Chapter 1: SIMPLE TRUSSES (Revision)
1.1 Definition
A simple truss is a planar truss which begins with a triangular element and can be expanded
by adding two members and a joint.
A truss is a structure composed of slender members joined together at their end points.
Fig. 1
1.2. The Method of Sections
The method of sections is a process used to solve for the unknown forces acting on members
of a truss. The method involves breaking the truss down into individual sections and analyzing
each section as a separate rigid body.
In this method, we will cut the truss into two sections by passing a cutting plane through the
members whose internal forces we wish to determine. This method permits us to solve directly
any member by analyzing the left or the right section of the cutting plane.
To remain each section in equilibrium, the cut members will be replaced by forces equivalent
to the internal load transmitted to the members.
Each section may constitute of non-concurrent force system from which three equilibrium
equations can be written.
NB: Because we can only solve up to three unknowns, it is important not to cut more than
three members of the truss.
Depending on the type of truss and which members to solve, one may have to repeat Method
of Sections more than once to determine all the desired forces.
Example 1
Using the method of sections, determine the force in members BD, CD, and CE of the roof
truss shown in Fig. 3
Fig. 3
Solutions
Step 1: Sketch a free-body diagram of the structure.
Step 2: Determine the reactions at the supports, using the two conditions for static equilibrium
if they are required or necessary for further calculations.
Step 3: Draw a section line through the structure, cutting those members in which the forces
are required. A maximum of three members should be cut by a section line.
Step 4:
- The external forces on only one side of the section line must be considered.
- Indicate this direction on the section line.
- Indicate the assumed force directions in the members that have been cut by the section
line.
- Take moments about any joint on the structure, considering only the external forces and
forces in the members that have been cut as indicated in steps 3. The selected joint must
have only one unknown force causing a moment about that joint. This force may then
be calculated using the second condition for static equilibrium.
NB: Tie = Tension and Strut = Compression
1.3. The Method of Tensions Coefficients
The tension coefficient for a member of a frame is defined as the pull or tension in that member
is divided by its length.
Consider the following truss member having coordinates of (x1, y1, z1) and (x2, y2, z2) and
force F acting on it:
Fig.4
Unit of tension coefficient (t) = kN/m
Similarly, FAB sinΘ = tAB (y2-y1)
-
Force in any member in any direction is equal to the product of tension coefficient in
that member and difference in the co-ordinates in that direction.
-
Using the above formula and the conditions ΣFx = 0, ΣFy = 0, ΣFz = 0 at any joints of
the truss three equations can be formed and hence three unknown values of tension
coefficient can be found.
-
After finding each of the tension coefficients in the truss it can be multiplied with its
length forces in that particular member can be found.
Example 2
Determine the forces in the members of the pin-jointed truss shown in Fig. 2.
Fig. 5
The equations of equilibrium at joint A are
Substituting the values of RA,H, RA,V and the joint coordinates into the above equations
tAC = 2 kN/m
and from Eq. (y)
tAB(1.5−0) + tAC(0−0)+1=0
tAB= −0.67kN/m
We can now proceed to joint B at which, since tBA (= tAB) has been calculated, there are two
unknowns
x - direction: tBA(xA−xB) + tBC(xC−xB) + tBD(xD–xB) +3 = 0
y - direction: tBA(yA−yB) + tBC(yC−yB) + tBD(yD−yB) = 0
Substituting the values of the joint coordinates and tBA in Eqs (x) and (y) we have, from Eq. (x)
−0.67(0−0) + tBC(1.5−0) + tBD(1.5−0) + 3=0
which simplifies to (x): 1.5tBC + 1.5tBD + 3 = 0
and from Eq. (y): −0.67(0−1.5) + tBC(0−1.5) + tBD(1.5−1.5)=0
tBC= 0.67 kN/m
Hence, from Eq. (y)
tBD= −2.67 kN/m
There are now just two unknown member forces at joint D. Hence, at D
x - direction: tDB(xB−xD) + tDF(xF−xD) + tDC(xC−xD) = 0
y - direction: tDB(yB−yD) + tDF(yF−yD) + tDC(yC−yD) – 5 = 0
Substituting values of joint coordinates and the previously calculated value of tDB (= tBD) in
Eqs (x) and (y) we obtain, from Eq. (x)
−2.67(0−1.5) + tDF(3.0−1.5) + tDC(1.5−1.5)−5 = 0
tDF= –2.67kN/m
and from Eq. (y) −2.67(1.5−1.5) + tDF(1.5−1.5) + tDC(0−1.5) = 0
tDC= –3.33kN/m
The solution then proceeds to joint C to obtain tCF and tCE or to joint F to determine tFC and tFE;
joint F would be preferable since fewer members meet at F than at C. Finally, the remaining
unknown tension coefficient (tEC or tEF) is found by considering the equilibrium of joint E.
Then, tFC=2.67kN/m, tFE=−2.67kN/m, tEC=0
The forces in the truss members are now calculated by multiplying the tension coefficients by
the member lengths, i.e.
TAB = tABLAB = −0.67×1.5 = −1 kN (compression)
TAC = tACLAC = 2×1.5 = 3 kN (tension)
TBC =tBCLBC
= 2.12 m
TBC = 0.67×2.12= 1.42kN (tension)
Note that in the calculation of member lengths it is immaterial in which order the joint
coordinates occur in the brackets since the brackets are squared. Also
TBD = tBDLBD = −2.67×1.5= −4 kN (compression)
TDF =−4 kN (compression)
TDC =−5.0kN (compression)
TFC = 5.67kN (tension)
TFE = −4 kN (compression)
TEC = 0