ≈ 3 C C D D F kx kh F kx kh θ θ = ≈ = ≈ 0: 3 0 A D C B M hF hF Px Σ = + − = 2 9 2 4 0, 5 2 khθ+ khθ− hP= P= kh
Data: k=2.0 kip/in. h=2ft 24 in.= 5
(2.0)(24) 2
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 10 psi× 6 .
SOLUTION Solid: 4 2 4 4 4 2 64 o o S o s d A = πd I = π = π d Hollow:
(
2 2)
3 3 2 4 4 4 4 H o i S o A = π d −d = A = πd 2 1 2 1 0.5 in. 4 2 i o i o d = d d = d = Solid rod: (1.0)4 0.049087 in4 64 S I = π = 2 2 6 3 cr 2 2 (10.6 10 )(0.049087) 12.839 10 lb (20) S EI P L π π × = = = × Hollow rod:(
)
4 4 4 (1)4 1 0.046019 in4 64 64 2 π π = − = − = H o i I d d 2 2 6 3 cr 2 2 (10.6 10 )(0.046019) 12.036 10 lb 12.04 kips (20) H EI P L π π × = = = × = (a) 3 3 3 12.839 10 12.036 10 0.0625 12.839 10 S H S P P P − = × − × = × Percent reduction = 6.25% (b) For the hollow rod, Pcr =12.04 kips
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 10.13
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). 4 1 12 a I = d 2 2 4 cr,a 2 12 2 e e EI Ed P L L π π = = Arrangement (b). 3 3 min 3 4 1 1 ( ) ( ) 12 3 12 3 1 19 ( ) 12 3 324 y d d I I d d d d d = = + + = 2 2 4 cr, 2 2 19 324 b e e EI Ed P L L π π = = cr, cr, 1 324 27 12 19 19 a b P P = ⋅ = cr, cr, 1.421 a b P P =
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: (2)(7.20 10 )6 14.40 10 mm6 4 x I = × = × 6 2 6 4 2[(2.59 10 ) (3060)(24.9) ] 8.975 10 mm y I = × + = × 6 4 6 4 min y 8.975 10 mm 8.975 10 m I = I = × = × − 2 2 9 6 3 cr 2 2 (200 10 )(8.975 10 ) 361.5 10 N 361.5 kN (7.0) e EI P L π π × × − = = = × = cr all 361.5 . . 2.2 P P F S = = Pall =164.0 kN
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 10.20
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,
5.2 sin 25 sin 20 sin 135
3.1079 kN (Comp) 2.5152 kN (Comp) BC AB AB BC F F F F = = ° ° ° = = Member AB: 4 4 3 4 9 4 18 5.153 10 mm 4 2 4 2 5.153 10 m AB d I π 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 AB AB AB AB AB EI F L F F S F π π × × − = = = × = = = = Member BC: 4 22 4 4 2 4 2 BC d I =π =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 BC BC BC BC BC L EI F L F F S F π π − − = × = × = + = × × = = = × = = = = Smallest F.S. governs. F S. . 2.27=
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 10× 6psi. SOLUTION 4 4 W10 22: 118 in 11.4 in x y I I × = = 3 3 3 cr 15 10 lb ( . .) (2.2)(15 10 ) 33 10 lb P P F S P = × = = × = × Buckling in xz-plane. Le=0.7L 2 cr 2 cr 0.7 (0.7 ) y y EI EI P L P L π π = = 6 3 (29 10 )(11.4) 449.2 in. 0.7 33 10 L= π × = × Buckling in yz-plane. Le=2L 2 cr 2x EI P =π
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 10.27
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 3 3 cr 2 cr 2 cr ( . .) (3.2)(3.8 10 ) 12.16 10 N e e P F S P EI EI P L P L π π = = × = × = = Buckling in xz-plane. cr e EI L L P 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 10 L=π × × − = × Buckling in yz-plane. cr 2 2 2 e e L EI L 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 L π − − = = = × = × × × = = ×
