SME
Advanced Solid Mechanics (Solid Mechanics II) · Problem type
Strain rosette
Use this for “A 45° rosette reads εₐ = 600 με, εb = 300 με and εc = −200 με. Find γₓᵧ and the principal strains.”
Diagram
Gauge a reads εₓ and gauge c reads εᵧ.
The
45°
gauge
reads
εₓ + εᵧ
2
+
γₓᵧ
2,
so
γₓᵧ
=
2
×
εb
−
εₐ
−
εc
=
2
×
300
−600
−
(−200)
=
200
με.
The
circle’s
center
is
εₓ + εᵧ
2
=
200
με
and
its
radius
√((
εₓ − εᵧ
2
)²
+
(
γₓᵧ
2
)
2
)
=
412.3
με:
ε₁
=
612.3
με,
ε₂
=
−212.3
με.
γ
2
is
plotted
positive
down,
as
τ
is
on
the
stress
circle.
This is an example. Type your own values to start a new problem.
γₓᵧ (με)
=
2
×
εb (με)
−
εₐ (με)
−
(
εc (με)
)
εₓ
Normal strain on x
calculated · με
εᵧ
Normal strain on y
calculated · με
ε₁
Larger principal strain
worked out (not typed) · με
ε₂
Smaller principal strain
worked out (not typed) · με
Formulas
εₓ = εₐ
+600 = 600
εᵧ = εc
−200 = −200
γₓᵧ = 2 × εb − εₐ − εc
+200 = 2 × 300 − 600 − (−200)
ε₁
=
εₓ + εᵧ
2
+
√((
εₓ − εᵧ
2
)²
+
(
γₓᵧ
2
)
2
)
+612.3106
=
600 + (−200)
2
+
√((
600 − (−200)
2
)²
+
(
200
2
)
2
)
ε₂
=
εₓ + εᵧ
2
−
√((
εₓ − εᵧ
2
)²
+
(
γₓᵧ
2
)
2
)
−212.3106
=
600 + (−200)
2
−
√((
600 − (−200)
2
)²
+
(
200
2
)
2
)
Assumptions
•
A 45° rectangular rosette: gauge a lies along x, b at 45°, c along y.
•
Each gauge reads normal strain only; γₓᵧ is the engineering shear strain (twice the tensor one).
•
ε₁ and ε₂ are the in-plane principal strains; stretching is +.
Step-by-step
WE KNOW
εₐ = +600 με, εb = +300 με, εc = −200 με
FIND
normal strain on x (εₓ), normal strain on y (εᵧ), shear strain (γₓᵧ), larger principal strain (ε₁), smaller principal strain (ε₂)
Step 1 · Find normal strain on x (εₓ)
Use εₓ = εₐ
Gauge a lies along x, so it reads εₓ directly.
εₓ = +600
εₓ = +600 με
Step 2 · Find normal strain on y (εᵧ)
Use εᵧ = εc
Gauge c lies along y, so it reads εᵧ directly.
εᵧ = −200 με
Step 3 · Find shear strain (γₓᵧ)
Use γₓᵧ = 2 × εb − εₐ − εc
At
45°,
εb
=
εₓ + εᵧ
2
+
γₓᵧ
2
;
solve
that
for
γₓᵧ.
γₓᵧ = 2 × 300 − 600 − (−200)
γₓᵧ = 600 − 600 − (−200)
γₓᵧ = +200 με
Step 4 · Find larger principal strain (ε₁)
Use ε₁ = (εₓ + εᵧ) ÷ 2 + √(((εₓ − εᵧ) ÷ 2)² + (γₓᵧ ÷ 2)²)
Mohr’s
circle
for
strain:
the
center
plus
the
radius,
with
γₓᵧ
2
on
the
vertical
axis.
ε₁
=
600 + (−200)
2
+
√((
600 − (−200)
2
)²
+
(
200
2
)
2
)
ε₁
=
600 + (−200)
2
+
√(
(
800
2
)
2
+
(
200
2
)
2
)
ε₁
=
600 + (−200)
2
+
√
400
2
+
100
2
ε₁
=
400
2
+
√
160,000 + 10,000
ε₁
=
400
2
+
√
170
,000
ε₁
=
400
2
+
412.3
ε₁
=
200
+
412.3
ε₁
=
+612.3106
με
Step 5 · Find smaller principal strain (ε₂)
Use ε₂ = (εₓ + εᵧ) ÷ 2 − √(((εₓ − εᵧ) ÷ 2)² + (γₓᵧ ÷ 2)²)
Mohr’s circle for strain: the center less the radius.
ε₂
=
600 + (−200)
2
−
√((
600 − (−200)
2
)²
+
(
200
2
)
2
)
ε₂
=
600 + (−200)
2
−
√(
(
800
2
)
2
+
(
200
2
)
2
)
ε₂
=
600 + (−200)
2
−
√
400
2
+
100
2
ε₂
=
400
2
−
√
160,000 + 10,000
ε₂
=
400
2
−
√
170
,000
ε₂
=
400
2
−
412.3
ε₂
=
200
−
412.3
ε₂
=
−212.3106
με
Check
Put the numbers back in. Both sides should match.
+600 = 600 ✓
−200 = −200 ✓
+200 = 2 × 300 − 600 − (−200) ✓
+612.3106
=
600 + (−200)
2
+
√((
600 − (−200)
2
)²
+
(
200
2
)
2
)
✓
−212.3106
=
600 + (−200)
2
−
√((
600 − (−200)
2
)²
+
(
200
2
)
2
)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.