978-0073398167 Chapter 11 Solution Manual Part 7

subject Type Homework Help
subject Pages 17
subject Words 1070
subject Authors David Mazurek, E. Johnston, Ferdinand Beer, John DeWolf

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page-pf1
PROBLEM 11.59
A short column is made by nailing two
1 4-in.×
planks to a
2 4-in.×
timber. Determine the largest compressive stress created in the column by a
12-kip load applied as shown in the center of the top section of the timber if
(a) the column is as described, (b) plank 1 is removed, (c) both planks are
removed.
SOLUTION
2
page-pf2
consent of McGraw-Hill Education.
PROBLEM 11.60
Knowing that the allowable stress in section ABD is 10 ksi,
determine the largest force P that can be applied to the bracket
shown.
SOLUTION
Statics:
2.45MP=
Cross section:
2
(0.9)(1.2) 1.08 inA= =
1(0.9) 0.45 in.
2
c= =
34
1(1.2)(0.9) 0.0729 in
12
I= =
At point B:
10 ksi
s
= −
(2.45 )(0.45)
10 16.049
1.08 0.0729
P Mc
AI
PP P
s
=−−
−=− − =−
0.623 kipsP=
623 lbP=
page-pf3
PROBLEM 11.61
A milling operation was used to remove a portion of a solid bar
of square cross section. Knowing that a = 30 mm, d = 20 mm,
and
all 60 MPa,
σ
=
determine the magnitude P of the largest
forces that can be safely applied at the centers of the ends of the
bar.
SOLUTION
3
3
22
11
, ,
12 2
22
6
3() 13()
where
A ad I ad c d
ad
e
P Mc P Ped
A I ad ad
P Pad ad
KP K
ad ad
ad ad
= = =
= −
=+=+
−−
=+= =+
σ
σ
Data:
32
2
30 mm 0.030 m 20 mm 0.020 m
1 (3)(0.010) 4.1667 10 m
(0.030)(0.020) (0.030)(0.020)
ad
K
= = = =
=+=×
63
3
60 10 14.40 10 N
4.1667 10
PK
σ
×
= = = ×
×
14.40 kNP=
page-pf4
PROBLEM 11.62
A milling operation was used to remove a portion of a solid bar of
square cross section. Forces of magnitude P = 18 kN are applied at
the centers of the ends of the bar. Knowing that a = 30 mm and
all 135 MPa,
σ
=
determine the smallest allowable depth d of the
milled portion of the bar.
SOLUTION
3
2
3
2
2
11
,,
12 2
22
11
() 3( )
22
1
12
32 2
or 3 0
A ad I ad c d
ad
e
Pad d
P Mc P Pec P P P a d
A I ad I ad ad ad
ad
PP P
d dP
ad a
d
= = =
= −
=+=+ =+ =+
= − + −=
σ
σσ
Solving for d,
2
12 2
12
2
PP
dP
aa



= +−





σ
σ
Data:
36
0.030 m, 18 10 N, 135 10 PaaP
σ
= =×=×
2
33
36
6
1 (2)(18 10 ) (2)(18 10 )
12(18 10 )(135 10 )
0.030 0.030
(2)(135 10 )
d


××

= + × ×−


×


3
16.04 10 m
= ×
16.04 mmd=
page-pf5
consent of McGraw-Hill Education.
PROBLEM 11.63
A vertical rod is attached at point A to the cast
iron hanger shown. Knowing that the allowable
stresses in the hanger are
all
5 ksi
s
= +
and
all 12 ksi= −
s
, determine the largest downward
force and the largest upward force that can be
exerted by the rod.
SOLUTION
3
2
2
all all
32
3 23 2
(1 3)(0.5) 2(3 0.25)(2.5)
(1 3) 2(3 0.75)
12.75 in 1.700 in.
7.5 in
7.5 in
5 ksi 12 ksi
1
12
11
(3)(1) (3 1)(1.70 0.5) (1.5)(3) (1.5 3)(2.5 1.70)
12 12
10.825 in
c
c
Ay
XA
X
A
I bh Ad
I
ss
× +×
= = ×+ ×
= =
=
=+=−

= +


= + + ×
=
4
Downward force.
(1.5 in.+1.70 in.) (3.20 in.)MP P= =
At D:
D
P Mc
AI
s
=++
(3.20) (1.70)
5 ksi 7.5 10.825
5 ( 0.6359)
PP
P
+=+
+= +
7.86 kips
= ↓P
At E:
E
P Mc
AI
s
=+−
(3.20) (2.30)
12 ksi 7.5 10.825
12 ( 0.5466)
PP
P
−=−
−=−
21.95 kips= ↓P
We choose the smaller value.
7.86 kips= ↓
P
page-pf6
consent of McGraw-Hill Education.
SOLUTION Continued
Upward force.
(1.5 in. 1.70 in.) (3.20 in.)MP P=+=
At D:
D
P Mc
AI
s
=+−
(3.20) (1.70)
12 ksi 7.5 10.825
12 ( 0.6359)
PP
P
− =−−
−=−
18.87 kips= ↑P
At E:
E
P Mc
AI
s
=++
(3.20) (2.30)
5 ksi 7.5 10.825
5 ( 0.5466)
PP
P
+=+
+= +
9.15 kips= ↑P
We choose the smaller value.
9.15 kips= ↑P
page-pf7
consent of McGraw-Hill Education.
PROBLEM 11.64
A steel rod is welded to a steel plate to form the machine
element shown. Knowing that the allowable stress is 135 MPa,
determine (a) the largest force P that can be applied to the
element, (b) the corresponding location of the neutral axis.
Given: The centroid of the cross section is at C and
4
4195 mm
z
I=
.
SOLUTION
(a)
2 2 62
(3)(18) (6) 82.27 mm 82.27 10 m
4
A
π
=+= =×
4 12 4
4195 mm 4195 10 m
I
= = ×
13.12 mm 0.01312 me= =
Based on tensile stress at
13.12 mm 0.01312 my=−=−
1P Pec ec P KP
A I AI
σ

=+=+ =


32
6 12
1 1 (0.01312)(0.01312) 53.188 10 m
82.27 10 4195 10
ec
KAI
−−
=+= + = ×
××
2.54 kNP=
(b) Location of neutral axis.
0
σ
=
1
0
P My P Pey ey
AI AI IA
σ
=−=− = =
12 3
6
4195 10 3.89 10 m
(82.27 10 )(0.01312)
I
yAe
×
= = = ×
×
3.89 mm
y=
of the loads.
page-pf8
consent of McGraw-Hill Education.
PROBLEM 11.65
The shape shown was formed by bending a thin steel plate. Assuming that the
thickness t is small compared to the length a of a side of the shape, determine the
stress (a) at A, (b) at B, (c) at C.
SOLUTION
Moment of inertia about centroid:
( )
3
3
122
12 2
1
12
a
It
ta

=

=
Area:
( )
22 2,
2 22
aa
A t at c

= = =


(a)
(
)( )
22 22
3
1
12
2
aa
A
P
P Pec P
A I at ta
σ
=−=−
2
A
P
at
σ
= −
(b)
( )( )
22
3
1
12
2
aa
B
P
P Pec P
A I at ta
σ
=+=+
2
B
P
at
σ
= −
(c)
CA
σσ
=
2
C
P
at
σ
= −
page-pf9
consent of McGraw-Hill Education.
PROBLEM 11.66
Knowing that the clamp shown has been tightened until
400 N,P=
determine (a) the stress at point A,
(b) the stress at point B, (c) the location of the neutral
axis of section a a.
SOLUTION
Cross section: Rectangle + Circle
2
1
1
22
2
2
(20 mm)(4 mm) 80 mm
1(20 mm) 10 mm
2
(2 mm) 4 mm
20 2 18 mm
A
y
A
y
ππ
= =
= =
= =
= −=
1
2
(80)(10) (4 )(18) 11.086 mm
80 4
20 8.914 mm
11.086 10 1.086 mm
18 11.086 6.914 mm
B
A
Ay
cy A
cy
d
d
π
π
+
= = = =
+
= −=
= −=
=−=
2 3 2 34
1 1 11
2 4 2 34
2 2 22
3 4 94
12
2 62
12
1(4)(20) (80)(1.086) 2.761 10 mm
12
(2) (4 )(6.914) 0.613 10 mm
4
3.374 10 mm 3.374 10 m
92.566 mm 92.566 10 m
I I Ad
I I Ad
III
AAA
ππ
=+= + = ×
=+= + =×
=+= × = ×
=+= = ×
page-pfa
consent of McGraw-Hill Education.
SOLUTION Continued
32 8.914 40.914 mm 0.040914 m
(400 N)(0.040914 m) 16.3656 N m
e
M Pe
=+= =
= = =
(a) Point A:
3
69
66 6
400 (16.3656)(8.914 10 )
92.566 10 3.374 10
4.321 10 43.23 10 47.55 10 Pa
AP Mc
AI
σ
−−
×
=+= +
××
= ×+ ×= ×
47.6 MPa
A
σ
=
(b) Point B:
69
66 6
400 (16.3656)(11.086)
92.566 10 3.374 10
4.321 10 53.72 10 49.45 10 Pa
B
P Mc
AI
σ
−−
=−= −
××
= ×− ×=− ×
49.4 MPa
B
σ
= −
(c) Neutral axis: By proportions,
20
47.55 47.55 49.45
9.80 mm
a
a
=+
=
9.80 mm below top of section
consent of McGraw-Hill Education.
PROBLEM 11.60
Knowing that the allowable stress in section ABD is 10 ksi,
determine the largest force P that can be applied to the bracket
shown.
SOLUTION
Statics:
2.45MP=
Cross section:
2
(0.9)(1.2) 1.08 inA= =
1(0.9) 0.45 in.
2
c= =
34
1(1.2)(0.9) 0.0729 in
12
I= =
At point B:
10 ksi
s
= −
(2.45 )(0.45)
10 16.049
1.08 0.0729
P Mc
AI
PP P
s
=−−
−=− − =−
0.623 kipsP=
623 lbP=
PROBLEM 11.61
A milling operation was used to remove a portion of a solid bar
of square cross section. Knowing that a = 30 mm, d = 20 mm,
and
all 60 MPa,
σ
=
determine the magnitude P of the largest
forces that can be safely applied at the centers of the ends of the
bar.
SOLUTION
3
3
22
11
, ,
12 2
22
6
3() 13()
where
A ad I ad c d
ad
e
P Mc P Ped
A I ad ad
P Pad ad
KP K
ad ad
ad ad
= = =
= −
=+=+
−−
=+= =+
σ
σ
Data:
32
2
30 mm 0.030 m 20 mm 0.020 m
1 (3)(0.010) 4.1667 10 m
(0.030)(0.020) (0.030)(0.020)
ad
K
= = = =
=+=×
63
3
60 10 14.40 10 N
4.1667 10
PK
σ
×
= = = ×
×
14.40 kNP=
PROBLEM 11.62
A milling operation was used to remove a portion of a solid bar of
square cross section. Forces of magnitude P = 18 kN are applied at
the centers of the ends of the bar. Knowing that a = 30 mm and
all 135 MPa,
σ
=
determine the smallest allowable depth d of the
milled portion of the bar.
SOLUTION
3
2
3
2
2
11
,,
12 2
22
11
() 3( )
22
1
12
32 2
or 3 0
A ad I ad c d
ad
e
Pad d
P Mc P Pec P P P a d
A I ad I ad ad ad
ad
PP P
d dP
ad a
d
= = =
= −
=+=+ =+ =+
= − + −=
σ
σσ
Solving for d,
2
12 2
12
2
PP
dP
aa



= +−





σ
σ
Data:
36
0.030 m, 18 10 N, 135 10 PaaP
σ
= =×=×
2
33
36
6
1 (2)(18 10 ) (2)(18 10 )
12(18 10 )(135 10 )
0.030 0.030
(2)(135 10 )
d


××

= + × ×−


×


3
16.04 10 m
= ×
16.04 mmd=
consent of McGraw-Hill Education.
PROBLEM 11.63
A vertical rod is attached at point A to the cast
iron hanger shown. Knowing that the allowable
stresses in the hanger are
all
5 ksi
s
= +
and
all 12 ksi= −
s
, determine the largest downward
force and the largest upward force that can be
exerted by the rod.
SOLUTION
3
2
2
all all
32
3 23 2
(1 3)(0.5) 2(3 0.25)(2.5)
(1 3) 2(3 0.75)
12.75 in 1.700 in.
7.5 in
7.5 in
5 ksi 12 ksi
1
12
11
(3)(1) (3 1)(1.70 0.5) (1.5)(3) (1.5 3)(2.5 1.70)
12 12
10.825 in
c
c
Ay
XA
X
A
I bh Ad
I
ss
× +×
= = ×+ ×
= =
=
=+=−

= +


= + + ×
=
4
Downward force.
(1.5 in.+1.70 in.) (3.20 in.)MP P= =
At D:
D
P Mc
AI
s
=++
(3.20) (1.70)
5 ksi 7.5 10.825
5 ( 0.6359)
PP
P
+=+
+= +
7.86 kips
= ↓P
At E:
E
P Mc
AI
s
=+−
(3.20) (2.30)
12 ksi 7.5 10.825
12 ( 0.5466)
PP
P
−=−
−=−
21.95 kips= ↓P
We choose the smaller value.
7.86 kips= ↓
P
consent of McGraw-Hill Education.
SOLUTION Continued
Upward force.
(1.5 in. 1.70 in.) (3.20 in.)MP P=+=
At D:
D
P Mc
AI
s
=+−
(3.20) (1.70)
12 ksi 7.5 10.825
12 ( 0.6359)
PP
P
− =−−
−=−
18.87 kips= ↑P
At E:
E
P Mc
AI
s
=++
(3.20) (2.30)
5 ksi 7.5 10.825
5 ( 0.5466)
PP
P
+=+
+= +
9.15 kips= ↑P
We choose the smaller value.
9.15 kips= ↑P
consent of McGraw-Hill Education.
PROBLEM 11.64
A steel rod is welded to a steel plate to form the machine
element shown. Knowing that the allowable stress is 135 MPa,
determine (a) the largest force P that can be applied to the
element, (b) the corresponding location of the neutral axis.
Given: The centroid of the cross section is at C and
4
4195 mm
z
I=
.
SOLUTION
(a)
2 2 62
(3)(18) (6) 82.27 mm 82.27 10 m
4
A
π
=+= =×
4 12 4
4195 mm 4195 10 m
I
= = ×
13.12 mm 0.01312 me= =
Based on tensile stress at
13.12 mm 0.01312 my=−=−
1P Pec ec P KP
A I AI
σ

=+=+ =


32
6 12
1 1 (0.01312)(0.01312) 53.188 10 m
82.27 10 4195 10
ec
KAI
−−
=+= + = ×
××
2.54 kNP=
(b) Location of neutral axis.
0
σ
=
1
0
P My P Pey ey
AI AI IA
σ
=−=− = =
12 3
6
4195 10 3.89 10 m
(82.27 10 )(0.01312)
I
yAe
×
= = = ×
×
3.89 mm
y=
of the loads.
consent of McGraw-Hill Education.
PROBLEM 11.65
The shape shown was formed by bending a thin steel plate. Assuming that the
thickness t is small compared to the length a of a side of the shape, determine the
stress (a) at A, (b) at B, (c) at C.
SOLUTION
Moment of inertia about centroid:
( )
3
3
122
12 2
1
12
a
It
ta

=

=
Area:
( )
22 2,
2 22
aa
A t at c

= = =


(a)
(
)( )
22 22
3
1
12
2
aa
A
P
P Pec P
A I at ta
σ
=−=−
2
A
P
at
σ
= −
(b)
( )( )
22
3
1
12
2
aa
B
P
P Pec P
A I at ta
σ
=+=+
2
B
P
at
σ
= −
(c)
CA
σσ
=
2
C
P
at
σ
= −
consent of McGraw-Hill Education.
PROBLEM 11.66
Knowing that the clamp shown has been tightened until
400 N,P=
determine (a) the stress at point A,
(b) the stress at point B, (c) the location of the neutral
axis of section a a.
SOLUTION
Cross section: Rectangle + Circle
2
1
1
22
2
2
(20 mm)(4 mm) 80 mm
1(20 mm) 10 mm
2
(2 mm) 4 mm
20 2 18 mm
A
y
A
y
ππ
= =
= =
= =
= −=
1
2
(80)(10) (4 )(18) 11.086 mm
80 4
20 8.914 mm
11.086 10 1.086 mm
18 11.086 6.914 mm
B
A
Ay
cy A
cy
d
d
π
π
+
= = = =
+
= −=
= −=
=−=
2 3 2 34
1 1 11
2 4 2 34
2 2 22
3 4 94
12
2 62
12
1(4)(20) (80)(1.086) 2.761 10 mm
12
(2) (4 )(6.914) 0.613 10 mm
4
3.374 10 mm 3.374 10 m
92.566 mm 92.566 10 m
I I Ad
I I Ad
III
AAA
ππ
=+= + = ×
=+= + =×
=+= × = ×
=+= = ×
consent of McGraw-Hill Education.
SOLUTION Continued
32 8.914 40.914 mm 0.040914 m
(400 N)(0.040914 m) 16.3656 N m
e
M Pe
=+= =
= = =
(a) Point A:
3
69
66 6
400 (16.3656)(8.914 10 )
92.566 10 3.374 10
4.321 10 43.23 10 47.55 10 Pa
AP Mc
AI
σ
−−
×
=+= +
××
= ×+ ×= ×
47.6 MPa
A
σ
=
(b) Point B:
69
66 6
400 (16.3656)(11.086)
92.566 10 3.374 10
4.321 10 53.72 10 49.45 10 Pa
B
P Mc
AI
σ
−−
=−= −
××
= ×− ×=− ×
49.4 MPa
B
σ
= −
(c) Neutral axis: By proportions,
20
47.55 47.55 49.45
9.80 mm
a
a
=+
=
9.80 mm below top of section

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