SME
Machine Design · Problem type
Peak stress at a hole
Use this for “A 50 mm × 10 mm plate with a 10 mm hole carries 20 kN; Kₜ = 2.5 from the chart. Find the nominal and peak stresses.”
Diagram
Aₙₑₜ
=
(w
−
d)t
=
400
mm².
σₙₒₘ
=
P
Aₙₑₜ
=
50
MPa,
spread
evenly
as
if
there
were
no
hole.
At the hole’s edge σₘₐₓ = Kₜ σₙₒₘ = 2.5 × 50 MPa = 125 MPa.
At
d
w
=
0.2
a
chart
reads
Kₜ
≈
2.506.
This is an example. Type your own values to start a new problem.
σₙₒₘ (MPa)
=
P (kN)
(
w (mm)
−
d (mm)
)
×
t (mm)
σₘₐₓ (MPa)
=
Kₜ
×
σₙₒₘ (MPa)
Aₙₑₜ
Net area
calculated
Formulas
Aₙₑₜ
=
(
w
−
d
)
×
t
400 = (50 − 10) × 10
σₙₒₘ
=
P
Aₙₑₜ
50
=
20,000
400
σₘₐₓ = Kₜ × σₙₒₘ
125 = 2.5 × 50
Formulas are worked in N. The step-by-step shows the conversions.
Assumptions
•
An axial load P on a flat plate, width w and thickness t, with a round hole d on its center line.
•
σₙₒₘ acts on the net section through the hole; read Kₜ for d ÷ w from a net-section chart.
•
Kₜ counts for brittle parts and fatigue; a ductile part under steady load yields locally instead.
Step-by-step
WE KNOW
P = 20 kN, w = 50 mm, d = 10 mm, t = 10 mm, Kₜ = 2.5
FIND
net area (Aₙₑₜ), nominal stress (σₙₒₘ), peak stress (σₘₐₓ)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to N).
P = 20 kN = 20,000 N (1 kN = 1000 N)
Step 2 · Find net area (Aₙₑₜ)
Use Aₙₑₜ = (w − d) × t
Across
the
hole
only
w
−
d
of
the
width
is
left,
all
of
it
t
thick.
Aₙₑₜ = (50 − 10) × 10
Aₙₑₜ = 40 × 10
Aₙₑₜ = 400 mm²
Step 3 · Find nominal stress (σₙₒₘ)
Use σₙₒₘ = P ÷ Aₙₑₜ
The
load
spread
evenly
over
the
net
section,
as
if
there
were
no
peak;
N
mm²
is
MPa.
σₙₒₘ
=
20,000
400
σₙₒₘ = 50 MPa
Step 4 · Find peak stress (σₘₐₓ)
Use σₘₐₓ = Kₜ × σₙₒₘ
The stress crowds round the hole: its edge carries Kₜ times the nominal stress.
σₘₐₓ = 2.5 × 50
σₘₐₓ = 125 MPa
Check
Put the numbers back in. Both sides should match.
400 = (50 − 10) × 10 ✓
50
=
20,000
400
✓
125 = 2.5 × 50 ✓
Topic check
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