Solucionario 1 - 1 A 1 - 28 [PDF]

Zitiervorschau

PROBLEM 1.1 Two solid cylindrical rods AB and BC are welded together at B and loaded as shown. Knowing that d1 30 mm and d 2 50 mm, find the average normal stress at the midsection of (a) rod AB, (b) rod BC.

SOLUTION (a)

Rod AB: Force:

P

Area:

A

Normal stress: (b)

AB

60 103 N tension 4

d12

4

(30 10 3 ) 2

60 103 706.86 10

P A

6

706.86 10 6 m 2 84.882 106 Pa

AB

84.9 MPa

Rod BC: Force:

P

Area:

A

Normal stress:

BC

60 103 4 P A

d 22

(2)(125 103 ) 4

(50 10 3 )2

190 103 1.96350 10

190 103 N 1.96350 10 3 m 2 96.766 106 Pa

3

BC

96.8 MPa

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PROBLEM 1.2 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.

SOLUTION (a)

Rod AB: Force:

P

Stress:

AB

Area:

A AB

4

d12

60 103 N 150 106 Pa d12

4 P A

AB

P AB

d12

(4)(60 103 ) (150 106 )

4P AB

(b)

P

A

d1

22.568 10 3 m

Force:

P

60 103

Stress:

BC

509.30 10 6 m 2 d1

22.6 mm

d2

40.2 mm

Rod BC:

Area:

A BC

d 22

190 103 N

150 106 Pa 4 P A

d 22 4P d 22

4P BC

d2

(2)(125 103 )

(4)( 190 103 ) ( 150 106 )

40.159 10 3 m

1.61277 10 3 m 2

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PROBLEM 1.3 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.

SOLUTION (a)

Rod AB: P A AB

(b)

12 4 P A

10 d12

22 kips

(1.25) 2 1.22718 in 2 4 22 17.927 ksi 1.22718

AB

17.93 ksi

Rod BC: P

10 kips

A

d 22

AB

4 P A

(0.75)2

4 10 0.44179

0.44179 in 2

22.635 ksi

AB

22.6 ksi

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PROBLEM 1.4 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.

SOLUTION (a)

Rod AB: P A

(b)

P

12 kips d2 4

4

(1.25 in.) 2

A

1.22718 in 2

AB

P 12 kips 1.22718 in 2

Rod BC: P A

P 4

d2

4

(0.75 in.)2

A

0.44179 in 2

BC

P 0.44179 in 2

AB

BC

P 12 kips 1.22718 in 2 5.3015

P 0.44179 in 2 0.78539P

P

6.75 kips

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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.

SOLUTION P A Geometry:

A

4

(d12

P

A

d 22 ) 4A

d 22

d12

d 22

(25 10 3 )2

4P

d12

(4)(1200) (3.80 106 )

222.92 10 6 m 2 d2

14.93 10 3 m

d2

14.93 mm

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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:

b

4 4

(15 mm) 2

176.715 mm 2

(10 mm) 2

78.54 mm 2

100

176.715 10 6 m 2 78.54 10 6 m 2

a

WAB

g AAB

AB

(8470)(9.81)(176.715 10 6 ) a

14.683 a

WBC

g ABC

BC

(8470)(9.81)(78.54 10 6 )(100

a)

652.59

6.526 a

Normal stresses: At A,

At B,

(a)

PA

WAB

WBC

652.59

A

PA AAB

3.6930 106

PB

WBC

652.59

B

PB ABC

8.3090 106

(1)

46.160 103a

6.526a

(2)

83.090 103a

Length of rod AB. The maximum stress in ABC is minimum when 4.6160 106 a

(b)

8.157a

129.25 103a

A

B

or

0

35.71 m

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 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.

MC

0:

(0.040) FBD FBD

MB

0:

(0.040) FCE

(0.025

0.040)(20 103 )

32.5 103 N

Link BD is in tension. 3

(0.025)(20 10 ) 12.5 103 N

FCE

0 Link CE is in compression.

Net area of one link for tension

(0.008)(0.036

For two parallel links,

320 10 6 m 2

(a)

(b)

BD

FBD Anet

A net

32.5 103 320 10 6

0

0.016)

101.563 106

BD

Area for one link in compression

(0.008)(0.036)

For two parallel links,

576 10 6 m 2

CE

FCE A

12.5 103 576 10 6

160 10 6 m 2

A

21.701 10

6

101.6 MPa

288 10 6 m 2

CE

21.7 MPa

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PROBLEM 1.8 Link AC has a uniform rectangular cross section

1 8

in. thick and 1 in. wide.

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

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 FAC

0:

(12

4)( FAC cos 30 )

1200 lb 16 cos 30 10 sin 30

Area of link AC: Stress in link AC:

(10)( FAC sin 30 )

1200 lb

0

135.500 lb

A

1 in.

AC

FAC A

1 in. 0.125 in 2 8 135.50 1084 psi 0.125

1.084 ksi

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PROBLEM 1.9 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:

MD C

Free Body AC:

Required area of BE:

0: 0.150P

0.250C

0.6 P

MA

0: 0.150FBE

FBE

1.07 P 0.150

BE

ABE

0

0.350P

7.1333 P

0.450P

0.450C

(7.133)(4 kN)

0

28.533 kN

FBE ABE FBE BE

28.533 103 100 10 6

285.33 10 6 m 2 ABE

285 mm 2

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PROBLEM 1.10 Link BD consists of a single bar 1 in. wide and 1 in. thick. Knowing that each pin has a 3 -in. 2 8 diameter, determine the maximum value of the average normal stress in link BD if (a) = 0, (b) = 90 .

SOLUTION Use bar ABC as a free body.

(a)

0. MA

0: (18 sin 30 )(4)

FBD

3.4641 kips (tension)

Area for tension loading:

A

(b

d )t

FBD A

Stress: (b)

(12 cos30 ) FBD

0

3 1 8 2 3.4641 kips 0.31250 in 2 1

0.31250 in 2 11.09 ksi

90 . MA FBD

0:

(18 cos30 )(4)

0

6 kips i.e. compression.

Area for compression loading: Stress:

(12 cos 30 ) FBD A

bt FBD A

(1)

1 0.5 in 2 2 6 kips 0.5 in 2

12.00 ksi

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PROBLEM 1.11 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. MH Ay

0: (9)(80)

(18)(80)

(27)(80)

36 Ay

0

120 kips

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

0: 120 FBE A

80

12 FBE 15

0

FBE

50 kips

50 kips 5.87 in 2 BE

8.52 ksi

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PROBLEM 1.12 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 12 -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, MA

0: 40Dx

(45

Dx

30)(480)

0

900 lb

Use member DEF as free body. Reaction at D must be parallel to FBE and FCF . 4 Dx 3

Dy MF

1200 lb

0:

(30)

FBE

2250 lb

ME

0: (30)

FCE

750 lb

4 FBE 5

4 FCE 5

(30

15) DY

(15) DY

0

0

Stress in compression member BE: Area:

A

(a)

BE

2 in.

4 in.

FBE A

8 in 2

2250 8

BE

281 psi

Minimum section area occurs at pin. Amin Stress in tension member CF:

(b)

CF

(2)(4.0 FCF Amin

0.5) 750 7.0

7.0 in 2 CF

107.1 psi

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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 )

MA

0: 850R

R

2654.5 N

1962.00 N

1150(1962.00 N)

0

FREE BODY – BOTH ARM & WHEEL UNITS:

tan

100 675

8.4270

ME

0: ( FCD cos )(550)

R(500)

FCD

500 (2654.5 N) 550 cos 8.4270

0

2439.5 N (comp.) CD

FCD ACD

2439.5 N (0.0125 m)2

4.9697 106 Pa

CD

4.97 MPa

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PROBLEM 1.14 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.

SOLUTION Use member ABC as free body.

MB

0: (0.150)

FAE

4 103 N

4 FAE 5

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

(0.600)(800)

A

4

d2

4

(20 10 3 ) 2

4 103 314.16 10

FAE A

AE

0

314.16 10 6 m 2 12.7324 106 Pa

6

(a)

AE

12.73 MPa

Use combined members ABC and BFD as free body.

Area of rod DG: Stress in rod DG:

A

MF

0: (0.150)

FDG

1500 N

4 DG

d2 FDG A

4

4 FAE 5

(20 10 3 ) 2 1500 3.1416 10

(0.200)

4 FDG 5

(1.050

0.350)(800)

0

314.16 10 6 m 2 6

4.7746 106 Pa (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

P A

Shearing stress: Therefore, Finally,

dt

P d

or

A

P

dt P t 45 103 N (0.006 m)(55 106 Pa) 43.406 10 3 m d

43.4 mm

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PROBLEM 1.16 Two wooden planks, each 12 in. thick and 9 in. 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 1.20 ksi, determine the magnitude P of the axial load that will cause the joint to fail.

SOLUTION Six areas must be sheared off when the joint fails. Each of these areas has dimensions

5 8

in.

1 2

in., its area

being A

5 8

1 2

5 2 in 16

0.3125 in 2

At failure, the force carried by each area is F

A

(1.20 ksi)(0.3125 in 2 )

0.375 kips

Since there are six failure areas, P

6F

(6)(0.375)

P

2.25 kips

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PROBLEM 1.17 When the force P reached 1600 lb, 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

3 in. 0.6 in.

Force:

P

1600 lb

Shearing stress:

P A

1600 lb 1.8 in 2

1.8 in 2

8.8889 102 psi

889 psi

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PROBLEM 1.18 A load P is applied to a steel rod supported as shown by an aluminum plate into which a 12-mm-diameter hole has been drilled. Knowing that the shearing stress must not exceed 180 MPa in the steel rod and 70 MPa in the aluminum plate, determine the largest load P that can be applied to the rod.

SOLUTION For steel:

A1

dt

(0.012 m)(0.010 m)

376.99 10 6 m 2 P A

1

P

A1

1

(376.99 10 6 m 2 )(180 106 Pa)

67.858 103 N For aluminum:

A2 2

P A2

P

dt A2

2

(0.040 m)(0.008 m)

1.00531 10 3 m 2

(1.00531 10 3 m 2 )(70 106 Pa)

Limiting value of P is the smaller value, so

70.372 103 N P

67.9 kN

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PROBLEM 1.19 The axial force in the column supporting the timber beam shown is P 20 kips. Determine the smallest allowable length L of the bearing plate if the bearing stress in the timber is not to exceed 400 psi.

SOLUTION Bearing area: Ab

Lw b

L

P Ab P bw

P Lw 20 103 lb (400 psi)(6 in.)

8.33 in.

L

8.33 in.

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PROBLEM 1.20 Three wooden planks are fastened together by a series of bolts to form a column. The diameter of each bolt is 12 mm and the inner diameter of each washer is 16 mm, which is slightly larger than the diameter of the holes in the planks. Determine the smallest allowable outer diameter d of the washers, knowing that the average normal stress in the bolts is 36 MPa and that the bearing stress between the washers and the planks must not exceed 8.5 MPa.

SOLUTION Bolt:

d2 4

ABolt

(0.012 m)2 4

P A

Tensile force in bolt:

P

1.13097 10 4 m 2

A (36 106 Pa)(1.13097 10 4 m 2 ) 4.0715 103 N

Bearing area for washer:

Aw

and

Aw

do2

4

di2

P BRG

Therefore, equating the two expressions for Aw gives 4

do2

di2

P BRG

do2

4P BRG

di2

do2

4 (4.0715 103 N) (8.5 106 Pa)

do2

8.6588 10 4 m 2

do

29.426 10 3 m

(0.016 m) 2

do

29.4 mm

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PROBLEM 1.21 A 40-kN axial load is applied to a short wooden post that is supported by a concrete footing resting on undisturbed soil. Determine (a) the maximum bearing stress on the concrete footing, (b) the size of the footing for which the average bearing stress in the soil is 145 kPa.

SOLUTION (a)

Bearing stress on concrete footing. 40 kN

A

(100)(120)

Footing area. P

12 103 mm 2

40 103 12 10 3

P A (b)

40 103 N

P

40 103 N

3.3333 106 Pa

A

P

3.33 MPa

45 103 Pa

145 kPa

P A

12 10 3 m 2

40 103 145 103

0.27586 m 2

b2

Since the area is square, A b

A

0.27586

0.525 m

b

525 mm

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PROBLEM 1.22 An axial load P is supported by a short W8 40 column of crosssectional area A 11.7 in 2 and is distributed to a concrete foundation by a square plate as shown. Knowing that the average normal stress in the column must not exceed 30 ksi and that the bearing stress on the concrete foundation must not exceed 3.0 ksi, determine the side a of the plate that will provide the most economical and safe design.

SOLUTION For the column,

P or A P

For the a

a plate,

A

(30)(11.7)

351 kips

3.0 ksi P

A

Since the plate is square, A

351 3.0

117 in 2

a2 a

A

117

a

10.82 in.

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PROBLEM 1.23 Link AB, of width b = 2 in. and thickness t = 14 in., is used to support the end of a horizontal beam. Knowing that the average normal stress in the link is 20 ksi and that the average shearing stress in each of the two pins is 12 ksi, determine (a) the diameter d of the pins, (b) the average bearing stress in the link.

SOLUTION Rod AB is in compression. A

bt

where b

P Pin:

P

and

AP

(a)

4 AP

d

4P P

(b)

b

P dt

A

2 in. and t

( 20)(2)

1 4

1 in. 4

10 kips

P AP 4

d2 (4)(10) (12)

10 (1.03006)(0.25)

1.03006 in. d

1.030 in.

b

38.8 ksi

38.833 ksi

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PROBLEM 1.24 Determine the largest load P which may be applied at A when 60°, knowing that the average shearing stress in the 10-mmdiameter pin at B must not exceed 120 MPa and that the average bearing stress in member AB and in the bracket at B must not exceed 90 MPa.

SOLUTION Geometry: Triangle ABC is an isoseles triangle with angles shown here.

Use joint A as a free body.

Law of sines applied to force triangle: P sin 30

FAB sin 120

FAC sin 30

P

FAB sin 30 sin 120

0.57735FAB

P

FAC sin 30 sin 30

FAC

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PROBLEM 1.24 (Continued) If shearing stress in pin at B is critical, A

4

FAB

d2

4

(0.010) 2

78.54 10 6 m 2

(2)(78.54 10 6 )(120 106 )

2A

18.850 103 N

If bearing stress in member AB at bracket at A is critical, Ab

td

FAB

Ab

(0.016)(0.010) b

160 10 6 m 2

(160 10 6 )(90 106 )

14.40 103 N

If bearing stress in the bracket at B is critical, Ab FAB

2td Ab

(2)(0.012)(0.010) b

Allowable FAB is the smallest, i.e., 14.40 Then from statics,

Pallow

240 10 6 m2

(240 10 6 )(90 106 )

21.6 103 N

103 N

(0.57735)(14.40 103 ) 8.31 103 N

8.31 kN

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PROBLEM 1.25 Knowing that 40° and P 9 kN, determine (a) the smallest allowable diameter of the pin at B if the average shearing stress in the pin is not to exceed 120 MPa, (b) the corresponding average bearing stress in member AB at B, (c) the corresponding average bearing stress in each of the support brackets at B.

SOLUTION Geometry: Triangle ABC is an isoseles triangle with angles shown here.

Use joint A as a free body.

Law of sines applied to force triangle: P sin 20 FAB

FAB FAC sin110 sin 50 P sin110 sin 20 (9)sin110 24.727 kN sin 20

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PROBLEM 1.25 (Continued)

(a)

Allowable pin diameter. FAB 2 AP d2

FAB 2 4 d2

2FAB where FAB d2

(2)(24.727 103 ) (120 106 )

2FAB

d (b)

131.181 10 6 m 2 11.4534 10 3 m

11.45 mm

Bearing stress in AB at A. Ab b

(c)

24.727 103 N

(0.016)(11.4534 10 3 )

td

24.727 103 183.254 10 6

FAB Ab

183.254 10 6 m 2

134.933 106 Pa

134.9 MPa

Bearing stress in support brackets at B. A b

(0.012)(11.4534 10 3 )

td 1 2

FAB A

(0.5)(24.727 103 ) 137.441 10 6

137.441 10 6 m 2 89.955 106 Pa

90.0 MPa

PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 29

PROBLEM 1.26 The hydraulic cylinder CF, which partially controls the position of rod DE, has been locked in the position shown. Member BD is 15 mm thick and is connected at C to the vertical rod by a 9-mm-diameter bolt. Knowing that P 2 kN and 75 , determine (a) the average shearing stress in the bolt, (b) the bearing stress at C in member BD.

SOLUTION Free Body: Member BD.

Mc

0:

40 FAB (100 cos 20 ) 41

9 FAB (100 sin 20 ) 4

(2 kN) cos 75 (175sin 20 ) 100 FAB (40cos 20 41

(2 kN)sin 75 (175cos 20 )

9sin 20 ) FAB

Fx

0: Cx

9 (4.1424 kN) 41

(2 kN)(175)sin(75

3

0: C y

(a)

ave

C A

5.9860 10 N (0.0045 m)2

(b)

b

C td

5.9860 103 N (0.015 m)(0.009 m)

40 (4.1424 kN) 41

94.1 106 Pa

20 )

4.1424 kN

(2 kN)cos 75 Cx

Fy

0

(2 kN)sin 75

0 0.39167 kN 0

Cy

5.9732 kN

C

5.9860 kN

86.2°

94.1 MPa

44.3 106 Pa

44.3 MPa

PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 30

PROBLEM 1.27 For the assembly and loading of Prob. 1.7, determine (a) the average shearing stress in the pin at B, (b) the average bearing stress at B in member BD, (c) the average bearing stress at B in member ABC, knowing that this member has a 10 50-mm uniform rectangular cross section. 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.

(a)

MC

0 : (0.040)FBD

FBD

32.5 103 N

(b)

Bearing: link BD.

A

A b

(c)

Bearing in ABC at B.

0

FBD for double shear 2A

Shear pin at B. where

0.040)(20 103 )

(0.025

4

d2

4

(0.016) 2

201.06 10 6 m 2

32.5 103 (2)(201.06 10 6 )

80.822 106 Pa

dt

128 10 6 m2

1 2

(0.016)(0.008)

FBD A

A

dt

b

FBD A

(0.5)(32.5 103 ) 128 10 6

126.95 106 Pa

(0.016)(0.010)

160 10 6 m 2

32.5 103 160 10 6

203.12 106 Pa

80.8 MPa

127.0 MPa

b

b

203 MPa

PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 31

PROBLEM 1.28 Two identical linkage-and-hydraulic-cylinder systems control the position of the forks of a fork-lift truck. The load supported by the one system shown is 1500 lb. Knowing that the thickness of member BD is 5 in., determine (a) the average shearing stress in the 12 -in.-diameter 8 pin at B, (b) the bearing stress at B in member BD.

SOLUTION Use one fork as a free body. MB

0: 24 E

(20)(1500)

E

(a)

0: E

Bx

E

Bx

Bx

1250 lb

Fy

0: By

B

Bx2

1500 By2

1250 lb

0

0 1250 2

By

1500 lb

15002

1952.56 lb

Shearing stress in pin at B. Apin

4

2 d pin

B Apin (b)

Fx

0

1 4 2

2

0.196350 in 2

1952.56 0.196350

9.94 103 psi

9.94 ksi

Bearing stress at B. B dt

1952.56 1 2

5 8

6.25 103 psi

6.25 ksi

PROPRIETARY MATERIAL. Copyright © 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 32