Approximate Analysis of Rectangular Building Frames
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Used in preliminary design to estimate internal
forces without computing deflections.
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Simplifies design by assuming deformation or
force distribution patterns.
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Saves time in early planning stages for
comparing design alternatives.
Applications of approximate analysis:
- To
obtain initial member sizes before exact analysis.
- To
roughly check results of complex exact analysis.
- To
understand and evaluate older structures (especially pre-1960 designs) as
these structures designed solely on the basis of approximate analysis.
Methods
Methods are commonly used for approximate analysis of
rectangular frames subjected to lateral loads:
(1) The Portal Method
(2) The Cantilever Method.
Analysis for Lateral Loads—Portal Method
Assumptions for approximate analysis of rigid frames
subjected to lateral loads are:
1.
Inflection points are at midspan of each beam.
Exception: Cantilevered beams have inflection points at their free ends.
2.
Inflection points are at mid height of each
column. Exception: The lowest level columns that are pin supported have
inflection points at the supports.
3.
On each story of the frame, the shears in
interior columns are twice as large as the shears in exterior columns.
Procedure for Analysis
1.
Simplified Frame with Internal Hinges
·
Insert an internal hinge at the midpoint of
each member (beam and column).
·
This converts the statically indeterminate
frame into a statically determinate system.
·
Draw a sketch of the resulting simplified
frame.
2.
Determine Column Shears
·
For each story of the frame:
a.
Pass a horizontal section through all
columns of the story, dividing the frame into upper and lower portions.
b.
Assume shears in interior columns are twice
those in exterior columns.
c.
Apply the horizontal equilibrium equation
to the upper portion:
d.
Solve to find the shears in all columns of
the story.
3.
Draw Free-Body Diagrams (FBDs)
·
Draw FBDs of all members and joints,
including:
o External
applied loads
o Column
end shears determined in Step 2
·
This helps visualize the force distribution
in the frame.
4.
Determine Column End Moments
·
Use the equations of condition for the mid-height
internal hinge (inflection point):
o Bending
moment at the hinge = 0
o Apply
to free body of the column:
·
The end moments (MC) of the column:
o Equal
in magnitude at top and bottom
o Same
sense (both clockwise or counterclockwise)
o Magnitude
formula:
where:
§ 
= column shear
§ 
= column height
5.
Compute End Moments for All Columns
·
Repeat Step 4 for every column in the
frame to complete the bending moment distribution.
6.
Determine girder axial forces, moments,
and shears (progressing story-by-story, left → right)
a)
Work from the top story downward. For
each story, start at the far-left joint and move across to the right.
b)
At a joint (left of a girder):
o Write
equilibrium for the joint to find the axial force and end moment
transmitted into the right-hand girder:
§ Use
for axial force.
§ Use
(about a convenient point) for the
girder end moment 
at the left end.
c)
Compute the girder shear at the left end
by enforcing the assumed inflection at the girder midspan (moment = 0 there):
where 
is the girder clear span (length
between mid-hinges).
d) Free
body of the entire girder: apply 
to obtain the axial, shear,
and moment at the right end of the girder.
o Axial
and shear at the two ends are equal in magnitude and opposite in sense.
o End
moments are equal in magnitude and sense (same sign).
e)
Advance to the next joint to the right
and repeat steps 2–4 until all girders in the story are processed.
o The
unused equilibrium equations at the right-most joint of the story provide a consistency
check.
f)
Proceed to the story below: start at its
far-left joint and repeat the whole sequence until girders of all stories have
been determined.
7.
Determine column axial forces (top →
bottom)
a)
Begin at the top story. For each joint,
take the free body of the joint and apply vertical equilibrium:
to
solve for the axial forces carried by the columns meeting at that joint.
b)
Advance story by story downward, using
the previously found girder end forces and tributary loads to determine axial
forces in the next lower story’s columns.
c)
Continue until axial forces for every column
in the frame are obtained.
8.
Global equilibrium and reaction check
a)
The lower ends of the bottom-story
columns (computed axial, shear, moment) represent the support reactions.
b)
Apply the three equilibrium equations to the entire
frame (external loads + computed support reactions):
∑F X =0, ∑F Y =0, ∑M=0.
c)
Satisfaction of these three equations is a
necessary check that the approximate analysis was carried out consistently. If
any of the three is not satisfied, re-check the steps and sign conventions.
Example:
Determine the approximate axial forces, shears, and moments
for all the members of the frame shown in Fig.(a) by using the portal method.
Solution:
Simplified Frame:
The simplified frame is obtained by inserting internal
hinges at the midpoints of all the members of the given frame.
Column Shears:
Insert an imaginary cut (section aa and bb) across the
second story and first story of the frame.
Portal Method Assumption
- Interior
column shear = 2 × exterior column shear
Thus:
Let shear in DG = S
Shear in FI = S
Shear in EH = 2S
By applying the equilibrium equation ∑ FX = 0, for second story, we obtain
10 – S2 – 2S2 – S2 = 0
Or, S2 = 2.5 k
Therefore, second-story column shears at the lower ends:
- DG
= 2.5 k ←
- FI
= 2.5 k ←
- EH
= 5 k ←
First Story Column Shears (Section bb)
Repeat the same steps for the lower story using section bb.
Given total story shear = 30 k, solving gives:
30 – S1 – 2S1 – S1 = 0
Or, S1 = 7.5 k
- Exterior
column shear = 7.5 k
- Interior
column shear = 15 k
Lower ends of first-story columns:
- AD
= 7.5 k ←
- CF
= 7.5 k ←
- BE
= 15 k ←
Upper-End Column Shears
To find the shear at the top of each column:
Use ΣFx = 0 for each individual column.
Example for column DG:
- Bottom
shear (at D) = 2.5 k ←
- So
top shear must be 2.5 k → (equal and opposite to satisfy
equilibrium)
Therefore:
For every column, upper-end shear = same magnitude but
opposite direction.
Column Moments
Once the column shears are known, finding the column
moments is very easy.
1. Formula
For any column:
Where:
= column shear- = column height
= distance from end to
internal hinge (portal simplification)
2. For Example — Column DG
Given:
- Column
height = 12 ft
- Shear
at each end = 2.5 k
Thus:
- Moment
at top of DG = 15 k-ft
- Moment
at bottom of DG = 15 k-ft
Both moments have the same magnitude but opposite
direction to the shears—
meaning both are counterclockwise on the column FBD.
3. For all other columns
Repeat the same formula:
All column heights are 12 ft → same “half-height” = 6 ft.
Girder Axial Forces, Moments & Shears:
Analyze each girder by starting at the upper left joint
(G) and moving across the frame.
At Joint G → Determine Axial Force & Moment for
Girder GH
From the column DG at joint G:
- Column
shear = 2.5 k ←
(on joint)
- Column
moment = 15 k-ft (on joint)
Girder Axial Force (QGH)
Sum of horizontal forces at joint G:
So, at the girder end G, it acts → (opposite
direction).
Girder Moment (MGH)
Moment equilibrium at joint G:
Direction:
- At
joint G: counterclockwise
- At
girder end G: clockwise
(Always opposite directions between joint and member.)
Girder GH → Find Shear SGH
To find shear, use moment equilibrium of half of the
girder (portal hinge at midspan).
Given:
- Girder
length = 30 ft
- Moment
at G = 15 k-ft
So:
- At
end G: 1 k ↓
- At
end H: 1 k ↑
Use same process to determine axial forces, moments and
shears for the other joints and members.
For each girder:
1.
Get forces at the joint (shear &
moment from column).
2.
Apply ΣFx = 0 → girder axial force.
3.
Apply ΣM = 0 → girder end moment.
4.
Girder shear = M / (L/2).
5.
Apply equilibrium to the girder → get forces
at the other end.
6.
Move to next joint and repeat.
Column Axial Forces
Column axial forces come from vertical equilibrium at
each joint.
Start at the Upper Left Joint (G)
From the girder GH, we already know the girder shear at
joint G = 1 k (downward on girder).
- On
the joint, the girder applies 1 k upward.
- Therefore,
the column DG must resist this with 1 k upward axial force.
So, at the top of column DG:
Find Axial Force at Bottom of Column DG
Apply vertical force equilibrium on the column:
So:
- Top:
1 k ↑
- Bottom:
1 k ↓
This means the column carries 1 k of tension.
Do the Same for Other Columns: Repeat the same steps
at all joints.
For each joint:
1.
Look at the vertical shears coming from
the girders.
2.
The column’s axial force must be equal and
opposite to that girder shear.
3.
Use ΣFy = 0 on the column to find the axial
force at the opposite end.
Reactions
The reactions at the supports are simply the lower-end actions
of the first-story columns AD, BE, CF.
Global equilibrium check
