재료역학+1장+솔루션 - 비어의 재료역학 1장 솔루션 PDF

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CCHHAAPPTTEERR PROPRIETARY MATERIAL. © 2013 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...

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CHAPTER 1

PROBLEM 1.1 Two solid cylindrical rods AB and BC are welded together at B and loaded as shown. Knowing that the average normal stress must not exceed 175 MPa in rod AB and 150 MPa in rod BC, determine the smallest allowable values of d1 and d2.

SOLUTION (a)

Rod AB 3 P = 40 + 30 = 70 kN = 70 × 10 N

σ AB = d1 =

(b)

P = AAB

P 4P = 2 d π d12 4 1 π

4P

=

πσ AB

(4)(70 × 103)

π (175 × 106 )

= 22.6 × 10−3 m

d1 = 22.6 mm 

= 15.96 × 10−3 m

d2 = 15.96 mm 

Rod BC P = 30 kN = 30 × 103 N

σ BC = d2 =

P = ABC 4P

πσ BC

P 4P = 2 d π d 22 4 2 π

=

(4)(30 × 103 ) 6

π (150 × 10 )

PROPRIETARY MATERIAL. © 2013 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 1.2 Two solid cylindrical rods AB and BC are welded together at B and loaded as shown. Knowing that d1 = 50 mm and d 2 = 30 mm, find the average normal stress at the midsection of (a) rod AB, (b) rod BC.

SOLUTION (a)

Rod AB P = 40 + 30 = 70 kN = 70 × 103 N A =

σ AB =

(b)

π 4

d 12 =

π 4

(50)2 = 1.9635 × 103 mm2 = 1.9635 × 10−3 m2

70 × 103 P = = 35.7 × 106 Pa A 1.9635 × 10 −3

σ AB = 35.7 MPa 

Rod BC P = 30 kN = 30 × 103 N A =

σ BC =

π 4

d 22 =

π 4

(30)2 = 706.86 mm2 = 706.86 × 10 −6 m2

P 30 × 103 = = 42.4 × 106 Pa A 706.86 × 10−6

σ BC = 42.4 MPa 

PROPRIETARY MATERIAL. © 2013 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 1.3 Two solid cylindrical rods AB and BC are welded together at B and loaded as shown. Determine the magnitude of the force P for which the tensile stress in rod AB is twice the magnitude of the compressive stress in rod BC.

SOLUTION

σ AB

π

(75)2 = 4417.9 mm2 4 2(120) − P = ABC

A BC =

σ BC

π

(50)2 = 1963.5 mm2 4 P P = = A AB 1963.5

AAB =

=

240 − P 4417.9

Equating σ AB to 2σBC P 2(240 − P ) = 1963.5 4417.9

P = 112.9 kN 

PROPRIETARY MATERIAL. © 2013 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 1.4 In Prob. 1.3, knowing that P = 160 kN, determine the average normal stress at the midsection of (a) rod AB, (b) rod BC. PROBLEM 1.3 Two solid cylindrical rods AB and BC are welded together at B and loaded as shown. Determine the magnitude of the force P for which the tensile stress in rod AB is twice the magnitude of the compressive stress in rod BC.

SOLUTION (a)

Rod AB P = 160 kN (tension)

AAB =

σ AB = (b)

2 π d AB

4

=

π 4

(50) 2 = 1963.5 mm 2

3 P 160 ×10 N = AAB 1963.5 × 10 −6 m2

σAB = 81.5 MPa 

Rod BC F = 160 − 2(120) = − 80 kN i.e., 80 kN compression. ABC =

σ BC =

πd 2BC 4

=

π 4

(75)2 = 4417.9 mm2

F −80 ×103 N = ABC 4417.9 × 10−6 m2

σBC = − 18.11 MPa 

PROPRIETARY MATERIAL. © 2013 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 1.5 Two steel plates are to be held together by means of 16-mmdiameter high-strength steel bolts fitting snugly inside cylindrical brass spacers. Knowing that the average normal stress must not exceed 200 MPa in the bolts and 130 MPa in the spacers, determine the outer diameter of the spacers that yields the most economical and safe design.

SOLUTION At each bolt location the upper plate is pulled down by the tensile force Pb of the bolt. At the same time, the spacer pushes that plate upward with a compressive force Ps in order to maintain equilibrium.

Pb = Ps

For the bolt,

σb =

Fb 4 Pb = Ab π d b2

or

Pb =

For the spacer,

σs =

Ps 4Ps = As π (ds 2 − db 2)

or

Ps =

π 4

π 4

σb db2 σs ( ds2 − db2 )

Equating Pb and Ps ,

π 4

σ bd 2b = ds =

π 4

σ s (d 2s − d 2b )

 σ  1 + b  db = σs  

200   1 +  (16) 130  

d s = 25.2 mm 

PROPRIETARY MATERIAL. © 2013 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 1.6 Two brass rods AB and BC, each of uniform diameter, will be brazed together at B to form a nonuniform rod of total length 100 m, which will be suspended from a support at A as shown. Knowing that the density of brass is 8470 kg/m3, determine (a) the length of rod AB for which the maximum normal stress in ABC is minimum, (b) the corresponding value of the maximum normal stress.

SOLUTION Areas:

AAB = ABC =

From geometry, Weights:

π 4

π 4

(15 mm) 2 = 176.71 mm 2 = 176.71 × 10− 6m 2 (10 mm) 2 = 78.54 mm2 = 78.54 × 10− 6 m2

b = 100 − a WAB = ρg AAB AB = (8470)(9.81)(176.71 × 10 −6 ) a = 14.683 a WBC = ρ g ABC BC = (8470)(9.81)(78.54 × 10 −6)(100 − a) = 652.59 − 6.526 a

Normal stresses: PA = W AB + W BC = 652.59 + 8.157a

At A,

σA

PB = W BC = 652.59 − 6.526a

At B,

σB = (a)

(1)

P = A = 3.6930 × 106 + 46.160 × 103 a AAB

(2)

PB = 8.3090 × 10 6 − 83.090 × 10 3a ABC

Length of rod AB. The maximum stress in ABC is minimum when σ A = σ B or 4.6160 × 106 − 129.25 × 103 a = 0 a = 35.71 m

(b)

 AB = a = 35.7 m 

Maximum normal stress.

σ A = 3.6930 × 10 6 + (46.160 × 10 3)(35.71) σ B = 8.3090 × 10 6 − (83.090 × 10 3)(35.71) σ A = σB = 5.34 × 106 Pa

σ = 5.34 MPa 

PROPRIETARY MATERIAL. © 2013 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 1.7 Each of the four vertical links has an 8 × 36-mm uniform rectangular cross section and each of the four pins has a 16-mm diameter. Determine the maximum value of the average normal stress in the links connecting (a) points B and D, (b) points C and E.

SOLUTION Use bar ABC as a free body.

ΣMC = 0 :

(0.040) FBD − (0.025 + 0.040)(20 × 10 3) = 0 3 FBD = 32.5 × 10 N

Link BD is in tension. 3

ΣMB = 0 : −(0.040) FCE − (0.025)(20 × 10 ) = 0 3 FCE = − 12.5 × 10 N

Link CE is in compression. .

For two parallel links, (a)

σ BD =

A net = 320 × 10 − 6m

2

FBD 32.5 × 103 = = 101.56 × 106 −6 Anet 320 × 10

σ BD = 101.6 MPa 

Area for one link in compression = (0.008)(0.036)= 288 × 10− 6m 2. For two parallel links, (b)

σ CE =

A = 576 × 10− 6m

FCE −12.5 × 10 3 = = −21.70 × 10 −6 −6 A 576 × 10

2

σ CE = −21.7 MPa 

PROPRIETARY MATERIAL. © 2013 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 1.8 Knowing that link DE is 25 mm wide and 3 mm thick, determine the normal stress in the central portion of that link when (a) θ = 0, (b) θ = 90°.

SOLUTION Use member CEF as a free body ΣM C = 0: − 0.3FDE − (0.2)(240 sinθ ) − (0.4)(240 cosθ ) = 0

FDE = − 160sin θ − 320 cos θ N ADE = (0.025)(0.003) = 75 × 10−6 m2

σ DE = (a)

(b)

FDE A DE

θ = 0: FDE = − 320 N σ DE =

− 320 = − 4.27 MPa  75 × 10−6

σ DE =

− 160 = − 2.13 MPa  75 × 10−6

θ = 90° : FDE = − 160 N

PROPRIETARY MATERIAL. © 2013 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 1.9 Knowing that the central portion of the link BD has a uniform cross-sectional area of 800 mm2, determine the magnitude of the load P for which the normal stress in that portion of BD is 50 MPa.

SOLUTION

F BD = σ A = (50 ×106 )(800 ×10 −6 ) = 40 ×103 N 2 2 BD = (0.56) + (1.92)

= 2.00 m

Use Free Body AC for statics. Σ MC = 0:

0.56 1.92 (40 × 103 )(1.4) + (40 × 103 )(1.4) − P (0.7 + 1.4) = 0 2.00 2.00 P = 33.1 × 10 3 N

P = 33.1 kN 

PROPRIETARY MATERIAL. © 2013 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 1.10 Three forces, each of magnitude P = 4 kN, are applied to the mechanism shown. Determine the cross-sectional area of the uniform portion of rod BE for which the normal stress in that portion is +100 MPa.

SOLUTION Draw free body diagrams of AC and CD.

Free Body CD:

ΣM D = 0: 0.150P − 0.250C = 0 C = 0.6P

Free Body AC:

Required area of BE:

M A = 0: 0.150FBE − 0.350P − 0.450P − 0.450C = 0 FBE =

1.07 P = 7.1333 P = (7.133)(4 kN) = 28.533 kN 0.150

σ BE =

FBE ABE

ABE =

FBE

σ BE

=

28.533 × 103 100 × 10 6

= 285.33 × 10−6 m2 ABE = 285 mm2 

PROPRIETARY MATERIAL. © 2013 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 1.11 The rigid bar EFG is supported by the truss system shown. Knowing that the member CG is a solid circular rod of 18 mm diameter, determine the normal stress in CG.

SOLUTION Using portion EFGCB as a free body 0.9 1.5 = 25 kN

Σ Fy = 0: F AE

FAB − 15 = 0

Using beam EFG as a free body M F = 0: − (1.2)

0.9  0.9  F + (1.2)  FCG  = 0 1.2 AE 1.2  

FCG = FAE = 25 kN

Cross sectional area of member CG ACG =

π 4

d2 =

π 4

(0.018)2 = 254.4 × 10−6 m2

Normal stress in CG.

σCG =

FCG 25 = = 98.3 MPa ACG 254.4 × 10−6



PROPRIETARY MATERIAL. © 2013 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 1.12 The rigid bar EFG is supported by the truss system shown. Determine the cross-sectional area of member AE for which the normal stress in the member is 105 MPa.

SOLUTION Using portion EFGCB as a free body 0.9 1.5 = 25 kN

Σ Fy = 0: F AE

FAE − 15 = 0

Stress in member AE

σ AE = A AE =

σ A E = 105 MPa

FAE AAE FAC

σA E

=

25 × 103 = 238.1 × 10 −6 m 2 105 × 10−6



PROPRIETARY MATERIAL. © 2013 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 1.13 An aircraft tow bar is positioned by means of a single hydraulic cylinder connected by a 25-mm-diameter steel rod to two identical arm-and-wheel units DEF. The mass of the entire tow bar is 200 kg, and its center of gravity is located at G. For the position shown, determine the normal stress in the rod.

SOLUTION 

FREE BODY – ENTIRE TOW BAR:

W = (200 kg)(9.81 m/s2 ) = 1962.00 N Σ M A = 0 : 850R − 1150(1962.00 N) = 0 R = 2654.5 N

 

FREE BODY – BOTH ARM & WHEEL UNITS:

tan α =

100 675

α = 8.4270°

Σ M E = 0 : (FCD cos α)(550) − R (500) = 0 FCD =

500 (2654.5 N) 550 cos 8.4270°

= 2439.5 N (comp.)



σCD = −

FCD 2439.5 N =− ACD π (0.0125 m)2

= −4.9697 × 10 6 Pa

σ CD = − 4.97 MPa 

PROPRIETARY MATERIAL. © 2013 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 1.14 A couple M of magnitude 1500 N ⋅ m is applied to the crank of an engine. For the position shown, determine (a) the force P required to hold the engine system in equilibrium, (b) the average normal stress in the connecting rod BC, which has a 450-mm2 uniform cross section.

SOLUTION

Use piston, rod, and crank together as free body. Add wall reaction H and bearing reactions Ax and Ay. Σ M A = 0 : (0.280 m)H − 1500 N ⋅ m = 0 3 H = 5.3571 × 10 N

Use piston alone as free body. Note that rod is a two-force member; hence the direction of force FBC is known. Draw the force triangle and solve for P and FBE by proportions. l =

2002 + 60 2 = 208.81 mm

P 200 = H 60



∴

P = 17.86 × 103 N

(a)

P = 17.86 kN 

208.81 FBC = ∴ FBC = 18.643× 103 N 60 H

Rod BC is a compression member. Its area is 450 mm2 = 450 × 10− 6 m2

Stress,

σ BC = 

−FBC −18.643 × 103 = = − 41.4 × 106 Pa A 450 × 10 −6

(b)

σ BC = − 41.4 MPa 

PROPRIETARY MATERIAL. © 2013 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 1.15 When the force P reached 8 kN, the wooden specimen shown failed in shear along the surface indicated by the dashed line. Determine the average shearing stress along that surface at the time of failure.

SOLUTION Area being sheared:

A = 90 mm × 15 mm = 1350 mm 2 = 1350× 10− 6 m 2

Force:

P = 8 × 103 N

Shearing stress:

τ =

8 × 103 P − = 5.93 × 106 Pa A 1350 × 10 −6

τ = 5.93 MPa 

PROPRIETARY MATERIAL. © 2013 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 1.16 Two wooden planks, each 12 mm thick and 225 mm wide, are joined by the dry mortise joint shown. Knowing that the wood used shears off along its grain when the average shearing stress reaches 8 MPa, determine the magnitude P of the axial load which will cause the joint to fail.

SOLUTION Six areas must be sheared off when the joint fails. Each of these areas has dimensions 16 mm × 12 mm, its area being A = (16)(12) = 192 mm2 = 192 × 10 −6 m2

At failure the force F carried by each of areas is F = τ A = (8 × 10 6 )(192 × 10− 6 ) = 1536 N = 1.536 kN

Since there are six failure areas

P = 6 F = (6)(1.536) = 9.22 kN



PROPRIETARY MATERIAL. © 2013 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.

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