Learn: Cantilever — UDL deflection
Find the maximum transverse deflection of a cantilever carrying a uniform load over its entire length. The maximum is at the free end; this is a displacement calculation, not a bending-strength check.
What the formula is saying
The beam-curvature relationship integrates to the coefficient 1/8 for these exact supports and this exact loading. The span appears to power 4, so length has a much stronger effect than a simple proportional change.
Read the symbols in plain language
- w
- UDL
UDL. Apply the coefficient that belongs to this specific support and load arrangement.
N/mUse N/m as the base unit shown here. Use w in N/m, L in m, E in Pa and I in m⁴. The result is m; multiplying by 1000 converts it to mm. A UDL is a force per length; do not substitute the total load wL as w.
- L
- Length/span
Length/span. The integrated curvature equation makes span a strong influence on displacement.
mMetres measure length; 1 m = 1000 mm.
- E
- Young’s modulus
Elastic stiffness: the stress change needed for a unit strain in the stated material model. It is not a strength limit.
PaOne pascal is one newton per square metre. 1 MPa = 10⁶ Pa.
- 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.
- δmax
- Result to find
Cantilever — UDL deflection. Divide to obtain the displacement magnitude at the location stated in the lesson.
m
Sort out the units first
Use w in N/m, L in m, E in Pa and I in m⁴. The result is m; multiplying by 1000 converts it to mm. A UDL is a force per length; do not substitute the total load wL as w.
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 δmax and explain the result in the stated output unit.
- w · UDL
- 10000 N/m
- L · Length/span
- 3 m
- E · Young’s modulus
- 200000000000 Pa
- I · Second moment of area
- 0.005 m⁴
Apply the span power
The integrated curvature equation makes span a strong influence on displacement.
(3)^4 = 81 m^4Form the load-and-span term
Apply the coefficient that belongs to this specific support and load arrangement.
1 × (10000) × (81) = 810000 N·m³Form the stiffness divisor
Bending rigidity EI reduces deflection; the support coefficient multiplies it.
8 × (200000000000) × (0.005) = 8000000000 N·m²Calculate maximum deflection
Divide to obtain the displacement magnitude at the location stated in the lesson.
(810000) ÷ (8000000000) = 0.00010125 m
Does this worked answer make sense?
Doubling span while holding all other inputs fixed multiplies deflection by 16. Doubling E or I halves it. Check that the result is small compared with the span before trusting small-deflection theory.
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.
- w · UDL
- 5000 N/m
- L · Length/span
- 4 m
- E · Young’s modulus
- 30000000000 Pa
- I · Second moment of area
- 0.003 m⁴
Apply the span power
The integrated curvature equation makes span a strong influence on displacement.
(4)^4 = 256 m^4Form the load-and-span term
Apply the coefficient that belongs to this specific support and load arrangement.
1 × (5000) × (256) = 1280000 N·m³Form the stiffness divisor
Bending rigidity EI reduces deflection; the support coefficient multiplies it.
8 × (30000000000) × (0.003) = 720000000 N·m²Calculate maximum deflection
Divide to obtain the displacement magnitude at the location stated in the lesson.
(1280000) ÷ (720000000) ≈ 0.001777777778 m
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.
- w · UDL
- 8000 N/m
- L · Length/span
- 5 m
- E · Young’s modulus
- 200000000000 Pa
- I · Second moment of area
- 0.002 m⁴
Find: Learn: Cantilever — UDL deflection
A hint, not the answer
The beam-curvature relationship integrates to the coefficient 1/8 for these exact supports and this exact loading. The span appears to power 4, so length has a much stronger effect than a simple proportional change.
Use w in N/m, L in m, E in Pa and I in m⁴. The result is m; multiplying by 1000 converts it to mm. A UDL is a force per length; do not substitute the total load wL as w.
Show the full practice solution
Compare the steps with your work; revealing a solution does not mark the lesson complete.
Apply the span power
The integrated curvature equation makes span a strong influence on displacement.
(5)^4 = 625 m^4Form the load-and-span term
Apply the coefficient that belongs to this specific support and load arrangement.
1 × (8000) × (625) = 5000000 N·m³Form the stiffness divisor
Bending rigidity EI reduces deflection; the support coefficient multiplies it.
8 × (200000000000) × (0.002) = 3200000000 N·m²Calculate maximum deflection
Divide to obtain the displacement magnitude at the location stated in the lesson.
(5000000) ÷ (3200000000) = 0.0015625 m
Avoid the common trap
Do not mix the simply supported and cantilever coefficients. Use the bending-axis I, not area or polar J, and do not lose the 4th power on L.
When this method applies — and when it does not
Only the stated support and load arrangement is included. Use a positive load magnitude for the downward-deflection magnitude; uplift, partial-span loading, support settlement, cracking, creep and allowable-deflection limits require additional treatment.
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.
One idea understood. Keep going.
This optional checkmark is saved only in this browser. It is your own progress note, not a certificate.
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: Cantilever — UDL deflection. See the Stresses in Beams and Beam Displacements modules; match the load and support conditions, not just the equation’s appearance.
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.