SME
Mechanics of Materials (Solid Mechanics I)
Beam deflection
Use this for “A 2 m steel cantilever with I = 8 × 10⁶ mm⁴ carries 5 kN at its tip. How far does the tip drop, and at what slope?”
Diagram
ΣF = 0: RA = 5,000 N, the total load.
The wall at A holds MA = 1 × 10⁷ N·mm, counterclockwise.
The bent shape is drawn bigger than life; it meets every support.
I is taken about the dashed bending axis.
EI is how stiff the beam is in bending: the larger it is, the less the beam bends.
This is an example. Type your own values to start a new problem.
δₘₐₓ (mm)
=
P (kN)
×
(
L (m)
)
3
3
×
E (GPa)
×
I (mm⁴)
θₘₐₓ (rad)
=
P (kN)
×
(
L (m)
)
2
2
×
E (GPa)
×
I (mm⁴)
Formulas
δₘₐₓ
=
P
×
L
3
3 ×
E
×
I
8.3333
=
5,000
×
2,
000
3
3 × 200,000 × 8,000,000
θₘₐₓ
=
P
×
L
2
2 ×
E
×
I
0.00625
=
5,000
×
2,
000
2
2 × 200,000 × 8,000,000
Formulas are worked in N, mm, MPa. The step-by-step shows the conversions.
Assumptions
•
Linear elastic, small deflections; the steps work in N, mm and MPa.
•
The beam is fixed at the wall and the load stands at the free tip.
•
One material and one section along the length: E and I stay the same.
Step-by-step
WE KNOW
P = 5 kN, L = 2 m, E = 200 GPa, I = 8,000,000 mm⁴
FIND
tip deflection (δₘₐₓ), tip slope (θₘₐₓ)
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 = 5 kN = 5,000 N (1 kN = 1000 N)
L = 2 m = 2,000 mm (1 m = 1000 mm)
E = 200 GPa = 200,000 MPa (1 GPa = 1000 MPa)
Step 2 · Find tip deflection (δₘₐₓ)
Use δₘₐₓ = P × L³ ÷ (3 × E × I)
Integrate
EIy″
=
M(x)
twice
with
y
=
0
and
y′
=
0
at
the
wall;
at
the
tip
this
gives
PL
³
3
EI
.
δₘₐₓ
=
5,000
×
2,
000
3
3 × 200,000 × 8,000,000
δₘₐₓ
=
5,000
×
2,
000
3
600,000 × 8,000,000
δₘₐₓ
=
5,000
×
2,
000
3
4,800,000,000,000
δₘₐₓ
=
5,000
×
8,000,000,000
4,800,000,000,000
δₘₐₓ
=
40,000,000,000,000
4,800,000,000,000
δₘₐₓ = 8.3333 mm
Step 3 · Find tip slope (θₘₐₓ)
Use θₘₐₓ = P × L² ÷ (2 × E × I)
Integrate
EIy″
=
M(x)
once
with
y′
=
0
at
the
wall;
at
the
tip
the
slope
is
PL
²
2
EI
,
in
radians.
θₘₐₓ
=
5,000
×
2,
000
2
2 × 200,000 × 8,000,000
θₘₐₓ
=
5,000
×
2,
000
2
400,000 × 8,000,000
θₘₐₓ
=
5,000
×
2,
000
2
3,200,000,000,000
θₘₐₓ
=
5,000
×
4,000,000
3,200,000,000,000
θₘₐₓ
=
20,000,000,000
3,200,000,000,000
θₘₐₓ = 0.00625 rad
Check
Put the numbers back in. Both sides should match.
8.3333
=
5,000
×
2,
000
3
3 × 200,000 × 8,000,000
✓
0.00625
=
5,000
×
2,
000
2
2 × 200,000 × 8,000,000
✓
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Sag of a simply supported beam under a uniform load
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