Member-AB is_____. options: a two-force member a multi-force member a zero-force member a torsional member b) Member-BC is _____. options: a two-force member a multi-force member a zero-force member a torsional member c) Member-DE is _____. options: a two-force member a multi-force member a zero-force member a torsional member
Member-AB is_____. options: a two-force member a multi-force member a zero-force member a torsional member b) Member-BC is _____. options: a two-force member a multi-force member a zero-force member a torsional member c) Member-DE is _____. options: a two-force member a multi-force member a zero-force member a torsional member
Member-AB is_____. options: a two-force member a multi-force member a zero-force member a torsional member b) Member-BC is _____. options: a two-force member a multi-force member a zero-force member a torsional member c) Member-DE is _____. options: a two-force member a multi-force member a zero-force member a torsional member
This solution contains _____ reaction(s) in the "overall system" FBDs.
HINT:
Think "component" form.
i)On any given 2-D FBD, at most _____ Equilibrium Equations (EQ EQ) will assist you in calculating the unknowns.
j)
ASIDE (but useful to realize):
On any given 3-D FBD, at most _____ Equilibrium Equations (EQ EQ) will assist you in calculating the unknowns.
k)
Using the "overall system" FBD ONLY, all of the reactions _____ be calculated using the Equilibrium Equations.
options:
can
cannot
l)This solution contains _____ FBDs.
m)This solution contains _____ FBDs within the "exploded diagram".
Transcribed Image Text:Solution. We note first that the frame is not a rigid unit when removed from
sequently the external reactions cannot be completely determined until the indi-
O its supports since BDEF is a movable quadrilateral and not a rigid triangle. Con-
Neglect the weight of the frame and compute the forces acting on all of its
components of the reactions at A and C from the free-body diagram of the frame
vidual members are analyzed. However, we can determine the vertical
O remaining unknown C, is found from
Article 4/6 Frames and Machines 207
SAMPLE PROBLEM 4/7
MAS
30 lb
members.
20"
12"
D
50 lb
a
20"
E
12"
Bel members are analyzed. However, we can determine the vertical
A
30"
(EMc = 0]
50(12) + 30(40) – 30A, = 0 A, = 60 lb
30 lb
3
%3D
-
Ans.
4
C, - 50(4/5) - 60 = 0
2F, = 0]
= 100 lb
Ans.
Next we dismember the frame and draw the free-body diagram of each part.
Since EF is a two-force member, the direction of the force at E on ED and at F on
AR is known. We assume that the 30-lb force is applied to the pin as a part of
O member BC. There should be no difficulty in assigning the correct directions for
forces E, F, D, and Bx. The direction of By, however, may not be assigned by inspec-
tion and therefore is arbitrarily shown as downward on AB and upward on BC.
50 lb
y
L--x
Ax
Ay
Cy
Member ED. The two unknowns are easily obtained by
1on Helpful Hints
1 We see that this frame corresponds to
the category illustrated in Fig. 4/14b.
2 The directions of A, and C, are not
obvious initially and can be assigned
arbitrarily, to be corrected later if
(EMp = 0]
50(12) – 12E = 0
E = 50 lb
Ans.
%3D
[EF = 0]
D - 50 – 50 = 0
D= 100 lb
Ans.
necessary.
Member EF. Clearly F is equal and opposite to E with the magnitude of 50 lb.
3 Alternatively the 30-lb force could
be applied to the pin considered a
part of BA, with a resulting change
in the reaction Bg.
Member AB. Since F is now known, we solve for Bx, Ax, and B, from
[ZMA = 0]
B, = 15 lb
Ans.
|B,
B,
50(3/5)(20) – B,(40) = 0
y
X.
[ZF, = 0]
A, + 15 50(3/5) = 0
A, = 15 lb
Ans.
B B
30 lb
3
y
Ans.
[ZF, = 0]
B, = -20 lb
50(4/5) – 60 – B, = 0
4
%3D
F
L--x
3
The minus sign shows that we assigned B, in the wrong direction.
4
D
F
Member BC. The results for B., By, and D are now transferred to BC, and the
remaining unknown C, is found from
E
50 lb
Cx
4
[EF, = 0]
C = 75 lb
Ans.
%3D
30 + 100(3/5)- 15 – C = 0
|
A, =
= 60 lb
E
C, = 100 lb
%3D
We may apply the remaining two equilibrium equations as a check. Thus,
(2F, = 0]
4 Alternatively we could have re-
turned to the free-body diagram of
the frame as a whole and found C.
100 + (-20) –- 100(4/5) = 0
%3D
IZMC = 0]
(30 – 15)(40) + (-20)(30) = 0
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![Solution. We note first that the frame is not a rigid unit when removed from
sequently the external reactions cannot be completely determined until the indi-
O its supports since BDEF is a movable quadrilateral and not a rigid triangle. Con-
Neglect the weight of the frame and compute the forces acting on all of its
components of the reactions at A and C from the free-body diagram of the frame
vidual members are analyzed. However, we can determine the vertical
O remaining unknown C, is found from
Article 4/6 Frames and Machines 207
SAMPLE PROBLEM 4/7
MAS
30 lb
members.
20"
12"
D
50 lb
a
20"
E
12"
Bel members are analyzed. However, we can determine the vertical
A
30"
(EMc = 0]
50(12) + 30(40) – 30A, = 0 A, = 60 lb
30 lb
3
%3D
-
Ans.
4
C, - 50(4/5) - 60 = 0
2F, = 0]
= 100 lb
Ans.
Next we dismember the frame and draw the free-body diagram of each part.
Since EF is a two-force member, the direction of the force at E on ED and at F on
AR is known. We assume that the 30-lb force is applied to the pin as a part of
O member BC. There should be no difficulty in assigning the correct directions for
forces E, F, D, and Bx. The direction of By, however, may not be assigned by inspec-
tion and therefore is arbitrarily shown as downward on AB and upward on BC.
50 lb
y
L--x
Ax
Ay
Cy
Member ED. The two unknowns are easily obtained by
1on Helpful Hints
1 We see that this frame corresponds to
the category illustrated in Fig. 4/14b.
2 The directions of A, and C, are not
obvious initially and can be assigned
arbitrarily, to be corrected later if
(EMp = 0]
50(12) – 12E = 0
E = 50 lb
Ans.
%3D
[EF = 0]
D - 50 – 50 = 0
D= 100 lb
Ans.
necessary.
Member EF. Clearly F is equal and opposite to E with the magnitude of 50 lb.
3 Alternatively the 30-lb force could
be applied to the pin considered a
part of BA, with a resulting change
in the reaction Bg.
Member AB. Since F is now known, we solve for Bx, Ax, and B, from
[ZMA = 0]
B, = 15 lb
Ans.
|B,
B,
50(3/5)(20) – B,(40) = 0
y
X.
[ZF, = 0]
A, + 15 50(3/5) = 0
A, = 15 lb
Ans.
B B
30 lb
3
y
Ans.
[ZF, = 0]
B, = -20 lb
50(4/5) – 60 – B, = 0
4
%3D
F
L--x
3
The minus sign shows that we assigned B, in the wrong direction.
4
D
F
Member BC. The results for B., By, and D are now transferred to BC, and the
remaining unknown C, is found from
E
50 lb
Cx
4
[EF, = 0]
C = 75 lb
Ans.
%3D
30 + 100(3/5)- 15 – C = 0
|
A, =
= 60 lb
E
C, = 100 lb
%3D
We may apply the remaining two equilibrium equations as a check. Thus,
(2F, = 0]
4 Alternatively we could have re-
turned to the free-body diagram of
the frame as a whole and found C.
100 + (-20) –- 100(4/5) = 0
%3D
IZMC = 0]
(30 – 15)(40) + (-20)(30) = 0](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6cf78908-a1f9-4330-9ece-f4021df10613%2F7079ce10-71f2-42d0-b65f-a4d4e478e61e%2Ft5hig3m_processed.jpeg&w=3840&q=75)
