Two-Force Members,
Three-Force Members,
Distributed Loads
Two-Force Members Examples
ME 202
2
Two-Force Members
Find the force in the strut BC.
Only two forces
act on the body.
The line of action
(LOA) of forces at
both A and B must
lie on the line
connecting A and
B. Why?
Hint: Resolve forces into components
parallel and perpendicular to line AB.
3
4
Three-Force Members
Pick two of the forces.
If their LOA intersect, the
LOA of the third force must
pass through the point of
intersection. Why?
Only two possibilities:
(a) concurrent
(b) parallel
!
∑M
O
No vertical support at A.
Find maximum reaction
magnitudes at A and B if
1.5 ft ≤ x ≤ 7.5 ft
=0
Pick two of the forces.
If their LOA don’t
intersect, the third LOA
must be parallel to the
others. Why?
!
∑F = 0
5
Distributed Force: Resultant
6
Distributed Force: Examples
Uniform
•
•
•
36 lb
=
Height of load diagram is force per unit
length.
Total force: area under load diagram.
3 lb/ft
12 ft
Linear
6 ft
=
8 ft
12 ft
7
18 lb
3 lb/ft
Location of line of action: centroid of
load diagram.
6 ft
8
4 ft
Example
1.4 kN
3 kN/m
FV
m
1.2
30 deg
M
4 kN-m
2.4 m
Resultant of distributed load:
1
( 2.4 m )( 3 kN/m ) = 3.6 kN
2
Location of resultant’s LOA:
2
( 2.4 m ) = 1.6 m (from left end)
3
9
FH
10
Two-Force and Three-Force Members and Distributed Loads
2
A two-force member can have any shape. The only requirement is that it have forces acting on it at only two
points.
3
For any two-force member of a structure, summing moments about either of the two points where a force is
applied will show that because the net moment about that point must be zero, the force at the other point
cannot have a component perpendicular to the line connecting the two points. It follows that the forces are
parallel to the line connecting the two points.
4
Since forces are applied at three points of member AB, AB is not a two-force member. Although the three
forces are coplanar, there is no reason to expect that they are concurrent. As we will see in subsequent notes
on non-concurrent, coplanar systems, we can write three independent equations (by summing forces and
moments), and solve for as many as three unknowns. Since we know neither the magnitude nor the direction
of the force at A, that force gives us two unknowns. Rather than use the magnitude and an angle as the unknowns, we will use the force’s horizontal and vertical components. If we did the same for the force at B, we
would then have four unknowns. But because we will not find more than three independent equations, we
must somehow reduce the number of unknowns to no more than three. The key to the problem is to recogPage 1 of 4
nize that member BC is a two-force member. This tells us the direction of the force at B, so that only its magnitude is unknown. Then we can sum moments about A to get the force in member BC. Finally, we can sum
forces horizontally and vertically to get the two unknown components of the force at A.
5
The answer to the first question on this slide is that because there can be no net moment about any point, and
because the lines of action of two of the forces pass through a common point, the moment about that point
can be zero only if the third force also passes through that point. The answer to the second question is that
because there can be no net force in any direction, and because the two forces whose lines of action are parallel have no net force perpendicular to their lines of action, the third force can have no component perpendicular to those lines of action and so must be parallel to the other two forces.
6
There are two unknown force components at B and one at A. Then If both x and θ are given, we have only
three unknowns and can solve this problem without knowing anything about three-force members. But the
solution will be simpler if we use the fact that the L-shaped structure is a three-force member. Since the force
at A must be horizontal and the applied load at T is vertical, the three forces are not parallel and so must be
concurrent. Therefore, the line of action of the force at B must pass through both B and the point where the
other two forces intersect. This leads to the FBD shown here. The force at B can then be found simply by
summing vertical forces.
Page 2 of 4
7
As mentioned in an earlier set of slides, a real force is not applied at a point, but rather is distributed over a
length, or an area, or a volume.
Gravity exerts a load on every point of your body, which is a force per unit volume. In this course, we only
consider the resultant of that load, which is a force that we call “weight.”
Think of the load beneath your feet that supports you when you stand. It is distributed over the bottoms of
your feet as a pressure, which is force per unit area. If you stand still, the resultant of all that pressure is a net
upward force equal to your weight.
Oftentimes it is convenient to treat a load as a force per unit length. This slide shows the only three things we
need to know about such loads. Think carefully about these things before moving to the next slide.
8
In general, the area under a diagram of a force per unit length can be computed by integral calculus. But in
this course, we will consider only distributions that are constant or change linearly. This slide shows distributions of both kinds. For these simple distributions, we need to compute the area of either a rectangle or a
right triangle. We also need to know that the centroid of a rectangle is midway between the sides, and the
centroid of a right triangle is one-third of the way from a side of the triangle to the opposite vertex. Given the
locations of the centroids and the areas, we can easily compute the resultant forces and locate their lines of
action.
Page 3 of 4
9
In applying Newton’s first law to a FBD of the structure shown here, we will find that summing forces and
moments is easier if we first replace the distributed load with its resultant.
10 Since the left end of the structure is fixed, there could be three unknown reactions there, as shown on this
FBD. Summing forces and moments will generate three independent equations from which we can solve for
these unknowns.
Page 4 of 4