SME
Mechanics of Materials (Solid Mechanics I)
Stress transformation, combined loading and pressure vessels
Use this for “A point carries σₓ = 60 MPa, σᵧ = −20 MPa and τₓᵧ = 30 MPa. Find the stresses on a plane turned 25° counterclockwise.”
Diagram
σavg
=
σₓ + σᵧ
2
=
60 + −20
2
σavg = 20 MPa
R
=
√((
σₓ − σᵧ
2
)²
+
τₓᵧ²)
=
50
MPa.
σ₁
=
σavg
+
R
=
70
MPa;
σ₂
=
σavg
−
R
=
−30
MPa;
τₘₐₓ
=
R.
Turned θ = 25°, the x′ faces carry σₓ′ = +68.69 MPa and τₓ′ᵧ′ = −11.36 MPa, the y′ faces σᵧ′ = −28.69 MPa. τ is plotted positive down, so X turns 2θ round the circle to X′ the way the element turns.
This is an example. Type your own values to start a new problem.
σₓ′ (MPa)
=
σₓ (MPa)
+
(
σᵧ (MPa)
)
2
+
σₓ (MPa)
−
(
σᵧ (MPa)
)
2
×
cos(2
×
θ
°)
+
τₓᵧ (MPa)
×
sin(2
×
θ
°)
τₓ′ᵧ′ (MPa)
=
τₓᵧ (MPa)
×
cos(2
×
θ
°)
−
σₓ (MPa)
−
(
σᵧ (MPa)
)
2
×
sin(2
×
θ
°)
σᵧ′
Normal stress on y′
calculated
Formulas
σₓ′
=
σₓ + σᵧ
2
+
(
σₓ − σᵧ
2
)
×
cos(2
×
θ
°)
+
τₓᵧ
×
sin(2
×
θ
°)
+68.6928
=
60 + (−20)
2
+
(
60 − (−20)
2
)
×
cos(2
×
25°)
+
30
×
sin(2
×
25°)
τₓ′ᵧ′
=
−(
σₓ − σᵧ
2
)
×
sin(2
×
θ
°)
+
τₓᵧ
×
cos(2
×
θ
°)
−11.3581
=
−(
60 − (−20)
2
)
×
sin(2
×
25°)
+
30
×
cos(2
×
25°)
σᵧ′ = σₓ + σᵧ − σₓ′
−28.6928 = 60 + (−20) − 68.6928
Assumptions
•
Plane stress: σz, τₓz and τyz are 0. Tension is +; τₓᵧ is + up on the right face.
•
θ turns the x′ axis from x, counterclockwise +; σᵧ′ acts on the face 90° further round.
•
σₓ′ + σᵧ′ = σₓ + σᵧ on every plane, a quick check on the answers.
Step-by-step
WE KNOW
θ = +25°, σₓ = +60 MPa, σᵧ = −20 MPa, τₓᵧ = +30 MPa
FIND
normal stress on x′ (σₓ′), shear stress on x′ (τₓ′ᵧ′), normal stress on y′ (σᵧ′)
Step 1 · Find normal stress on x′ (σₓ′)
Use σₓ′ = (σₓ + σᵧ) ÷ 2 + ((σₓ − σᵧ) ÷ 2) × cos(2 × θ°) + τₓᵧ × sin(2 × θ°)
Balance
the
forces
on
a
wedge
cut
at
θ
:
the
mean
stress,
plus
the
half
difference
and
the
shear
turned
through
2
θ
.
σₓ′
=
60 + (−20)
2
+
(
60 − (−20)
2
)
×
cos(2
×
25°)
+
30
×
sin(2
×
25°)
σₓ′
=
40
2
+
80
2
×
cos(50°)
+
30
×
sin(50°)
σₓ′
=
40
2
+
80
2
×
0.6428
+
30
×
0.766
σₓ′ = 20 + 40 × 0.6428 + 22.98
σₓ′ = 20 + 25.712 + 22.98
σₓ′ = +68.6928 MPa
Step 2 · Find shear stress on x′ (τₓ′ᵧ′)
Use τₓ′ᵧ′ = −((σₓ − σᵧ) ÷ 2) × sin(2 × θ°) + τₓᵧ × cos(2 × θ°)
Balance
the
wedge
along
the
cut
face:
the
half
difference
and
the
shear,
each
turned
through
2
θ
.
τₓ′ᵧ′
=
−(
60 − (−20)
2
)
×
sin(2
×
25°)
+
30
×
cos(2
×
25°)
τₓ′ᵧ′
=
−(
80
2
)
×
sin(50°)
+
30
×
cos(50°)
τₓ′ᵧ′
=
−(
80
2
)
×
0.766
+
30
×
0.6428
τₓ′ᵧ′ = −40 × 0.766 + 19.284
τₓ′ᵧ′ = −30.64 + 19.284
τₓ′ᵧ′ = −11.3581 MPa
Step 3 · Find normal stress on y′ (σᵧ′)
Use σᵧ′ = σₓ + σᵧ − σₓ′
The sum of the normal stresses is the same on every pair of faces at right angles.
σᵧ′ = 60 + (−20) − 68.6928
σᵧ′ = −28.6928 MPa
Check
Put the numbers back in. Both sides should match.
+68.6928
=
60 + (−20)
2
+
(
60 − (−20)
2
)
×
cos(2
×
25°)
+
30
×
sin(2
×
25°)
✓
−11.3581
=
−(
60 − (−20)
2
)
×
sin(2
×
25°)
+
30
×
cos(2
×
25°)
✓
−28.6928 = 60 + (−20) − 68.6928 ✓
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Bending, torsion and axial load on a shaft
Topic check
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