Learn: Beam shear stress
Shear force does not normally produce a uniform stress over a beam cross-section. This formula finds the local shear stress at a chosen level by using the area on one side of that level.
What the formula is saying
Q is the first moment of that partial area about the neutral axis. VQ describes the change in bending force that must be transferred; dividing by I and the local width converts it to shear stress.
Read the symbols in plain language
- V
- Shear force
Shear force. Multiply the section shear force by the first moment for the chosen level.
NNewtons measure force; 1000 N = 1 kN.
- Q
- First moment of area
Area on one side of the shear-evaluation level multiplied by its centroid’s distance from the neutral axis; this is Q, not I.
m³Cubic metres measure volume, or a first moment of area where explicitly identified.
- I
- Second moment of area
The area-weighted square of distance from the stated axis. It measures the spread of the cross-section, not its area or mass.
m⁴The fourth power of metres is used for a second moment of area; 1 m⁴ = 10¹² mm⁴.
- b
- Width at evaluation point
Width at evaluation point. The full-section second moment and local width control the stress distribution.
mMetres measure length; 1 m = 1000 mm.
- τ
- Result to find
Beam shear stress. Divide to obtain force per unit area at the selected location.
Pa
Sort out the units first
Use N, m³, m⁴ and m for V, Q, I and b. The units reduce to N/m² = Pa. Q is not a force or the full section’s second moment.
Assumptions before calculating
The beam is straight, slender and prismatic; E and I are constant. Deflections are small and Euler–Bernoulli bending applies: shear deformation, joint flexibility and geometric nonlinearity are neglected.
Let’s solve one together
Read the given values, follow each operation, then check what the result means.
Read the supplied values as one complete study case. Find τ and explain the result in the stated output unit.
- V · Shear force
- 100000 N
- Q · First moment of area
- 0.002 m³
- I · Second moment of area
- 0.005 m⁴
- b · Width at evaluation point
- 0.3 m
Form the shear-transfer numerator
Multiply the section shear force by the first moment for the chosen level.
(100000) × (0.002) = 200 N·m³Build the geometric divisor
The full-section second moment and local width control the stress distribution.
(0.005) × (0.3) = 0.0015 m⁵Calculate local shear stress
Divide to obtain force per unit area at the selected location.
(200) ÷ (0.0015) ≈ 133333.3333 Pa
Does this worked answer make sense?
At a free top or bottom surface Q is zero, so this ideal shear stress is zero. Increasing local width with all other inputs fixed reduces the stress.
A second worked example — different values
A second case uses different data. Predict which way the answer will change, then calculate it without reusing the first answer.
- V · Shear force
- 50000 N
- Q · First moment of area
- 0.001 m³
- I · Second moment of area
- 0.002 m⁴
- b · Width at evaluation point
- 0.2 m
Form the shear-transfer numerator
Multiply the section shear force by the first moment for the chosen level.
(50000) × (0.001) = 50 N·m³Build the geometric divisor
The full-section second moment and local width control the stress distribution.
(0.002) × (0.2) = 0.0004 m⁵Calculate local shear stress
Divide to obtain force per unit area at the selected location.
(50) ÷ (0.0004) = 125000 Pa
Now try your own values
Change a value or its unit. The same method will show your calculation, step by step.
Results update only when you calculate. The lesson example above stays unchanged.
Your turn — check your understanding
Solve this separate case yourself. Use only the values below; the two worked examples use different data. Give the requested result in the selected unit.
- V · Shear force
- 80000 N
- Q · First moment of area
- 0.0015 m³
- I · Second moment of area
- 0.004 m⁴
- b · Width at evaluation point
- 0.25 m
Find: Learn: Beam shear stress
A hint, not the answer
Q is the first moment of that partial area about the neutral axis. VQ describes the change in bending force that must be transferred; dividing by I and the local width converts it to shear stress.
Use N, m³, m⁴ and m for V, Q, I and b. The units reduce to N/m² = Pa. Q is not a force or the full section’s second moment.
Show the full practice solution
Compare the steps with your work; revealing a solution does not mark the lesson complete.
Form the shear-transfer numerator
Multiply the section shear force by the first moment for the chosen level.
(80000) × (0.0015) = 120 N·m³Build the geometric divisor
The full-section second moment and local width control the stress distribution.
(0.004) × (0.25) = 0.001 m⁵Calculate local shear stress
Divide to obtain force per unit area at the selected location.
(120) ÷ (0.001) = 120000 Pa
Avoid the common trap
Use width at the evaluation point, not automatically the flange width. Build Q from the correct partial area and centroid distance; do not replace it with I.
When this method applies — and when it does not
This elementary formula applies where beam theory describes the stress distribution. Local concentrations near supports, holes, load introduction and abrupt thickness changes need separate assessment.
For study and understanding, not approval of a real structure, site operation or design. Apply the correct standard, National Annex and professional review to actual engineering work.
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Sources & further reading
References open in a new tab and explain the underlying principles. The teaching text and examples here are SimpleFlick’s own; the source organisations have not endorsed this calculator.
Reading focus: Beam shear stress. Use the relevant elastic-response, equilibrium, constitutive-relations or shear-and-torsion module; the lesson states its particular sign convention.
Lesson updated: · Both examples and the separate practice case are checked against an independent high-precision numerical implementation. This verifies arithmetic for the stated model, not engineering certification.