consent of McGraw–Hill Education.
Structure (c):
Nonsimple truss with
3,r=
13,m=
8n=
so
16 2 .rm n+= =
To further examine, follow procedure in
part (a) above to get truss at left.
Since
10
≠
F
(from solution of joint F),
1A
M aFΣ=
0≠
and there is no equilibrium.
Structure is improperly constrained.
consent of McGraw–Hill Education.
PROBLEM 6.48
Classify each of the structures shown as completely, partially, or improperly constrained; if completely
constrained, further classify as determinate or indeterminate. (All members can act both in tension and in
compression.)
SOLUTION
Structure (a):
consent of McGraw–Hill Education.
Structure (c):
Simple truss with
3,r=
17,m=
10,n=
20 2 ,mr n+= =
but the horizontal reaction forces
and
xx
AE
are collinear and no equilibrium equation will resolve them, so the structure is improperly constrained
and indeterminate.
consent of McGraw–Hill Education.
PROBLEM 6.49
Determine the force in member BD and the components of the
reaction at C.
SOLUTION
We note that BD is a two–force member. The force it exerts on ABC, therefore, is directed along line BD.
Free body: ABC:
BD
0:
C
M
Σ=
450
(400 N)(135 mm) (240 mm) 0
510
BD
F

+=


255 N
BD
F= −
255 N
BD
FC=
240
0: ( 255 N) 0
510
xx
FCΣ= + − =
120.0 N
x
C= +
120.0 N
x=C
450
0: 400 N ( 255 N) 0
510
yy
FCΣ= − + − =
625 N
y
C= +
625 N
y=C
consent of McGraw–Hill Education.
PROBLEM 6.50
Determine the force in member BD and the components
of the reaction at C.
SOLUTION
We note that BD is a two–force member. The force it exerts on ABC, therefore, is directed along line BD.
PROBLEM 6.51
Determine the force in member BD and the components of the reaction
at C.
SOLUTION
We note that BD is a two–force member. The force it exerts on ABC, therefore, is directed along line BD.
PROBLEM 6.52
Determine the components of all forces acting on member ABCD of
the assembly shown.
SOLUTION
Free body: Entire assembly:
0: (6 in.) (120 lb)(4 in.) 0
B
MDΣ= − =
80.0 lb=D
0: 120 lb 0
xx
FBΣ= + =
120.0 lb
x
=B
0: 80 lb 0
yy
FBΣ= + =
80.0 lb
y
=B
ABCD:
0: (80 lb)(10 in.) (8 in.) (120 lb)(2 in.)
(80 lb)(4 in.) 0
A
MCΣ= − −
−=
30.0 lb=C
0: 120 lb 0
xx
FAΣ= − =
120.0 lb
x
=
A
0: 80 lb 30 lb 80 lb 0
yy
FAΣ= − − + =
30.0 lb
y=A
consent of McGraw–Hill Education.
PROBLEM 6.53
Determine the components of all forces acting on member
ABCD when
0.
θ
=
SOLUTION
Free body: Entire assembly
0: (8 in.) (60 lb)(20 in.) 0
B
MA
Σ= − =
150 lbA=
150.0 lb=A
0: 150.0 lb 0 150 lb
xx x
FB BΣ= + = =−
150.0 lb
x
=B
0: 60.0 lb 0 60.0 lb
yy y
FB BΣ= − = =+
60.0 lb
y
=B
Free body: Member ABCD We note that D is directed along DE, since DE is a two–force member.
0: (12) (60 lb)(4) (150 lb)(8) 0
C
MDΣ= − + =
80 lbD= −
80.0 lb=D
0: 150.0 150.0 0 0
xx x
FC C
Σ= + − = =
0: 60.0 80.0 0 20.0 lb
yy y
FC CΣ= + − = =+
20.0 lb=C
consent of McGraw–Hill Education.
PROBLEM 6.54
Determine the components of all forces acting on member
ABCD when
90 .
θ
= °
SOLUTION
Free body: Entire assembly
0: (8 in.) (60 lb)(8 in.) 0
B
MAΣ= − =
60.0 lbA= +
60.0 lb=A
0: 60 lb 60 lb 0 0
xx x
FB BΣ= + − = =
0: 0
yy
FBΣ= =
0=B
Free body: Member ABCD We note that D is directed along DE, since DE is a two–force member.
0: (12 in.) (60 lb)(8 in.) 0
C
MDΣ= + =
40.0 lbD= −
40.0 lb=D
0: 60 lb 0
xx
FCΣ= + =
60 lb
x
C= −
60.0 lb
x
=
C
0: 40 lb 0
yy
FCΣ= − =
40 lb
y
C= +
40.0 lb
y
=C
consent of McGraw–Hill Education.
Structure (c):
Nonsimple truss with
3,r=
13,m=
8n=
so
16 2 .rm n+= =
To further examine, follow procedure in
part (a) above to get truss at left.
Since
10
≠
F
(from solution of joint F),
1A
M aFΣ=
0≠
and there is no equilibrium.
Structure is improperly constrained.
consent of McGraw–Hill Education.
PROBLEM 6.48
Classify each of the structures shown as completely, partially, or improperly constrained; if completely
constrained, further classify as determinate or indeterminate. (All members can act both in tension and in
compression.)
SOLUTION
Structure (a):
consent of McGraw–Hill Education.
Structure (c):
Simple truss with
3,r=
17,m=
10,n=
20 2 ,mr n+= =
but the horizontal reaction forces
and
xx
AE
are collinear and no equilibrium equation will resolve them, so the structure is improperly constrained
and indeterminate.
consent of McGraw–Hill Education.
PROBLEM 6.49
Determine the force in member BD and the components of the
reaction at C.
SOLUTION
We note that BD is a two–force member. The force it exerts on ABC, therefore, is directed along line BD.
Free body: ABC:
BD
0:
C
M
Σ=
450
(400 N)(135 mm) (240 mm) 0
510
BD
F

+=


255 N
BD
F= −
255 N
BD
FC=
240
0: ( 255 N) 0
510
xx
FCΣ= + − =
120.0 N
x
C= +
120.0 N
x=C
450
0: 400 N ( 255 N) 0
510
yy
FCΣ= − + − =
625 N
y
C= +
625 N
y=C
consent of McGraw–Hill Education.
PROBLEM 6.50
Determine the force in member BD and the components
of the reaction at C.
SOLUTION
We note that BD is a two–force member. The force it exerts on ABC, therefore, is directed along line BD.
PROBLEM 6.51
Determine the force in member BD and the components of the reaction
at C.
SOLUTION
We note that BD is a two–force member. The force it exerts on ABC, therefore, is directed along line BD.
PROBLEM 6.52
Determine the components of all forces acting on member ABCD of
the assembly shown.
SOLUTION
Free body: Entire assembly:
0: (6 in.) (120 lb)(4 in.) 0
B
MDΣ= − =
80.0 lb=D
0: 120 lb 0
xx
FBΣ= + =
120.0 lb
x
=B
0: 80 lb 0
yy
FBΣ= + =
80.0 lb
y
=B
ABCD:
0: (80 lb)(10 in.) (8 in.) (120 lb)(2 in.)
(80 lb)(4 in.) 0
A
MCΣ= − −
−=
30.0 lb=C
0: 120 lb 0
xx
FAΣ= − =
120.0 lb
x
=
A
0: 80 lb 30 lb 80 lb 0
yy
FAΣ= − − + =
30.0 lb
y=A
consent of McGraw–Hill Education.
PROBLEM 6.53
Determine the components of all forces acting on member
ABCD when
0.
θ
=
SOLUTION
Free body: Entire assembly
0: (8 in.) (60 lb)(20 in.) 0
B
MA
Σ= − =
150 lbA=
150.0 lb=A
0: 150.0 lb 0 150 lb
xx x
FB BΣ= + = =−
150.0 lb
x
=B
0: 60.0 lb 0 60.0 lb
yy y
FB BΣ= − = =+
60.0 lb
y
=B
Free body: Member ABCD We note that D is directed along DE, since DE is a two–force member.
0: (12) (60 lb)(4) (150 lb)(8) 0
C
MDΣ= − + =
80 lbD= −
80.0 lb=D
0: 150.0 150.0 0 0
xx x
FC C
Σ= + − = =
0: 60.0 80.0 0 20.0 lb
yy y
FC CΣ= + − = =+
20.0 lb=C
consent of McGraw–Hill Education.
PROBLEM 6.54
Determine the components of all forces acting on member
ABCD when
90 .
θ
= °
SOLUTION
Free body: Entire assembly
0: (8 in.) (60 lb)(8 in.) 0
B
MAΣ= − =
60.0 lbA= +
60.0 lb=A
0: 60 lb 60 lb 0 0
xx x
FB BΣ= + − = =
0: 0
yy
FBΣ= =
0=B
Free body: Member ABCD We note that D is directed along DE, since DE is a two–force member.
0: (12 in.) (60 lb)(8 in.) 0
C
MDΣ= + =
40.0 lbD= −
40.0 lb=D
0: 60 lb 0
xx
FCΣ= + =
60 lb
x
C= −
60.0 lb
x
=
C
0: 40 lb 0
yy
FCΣ= − =
40 lb
y
C= +
40.0 lb
y
=C
consent of McGraw–Hill Education.