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PROBLEM 9.65
Two holes have been drilled through a long steel bar that is subjected to a centric
axial load as shown. For
6.5P=
kips, determine the maximum value of the
stress (a) at A, (b) at B.
SOLUTION
(a) At hole A:
11 1
in.
22 4
r=⋅=
1
3 2.50 in.
2
=−=d
2
net
1
(2.50) 1.25 in
2
= = =
A dt
non net
6.5 5.2 ksi
1.25
P
A
s
= = =
( )
1
4
2
20.1667
3
r
D= =
From Fig. 9.49,
2.56K=
max non
(2.56)(5.2)K
ss
= =
max 13.31ksi
s
=
(b) At hole B:
1(1.5) 0.75, 3 1.5 1.5 in.
2
rd
= = =−=
2
net non net
1 6.5
(1.5) 0.75 in , 8.667 ksi
2 0.75
s
= = = = = =
P
A dt A
2 2(0.75) 0.5
3
r
D= =
From Fig. 9.49,
2.16K=
max non
(2.16)(8.667)
ss
= =K
max 18.72 ksi
s
=
consent of McGraw-Hill Education.
PROBLEM 9.66
Knowing that
all
16 ksi,
s
=
determine the maximum allowable value of the
consent of McGraw-Hill Education.
PROBLEM 9.67
Knowing that, for the plate shown, the allowable stress is 125 MPa, determine
consent of McGraw-Hill Education.
PROBLEM 9.68
Knowing that P = 38 kN, determine the maximum stress when (a) r = 10 mm,
consent of McGraw-Hill Education.
PROBLEM 9.69
(a) Knowing that the allowable stress is 20 ksi, determine the maximum
allowable magnitude of the centric load P. (b) Determine the percent
change in the maximum allowable magnitude of P if the raised portions
consent of McGraw-Hill Education.
PROBLEM 9.70
A centric axial force is applied to the steel bar shown. Knowing that
consent of McGraw-Hill Education.
PROBLEM 9.71
Knowing that the hole has a diameter of 9 mm, determine (a) the radius
rf of the fillets for which the same maximum stress occurs at the hole A
and at the fillets, (b) the corresponding maximum allowable load P if
the allowable stress is 100 MPa.
SOLUTION
For the circular hole,
1(9) 4.5 mm
2
r
= =
2 2(4.5)
96 9 87 mm 0.09375
96
r
dD
= −= = =
62
net
(0.087 m)(0.009 m) 783 10 mA dt
−
= = = ×
From Fig. 9.49,
hole
2.72K=
hole
max net
KP
A
σ
=
66 3
net max
hole
(783 10 )(100 10 ) 28.787 10 N
2.72
A
PK
σ
−
××
= = = ×
(a) For fillet,
96 mm, 60 mmDd
= =
96 1.60
60
D
d= =
62
min
(0.060 m)(0.009 m) 540 10 mA dt
−
= = = ×
66
fillet min max
max fillet 3
min
(5.40 10 )(100 10 )
28.787 10
1.876
KP A
AP
K
σ
σ
−
××
= = = ×
=
∴
From Fig. 9.49,
0.19 0.19 0.19(60)≈ ∴≈ =
ff
rrd
d
11.4 mm
f
r=
(b)
28.8 kNP=
consent of McGraw-Hill Education.
PROBLEM 9.72
For
100 kN,P=
determine the minimum plate thickness t required if the
allowable stress is 125 MPa.
SOLUTION
At the hole:
20 mm 88 40 48 mm
AA
rd
= =−=
2 2(20) 0.455
88
A
A
r
D= =
From Fig. 9.49,
2.20K=
max net max
33
6
(2.20)(100 10 N) 36.7 10 m 36.7 mm
(0.048 m)(125 10 Pa)
AA
KP KP KP
t
A dt d
t
σσ
−
= = ∴=
×
= =×=
×
At the fillet:
88
88 mm, 64 mm 1.375
64
BB
D
Dd d
= = = =
15
15 mm 0.2344
64
B
BB
r
rd
= = =
From Fig. 9.49,
1.70K=
max min B
KP KP
A dt
σ
= =
33
6
max
(1.70)(100 10 N) 21.25 10 m 21.25 mm
(0.064 m)(125 10 Pa)
B
KP
td
σ
−
×
= = =×=
×
The larger value is the required minimum plate thickness.
36.7 mmt=
consent of McGraw-Hill Education.
PROBLEM 9.73
The aluminum rod ABC (
6
10.1 10 psi),E= ×
which consists of two
cylindrical portions AB and BC, is to be replaced with a cylindrical steel
rod DE (
6
29 10 psi)E= ×
of the same overall length. Determine the
minimum required diameter d of the steel rod if its vertical deformation is
SOLUTION
Deformation of aluminum rod.
3
62 2
44
28 10 12 18
10.1 10 (1.5) (2.25)
0.031376 in.
BC
AB
AAB BC
BC
AB
AB BC
PL
PL
AE AE
L
L
P
EA A
ππ
δ
= +
= +
×
= +
×
=
Steel rod.
0.031376 in.
δ
=
32
6
32
3
(28 10 )(30) 0.92317 in
(29 10 )(0.031376)
28 10 1.16667 in
24 10
PL PL
A
EA E
PP
A
A
δδ
σσ
×
= ∴= = =
×
×
= ∴== =
×
Required area is the larger value.
2
1.16667 inA=
Diameter:
4 (4)(1.16667)A
d
ππ
= =
1.219 in.=
d
PROBLEM 9.65
Two holes have been drilled through a long steel bar that is subjected to a centric
axial load as shown. For
6.5P=
kips, determine the maximum value of the
stress (a) at A, (b) at B.
SOLUTION
(a) At hole A:
11 1
in.
22 4
r=⋅=
1
3 2.50 in.
2
=−=d
2
net
1
(2.50) 1.25 in
2
= = =
A dt
non net
6.5 5.2 ksi
1.25
P
A
s
= = =
( )
1
4
2
20.1667
3
r
D= =
From Fig. 9.49,
2.56K=
max non
(2.56)(5.2)K
ss
= =
max 13.31ksi
s
=
(b) At hole B:
1(1.5) 0.75, 3 1.5 1.5 in.
2
rd
= = =−=
2
net non net
1 6.5
(1.5) 0.75 in , 8.667 ksi
2 0.75
s
= = = = = =
P
A dt A
2 2(0.75) 0.5
3
r
D= =
From Fig. 9.49,
2.16K=
max non
(2.16)(8.667)
ss
= =K
max 18.72 ksi
s
=
consent of McGraw-Hill Education.
PROBLEM 9.66
Knowing that
all
16 ksi,
s
=
determine the maximum allowable value of the
consent of McGraw-Hill Education.
PROBLEM 9.67
Knowing that, for the plate shown, the allowable stress is 125 MPa, determine
consent of McGraw-Hill Education.
PROBLEM 9.68
Knowing that P = 38 kN, determine the maximum stress when (a) r = 10 mm,
consent of McGraw-Hill Education.
PROBLEM 9.69
(a) Knowing that the allowable stress is 20 ksi, determine the maximum
allowable magnitude of the centric load P. (b) Determine the percent
change in the maximum allowable magnitude of P if the raised portions
consent of McGraw-Hill Education.
PROBLEM 9.70
A centric axial force is applied to the steel bar shown. Knowing that
consent of McGraw-Hill Education.
PROBLEM 9.71
Knowing that the hole has a diameter of 9 mm, determine (a) the radius
rf of the fillets for which the same maximum stress occurs at the hole A
and at the fillets, (b) the corresponding maximum allowable load P if
the allowable stress is 100 MPa.
SOLUTION
For the circular hole,
1(9) 4.5 mm
2
r
= =
2 2(4.5)
96 9 87 mm 0.09375
96
r
dD
= −= = =
62
net
(0.087 m)(0.009 m) 783 10 mA dt
−
= = = ×
From Fig. 9.49,
hole
2.72K=
hole
max net
KP
A
σ
=
66 3
net max
hole
(783 10 )(100 10 ) 28.787 10 N
2.72
A
PK
σ
−
××
= = = ×
(a) For fillet,
96 mm, 60 mmDd
= =
96 1.60
60
D
d= =
62
min
(0.060 m)(0.009 m) 540 10 mA dt
−
= = = ×
66
fillet min max
max fillet 3
min
(5.40 10 )(100 10 )
28.787 10
1.876
KP A
AP
K
σ
σ
−
××
= = = ×
=
∴
From Fig. 9.49,
0.19 0.19 0.19(60)≈ ∴≈ =
ff
rrd
d
11.4 mm
f
r=
(b)
28.8 kNP=
consent of McGraw-Hill Education.
PROBLEM 9.72
For
100 kN,P=
determine the minimum plate thickness t required if the
allowable stress is 125 MPa.
SOLUTION
At the hole:
20 mm 88 40 48 mm
AA
rd
= =−=
2 2(20) 0.455
88
A
A
r
D= =
From Fig. 9.49,
2.20K=
max net max
33
6
(2.20)(100 10 N) 36.7 10 m 36.7 mm
(0.048 m)(125 10 Pa)
AA
KP KP KP
t
A dt d
t
σσ
−
= = ∴=
×
= =×=
×
At the fillet:
88
88 mm, 64 mm 1.375
64
BB
D
Dd d
= = = =
15
15 mm 0.2344
64
B
BB
r
rd
= = =
From Fig. 9.49,
1.70K=
max min B
KP KP
A dt
σ
= =
33
6
max
(1.70)(100 10 N) 21.25 10 m 21.25 mm
(0.064 m)(125 10 Pa)
B
KP
td
σ
−
×
= = =×=
×
The larger value is the required minimum plate thickness.
36.7 mmt=
consent of McGraw-Hill Education.
PROBLEM 9.73
The aluminum rod ABC (
6
10.1 10 psi),E= ×
which consists of two
cylindrical portions AB and BC, is to be replaced with a cylindrical steel
rod DE (
6
29 10 psi)E= ×
of the same overall length. Determine the
minimum required diameter d of the steel rod if its vertical deformation is
SOLUTION
Deformation of aluminum rod.
3
62 2
44
28 10 12 18
10.1 10 (1.5) (2.25)
0.031376 in.
BC
AB
AAB BC
BC
AB
AB BC
PL
PL
AE AE
L
L
P
EA A
ππ
δ
= +
= +
×
= +
×
=
Steel rod.
0.031376 in.
δ
=
32
6
32
3
(28 10 )(30) 0.92317 in
(29 10 )(0.031376)
28 10 1.16667 in
24 10
PL PL
A
EA E
PP
A
A
δδ
σσ
×
= ∴= = =
×
×
= ∴== =
×
Required area is the larger value.
2
1.16667 inA=
Diameter:
4 (4)(1.16667)A
d
ππ
= =
1.219 in.=
d
