SME
Structural Analysis · Problem type
Cantilever tip deflection, load part-way along
Use this for “A 3 m steel cantilever (I = 40 × 10⁶ mm⁴) carries 25 kN 2 m from the wall. How far does the free end drop?”
Diagram
ΣF = 0: RA = 25,000 N, the total load.
The wall at A holds MA = 5 × 10⁷ N·mm, counterclockwise.
The bent shape is drawn bigger than life; it meets every support.
This is an example. Type your own values to start a new problem.
θₐ (rad)
=
P (kN)
×
(
a (m)
)
2
÷
(2
×
E (GPa)
×
I (mm⁴)
)
δₐ (mm)
=
P (kN)
×
(
a (m)
)
3
÷
(3
×
E (GPa)
×
I (mm⁴)
)
δₜᵢₚ (mm)
=
δₐ (mm)
+
θₐ
×
(
L (m)
−
a (m)
)
Formulas
θₐ
=
P
×
a
2
2 ×
E
×
I
0.00625
=
25,000
×
2,
000
2
2 × 200,000 × 40,000,000
δₐ
=
P
×
a
3
3 ×
E
×
I
8.3333
=
25,000
×
2,
000
3
3 × 200,000 × 40,000,000
δₜᵢₚ
=
δₐ
+
θₐ
×
(
L
−
a
)
14.5833 = 8.3333 + 0.00625 × (3,000 − 2,000)
Formulas are worked in N, mm, MPa. The step-by-step shows the conversions.
Assumptions
•
Fixed at the wall, constant EI, linear elastic; the steps work in N, mm and MPa.
•
Small slopes: tan θ ≈ θ in radians, so the straight end drops θₐ(L − a).
•
The load stands on the beam: a ≤ L.
Step-by-step
WE KNOW
P = 25 kN, a = 2 m, L = 3 m, E = 200 GPa, I = 40,000,000 mm⁴
FIND
slope at the load (θₐ), deflection at the load (δₐ), tip deflection (δₜᵢₚ)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to N, mm, MPa).
P = 25 kN = 25,000 N (1 kN = 1000 N)
a = 2 m = 2,000 mm (1 m = 1000 mm)
L = 3 m = 3,000 mm (1 m = 1000 mm)
E = 200 GPa = 200,000 MPa (1 GPa = 1000 MPa)
Step 2 · Find slope at the load (θₐ)
Use θₐ = P × a² ÷ (2 × E × I)
Up
to
the
load
the
beam
bends
like
one
cantilever
of
length
a
with
P
at
its
tip:
θ
=
Pa
²
2
EI
,
in
radians.
θₐ
=
25,000
×
2,
000
2
2 × 200,000 × 40,000,000
θₐ
=
25,000
×
2,
000
2
400,000 × 40,000,000
θₐ
=
25,000
×
2,
000
2
16,000,000,000,000
θₐ
=
25,000
×
4,000,000
16,000,000,000,000
θₐ
=
100,000,000,000
16,000,000,000,000
θₐ = 0.00625 rad
Step 3 · Find deflection at the load (δₐ)
Use δₐ = P × a³ ÷ (3 × E × I)
The
same
short
cantilever
of
length
a
:
its
tip,
under
the
load,
drops
Pa
³
3
EI
.
δₐ
=
25,000
×
2,
000
3
3 × 200,000 × 40,000,000
δₐ
=
25,000
×
2,
000
3
600,000 × 40,000,000
δₐ
=
25,000
×
2,
000
3
24,000,000,000,000
δₐ
=
25,000
×
8,000,000,000
24,000,000,000,000
δₐ
=
200,000,000,000,000
24,000,000,000,000
δₐ = 8.3333 mm
Step 4 · Find tip deflection (δₜᵢₚ)
Use δₜᵢₚ = δₐ + θₐ × (L − a)
Past
the
load
the
beam
carries
no
moment,
so
it
runs
on
straight
at
slope
θₐ
for
the
last
L
−
a
(moment-area:
no
M
EI
there).
δₜᵢₚ = 8.3333 + 0.00625 × (3,000 − 2,000)
δₜᵢₚ = 8.3333 + 0.00625 × 1,000
δₜᵢₚ = 8.3333 + 6.25
δₜᵢₚ = 14.5833 mm
Check
Put the numbers back in. Both sides should match.
0.00625
=
25,000
×
2,
000
2
2 × 200,000 × 40,000,000
✓
8.3333
=
25,000
×
2,
000
3
3 × 200,000 × 40,000,000
✓
14.5833 = 8.3333 + 0.00625 × (3,000 − 2,000) ✓
Topic check
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