Unlock document.
This document is partially blurred.
Unlock all pages and 1 million more documents.
Get Access
PROBLEM 8.12
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.
3
4
0: (0.150) (0.600)(800) 0
5
4 10 N
Σ= − =
= ×
B AE
AE
MF
F
Area of rod in member AE is
2 32 6 2
(20 10 ) 314.16 10 m
44
Ad
ππ
−−
==×= ×
Stress in rod AE:
36
6
4 10 12.7324 10 Pa
314.16 10
AE
AE
F
A
σ
−
×
= = = ×
×
(a)
12.73 MPa
AE
σ
=
Use combined members ABC and BFD as free body.
44
0: (0.150) (0.200) (1.050 0.350)(800) 0
55
1500 N
Σ= − − − =
= −
F AE DG
DG
M FF
F
Area of rod DG:
2 32 6 2
(20 10 ) 314.16 10 m
44
Ad
ππ
−−
==×= ×
Stress in rod DG:
6
6
1500 4.7746 10 Pa
3.1416 10
DG
DG
F
A
σ
−
−
== =−×
×
(b)
4.77 MPa
DG
σ
= −
consent of McGraw-Hill Education.
PROBLEM 8.13
The wooden members A and B are to be joined by plywood splice
plates which will be fully glued on the surfaces in contact. As part of
the design of the joint, and knowing that the clearance between the
ends of the members is to be 8 mm, determine the smallest allowable
length L if the average shearing stress in the glue is not to exceed 800
kPa.
SOLUTION
PROBLEM 8.14
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 dt
π
=
Shearing stress:
or
PP
A
A
ττ
= =
Therefore,
Pdt
π
τ
=
Finally,
3
6
3
45 10 N
(0.006 m)(55 10 Pa)
43.406 10 m
P
dt
πτ
π
−
=
×
=×
= ×
43.4 mmd=
consent of McGraw-Hill Education.
PROBLEM 1.15
Two wooden planks, each
1
2
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
51
PROBLEM 8.16
A load P is applied to a steel rod supported as shown by an
aluminum plate into which a 0.6-in.-diameter hole has been
drilled. Knowing that the shearing stress must not exceed
18 ksi in the steel rod and 10 ksi in the aluminum plate,
determine the largest load P that can be applied to the rod.
SOLUTION
For steel:
1
2
(0.6)(0.4)
0.7540 in
ππ
= =
=
A dt
1 11 (0.7540)(18)
13.57 kips
ττ
= ∴= =
=
PPA
A
For aluminum:
2
2
(1.6)(0.25) 1.2566 in
ππ
= = =A dt
2 22
2
(1.2566)(10) 12.57 kips
ττ
= ∴= = =
PPA
A
Limiting value of P is the smaller value, so
12.57 kips=P
consent of McGraw-Hill Education.
PROBLEM 8.17
An axial load P is supported by a short
W8 40×
column of cross-
sectional area
2
11.7 in=A
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
PROBLEM 8.18
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:
=
b
A Lw
3
20 10 lb 8.33 in.
(400 psi)(6 in.)
s
s
= =
×
= = =
bb
b
PP
A Lw
P
Lw
8.33 in.=L
consent of McGraw-Hill Education.
PROBLEM 8.19
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:
22 42
Bolt
(0.012 m) 1.13097 10 m
44
d
A
ππ
−
= = = ×
Tensile force in bolt:
PPA
A
σσ
= ⇒=
6 42
3
(36 10 Pa)(1.13097 10 m )
4.0715 10 N
−
=××
= ×
Bearing area for washer:
( )
22
4
π
= −
w oi
A dd
and
wBRG
P
A
σ
=
Therefore, equating the two expressions for Aw gives
()
22
22
3
22
6
2 42
3
4
4
4 (4.0715 10 N) (0.016 m)
(8.5 10 Pa)
8.6588 10 m
29.426 10 m
oi BRG
oi
BRG
o
o
o
P
dd
P
dd
d
d
d
π
σ
πσ
π
−
−
−=
= +
×
= +
×
= ×
= ×
29.4 mm
=
o
d
consent of McGraw-Hill Education.
PROBLEM 8.20
Link AB, of width b = 50 mm and thickness t = 6 mm, is used to support
the end of a horizontal beam. Knowing that the average normal stress in
the link is − 140 MPa, and that the average shearing stress in each of the
two pins is 80 MPa, 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
50 mm=b
and
6 mm=t
62
66
3
(0.050)(0.006) 300 10 m
( 140 10 )(300 10 )
42 10 N
σ
−
−
= = ×
=−=−−× ×
= ×
A
PA
For the pin,
2
4
p
Ad
p
=
and
τ
=
p
P
A
362
6
42 10 525 10 m
80 10
τ
−
×
= = = ×
×
p
P
A
(a) Diameter d
63
4(4)(525 10 ) 2.585 10 m
p
A
d
pp
−−
×
= = = ×
25.9 mm=d
(b) Bearing stress
36
3
42 10 271 10 Pa
(25.85 10 )(0.006)
σ
−
×
= = = ×
×
b
P
dt
271 MPa
σ
=
b
consent of McGraw-Hill Education.
PROBLEM 8.12
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.
3
4
0: (0.150) (0.600)(800) 0
5
4 10 N
Σ= − =
= ×
B AE
AE
MF
F
Area of rod in member AE is
2 32 6 2
(20 10 ) 314.16 10 m
44
Ad
ππ
−−
==×= ×
Stress in rod AE:
36
6
4 10 12.7324 10 Pa
314.16 10
AE
AE
F
A
σ
−
×
= = = ×
×
(a)
12.73 MPa
AE
σ
=
Use combined members ABC and BFD as free body.
44
0: (0.150) (0.200) (1.050 0.350)(800) 0
55
1500 N
Σ= − − − =
= −
F AE DG
DG
M FF
F
Area of rod DG:
2 32 6 2
(20 10 ) 314.16 10 m
44
Ad
ππ
−−
==×= ×
Stress in rod DG:
6
6
1500 4.7746 10 Pa
3.1416 10
DG
DG
F
A
σ
−
−
== =−×
×
(b)
4.77 MPa
DG
σ
= −
consent of McGraw-Hill Education.
PROBLEM 8.13
The wooden members A and B are to be joined by plywood splice
plates which will be fully glued on the surfaces in contact. As part of
the design of the joint, and knowing that the clearance between the
ends of the members is to be 8 mm, determine the smallest allowable
length L if the average shearing stress in the glue is not to exceed 800
kPa.
SOLUTION
PROBLEM 8.14
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 dt
π
=
Shearing stress:
or
PP
A
A
ττ
= =
Therefore,
Pdt
π
τ
=
Finally,
3
6
3
45 10 N
(0.006 m)(55 10 Pa)
43.406 10 m
P
dt
πτ
π
−
=
×
=×
= ×
43.4 mmd=
consent of McGraw-Hill Education.
PROBLEM 1.15
Two wooden planks, each
1
2
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
51
PROBLEM 8.16
A load P is applied to a steel rod supported as shown by an
aluminum plate into which a 0.6-in.-diameter hole has been
drilled. Knowing that the shearing stress must not exceed
18 ksi in the steel rod and 10 ksi in the aluminum plate,
determine the largest load P that can be applied to the rod.
SOLUTION
For steel:
1
2
(0.6)(0.4)
0.7540 in
ππ
= =
=
A dt
1 11 (0.7540)(18)
13.57 kips
ττ
= ∴= =
=
PPA
A
For aluminum:
2
2
(1.6)(0.25) 1.2566 in
ππ
= = =A dt
2 22
2
(1.2566)(10) 12.57 kips
ττ
= ∴= = =
PPA
A
Limiting value of P is the smaller value, so
12.57 kips=P
consent of McGraw-Hill Education.
PROBLEM 8.17
An axial load P is supported by a short
W8 40×
column of cross-
sectional area
2
11.7 in=A
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
PROBLEM 8.18
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:
=
b
A Lw
3
20 10 lb 8.33 in.
(400 psi)(6 in.)
s
s
= =
×
= = =
bb
b
PP
A Lw
P
Lw
8.33 in.=L
consent of McGraw-Hill Education.
PROBLEM 8.19
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:
22 42
Bolt
(0.012 m) 1.13097 10 m
44
d
A
ππ
−
= = = ×
Tensile force in bolt:
PPA
A
σσ
= ⇒=
6 42
3
(36 10 Pa)(1.13097 10 m )
4.0715 10 N
−
=××
= ×
Bearing area for washer:
( )
22
4
π
= −
w oi
A dd
and
wBRG
P
A
σ
=
Therefore, equating the two expressions for Aw gives
()
22
22
3
22
6
2 42
3
4
4
4 (4.0715 10 N) (0.016 m)
(8.5 10 Pa)
8.6588 10 m
29.426 10 m
oi BRG
oi
BRG
o
o
o
P
dd
P
dd
d
d
d
π
σ
πσ
π
−
−
−=
= +
×
= +
×
= ×
= ×
29.4 mm
=
o
d
consent of McGraw-Hill Education.
PROBLEM 8.20
Link AB, of width b = 50 mm and thickness t = 6 mm, is used to support
the end of a horizontal beam. Knowing that the average normal stress in
the link is − 140 MPa, and that the average shearing stress in each of the
two pins is 80 MPa, 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
50 mm=b
and
6 mm=t
62
66
3
(0.050)(0.006) 300 10 m
( 140 10 )(300 10 )
42 10 N
σ
−
−
= = ×
=−=−−× ×
= ×
A
PA
For the pin,
2
4
p
Ad
p
=
and
τ
=
p
P
A
362
6
42 10 525 10 m
80 10
τ
−
×
= = = ×
×
p
P
A
(a) Diameter d
63
4(4)(525 10 ) 2.585 10 m
p
A
d
pp
−−
×
= = = ×
25.9 mm=d
(b) Bearing stress
36
3
42 10 271 10 Pa
(25.85 10 )(0.006)
σ
−
×
= = = ×
×
b
P
dt
271 MPa
σ
=
b
consent of McGraw-Hill Education.
