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PROBLEM 8.2
Two solid cylindrical rods AB and BC are welded together at B and
loaded as shown. Knowing that the average normal stress must not
exceed 175 MPa in rod AB and 150 MPa in rod BC, determine the
smallest allowable values of d1 and d2.
SOLUTION
(a) Rod AB
3
22
11
4
33
16
40 30 70 kN 70 10 N
4
4 (4)(70 10 ) 22.6 10 m
(175 10 )
AB AB
AB
P
PP P
Add
P
d
π
σπ
πσ π
−
=+= =×
= = =
×
= = = ×
×
1
22.6 mmd=
(b) Rod BC
3
22
22
4
33
26
30 kN 30 10 N
4
4 (4)(30 10 ) 15.96 10 m
(150 10 )
BC BC
BC
P
PP P
Add
P
d
π
σπ
πσ π
−
= = ×
= = =
×
= = = ×
×
2
15.96 mmd=
consent of McGraw-Hill Education.
PROBLEM 8.3
Two solid cylindrical rods
AB and BC are welded together at
B
and loaded as shown. Knowing that P = 40 kips, determine
the average normal stress at the midsection of (
a) rod AB, (b)
rod
BC.
SOLUTION
(a) Rod AB
40 kips (tension)P=
22 2
(2) 3.1416 in
44
40
3.1416
AB
AB
AB AB
d
A
P
A
ππ
σ
= = =
= =
12.73 ksi
AB
σ
=
(b) Rod BC
40 (2)(30) 20 kips, i.e., 20 kips compression.F=−=−
22 2
(3) 7.0686 in
44
20
7.0686
BC
BC
BC BC
d
A
F
A
ππ
σ
= = =
−
= =
2.83 ksi
BC
σ
= −
consent of McGraw-Hill Education.
PROBLEM 8.4
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 stress in rod AB is twice the
magnitude of the compressive stress in rod BC.
SOLUTION
22
22
(2) 3.1416 in
4
3.1416
0.31831
(3) 7.0686 in
4
(2)(30)
60 8.4883 0.14147
7.0686
AB
AB AB
BC
BC AB
A
PP
A
P
A
P
A
PP
π
σ
π
σ
= =
= =
=
= =
−
=
−
= = −
Equating
AB
σ
to
2
BC
σ
0.31831 2(8.4883 0.14147 )PP= −
28.2 kipsP=
PROBLEM 8.5
Link
BD consists of a single wooden member 36
mm wide and 18 mm thick. Knowing that each pin
has a 12
-mm diameter, determine the maximum
value of the average normal stress in link
BD if (a)
ϴ
= 0, (b) ϴ = 90°.
SOLUTION
Use bar ABC as a free body.
(a)
0.
θ
=
0: (450 sin30 )(24) (300 cos30 ) 0
20.7846 kN (tension)
A BD
BD
MF
F
Σ= ° − ° =
=
Area for tension loading:
( )( )
2
( ) 36 12 18 432 mmA b dt=−=− =
Stress:
62
20.7846 kN
432 10 m
BD
F
A
σ
−
= = ×
48.1 MPa
σ
=
(b)
90 .
θ
= °
0: (450 cos30 )(24) (300 cos30 ) 0
36 kN i.e. compression.
A BD
BD
MF
F
Σ= − ° − ° =
= −
Area for compression loading:
( )
2
(36) 18 648 mmA bt= = =
Stress:
62
36 kN
648 10 m
BD
F
A
σ
−
−
= = ×
55.6 MPa
σ
=
consent of McGraw-Hill Education.
PROBLEM 8.6
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.
3
3
3
3
0 : (0.040) (0.025 0.040)(20 10 ) 0
32.5 10 N Link is in tension.
0 : (0.040) (0.025)(20 10 ) 0
12.5 10 N Link is in compression.
Σ= − + × =
= ×
Σ= − − × =
=−×
C BD
BD
B CE
CE
MF
F BD
MF
F CE
Net area of one link for tension
(0.008)(0.036 0.016)= −
62
160 10 m
−
= ×
For two parallel links,
62
net 320 10 mA−
= ×
(a)
36
6
net
32.5 10 101.563 10
320 10
BD
BD
F
A
σ
−
×
= = = ×
×
101.6 MPa
σ
=
BD
Area for one link in compression
(0.008)(0.036)=
62
288 10 m
−
= ×
For two parallel links,
62
576 10 mA
−
= ×
(b)
36
6
12.5 10 21.701 10
576 10
CE
CE
F
A
σ
−
−
−×
== =−×
×
21.7 MPa
σ
= −
CE
PROBLEM 8.7
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.
0: (12 4)( cos30 ) (10)( sin30 ) 1200 lb 0
1200 lb 135.500 lb
16 cos30 10 sin30
Σ = − + °+ °− =
=−=−
°− °
B AC AC
AC
M FF
F
Area of link AC:
2
1
1 in. in. 0.125 in
8
=×=A
Stress in link AC:
135.50 1084 psi 1.084 ksi
0.125
AC
AC F
A
s
==−= =
consent of McGraw-Hill Education.
PROBLEM 8.8
Two horizontal 5-kip forces are applied to pin B of the assembly shown.
Knowing that a pin of 0.8-in. diameter is used at each connection,
determine the maximum value of the average normal stress (a) in link
AB, (b) in link BC.
SOLUTION
Use joint B as free body.
PROBLEM 8.9
For the Pratt bridge truss and loading shown, determine the
average normal stress in member BE, knowing that the cross-
sectional area of that member is 5.87 in2.
SOLUTION
Use entire truss as free body.
0: (9)(80) (18)(80) (27)(80) 36 0
120 kips
Hy
y
MA
A
Σ= + + − =
=
Use portion of truss to the left of a section cutting members
BD, BE, and CE.
12
0: 120 80 0 50 kips
15
y BE BE
F FF+Σ = − − = ∴ =
↑
2
50 kips
5.87 in
BE
BE
F
A
s
= =
8.52 ksi
s
=
BE
consent of McGraw-Hill Education.
PROBLEM 8.10
Knowing that the average normal stress in member CE of the
Pratt bridge truss shown must not exceed 21 ksi for the given
loading, determine the cross-sectional area of that member that
will yield the most economical and safe design. Assume that both
ends of the member will be adequately reinforced.
SOLUTION
Use entire truss as free body.
0: (9)(80) (18)(80) (27)(80) 36 0
120 kips
Hy
y
MA
A
Σ= + + − =
=
Use portion of truss to the left of a section cutting members
BD, BE, and CE.
0: 12 (9)(120) 0 90 kips
B CE CE
MF FΣ= − =∴ =
CE
CE CE
F
GA
=
2
90 4.29 in
21
CE
CE CE
F
AG
= = =
consent of McGraw-Hill Education.
PROBLEM 8.2
Two solid cylindrical rods AB and BC are welded together at B and
loaded as shown. Knowing that the average normal stress must not
exceed 175 MPa in rod AB and 150 MPa in rod BC, determine the
smallest allowable values of d1 and d2.
SOLUTION
(a) Rod AB
3
22
11
4
33
16
40 30 70 kN 70 10 N
4
4 (4)(70 10 ) 22.6 10 m
(175 10 )
AB AB
AB
P
PP P
Add
P
d
π
σπ
πσ π
−
=+= =×
= = =
×
= = = ×
×
1
22.6 mmd=
(b) Rod BC
3
22
22
4
33
26
30 kN 30 10 N
4
4 (4)(30 10 ) 15.96 10 m
(150 10 )
BC BC
BC
P
PP P
Add
P
d
π
σπ
πσ π
−
= = ×
= = =
×
= = = ×
×
2
15.96 mmd=
consent of McGraw-Hill Education.
PROBLEM 8.3
Two solid cylindrical rods
AB and BC are welded together at
B
and loaded as shown. Knowing that P = 40 kips, determine
the average normal stress at the midsection of (
a) rod AB, (b)
rod
BC.
SOLUTION
(a) Rod AB
40 kips (tension)P=
22 2
(2) 3.1416 in
44
40
3.1416
AB
AB
AB AB
d
A
P
A
ππ
σ
= = =
= =
12.73 ksi
AB
σ
=
(b) Rod BC
40 (2)(30) 20 kips, i.e., 20 kips compression.F=−=−
22 2
(3) 7.0686 in
44
20
7.0686
BC
BC
BC BC
d
A
F
A
ππ
σ
= = =
−
= =
2.83 ksi
BC
σ
= −
consent of McGraw-Hill Education.
PROBLEM 8.4
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 stress in rod AB is twice the
magnitude of the compressive stress in rod BC.
SOLUTION
22
22
(2) 3.1416 in
4
3.1416
0.31831
(3) 7.0686 in
4
(2)(30)
60 8.4883 0.14147
7.0686
AB
AB AB
BC
BC AB
A
PP
A
P
A
P
A
PP
π
σ
π
σ
= =
= =
=
= =
−
=
−
= = −
Equating
AB
σ
to
2
BC
σ
0.31831 2(8.4883 0.14147 )PP= −
28.2 kipsP=
PROBLEM 8.5
Link
BD consists of a single wooden member 36
mm wide and 18 mm thick. Knowing that each pin
has a 12
-mm diameter, determine the maximum
value of the average normal stress in link
BD if (a)
ϴ
= 0, (b) ϴ = 90°.
SOLUTION
Use bar ABC as a free body.
(a)
0.
θ
=
0: (450 sin30 )(24) (300 cos30 ) 0
20.7846 kN (tension)
A BD
BD
MF
F
Σ= ° − ° =
=
Area for tension loading:
( )( )
2
( ) 36 12 18 432 mmA b dt=−=− =
Stress:
62
20.7846 kN
432 10 m
BD
F
A
σ
−
= = ×
48.1 MPa
σ
=
(b)
90 .
θ
= °
0: (450 cos30 )(24) (300 cos30 ) 0
36 kN i.e. compression.
A BD
BD
MF
F
Σ= − ° − ° =
= −
Area for compression loading:
( )
2
(36) 18 648 mmA bt= = =
Stress:
62
36 kN
648 10 m
BD
F
A
σ
−
−
= = ×
55.6 MPa
σ
=
consent of McGraw-Hill Education.
PROBLEM 8.6
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.
3
3
3
3
0 : (0.040) (0.025 0.040)(20 10 ) 0
32.5 10 N Link is in tension.
0 : (0.040) (0.025)(20 10 ) 0
12.5 10 N Link is in compression.
Σ= − + × =
= ×
Σ= − − × =
=−×
C BD
BD
B CE
CE
MF
F BD
MF
F CE
Net area of one link for tension
(0.008)(0.036 0.016)= −
62
160 10 m
−
= ×
For two parallel links,
62
net 320 10 mA−
= ×
(a)
36
6
net
32.5 10 101.563 10
320 10
BD
BD
F
A
σ
−
×
= = = ×
×
101.6 MPa
σ
=
BD
Area for one link in compression
(0.008)(0.036)=
62
288 10 m
−
= ×
For two parallel links,
62
576 10 mA
−
= ×
(b)
36
6
12.5 10 21.701 10
576 10
CE
CE
F
A
σ
−
−
−×
== =−×
×
21.7 MPa
σ
= −
CE
PROBLEM 8.7
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.
0: (12 4)( cos30 ) (10)( sin30 ) 1200 lb 0
1200 lb 135.500 lb
16 cos30 10 sin30
Σ = − + °+ °− =
=−=−
°− °
B AC AC
AC
M FF
F
Area of link AC:
2
1
1 in. in. 0.125 in
8
=×=A
Stress in link AC:
135.50 1084 psi 1.084 ksi
0.125
AC
AC F
A
s
==−= =
consent of McGraw-Hill Education.
PROBLEM 8.8
Two horizontal 5-kip forces are applied to pin B of the assembly shown.
Knowing that a pin of 0.8-in. diameter is used at each connection,
determine the maximum value of the average normal stress (a) in link
AB, (b) in link BC.
SOLUTION
Use joint B as free body.
PROBLEM 8.9
For the Pratt bridge truss and loading shown, determine the
average normal stress in member BE, knowing that the cross-
sectional area of that member is 5.87 in2.
SOLUTION
Use entire truss as free body.
0: (9)(80) (18)(80) (27)(80) 36 0
120 kips
Hy
y
MA
A
Σ= + + − =
=
Use portion of truss to the left of a section cutting members
BD, BE, and CE.
12
0: 120 80 0 50 kips
15
y BE BE
F FF+Σ = − − = ∴ =
↑
2
50 kips
5.87 in
BE
BE
F
A
s
= =
8.52 ksi
s
=
BE
consent of McGraw-Hill Education.
PROBLEM 8.10
Knowing that the average normal stress in member CE of the
Pratt bridge truss shown must not exceed 21 ksi for the given
loading, determine the cross-sectional area of that member that
will yield the most economical and safe design. Assume that both
ends of the member will be adequately reinforced.
SOLUTION
Use entire truss as free body.
0: (9)(80) (18)(80) (27)(80) 36 0
120 kips
Hy
y
MA
A
Σ= + + − =
=
Use portion of truss to the left of a section cutting members
BD, BE, and CE.
0: 12 (9)(120) 0 90 kips
B CE CE
MF FΣ= − =∴ =
CE
CE CE
F
GA
=
2
90 4.29 in
21
CE
CE CE
F
AG
= = =
consent of McGraw-Hill Education.
