Mechanics of Materials 7th Edition Beer Solution Manual chapter 1 PDF

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CHAPTER 1

d1

d2

125 kN B

PROBLEM 1.1 C

A

Two solid cylindrical rods AB and BC are welded together at B and loaded as shown. Knowing that d1  30 mm and d2  50 mm, find the average normal stress at the midsection of (a) rod AB, (b) rod BC.

60 kN 125 kN 0.9 m

1.2 m

SOLUTION (a)

Rod AB: Force:

P  60  10 3 N tension

Area:

A

Normal stress: (b)

 AB 

 4

d12 

 4

(30  10 3) 2  706.86  10 6 m 2

P 60  103   84.882  106 Pa A 706.86  10 6

AB  84.9 MPa 

Rod BC: Force:

P  60  10 3  (2)(125  10 3)   190  10 3 N

Area:

A

Normal stress:

BC 

 4

d 22 

 4

(50  10 3) 2  1.96350  10 3 m 2

P 190  103   96.766  106 Pa A 1.96350  10 3

BC  96.8 MPa 

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d1

d2

125 kN B

PROBLEM 1.2 C

A

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 150 MPa in either rod, determine the smallest allowable values of the diameters d1 and d2.

60 kN 125 kN 0.9 m

1.2 m

SOLUTION (a)

Rod AB: Force:

P  60  10 3 N

Stress:

 AB  150  10 6 Pa  2 A 

Area:

 AB  4

d1 4 P P   A  AB A

d12  d12 

P

 AB 4P

 AB



(4)(60  103 )  509.30  106 m2  (150  106 )

3 d1  22.568  10  m

(b)

d1  22.6 mm 

Rod BC: Force: Stress: Area:

P  60  10 3  (2)(125  10 3)  190  10 3 N

 BC  150  106 Pa  2 A

 BC

d2 4 P 4P   A  d22

d 22 

4P

 BC



(4)( 190  103 )  1.61277  103 m2 6  ( 150  10 )

d2  40.159  10 3 m

d2  40.2 mm 

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PROBLEM 1.3 A

Two solid cylindrical rods AB and BC are welded together at B and loaded as shown. Knowing that P = 10 kips, find the average normal stress at the midsection of (a) rod AB, (b) rod BC.

30 in. 1.25 in. B 12 kips 25 in. 0.75 in. C

P

SOLUTION (a)

Rod AB: P  12  10  22 kips A 

 AB (b)



d 12 



(1.25)2  1.22718 in 2 4 4 P 22    17.927 ksi A 1.22718

AB  17.93 ksi 

Rod BC: P  10 kips A

 AB





d 2  (0.75) 2  0.44179 in 2 4 2 4 P 10    22.635 ksi A 0.44179

 AB  22.6 ksi 

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PROBLEM 1.4 A

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 stresses in rods AB and BC are equal.

30 in. 1.25 in. B 12 kips 25 in. 0.75 in. C

P

SOLUTION (a)

Rod AB: P  P  12 kips A

d 2 4



 4

(1.25 in.) 2

A  1.22718 in2

 AB  (b)

P  12 kips 1.22718 in2

Rod BC: P P A

 4

d2 

 4

(0.75 in.)2

A  0.44179 in 2

 BC 

P 0.44179 in2

 AB   BC P  12 kips P  2 1.22718 in 0.44179 in2 5.3015  0.78539P

P  6.75 kips 

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1200 N

PROBLEM 1.5 A strain gage located at C on the surface of bone AB indicates that the average normal stress in the bone is 3.80 MPa when the bone is subjected to two 1200-N forces as shown. Assuming the cross section of the bone at C to be annular and knowing that its outer diameter is 25 mm, determine the inner diameter of the bone’s cross section at C.

A

C

B 1200 N

SOLUTION

  Geometry:

A

 4

P P  A  A

(d12  d 22) d 22  d 12 

4A



 d 12 

4P



(4)(1200) d 22  (25  10  3) 2  (3.80  10 6)  222.92  10 6 m 2 d 2  14.93  103 m

d2  14.93 mm 

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PROBLEM 1.6

A a 15 mm B 100 m

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.

b 10 mm

C

SOLUTION AAB 

Areas:

ABC 

4

 4

(15 mm)2  176.715 mm2  176.715 10 6 m2 (10 mm)2  78.54 mm2  78.54 10 6 m2

b  100  a

From geometry, Weights:



WAB  g AAB AB  (8470)(9.81)(176.715  10 6)a  14.683a WBC   g ABC BC  (8470)(9.81)(78.54 10 6)(100  a)  652.59  6.526 a

Normal stresses: At A,

PA  WAB  WBC  652.59  8.157a

A  At B,

(a)

PA  3.6930  106  46.160  103 a A AB

PB  WBC  652.59  6.526a

B 

(1)

(2)

PB  8.3090  106  83.090  103a ABC

Length of rod AB. The maximum stress in ABC is minimum when  A   B or 4 .6160  10 6  129.25  10 3a  0

a  35.71 m (b)

 AB  a  35.7 m 

Maximum normal stress.

 A  3.6930  106  (46.160  103 )(35.71)  B  8.3090  106  (83.090  103 )(35.71)  A   B  5.34  106 Pa

  5.34 MPa 

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PROBLEM 1.7 0.4 m C 0.25 m

0.2 m

B

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.

E

20 kN D A

SOLUTION Use bar ABC as a free body.

MC  0 :

(0.040) FBD  (0.025  0.040)(20  103 )  0 FBD  32.5  103 N

Link BD is in tension.

MB  0 :  (0.040) FCE  (0.025)(20  10 )  0 3

FCE   12.5  10 3 N

Link CE is in compression.

Net area of one link for tension  (0.008)(0.036  0.016) 160  10 6 m 2 For two parallel links, (a)

 BD 

A net  320  10 6 m2

FBD 32.5  103   101.563  106 Anet 320  106

 BD  101.6 MPa 

Area for one link in compression  (0.008)(0.036)  288  10 6 m 2 For two parallel links, (b)

CE 

A  576  10  6 m 2

12.5  10 3 FCE   21.701  10  6 A 576  106

 CE  21.7 MPa 

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PROBLEM 1.8 B

2 in.

Link AC has a uniform rectangular cross section

12 in.

1 8

in. thick and 1 in. wide.

Determine the normal stress in the central portion of the link.

120 lb 4 in. 30⬚

120 lb

A C 10 in.

8 in.

SOLUTION Use the plate together with two pulleys as a free body. Note that the cable tension causes at 1200 lb-in. clockwise couple to act on the body.

MB  0:  (12  4)(FAC cos30 )  (10)(FAC sin 30 )  1200 lb  0 FAC  

1200 lb  135.500 lb 16 cos30  10 sin 30

Area of link AC: Stress in link AC:

1 in.  0.125 in 2 8 135.50 F  AC    1084 psi  1.084 ksi A 0.125

A  1 in. 

AC



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0.100 m

PROBLEM 1.9

E P

P

P D

A

0.150 m

B

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.

C

0.300 m

0.250 m

SOLUTION Draw free body diagrams of AC and CD.

Free Body CD:

MD  0: 0.150 P  0.250C  0 C  0.6 P

Free Body AC:

Required area of BE:

MA  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

 BE 

FBE ABE

ABE 

FBE

 BE



28.533  103  285.33  106 m2 100  106 2 ABE  285 mm 

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4 kips C

6i

n.

Link BD consists of a single bar 1 in. wide and 1 in. thick. Knowing that each pin has a 83 -in. 2

B

i 12

A

PROBLEM 1.10

u

diameter, determine the maximum value of the average normal stress in link BD if (a)  = 0, (b)  = 90.

n.

308

D

SOLUTION Use bar ABC as a free body.

(a)

  0.  M A  0: (18 sin 30 )(4)  (12 cos30 )FBD  0 FBD  3.4641 kips (tension)

Area for tension loading: Stress:

(b)

3 1   A  ( b  d ) t   1     0.31250 in 2 8 2  FBD 3.4641 kips    A 0.31250 in 2

  11.09 ksi 

  90 .  M A  0:  (18 cos30 )(4)  (12 cos30 )FBD  0 FBD   6 kips i.e. compression.

Area for compression loading: Stress:

1  A  bt  (1)    0.5 in 2  2 6 kips F   BD  A 0.5 in 2

  12.00 ksi 

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B

D

PROBLEM 1.11

F

12 ft H

A C 9 ft

E 9 ft

80 kips

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.

G 9 ft

80 kips

9 ft 80 kips

SOLUTION 

Use entire truss as free body.  M H  0: (9)(80)  (18)(80)  (27)(80)  36 Ay  0 Ay  120 kips Use portion of truss to the left of a section cutting members BD, BE, and CE.

 

   Fy  0: 120  80 

 BE 

12 FBE  0 15

 FBE  50 kips

FBE 50 kips  A 5.87 in2

 BE  8.52 ksi 



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PROBLEM 1.12 45 in.

A

B

30 in. C

480 lb 4 in.

4 in.

40 in.

D

E 15 in.

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 21-in. diameter, determine the maximum value of the average normal stress (a) in member BE, (b) in member CF.

F 30 in.

SOLUTION 

Add support reactions to figure as shown. 

Using entire frame as free body,  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 F CF. Dy 

4 D  1200 lb 3 x

4   MF  0:  (30)  FBE   (30  15) DY  0 5  FBE   2250 lb 4   ME  0: (30) FCE   (15) DY  0 5   FCE  750 lb Stress in compression member BE: A  2 in.  4 in.  8 in 2

Area: (a)

 BE 

FBE 2250  A 8

 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)

 CF 

FCF 750  7.0 A min

 CF 107.1 psi 

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PROBLEM 1.13

Dimensions in mm 1150

D 100 C

G A

B F

850

250

E

500

450

675

825

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/s 2)  1962.00 N  M A  0: 850R  1150(1962.00 N)  0 R  2654.5 N

 

FREE BODY – BOTH ARM & WHEEL UNITS:

tan  

100 675

  8.4270

M E  0: ( FCD cos  )(550)  R(500)  0 FCD 

500 (2654.5 N) 550 cos 8.4270

 2439.5 N (comp.)



CD  

FCD 2439.5 N  ACD  (0.0125 m)2

 4.9697  106 Pa

 CD   4.97 MPa 

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PROBLEM 1.14

150 mm 300 mm A

Two hydraulic cylinders are used to control the position of the robotic arm ABC. Knowing that the control rods attached at A and D each have a 20-mm diameter and happen to be parallel in the position shown, determine the average normal stress in (a) member AE, (b) member DG.

C B

400 mm E

800 N

600 mm

D F

150 mm

G

200 mm

SOLUTION Use member ABC as free body.

 M B  0: (0.150)

4 F AE  (0.600)(800)  0 5

FAE  4  103 N

Area of rod in member AE is A  Stress in rod AE:

 AE 

 4

d2

 4

(20  10  3) 2  314.16  10  6m 2

FAE 4  10 3   12.7324  10 6 Pa A 314.16  10 6

(a)

 AE  12.73 MPa 

Use combined members ABC and BFD as free body. 4  4   M F  0: (0.150)  F AE   (0.200) F DG   (1.050  0.350)(800)  0 5 5     FDG   1500 N

Area of rod DG: Stress in rod DG:

A 

 4

d2 

 DG 

 4

(20  10  3) 2  314.16  10 6 m 2

FDG 1500    4.7746  10 6 Pa A 3.1416  106 (b)

 DG   4.77 MPa 

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PROBLEM 1.15 Determine the diameter of the largest circular hole that can be punched into a sheet of polystyrene 6 mm thick, knowing that the force exerted by the punch is 45 kN and that a 55-MPa average shearing stress is required to cause the material to fail.

SOLUTION For cylindrical failure surface:

A   dt

Shearing stress:



Therefore, Finally,

P



P A

or

A

P



  dt

d 

P  t 45  103 N (0.006 m)(55  106 Pa)

 43.406  10 3 m d  43.4 mm 

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