CELE Structural Theory & Analysis — Analysis of Determinate StructuresRevision Notes

Formulas

Exam Tips

• Memorise both formulas: DI = (3m + r) − (3n + ec) for frames, DI = r − (3 + ec) for beams. Board exams test both.
• Draw the structure, label all supports and their reaction components, then substitute directly into the formula.
• When asked 'Is the structure stable and determinate?', answer two separate questions: (1) Is DI = 0? (2) Are reactions non-concurrent and non-parallel?
• For trusses (not covered here but related): DI = m + r − 2j; know which formula applies to which structure type.
• When in doubt about stability, check: can the structure translate horizontally? vertically? rotate? If yes to any, it is unstable.

Key Points

• A structure is statically determinate if all unknown reactions and internal forces can be found using the three equilibrium equations alone: ΣFx = 0, ΣFy = 0, ΣM = 0.
• Degree of Indeterminacy (DI) for planar frames: DI = (3m + r) − (3n + ec), where m = number of members, r = total reaction components, n = number of joints (nodes), ec = equations of condition (internal releases).
• Shortcut for beams: DI = r − (3 + ec). Each internal hinge contributes exactly one equation of condition (ec = 1 per hinge).
• DI = 0 → statically determinate; DI > 0 → indeterminate to degree DI; DI < 0 → geometrically unstable (mechanism).
• A pin connecting k members at one joint contributes (k − 1) equations of condition, not just one.
• Stability is a separate check from determinacy. A structure with exactly enough reactions can still be unstable if all reaction lines are concurrent (pass through one point) or parallel (no resistance to one direction of translation).
• External determinacy focuses on reactions; internal determinacy focuses on internal forces — both must be checked independently.
• Typical support conditions: pin = 2 reactions (Ax, Ay); roller = 1 reaction (Ay); fixed = 3 reactions (Ax, Ay, MA).

Definitions

Section Title

1. Determinacy and Stability

Common Mistakes

• Confusing DI < 0 (mechanism/unstable) with DI > 0 (indeterminate) — negative DI means the structure is a mechanism, not just under-supported.
• Counting roller supports as contributing 2 reactions instead of 1 — a standard roller provides only 1 reaction perpendicular to the rolling surface.
• Forgetting that a fixed support contributes 3 reactions (Fx, Fy, and moment MA).
• Using the beam formula DI = r − (3 + ec) for frames — always use the general formula DI = (3m + r) − (3n + ec) for frames.
• Assuming determinacy implies stability — concurrent or parallel reactions make a determinate structure geometrically unstable.
• Counting ec = 1 for a hinge connecting 3+ members — the correct count is ec = k − 1 per multi-member hinge.

Formulas

Exam Tips

• For beam problems, always take ΣM = 0 at the pin/roller support farthest from where you want to solve — this gives the reaction at the other support directly.
• Label assumed positive directions for reactions before solving — if an answer is negative, the reaction is opposite to your assumed direction.
• For Gerber beams, redraw each segment as a separate FBD after identifying the hinge forces from the suspended span.
• Memorise: roller on horizontal surface → vertical reaction only; roller on vertical wall → horizontal reaction only; pin → two orthogonal reactions.
• In timed board exam conditions, write ΣM = 0 first (one unknown), then ΣFy = 0, then ΣFx = 0 — this sequential approach minimises simultaneous equation solving.

Key Points

• Apply the three global equilibrium equations to the entire free body to find unknown reactions.
• For compound (Gerber) beams, isolate the suspended span first, solve its reactions, then transfer those forces as loads onto the next supporting span — work from the free end inward.
• Moment equations are most efficient: take moments about the point where the most unknowns intersect, reducing the algebra.
• For beams with internal hinges, the condition ΣM = 0 at the hinge (for the sub-body on one side) gives an additional equation to close the system.
• Distributed loads must be converted to their resultant (magnitude = w × length, acting at the centroid) before applying equilibrium equations.
• Inclined loads must be resolved into horizontal and vertical components before substitution into ΣFx and ΣFy.
• All reactions should be checked by substituting back into all three equilibrium equations — an error in one equation propagates everywhere.

Definitions

Section Title

2. Reactions of Determinate Structures

Common Mistakes

• Taking moments about the wrong point — always take moments about a point where the most unknowns are concurrent (eliminating them from the equation).
• Forgetting to include the moment of a fixed support (MA) when writing ΣM = 0 for the whole beam.
• Solving the main span of a Gerber beam first instead of the suspended span — always work from the free end toward the fixed end.
• Applying UDL resultant at the wrong location — the resultant acts at the centroid of the load diagram, which is the midpoint for a UDL.
• Not resolving inclined loads into components before applying ΣFx and ΣFy.
• Forgetting to check the solution by substituting back into all three equilibrium equations.

Formulas

Exam Tips

• Use area method: ΔV = −(area of load diagram); ΔM = area of SFD. This is far faster than writing equations at every point.
• Sketch the qualitative shape first (constant, linear, parabolic) based on the load type, then compute key values (at reactions, load points, and mid-span).
• At a free end with no applied load: V = 0 and M = 0 — use these as boundary conditions.
• For frame problems, start from the free ends of cantilever members and work toward fixed supports.
• Board exams often ask only for the maximum moment and its location — set V(x) = 0 and compute M at that x.

Key Points

• Internal forces at any section are found by cutting the structure and applying equilibrium to one side of the cut (the simpler side is preferred).
• Sign convention (standard): Positive shear — left portion has upward transverse resultant on cut face; positive moment — sagging (tension at bottom).
• Shear and moment are functions of position x along the member and can be expressed as equations or shown graphically as diagrams (SFD and BMD).
• Key relationships: dV/dx = −w(x) (rate of change of shear equals negative distributed load intensity); dM/dx = V (rate of change of moment equals shear).
• These differential relationships enable rapid diagram sketching: where w = 0, shear is constant; where V = 0, moment has an extremum (maximum or minimum).
• Point loads cause step changes in the SFD; point moments cause step changes in the BMD.
• For frames, draw N, V, and M diagrams for each member separately, maintaining a consistent global sign convention throughout.

Definitions

Section Title

3. Internal Forces — Shear and Moment Diagrams

Common Mistakes

• Inconsistent sign convention — pick one and stick to it throughout the entire problem; mixing conventions yields wrong diagrams.
• Forgetting that the BMD is the integral of the SFD — the area of the SFD between two points equals the change in moment between those points.
• Missing step changes in BMD at points of applied moments (concentrated couples).
• Drawing the BMD on the wrong side — by convention, positive (sagging) moments are plotted below the beam axis (on the tension side) in the Philippines.
• Computing M by summing moments of forces on the right side but using the left-side sign convention — choose one side and apply that side's convention consistently.
• Neglecting axial force N in frame members — frames carry combined N, V, and M.

Formulas

Exam Tips

• For a three-hinged arch, always solve in this order: (1) ΣMA = 0 for VB, (2) ΣFy = 0 for VA, (3) ΣMC_left = 0 for H, (4) check with ΣMC_right = 0.
• Memorise: M_arch = M_beam − H·y. This formula alone answers most arch internal force questions.
• For a symmetric parabolic arch under full UDL: H = wL²/(8h) (analogous to the cable formula; derived from the crown condition).
• Know the difference: two-hinged arch = indeterminate to 1st degree; three-hinged arch = determinate; fixed arch = indeterminate to 3rd degree.
• Board exam shortcut: at the crown of a symmetric arch, dM/dx = 0 by symmetry — maximum arch moment is NOT at the crown under non-symmetric loading.

Key Points

• A three-hinged arch has supports at A and B (hinges providing Ax, Ay, Bx, By — four unknowns) plus a third hinge at the crown C, giving four equations: ΣFx = 0, ΣFy = 0, ΣMA = 0, and ΣMC = 0 (for one side of the arch). The structure is therefore statically determinate.
• The horizontal thrust H (inward at both supports) is the defining feature of arch action. It creates a compression-dominant state that reduces bending moments significantly compared with an equivalent simply supported beam.
• The moment at any section of an arch: M_arch = M_beam − H·y, where M_beam is the moment in an equivalent simply supported beam and y is the arch rise at that section.
• A parabolic arch under a uniformly distributed load carries zero bending moment throughout — it is in pure compression (funicular shape for UDL).
• The crown hinge condition: take either the left or right sub-arch as a free body, apply ΣMC = 0 — the moment at C is zero because it is an internal hinge.
• When supports are at the same level, the horizontal reactions at A and B are equal in magnitude (HA = HB = H) and directed inward.
• When supports are at different levels (skewed arch), the two horizontal reactions are still equal (by ΣFx = 0 if no horizontal loads), but the geometry of the moment arm changes.

Definitions

Section Title

4. Three-Hinged Arches

Common Mistakes

• Forgetting to apply the crown-hinge moment condition (ΣMC = 0 for one side) — without this equation, H cannot be found.
• Including the crown hinge reaction in the sub-arch FBD — at the cut through the crown hinge, internal forces are N (axial) and V (shear) but NOT moment.
• Using the wrong arch height y when computing H — y must be measured at the crown (the hinge location), not at an arbitrary section.
• Assuming both horizontal reactions are inward for all load conditions — with inclined or horizontal loads, check ΣFx = 0 globally first.
• Computing M_arch without subtracting H·y — the arch moment is always less than the beam moment due to thrust action.

Formulas

Exam Tips

• For an L-frame with a fixed base and loads only at the free end: reactions equal the applied loads (Ax = Px, Ay = Py) and MA = sum of moments of loads about A.
• At the knee of an L-frame: M is the same for both the column top and beam end (moment is continuous through a rigid joint).
• When sketching the BMD for an L-frame, the moment is maximum at the fixed base A and zero at the free end C — linear variation along each straight member under point loads.
• Know by heart: for a simply supported portal frame (pins at both column bases) under horizontal load H at the beam level, each column base carries H/2 horizontal reaction.

Key Points

• A determinate frame is analyzed the same way as a beam — FBD of the entire frame, three equilibrium equations for reactions, then internal forces member by member.
• An L-frame (right-angle frame) typically has a vertical column and a horizontal beam, with a fixed or pinned base and loads applied at the free end or along the members.
• Internal forces at any section of a frame member are: N (along member axis), V (perpendicular to axis), and M (moment about the section).
• At the connection (knee) of an L-frame, forces are transferred between members — the shear in the beam becomes the axial force in the column at the joint, and vice versa.
• Draw N, V, and M diagrams for each member separately. Use consistent sign conventions — define tension as positive N, and define positive V and M directions at the start.
• For portal frames under horizontal load, the column carries shear and moment; the beam carries axial force (the horizontal thrust transmitted between columns).

Definitions

Section Title

5. Determinate Frames — L-Frames and Portal Frames

Common Mistakes

• Forgetting to include the fixed-end moment MA when summing moments for the frame FBD.
• Misidentifying the direction of axial force in the column — the vertical reaction Ay at the base equals the axial compression in the column (not the shear).
• Drawing the BMD on the wrong side of the member — convention: plot M on the tension side of each member.
• Not checking equilibrium at the knee joint — forces and moments must balance at every joint in a frame.
• Treating an L-frame with a pin at both base and knee as determinate — a pin base + free end frame has only 2 reactions (pin), which is less than the 3 needed for a general planar system, making it a mechanism unless a roller is added.