SME
Statics · Problem type
I of a composite T-section
Use this for “Find I about the centroid of a T with a 100 × 20 mm flange on an 80 × 20 mm web.”
Diagram
ȳ
=
ΣAy
ΣA
=
244,000
3,600
=
67.78
mm
I = 3,142,000 mm⁴ about the centroidal axis
This is an example. Type your own values to start a new problem.
bf
Flange width
example · mm
tf
Flange thickness
example · mm
hw
Web height
example · mm
tw
Web thickness
example · mm
A₁
Flange area
worked out (not typed) · mm²
A₂
Web area
worked out (not typed) · mm²
ȳ
Centroid height
calculated · mm
d₁
Flange offset
calculated · mm
d₂
Web offset
calculated · mm
I
Moment of inertia about the centroid
calculated · mm⁴
Formulas
A₁
=
bf
×
tf
2,000 = 100 × 20
A₂
=
tw
×
hw
1,600 = 20 × 80
ȳ
=
(
A₁
×
(hw
+
tf
÷
2)
+
A₂
×
hw
÷
2)
÷
(
A₁
+
A₂
)
67.7778 = (2,000 × (80 + 20 ÷ 2) + 1,600 × 80 ÷ 2) ÷ (2,000 + 1,600)
d₁
=
hw
+
tf
2
−
ȳ
22.2222
=
80
+
20
2
−
67.7778
d₂
=
hw
2
−
ȳ
−27.7778
=
80
2
−
67.7778
I
=
bf
×
tf
3
12
+
bf
×
tf
×
d₁
2
+
tw
×
hw
3
12
+
tw
×
hw
×
d₂
2
3.1422 ×
10
6
=
100
×
20
3
12
+
100
×
20
×
22.2222
2
+
20
×
80
3
12
+
20
×
80
×
(−27.7778)
2
Assumptions
•
The T is two rectangles: the flange centered on top of the web.
•
Each part adds its own bh³ ÷ 12 plus its area times d² (the parallel-axis theorem).
•
ȳ is up from the web’s bottom; d₁ and d₂ run from ȳ to each part’s center, up positive.
Step-by-step
WE KNOW
bf = 100 mm, tf = 20 mm, hw = 80 mm, tw = 20 mm
FIND
flange area (A₁), web area (A₂), centroid height (ȳ), flange offset (d₁), web offset (d₂), moment of inertia about the centroid (I)
Step 1 · Find flange area (A₁)
Use A₁ = bf × tf
The flange is a rectangle bf wide and tf thick.
A₁
=
100
×
20
A₁
=
2,000
mm²
Step 2 · Find web area (A₂)
Use A₂ = tw × hw
The web is a rectangle tw wide and hw tall.
A₂
=
20
×
80
A₂
=
1,600
mm²
Step 3 · Find centroid height (ȳ)
Use ȳ = (A₁ × (hw + tf ÷ 2) + A₂ × hw ÷ 2) ÷ (A₁ + A₂)
Each
area
times
the
height
of
its
own
center
(the
flange’s
at
hw
+
tf
2,
the
web’s
at
hw
2
),
added,
then
divided
by
the
whole
area.
ȳ
=
(2,000
×
(80
+
20
÷
2)
+
1,600
×
80
÷
2)
÷
(2,000
+
1,600)
ȳ
=
(2,000
×
(80
+
10)
+
1,600
×
80
÷
2)
÷
(2,000
+
1,600)
ȳ
=
(2,000
×
90
+
1,600
×
80
÷
2)
÷
(2,000
+
1,600)
ȳ
=
180,000 + 128,000 ÷ 2
3,600
ȳ
=
180,000 + 64,000
3,600
ȳ
=
244,000
3,600
ȳ
=
67.7778
mm
Step 4 · Find flange offset (d₁)
Use d₁ = hw + tf ÷ 2 − ȳ
The
flange’s
center
is
at
hw
+
tf
2
;
take
away
ȳ
.
d₁
=
80
+
20
2
−
67.7778
d₁
=
80
+
10
−
67.7778
d₁
=
22.2222
mm
Step 5 · Find web offset (d₂)
Use d₂ = hw ÷ 2 − ȳ
The
web’s
center
is
at
hw
2
;
take
away
ȳ
.
d₂
=
80
2
−
67.7778
d₂
=
40
−
67.7778
d₂
=
−27.7778
mm
Step 6 · Find moment of inertia about the centroid (I)
Use I = bf × tf³ ÷ 12 + bf × tf × d₁² + tw × hw³ ÷ 12 + tw × hw × d₂²
Each
rectangle
about
its
own
center,
bh³
12,
plus
its
area
times
its
offset
squared,
for
both
parts.
I
=
100
×
20
3
12
+
100
×
20
×
22.2222
2
+
20
×
80
3
12
+
20
×
80
×
(−27.7778)
2
I
=
66,666.7
+
987,654
+
853,333
+
1,234,568
I
=
3.1422 ×
10
6
mm⁴
Check
Put the numbers back in. Both sides should match.
2,000 = 100 × 20 ✓
1,600 = 20 × 80 ✓
67.7778 = (2,000 × (80 + 20 ÷ 2) + 1,600 × 80 ÷ 2) ÷ (2,000 + 1,600) ✓
22.2222
=
80
+
20
2
−
67.7778
✓
−27.7778
=
80
2
−
67.7778
✓
3.1422 ×
10
6
=
100
×
20
3
12
+
100
×
20
×
22.2222
2
+
20
×
80
3
12
+
20
×
80
×
(−27.7778)
2
✓
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