Analysis of Determinate Structures - Cheat Sheet
Your 30-minute exam companion covering static determinacy, reactions, internal forces, and three-hinged arches. Focus on formula application, sign conventions, and equation-of-condition logic.
Static Determinacy & Stability
Formulas
DI = (3m + r) - (3n + e_c)
Meaning
m = members, r = reaction components, n = joints (nodes), e_c = equations of condition (internal releases)
Watch out
Count ALL members (including those that appear as one visual element) and distinguish joints from members. Internal hinge = +1 e_c only; pin connecting k members = (k-1) e_c
When to use
Classify any planar truss, beam, or frame structure
DI = r - (3 + e_c)
Meaning
Shortcut for simple beams only: r = total reaction unknowns, 3 = global equilibrium equations, e_c = internal hinges
Watch out
Only valid for beams, NOT frames. Each internal hinge adds exactly +1 to e_c. Do NOT use for trusses or complex frames.
When to use
Quick check for determinate beam (DI = 0) or indeterminate (DI > 0) or unstable (DI < 0)
Important facts
- Determinacy ≠ stability: DI = 0 is necessary but NOT sufficient; always check reaction geometry
- Three concurrent or parallel reactions → unstable, even if DI = 0
- Internal hinge ALWAYS adds exactly one equation of condition
- For cantilever: r = 3 (A_x, A_y, M_A); can be determinate with one load
- Continuous beam with r = 3 + number of interior supports; each interior support adds +1 indeterminacy
- Compound (Gerber) beams: solve suspended spans first (lowest DI), then propagate reactions upward
- Member count includes all bars/elements; rigid joints between two members count as ONE joint
Key definitions
Statically Determinate
Example
Simple supported beam, cantilever beam with one load, three-hinged arch
DI = 0; reactions and internal forces solvable by equilibrium equations alone (ΣF_x, ΣF_y, ΣM) without using member compatibility.
Statically Indeterminate
Example
Continuous beam, fixed-fixed beam, portal frame with multiple fixed supports
DI > 0; equilibrium alone is insufficient; compatibility conditions and elastic deformation required to solve.
Unstable Structure
Example
Three collinear supports, reactions parallel to a load direction
DI < 0 or reactions are concurrent/parallel; cannot resist all possible load directions despite reaction count.
Equation of Condition (e_c)
Example
Internal hinge: ΣM = 0 on one side of hinge; roller release: V = 0 at that point
Extra equilibrium relation from internal release (hinge, pin, or member release); adds one unknown but also one equation.
Diagrams to know
- Free body diagram showing all reactions and external loads
- Support symbols: fixed (3 reactions), pin/roller (2 and 1 respectively), internal hinge (dashed line or circle)
Reactions of Determinate Structures
Formulas
ΣF_x = 0
Meaning
Sum of all horizontal forces (and reaction components) equals zero
Watch out
Don't forget distributed loads over a length; replace with concentrated equivalent at the centroid. Watch sign: rightward = +, leftward = −
When to use
Always start here for any structure; rearrange to find horizontal reactions
ΣF_y = 0
Meaning
Sum of all vertical forces (and reaction components) equals zero
Watch out
Common error: forgetting that a reaction has BOTH vertical and horizontal components even if only one load direction is given
When to use
Find vertical reaction components; include weight of members if given
ΣM = 0 (about any point)
Meaning
Sum of moments about a chosen point equals zero; moment arm = perpendicular distance from force line to point
Watch out
Choose the moment point wisely to eliminate multiple unknowns. Counterclockwise = + (or reverse, but BE CONSISTENT). Moment of a couple is the same about any point.
When to use
Take moments about support A to eliminate A's reactions and solve for reactions at other supports
ΣM_C = 0 (at internal hinge C)
Meaning
At an internal hinge, the net moment on one side = 0 (no moment transfer across hinge)
Watch out
MUST take moments on ONE side of the hinge only. Apply to the ENTIRE left (or right) portion from the support to the hinge.
When to use
Three-hinged arch or Gerber beam with internal hinge; use as the 4th equation for determinate structures
Important facts
- Order of solving: (1) global ΣF_x, ΣF_y, ΣM, (2) internal hinge conditions ΣM_C = 0, (3) check with alternate moment point
- Always draw a complete free body diagram (FBD) of the entire structure before writing equations
- Distributed load w (kN/m) over length L → equivalent force = w × L, acting at centroid (L/2 for uniform)
- For inclined loads: resolve into components parallel and perpendicular to chosen axes
- Moment of a couple (M) is constant; moment arm irrelevant — it contributes equally to ΣM about any point
- Support reactions are EXTERNAL; internal forces at sections are found separately
Key definitions
Support Reaction
Example
Fixed support: 3 reactions (R_x, R_y, M); pin: 2 reactions (R_x, R_y); roller: 1 reaction perpendicular to surface
Force (or moment) exerted by a support on the structure to maintain equilibrium; direction opposite to movement prevented.
Simple Support (Pin or Roller)
Example
Beam on knife-edge at each end; bridge simply supported on bearings
Two reactions (pin) or one reaction perpendicular to surface (roller); no moment transfer.
Fixed Support (Cantilever Base)
Example
Beam rigidly connected to a column base; wall-mounted bracket
Three reactions: two force components and one moment; prevents all translation and rotation.
Diagrams to know
- Free body diagram of entire structure with all loads and reactions labeled
- Isolated free body of a segment (used for internal forces at a section)
Reactions or equations
R_A = ΣF / 2 (for symmetric load on simply supported beam, both ends at same level)
Note
Quick check: R_A = R_B = (sum of all vertical loads) / 2
Conditions
Load is symmetric about midspan; no horizontal load; reactions are vertical only
ΣM_A = 0 ⟹ R_B × L = Σ(P_i × x_i) + w(centroid location)
Note
Most common method for finding support reactions in determinate beams
Conditions
Taking moments about A to solve for B; L = span length
Internal Forces (N, V, M) in Determinate Structures
Formulas
N (Axial Force) = ΣF_parallel (on one side of section)
Meaning
Sum of forces parallel to member axis; positive = tension, negative = compression (convention may vary — state yours)
Watch out
Sign convention: establish clearly at the start. Many textbooks use opposite signs for N. Consistency across the diagram is critical.
When to use
Cut the member at desired section; sum parallel forces on the left (or right, CONSISTENTLY) side
V (Shear Force) = ΣF_perpendicular (on one side of section)
Meaning
Sum of forces perpendicular to member axis; positive = shear causing clockwise rotation on left face
Watch out
Shear jumps by the magnitude of the point load at that location. At a distributed load, shear varies linearly.
When to use
At any section, sum transverse loads on one side. Used to draw shear force diagram (SFD)
M (Bending Moment) = ΣM_about_section (on one side of section)
Meaning
Sum of moments about the section; positive = tension in bottom fiber (sagging, ∪ shape)
Watch out
Sign: upward load causes positive (sagging) moment. Negative moment (hogging, ∩) occurs over supports in continuous beams. Slope of M-diagram = shear force (dM/dx = V).
When to use
Take moments on one side of the cut section to find internal moment. Used to draw moment diagram (BMD)
Important facts
- dV/dx = −w (rate of change of shear = negative of distributed load intensity)
- dM/dx = V (slope of moment diagram = shear force at that section)
- At a point load: shear jumps; at a concentrated moment: moment jumps (no change in V)
- Free end: V = 0 and M = 0 (or M = applied moment if present)
- Pinned support: M = 0 (hinge cannot carry moment); V ≠ 0 (can carry shear)
- Fixed support: M ≠ 0 and V ≠ 0 (both can be carried by a fixed base)
- Maximum moment often occurs where V = 0 (inflection point if M changes sign)
Key definitions
Axial Force (N)
Example
In a truss member carrying load along its length; in the vertical column of an L-frame
Internal force parallel to member axis; positive = tension (pulling apart), negative = compression (pushing together).
Shear Force (V)
Example
Maximum at supports of a simply supported beam; zero at midspan of uniformly loaded beam
Internal force perpendicular to member axis; causes one part to slide past the other; jumps at point loads.
Bending Moment (M)
Example
Maximum at center of simply supported uniformly loaded beam; zero at free ends and over pinned supports
Internal couple (moment) that resists bending; positive moment creates tension in bottom fiber (sagging parabola).
Sign Convention (Standard)
Example
Uniformly loaded simply supported beam: V positive to the left of center, negative to the right; M is everywhere positive (sagging)
Positive shear: right face moves up, left face moves down (clockwise on left face). Positive moment: tension in bottom (∪ shape, sagging).
Diagrams to know
- Axial Force Diagram (AFD): shows N variation along member length
- Shear Force Diagram (SFD): shows V variation; jumps at point loads, linear under distributed loads
- Bending Moment Diagram (BMD): shows M variation; parabolic under uniform load, linear under point loads, zero at hinges
Reactions or equations
V_left = R_A − (sum of loads between A and section) = R_A − ∫w dx
Note
Positive V (upward) on left face means the shear is pulling the left section down and right section up
Conditions
For a cantilever or simple beam, taking left-side equilibrium
M_section = R_A × x − ∫∫w dx² (for cantilever; add concentrated moment loads where applicable)
Note
For cantilever, M increases as you move from free end to fixed end (M_max at fixed support)
Conditions
x measured from left support; integrate distributed load to get moment contribution
Three-Hinged Arches
Formulas
ΣM_crown = 0 (on one side of the crown hinge C)
Meaning
At the crown hinge, net moment on left (or right) side = 0; this gives the horizontal thrust H
Watch out
Must apply to ONE side of the crown ONLY. Include the weight and all loads on that side, measured perpendicular to the centroidal axis. Forget this condition and you cannot solve the arch.
When to use
Determine the horizontal thrust in a three-hinged arch (symmetrical or not)
H = (ΣM_crown_left) / (rise h)
Meaning
H (horizontal thrust) = moment about crown from left side divided by the vertical distance from support to crown
Watch out
H is the same throughout the arch (any vertical slice). Moment includes the lever arm from the crown to the load.
When to use
After ΣM_C = 0 and solving for one reaction, use this to find H explicitly
V_A + V_B = ΣP_vertical (total vertical load)
Meaning
Sum of vertical reactions equals total vertical load (global equilibrium)
Watch out
Often V_A ≠ V_B even for symmetric arch under point load if load is not at center
When to use
Check; find V_A or V_B given the other and total load
H_A + H_B = 0 (for an arch with no external horizontal load at supports)
Meaning
Horizontal reactions are equal and opposite (equilibrium in x-direction)
Watch out
If there is an external horizontal load P_x, then H_A + H_B = P_x (not zero)
When to use
Verify that H_A = H_B = H (the thrust)
Important facts
- A three-hinged arch is ALWAYS statically determinate (DI = 0) by definition
- The presence of H (horizontal thrust) is what makes an arch efficient — it lowers bending moment vs. a simple beam
- For a symmetrical arch under symmetrical load: V_A = V_B, H_A = H_B = H, and both by symmetry
- Arch moment at any point = (moment if it were a beam) − (H × rise at that point); thus M_arch ≤ M_beam
- Crown condition ΣM_C = 0 is the key; without it, 4 unknowns (H_A, V_A, H_B, V_B) with only 3 global equations → indeterminate
- At supports (A, B): both vertical and horizontal reactions exist; at crown C: moment discontinuity (jump from one side to other) = 0
Key definitions
Three-Hinged Arch
Example
Symmetrical parabolic arch with support hinges at same level and crown hinge at midspan
Determinate arch with hinged supports at A and B and an internal hinge at the crown C; defines 4 unknowns and has 4 equations (3 global + crown condition).
Horizontal Thrust (H)
Example
In a symmetrical arch under vertical load: H = ΣM_C (at crown from left side) / h, where h = rise
The horizontal reaction at arch supports; resists the lateral push from the arch curvature; reduces bending compared with a straight beam of same span.
Crown Hinge
Example
In a symmetric arch, often at midspan; allows rotation without moment transfer between left and right halves
The internal hinge at the top of the arch (or at the apex); enforces the condition ΣM = 0 on each side independently.
Arch Rise (h)
Example
For a parabolic arch spanning 20 m with rise 5 m: h = 5 m
Vertical distance from support level to the crown hinge; determines the relationship between moment at crown and horizontal thrust.
Diagrams to know
- Arch with three hinges labeled: supports A, B and crown C
- Free body of left half of arch, showing V_A, H_A, loads, and moment equilibrium at C
- Thrust diagram showing H constant along the arch
- Moment diagram: parabolic shape under uniform load, much lower than a simple beam
Reactions or equations
ΣM_A (global) = 0 ⟹ V_B × L = ΣP_i × x_i
Note
First step: find vertical reactions using global moment equilibrium
Conditions
L = span length; x_i = horizontal distance of load i from A
ΣM_C^left = 0 ⟹ V_A × (L/2) − H × h − (loads on left side with their moments) = 0
Note
This is the crown condition; rearrange to solve for H
Conditions
Left side extends from A to C; h = rise (vertical distance from A to C)
ΣF_x = 0 ⟹ H_A = H (from left side condition); H_B = H (by symmetry of moment equilibrium)
Note
Both thrust components are equal; this is the defining characteristic of the three-hinged arch
Conditions
No external horizontal load at supports
Compound (Gerber) Beams & Internal Hinges
Formulas
Analyze suspended spans FIRST; then propagate reactions upward
Meaning
In a Gerber beam, identify the suspended (cantilevered) portion; solve it independently, then add its reaction to the next support below
Watch out
Many students solve the whole beam at once; you'll get contradictions. Always start with the span that has the fewest supports.
When to use
Multi-span determinate beams with internal hinges (roller or pin connections between spans)
e_c (at internal hinge) = number of releases = 1 (for a single hinge between two parts)
Meaning
Each hinge adds one equation of condition (ΣM = 0 at hinge for moment on one side)
Watch out
Do NOT confuse a simple hinge with a pin connecting k members: a simple internal hinge between two beam segments = 1 e_c; a pin at k members = (k − 1) e_c
When to use
Count to verify DI = 0 for the compound beam: DI = r − (3 + e_c)
Important facts
- Compound beam is determinate (DI = 0) by design; internal hinges reduce indeterminacy
- Solve bottom-up or left-to-right, starting with the most-dependent (most-cantilever-like) span
- At an internal hinge: shear V may be non-zero, but bending moment M = 0
- Reaction of a suspended span acts as an external downward load on the supporting span below
- Each internal hinge adds exactly +1 to e_c and allows one additional equilibrium relation (ΣM_hinge = 0)
Key definitions
Compound (Gerber) Beam
Example
Main beam rests on three supports; a suspended (secondary) beam connects between two main supports via a hinge
A determinate multi-span beam with internal hinges (roller or pin) connecting the spans; allows independent analysis of cantilever portions.
Suspended Span
Example
In a three-span beam with a central suspended span: the left and right outer spans are main; the center is suspended on the left support
The portion of a Gerber beam that is cantilever-like, supported only at one end by another beam; analyzed first, independently.
Internal Hinge (or Pin)
Example
Roller connection between main and secondary beam; hinge pin joining two structural members
A connection between two adjacent beam segments that allows relative rotation; moment = 0 at the hinge interface.
Diagrams to know
- Compound beam with internal hinge clearly marked (usually shown as a circle or small gap with a pin)
- Free body of suspended span isolated from rest of structure
- Free body of main span with the suspended span reaction shown as an external load
Reactions or equations
Solve suspended span first: ΣM = 0, ΣF = 0 independently
Note
The reaction of the suspended span becomes a load on the main beam
Conditions
Suspended span has loads; find its reactions
Add suspended span reaction to main beam; then solve main beam reactions
Note
This is the most common approach in Philippine board exams
Conditions
Main beam now has the suspended span reaction as an additional downward load
Sign Conventions & Consistency
Important facts
- Choose ONE sign convention at the start and apply it consistently throughout all calculations and diagrams
- Most common: counterclockwise = +, clockwise = − (for moments); rightward = +, upward = + (for forces)
- Shear force sign: on the LEFT face of a cut, upward = +; on the RIGHT face, upward = −
- Internal moment diagram sign: sagging (∪) = +, hogging (∩) = − (per ACI and NSCP standard)
- If you reverse direction of equilibrium (left side vs. right side of a cut), reverse the sign of the result
Key definitions
Right-Hand Rule (for moments)
Example
For a horizontal beam, counterclockwise moment (viewed from above) is positive; causes upward curvature in left overhang
Thumb in direction of axis (along member); fingers curl from positive to negative; counterclockwise viewed from that direction = positive.
Sagging Moment (Positive)
Example
At midspan of a simply supported uniformly loaded beam
Bending moment that creates tension in the bottom fiber and compression in the top; forms a ∪ shape (smile).
Hogging Moment (Negative)
Example
Over intermediate supports in a continuous beam
Bending moment that creates compression in bottom fiber and tension in top; forms a ∩ shape (frown).
Diagrams to know
- Shear and moment conventions: diagram showing positive and negative directions on cut faces
Quick Checks & Relationships
Formulas
dV/dx = −w (distributed load intensity)
Meaning
Slope of shear diagram = negative of the distributed load; shear decreases if load acts downward
Watch out
If load is upward (negative w), then dV/dx is positive (shear increases to the right). Watch sign!
When to use
Verify SFD shape under distributed loads; check for discontinuities at point loads
dM/dx = V (shear force)
Meaning
Slope of moment diagram = shear force at that section; moment is constant where V = 0
Watch out
Linear shear → parabolic moment; zero shear → horizontal (constant) moment; positive shear → moment increasing to the right
When to use
Verify BMD shape; find maximum moment (occurs where V = 0)
ΔM = ∫V dx (area under SFD)
Meaning
Change in moment between two sections = area under shear force diagram
Watch out
If SFD is negative, area is negative → moment decreases to the right
When to use
Quick check: calculate moment at two points and verify difference equals area under SFD
ΔV = −∫w dx (area under load diagram)
Meaning
Change in shear = negative area under distributed load
Watch out
Downward load (positive w) → negative change in V as you move right; SFD slopes downward
When to use
Verify shear diagram shape and values
Important facts
- At a concentrated load: shear diagram has a vertical jump equal to the load magnitude
- At a concentrated moment: moment diagram has a vertical jump equal to the moment magnitude; shear does NOT change
- Under a uniformly distributed load: shear is linear (slope = −w), moment is parabolic (second-degree curve)
- At a free end: both V and M must be zero (unless a moment is applied at that end, then M ≠ 0)
- At a pin support: M = 0 and V = reaction (shear is non-zero but moment must be zero)
- At a fixed support: both V and M = reaction values; moment can be very large at fixed base
Key definitions
Inflection Point
Example
Over an interior support of a continuous beam (transition from sagging to hogging)
Location where bending moment changes sign (from + to − or vice versa); curvature of the beam changes direction.
Critical Section
Example
Midspan of simply supported beam (max M); just left of a support (max V); at a point load (shear jump)
A section where internal forces reach local maximum or minimum; typically where V = 0 for moment, or at supports and point loads.
Diagrams to know
- Load diagram → Shear Force Diagram → Bending Moment Diagram (relationship of slopes)
Frames & Portal Structures
Formulas
DI = (3m + r) − (3n + e_c)
Meaning
For frames: m = number of members, r = reaction components, n = joints, e_c = internal releases
Watch out
Count EVERY member (horizontal, vertical, diagonal); a single L-corner counts as TWO members meeting at a joint. A pin at a joint where 3 members meet = 2 e_c, not 1.
When to use
Determine if a multi-member frame is determinate, indeterminate, or unstable
Important facts
- In a frame, internal forces (N, V, M) must be found member by member using equilibrium of joints or sections
- At a rigid joint (no hinge): moment continuity is enforced; M_in one member = −M_out on the other (internal equilibrium)
- At an internal hinge in a frame: M = 0 on both sides at that joint; forces can still exist
- Portal frame base reactions: if base is fixed, 3 reactions; if pinned, 2 reactions; if supported on two column bases (spread), each base has its own reactions
- Typical portal under horizontal load (wind): lateral deflection is resisted by bending of columns; moment largest at base of columns
- By symmetry (if structure and loads are symmetric): half-frame analysis often suffices
Key definitions
Portal Frame (Rectangular Frame)
Example
Building frame: four members (two columns, one beam at top, base connections); base is fixed or pinned
A rigid frame with vertical columns and a horizontal beam, typically cantilevered or supported on a spread foundation.
Knee (or Corner Joint)
Example
Where a column meets a beam in a portal; moment at knee on column = moment at knee on beam (internal equilibrium)
A rigid joint where two members meet at an angle (typically 90° in portal frames); moment transfer occurs.
Member Axis (Local Coordinates)
Example
In a vertical column: axial = vertical, shear = horizontal; in a horizontal beam: axial = horizontal, shear = vertical
For each member of a frame, a local coordinate system: axial force along the member, shear perpendicular, moment about the member.
Diagrams to know
- Frame outline with supports, joints labeled, loads shown
- Free body of individual joints (pin representations at each node)
- Axial force diagram on frame (showing N in each member)
- Shear force diagram on frame (showing V in each member)
- Bending moment diagram on frame (showing M in each member, often drawn to a specific side convention)
Reactions or equations
Joint equilibrium: ΣF_x = 0, ΣF_y = 0, ΣM = 0 (at each joint of a frame)
Note
Careful with sign: forces on the member face pointing outward from joint are INTERNAL forces, equal and opposite on the joint free body
Conditions
Must apply to every joint; internal forces on cut members balance with applied loads and reactions
Section method: cut the frame and apply equilibrium to one part
Note
Often simpler than joint-by-joint analysis; typical in board exams
Conditions
Choose a section that cuts through the member where you want internal forces
Must remember
DI = (3m + r) − (3n + e_c) for frames; DI = r − (3 + e_c) for beams. DI = 0 → determinate (solvable by equilibrium); DI > 0 → indeterminate (need compatibility); DI < 0 → unstable (cannot be solved).
Determinacy ≠ stability. A structure can have DI = 0 yet be unstable if reactions are concurrent (all meet at one point) or parallel (all point in same direction). Always check reaction geometry.
Three equilibrium equations only (ΣF_x = 0, ΣF_y = 0, ΣM = 0 at any point). Each internal hinge or release adds EXACTLY one equation of condition (e_c). This is how you get 4 equations for a three-hinged arch (3 + 1).
Three-hinged arch: use crown condition ΣM_C = 0 on ONE side of the crown ONLY. This gives the horizontal thrust H = moment_at_crown / rise. Without this, you cannot solve the arch (you'll have 4 unknowns and only 3 global equations).
Internal forces (N, V, M) come from equilibrium of a cut free body. Choose to cut and analyze the LEFT side (or RIGHT, but be consistent). Sum forces and moments on that one side; the internal forces are what you need to balance equilibrium.
Sign convention must be CONSISTENT throughout. Standard: counterclockwise moment = +, rightward force = +, upward force = +. Sagging moment (∪, tension bottom) = +; hogging moment (∩, tension top) = −.
dV/dx = −w and dM/dx = V. Use these relationships to verify SFD and BMD shapes. Zero shear (V = 0) is where moment is often maximum. Shear jumps at point loads; moment jumps at concentrated moments.
In a Gerber (compound) beam, ALWAYS analyze suspended (cantilever-like) spans first, then add their reactions to the supporting spans below. This is the correct sequential approach; solving the whole beam at once will lead to error.
At an internal hinge: M = 0 (no moment transfer across hinge), but V ≠ 0 (shear can exist). At a pin or roller support: M = 0 and V = reaction. At a fixed support: M ≠ 0 and V ≠ 0.
Always draw a complete FREE BODY DIAGRAM (FBD) before writing any equation. Include ALL external loads, ALL reaction components, and indicate which direction is positive. An error in the FBD will corrupt all subsequent work.
Last minute tips
Count members, joints, reactions, and releases CAREFULLY for determinacy. In a frame, a rigid joint where two members meet counts as ONE joint, not two. An internal pin at that joint = 1 e_c. A pin connecting THREE members = 2 e_c (not 3).
For three-hinged arches: immediately recognize the structure type and write ΣM_C = 0 on ONE side as your 4th equation. If you forget this, you'll be stuck with 4 unknowns and 3 equations. This is a common mistake that loses exam points fast.
When solving internal forces at a section, ALWAYS cut one member or joint, isolate the free body on ONE side, and apply equilibrium. Write the equations in a logical order (ΣF_x, ΣF_y, then ΣM) to minimize coupling and risk of algebraic error.
Check your reactions by verifying ΣF_y (total) and ΣM (about a different point, NOT where you solved). If these don't balance, your reactions are wrong; find the error before moving to internal forces.
For complex frames, draw internal force diagrams (N, V, M) member by member, not as one overall diagram. Label each member clearly and apply local member coordinates consistently. This prevents sign-convention errors and makes checking easier.
Comparison tables
Determinacy, Stability & Indeterminacy Quick Reference
| Classification | DI Value | Solvable by Equilibrium Alone? | Stability Check Required? | Example |
|---|---|---|---|---|
| Statically Determinate | DI = 0 | Yes, use ΣF = 0, ΣM = 0 only | Yes; concurrent/parallel reactions = unstable | Simple beam, cantilever, three-hinged arch |
| Statically Indeterminate | DI > 0 | No; need compatibility + elastic constants | Yes; indeterminate degree must match equation count | Continuous beam, fixed-fixed beam, two-story frame |
| Unstable (Mechanism) | DI < 0 OR reactions concurrent/parallel | Cannot solve; structure can move freely | Structure will fail or displace without support | Three collinear supports, or all reactions parallel to a load |
| Compound Beam | DI = 0 (by design with internal hinges) | Yes, but use internal hinge condition first | Yes; hinges allow cantilever action safely | Gerber beam: suspended span + main spans with hinges |
Support Types: Reactions & Moments
| Support Type | Number of Reactions | Reaction Components | Moment Carried? | Symbol / Diagram |
|---|---|---|---|---|
| Pin (Hinge) | 2 | R_x (horizontal), R_y (vertical) | No (M = 0) | Circle or • |
| Roller | 1 | R perpendicular to surface (usually vertical) | No (M = 0) | Circle on wheels or ▲ with wheels |
| Fixed (Cantilever) | 3 | R_x, R_y, and M (moment) | Yes; can be large | Thick line or filled square |
| Internal Hinge | Counted as e_c = 1 (adds one equation) | Forces transmitted; moment BLOCKED | No; ΣM = 0 on each side | Small circle, dashed line, or pin symbol between members |
Internal Forces: Sign Convention & Typical Values
| Internal Force | Positive Direction | On Left Face | On Right Face | Typical Location (Simply Supported Beam) | |
|---|---|---|---|---|---|
| Axial Force (N) | Tension (pulling) | Pointing right (+) | Pointing left, magnitude N | Pointing right, magnitude N | Often zero in beams; non-zero in trusses & frames |
| Shear Force (V) | Upward on left face, downward on right (clockwise) | Upward, magnitude V (internal shear resisting slip) | Downward, magnitude V | Positive from left support toward midspan; negative from midspan toward right support | |
| Bending Moment (M) | Sagging (∪ shape, bottom in tension) | Counterclockwise at cut, magnitude M | Clockwise at cut, magnitude M | Zero at supports; maximum positive at midspan under uniform load |
Distributed Load → Internal Force & Moment Relationships
| Load Type | Shear Change (dV/dx) | Moment Change (dM/dx) | SFD Shape | BMD Shape |
|---|---|---|---|---|
| No distributed load (w = 0) | dV/dx = 0 (constant) | dM/dx = V (linear if V ≠ 0) | Horizontal (constant) | Linear (sloped line) |
| Uniform downward load (w > 0) | dV/dx = −w (decreases left to right) | dM/dx = V (linear shear → parabolic moment) | Sloped downward (linear decrease) | Parabola opening down (concave) |
| Concentrated point load P downward | Jump down by P (discontinuity) | Slope changes abruptly (corner in BMD) | Vertical drop equal to P | Kink (change in slope) at load location |
| Concentrated moment M_applied | No change (ΔV = 0) | Jump up by M (discontinuity) | No change (constant between loads) | Vertical jump equal to M |
Three-Hinged Arch vs. Simple Beam: Comparison
| Aspect | Simple Supported Beam | Three-Hinged Arch |
|---|---|---|
| Determinacy | DI = 0 (determinate) | DI = 0 (determinate; 4 unknowns, 4 equations including crown condition) |
| Vertical Reactions | V_A = ΣP · (L − x) / L; V_B = ΣP · x / L (beam equation) | V_A and V_B found from global ΣM = 0, same as beam |
| Horizontal Reactions | None (H = 0) if no horizontal load | H_A = H_B = H (non-zero thrust due to curvature); found from crown condition |
| Maximum Bending Moment | M_max = P · L / 4 (for point load at center) | M_max ≤ M_beam − H × h (reduced by the thrust effect) |
| Primary Advantage | Simple design, easy to analyze | Lower internal bending moment; more efficient for long spans (less material, less deflection) |
| Crown Condition | No crown; simply ΣM = 0 everywhere for equilibrium | ΣM_C = 0 (on one side of crown hinge); essential for solving H |
Diagrams
Static Determinacy & Structure Classification
Overview of how to classify structures: the determinacy formula differs between frames and beams, and determinacy alone does not guarantee stability. Always verify reaction geometry.
Step-by-Step Solution Procedure for Determinate Structures
Complete procedure from structure classification through internal force diagrams, including key verification steps. Emphasizes importance of FBD and checking reactions before moving to internal forces.
Three-Hinged Arch Analysis — Finding Reactions & Thrust
Focused workflow for three-hinged arch problems. Highlights the critical crown condition (ΣM_C = 0 on one side) that distinguishes arch from beam and yields the horizontal thrust H.
Load-to-Moment-Diagram Evolution Under Distributed Load
State progression showing how distributed load generates shear pattern, which then determines moment diagram shape. Illustrates the differential relationships dV/dx = −w and dM/dx = V.
Gerber (Compound) Beam Analysis — Sequential Solving Order
Sequential timeline emphasizing the critical approach: solve suspended/cantilevered portions first, then work upward. This prevents inconsistency errors and matches the typical board-exam workflow.