Cantilever Method – Analysis for Lateral Loads
Origin and Applicability
- Developed
by: A. C. Wilson (1908)
- Suitable
for:
- Tall,
slender buildings (height >> width)
- Frames
where lateral deflection dominates (wind/earthquake loads).
- The
structure behaves like a vertical cantilever beam fixed at the
base.
Basic Concept
- The
entire frame is idealized as a cantilever beam subjected to lateral
loads.
- The columns
represent the longitudinal fibers of the cantilever beam.
- The axial
stress in columns varies linearly with their distance from the centroidal
axis (neutral axis).
- Columns
on one side → Compression
- Columns
on the other side → Tension
Assumptions
1.
Inflection Points:
o Each
beam and column has an inflection point (zero bending moment) at its mid-height
or midspan — same as in the Portal Method.
2.
Axial Force Distribution:
o On
any story, axial forces in the columns vary linearly with their horizontal
distance from the centroid of all column areas on that story.
o If
all columns have equal cross-sectional areas, then:
where 
= axial force in column i,
and 
= distance from centroid.
3.
Direction of Forces:
o When
the frame sways to the right:
§ Right-side
columns → Compression
§ Left-side
columns → Tension
o When
it sways to the left, the condition reverses.
Procedure for Analysis — Cantilever Method
Step 1: Simplify the Frame
- Draw a
simplified model by inserting an internal hinge at the midpoint
of each beam and column.
→ This represents the inflection
points (where bending moment ≈ 0).
Step 2: Determine Column Axial Forces
For each story:
(a) Pass a horizontal
section through all column mid-heights, dividing the frame into two
parts — the upper portion (free body) and the lower portion.
(b) Draw the free-body
diagram (FBD) of the upper portion.
At the hinge cuts (column mid-heights), only axial forces and shears act
— no moments, since the hinge is a zero-moment point.
(c) Find the centroid
of the cross-sectional areas of all columns on that story:
(d) Assuming that the axial forces in the columns
are proportional to their distances from the centroid, determine the
column axial forces by applying the moment equilibrium equation, ∑ M = 0, to
the free body of the frame above the section. To eliminate the unknown
column shears from the equilibrium equation, the moments should be summed about
one of the internal hinges at the column mid- heights through which the section
has been passed.
🔸 Columns on one side of
the centroid → Compression
🔸
Columns on the other side → Tension
Step 3: Free-Body Diagrams of Members and Joints
- Show all
external loads, column axial forces, and internal forces
on an FBD of each joint and member.
- This
setup is used for shear and moment calculations in the next step.
Step 4: Determine Girder (Beam) Shears and Moments
For each story:
(a) Start at the leftmost
joint, apply vertical equilibrium (∑Fy = 0) to find the shear
at the left end of the next girder.
(b) Determine the moment
at the left end of the girder using:
(since moment at girder midspan
= 0, due to internal hinge)
(c) Using ∑Fy = 0 and
∑M = 0, compute the shear and moment at the right end
of the girder.
(d) Move to the next
joint on the right and repeat (a)–(c).
At the last joint, check equilibrium (∑Fy = 0) to ensure correctness.
Step 5: Determine Column Moments and Shears
Start from the top story:
(a) Apply moment
equilibrium (∑M = 0) at each joint to find the moment at the upper end
of the column below that joint.
(b) Compute column shear
at the upper end:
(where 
= column height, moment is
assumed zero at midheight hinge).
(c) Apply ∑Fx = 0 and
∑M = 0 to the column’s FBD to find moment and shear at its
lower end.
(d) Repeat for all columns,
story by story, moving downward.
Step 6: Determine Girder Axial Forces
For each story:
- Starting
from the far left joint, apply horizontal equilibrium (∑Fx = 0)
successively to each joint to find axial forces in girders.
Step 7: Check Support Reactions (Verification)
At the base of the frame:
- The forces
and moments at the bottom of the lowest columns are the support
reactions.
- Check overall
equilibrium of the entire frame:
If satisfied → the analysis is consistent and correct.
Example:
Determine the approximate axial forces, shears, and moments
for all the members of the frame shown in Fig.(a) by using the cantilever
method.
Solution:
Simplified Frame:
The simplified frame,
obtained by inserting internal hinges at midpoints of all the members of the
given frame, is shown in Fig.(b).
Column axial forces:
Assuming that the
cross-sectional areas of the columns are equal, we determine the location of
the centroid of the three columns from the left column DG by using the
relationship
The lateral loads are acting on the frame to the right, so
the axial force in column DG, which is to the left of the centroid, must
be tensile, whereas the axial forces in the columns EH and FI,
located to the right of the centroid, must be compressive.
Since the axial forces in the columns are assumed to vary
linearly with their horizontal distances from the centroid, the ratios between
the column axial forces can be obtained by using similar triangles drawn
between the force distribution diagram and the column layout.
By summing moments about
the left internal hinge J, we eliminate the unknown column shears and obtain an
equilibrium equation involving only the axial forces.
From the relationships
and above equation:
The axial forces in the first-story columns can be
determined in a similar manner by introducing section bb through the
internal hinges of the lower-story columns. Because the column layout is
identical in both stories, the horizontal location of the centroid—and
therefore the proportional relationships between the axial forces—remains the
same as in the second story. Thus, the same distance ratios used previously
apply directly to the first-story columns.
By summing
moments about the internal hinge K,
From the relationships
and above equation:
Girder Shears and Moments:
With the column axial forces known, the shears in the girders can be determined
by applying vertical equilibrium at the joints. Beginning at the upper-left
joint G, we apply the equilibrium equation
to the free body of the joint (see Fig. (e)), which gives
the shear at the left end of girder GH as
The corresponding moment at the left end of the girder is
found by multiplying this shear by half the girder length:
Next, the shear and moment at the right end of girder GH
(joint H) are obtained by applying
to the free body of the girder. These equilibrium equations
give
Note that the end moments 
and 
have equal magnitude and the same
rotational sense.
The shears and moments for other girders both in second and
first story are then determined in the same manner.
Column Moments and Shears:
With the girder end moments now known, the column end moments can be found by
enforcing moment equilibrium at the joints. Starting at the second story, we
apply
to the free body of joint G (see Fig.(e)), which
gives the moment at the top of column DG as
The corresponding shear at the upper end of column DG is
determined by dividing the end moment by half the column height:
→
This shear must act to the right in order to produce
a clockwise moment that balances the counterclockwise girder moment at joint G.
The moment and shear at the lower end of the column (joint D)
are then obtained by applying the equilibrium equations to the free body of
column DG (Fig. (e)).
The same procedure is repeated for other columns in both
second and first story.
Girder Axial Forces:
The computation of girder axial forces begins at the upper-left joint G.
Applying the horizontal equilibrium equation
to the free-body diagram of joint G (Fig. (e)) gives the
axial force in girder GH as
The axial force in girder HI is then obtained in the
same manner by considering the horizontal equilibrium of joint H.
The axial forces in the first-story girders DE and EF
are evaluated next by applying horizontal equilibrium at joints D and E,
respectively.
Reactions:
The forces and moments acting at the lower ends of the first-story columns AD,
BE, and CF represent the support reactions at the fixed bases A,
B, and C, respectively.
Checking Computations:
To verify the results, we apply the three equilibrium equations to the free
body of the entire frame. This ensures that all computed column and girder
forces, shears, and moments satisfy overall static equilibrium.
Home Work:
Determine the approximate axial forces, shears, and moments for all the members of the frames shown in Fig. 1 through Fig. 10 by using
Portal Method and
Cantilever Method.
Figure:1
Figure:2
Figure:3
Figure:4
Figure:5
Figure:6
Figure:7
