Three-Hinged Arch

Q: Why is a three-hinged arch statically determinate while a two-hinged arch is not?

A three-hinged arch has 4 unknown reaction components (V A , H A , V B , H B ) and 4 equations (ΣFx = 0, ΣFy = 0, ΣM A = 0, and M C = 0 at crown). The four equations exactly equal the four unknowns — DSI = 0, determinate. A two-hinged arch also has 4 unknowns but only 3 equilibrium equations (no internal hinge = no crown condition equation). DSI = 4 − 3 = 1 — indeterminate to the 1st degree. The extra equation needed comes from compatibility: the horizontal displacement of support B (which is a

Q: What happens to bending moments in an arch when the supports settle?

For a three-hinged arch (determinate), settlement of supports causes the arch to move rigidly without developing additional stresses — just as settlement has no effect on any determinate structure. This is a major practical advantage of the three-hinged arch over the two-hinged and fixed arches. For a two-hinged or fixed arch (indeterminate), support settlement changes the horizontal thrust H, which changes the bending moments throughout the arch. This sensitivity to settlement is why three-hing

Q: What is the difference between a tied arch and a regular arch?

In a regular arch, the horizontal thrust is transferred to the foundations through the supports — the foundations must resist the outward thrust. This requires strong, stiff foundations (rock, piles) that can take horizontal loads. In a tied arch , the horizontal thrust is resisted internally by a tie rod or tie beam connecting the two support points. The tie takes the tension that balances the arch's compression, so the net force transferred to the foundations is purely vertical. Tied arches ar

Q: How does the rise-to-span ratio affect arch behaviour?

The rise-to-span ratio (h/L) significantly affects the horizontal thrust and the arch's structural efficiency. A higher h/L ratio (tall, steep arch) gives a lower horizontal thrust H = wL²/8h for the same span and loading — the arch transfers load more vertically, reducing the thrust on foundations. However, a very tall arch has longer members and more self-weight, which can offset the structural advantage. A lower h/L ratio (flat arch) gives a much higher horizontal thrust — flat arches develop

Horizontal Thrust, Bending Moment, Normal Thrust & Radial Shear — Parabolic & Circular Arches with Solved GATE Examples

Last Updated: March 2026

📌 Key Takeaways

An arch is a curved structural member that carries loads primarily through axial compression, with the horizontal thrust at the supports allowing the arch to resist loads more efficiently than a beam over the same span.
A three-hinged arch has hinges at both supports (A and B) and one internal hinge at the crown (C) — making it statically determinate (DSI = 0).
The horizontal thrust H is found by applying the condition M = 0 at the crown hinge: H = M₀_C / h, where M₀_C is the free bending moment at the crown section and h is the rise of the arch.
The bending moment at any section of an arch: M = M₀ − Hy, where M₀ = free BM (as if simply supported beam) and y = height of arch at that section.
For a parabolic arch under full-span UDL, the BM is zero everywhere — the arch is in pure compression. This is the most important result in arch theory.
The normal thrust (axial compression along the arch axis) and radial shear (perpendicular to the arch axis) at any section require resolving the forces using the arch slope angle φ.
Three-hinged arch problems appear regularly in GATE CE — typically as 2-mark NAT questions asking for horizontal thrust, BM at a specific section, or normal thrust.

1. What is an Arch? — Structural Behaviour

A beam spanning a large opening must resist the applied loads through bending — the bottom fibre is in tension and the top fibre is in compression (for a sagging beam). For long spans, the bending moments become very large, requiring deep, heavy beams. This is inefficient.

An arch achieves the same task — spanning an opening and carrying loads — but does so primarily through axial compression along its curved axis. Instead of bending, the arch material is compressed, which is far more efficient for materials like stone, brick, and concrete that are strong in compression but weak in tension. The key to this efficiency is the horizontal thrust at the supports.

When an arch carries a vertical load, the support reactions have both vertical and horizontal components. The horizontal component — the thrust H — is directed inward (toward the arch) at both supports. This horizontal thrust, acting at the level of the supports, creates a hogging moment that counteracts the sagging moment from the vertical loads. The net result is a much smaller bending moment throughout the arch than would occur in an equivalent straight beam.

For the ideal case of a parabolic arch under a uniformly distributed load, the horizontal thrust exactly cancels the free bending moment at every section — the bending moment is zero throughout the arch, and the arch carries the load purely in compression. This is why parabolic arches are used for bridges and buildings where the dominant load is distributed gravity load.

In real structures, loads are rarely perfectly distributed — there are point loads (vehicles on bridges), asymmetric loads (wind, eccentric live loads), and temperature effects. These produce non-zero bending moments in the arch, which must be designed for. But the bending moments in an arch are always much smaller than in an equivalent beam, making arches highly material-efficient for long spans.

2. Types of Arches

Arch Type	Number of Hinges	DSI	Analysis Method	Use Case
Three-hinged arch	3 (both supports + crown)	0 (Determinate)	Equilibrium + crown hinge condition	Bridges, roofs — when differential settlement is possible; exam standard
Two-hinged arch	2 (supports only)	1	Force method (compatibility)	Long-span bridges — more rigid than three-hinged
Fixed arch (hingeless)	0	3	Force method or FEM	Short, stiff bridges and tunnels — highest rigidity
Tied arch	Varies	Varies	Depends on hinge count	When horizontal thrust cannot be transferred to supports (e.g., on soft ground)

Why study the three-hinged arch? Being statically determinate, it can be completely analysed using the three equilibrium equations plus the one crown-hinge condition (M = 0 at crown). This makes it the standard arch type for hand calculation and for GATE CE examination questions. The two-hinged and fixed arches require compatibility equations and are more complex.

3. Three-Hinged Arch — Determinacy & Reactions

A three-hinged arch has pin supports at A (left) and B (right) and an internal hinge at the crown C. The pin supports provide 2 reactions each (vertical + horizontal), giving a total of 4 reaction components: V_A, H_A, V_B, H_B. With only 3 equilibrium equations, the structure would appear to have DSI = 1. However, the internal hinge at C provides one additional condition equation (M_C = 0), making the total number of equations equal to the number of unknowns — DSI = 0.

Reaction Components — Three-Hinged Arch

4 unknowns: V_A (vertical at A), H_A (horizontal at A), V_B (vertical at B), H_B (horizontal at B)

4 equations:

1. ΣF_x = 0: H_A = H_B = H (for vertical loads only — horizontal thrusts are equal and opposite)

2. ΣF_y = 0: V_A + V_B = total vertical load

3. ΣM_A = 0: finds V_B (moments about A eliminate H_A if supports are at the same level)

4. M_C = 0 (crown hinge condition): used to find H

For supports at the same level (most common case):

H_A = H_B = H (equal, opposite, horizontal — the thrust)

V_A and V_B are found from ΣM_A = 0 and ΣF_y = 0, exactly as for a simply supported beam.

H is found from M_C = 0 (see next section).

Note on direction of thrust: The horizontal thrust H acts inward at both supports — toward the arch. At support A, H_A acts to the right (into the arch). At support B, H_B acts to the left (into the arch). This inward thrust is what makes the arch work — it is what distinguishes an arch from a curved beam.

4. Horizontal Thrust — Derivation & Formula

The horizontal thrust H is found by applying the crown hinge condition: the bending moment at the crown hinge C must be zero. This is because a hinge cannot transmit moment — if a moment existed at C, the two halves of the arch would rotate relative to each other, which would be a mechanism.

Horizontal Thrust H — General Formula

Consider the left half of the arch (from A to C). Taking moments about C:

M_C = V_A × (L/2) − H × h − (moments of loads on left half about C) = 0

Where h = rise of arch at crown (height of C above the springing line AB).

Rearranging: H = M₀_C / h

Where M₀_C = free bending moment at the crown section = the BM at the midspan of an equivalent simply supported beam under the same loading.

For supports at the same level: M₀_C = V_A × (L/2) − (sum of moments of loads on left half about C)

Memory rule: H = (Free BM at crown) / (Rise of arch at crown)

Key Results for Common Loading — Horizontal Thrust

Full-span UDL w (symmetric arch, rise h, span L):

Free BM at crown = wL²/8

H = wL²/8h

Central point load W (symmetric arch):

Free BM at crown = WL/4

H = WL/4h

Point load W at distance a from A (crown at L/2):

V_A = W(L−a)/L, V_B = Wa/L

M₀_C = V_A×(L/2) − W×(L/2−a) if a < L/2 (load on left of crown)

= W(L−a)/L × L/2 − W(L/2−a) = WL(L−a)/2L − W(L/2−a)

= W(L−a)/2 − W(L/2−a) = WL/2 − Wa/2 − WL/2 + Wa = Wa/2…

Correction: M₀_C = V_A×(L/2) for load to right of crown: M₀_C = W(L−a)/L × L/2 = W(L−a)/2

For load at distance a from A (a ≤ L/2): M₀_C = Wa(L−a)/L × … standard result: M₀_C = Wab/L × …

Most directly: M₀_C = free BM at midspan for asymmetric load = V_A×(L/2) − W×(L/2 − a) for a < L/2

= [W(L−a)/L](L/2) − W(L/2−a) = W(L−a)/2 − WL/2 + Wa = W(L−a)/2 + Wa − WL/2

= WL/2 − Wa/2 + Wa − WL/2 = Wa/2

So M₀_C = Wa/2 (for load at distance a from A, a < L/2)

H = Wa/(2h) for point load at distance a < L/2 from A

5. Bending Moment at Any Section

Once the reactions (V_A, V_B, H) are known, the bending moment at any section of the arch can be found by taking moments of all forces on one side of that section about the section point, treating the arch as a curved beam.

Bending Moment at Section D (at horizontal distance x from A)

M_D = M₀_D − H × y_D

Where:

M₀_D = free bending moment at section D = BM at distance x in the equivalent simply supported beam under the same loading (calculated ignoring the arch shape, as if it were a straight beam)

H = horizontal thrust

y_D = height of the arch axis above the springing line at section D

The term Hy is the “relieving moment” — the horizontal thrust acting at the height y creates a hogging moment that reduces the sagging free BM.

Sign convention: M₀ is positive (sagging). Hy is always positive (hogging effect reduces net BM). Therefore M = M₀ − Hy: if M is positive the section is sagging; if negative it is hogging.

Physical insight: The arch shape y(x) determines how much of the free BM is “relieved” at each section. If the arch shape perfectly matches the free BM diagram (divided by H), then M = M₀ − Hy = 0 everywhere — the arch is in pure compression. This is why the parabolic arch is ideal for UDL (since the free BMD for a simply supported beam under UDL is also parabolic).

6. Normal Thrust and Radial Shear

In a beam, the internal forces at any section are bending moment M and shear force V. In an arch, the internal forces are bending moment M, normal thrust N (compression along the arch axis), and radial shear Q (perpendicular to the arch axis). The normal thrust and radial shear are found by resolving the net force at the section along and perpendicular to the arch tangent.

Normal Thrust and Radial Shear — Formulae

Let φ = angle of the arch tangent with the horizontal at the section (the slope angle of the arch at that point).

Let V = net vertical force on one side of the section (V_A minus any vertical loads to the left)

H = horizontal thrust

Normal thrust (axial compression along arch axis):

N = V sinφ + H cosφ

Radial shear (perpendicular to arch axis):

Q = V cosφ − H sinφ

Where V and H are the algebraic resultants of all forces to the left of the section:

V = V_A − (sum of vertical loads to left of section)

H = horizontal thrust (acts horizontally at the supports)

For a parabolic arch under full UDL: Q = 0 everywhere (no radial shear) and N = H/cosφ (pure axial compression — confirming zero bending and zero shear).

Slope of Parabolic Arch at Any Section

For a parabolic arch: y = 4h·x(L−x)/L²

Slope: dy/dx = 4h(L−2x)/L²

tan φ = dy/dx = 4h(L−2x)/L²

At the crown (x = L/2): dy/dx = 0 → φ = 0 (arch is horizontal at crown)

At the springing (x = 0 or x = L): dy/dx = ±4h/L → tan φ = 4h/L → φ = arctan(4h/L)

sin φ = (4h/L)/√(1 + 16h²/L²) | cos φ = 1/√(1 + 16h²/L²)

7. Parabolic Arch — Geometry & Key Results

The parabolic arch is the most important arch shape in structural engineering because it is the funicular shape for uniformly distributed load — meaning it carries UDL in pure compression with zero bending moment throughout. This makes it the optimal shape for bridges and roofs carrying distributed gravity loads.

Equation of a Parabolic Arch

For a symmetric parabolic arch with span L and rise h (both supports at the same level, origin at left support A):

y = 4hx(L−x)/L²

At x = 0 (support A): y = 0 ✓

At x = L/2 (crown): y = 4h(L/2)(L/2)/L² = h ✓

At x = L (support B): y = 0 ✓

Key property: The parabolic shape matches the free BMD of a simply supported beam under UDL (which is also parabolic with maximum = wL²/8 at midspan). Since M = M₀ − Hy, and M₀/H = (wL²/8 × x(L−x)/(L²/8)) … the arch height y is proportional to M₀ at every point, making M = 0 everywhere under UDL.

Summary of Key Results — Parabolic Arch Under Full-Span UDL

Horizontal thrust: H = wL²/8h

Bending moment everywhere: M = 0 (pure compression)

Radial shear everywhere: Q = 0

Normal thrust at crown (φ = 0): N_crown = H = wL²/8h

Normal thrust at springing (x = 0):

V_A = wL/2 | H = wL²/8h | tanφ = 4h/L

N_spring = √(V_A² + H²) = √((wL/2)² + (wL²/8h)²)

= (wL/2)√(1 + L²/16h²)

This is the maximum normal thrust — the arch is most heavily compressed at the springing.

8. Worked Example 1 — Point Load on Parabolic Three-Hinged Arch

Problem: A parabolic three-hinged arch has span L = 20 m and rise h = 4 m. It carries a point load of 60 kN at 5 m from the left support A. Find: (a) horizontal thrust H, (b) bending moment at the load point D (5 m from A), and (c) bending moment at the crown C.

Step 1 — Vertical Reactions

Taking moments about A: V_B × 20 = 60 × 5 → V_B = 15 kN

ΣF_y = 0: V_A = 60 − 15 = 45 kN

Step 2 — Horizontal Thrust H

Crown is at x = L/2 = 10 m from A, height y = h = 4 m.

Free BM at crown (M₀_C) = BM at x = 10 m in equivalent SS beam:

M₀_C = V_A × 10 − 60 × (10 − 5) = 45 × 10 − 60 × 5 = 450 − 300 = 150 kN·m

(Load is at 5 m from A, which is to the left of crown at 10 m — so the load is included in the left side moment.)

H = M₀_C / h = 150 / 4 = 37.5 kN

Step 3 — Arch Height at D (x = 5 m)

y_D = 4h × x(L−x)/L² = 4×4×5×15/400 = 4×4×75/400 = 1200/400 = 3.0 m

Step 4 — Bending Moment at D (x = 5 m)

Free BM at D (M₀_D) = V_A × 5 − 0 (no load to left of D except reaction) = 45 × 5 = 225 kN·m

Wait — the load IS at D (x = 5 m). For BM just to the right of D:

M₀_D = V_A × 5 = 45 × 5 = 225 kN·m (considering forces to left of D, before the load)

Arch BM: M_D = M₀_D − H × y_D = 225 − 37.5 × 3.0 = 225 − 112.5 = +112.5 kN·m

(Positive = sagging — the arch has not fully relieved the free BM because the load is not at the crown.)

Step 5 — Bending Moment at Crown C (x = 10 m)

Free BM at C: M₀_C = 150 kN·m (calculated above)

y_C = h = 4 m

M_C = 150 − 37.5 × 4 = 150 − 150 = 0 kN·m

BM at crown = 0 ✓ (as required by the crown hinge condition — this confirms the calculation is correct)

9. Worked Example 2 — Full-Span UDL on Parabolic Arch

Problem: A parabolic three-hinged arch has span L = 24 m and rise h = 6 m. It carries a UDL of 20 kN/m over the entire span. Find: (a) horizontal thrust H, (b) bending moment at x = 6 m from A, (c) bending moment at crown, and (d) normal thrust at x = 6 m.

(a) Horizontal Thrust H

By symmetry: V_A = V_B = wL/2 = 20×24/2 = 240 kN

Free BM at crown: M₀_C = wL²/8 = 20×576/8 = 1440 kN·m

H = 1440/6 = 240 kN

(b) BM at x = 6 m from A

Free BM at x = 6: M₀ = V_A×6 − w×6×3 = 240×6 − 20×6×3 = 1440 − 360 = 1080 kN·m

Arch height at x = 6: y = 4×6×6×(24−6)/24² = 4×6×6×18/576 = 2592/576 = 4.5 m

Arch BM: M = 1080 − 240×4.5 = 1080 − 1080 = 0 kN·m

BM = 0 everywhere under full UDL on a parabolic arch ✓

(c) BM at Crown

M = M₀_C − H×h = 1440 − 240×6 = 1440 − 1440 = 0 kN·m ✓

(d) Normal Thrust at x = 6 m

Since M = 0 and Q = 0 everywhere (pure compression under UDL), only N exists.

Net vertical force on left of section: V = V_A − w×6 = 240 − 120 = 120 kN

Arch slope at x = 6: tanφ = dy/dx = 4h(L−2x)/L² = 4×6×(24−12)/576 = 4×6×12/576 = 288/576 = 0.5

φ = arctan(0.5) → sinφ = 0.5/√1.25 = 0.4472, cosφ = 1/√1.25 = 0.8944

N = V×sinφ + H×cosφ = 120×0.4472 + 240×0.8944 = 53.67 + 214.66 = 268.3 kN (compression)

Check Q = V×cosφ − H×sinφ = 120×0.8944 − 240×0.4472 = 107.33 − 107.33 = 0 ✓

10. Worked Example 3 — Unsymmetric Loading

Problem: A parabolic three-hinged arch, span L = 16 m, rise h = 4 m, crown at midspan. Loads: UDL of 10 kN/m on the LEFT half only (0 to 8 m). Find: (a) vertical reactions, (b) horizontal thrust H, (c) bending moment at x = 4 m, and (d) bending moment at crown.

(a) Vertical Reactions

Total load = 10×8 = 80 kN, acting at 4 m from A (centroid of left-half UDL).

ΣM_A = 0: V_B×16 = 80×4 → V_B = 20 kN

ΣF_y: V_A = 80 − 20 = 60 kN

(b) Horizontal Thrust H

Free BM at crown (x = 8 m) in equivalent SS beam:

M₀_C = V_A×8 − 10×8×(8−4) = 60×8 − 10×8×4 = 480 − 320 = 160 kN·m

(The UDL covers the entire left half — its resultant of 80 kN is at 4 m from A, so its moment about crown = 80×(8−4) = 320 kN·m)

H = M₀_C/h = 160/4 = 40 kN

(c) BM at x = 4 m (within the loaded half)

Free BM at x = 4: M₀ = V_A×4 − 10×4×2 = 60×4 − 80 = 240 − 80 = 160 kN·m

Arch height at x = 4: y = 4×4×4×(16−4)/16² = 4×4×4×12/256 = 768/256 = 3.0 m

Arch BM: M = 160 − 40×3.0 = 160 − 120 = +40 kN·m (sagging)

The arch has a sagging moment here — the partial UDL creates bending in the arch.

(d) BM at Crown (x = 8 m)

M_C = M₀_C − H×h = 160 − 40×4 = 160 − 160 = 0 kN·m ✓

BM at crown = 0 (as required by hinge condition) ✓

Note: For the right half (x = 8 to 16 m), with no load, the arch BM will be negative (hogging) at some sections — this is the consequence of unsymmetric loading on an arch.

11. Worked Example 4 — Normal Thrust and Radial Shear at a Section

Problem: For the arch in Example 3 (span 16 m, rise 4 m, partial UDL on left half), find the normal thrust N and radial shear Q at x = 4 m.

Solution

From Example 3: V_A = 60 kN, H = 40 kN, UDL = 10 kN/m on left half.

Net vertical force V at x = 4 m (considering left side):

V = V_A − 10×4 = 60 − 40 = 20 kN (upward on left)

Arch slope at x = 4 m:

dy/dx = 4h(L−2x)/L² = 4×4×(16−8)/256 = 4×4×8/256 = 128/256 = 0.5

tanφ = 0.5 → sinφ = 0.4472, cosφ = 0.8944 (same as Example 2 — same geometry)

Normal thrust:

N = V×sinφ + H×cosφ = 20×0.4472 + 40×0.8944 = 8.944 + 35.777 = 44.72 kN (compression)

Radial shear:

Q = V×cosφ − H×sinφ = 20×0.8944 − 40×0.4472 = 17.889 − 17.889 = 0 kN

Wait — Q = 0 here despite unsymmetric loading? This is because at x = 4 m under UDL, the resultant force direction happens to align with the arch tangent at this section for this particular combination of V and H. This is a coincidence of the numbers — in general, Q ≠ 0 for unsymmetric loading.

Let us verify at x = 2 m instead (to show Q ≠ 0 in general):

V at x=2: 60 − 10×2 = 40 kN | H = 40 kN

dy/dx at x=2: 4×4×(16−4)/256 = 4×4×12/256 = 192/256 = 0.75 → tanφ = 0.75

sinφ = 0.6, cosφ = 0.8

N = 40×0.6 + 40×0.8 = 24 + 32 = 56 kN

Q = 40×0.8 − 40×0.6 = 32 − 24 = 8 kN ≠ 0 ✓ (radial shear is non-zero under partial UDL)

12. Influence Lines for Three-Hinged Arch

For a three-hinged arch, influence lines for H, BM at a section, and normal thrust can be constructed by placing a unit load at various positions and computing the response each time. For a parabolic arch, the ILD for horizontal thrust H is particularly elegant.

ILD for Horizontal Thrust H — Parabolic Arch

When a unit load is at position x from A:

V_A = (L−x)/L | M₀_C = V_A×(L/2) for load to right of crown, or M₀_C = V_B×(L/2) for load to left

For unit load at position x (x ≤ L/2):

M₀_C = V_A×(L/2) − 1×(L/2−x) = (L−x)/L×(L/2) − (L/2−x)

= (L−x)/2 − L/2 + x = x/2

H = M₀_C/h = x/(2h)

For unit load at position x (x ≥ L/2):

By symmetry: H = (L−x)/(2h)

ILD for H: Triangle with peak H = L/(8h) × 2… No: at x = L/2, H = (L/2)/(2h) = L/(4h)

Peak ordinate of ILD for H = L/(4h) at midspan (crown position)

ILD for H is a triangle: 0 at A, rises to L/(4h) at crown, falls to 0 at B.

ILD for BM at Section D

M_D = M₀_D − H×y_D

ILD for M_D = ILD for M₀_D − y_D × ILD for H

= (triangular ILD for free BM at D) − y_D × (triangular ILD for H)

Since both are triangles, the ILD for M_D is the algebraic difference of two triangles — which produces a shape that can have both positive and negative ordinates, even for a simply supported arch under gravity loads.

13. Common Mistakes Students Make

Finding V_A and V_B using the arch geometry instead of equilibrium: The vertical reactions V_A and V_B in a three-hinged arch are found exactly as for an equivalent simply supported beam under the same vertical loads — using moment equilibrium about A and B. The arch shape does not affect the vertical reactions (for vertical loads and supports at the same level). Many students try to involve H or the arch height in the vertical reaction calculation, which is unnecessary and wrong.
Using the wrong section for the crown hinge condition: The crown hinge condition M_C = 0 applies at the internal crown hinge, not at some other point on the arch. When applying this condition to find H, take moments of all forces on ONE side of the crown (left half or right half) about the crown point C. Including forces from both sides in a single moment equation is an error.
Forgetting to subtract Hy when calculating arch BM: The bending moment at any arch section is M = M₀ − Hy, not simply M₀. Students who forget the Hy term calculate the free BM (as for a straight beam) and miss the arch action entirely. The arch BM is always less than the free BM — by the amount Hy — because the thrust relieves the sagging moment.
Using H = wL²/8h for partial UDL or point loads: The formula H = wL²/8h applies only for a full-span UDL on a symmetric parabolic arch. For any other loading, H must be found from M₀_C/h where M₀_C is the free BM at the crown for the actual loading — not a formula lookup. Applying the UDL formula to partial loads or point loads is a very common error.
Wrong angle φ for normal thrust and radial shear: The angle φ is the slope of the arch tangent at the section — not the angle of the chord from support to crown. For a parabolic arch, φ varies along the span and must be computed from dy/dx = 4h(L−2x)/L² at the specific x of interest. Using a fixed angle (such as the angle at the springing) for all sections is incorrect.

14. Frequently Asked Questions

Why is a three-hinged arch statically determinate while a two-hinged arch is not?

A three-hinged arch has 4 unknown reaction components (V_A, H_A, V_B, H_B) and 4 equations (ΣFx = 0, ΣFy = 0, ΣM_A = 0, and M_C = 0 at crown). The four equations exactly equal the four unknowns — DSI = 0, determinate. A two-hinged arch also has 4 unknowns but only 3 equilibrium equations (no internal hinge = no crown condition equation). DSI = 4 − 3 = 1 — indeterminate to the 1st degree. The extra equation needed comes from compatibility: the horizontal displacement of support B (which is a roller in a two-hinged arch) must be zero, relating H to the arch deformations through Castigliano’s theorem.

What happens to bending moments in an arch when the supports settle?

For a three-hinged arch (determinate), settlement of supports causes the arch to move rigidly without developing additional stresses — just as settlement has no effect on any determinate structure. This is a major practical advantage of the three-hinged arch over the two-hinged and fixed arches. For a two-hinged or fixed arch (indeterminate), support settlement changes the horizontal thrust H, which changes the bending moments throughout the arch. This sensitivity to settlement is why three-hinged arches are preferred in situations where foundation movement is likely — for example, on compressible soils or in seismic regions.

What is the difference between a tied arch and a regular arch?

In a regular arch, the horizontal thrust is transferred to the foundations through the supports — the foundations must resist the outward thrust. This requires strong, stiff foundations (rock, piles) that can take horizontal loads. In a tied arch, the horizontal thrust is resisted internally by a tie rod or tie beam connecting the two support points. The tie takes the tension that balances the arch’s compression, so the net force transferred to the foundations is purely vertical. Tied arches are used where horizontal foundation reactions are difficult — for example, on soft ground or for bridges where the deck serves as the tie. The arch-tie system is self-contained and eliminates the need for thrust-resistant foundations.

How does the rise-to-span ratio affect arch behaviour?

The rise-to-span ratio (h/L) significantly affects the horizontal thrust and the arch’s structural efficiency. A higher h/L ratio (tall, steep arch) gives a lower horizontal thrust H = wL²/8h for the same span and loading — the arch transfers load more vertically, reducing the thrust on foundations. However, a very tall arch has longer members and more self-weight, which can offset the structural advantage. A lower h/L ratio (flat arch) gives a much higher horizontal thrust — flat arches develop enormous thrust forces that require very strong foundations. In practice, rise-to-span ratios for bridges typically range from 1:3 to 1:7. Masonry and stone arch bridges often use 1:4 to 1:6. For steel arches, 1:5 to 1:8 is common. The minimum practical rise-to-span ratio is about 1:8 below which the thrust becomes prohibitively large.