SME
Mechanics of Materials (Solid Mechanics I) · Problem type
Size a beam by its section modulus
Use this for “A beam twice as deep as it is wide carries 45 kN·m at 150 MPa allowed. Find its depth.”
Diagram
σ
=
Mc
I
:
zero
at
the
neutral
axis
through
the
centroid,
largest
at
the
far
edge.
This is an example. Type your own values to start a new problem.
S (mm³)
=
M (kN·m)
σallow (MPa)
=
(
h (mm)
)
3
12
A rectangle’s section modulus is S = bh² ÷ 6; with b = h ÷ 2 it is h³ ÷ 12.
b
Width
calculated
Formulas
S
=
M
σallow
300,000
=
45,000,000
150
S
=
h
3
12
300,000
=
153.3
3
12
b
=
h
2
76.63
=
153.3
2
Formulas are worked in N·mm. The step-by-step shows the conversions.
Assumptions
•
A rectangle twice as deep as it is wide (b = h ÷ 2), bent about its strong axis.
•
Linear elastic: the largest stress M ÷ S may reach σallow but not pass it.
•
Worked in N, mm and MPa: N·mm ÷ MPa is mm³.
Step-by-step
WE KNOW
M = 45 kN·m, σallow = 150 MPa
FIND
section modulus needed (S), depth (h), width (b)
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).
M = 45 kN·m = 45,000,000 N·mm (1 kN·m = 10⁶ N·mm)
Step 2 · Find section modulus needed (S)
Use S = M ÷ σallow
The
flexure
formula
σ
=
Mc
I
=
M
S
,
with
σ
at
its
allowed
value.
S
=
45,000,000
150
S
=
300,000
mm³
Step 3 · Find depth (h)
Use S = h³ ÷ 12
Multiply
S
by
12,
then
take
the
cube
root.
h
=
∛(12
×
S
)
h
=
∛(12
×
300,000)
h
=
∛3,600,000
h
=
153.26
mm
Step 4 · Find width (b)
Use b = h ÷ 2
The width is half the depth.
b
=
153.26
2
b
=
76.63
mm
Check
Put the numbers back in. Both sides should match.
300,000
=
45,000,000
150
✓
300,000
=
153.2619
3
12
✓
76.63
=
153.26
2
✓
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