Chapter 6 - Analysis of Structures
Sections 6.9 - 6.11
Sections 6.9 - 6.11 - Frames and
Machines



Frames differ from trusses because
they contain at least one multiforce
member.
A multiforce has at least 3 forces acting
on it. Remember that a truss is made
entirely of 2 force members!
In Fig 6.20, the frame shown contains
two multiforce members: AD and CF.
Sections 6.9 - 6.11 - Frames and
Machines cont.


With any multiforce member it can no
longer be assumed that all forces follow
the axis line of the member. This makes
the analysis more difficult.
The procedure for analyzing the forces
in a frame is usually to first solve for the
reactions by considering the frame as a
single rigid body.
Sections 6.9 - 6.11 - Frames and
Machines cont.


Once you have the external reactions
you must separate the members and
examine each with a free body diagram.
In the free body diagram, show all
forces acting on the pins in each
member, all external support reactions
and all external applied loads.
Sections 6.9 - 6.11 - Frames and
Machines cont.


When you solve for an unknown force
acting on a pin, you can apply an equal
and opposite force acting on the same
pin in a connected member. (See Fig
6.20)
For a multiforce member you have three
equilibrium equations to work with.
– Usually MP, Fx, Fy but you may use
more than one moment equation.
Sections 6.9 - 6.11 - Frames and
Machines cont.

To determine whether a frame is
statically determinate, break it into
members and count the unknowns.

If there are more unknowns than there
are independent equations, the frame is
indeterminate.
Sample Problem 6.4
Determine the force in DE and the force at pin C.
Ay
Ax
160 mm
B
480 kN
B
Fx = 0 = 300 N - Ax
Ax = 300 N
100 mm
MA = 0 = B x 160 - 480 x 100
B = (480 x 100)/160 = 300 N
Fy = 0 = - Ay - 480
Ay = - 480 N
Sample Problem 6.4 cont.
Ay = 480 N
Ax = 300 N
220 mm
Cx
DE
80 mm
170
Cy
80
150
100 mm
MC = 0 = 300 N x 220 mm + (150/170)DE x 80 mm
+ (80/170)DE x 100 mm
0 = 300 x 220 + 70.6 DE + 47 DE
DE = -66,000/117.6 = -561.2 N (Compression)
Sample Problem 6.4 cont.
At pin C:
Fx = 0 = - Cx - 300 N - (150/170)(561)
Cx = 795 N
Fy = 0 = - Cy + 480 N - (80/170)(DE)
Cy = 216 N
•Now these are the forces exerted at C on member ACE.
•However for member BCD these forces would have to be
turned around.
Cy = 216 N
B
Cx = 795 N
480
Sample Problem 6.5
Determine the components of the forces acting on each
member of the frame shown.
3.6 m
A
2.7 m
C
D
B
2.7 m
E
F
4.8 m
Sample Problem 6.5 cont.
Free Body: Entire Frame
3.6 m
A
C
2400 N
D
B
Ex
E
F
Ey
F
4.8 m
Since the external reactions involve
only three unknowns, we compute
the reactions by considering the
free-body diagram of the entire
frame.
Sample Problem 6.5 cont.
+
ME = 0: -(2400 N)(3.6 m) + F(4.8 m) = 0
F = + 1800 N
+
Fy = 0: -2400 N + 1800 N + Ey = 0
Ey = + 600 N
+
F = 1800 N
Fx = 0:
Ey = 600 N
Ex = 0
Members. The frame is now dismembered; since only two
members are connected at each joint, equal and opposite
components are shown on each member at each joint.
Sample Problem 6.5 cont.
By
Free Body: Member BCD
+
Bx
2.4 m
Cy
B
1.2 m 2400 N
C Cx
MB = 0: -(2400 N)(3.6 m) + Cy(2.4 m) = 0
Cy = + 3600 N
+
MC = 0: -(2400 N)(1.2 m) + By(2.4 m) = 0
By = + 1200 N
+
Fx = 0: -Bx + Cx = 0
We note that neither Bx nor Cx can be obtained by considering only
member BCD. The positive values obtained for By and Cy indicate
that the force components By and Cy are directed as assumed.
D
Sample Problem 6.5 cont.
Free Body: Member ABE
+
MB = 0: Bx(2.7 m) = 0
Bx = 0
+
Fx = 0: +Bx - Ax = 0
Ax = 0
+
Fy = 0: -Ay + By + 600 N = 0
-Ay + 1200 N + 600 N = 0
A
Ay = +1800 N
Ax
Ay
2.7 m
B
Free Body: Member BCD
By
Bx
2.7 m
E
600 N
Returning now to member BCD, we write
+
Fx = 0: -Bx + Cx = 0; 0 + Cx = 0
Cx = 0
Sample Problem 6.5 cont.
Free Body: Member ACF (check)
All unknown components have now been found; to
check the results, we verify that member ACF is in
equilibrium.
MC = (1800 N)(2.4 m) - Ay(2.4 m) - Ax(2.7 m)
+
= (1800 N)(2.4 m) - (1800 N)(2,4 m) - 0 = 0
Ay
A
Ax
0 = 0 (checks)
C
Cx
Cy
2.4 m
F
1800 N
Sample Problem 6.6
Determine the forces acting on the vertical members of the frame.
600 lb
(1)
Since this frame has 2 pin
supports, it is technically
indeterminate.
10'
E
F
Ex
Ey
6'
Fy
ME = 0 = - 600 x 10' + Fy x 6'
Fy = 6000/6 = 1000 lb (1)
Fy = 0 = 1000 + Ey
Ey = -1000 lb (1)
Fx
However you can still
solve for Ey and Fy, which
will get you started.
Sample Problem 6.6 cont.
600 lb
(2)
2.5 6.5
6
AB
7.5
Now write a FBD for AE.
CD
Fy = 0 = -1000 + (2.5/6.5)CD
+ (2.5/6.5)AB
ME = 0 = (6/6.5)AB x 10'
-(6/6.5)CD x 2.5 - 600 x 10
CD = (1000 - .385)AB/.385
CD = 2597 - AB
6.5 2.5
6
2.5
Ex
Ey = 1000 lb
Sample Problem 6.6 cont.
0 = 9.23 AB - 2.31 CD - 6000
0 = 9.23 AB - 2.31(2597 - AB) - 6000
0 = 11.54 AB - 12,000
AB = 12000/11.54 = 1040 lb (force as drawn in FBD (2))
CD = 2597 - 1040 = 1557 lb (force as drawn in FBD (2))
and Fx = 0 = 600 lb - (6/6.5)AB + (6/6.5)CD - Ex
0 = 600 - 960 + 1437 - Ex
Ex = 1077 lb
(2)
Sample Problem 6.6 cont.
Now for BF
AB
Fy = 0 = -((6/6.5)AB x 7.5)
+ ((6/6.5)CD x 5.0)
2.5
6
6.92 AB = 4.62 CD
2.5
Fy = 0 = -(2.5/6.5)AB - (2.5/6.5)CD
+1000
D
6.5
CD
1000 = .385 AB + .385 CD
5.0
F
1000 = .258 CD + .385 CD
AB = 1040 lb
2.5
6
AB = (4.62/6.92)CD = .67 CD
CD = 1556 lb
B
6.5
Fx
Fy
Fx = 0 = (6/6.5)AB - (6/6.5)CD + Fx
Fx = 480 lb