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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 P = 40 + 30 = 70 kN = 70 × 103 N

(b)

σ AB =

P = AAB

d1 =

4P

P 4P = 2 d π d12 4 1 π

=

πσ AB

(4)(70 × 103) = 22.6 × 10−3 m 6 × π (175 10 )

d1 = 22.6 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 ) = 15.96 × 10 −3 m 6 × π (150 10 )

d2 = 15.96 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.

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

d1 2 =

π 4

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

P 70 × 103 = = 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. © 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 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 AAB =

σ AB

π

(2)2 = 3.1416 in2 4 P P = = AAB 3.1416 = 0.31831P

ABC =

σ BC

π

(3)2 = 7.0686 in2 4 (2)(30) − P = AAB =

60 − P = 8.4883 − 0.14147 P 7.0686

Equating σ AB to 2σ BC 0.31831P = 2(8.4883 − 0.14147 P )

P = 28.2 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 1.4 In Prob. 1.3, knowing that P = 40 kips, 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 = 40 kips (tension)

A AB =

σ AB (b)

πd 2AB

=

π (2)2

4 4 P 40 = = 3.1416 AAB

= 3.1416 in2

σ AB = 12.73 ksi 

Rod BC F = 40 − (2)(30) = −20 kips, i.e., 20 kips compression.

ABC =

σ BC

2 π d BC

=

π (3) 2

4 4 −20 F = = ABC 7.0686

= 7.0686 in 2

σ BC = −2.83 ksi 

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 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 (d s2 − db2 )

or

Ps =

π 4

π 4

σ b db2 σ s (d 2s − d b2 )

Equating Pb and Ps ,

π 4

σ bd b2 = ds =

π 4

σ s (d s2 − d b2 )  σb  1 + d = σs  b 

200   1 + 130 (16)  

d s = 25.2 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.

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 AAB =

Areas:

ABC =

From geometry, Weights:

π 4

π 4

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

(10 mm) 2 = 78.54 mm2 = 78.54 × 10 6 m 2

b = 100 − a − WAB = ρ g 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: At A,

P A = W AB + W BC = 652.59 + 8.157a

σA = At B,

(a)

PA = 3.6930 × 106 + 46.160 × 103 a AAB

PB = W BC = 652.59 − 6.526a

σB =

(1)

(2)

PB = 8.3090 × 106 − 83.090 × 103 a 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 × 106 + (46.160 × 103 )(35.71) σ B = 8.3090 × 106 − (83.090 × 103 )(35.71) σ A = σ B = 5.34 × 106 Pa

σ = 5.34 MPa 

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 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 × 103 ) = 0 FBD = 32.5 × 103 N

Link BD is in tension.

ΣMB = 0 : −(0.040) FCE − (0.025)(20 × 103 ) = 0 FCE = − 12.5 × 103 N

Link CE is in compression.

Net area of one link for tension = (0.008)(0.036 − 0.016) = 160 × 10− 6m 2. For two parallel links, (a)

σ BD =

A net = 320 × 10 −6 m 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 2

−12.5 × 103 FCE = = −21.70 × 10−6 −6 × 576 10 A

σ CE = −21.7 MPa 

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 1.8 Knowing that the link DE is 81 in. thick and 1 in. wide, 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.

Σ M C = 0 : − 12 F DE − (8)(60 sinθ ) − (16)(60 cosθ ) = 0

FDE = − 40 sinθ − 80 cosθ lb.

1 ADE = (1)   = 0.125 in.2 8 F σ DE = DE A DE

(a)

θ = 0: FDE = −80 lb. σDE =

(b)

−80 0.125

σ DE = −640 psi 

θ = 90 °: FDE = −40 lb. σ DE =

−40 0.125

σ DE = −320 psi 

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 1.9 Link AC has a uniform rectangular cross section 161 in. thick and 41 in. wide. Determine the normal stress in the central portion of the link.

SOLUTION Free Body Diagram of Plate

Note that the two 240-lb forces form a couple of moment (240 lb)(6 in.) = 1440 lb ⋅ in.

Σ M B = 0 : 1440 lb ⋅ in − (F AC cos 30° )(10 in.) = 0

FAC = 166.277 lb.

Area of link:

1  1  in.  in. = 0.015625 in.2 AAC =  16 4   

Stress:

σ AC =

FAC 166.277 = = 10640 psi 0.015625 AAC

σ AC = 10.64 ksi 

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 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:

Σ M D = 0: 0.150P − 0.250C = 0

C = 0.6P

Free Body AC:

Required area of BE:

M A = 0: 0.150 FBE − 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 − = 285.33 × 10 6 m2 100 × 10 6

ABE = 285 mm2 

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 1.11 The frame shown consists of four wooden members, ABC, DEF, BE, and CF. Knowing that each member has a 2 × 4-in. rectangular cross section and that each pin has a 1/2-in. diameter, determine the maximum value of the average normal stress (a) in member BE, (b) in member CF.

SOLUTION 

Add support reactions to figure as shown. 

Using entire frame as free body, Σ M A = 0: 40 Dx − (45 + 30)(480) = 0

Dx = 900 lb.



Use member DEF as free body. Reaction at D must be parallel to F BE and FCF . Dy =

4 Dx = 1200 lb. 3

4  ΣM F = 0: − (30)  FBE  − (30 + 15)DY = 0 5  F BE = − 2250 lb. 4  ΣM E = 0: (30) FCE  − (15)DY = 0 5   FCE = 750 lb.

Stress in compression member BE Area: (a)

A = 2 in × 4 in = 8 in 2

σ BE =

− 2250 FBE = 8 A

σ BE = −281 psi 

Minimum section area occurs at pin. Amin = (2)(4.0 − 0.5) = 7.0 in 2

Stress in tension member CF

(b)

σ CF =

FCF 750 = Amin 7.0

σCF =107.1 psi 

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 1.12 For the Pratt bridge truss and loading shown, determine the average normal stress in member BE, knowing that the crosssectional area of that member is 5.87 in2.

SOLUTION 

Use entire truss as free body. ΣM H = 0: (9)(80) + (18)(80) + (27)(80) − 36Ay = 0 A y = 120 kips

Use portion of truss to the left of a section cutting members BD, BE, and CE.

 

+ ↑ Σ F y = 0: 120 − 80 −

σBE =

12 F =0 15 BE

∴ FBE = 50 kips

FBE 50 kips = A 5.87 in 2

σ BE = 8.52 ksi 



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 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 ME = 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 × 106 Pa

σ CD = − 4.97 MPa 

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 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 H = 5.3571 × 10 3 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 × 10 3N

(a)

P = 17.86 kN 

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

Rod BC is a compression member. Its area is 450 mm 2 = 450 × 10− 6m 2

Stress,

σ BC =

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

 (b)

σ BC = −41.4 MPa 

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 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 mm2 = 1350 × 10− 6 m2

Force:

P = 8 × 103 N

Shearing stress:

τ =

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

τ = 5.93 MPa 

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 1.16 The wooden members A and B are to be joined by plywood splice plates, that will be fully glued on the surfaces in contact. As part of the design of the joint, and knowing that the clearance between the ends of the members is to be 41 in., determine the smallest allowable length L if the average shearing stress in the glue is not to exceed 120 psi.

SOLUTION There are four separate areas that are glued. Each of these areas transmits one ha...