Solucionario 1_1 a 1_28
Short Description
resitencia materias...
Description
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
PROPRIETARY MATERIAL . Copyright
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)
P
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
PROPRIETARY MATERIAL. Copyright
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)
P
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
PROPRIETARY MATERIAL. Copyright
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
PROPRIETARY MATERIAL . Copyright
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
PROPRIETARY MATERIAL. Copyright
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 bones cross section at C .
SOLUTION
P A Geometry:
A
4
(d12
P
A
d 22 ) d 22
d12
4 A
d 22
(25
10 3 ) 2
222.92 d 2
PROPRIETARY MATERIAL . Copyright
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
PROPRIETARY MATERIAL. Copyright
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
For two parallel links,
320
(a)
(b)
BD
A net
F BD
32.5 103
Anet
6
320
10
0
Link BD is in tension.
12.5 103 N
FCE
103 )
0.040)(20
10
6
0.016)
For two parallel links,
576
CE
A
A
576
10
6
PROPRIETARY MATERIAL . Copyright
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
Area of link AC : Stress in link AC :
AC
PROPRIETARY MATERIAL. Copyright
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
PROPRIETARY MATERIAL . Copyright
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)
Area for tension loading:
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.
Area for compression loading: Stress: PROPRIETARY MATERIAL. Copyright
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
PROPRIETARY MATERIAL . Copyright
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 :
PROPRIETARY MATERIAL. Copyright
(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:
W
(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
PROPRIETARY MATERIAL . Copyright
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)
PROPRIETARY MATERIAL. Copyright
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
d
P
or A
dt P t 45
103 N
(0.006 m)(55 106 Pa ) 43.406
10
3
m d
PROPRIETARY MATERIAL . Copyright
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
PROPRIETARY MATERIAL. Copyright
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
PROPRIETARY MATERIAL . Copyright
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
PROPRIETARY MATERIAL. Copyright
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
PROPRIETARY MATERIAL . Copyright
P
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
PROPRIETARY MATERIAL. Copyright
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
P
103
40
3
145
10
3.33 MPa
0.27586 m 2
b2
Since the area is square, A b
PROPRIETARY MATERIAL . Copyright
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
PROPRIETARY MATERIAL. Copyright
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
d
4 A P
4 P
A P d 2
4
(4)(10) (12)
P
(b)
( 20)(2)
t
P
Pin:
(a)
A
2 in. and
b
P
10
dt
(1.03006)(0.25)
PROPRIETARY MATERIAL . Copyright
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:
PROPRIETARY MATERIAL. Copyright
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
PROPRIETARY MATERIAL . Copyright
10
8.31 kN
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Not authorized for sale or distributi on in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 27
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
PROPRIETARY MATERIAL. Copyright
24.727 kN
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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. 28
PROBLEM 1.25
(a)
(Continued)
Allowable pin diameter. F AB
FAB
2 A P 2
d
24d
2 F AB
2 F AB 2
2
d
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
PROPRIETARY MATERIAL . Copyright
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
© 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distributi on 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.
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)
C
5.9860
td
0: C y
2
41
(4.1424 kN)
94.1 106 Pa
103 N
(0.015 m)(0.009 m)
PROPRIETARY MATERIAL. Copyright
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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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)
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)
Bearing: link BD.
A b
(c)
0
Bearing in ABC at B.
A b
PROPRIETARY MATERIAL . Copyright
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
© 2015 McGraw-Hill Education. This is proprietary material solely for authorized instructor use.
Not authorized for sale or distributi on in any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part. 31
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