SME
Structural Analysis
Influence lines
Use this for “A 10 m simple span: what largest moment can a 50 kN moving load cause 4 m from A? And 12 kN/m over the whole span?”
Diagram
Each ordinate is the moment a load of 1 makes standing there; the largest is 2.4 at x = 4 m.
This is an example. Type your own values to start a new problem.
yC (m)
=
c (m)
×
(
L (m)
−
c (m)
)
÷
L (m)
MP (kN·m)
=
P (kN)
×
yC (m)
Mw (kN·m)
=
w (kN/m)
×
L (m)
×
yC (m)
÷
2
Formulas
yC
=
c
×
L
−
c
L
2.4
=
4
×
10 − 4
10
MP
=
P
×
yC
120 = 50 × 2.4
Mw
=
w
×
L
×
yC
2
144
=
12
×
10
×
2.4
2
Assumptions
•
Simply supported; the influence line is the moment at C made by a load of 1 standing anywhere.
•
The point load does most at C itself, where the triangle peaks.
•
The uniform load covers the whole span: its moment is w times the triangle’s area.
Step-by-step
WE KNOW
L = 10 m, c = 4 m, P = 50 kN, w = 12 kN/m
FIND
peak ordinate (yC), moment at C from P (MP), moment at C from w (Mw)
Step 1 · Find peak ordinate (yC)
Use yC = c × (L − c) ÷ L
A
load
of
1
at
C
makes
RA
=
L
−
c
L
,
and
the
moment
at
C
is
RA
×
c
.
yC
=
4
×
10 − 4
10
yC
=
4
×
6
10
yC
=
24
10
yC = 2.4 m
Step 2 · Find moment at C from P (MP)
Use MP = P × yC
A point load’s effect is its size times the ordinate where it stands.
MP = 50 × 2.4
MP = 120 kN·m
Step 3 · Find moment at C from w (Mw)
Use Mw = w × L × yC ÷ 2
A
uniform
load’s
effect
is
w
times
the
area
under
the
line,
1
2
×
L
×
yC.
Mw
=
12
×
10
×
2.4
2
Mw
=
120
×
2.4
2
Mw
=
288
2
Mw = 144 kN·m
Check
Put the numbers back in. Both sides should match.
2.4
=
4
×
10 − 4
10
✓
120 = 50 × 2.4 ✓
144
=
12
×
10
×
2.4
2
✓
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Determinate, indeterminate or unstable
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Influence line for the shear at a section
Topic check
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More problem types
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