SME
Structural Analysis · Problem type
Influence line for the shear at a section
Use this for “A 10 m simple span: what largest positive and negative shears can a 50 kN moving load cause 4 m from A?”
Diagram
Each ordinate is the shear a load of 1 makes standing there; the largest is 0.6 at x = 4 m.
Largest negative shear at C Vₘᵢₙ = −20 kN
This is an example. Type your own values to start a new problem.
yL
=
−
c (m)
÷
L (m)
yR
=
(
L (m)
−
c (m)
)
÷
L (m)
Vₘₐₓ (kN)
=
P (kN)
×
yR
,
Vₘᵢₙ (kN)
=
P (kN)
×
(
yL
)
Formulas
yL
=
−
c
L
−0.4
=
−
4
10
yR
=
L
−
c
L
0.6
=
10 − 4
10
Vₘₐₓ
=
P
×
yR
30 = 50 × 0.6
Vₘᵢₙ
=
P
×
yL
−20 = 50 × (−0.4)
Assumptions
•
Simply supported; shear at C sums the forces left of C, up positive.
•
The line jumps by 1 at C: the load just right of C gives the largest positive shear, just left the most negative.
Step-by-step
WE KNOW
L = 10 m, c = 4 m, P = 50 kN
FIND
ordinate just left of C (yL), ordinate just right of C (yR), largest positive shear at C (Vₘₐₓ), largest negative shear at C (Vₘᵢₙ)
Step 1 · Find ordinate just left of C (yL)
Use yL = −c ÷ L
A
load
of
1
just
left
of
C:
the
shear
at
C
is
RA
−
1,
which
is
−RB
=
−
c
L
.
yL
=
−
4
10
yL = −0.4
Step 2 · Find ordinate just right of C (yR)
Use yR = (L − c) ÷ L
A
load
of
1
just
right
of
C:
the
shear
at
C
is
RA
=
L
−
c
L
.
yR
=
10 − 4
10
yR
=
6
10
yR = 0.6
Step 3 · Find largest positive shear at C (Vₘₐₓ)
Use Vₘₐₓ = P × yR
The
load
just
right
of
C
gives
the
largest
positive
shear:
P
times
that
ordinate.
Vₘₐₓ = 50 × 0.6
Vₘₐₓ = 30 kN
Step 4 · Find largest negative shear at C (Vₘᵢₙ)
Use Vₘᵢₙ = P × yL
Load
just
left
of
C
for
the
most
negative
shear:
P
times
that
ordinate.
Vₘᵢₙ = 50 × (−0.4)
Vₘᵢₙ = −20 kN
Check
Put the numbers back in. Both sides should match.
−0.4
=
−
4
10
✓
0.6
=
10 − 4
10
✓
30 = 50 × 0.6 ✓
−20 = 50 × (−0.4) ✓
Topic check
See what you know: 5 questions on this topic
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