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Structures 1 (Civil & Aerospace)
MACE10010 & MACE11001 13
7) Forces within Structures – Axial Members
So far we have looked at how to calculate the forces a structure exerts on its supports (or,
equivalently, the forces supports exert on a structure). Next we need to determine the forces that
are developed within a structure, so we can then check if its members have sufficient strength. How
to do this depends on the type of structural element. In this section will examine forces in axial
members (columns and ties), and trusses. In later sections we will look at forces and moments in
beams.
The forces in axial members are easy to analyse intuitively. However, it is worth looking at a
rigorous approach as the concepts used are applicable when it comes to the more complex methods
needed for trusses in the next section. Two key ideas are:
• Taking an imaginary cut through a member and asking what internal forces must be acting
for equilibrium to be maintained, and
• Using positive and negative signs to distinguish between tensile and compressive internal
forces
These ideas are demonstrated in the following example and will be used in later sections too.
Figure 10a) shows an axial member with a force of
10kN applied to its end. We know how to find the
force at the support but are now asking how to find
the force within the length of the member.
Intuitively this is going to be a tensile force of 10 kN, but how do we show this formally? The
approach is to imagine cutting the member at some
point along its length as in Figure 10b) and
considering the equilibrium of the “cut off” piece. If
we consider the cut part of the beam in a free body
diagram as in Figure 10c) and apply equilibrium
equations we confirm the intuitive answer: 0 = 10 − (+) ( = 0)
= 10kN (tension)
With internal forces, positive and negative signs are
used to indicate whether a force is tensile or compressive, rather than to indicate its direction. A
tensile force (a force that tends to make a member longer) is normally considered positive while a
compressive force (one that tends to make a member shorted) is normally taken as compressive.
This not a universal convention however, so always be clear about the convention being used.
To ensure internal forces are determined as tensile or compressive correctly in analyses, assume an
unknown internal force is tensile (i.e. positive). If it is in fact compressive, the answer will come out
negative.
In this example the sign of has been explicitly written as positive within brackets to show it is
initially being assumed positive. This is not normally done but great care must be taken to get signs
of forces right in analyses.
8) Forces in Structures - Trusses
Analysis of the forces in a truss is simply analysing the forces in a collection of axial members so
broadly the same approach can be used as for axial members (taking imaginary cuts and looking at
10kN
10kN F
a)
b)
c)
10kN
Figure 10 a) An axial member with a force acting at its
end. b) The same member with an imaginary cut at an
arbitrary point. c) Since the part to left of the cut is in
equilibrium, there must be an internal force F acting.
Structures 1 (Civil & Aerospace)
MACE10010 & MACE11001 14
equilibrium of “pieces” of a structure). However, calculations can quickly become confusing if a
methodical approach isn’t adopted. Two such approaches are common and will be presented here
by reference to a simple example truss. These are called “The Method of Joints” and the “Method of
Sections”.
Method of Joints
This approach looks at the equilibrium of each joint of a truss to determine the member forces. It is
the best method if the forces in all the members of a truss are needed. As an example we will
consider the truss in Figure 11.
Figure 12 Free-body diagram of the truss in Figure 11
The first step is to obtain the support reactions as before. There are three of these - vertical and
horizontal at A, and vertical at E - as shown in the free-body diagram in Figure 12. Applying
equilibrium equations horizontally, rotationally and vertically to the whole truss gives:
0=15+HA (Fx=0)
HA=-15kN
0=(15x1xcos30)-(VEx2) (MA=0)
VE=6.5kN
0=VA+ VE=VA+6.5 (Fy=0)
VA=-6.5kN
Next, apply equilibrium at each joint having made imaginary cuts through all the members that
connect to the joint. Assume any unknown member forces are positive (if they in fact aren’t, the
answer will come out negative). Although in principle joints can be considered in any order, it is best
to choose an order so that only two unknown forces are present at each joint being analysed to
avoid having to solve simultaneous equations. Always draw an FBD for each joint.
15kN
A
B
C
D
E
All internal angles 600
All members 1m
HA
V VE
15kN
A
B
C
D
E
All internal angles 600
All members 1m
Figure 11 An example pin-jointed truss.
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Starting with joint A and “cutting” AB and AC we have forces as
shown in Figure 13. We know the joint is in equilibrium so we
can use the usual equations. Resolving forces vertically gives:
0=-6.5+FABcos30
FAB=7.5kN (tension)
and horizontally
0=-15+7.5cos60+FAC
FAC=11.25kN (tension)
Similarly, at joint B resolving vertically (and deliberately
emphasising signs for the example) gives
0=-(+7.5)cos30-(+FBC)cos30
FBC=-7.5kN (compression)
and horizontally:
0=15-(+7.5cos60)+(-7.5cos60)+FBD
FBD= -7.5kN (compression)
At joint E resolving vertically gives
0=6.5+FDEcos30
FDE=-7.5kN (compression)
and horizontally:
0=-(+FCE)-(-7.5)cos60
FCE=3.8kN (tension)
Finally at joint D resolving vertically gives
0=-(-7.5)cos30-(+FDC)cos30
FDC=7.5kN (tension)
And we have the forces in all the members.
Method of Sections
One drawback with the method of joints is that to get the force in a specific member of a truss often
requires a lot of time consuming calculation about forces in all the other members that may not be
of interest. The method of sections allows forces in specific members to be determined directly.
A
-15kN
-6.5kN
FAB
FAC
Figure 13 FBD of forces at joint A
B
15kN
FBC
FBD
7.5kN
Figure 14 FBD of forces at joint B
D
-7.5kN
FDE FDC
Figure 16 FBD of forces at joint D
FCE
FDE
6.5kN
E
Figure 15 FBD of forces at joint B
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The approach is to consider the equilibrium of a part of the truss after making an imaginary cut
through a section of it. We will use the same example as before to demonstrate the method.
If in the truss in Figure 11 we wanted to know the forces in members BD, BC and AC but weren’t
interested in the forces in other members, we could use the method of sections. As before, the first
step is to calculate the support reactions. Here we know these from the previous example (HA=-
15kN, VA=-6.5kN and VB=6.5kN). Next we imagine a cut through the truss that passes through each
member we are interested in, as in Figure 17. We then consider the equilibrium of a piece of truss.
As before, we assume that the forces in each cut member are tensile.
Taking moments about B (clockwise positive) in Figure 17 gives
0=(FACx1cos30)+(-15x1cos30)-(-6.5x1cos60)
FAC=11.3kN (tension)
Resolving forces vertically gives
0=-6.5-(+FBCcos30)
FBC=-7.5kN (compression)
And horizontally
0=-15+15+11.25+(-7.5)cos60+FBD
FBD=-7.5kN (compression)
These three results are the same as when analysed using the method of joints. Note that moments
were taken around B which had two of the unknown forces passing through it. This means they do
not appear in the moment equation which can then be solved without using simultaneous
equations.
Shortcuts and Limitations of Truss Analysis
Zero Force Members
Many trusses will have members that carry no load. This may be
because the truss is designed for several load cases and the one being
analysed does not load some members, or it may be that some
members act to prevent compression members buckling (more on this
next year).
Sometimes zero force members can be identified without calculation.
This can make the analysis of a truss much quicker. A simple example
of a truss with zero force members is given in Figure 18 . Imagine
using the method of joints at B. Since there are only two members
here and no external forces acting, the forces in AB and BC can be
seen by inspection to be zero.
15kN
A
B
All internal angles 600
All members 1m
-6.5kN
FBD
FBC
FAC
-15kN
Figure 17 Free-body diagram after "cutting" the truss in Figure 11 for analysis by the Method of Sections.
A B
C
Figure 18 A truss with two zero force
members
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A less obvious example is the truss shown in Figure 19. Here all the internal members carry zero
force. This can be seen by imagining using the method of joints at joint A. Resolving forces
perpendicular to the upper chord gives the force in member AB as zero. With this established it is
possible to use the method of joints at B and resolve forces perpendicular to the bottom chord,
which gives the force in member BC as zero. Continuing in this way, all the internal forces are found
to be zero. With certain forms of truss and loading it thus possible to establish many member forces
very quickly.
Redundant Trusses
Not all trusses can be analysed using the equations of equilibrium and the methods discussed. If
there are more unknown forces than can be determined using equilibrium alone, a truss (or any
structure) is said to be “redundant” or “statically indeterminate”. This is in contrast to the “statically
determinate” trusses analysed in this course. Figure 20 shows three trusses; one is determinate the
other two indeterminate. It is worth trying to analyse all three using the methods above to see why
the second two cannot be solved (there are not enough equations of equilibrium). Note, using the
shortcut discussed above, that the first one has two zero force members before doing any
calculations.
Figure 20 Three trusses. The left-hand one is determinate and can be analysed using the method of joints. The
second is indeterminate due to the change in support condition and the third is indeterminate due to an additional
member. These two cannot be analysed using equilibrium alone. Techniques for analysing indeterminate trusses
will be discussed in later years and require knowledge of the materials used in the members.
A
B
C
Figure 19 Truss with all internal members carrying zero force (The inclined internal members are perpendicular to the
chord)
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Summary
The following steps are applicable to all the trusses that will be met on this course:
0. (Not always needed or possible). Identify by inspection any members with zero force.
1. Use the methods in Section 6) to determine the support reactions
2. Either
a. Use the method of joints (if all member forces are needed), or
b. Use the method of sections (if only certain member forces are needed).
