CELE Structural Theory & Analysis — Analysis of Determinate StructuresExam Answer Templates

Marks

1

Topic

Determinacy and Stability

Difficulty

easy

Template Id

T1

Examiner Tip

This is a definition question — examiners award the mark ONLY if both elements are present: the qualitative definition AND the quantitative criterion. One without the other earns zero.

Model Answer

A statically determinate structure is one in which all reaction components and internal forces can be found using the three equations of static equilibrium alone (ΣFx = 0, ΣFy = 0, ΣM = 0). For a planar beam: DI = r − (3 + e_c) = 0.

Question Type

very_short_answer

Answer Structure

• Sentence 1: Definition of statically determinate structure (mention equilibrium equations) [0.5 mark]
• Sentence 2: Determinacy condition — DI = r − (3 + e_c) = 0 [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Writing 'solvable by inspection' without mentioning the three equilibrium equations
• Omitting the DI = 0 condition or writing DI < 3 (wrong threshold)
• Not mentioning r, e_c variables

Key Phrases To Include

• three equations of static equilibrium
• DI = r − (3 + e_c) = 0
• reaction components
• equilibrium alone

Marks

2

Topic

Determinacy and Stability

Difficulty

easy

Template Id

T2

Examiner Tip

Always show the substitution step separately. Writing 'DI = 4 − 4 = 0' without showing '3 + e_c = 4' loses the intermediate mark in a 2-mark question.

Model Answer

Given: r = 4, e_c = 1 (one internal hinge). Using the beam determinacy formula: DI = r − (3 + e_c) DI = 4 − (3 + 1) DI = 4 − 4 = 0 ∴ The beam is STATICALLY DETERMINATE.

Question Type

short_answer

Answer Structure

• Line 1: Identify given values — r = 4, e_c = 1 [0.5 mark]
• Line 2: Write formula: DI = r − (3 + e_c) [0.5 mark]
• Line 3: Substitute and compute DI = 0 [0.5 mark]
• Line 4: State conclusion — statically determinate [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Setting e_c = 0 (ignoring the internal hinge) and getting DI = 1 (wrong conclusion)
• Writing the general frame formula DI = (3m + r) − (3n + e_c) instead of the beam shortcut
• Computing correctly but not stating the classification explicitly

Key Phrases To Include

• e_c = 1
• DI = r − (3 + e_c)
• DI = 0
• statically determinate

Marks

2

Topic

Determinacy and Stability

Difficulty

medium

Template Id

T3

Examiner Tip

For frame problems, the choice of formula — beam shortcut vs. general — is itself a tested skill. Using the wrong formula, even with correct arithmetic, shows conceptual error and loses the first mark.

Model Answer

Given: m = 3, n = 4, r = 6, e_c = 0. Using the general frame formula: DI = (3m + r) − (3n + e_c) DI = (3×3 + 6) − (3×4 + 0) DI = (9 + 6) − (12 + 0) DI = 15 − 12 = 3 ∴ The frame is STATICALLY INDETERMINATE to the 3rd degree.

Question Type

short_answer

Answer Structure

• Line 1: List all given parameters — m, n, r, e_c [0.5 mark]
• Line 2: Write general frame formula [0.5 mark]
• Line 3: Substitute numerical values step by step [0.5 mark]
• Line 4: State DI = 3 and conclude '3rd degree indeterminate' [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using beam shortcut formula r − (3 + e_c) for a frame
• Counting nodes incorrectly (including supports vs. not including supports)
• Not stating the degree explicitly — writing 'indeterminate' without 'to the 3rd degree'

Key Phrases To Include

• DI = (3m + r) − (3n + e_c)
• m = 3, n = 4, r = 6
• DI = 3
• indeterminate to the 3rd degree

Marks

3

Topic

Reactions of Determinate Structures

Difficulty

easy

Template Id

T4

Examiner Tip

The verification step (Step 3) is worth its weight in gold — it takes 10 seconds, proves your answer is self-consistent, and shows examiner you understand equilibrium fully. Never skip it.

Model Answer

Sign convention: +↑ positive vertical, + counterclockwise moment positive. FBD: Beam AB, span = 8 m. Supports: pin at A (R_A ↑), roller at B (R_B ↑). Load: 40 kN ↓ at 3 m from A. Step 1 — ΣM_A = 0: R_B(8) − 40(3) = 0 R_B = 120/8 R_B = 15 kN ↑ Step 2 — ΣF_y = 0: R_A + R_B = 40 R_A = 40 − 15 R_A = 25 kN ↑ Step 3 — Check ΣM_B = 0: R_A(8) − 40(8 − 3) = 25(8) − 40(5) = 200 − 200 = 0 ✓ ∴ R_A = 25 kN ↑, R_B = 15 kN ↑

Question Type

numerical

Answer Structure

• State sign convention and sketch FBD [0.5 mark]
• Apply ΣM_A = 0 to solve R_B [1 mark]
• Apply ΣF_y = 0 to solve R_A [1 mark]
• Verify with check equation (ΣM_B = 0 or ΣF_y check) [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• No FBD drawn — automatic deduction of 0.5 to 1 mark
• Moment arm error — using 3 m instead of 5 m for the 40 kN load in ΣM_B
• Missing units on final answer
• No verification check (check equation skipped)

Key Phrases To Include

• ΣM_A = 0
• ΣF_y = 0
• sign convention stated
• R_B = 15 kN
• R_A = 25 kN
• verification check

Marks

2

Topic

Determinacy and Stability

Difficulty

medium

Template Id

T5

Examiner Tip

Determinacy and stability are INDEPENDENT concepts. A structure can be indeterminate and stable (which is the normal case for propped cantilevers). Mixing them up is a classic board exam trap.

Model Answer

Given: r = 4, e_c = 0 (no internal hinges). DI = r − (3 + e_c) = 4 − (3 + 0) = 1 ∴ Indeterminate to the 1st degree. Stability: The reactions are neither all parallel nor all concurrent, so the beam is STABLE. Conclusion: Stable and statically indeterminate to the 1st degree.

Question Type

short_answer

Answer Structure

• State r = 4, e_c = 0 and apply formula [0.5 mark]
• Compute DI = 1 and state '1st degree indeterminate' [0.5 mark]
• Address stability separately — reactions not parallel or concurrent [1 mark]

Scoring Breakdown

Common Mark Deductions

• Confusing determinacy with stability — writing 'DI > 0 therefore unstable' (wrong!)
• Not addressing stability at all — the question explicitly asks for it
• Omitting the e_c = 0 identification

Key Phrases To Include

• DI = 1
• 1st degree indeterminate
• not parallel
• not concurrent
• stable

Marks

1

Topic

Determinacy and Stability

Difficulty

easy

Template Id

T6

Examiner Tip

The key technical term is 'moment release' — an internal hinge is a moment release, meaning M = 0 there. Write that exact phrase and you will earn the mark.

Model Answer

An equation of condition (e_c) is an additional equilibrium condition provided by a structural release such as an internal hinge. At an internal hinge, the bending moment is zero; taking ΣM = 0 for the portion on either side of the hinge gives one independent equation, increasing the number of equilibrium equations available to solve the structure.

Question Type

very_short_answer

Answer Structure

• Sentence 1: Define equation of condition — additional equilibrium condition from structural release [0.5 mark]
• Sentence 2: Explain hinge contribution — M = 0 at hinge, ΣM = 0 on one side gives extra equation [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Saying 'shear is zero at a hinge' — wrong; it is moment that is zero
• Vague answer like 'hinge adds freedom' without stating M = 0 and the extra equation

Key Phrases To Include

• equation of condition
• internal hinge
• bending moment is zero
• ΣM = 0 on one side
• additional equilibrium equation

Marks

5

Topic

Three-Hinged Arches

Difficulty

medium

Template Id

T7

Examiner Tip

The crown condition is the heart of every three-hinged arch problem. Write it clearly as 'ΣM_C = 0 for left portion [or right portion]' and show the free body of ONLY that portion. This is what distinguishes students who score 5/5 from those who score 2/5.

Model Answer

Given: Span L = 20 m, rise h = 5 m, crown at midspan (x_C = 10 m), P = 60 kN at x = 5 m from A. Sign convention: +↑, +→, + counterclockwise. ── PART (a): Vertical Reactions ── Global ΣM_A = 0: V_B(20) − 60(5) = 0 V_B = 300/20 = 15 kN ↑ Global ΣF_y = 0: V_A + V_B = 60 V_A = 60 − 15 = 45 kN ↑ ── PART (b): Horizontal Thrust H ── Crown condition — isolate left portion AC (x = 0 to 10 m): At C (x = 10, y = 5): ΣM_C (left portion) = 0: V_A(10) − H(5) − 60(10 − 5) = 0 45(10) − 5H − 60(5) = 0 450 − 5H − 300 = 0 5H = 150 H = 30 kN Global ΣF_x = 0: H_A = H_B = 30 kN (horizontal thrusts pointing inward at both supports) ── SUMMARY ── V_A = 45 kN ↑ V_B = 15 kN ↑ H = 30 kN (horizontal thrust at each support) [Verification: ΣM_B = 0: 60(15) − V_A(20) + H(0) = 900 − 900 = 0 ✓]

Question Type

numerical

Answer Structure

• Sketch arch with labels: A, B, C, span 20 m, rise 5 m, load position [0.5 mark]
• Apply ΣM_A = 0 globally to find V_B = 15 kN [1 mark]
• Apply ΣF_y = 0 to find V_A = 45 kN [0.5 mark]
• State crown condition: ΣM_C = 0 for left portion, identify moment arm of H as rise = 5 m [1 mark]
• Substitute and solve H = 30 kN [1 mark]
• State global ΣF_x = 0 confirming H_A = H_B = 30 kN, verification check [1 mark]

Scoring Breakdown

Common Mark Deductions

• Applying crown condition to the entire arch (ΣM_C = 0 globally) — this is trivially satisfied and gives H = 0
• Using the wrong moment arm for H — using span/2 = 10 m instead of rise = 5 m
• Not isolating the correct portion (must take only the left or only the right portion at crown)
• Forgetting that 60 kN load is within the left portion — its moment arm from C is (10 − 5) = 5 m, not 5 m from B
• No arch sketch — loses the diagram mark

Key Phrases To Include

• three-hinged arch
• crown condition
• ΣM_C = 0 for left portion
• horizontal thrust H
• rise h = 5 m
• V_A = 45 kN
• V_B = 15 kN
• H = 30 kN

Marks

5

Topic

Internal Forces in Frames

Difficulty

medium

Template Id

T8

Examiner Tip

For frame internal forces, always state which direction is the member axis. For a vertical column: axial is parallel to the vertical axis (so vertical loads cause N), and shear is perpendicular (so horizontal loads cause V). Getting this right is the difference between 3/5 and 5/5.

Model Answer

Sign convention: +↑, +→, + counterclockwise. ── REACTIONS AT FIXED SUPPORT A ── ΣF_x = 0: A_x − 10 = 0 → A_x = 10 kN (←) ΣF_y = 0: A_y − 15 = 0 → A_y = 15 kN (↑) ΣM_A = 0: M_A − 10(4) − 15(3) = 0 M_A = 40 + 45 = 85 kN·m (counterclockwise) ── INTERNAL FORCES AT BASE (SECTION AT A) ── Cut at base of column and analyze the free body ABOVE the cut (the entire L-frame minus the support): Axial force N (along column axis = vertical): N = 15 kN (compression, because 15 kN load acts downward) Shear force V (transverse to column = horizontal): V = 10 kN (due to the 10 kN horizontal load at C) Bending moment M (at base, about cut section): M = 10(4) + 15(3) = 40 + 45 = 85 kN·m ── SUMMARY ── Reactions: A_x = 10 kN ←, A_y = 15 kN ↑, M_A = 85 kN·m CCW At base of column: N = 15 kN (compression), V = 10 kN, M = 85 kN·m

Question Type

numerical

Answer Structure

• Sketch the L-frame with all loads and support reactions labeled [0.5 mark]
• Apply ΣF_x = 0 → A_x = 10 kN [0.5 mark]
• Apply ΣF_y = 0 → A_y = 15 kN [0.5 mark]
• Apply ΣM_A = 0 → M_A = 85 kN·m [1 mark]
• Identify N = 15 kN (compression along column) with reasoning [0.5 mark]
• Identify V = 10 kN (horizontal shear at base) [0.5 mark]
• Identify M = 85 kN·m at base with moment arm computation [0.5 mark]
• State all results with correct units and sense [1 mark]

Scoring Breakdown

Common Mark Deductions

• Not showing the fixed-end moment M_A in the FBD
• Confusing axial and shear — for a vertical column, axial is VERTICAL and shear is HORIZONTAL
• Moment arm error — using only one load's moment contribution, missing the other
• Forgetting to indicate sense of internal forces (tension vs. compression for N)

Key Phrases To Include

• fixed support
• ΣM_A = 0
• M_A = 85 kN·m
• axial N = 15 kN compression
• shear V = 10 kN
• moment M = 85 kN·m
• moment arm

Marks

3

Topic

Internal Forces — SFD and BMD

Difficulty

easy

Template Id

T9

Examiner Tip

For UDL problems, always write V(x) as a function. This shows the examiner you understand integration of loading — a UDL (zero-th degree load) produces a linear SFD (first-degree) and a parabolic BMD (second-degree). State this explicitly for bonus impression marks.

Model Answer

Given: Span L = 6 m, w = 10 kN/m (UDL over full span). Step 1 — Reactions: By symmetry: R_A = R_B = wL/2 = 10(6)/2 = 30 kN ↑ Step 2 — Shear function V(x) from A: V(x) = R_A − wx = 30 − 10x (0 ≤ x ≤ 6 m) Step 3 — Key values: At x = 0 (just right of A): V = +30 kN At x = 3 m (midspan): V = 30 − 10(3) = 0 ← zero crossing (location of max moment) At x = 6 m (just left of B): V = 30 − 10(6) = −30 kN Step 4 — SFD Description: The SFD is a STRAIGHT LINE from +30 kN at A, crossing zero at midspan (x = 3 m), to −30 kN at B. The diagram is linear because the load is uniformly distributed. [SKETCH: Horizontal baseline; upward block at A = +30 kN; straight line sloping down to −30 kN at B; zero crossing at center marked; shaded area above baseline on left, below on right]

Question Type

diagram_based

Answer Structure

• Compute R_A = R_B = 30 kN using symmetry or equilibrium [0.5 mark]
• Write shear function V(x) = 30 − 10x [0.5 mark]
• Compute key SFD values: +30 kN at A, 0 at midspan, −30 kN at B [1 mark]
• Draw/describe correct SFD shape — linear, noting zero crossing at center [1 mark]

Scoring Breakdown

Common Mark Deductions

• Drawing a curved SFD for a UDL — it must be linear (first-degree polynomial in x)
• Omitting the zero crossing location from the SFD
• Not labeling numerical values on the SFD at key points
• Missing units on the SFD ordinates

Key Phrases To Include

• V(x) = R_A − wx
• V = +30 kN at A
• V = 0 at midspan
• V = −30 kN at B
• linear SFD
• zero crossing at x = 3 m

Marks

3

Topic

Three-Hinged Arches

Difficulty

medium

Template Id

T10

Examiner Tip

Board examiners particularly reward the formula M_arch = M_beam − H·y because it quantitatively links arch geometry (rise y) to moment reduction. Write it clearly and define each term.

Model Answer

In a three-hinged arch, the supports develop a horizontal inward thrust H in addition to vertical reactions. This thrust acts as an internal couple that reduces the net bending moment at any section of the arch. For a simply supported beam under the same loading, the bending moment at a section at horizontal distance x from the left support is: M_beam = V_A(x) − (loading moment terms) For the arch at the same horizontal distance x, the arch has a height y above the baseline; the bending moment is: M_arch = M_beam − H·y The term (H × y) is the moment reduction due to thrust. Because y is positive for all interior points of the arch, M_arch < M_beam throughout the span. This is why arches require less material than beams for the same span and load — the arch form converts bending into primarily compressive axial forces, which concrete and masonry resist efficiently. This structural principle is the basis for the efficiency of arch bridges in Philippine infrastructure (e.g., Guadalupe Bridge, historic stone arches).

Question Type

short_answer

Answer Structure

• State that H is the horizontal inward thrust at supports [0.5 mark]
• Write M_arch = M_beam − H·y and identify the reduction term [1 mark]
• Explain that y > 0 at interior points → M_arch < M_beam always [1 mark]
• State physical implication: arch converts bending to compression, less material needed [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Stating 'H reduces shear' instead of 'H reduces moment'
• Not writing the formula M_arch = M_beam − H·y — just a verbal description earns maximum 1/3
• Not explaining WHY H·y reduces moment (because y > 0 at interior points)

Key Phrases To Include

• horizontal thrust H
• M_arch = M_beam − H·y
• H × y reduction
• compressive axial forces
• structural efficiency
• arch rise y

Marks

5

Topic

Compound (Gerber) Beams

Difficulty

hard

Template Id

T11

Examiner Tip

The rule for Gerber beams is absolute: ALWAYS start with the SUSPENDED span. It is the only span that is fully free to analyze. Then use its reactions as imposed loads on the next span. This order is a non-negotiable technique in board exams.

Model Answer

Sign convention: +↑, + counterclockwise. ── STEP 1: Analyze suspended span CD first ── FBD of span CD (3 m): supported at C (internal hinge reaction ↑) and D (roller ↑). Load: 24 kN at midpoint of CD (1.5 m from C). ΣM_C = 0 (for CD): R_D(3) − 24(1.5) = 0 R_D = 36/3 = 12 kN ↑ ΣF_y = 0 (for CD): R_C' + R_D = 24 R_C' = 24 − 12 = 12 kN ↑ [reaction from main beam onto CD at C] ∴ Force on main beam at C (by Newton's 3rd law) = 12 kN ↓ ── STEP 2: Analyze main span AC ── FBD of AC (6 m): pin at A (R_A ↑), roller at C_main (R_Cmain ↑). External load at C from suspended span = 12 kN ↓ (transferred from CD at 6 m from A = at point C). ΣM_A = 0 (for AC): R_Cmain(6) − 12(6) = 0 R_Cmain = 12 kN ↑ ΣF_y = 0 (for AC): R_A + R_Cmain = 12 R_A = 12 − 12 = 0 kN Note: R_Cmain is the reaction at the roller directly under C on the main span. The total upward reaction at hinge C = R_Cmain = 12 kN. ── SUMMARY ── R_A = 0 kN R_C = 12 kN ↑ (roller under main beam at C) R_D = 12 kN ↑ [Verification: Total vertical reactions = 0 + 12 + 12 = 24 kN = total applied load ✓]

Question Type

numerical

Answer Structure

• Draw FBD of suspended span CD with labels [0.5 mark]
• Apply ΣM_C = 0 for CD → R_D = 12 kN [1 mark]
• Apply ΣF_y = 0 for CD → reaction at C on main beam = 12 kN ↓ [1 mark]
• Draw FBD of main span AC with transferred load 12 kN ↓ at C [0.5 mark]
• Apply ΣM_A = 0 for AC → R_Cmain = 12 kN [1 mark]
• Apply ΣF_y = 0 for AC → R_A = 0 kN with global verification [1 mark]

Scoring Breakdown

Common Mark Deductions

• Analyzing the main span first — wrong order, leads to wrong transfer force
• Forgetting to reverse the direction of the hinge reaction when transferring to the main beam (Newton's 3rd law error)
• Treating the entire compound beam as one FBD without separating at the hinge

Key Phrases To Include

• analyze suspended span first
• internal hinge at C
• Newton's 3rd law transfer
• 12 kN ↓ at C on main beam
• R_D = 12 kN
• R_A = 0 kN
• global verification

Marks

3

Topic

Determinacy and Stability

Difficulty

medium

Template Id

T12

Examiner Tip

Board exams frequently test the distinction between DI < 0 (arithmetic instability) and geometric instability (arrangement). A structure can have DI = 0 and still be geometrically unstable. Make this point explicitly in your answer.

Model Answer

External instability occurs when the support reactions are insufficient or improperly arranged to prevent rigid-body motion of the entire structure, even though internal members may be adequate. Example: A beam with three parallel roller supports (no horizontal reaction available) is externally unstable — any horizontal load will cause sliding. Internal instability (or geometric instability) occurs when the internal members or joints of the structure cannot maintain its shape under load, even if external support conditions are adequate. Example: A four-bar frame without diagonal bracing will collapse into a mechanism (sway) under lateral load — a single diagonal member would stabilize it. Key distinction: External instability is checked by examining support reaction arrangement; internal instability is checked by examining the arrangement of members and joints.

Question Type

short_answer

Answer Structure

• Define external instability with example (parallel/concurrent supports) [1 mark]
• Define internal instability with example (mechanism, missing bracing) [1 mark]
• State the key distinction — how to check each type [1 mark]

Scoring Breakdown

Common Mark Deductions

• Confusing instability with indeterminacy (DI < 0 vs. arrangement problem)
• Giving only one type without the other
• Example not clearly linked to the type of instability described

Key Phrases To Include

• external instability
• parallel reactions
• concurrent reactions
• rigid-body motion
• internal instability
• mechanism
• member arrangement

Marks

3

Topic

Internal Forces in Beams

Difficulty

medium

Template Id

T13

Examiner Tip

Always compute ΣM about the CUT POINT (not about A or B) when finding the internal moment. This eliminates V from the moment equation entirely, making M solvable in one step.

Model Answer

Given: R_A = 16 kN ↑, 20 kN load at x = 1 m from A. Section cut at x = 2 m. FBD of left portion (0 to 2 m from A): Forces acting on left portion: R_A = 16 kN ↑ at x = 0 20 kN ↓ at x = 1 m Internal shear V and moment M act at the cut face (at x = 2 m) Sign convention for internal forces: V positive when left portion tends to move upward relative to right (+↑ on left face). M positive (sagging) when it causes tension on the bottom fiber. ΣF_y = 0 (left portion): R_A − 20 − V = 0 16 − 20 − V = 0 V = −4 kN [negative = downward shear on left face] ΣM_cut = 0 (take moments about cut at x = 2 m): R_A(2) − 20(2 − 1) − M = 0 16(2) − 20(1) − M = 0 32 − 20 = M M = +12 kN·m [positive = sagging] ∴ V = −4 kN (net downward on left face), M = +12 kN·m (sagging)

Question Type

numerical

Answer Structure

• Draw FBD of left portion with all forces labeled at correct positions [0.5 mark]
• Apply ΣF_y = 0 on left portion → V = −4 kN [1 mark]
• Apply ΣM_cut = 0 on left portion about the section → M = 12 kN·m [1 mark]
• State sense (sign/direction) of V and M explicitly [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Using right portion without adjusting — causes sign errors because R_B is needed
• Incorrect moment arm — using x = 2 m for the 20 kN load instead of (2 − 1) = 1 m
• Not stating sign convention before computing

Key Phrases To Include

• FBD of left portion
• section cut at x = 2 m
• ΣF_y = 0 on left portion
• ΣM_cut = 0
• V = −4 kN
• M = +12 kN·m
• moment arm

Marks

1

Topic

Three-Hinged Arches

Difficulty

easy

Template Id

T14

Examiner Tip

The phrase 'applied to one isolated portion' is non-negotiable. The global crown moment ΣM_C is always zero for any structure in equilibrium — it provides no information. It is the PARTIAL STRUCTURE moment that creates the useful fourth equation.

Model Answer

The fourth condition is the crown (hinge) condition: ΣM_C = 0 applied to one isolated portion of the arch (either left or right of crown hinge C). This equation is valid because at an internal hinge the bending moment is zero by definition (a hinge cannot transmit moment); therefore, the sum of moments about C of all forces on either side must equal zero.

Question Type

very_short_answer

Answer Structure

• Name the equation: ΣM_C = 0 for one isolated portion (crown condition) [0.5 mark]
• State why it is valid: internal hinge → M = 0 at C → extra equation [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Writing ΣM_C = 0 for the ENTIRE arch — this is trivially satisfied and does not provide extra information
• Saying 'hinge means shear is zero' — wrong; moment is zero at a hinge, not shear

Key Phrases To Include

• ΣM_C = 0
• one isolated portion
• internal hinge
• bending moment is zero at hinge
• crown condition

Marks

3

Topic

Reactions — Distributed Loads

Difficulty

medium

Template Id

T15

Examiner Tip

The centroid of a triangle is always at 1/3 of the base from the heavy side, or equivalently 2/3 of the base from the zero side. For a triangular load from 0 at A to w_max at B: centroid is at (2/3)L from A. Memorize this — it appears in almost every PRC exam cycle.

Model Answer

Given: Span L = 10 m. Load: triangular, 0 at A, 30 kN/m at B. Step 1 — Total load resultant W: W = (1/2)(base)(height) = (1/2)(10)(30) = 150 kN Acts at 2/3 of span from A = (2/3)(10) = 6.667 m from A [centroid of triangle is at 2/3 from zero end] Step 2 — ΣM_A = 0: R_B(10) − 150(6.667) = 0 R_B = 1000/10 = 100 kN ↑ Step 3 — ΣF_y = 0: R_A + R_B = 150 R_A = 150 − 100 = 50 kN ↑ Step 4 — Verification ΣM_B = 0: R_A(10) − 150(10 − 6.667) = 50(10) − 150(3.333) = 500 − 500 = 0 ✓ ∴ R_A = 50 kN ↑, R_B = 100 kN ↑

Question Type

numerical

Answer Structure

• Compute total resultant W = 150 kN and locate its centroid at 2/3 L from A [1 mark]
• Apply ΣM_A = 0 → R_B = 100 kN [1 mark]
• Apply ΣF_y = 0 → R_A = 50 kN with verification check [1 mark]

Scoring Breakdown

Common Mark Deductions

• Placing the centroid at 1/3 L from A instead of 2/3 L (confusing which end has the zero load)
• Placing the centroid at L/2 (midspan) — that is for uniform, not triangular load
• Not computing the total resultant W first — integrating directly is error-prone without showing W
• R_A = R_B = 75 kN (assuming symmetry, which is wrong for a triangular load)

Key Phrases To Include

• W = (1/2)(L)(w_max) = 150 kN
• centroid at 2/3 L from zero end
• 6.667 m from A
• R_B = 100 kN
• R_A = 50 kN
• verification check