Influence Lines
ü Concept introduced by Emil
Winkler in 1867.
ü Definition:
An influence line is a graph showing how a response function
(like reaction, shear, moment, or member force) varies with the position
of a unit moving load across a structure.
ü Purpose:
Influence lines help determine where to place loads to produce maximum
effects in structures.
Example: Bridge Truss with a Moving Vehicle
- When a
car moves across a bridge truss:
- The forces
in the truss members change depending on the position (x) of
the car.
- Different
members reach their maximum internal force at different load
positions.
- Example:
- Member
AB may experience its maximum force when the car is at position
x=x₁.
- Member
CH may reach its maximum force at position x=x₂.
- Therefore,
each member must be designed for the maximum force it can
experience as the load moves.
Two-Step Process for Analyzing Variable Loads
1.
Find the critical load position — where
the response (reaction, shear, bending moment, or member force) is maximum.
2.
Calculate the maximum value of that
response.
Influence Lines for Beams and Frames by Equilibrium
Method
ü
Consider a simply supported beam
The following scenario is to be analyzed:
- A beam
is simply supported at points A and C.
- A
single, downward unit load moves along the beam's length.
- The
load's position is defined by the variable x, measured from support A.
- The
objective is to determine and draw the influence lines for:
- The
vertical reaction at support A.
- The
vertical reaction at support C.
- The
shear force at a specific point B (a distance 'a' from A).
- The
bending moment at that same point B.
Influence Lines for
Reactions
The influence line for the vertical reaction Ay
Take moments about support 
(clockwise positive). For a
downward unit load at distance 
from 
,
Solve for 
:
Checks:
(unit load at is fully carried by ).
(unit load at 
is fully carried by ).- Linear
in (a straight line from
to 
) — this is the
influence line for .
The influence line for the vertical reaction C y
Take moments about support :
Solve for 
:
Checks:
- Linear
in (a straight line from
to 
) — this is the
influence line for .
Influence Line for Shear at
B
1. Derivation Using Free-Body Diagrams and Reaction
Influence Lines
The most efficient method uses the previously established
influence lines for the reactions, Ay and Cy.
Case 1: Unit Load is to the LEFT of B (0 ≤ x < a)
- Approach: To
simplify calculations, we analyze the free-body diagram of the beam
segment to the right of B (segment BC).
- Analysis: On
this segment, the only external force related to the reactions is Cy,
which acts upward. According to the beam sign convention, an upward force
to the right of a section causes negative shear.
- Result: Therefore,
the shear at B is equal to the negative of the reaction at C.
Case 2: Unit Load is to the RIGHT of B (a < x ≤ L)
- Approach: In
this case, it's simpler to analyze the free-body diagram of the beam
segment to the left of B (segment AB).
- Analysis: On
this segment, the reaction Ay acts upward. According to the beam sign
convention, an upward force to the left of a section causes positive
shear.
- Result: Therefore,
the shear at B is equal to the reaction at A.
Summary Equation:
The influence line for the shear at B is defined by a
piecewise function based on the load's position:
Graphical Construction:
This equation gives us a direct way to plot the influence
line using the graphs we already have for Ay and Cy.
- Segment
AB (from x=0 to x=a): The influence line for SB is the mirror
image (negative) of the influence line for Cy over this range.
- Segment BC (from x=a to x=L): The influence line for SB is identical to the influence line for Ay over this range.
2. Explicit Equations in Terms of x (Alternative Method)
By substituting the formulas for Ay and Cy, we can also
express the shear at B directly as a function of the load position x.
3. Description and
Interpretation of the Final Influence Line
The resulting influence line for SB [shown in Fig.(e)] has a
distinct shape with a sudden jump at point B:
- At
x = 0 (Load at A): The shear at B is
. - As
the load moves from A to B: The shear at B becomes increasingly
negative, decreasing linearly.
- Just
to the LEFT of B (x → a⁻): The shear value is
. - Just
to the RIGHT of B (x → a⁺): The shear value jumps to
. This discontinuity
(jump of magnitude 1) occurs because the unit load passes directly over
point B. - As
the load moves from B to C: The shear at B decreases linearly.
- At
x = L (Load at C): The shear at B returns to .
This influence line provides a powerful visual tool: for any
load position x, the corresponding ordinate (y-value) on the graph gives
the exact value of the shear force at point B.
Influence Line for Bending
Moment at B
To construct the influence line for the bending moment at a
specific point B, we analyze how the moment at B changes as a unit load moves
across the beam. The analysis is split based on whether the load is to the left
or right of point B.
1. Derivation Using Free-Body Diagrams and Reaction
Influence Lines
The most efficient method builds upon the known influence
lines for the reactions, Ay and Cy.
Case 1: Unit Load is to the LEFT of B (0 ≤ x ≤ a)
- Approach: We
analyze the free-body diagram of the beam segment to the right of
B (segment BC).
- Analysis: The
only significant force on this segment is the vertical reaction Cy,
located a distance (L - a) from point B. For the right segment,
a counter-clockwise moment from Cy is considered positive.
- Result: The
bending moment at B is the product of Cy and its lever arm.
Case 2: Unit Load is to the RIGHT of B (a ≤ x ≤ L)
- Approach: We
analyze the free-body diagram of the beam segment to the left of
B (segment AB).
- Analysis: The
significant force is the vertical reaction Ay, located a
distance a from point B. For the left segment, a clockwise
moment from Ay is considered positive.
- Result: The
bending moment at B is the product of Ay and its lever arm.
Summary Equation:
The influence line for the bending moment at B is defined by
a piecewise function:
Graphical Construction:
This equation provides a direct way to plot the influence
line:
- Segment
AB (from x=0 to x=a): The influence line for MB is obtained by
scaling the influence line for Cy by the constant factor (L - a).
- Segment
BC (from x=a to x=L): The influence line for MB is obtained by
scaling the influence line for Ay by the constant factor a.
2. Explicit Equations in Terms of x (Alternative Method)
Substituting the formulas for Ay and Cy gives the bending
moment at B directly as a function of the load position x.
3. Crucial Distinction: Influence Line vs. Bending Moment
Diagram
It is vital to understand the difference between these two
concepts:
- Bending
Moment Diagram: Shows the variation of bending moment at
all sections along the beam for a fixed load position.
- Influence
Line for Bending Moment: Shows the variation of bending
moment at one specific section (point B) as a single
unit load moves across the beam.
While the influence line for MB in a simply supported beam
might look similar to the bending moment diagram for a load at B, they convey
fundamentally different information.
4. Important Property of Statically Determinate
Structures
A key observation from this analysis is that the influence
lines for forces and moments (reactions, shear, bending
moment) in statically determinate structures are always
composed of straight-line segments.
- Contrast
with Deflections: Influence lines for deflections in
statically determinate structures, however, are composed of curved
lines. This highlights a fundamental difference in how these response
functions behave under a moving load.
Procedure for Analysis
A. Setup
1.
Choose origin and direction for the moving unit
downward load (convention: from left → right).
o Position
= measured from left end .
2. Know the sign conventions for reactions, shear and moment (use your beam sign convention).
B. Influence line for a reaction (e.g., 
or 
)
1.
Place unit load at an arbitrary position .
2.
Write equilibrium (moments or forces) to solve
the reaction algebraically in terms of .
o If
structure has internal hinges/rollers, do this separately for each rigid
segment the load can occupy.
3.
Repeat for all relevant ranges of (piecewise expressions).
4.
Plot reaction ordinate vs. . Positive ordinate →
reaction acts in the assumed positive sense.
Reminder (simply supported beam):
C. Influence line for shear at a point (e.g., 
)
1.
Ensure reaction ILs on the relevant side(s) of
the section are available.
2.
Place unit load to the left of the
section; take the free body of the portion to the right and write
equilibrium for shear 
(use reaction ILs if
present).
3.
Place unit load to the right of the
section; take the free body of the portion to the left and write
equilibrium for shear.
4.
Combine the two piecewise expressions and plot
(expect a jump of magnitude equal to the unit load when it crosses the
section).
Reminder (simply supported beam at 
):
Jump at 
equals 
.
D. Influence line for bending moment at a point (e.g., 
)
1.
Make sure reaction ILs on one side or the other
are available.
2.
For load left of the section take the
free body of the right portion and write moment about the section (use
reaction IL if convenient).
3.
For load right of the section take the
free body of the left portion and write moment about the section (use
reaction IL if convenient).
4.
Combine the piecewise expressions and plot.
Expect a continuous, piecewise-linear IL for statically determinate structures.
Reminder (simply supported beam at ):
E. Useful tips / checks
- Sum of
reaction ILs at any should equal 1 (vertical
equilibrium).
- Shear
IL has a jump of magnitude equal to the moving load when the load crosses
the section.
- Moment
IL is continuous at the section.
- For
structures with hinges/roller segments, treat each segment separately (ILs
become piecewise accordingly).
- For
statically determinate structures: ILs for forces/moments → straight-line
segments. Deflection ILs → curved.
Example
A simply supported beam AC has total length 
ft. Point is located at distance 
ft from the left support . A downward unit load
(magnitude = 1) moves along the beam; its position measured from is with 
ft. Construct the influence
lines (ILs) for:
- the
vertical reaction at ,
, - the
vertical reaction at ,
, - the
shear at ,
(use sign: positive upward on
left), - the
bending moment at ,
(positive sagging).
Free body diagram of the beam
1) Reaction at 
( 
)
Take moments about for the beam with the unit
load at . Vertical equilibrium gives
Substitute ft:
Key values:
(load at ),
,
(load at ).
2) Reaction at 
( 
)
Take moments about :
With :
Key values:
,
,
.
Note: 
for every (vertical equilibrium).
3) Shear at 
( 
)
Use free-body portions and the sign convention (positive
upward on the left of the section).
Case A — load left of 
(
):
Take the right portion (BC). On that right portion, the only
vertical reaction is (the unit load is on the
left piece), and the internal shear at (on the right face) must
balance . With the sign conventions
this leads to
Case B — load right of (
):
Use the left portion (AB). The shear at equals the left reaction,
Combine:
Important numeric checks:
.- Just
left of :
. - Just
right of :
. - Jump
across
: 
(equals unit load magnitude). 
.
So, the IL is two straight lines with a jump of +1 at .
4) Bending moment at ( 
)
Take moments about point . The moment IL is
continuous (no jump).
Case A — load left of (
):
Use the right portion (BC). The moment at is produced by acting at a lever arm 
:
With numbers 
:
Case B — load right of (
):
Use the left portion (AB). The moment at is 
times lever arm 
:
With numbers:
Combine:
Check continuity at :
- Left
formula:
(ft·k per unit load), - Right
formula:
. - End
values:
. The IL is two
straight segments meeting at with peak value 
(ft·k per unit load).
5) Short summary of numerical points (useful for plotting
or checks)
- Reactions:
→ at 
→ at 
- Shear
at :
- For
.
So, at 
- For
.
So, at 
.
- Moment
at :
- For
.
Peak at 
is (ft·k per unit load).
- For
.
Problem
Beam A–B–C–D–E, total length 60 ft. The spans from left to
right are:
- AB =
10 ft (B at 10 ft from A)
- BC =
10 ft (C at 20 ft from A)
- CD =
20 ft (D at 40 ft from A)
- DE =
20 ft (E at 60 ft from A)
There is an internal hinge at D that separates the rigid
left part AD (A→D, 40 ft) and rigid right part DE (D→E, 20 ft). Vertical
reactions are at A, C, and E (call them 
).
Draw influence lines for:
,- ,
- ,
- shear
just to the right of C (call
), - bending
moment at B (call
).
Solution
Place a unit vertical load (1 k) at a variable position measured from A (so 
) and determine how the
quantity of interest changes with . Because of the hinge at D,
treat cases with the load on the left rigid part (AD: 
) and on the right rigid
part (DE: 
) separately.
1) Influence line for 
(reaction at E)
Take moments about the hinge D on the right rigid piece DE
(length 20 ft). For equilibrium of the right piece, the only vertical support
in that piece is .
- Case
1: Unit load is between A and D (0 ≤ x ≤ 40 ft)
- Case 2: Unit load is between D and E (40 ft ≤ x ≤ 60 ft)
The unit load is now on segment D-E. Taking moments about D for segment DE:
Reaction at 
, 
Influence Line for
We now consider the entire beam and sum moments about point
A.
- Case
1: Unit load is between A and D (0 ≤ x ≤ 40 ft)
Substituting 
for this range:
- Case
2: Unit load is between D and E (40 ft ≤ x ≤ 60 ft)
Substituting 
:
Summary for :
Influence Line for
We use the vertical force equilibrium for the entire beam.
- Case
1: Unit load is between A and D (0 ≤ x ≤ 40 ft)
Substituting
and :
- Case
2: Unit load is between D and E (40 ft ≤ x ≤ 60 ft)
Substituting
and :
Summary for :
(Note: 
— with unit at A the
three reactions are
.)
4. Influence Line for Shear just to the right of C (
)
Step 1 – Cases for load position
Case 1: Unit load between A and just left of C (
)
On the right segment (C to E), no unit load.
Sum vertical forces on right segment:
From earlier: for .
So:
Case 2: Unit load between just right of C and D (
)
Now the unit load is on the right segment (between C and E).
Sum vertical forces on right segment:
For , .
So:
Case 3: Unit load between D and E ()
Still, the unit load is on the right segment (portion DE actually, but it's to
the right of C).
Equation: 
still holds.
For , .
So:
Finally:
Step 4 – Check values at key points
- At
to 
: 
- At
: 
- At
: from second
case, ; from third
case, 
→ matches. - At
: 
(reasonable, load
at E, shear just right of C is 0).
5. Influence Line for Bending Moment at B ()
Given:
Point B is 10 ft from support A.
We will determine using the left-hand
section (from A to B) and follow the sign convention: positive
moment = sagging (tension at bottom).
Case 1: Unit load between A and B (
)
Consider segment AB.
The unit load is located at (between A and B).
Take moments about B:
From earlier: 
for .
Substitute :
So, for 
:
Case 2: Unit load between B and E (
)
Now the unit load is to the right of B, so it
does not appear in the left-hand free body (A to B).
Moment at B:
Now substitute for different ranges:
- For
:
- For :
Check continuity at 
:
From Case 1: 
From Case 2 (): 
Matches.
At :
From : 
From : 
Matches.
At : 
✓
Final Piecewise Function for
Values at key points:
- :
: 
: 
- :
- :
This matches the expected behavior:
- Moment
at B is zero when load is at A, C, or E.
- Maximum
positive moment at B when load is at B.
- Negative
moment when load is between D and E, because hinge at D changes beam
behavior.
Extra
Influence Line for Shear just to the Left of
C (
)
Case 1: Load between A and just left of C (
)
The unit load is on the left segment.
Forces on left segment (A to just left of C):
- Reaction (upward)
- Unit
load 1 kip (downward) at
- Shear
at cut (upward
positive) - No because cut is
just left of C
Vertical equilibrium:
For , :
So, for :
Case 2: Load between just right of C and E (
)
The unit load is not on the left segment.
Vertical equilibrium of left segment:
- For :
- For :
Case 3: Load exactly at C (
)
The load is at the support, so for the left segment,
the load is not present.
Use equation from Case 2 (load not on left
segment):
At , 
:
Final Piecewise Function for
Check Values:
- At : ✓
- At :
✓ - At
: 
✓ - At : (jump occurs)
- At :
✓ - At :
✓ - At :
✓
Comparison with :
Now they are different, as expected. The jump at
C is consistent:
- jumps from 1 to 0
jumps from 0 to 1
The sum of jumps = 1, which equals the unit load magnitude.
Final Answer:
Here is the calculation only showing
how for .
We start from the vertical equilibrium equation for the
entire beam:
For , we have:
Substitute into the equation for :
Final:
