Shear for a simple rectangular beam: τ = 3F/(4·b·h). Add internal support (an extra bridge) to shorten the span if the check fails.
Load-case diagrams (from your inputs)
The shear-force, bending-moment and deflected-shape curves redraw every time you press Calculate.
Shear force V(x)
Bending moment M(x)
Deflected shape y(x)
Section modulus Wx–
Moment of inertia I–
Max bending moment M–
Bending stress σb–
Shear stress τ–
Max deflection δ–
Safety factor–
Midspan point load: Mmax = F·L/4
Deflection (point load): δ = F·L³ / (48·E·I)
Shear for a simple rectangular beam: τ = 3F/(4·b·h). Add internal support (an extra bridge) to shorten the span if the check fails.
Load-case diagrams (from your inputs)
The shear-force, bending-moment and deflected-shape curves redraw every time you press Calculate.
Shear force V(x)
Bending moment M(x)
Deflected shape y(x)
Bridge bending check — calculation guide
During extrusion the welding-chamber bridges carry the full extrusion pressure on the metal behind the profile. This module checks the bridge — treated as a simply supported rectangular beam spanning between two support pillars — against bending stress and transverse shear, using the selected allowable stress of the die steel.
Uniform load assumes the metal fills the whole span; the point-load case is a conservative estimate when the load concentrates over a central orifice. Pick the load case that best matches the bridge and the pressure-front behind it.
Uniform loadw = F/L — fills the whole span; Mmax at midspan
Midspan point loadF concentrated over the central orifice; conservative case
Derivation schematic
Beam model & coordinatesSimply supported Euler–Bernoulli beam; coordinate x along the span, curvature EI·y″ = M(x) at every section
Section & bending stressLinear stress σ = M·y/I across b×h, zero at the neutral axis, maximum at the outer fibre y = h/2; section modulus Wx = b·h²/6
Integration & boundary conditionsIntegrating the curvature twice gives the slope y′ then the deflection y; the supports set y(0)=0 and y(L)=0, maximum δmax at midspan
Bending derivation — moment diagrams
Read the two diagrams as envelopes of |M| along the span: the bending moment starts at zero at the supports and reaches its largest magnitude at midspan under both loads. Because σb = Mmax/Wx is evaluated where |M| is greatest, the midspan section is the critical one for the bending check in either case.
Uniform loadM(x) is a parabola peaking at midspan; Mmax = F·L/8 — σb is checked at this section
① Load: the arrows show a total load w = F/L spread evenly along the span.
② Shear & reactions: each support takes F/2; V starts +F/2, falls linearly through zero at midspan to −F/2.
③ Moment curve: M(x) = (w/2)·x·(L−x) is the area under the shear diagram — a parabola, zero at both ends, peaking at w·L²/8 = F·L/8 at midspan.
④ Checked section: |M| is greatest at midspan, so σb = Mmax/Wx is checked there.
Point loadM(x) is linear, rising to a peak at midspan; Mmax = F·L/4 — σb is checked at this section
① Load: a single force F acts at midspan, supported by F/2 at each end.
② Shear: constant +F/2 left of midspan, constant −F/2 right of midspan (step change at load).
③ Moment triangle: M(x) = (F/2)·x rises linearly to F·L/4 at midspan, then falls back to 0 — area under the shear rectangle.
④ Checked section: Mmax = F·L/4 is twice the uniform case, so σb doubles; critical at midspan.
Symbols
Symbol
Unit
Meaning
F
kN
Total load on the bridge (from extrusion pressure)
L
mm
Span — distance between the bridge supports
b
mm
Bridge width (cross-section)
h
mm
Bridge height (direction that resists bending)
[σ]
MPa
Allowable stress of the die steel at operating temperature
Wx
mm³
Section modulus about the bending axis
M
N·mm
Maximum bending moment on the span
σb
MPa
Bending stress
τ
MPa
Maximum transverse shear stress
E
MPa
Elastic modulus of the die steel (H13 ≈ 210 GPa)
I
mm⁴
Second moment of area of the cross-section
δ
mm
Maximum deflection at midspan
[δ]
mm
Max. allowable deflection (based on die gap and tolerance)
Key formulas
Wx = b·h² / 6Section modulus of a rectangle
I = b·h³ / 12Second moment of area about the bending axis
Mmax = F·L / 8Uniform load over the span
Mmax = F·L / 4Midspan point load
δ = 5·F·L³ / (384·E·I)Uniform load — max deflection at midspan
δ = F·L³ / (48·E·I)Midspan point load — max deflection
σb = Mmax / WxBending stress from the maximum moment
τ = 1.5·(V/A) ; V = F/2 → τ = 3F / (4·b·h)Max shear in a rectangular section (at mid-height)
n = [σ] / σbBending safety factor
Procedure
Get Wx = b·h²/6 and I = b·h³/12 from the bridge cross-section b × h.
Set the span L between the support pillars (or the full unsupported bridge width).
Estimate the total load F = p·A from the extrusion pressure p and the loaded area A behind the profile.
Compute Mmax (F·L/8 uniform, or F·L/4 point), then σb = Mmax/Wx and τ = 3F/(4·b·h).
Compare σb with [σ] and τ with 0.6·[σ]; keep the bending safety factor n ≥ 1.5.
Compute the deflection δ (5·F·L³/(384·E·I) uniform, or F·L³/(48·E·I) point) and compare it with the allowable [δ].
A total load F spread evenly gives each support F/2 and the peak moment Mmax = F·L/8 = 4.5×10⁶ N·mm at midspan, so σb = Mmax/Wx ≈ 138 MPa ≤ [σ] and n = [σ]/σb ≈ 3.2 ≥ 1.5.
The largest transverse shear acts at the supports; for a rectangle τ = 3F/(4·b·h) ≈ 80 MPa, far below 0.6·[σ] = 264 MPa, so shear is not critical.
Detailed derivation…
④ Deflection & verdict
δ = 5·F·L³/(384·E·I) ≈ 0.028 mm ≤ [δ] 0.2 mm → bridge OK
Twice integrating EI·y″ = M(x) with y = 0 at both supports gives δ = 5·F·L³/(384·E·I) ≈ 0.028 mm ≤ [δ] = 0.2 mm. All four criteria pass with comfortable margins; deflection is small because the bridge stays stiff.
Detailed derivation…
Worked example — point load case
Same bridge b×h = 40×70 mm, span L = 120 mm, but the 300 kN load concentrates at midspan:
A midspan point load makes the moment rise linearly from 0 at each support (the slope equals the shear, F/2) to a peak Mmax = F·L/4 = 9×10⁶ N·mm at midspan. Then σb = Mmax/Wx ≈ 275 MPa ≤ [σ], and n = [σ]/σb ≈ 1.60 ≥ 1.5.
Transverse shear is taken by the whole section: for a rectangle τ = 3F/(4·b·h) ≈ 80 MPa, far below 0.6·[σ] = 264 MPa, so shear never governs here.
Detailed derivation…
④ Deflection & verdict
δ = F·L³/(48·E·I) ≈ 0.045 mm ≤ [δ] 0.2 mm → bridge OK; the point load sets the bending limit
Twice integrating EI·y″ = M(x) with y = 0 at both supports gives the midspan deflection δ = F·L³/(48·E·I) ≈ 0.045 mm ≤ [δ] = 0.2 mm. All four criteria pass; bending stress (≈ 275 MPa, about double the uniform case) is the controlling margin under the point load.
Detailed derivation…
Practical remarks
A failing bridge is most often corrected by deepening h (bending strength grows as h² and stiffness as h³) or by adding an intermediate support bridge to shorten the span L. Always verify the final die with FEA and shop trials.
Where the formulas come from
The bridge is treated as a simply supported beam using Euler–Bernoulli beam theory, EI·y″ = M(x), where y is the lateral deflection and M the local bending moment. Integrating twice along the span and applying zero deflection at both supports yields each deflection formula below.
Bending stress — σ = M·y/I peaks at y = h/2, so σb = M/Wx ; Wx = (b·h³/12)·2/h = b·h²/6
Point load — linear moment diagram, Mmax = F·L/4 ; integrate twice → δ = F·L³/(48·E·I)
Shear — τmax = (3/2)·(V/A) with V = F/2 (support reaction) → 3F/(4·b·h)
Deflection check — the full derivation
Unlike the force-based bending and shear checks, deflection comes from integrating the beam-curvature equation twice (Euler–Bernoulli), EI·y″ = M(x), then applying the boundary conditions. Here it is shown step by step for both load cases.
Uniform load — w = F/L
Moment: M(x) = (w·L/2)·x − (w/2)·x² (a parabola peaking at w·L²/8). Integrate once → EI·y′ = (w·L/4)·x² − (w/6)·x³ + C₁.
Deflection peaks at midspan: δmax = |y(L/2)| = 5·w·L⁴/(384·E·I) → with w = F/L → δ = 5·F·L³/(384·E·I)
Midspan point load — F at L/2
For 0 ≤ x ≤ L/2 the moment is linear, M(x) = (F/2)·x, so EI·y = (F/12)·x³ + C₁·x + C₂ with C₂=0 from y(0)=0.
By symmetry the slope is zero at midspan: y′(L/2)=0 → C₁ = −F·L²/16.
At midspan: δmax = |y(L/2)| = F·L³/(48·E·I)
Flexural rigidity E·I — E is material stiffness (H13 ≈ 210 GPa) and I = b·h³/12 is section stiffness; they act as a product, so doubling I halves the deflection. Because deflection scales with h³, deepening the bridge is by far the most effective fix.
Why [δ] matters — bridge deflection changes the welding-chamber gap and therefore the profile dimensions. Set [δ] to a small fraction of the die-gap / profile tolerance so the loaded deflection stays comfortably inside tolerance.
Validity — the exact beam equation holds only while the bridge stays elastic and the deflections are small (linear theory), which is the normal operating range of an extrusion die.
Units & design assumptions
Units — load F in kN, lengths b, h, L in mm, modulus E in MPa (N/mm²). Internally F is converted to N, so moments come out in N·mm, stresses in MPa, deflection in mm.
Beam model — simply supported, uniform rectangular cross-section, isotropic tool steel at operating temperature, small elastic deflections (linear theory).
Load cases — the uniform case assumes the metal fills the whole span; the point case is a conservative bound for a pressure front concentrated over a central orifice.
Boundaries — the bridge stays elastic (σb ≤ [σ]), the deflection keeps the die gap and profile tolerance, and E ≈ 210 GPa for H13 (a little lower while hot).
If a check fails — what to do
Bending too high (σb > [σ] or n < 1.5) — deepen h: resistance grows as h², the most effective fix. Or widen b, or add a short auxiliary span.
Shear too high (τ > 0.6·[σ]) — enlarge the section area b·h (widening helps shear most) or reduce the pressure load.
Deflection too high (δ > [δ]) — stiffness grows as h³, so deepening h is very effective; otherwise shorten L with an intermediate support pillar or a second bridge.
After any change re-run bending, shear and deflection; re-check the die gap and land. Confirm the final die with FEA and a shop trial.