Global units
These selections apply to all inputs and results below. Choose the line load unit for w here.
- Use dot “.” as decimal separator.
- Positive w acts downward (as drawn).
- Load varies linearly from wa at x = a to
wb at x = a + b. The loaded region spans
[a, a+b].
- Tip: Enter c in its own unit (often inches) regardless of the length unit for L, a, b, x.
Numbers and units update together.
Need I for common shapes? Try the sectional properties calculators.
Results
| Parameter | Value |
|---|
| Reaction Force R₁ | --- lbf |
| Reaction Force R₂ | --- lbf |
| Shear @ x (Vₓ) | --- lbf |
| Max Shear (Vmax) | --- lbf |
| Reaction Moment Left (M₁) | --- lbf·in |
| Reaction Moment Right (M₂) | --- lbf·in |
| Moment @ x (Mₓ) | --- lbf·in |
| Max Moment (Mmax) | --- lbf·in |
| Slope @ x (θₓ) | --- radian |
| Max Slope (θmax) | --- radian |
| End Slope Left (θ₁) | --- radian |
| End Slope Right (θ₂) | --- radian |
| Deflection @ x (yₓ) | --- ft |
| Max Deflection (ymax) | --- ft |
| End Deflection Left (y₁) | --- ft |
| End Deflection Right (y₂) | --- ft |
| Bending Stress @ x (σₓ) | --- psi |
| Max Bending Stress (σmax) | --- psi |
About this load case
This calculator solves a beam fixed at both ends (zero rotation and zero deflection at
x = 0 and x = L) subjected to a
linearly varying distributed load acting only on the interval
[a, a+b].
The load intensity varies linearly from wa at
x = a to wb at
x = a + b:
w(x) = wa + (wb − wa) · (x − a) / b for a ≤ x ≤ a + b
- For x < a: the beam is unloaded by this line load (w(x) = 0).
- For a ≤ x ≤ a+b: the load varies linearly from wa to wb.
- For x > a+b: w(x) = 0 again.
- If a+b = L, the load reaches the right end; otherwise it is intermediate.
Sign convention
- Axes:
- x-axis: from left fixed end (x = 0) to right fixed end (x = L), positive to the right.
- y-axis: positive upward.
- Distributed load w(x):
- Positive w acts downward on the beam (as drawn in the diagram).
- Support reactions (at the fixed ends):
- R₁, R₂ > 0 → forces acting upward at the left and right ends.
- M₁, M₂ > 0 → moments acting clockwise on the beam
(compression at the top fibers near the support).
- Internal shear V(x):
- Defined on the left face of the cut.
- V(x) > 0 → internal shear acting upward on the left-hand segment.
- Internal moment M(x):
- M(x) > 0 corresponds to a sagging moment
(concave up, top fibers in compression).
- Slope θ(x):
- θ(x) > 0 means the cross-section rotates so that the beam tilts
upward to the right (counter-clockwise rotation).
- Because the beam is fixed–fixed, θ(0) = 0 and θ(L) = 0.
- Deflection y(x):
- y(x) > 0 is deflection upward.
- With positive w acting downward, typical deflections are negative (downwards).
- Bending stress σ(x):
- Computed as σ(x) = M(x) · c / I, where c is the distance to the extreme fiber and I is the second moment of area.
- Positive M with positive c corresponds to tension on the bottom fiber and compression on the top fiber.
Field equations (piecewise behavior)
The code uses closed-form polynomials consistent with these conventions and superposition. Conceptually:
- Before the load (0 ≤ x < a):
V(x) = R₁, M(x) = M₁ + R₁ x
- Within the load (a ≤ x ≤ a+b):
w(x) = wa + (wb − wa) · (x − a)/b, and
deflection/slope are obtained by integrating M(x)/(E·I) twice, with
fixed–fixed boundary conditions.
- After the load (a+b < x ≤ L):
the shear remains constant and the moment is linear, matching
M(a+b) and V(a+b).
All displayed reactions, internal forces, slopes and deflections are consistent with this
sign convention and the fixed–fixed boundary conditions.
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