# Solucionario 1_1 a 1_28

August 26, 2018 | Author: Lucas Warley | Category: Stress (Mechanics), Tension (Physics), Truss, Screw, Copyright

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PROBLEM 1.1

Two solid cylindrical rods  AB  AB   and  BC   are welded together at  B  B   and loaded as shown. Knowing that d 1 30 mm  and d 2 50 mm, find the average normal stress at the midsection of (a (a) rod AB rod AB,, (b (b) rod BC  rod BC .

SOLUTION

(a)

Rod AB:: Rod AB Force:

P

Area:

A

Normal stress: (b)

103 N

60

4

d 12

4

P   AB

A

Force:

P

60

Area:

A

tension

(30

10 3 ) 2

103

60 706.86

10

6

706.86 84.882

10

6

m2

106 Pa

AB

84.9 MPa

Rod BC : Rod BC

Normal stress:

4

1 03

d 22

P   BC

A

(2)(125 103) 4

(50

190 1.96350

10 3 )2

1.96350

103 10

190

96.766

3

103 N 10

3

m2

106 Pa  BC

96.8 MPa

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PROBLEM 1.2

Two solid cylindrical rods  AB  AB   and  BC   are welded together at  B  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 d 1 and d 2.

SOLUTION

( a)

Rod AB:: Rod AB Force:

P

Stress:

AB

Area:

A  AB

4

d 12

103 N

60

106 Pa

150 4  P

d 12

A

AB

P   AB

d 12

4 P

(4)(60 (150

AB

( b)

A

d 1

22.568 10 3 m

Force:

P

60

Stress:

BC

150

A

d 22

103 )

509.30

6

10 )

10

6

m2 d 1

22.6 mm

d 2

40.2 mm

Rod BC : Rod BC

Area:

BC

d 22

4  P

103

103 N

4P  d 22

A 4 P  40.159

190

106 Pa

BC

d 2

(2)(125 103)

(4)( 190 ( 150 10

3

103 ) 6

10 )

m

1.61277

10

3

m2

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PROBLEM 1.2

Two solid cylindrical rods  AB  AB   and  BC   are welded together at  B  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 d 1 and d 2.

SOLUTION

( a)

Rod AB:: Rod AB Force:

P

Stress:

AB

Area:

A  AB

4

d 12

103 N

60

106 Pa

150 4  P

d 12

A

AB

P   AB

d 12

4 P

(4)(60 (150

AB

( b)

A

d 1

22.568 10 3 m

Force:

P

60

Stress:

BC

150

A

d 22

103 )

509.30

6

10 )

10

6

m2 d 1

22.6 mm

d 2

40.2 mm

Rod BC : Rod BC

Area:

BC

d 22

4  P

103

103 N

4P  d 22

A 4 P  40.159

190

106 Pa

BC

d 2

(2)(125 103)

(4)( 190 ( 150 10

3

103 ) 6

10 )

m

1.61277

10

3

m2

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PROBLEM 1.3

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

SOLUTION

(a)

Rod AB Rod  AB::  P   A

AB

(b)

12

10 d 12

22 kips (1.25)2

4  P

4 22

A

1.22718

1.22718 in 2

17.927 ksi

AB

17.93 ksi

Rod BC : Rod BC   P

10 kips

A

d 22

AB

(0.75)2

4  P

4 10

A

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

P  12 kips d 2

A  A

4

4

(1.25 in.) 2

1.22718 in 2  P  12 kips

AB

(b)

1.22718 in 2

Rod BC :  P  A  A

P  4

d 2

4

(0.75 in.)2

0.44179 in 2  P

BC

AB

0.44179 in 2 BC

P

12 kips 2

1.22718 in 5.3015

P  0.44179 in 2 0.78539 P

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

A

d 22 ) d 22

d12

4 A

d 22

(25

10 3 ) 2

222.92 d 2

4 P

d 12

10

14.93 10

3

(4)(1200) (3.80 6

106 )

m2

m

d 2

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 3 from a support at A as shown. Knowing that the density of brass is 8470 kg/m , 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:

A AB  A BC

From geometry, Weights:

4 4

(15 mm) 2

176.715 mm 2

(10 mm) 2

78.54 mm 2

b  100

176.715

10 6 m 2

78.54

10 6 m 2

10 6 ) a

a

W AB

g AAB

AB

(8470)(9.81)(176.715

W BC

g ABC

BC

(8470)(9.81)(78.54

10 6 )(100

14.683 a a)

652.59

6.526 a

Normal stresses: At A,

At B,

P A

WBC

652.59

P   A

3.6930

106

46.160

A AB

P B

WBC

652.59

6.526a

P   B

8.3090

106

A BC

83.090

(1)

(2)

103a

103a

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

(b)

8.157a

A

B

(a)

WAB

106

129.25

103a

A

B

or

0

35.71m

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.

M C

0:

(0.040) F BD  F BD

M B

0:

(0.040) F CE

(0.025

32.5 103 N (0.025)(20

103 )

0 Link CE  is in compression.

Net area of one link for tension

(0.008)(0.036

320

(a)

(b)

BD

A net

F   BD

32.5 103

Anet

6

320

10

0

12.5 103 N

FCE

103 )

0.040)(20

10

6

0.016)

576

CE

A

A

576

10

6

6

m2

101.563 106 (0.008)(0.036)

12.5 103

10

m2

Area for one link in compression

F CE

160

10

21.701 10

6

288

10

6

BD

101.6 MPa

CE

21.7 MPa

m2

m2

6

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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.

B

F  AC

0:

(12

4)( FAC cos 30 )

1200 lb 16 cos 30

AC

1200 lb

0

135.500 lb

10 sin 30

A

(10)( F AC sin 30 )

1 in.

1 8

in.

0.125 in 2

F   AC

135.50

A

0.125

1084 psi

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:

M D C

Free Body AC :

M A  F BE

Required area of BE :

BE

A BE

0: 0.150P

0.250C

0

0.6 P  0: 0.150FBE  1.07 0.150

P

0.350 P

7.1333 P

0.450P

0.450C

(7.133)(4 kN)

0

28.533 kN

F   BE   A BE   F   BE   BE

28.533 103 100

10

6

285.33 10 6 m 2  A BE

285 mm2

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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 38 -in. 2 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.  A

F   BD

0: (18 sin 30 )(4)

(12 cos30 ) F BD

3.4641 kips (tension)

A

(b

d )t

3

1

1

8 2 3.4641 kips

F   BD

Stress:

(b)

0

0.31250 in 2

0.31250 in 2

A

11.09 ksi

90 .  A

F   BD

0:

(18 cos30 )(4)

(12 cos 30 ) F BD

0

6 kips i.e. compression.

A

bt

1

(1)

2 6 kips

F   BD  A

0.5 in 2

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 cross2 sectional area of that member is 5.87 in .

SOLUTION

Use entire truss as free body.  M H  A y

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 .  F y  BE

0: 120

80

12 15

F   BE

50 kips

A

5.87 in2

FBE

0

F BE

BE

50 kips

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, 0: 40 Dx

A

(45

D x

30)(480)

0

900 lb

Use member DEF as free body. Reaction at D must be parallel to  F   BE  and  F  CF . 4

D y

3

Dx

M F

1200 lb

0:

F   BE

4

(30)

5

FBE

(30

15) DY

(15) DY

0

0

2250 lb

M E

0: (30)

F CE

750 lb

4 5

FCE

Stress in compression member BE : Area:

A

(a)

BE

2 in.

8 in 2

4 in.

F   BE   A

2250 8

BE

281psi

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

(b)

CF

(2)(4.0

0.5)

F CF

750

Amin

7.0

7.0 in 2 CF

107.1psi

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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:

(200 kg)(9.81 m/s2 )

M A

0: 850 R

R

2654.5 N

1962.00 N

1150(1962.00 N)

0

FREE BODY  BOTH ARM & WHEEL UNITS:

100

tan

8.4270

675  E

F CD

0: ( FCD cos )(550) 500 550 cos 8.4270 2439.5 N

CD

0

(2654.5 N)

(comp.)

F CD

2439.5 N

ACD

(0.0125 m)2

4.9697

R(500)

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 s hown, determine the average normal stress in (a) member AE , (b) member DG.

SOLUTION

Use member ABC  as free body.

M B

0: (0.150)

F  AE

4

4 5

F AE

(0.600)(800)

0

103 N

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

4

d 2

4

F   AE

AE

103

4

A

10 3 )2

(20

314.16

6

10

314.16

10 6 m 2

12.7324

106 Pa (a)

AE

12.73 MPa

Use combined members ABC  and BFD as free body.

Area of rod DG: Stress in rod DG:

A

M F

0: (0.150)

F   DG

1500 N

4  DG

d 2

4

F   DG  A

(20

4 5

FAE

10 3) 2

314.16

1500 3.1416

10

(0.200)

6

10

4.7746

6

4 5

F DG

(1.050

0

m2 106 Pa (b)

0.350)(800)

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

Shearing stress: Therefore, Finally,

dt   A

P

or   A

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

1

5

8

2

16

in 2

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

6 F

(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.

Force:

P

1600 lb

Shearing stress:

P   A

0.6 in.

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  376.99

P  1

A

P

A1

1

(0.012 m)(0.010 m) 10

6

m2

(376.99

10

6

m 2 )(180

106 Pa)

67.858 103 N For aluminum:

A2  P  2

A2

P

A2

dt

(0.040 m)(0.008 m)

2

(1.00531 10 3 m 2 )(70

Limiting value of P is the smaller value, so

1.00531 10 106 Pa)

3

m2

70.372  P

103 N 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

Lw

P  bw

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:

d 2

ABolt

4  P

Tensile force in bolt:

(0.012 m) 2

A

1.13097

10

4

m2

106 Pa)(1.13097

10

4

m2)

4  P

A (36

4.0715 103 N Bearing area for washer:

Aw

and

Aw

4

do2

d i2

P   BRG

Therefore, equating the two expressions for Aw gives 4

do2

d i2

P   BRG

do2

4 P

d i2

BRG

d o2

4 (4.0715 103 N) (8.5

6

10 Pa)

d o2

8.6588 10

4

m2

d o

29.426

3

m

10

(0.016 m) 2

d o

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.  P

40 kN

A

(100)(120)  P   A

(b)

Footing area. P

40

40 12

103 N  P  A

103 mm 2

12 103

10

12

10 3 m 2

3.3333 106 Pa

3

145 kPa

45 103 Pa

103

40

3

145

10

3.33 MPa

0.27586 m 2

b2

Since the area is square,  A b

A

103 N

40

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 in2  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   A

or  P

For the a

a plate,

A

(30)(11.7)

351 kips

3.0 ksi  A

Since the plate is square,  A

P

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

P

and

A P

1 4

1 4

in.

10 kips

4 A P

4 P

A P  d 2

4

(4)(10) (12)

P

(b)

( 20)(2)

P

Pin:

(a)

A

2 in. and

b

P

10

dt

(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

F AB

F AC

sin 30

sin 120

sin 30

P

F   AB sin30

P

F   AC  sin 30

sin 120 sin 30

0.57735F   AB F   AC

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PROBLEM 1.24

(Continued)

If shearing stress in pin at B is critical,  A

4

F AB

d 2

4

2A

(0.010) 2

(2)(78.54

78.54

10

10 6 )(120

6

m2

106 )

18.850

103 N

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

td

F AB

Ab

(0.016)(0.010) b

(160

160

10 6 )(90

10

106 )

6

m2 14.40

103 N

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

2td  Ab

b

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

P allow

(2)(0.012)(0.010)

240

10 6 )(90

106 )

(240

6

m2

21.6

103 N

3

10  N

(0.57735)(14.40 103 ) 8.31 103 N

10

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

F AB

F AC

sin 20

sin110  P sin110

sin 50

F   AB

sin 20 (9)sin110 sin 20

24.727 kN

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PROBLEM 1.25

(a)

(Continued)

Allowable pin diameter.  F AB

FAB

2 A P  2

24d

2 F   AB

2 F AB 2

2

103)

(2)(24.727 (120

where F   AB

6

10 ) d

(b)

6

m2

10 3 m

11.45 mm

Bearing stress in AB at A.  Ab b

(c)

131.181 10 11.4534

103 N

24.727

td

10 3 )

(0.016)(11.4534

F  AB

24.727

Ab

183.254

103 10

183.254

10

6

m2

134.933 106 Pa

6

134.9 MPa

Bearing stress in support brackets at B.  A b

td

(0.012)(11.4534

1  F  2  AB

A

(0.5)(24.727

10 3 ) 103 )

137.441 10

6

137.441 10

6

m2

89.955 106 Pa

90.0 MPa

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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.

M c

0:

40 41

9

FAB (100 cos 20 )

4

(2 kN) cos 75 (175 sin 2 0 ) 100  F   AB (40 cos 20 41

F AB (100 sin 20 ) (2 kN) sin 75 (175 cos 20 )

9sin 20 )  F   AB

F x

0: C x

9 41

(4.1424 kN)

(2 kN)(175)sin(75

(a)

(b)

C  ave

b

5.9860

103 N

A

(0.0045 m)

5.9860

td

0: C y

2

41

(4.1424 kN)

94.1 106 Pa

103 N

(0.015 m)(0.009 m)

40

20 )

4.1424 kN

(2 kN)cos 75 C  x

F y

0

(2 kN)sin 75

0 0.39167 kN 0

C  y

5.9732 kN

C

5.9860 kN

86.2°

94.1 MPa

44.3 106 Pa

44.3 MPa

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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)

M C

0 : (0.040)F BD

F   BD

32.5

(0.025

103 )

0.040)(20

103 N

F   BD

Shear pin at B. where

for double shear

2 A  A

4

d 2

4

(0.016)2

32.5 103 6

(2)(201.06 10 ) (b)

A b

(c)

0

Bearing in ABC  at B.

A b

dt

F   BD  A

128

10

(0.016)(0.010) 32.5 160

103 10

6

160 203.12

6

80.8 MPa

128 10 6 m2

(0.5)(32.5 103 )

A

10 6 m2

80.822 10 6 Pa

(0.016)(0.008)

1  F  2  BD

dt

201.06

126.95 106 Pa

127.0 MPa

b

10 6 m 2 106 Pa

b

203 MPa

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