SME
Mechanics of Materials (Solid Mechanics I) · Problem type
Shear stress at the neutral axis
Use this for “A 100 × 200 mm beam carries a 30 kN shear force. Find the largest shear stress.”
Diagram
A = 20,000 mm²
τ
=
VQ
It
:
largest
at
the
neutral
axis,
zero
at
the
top
and
bottom.
This is an example. Type your own values to start a new problem.
A (mm²)
=
b (mm)
×
h (mm)
τₘₐₓ (MPa)
=
1.5
×
V (kN)
A (mm²)
Formulas
A
=
b
×
h
20,000 = 100 × 200
τₘₐₓ
=
1.5
×
V
A
2.25
=
1.5
×
30,000
20,000
Formulas are worked in N. The step-by-step shows the conversions.
Assumptions
•
A solid rectangle, linear elastic; τ = VQ ÷ (It) is spread evenly across the width.
•
τ is a parabola over the depth: 0 at the top and bottom, largest at the neutral axis.
•
Worked in N, mm and MPa: N ÷ mm² is MPa.
Step-by-step
WE KNOW
V = 30 kN, b = 100 mm, h = 200 mm
FIND
area (A), largest shear stress (τₘₐₓ)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to N).
V = 30 kN = 30,000 N (1 kN = 1000 N)
Step 2 · Find area (A)
Use A = b × h
A rectangle’s area is its width times its depth.
A
=
100
×
200
A
=
20,000
mm²
Step 3 · Find largest shear stress (τₘₐₓ)
Use τₘₐₓ = 1.5 × V ÷ A
At
the
neutral
axis
Q
=
bh
²
8,
I
=
bh
³
12
and
t
=
b
,
so
τ
=
VQ
It
is
1.5
times
the
average
V
A
.
τₘₐₓ
=
1.5
×
30,000
20,000
τₘₐₓ
=
45,000
20,000
τₘₐₓ = 2.25 MPa
Check
Put the numbers back in. Both sides should match.
20,000 = 100 × 200 ✓
2.25
=
1.5
×
30,000
20,000
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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Bending and shear
Main lesson
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