Engr210 – Spring 2005
Lesson # 20: Frames
Today’s Objectives:
a) Draw the free body diagram of a
frame and its members.
b) Determine the forces acting at
the joints and supports of a
frame or machine.
Instructor: Ahmed Abdel-Rahim
Page 1 of 2
Example 1
Frames are commonly used to support
various external loads.
Frames are structures that have
at least one multi-force member.
(Recall that trusses have nothing
but two-force members
How can you determine the forces at
the joints and supports of a frame?
Draw FBD for the Frame:
a) Identify any two-force members,
b) Forces on contacting surfaces (usually
between a pin and a member) are equal and
opposite, and,
2. Develop a strategy to apply the equations
of equilibrium to solve for the unknowns.
Example 1
Example 2
How is a frame different than a truss?
Pin B
FAB
Engr210 – Spring 2005
Lesson # 20: Frames
Instructor: Ahmed Abdel-Rahim
Page 2 of 2
FBD for each Element
TBD
FBD of the Pulley E
350 lb
T
T
BY
BX
B
45°
4 ft
B
175 lb
303.11 lb
4 ft
350 lb
700 lb
A FBD of member ABC
350 lb
A FBD of pulley R
700 lb
A
AY
30°
E
∑ MA = TBD sin 45° (4) – 303.1 (4) – 700 (8) = 0
TBD = 2409 lb
350 lb
C
Y
C
∑ FY
CX
350 lb
∑ FY = 2 T – 700 = 0, T = 350 lb [E]
∑ FX = CX – 350 = 0, CX = 350 lb [C]
∑ FY = CY – 350 = 0, CY = 350 lb [C]
∑ FX = – BX + 350 – 350 sin 30° = 0 [B]
∑ FY = BY – 350 cos 30° = 0 [B]
BY = 303.1 lb
BX = 175 lb
A FBD of member BD
2409 lb
D
45°
= AY + 2409 sin 45° – 303.1 – 700 = 0
AY = – 700 lb
∑ FX = AX – 2409 cos 45° + 175 – 350 = 0
AX = 1880 lb
A FBD of pulley C
+
+
+
+
+
AX
B
2409 lb
At D, the X and Y component are
+ DX = –2409 cos 45° = –1700 lb
+ DY = 2409 sin 45° = 1700 lb