SME
Mechanics of Materials (Solid Mechanics I)
Axial loading
Use this for “An aluminum bar, 400 mm of 400 mm² then 300 mm of 200 mm², carries 20 kN. How much does it stretch?”
Diagram
Each
segment
stretches
by
δᵢ
=
PᵢLᵢ
AᵢE
;
the
bar
stretches
by
their
sum.
δ = 0.286 + 0.429 = 0.714 mm.
The stretch is drawn bigger than life.
This is an example. Type your own values to start a new problem.
δ (mm)
=
P (kN)
×
L₁ (mm)
A₁ (mm²)
×
E (GPa)
+
P (kN)
×
L₂ (mm)
A₂ (mm²)
×
E (GPa)
δ₁
Stretch of segment 1
calculated · mm
δ₂
Stretch of segment 2
calculated · mm
Formulas
δ₁
=
P
×
L₁
A₁
×
E
0.2857
=
20
×
400
400 × 70
δ₂
=
P
×
L₂
A₂
×
E
0.4286
=
20
×
300
200 × 70
δ
=
δ₁
+
δ₂
0.7143 = 0.2857 + 0.4286
Assumptions
•
The load P acts along the axis at the free end, so both segments carry all of it.
•
One material throughout: E = 70 GPa for aluminum, 200 GPa for steel.
•
Linear elastic with small stretches; kN with GPa, like N with MPa, gives the stretch in mm.
Step-by-step
WE KNOW
P = 20 kN, E = 70 GPa, L₁ = 400 mm, A₁ = 400 mm², L₂ = 300 mm, A₂ = 200 mm²
FIND
stretch of segment 1 (δ₁), stretch of segment 2 (δ₂), total stretch (δ)
Step 1 · Find stretch of segment 1 (δ₁)
Use δ₁ = P × L₁ ÷ (A₁ × E)
Segment
1
carries
the
whole
load,
so
it
stretches
by
PL
AE
with
its
own
length
and
area.
δ₁
=
20
×
400
400 × 70
δ₁
=
20
×
400
28,000
δ₁
=
8,000
28,000
δ₁
=
0.2857
mm
Step 2 · Find stretch of segment 2 (δ₂)
Use δ₂ = P × L₂ ÷ (A₂ × E)
Segment
2
carries
the
whole
load,
so
it
stretches
by
PL
AE
with
its
own
length
and
area.
δ₂
=
20
×
300
200 × 70
δ₂
=
20
×
300
14,000
δ₂
=
6,000
14,000
δ₂
=
0.4286
mm
Step 3 · Find total stretch (δ)
Use δ = δ₁ + δ₂
The segments stretch one after the other, so their stretches add.
δ
=
0.2857
+
0.4286
δ
=
0.7143
mm
Check
Put the numbers back in. Both sides should match.
0.2857
=
20
×
400
400 × 70
✓
0.4286
=
20
×
300
200 × 70
✓
0.7143 = 0.2857 + 0.4286 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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