Structural Analysis Ch 6
APPLICATIONS
Long trusses are often used to construct bridges.
The method of joints requires that many joints be
analyzed before we can determine the forces in the middle
part of the truss.
THE METHOD OF SECTIONS
In the method of sections, a truss is divided into two parts by
taking an imaginary “cut” (shown here as a-a) through the truss.
Since truss members are subjected to only tensile or compressive
forces along their length, the internal forces at the cut member will
also be either tensile or compressive with the same magnitude.
This result is based on the equilibrium principle and Newton’s
third law.
STEPS FOR ANALYSIS
1. Decide how you need to “cut” the truss. This is based on:
a) where you need to determine forces, and, b) where the total
number of unknowns does not exceed three (in general).
2. Decide which side of the cut truss will be easier to work
with (minimize the number of reactions you have to find).
3. If required, determine the necessary support reactions by
drawing the FBD of the entire truss and applying the E-of-E.
STEPS FOR ANALYSIS (continued)
4. Draw the FBD of the selected part of the cut truss. We need to
indicate the unknown forces at the cut members. Initially we
may assume all the members are in tension, as we did when
using the method of joints. Upon solving, if the answer is
positive, the member is in tension as per our assumption. If the
answer is negative, the member must be in compression.
STEPS FOR ANALYSIS (continued)
5. Apply the equations of equilibrium (E-of-E) to the selected cut
section of the truss to solve for the unknown member forces.
Please note that in most cases it is possible to write one
equation to solve for one unknown directly.
PROBLEM
Given: Loading on the
truss as shown.
Find: The force in
members BC, BE,
and EF.
Plan:
a) Take a cut through the members BC, BE, and EF.
b) Analyze the top section (no support reactions!).
c) Draw the FBD of the top section.
d) Apply the equations of equilibrium such that every
equation yields answer to one unknown.
SOLUTION
+ FX = 5 + 10 – FBE cos 45º = 0
FBE = 21.2 kN (T)
+ ME = – 5(4) + FCB (4) = 0
FCB = 5 kN (T)
+ MB = – 5 (8) – 10 (4) – 5 (4) – FEF (4) = 0
FEF = – 25 kN or 25 kN (C)
PROBLEM
Determine the force in members GE,GC, and BC of the truss shown
in Fig. Indicate whether the members are in tension or compression.
PROBLEM
Plan:
a) Take a cut through the members GE, GC, and BC.
b) Determine the support reactions.
c) Work with the left part of the cut section.
d) Apply the Equilibrium Equations to find the forces in GE, GC, and
BC.
PROBLEM
PROBLEM
PROBLEM
EXAMPLE
Given: Loads as shown on the
roof truss.
Find: The force in members
DE, DL, and ML.
Plan:
a) Take a cut through the members DE, DL, and ML.
b) Work with the left part of the cut section.
c) Determine the support reaction at A.
d) Apply the Equilibrium Equations to find the forces in DE,
DL, and ML.
EXAMPLE (continued)
Analyzing the entire truss, we get FX = AX = 0. By
symmetry, the vertical support reactions are
AY = IY = 36 kN
+ MD = – 36 (8) + 6 (8) + 12 (4) + FML (5) = 0
FML = 38.4 kN ( T )
EXAMPLE (continued)
+ ML = –36 (12) + 6 (12) + 12 (8) + 12 (4) – FDE ( 4/17)(6) = 0
FDE = –37.11 kN or 37.1 kN (C)
+ FX = 38.4 + (4/17) (–37.11) + (4/41) FDL = 0
FDL = –3.84 kN or 3.84 kN (C)
