consent of McGraw–Hill Education.
SOLUTION Continued
(b) Bearing stress at C in member BC.
σ
=
BC
b
F
A
2
(0.5)(0.8) 0.4 in
8.9658 22.4
0.4
σ
= = =
= =
b
A td
(c) Bearing stress at B in member BC.
σ
=
BC
b
F
A
2
2 2(0.5)(0.8) 0.8 in
8.9658 11.21
0.8
σ
= = =
= =
b
A td
11.21 ksi
σ
=
b
consent of McGraw–Hill Education.
PROBLEM 8.22
The hydraulic cylinder CF, which partially controls the position of rod
DE, has been locked in the position shown. Member BD is
5
8
in. thick
and is connected to the vertical rod by a
3
8
–in.–diameter bolt. Determine
(a) the average shearing stress in the bolt, (b) the bearing stress at C in
member BD.
SOLUTION
Use member BCD as a free body, and note that AB is a two force member.
22
8 1.8 8.2 in.=+=
AB
l
8 1.8
0: (4cos20 ) (4sin 20 )
8.2 8.2
(7cos20 )(400sin 75 ) (7sin 20 )(400cos75 ) 0
3.36678 2789.35 0 828.49 lb
 
Σ= ° − °
 
 
− ° °− ° °=
− = ∴=
C AB AB
AB AB
M FF
FF
1.8
0: 400cos75 0
8.2
(1.8)(828.49) 400cos75 78.34 lb
8.2
Σ = − + + °=
= − °=
x AB x
x
F FC
C
22
8
0: 400sin75 0
8.2
(8)(828.49) 400sin75 1194.65 lb
8.2
1197.2 lb
Σ = − + − °=
= + °=
= +=
y AB y
y
xy
F FC
C
C CC
consent of McGraw–Hill Education.
SOLUTION Continued
(a) Shearing stress in the bolt:
1197.2 lb=P
2
22
30.11045 in
4 48
ππ

= = =


Ad
3
1197.2 10.84 10 psi
0.11045
τ
== =×=
P
A
10.84 ksi
(b) Bearing stress at C in member BCD:
1197.2 lb=P
2
35 0.234375 in
88
  
= = =
  
  
b
A dt
3
1197.2 5.11 10 psi
0.234375
s
== =×=
bb
P
A
5.11ksi
consent of McGraw–Hill Education.
PROBLEM 8.23
Knowing that
θ
= 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
SOLUTION Continued
(a) Allowable pin diameter.
22
4
2
22
π
τπ
= = =
AB AB AB
P
FF F
Add
where
3
24.727 10 N= ×
AB
F
3
2 62
6
2 (2)(24.727 10 ) 131.181 10 m
(120 10 )
πτ π
−
×
= = = ×
×
AB
F
d
3
11.4534 10 m
−
= ×d
11.45 mm
(b) Bearing stress in AB at A.
3 62
36
6
(0.016)(11.4534 10 ) 183.254 10 m
24.727 10 134.933 10 Pa
183.254 10
σ
−−
−
== ×= ×
×
= = = ×
×
b
AB
bb
A td
F
A
134.9 MPa
(c) Bearing stress in support brackets at B.
3 62
136
26
(0.012)(11.4534 10 ) 137.441 10 m
(0.5)(24.727 10 ) 89.955 10 Pa
137.441 10
−−
−
== ×= ×
×
×
AB
b
A td
F
A
consent of McGraw–Hill Education.
PROBLEM 8.24
Determine the largest load P which may be applied at A when
θ
= 60°, knowing that the average shearing stress in the 10–mm–
diameter 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
Use joint A as a free body.
Law of sines applied to force triangle:
sin 30 sin 120 sin 30
sin30 0.57735
sin 120
sin 30
sin 30
= =
° °°
°
= =
°
°
= =
°
AB AC
AB AB
AC AC
PF F
F
PF
F
PF
consent of McGraw–Hill Education.
SOLUTION Continued
If shearing stress in pin at B is critical,
2 2 62
66 3
(0.010) 78.54 10 m
44
2 (2)(78.54 10 )(120 10 ) 18.850 10 N
ππ
τ
−
−
= = = ×
== × ×= ×
AB
Ad
FA
62
66 3
(0.016)(0.010) 160 10 m
(160 10 )(90 10 ) 14.40 10 N
b
AB b b
A td
FA
σ
−
−
= = = ×
= =× ×= ×
62
66 3
2 (2)(0.012)(0.010) 240 10 m
(240 10 )(90 10 ) 21.6 10 N
σ
−
−
= = = ×
= =× ×=×
b
AB b b
A td
FA
Then from statics,
3
allow (0.57735)(14.40 10 )= ×P
3
8.31 10 N= ×
8.31 kN
consent of McGraw–Hill Education.
PROBLEM 8.25
Two wooden members of uniform rectangular cross section are joined
by the simple glued scarf splice shown. Knowing that P = 11 kN,
determine the normal and shearing stresses in the glued splice.
SOLUTION
3
3 2 32
0
2 32 3
3
0
90 45 45
11 kN 11 10 N
(150)(75) 11.25 10 mm 11.25 10 m
cos (11 10 )cos 45
489 10 Pa
11.25 10
θ
θ
s
−
−
= °− °= °
= = ×
= =×=×
×°
= = = ×
×
P
A
P
A
489 kPa
s
=
33
3
0
sin 2 (11 10 )(sin90 ) 489 10 Pa
2(2)(11.25 10 )
θ
τ
−
×°
= = = ×
×
P
A
489 kPa
τ
=
PROBLEM 8.26
Two wooden members of uniform rectangular cross section are joined
by the simple glued scarf splice shown. Knowing that the maximum
allowable shearing stress in the glued splice is 620 kPa, determine
(a) the largest load P that can be safely applied, (b) the corresponding
tensile stress in the splice.
SOLUTION
3 2 32
0
3
0
90 45 45
(150)(75) 11.25 10 mm 11.25 10 m
620 kPa 620 10 Pa
sin 2
2
A
P
A
θ
τ
θ
τ
−
= °− °= °
= =×=×
= = ×
=
(a)
33
0
2 (2)(11.25 10 )(620 10 )
sin2 sin90
τ
θ
−
××
= = °
A
P
3
13.95 10 N= ×
13.95 kNP=
(b)
2 32
3
0
cos (13.95 10 )(cos45 )
11.25 10
θ
s
−
×°
= = ×
P
A
3
620 10 Pa
= ×
620 kPa
s
=
SOLUTION Continued
(b) Bearing stress at C in member BC.
σ
=
BC
b
F
A
2
(0.5)(0.8) 0.4 in
8.9658 22.4
0.4
σ
= = =
= =
b
A td
(c) Bearing stress at B in member BC.
σ
=
BC
b
F
A
2
2 2(0.5)(0.8) 0.8 in
8.9658 11.21
0.8
σ
= = =
= =
b
A td
11.21 ksi
σ
=
b
consent of McGraw–Hill Education.
PROBLEM 8.22
The hydraulic cylinder CF, which partially controls the position of rod
DE, has been locked in the position shown. Member BD is
5
8
in. thick
and is connected to the vertical rod by a
3
8
–in.–diameter bolt. Determine
(a) the average shearing stress in the bolt, (b) the bearing stress at C in
member BD.
SOLUTION
Use member BCD as a free body, and note that AB is a two force member.
22
8 1.8 8.2 in.=+=
AB
l
8 1.8
0: (4cos20 ) (4sin 20 )
8.2 8.2
(7cos20 )(400sin 75 ) (7sin 20 )(400cos75 ) 0
3.36678 2789.35 0 828.49 lb
 
Σ= ° − °
 
 
− ° °− ° °=
− = ∴=
C AB AB
AB AB
M FF
FF
1.8
0: 400cos75 0
8.2
(1.8)(828.49) 400cos75 78.34 lb
8.2
Σ = − + + °=
= − °=
x AB x
x
F FC
C
22
8
0: 400sin75 0
8.2
(8)(828.49) 400sin75 1194.65 lb
8.2
1197.2 lb
Σ = − + − °=
= + °=
= +=
y AB y
y
xy
F FC
C
C CC
consent of McGraw–Hill Education.
SOLUTION Continued
(a) Shearing stress in the bolt:
1197.2 lb=P
2
22
30.11045 in
4 48
ππ

= = =


Ad
3
1197.2 10.84 10 psi
0.11045
τ
== =×=
P
A
10.84 ksi
(b) Bearing stress at C in member BCD:
1197.2 lb=P
2
35 0.234375 in
88
  
= = =
  
  
b
A dt
3
1197.2 5.11 10 psi
0.234375
s
== =×=
bb
P
A
5.11ksi
consent of McGraw–Hill Education.
PROBLEM 8.23
Knowing that
θ
= 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
SOLUTION Continued
(a) Allowable pin diameter.
22
4
2
22
π
τπ
= = =
AB AB AB
P
FF F
Add
where
3
24.727 10 N= ×
AB
F
3
2 62
6
2 (2)(24.727 10 ) 131.181 10 m
(120 10 )
πτ π
−
×
= = = ×
×
AB
F
d
3
11.4534 10 m
−
= ×d
11.45 mm
(b) Bearing stress in AB at A.
3 62
36
6
(0.016)(11.4534 10 ) 183.254 10 m
24.727 10 134.933 10 Pa
183.254 10
σ
−−
−
== ×= ×
×
= = = ×
×
b
AB
bb
A td
F
A
134.9 MPa
(c) Bearing stress in support brackets at B.
3 62
136
26
(0.012)(11.4534 10 ) 137.441 10 m
(0.5)(24.727 10 ) 89.955 10 Pa
137.441 10
−−
−
== ×= ×
×
×
AB
b
A td
F
A
consent of McGraw–Hill Education.
PROBLEM 8.24
Determine the largest load P which may be applied at A when
θ
= 60°, knowing that the average shearing stress in the 10–mm–
diameter 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
Use joint A as a free body.
Law of sines applied to force triangle:
sin 30 sin 120 sin 30
sin30 0.57735
sin 120
sin 30
sin 30
= =
° °°
°
= =
°
°
= =
°
AB AC
AB AB
AC AC
PF F
F
PF
F
PF
consent of McGraw–Hill Education.
SOLUTION Continued
If shearing stress in pin at B is critical,
2 2 62
66 3
(0.010) 78.54 10 m
44
2 (2)(78.54 10 )(120 10 ) 18.850 10 N
ππ
τ
−
−
= = = ×
== × ×= ×
AB
Ad
FA
62
66 3
(0.016)(0.010) 160 10 m
(160 10 )(90 10 ) 14.40 10 N
b
AB b b
A td
FA
σ
−
−
= = = ×
= =× ×= ×
62
66 3
2 (2)(0.012)(0.010) 240 10 m
(240 10 )(90 10 ) 21.6 10 N
σ
−
−
= = = ×
= =× ×=×
b
AB b b
A td
FA
Then from statics,
3
allow (0.57735)(14.40 10 )= ×P
3
8.31 10 N= ×
8.31 kN
consent of McGraw–Hill Education.
PROBLEM 8.25
Two wooden members of uniform rectangular cross section are joined
by the simple glued scarf splice shown. Knowing that P = 11 kN,
determine the normal and shearing stresses in the glued splice.
SOLUTION
3
3 2 32
0
2 32 3
3
0
90 45 45
11 kN 11 10 N
(150)(75) 11.25 10 mm 11.25 10 m
cos (11 10 )cos 45
489 10 Pa
11.25 10
θ
θ
s
−
−
= °− °= °
= = ×
= =×=×
×°
= = = ×
×
P
A
P
A
489 kPa
s
=
33
3
0
sin 2 (11 10 )(sin90 ) 489 10 Pa
2(2)(11.25 10 )
θ
τ
−
×°
= = = ×
×
P
A
489 kPa
τ
=
PROBLEM 8.26
Two wooden members of uniform rectangular cross section are joined
by the simple glued scarf splice shown. Knowing that the maximum
allowable shearing stress in the glued splice is 620 kPa, determine
(a) the largest load P that can be safely applied, (b) the corresponding
tensile stress in the splice.
SOLUTION
3 2 32
0
3
0
90 45 45
(150)(75) 11.25 10 mm 11.25 10 m
620 kPa 620 10 Pa
sin 2
2
A
P
A
θ
τ
θ
τ
−
= °− °= °
= =×=×
= = ×
=
(a)
33
0
2 (2)(11.25 10 )(620 10 )
sin2 sin90
τ
θ
−
××
= = °
A
P
3
13.95 10 N= ×
13.95 kNP=
(b)
2 32
3
0
cos (13.95 10 )(cos45 )
11.25 10
θ
s
−
×°
= = ×
P
A
3
620 10 Pa
= ×
620 kPa
s
=