CELE Engineering Mechanics — Analysis of TrussesExam Answer Templates

Marks

1

Topic

Truss Determinacy

Difficulty

easy

Template Id

T1

Examiner Tip

Examiners award the mark for the equation alone if it is clearly written; variable definitions are a bonus but always include them to avoid ambiguity.

Model Answer

A plane truss is statically determinate when m + r = 2j, where m is the number of members, r is the number of reaction components, and j is the number of joints.

Question Type

very_short_answer

Answer Structure

• Line 1: Write the determinacy equation with all variables defined [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing m + r > 2j (which is the indeterminate condition, not determinate)
• Stating the equation without defining any variable
• Confusing j (joints) with m (members)

Key Phrases To Include

• m + r = 2j
• members
• reactions
• joints
• statically determinate

Marks

2

Topic

Truss Assumptions

Difficulty

easy

Template Id

T2

Examiner Tip

The link between the definition and the assumptions is what earns the second mark. A definition alone is worth only one mark regardless of how well it is written.

Model Answer

A two-force member is a structural member loaded by forces at only two points (its two end joints), with no intermediate loads or moments. Because the resultant at each end must be equal, opposite, and collinear, the force is directed along the member — producing pure tension or pure compression with no bending. In an ideal truss, all loads are applied only at the joints, members are connected by frictionless pins (which transmit no moment), and members are straight. These three assumptions ensure that no transverse loads or couples act on any member, satisfying the definition of a two-force member exactly.

Question Type

short_answer

Answer Structure

• Sentence 1: Define two-force member — loaded at two points only, force directed along the member [1 mark]
• Sentence 2: Link the three truss assumptions (pin joints, loads at joints, straight members) to why every truss member qualifies [1 mark]

Scoring Breakdown

Common Mark Deductions

• Defining two-force member correctly but failing to connect it to the truss assumptions
• Mentioning only one truss assumption instead of all three
• Saying 'no bending' without explaining why (the pin connection)

Key Phrases To Include

• two-force member
• frictionless pin
• loads at joints only
• pure tension or compression
• directed along the member

Marks

2

Topic

Zero-Force Members

Difficulty

easy

Template Id

T3

Examiner Tip

Always state the rule before applying it. Examiners follow a two-step marking: rule identification + correct application. Skipping the rule costs you one mark even if the answer is correct.

Model Answer

Applying the zero-force member rule for a three-member joint: Rule: At a joint with three members where two members are collinear and no external load is applied, the third (non-collinear) member carries zero force. At joint K: members KL and KM are collinear (bottom chord), and no external load acts at K. Therefore, the vertical web member KN is a zero-force member (F_KN = 0). Members KL and KM are not zero-force; they transmit the chord force through the joint.

Question Type

short_answer

Answer Structure

• Line 1: State the applicable zero-force member rule [1 mark]
• Line 2: Apply the rule to the given joint and identify KN as zero-force [1 mark]

Scoring Breakdown

Common Mark Deductions

• Declaring all three members as zero-force (a common but serious error)
• Stating the answer without citing the rule
• Confusing which member is non-collinear

Key Phrases To Include

• zero-force member
• collinear
• no external load
• non-collinear member
• F = 0

Marks

2

Topic

Truss Determinacy

Difficulty

easy

Template Id

T4

Examiner Tip

The phrase 'assuming proper arrangement of members' or 'necessary but not sufficient' signals to the examiner that you understand the limitation of the equation — this phrase can earn you a partial bonus mark.

Model Answer

Apply the determinacy criterion: m + r vs. 2j 21 + 3 vs. 2(12) 24 vs. 24 Since m + r = 2j = 24, the truss is statically determinate (assuming proper arrangement of members). Note: The equation is a necessary but not sufficient condition. Geometric stability must also be verified separately.

Question Type

numerical

Answer Structure

• Line 1: Write the criterion equation [0.5 mark]
• Line 2: Substitute and compute both sides [0.5 mark]
• Line 3: Compare and state the classification [1 mark]

Scoring Breakdown

Common Mark Deductions

• Omitting the caveat about geometric stability (partial deduction)
• Arithmetic error in computing 2j
• Stating 'stable' instead of 'statically determinate' — these are different concepts

Key Phrases To Include

• m + r = 2j
• 24 = 24
• statically determinate
• necessary but not sufficient
• geometric stability

Marks

5

Topic

Method of Joints

Difficulty

medium

Template Id

T5

Examiner Tip

A full 5-mark truss solution should fill approximately one-half to one full exam page. Show every equilibrium equation explicitly — examiners award partial marks for correct equation setup even if arithmetic is wrong.

Model Answer

GIVEN: Pin at A(0,0), roller at B(8,0), load P = 30 kN ↓ at C(4,3). STEP 1 — DETERMINACY CHECK: m = 3, r = 3, j = 3 → m + r = 6 = 2(3) ✓ Statically determinate. STEP 2 — SUPPORT REACTIONS (whole truss): ΣM_A = 0: R_B(8) − 30(4) = 0 → R_B = 15 kN ↑ ΣF_y = 0: R_A_y + 15 − 30 = 0 → R_A_y = 15 kN ↑ ΣF_x = 0: R_A_x = 0 VERIFICATION: ΣF_y = 15 + 15 − 30 = 0 ✓ STEP 3 — MEMBER GEOMETRY: AC: Δx = 4, Δy = 3 → L_AC = 5 m; cos θ_AC = 4/5 = 0.800, sin θ_AC = 3/5 = 0.600 BC: Δx = −4, Δy = 3 → L_BC = 5 m; cos θ_BC = 4/5 = 0.800, sin θ_BC = 3/5 = 0.600 AB: horizontal, L = 8 m CONVENTION: Assume all unknown member forces in tension (+). Negative result → compression. STEP 4 — METHOD OF JOINTS: Joint A (unknowns: F_AC, F_AB; known: R_Ay = 15 kN ↑, R_Ax = 0): ΣF_y = 0: F_AC sin θ_AC + 15 = 0 F_AC (0.600) + 15 = 0 F_AC = −25.0 kN → COMPRESSION (25.0 kN C) ΣF_x = 0: F_AC cos θ_AC + F_AB = 0 (−25.0)(0.800) + F_AB = 0 F_AB = 20.0 kN → TENSION (20.0 kN T) Joint B (unknowns: F_BC; known: R_B = 15 kN ↑, F_AB = 20.0 kN T): ΣF_y = 0: F_BC sin θ_BC + 15 = 0 F_BC (0.600) + 15 = 0 F_BC = −25.0 kN → COMPRESSION (25.0 kN C) VERIFICATION at Joint C (ΣF_y): (−25.0)(0.600) + (−25.0)(0.600) − 30 = −15 − 15 − 30 ≠ 0 CORRECTION — AC and BC slope upward toward C, so at Joint C: ΣF_y = F_AC_y(pointing down at C) + F_BC_y(pointing down at C) − 30 = 0 Both members are in compression → they push outward on C. F_AC sin θ + F_BC sin θ = 30 → (25)(0.6)+(25)(0.6) = 15+15 = 30 ✓ FINAL ANSWERS: F_AC = 25.0 kN (Compression) F_BC = 25.0 kN (Compression) F_AB = 20.0 kN (Tension)

Question Type

numerical

Answer Structure

• Step 1: Determinacy check — m + r = 2j [0.5 mark]
• Step 2: Support reactions using global ΣM and ΣF with verification [1 mark]
• Step 3: Member geometry — compute angles/trig ratios for each member [0.5 mark]
• Step 4: Joint A equilibrium — solve F_AC and F_AB [1.5 marks]
• Step 5: Joint B equilibrium — solve F_BC [1 mark]
• Step 6: Verification at Joint C and state T/C for all members [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Skipping the determinacy check at the start (−0.5 mark)
• Not verifying support reactions with a global check
• Omitting T (Tension) or C (Compression) designation on final answers (−1 mark)
• Using degrees for angle but computing trig ratios incorrectly due to wrong reference direction
• Starting at a joint with more than two unknowns

Key Phrases To Include

• assume tension
• ΣF_x = 0
• ΣF_y = 0
• negative result → compression
• Tension
• Compression
• kN
• determinacy check

Marks

3

Topic

Truss Assumptions

Difficulty

easy

Template Id

T6

Examiner Tip

Each assumption is worth 1 mark, but the mark is split: 0.5 for the statement and 0.5 for the implication. Always explain what each assumption means for the member forces.

Model Answer

The three fundamental assumptions for an ideal plane truss are: 1. Frictionless pin connections: All joints are idealized as frictionless pins that transmit force but not moment. This ensures that no bending moment is induced at the ends of any member. 2. Loads applied at joints only: All external loads (including support reactions) act exclusively at the joint nodes. No load is applied along the length of any member, eliminating transverse shear and bending in the members. 3. Straight, weightless members: All members are perfectly straight. Self-weight is either neglected or, if considered, is distributed equally to the two end joints of each member. This ensures the internal force in each member is purely axial.

Question Type

short_answer

Answer Structure

• Point 1: Frictionless pins — transmit force, not moment [1 mark]
• Point 2: Loads at joints only — no intermediate loading [1 mark]
• Point 3: Straight, weightless members — pure axial force [1 mark]

Scoring Breakdown

Common Mark Deductions

• Listing assumptions without explaining their structural implication (earns only 0.5 per point)
• Writing 'rigid connections' instead of 'pin connections' — this is the opposite assumption
• Omitting the consequence of self-weight (split to end joints)

Key Phrases To Include

• frictionless pin
• no moment
• loads at joints
• straight members
• pure axial
• tension or compression

Marks

3

Topic

Method of Sections

Difficulty

medium

Template Id

T7

Examiner Tip

The examiner is testing whether you know WHY the three-member limit exists (only three equilibrium equations in 2-D). Stating this reason earns the second mark more reliably than just saying 'at most three.'

Model Answer

The Method of Sections is preferred when only the forces in a few specific members deep inside the truss are required, making it inefficient to analyze all preceding joints one by one with the Method of Joints. Procedure advantage: By passing a cutting plane through the truss and applying equilibrium to one portion as a rigid body (ΣFx = 0, ΣFy = 0, ΣM = 0), the forces in the cut members are found directly without analyzing every joint from the support. Key restriction: The section cut must pass through at most three members with unknown forces. This is because a free body in 2-D has only three independent equilibrium equations; cutting more than three unknowns yields an unsolvable system (unless additional conditions exist). Moment trick: Taking moments about the intersection point of two cut members eliminates those two unknowns, allowing direct solution of the third — a powerful technique for the PRC board exam.

Question Type

short_answer

Answer Structure

• Sentence 1–2: When to prefer Sections over Joints [1 mark]
• Sentence 3: Key restriction — at most 3 unknown members cut [1 mark]
• Sentence 4: The moment trick for direct solution [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating 'cut the truss in half' without specifying the three-member restriction
• Not mentioning the moment trick at all
• Confusing Method of Sections with Method of Joints in the explanation

Key Phrases To Include

• specific interior members
• at most three unknowns
• rigid body equilibrium
• ΣM about intersection
• three equilibrium equations

Marks

5

Topic

Method of Sections

Difficulty

hard

Template Id

T8

Examiner Tip

The method of sections is entirely about the smart choice of moment center. Explicitly state 'Taking ΣM about C(8,0) to eliminate F_EC and F_BC' — this one sentence demonstrates your analytical reasoning and typically earns a dedicated half-mark.

Model Answer

GIVEN: Span = 12 m, panel width = 4 m, truss height = 3 m. Loads: 24 kN ↓ at B(4,0) and 24 kN ↓ at C(8,0). Supports: Pin at A(0,0), Roller at D(12,0). STEP 1 — DETERMINACY: m = 9, r = 3, j = 6 → 9 + 3 = 12 = 2(6) ✓ Statically determinate. STEP 2 — SUPPORT REACTIONS: ΣM_A = 0: R_D(12) − 24(4) − 24(8) = 0 12 R_D = 96 + 192 = 288 R_D = 24 kN ↑ ΣF_y = 0: R_Ay + 24 − 24 − 24 = 0 → R_Ay = 24 kN ↑ ΣF_x = 0: R_Ax = 0 VERIFICATION: ΣF_y = 24 + 24 − 24 − 24 = 0 ✓ STEP 3 — SECTION CUT: Pass a vertical cutting plane between panels BE and CF, cutting through: • Top chord EF (horizontal, at y = 3 m) • Diagonal EC (from E(4,3) to C(8,0)) • Bottom chord BC (horizontal, at y = 0 m) Consider the LEFT portion (A–B–E side) with: • R_Ay = 24 kN ↑ at A • 24 kN ↓ at B • Unknown F_EF (horizontal, at y = 3) • Unknown F_EC (diagonal) • Unknown F_BC (horizontal, at y = 0) STEP 4 — SOLVE FOR F_EF: To eliminate F_EC and F_BC, take ΣM about their intersection point. F_BC acts along y = 0; F_EC, if extended, passes through C(8,0). Both forces pass through point C(8,0). ΣM_C = 0 (about C, taking counterclockwise positive): R_Ay(8) − 24(8−4) + F_EF(3) = 0 24(8) − 24(4) + 3 F_EF = 0 192 − 96 + 3 F_EF = 0 3 F_EF = −96 F_EF = −32 kN F_EF = 32 kN COMPRESSION (The negative sign confirms compression — expected for the top chord of a simply supported truss under gravity loading.) FINAL ANSWER: F_EF = 32 kN (Compression)

Question Type

numerical

Answer Structure

• Step 1: Determinacy check [0.5 mark]
• Step 2: Support reactions with ΣM_A and ΣF_y, verified [1 mark]
• Step 3: Identify and describe the section cut and the three members cut [1 mark]
• Step 4: Identify the moment center C and justify its choice [0.5 mark]
• Step 5: ΣM_C equation set up and solved for F_EF [1.5 marks]
• Step 6: Final answer with correct value, unit, and T/C designation [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Taking moments about the wrong point (not the intersection of the other two cut forces) — this gives an equation with two unknowns instead of one
• Including the roller reaction at D in the left-portion FBD instead of R_Ay
• Forgetting to include the 24 kN load at B in the moment equation for the left portion
• Incorrect moment arm — using 4 m instead of 8 m for R_Ay about point C

Key Phrases To Include

• section cut
• ΣM about C
• eliminate unknowns
• left portion
• F_EF = 32 kN
• Compression
• top chord compression

Marks

2

Topic

Truss Assumptions

Difficulty

easy

Template Id

T9

Examiner Tip

This is a conceptual 2-mark question. Examiners allocate one mark per concept. A sketch of a joint with arrows labeled 'T' and 'C' can substitute for words and often earns both marks faster.

Model Answer

Tension vs. Compression in truss members: A member in TENSION is being pulled apart — it pulls its two end joints toward the member (the member force arrow points away from the joint on an FBD). A member in COMPRESSION is being squeezed — it pushes its two end joints away from the member (the member force arrow points toward the joint on an FBD). Sign convention for the Method of Joints: Assume all unknown member forces are in TENSION (arrows pointing away from the joint). After solving: • Positive result → member is in Tension (assumption was correct) • Negative result → member is in Compression (force acts opposite to assumed direction)

Question Type

short_answer

Answer Structure

• Lines 1–2: Define tension (pulling joint) and compression (pushing joint) with FBD arrow direction [1 mark]
• Lines 3–4: State the 'assume tension' convention and interpretation of positive/negative results [1 mark]

Scoring Breakdown

Common Mark Deductions

• Reversing the arrow direction for tension vs compression
• Not explaining what a negative result means
• Stating 'assume compression' as the starting convention (non-standard)

Key Phrases To Include

• tension pulls
• compression pushes
• assume tension
• positive = tension
• negative = compression
• FBD arrow direction

Marks

5

Topic

Method of Joints

Difficulty

hard

Template Id

T10

Examiner Tip

Invoking symmetry for a symmetric truss under symmetric loading is not a shortcut — it is correct engineering. State 'By symmetry, F_CE = F_AD and F_BC = F_AB' explicitly. This earns full marks and saves 5 minutes.

Model Answer

GIVEN: Equilateral triangles, side = 3 m → all angles = 60°, height = 3 sin60° = 2.598 ≈ 2.6 m. Loads: 18 kN ↓ at D(1.5,2.6) and 18 kN ↓ at E(4.5,2.6). Supports: Pin A(0,0), Roller C(6,0). Member geometry (cos60° = 0.5, sin60° = 0.866): AD: from A(0,0) to D(1.5,2.6) → rises right at 60° BD: from B(3,0) to D(1.5,2.6) → rises left at 60° BE: from B(3,0) to E(4.5,2.6) → rises right at 60° CE: from C(6,0) to E(4.5,2.6) → rises left at 60° DE: horizontal top chord AB, BC: horizontal bottom chord STEP 1 — DETERMINACY: m = 7, r = 3, j = 5 → 7 + 3 = 10 = 2(5) ✓ STEP 2 — REACTIONS (by symmetry, symmetric loading): ΣM_A = 0: R_C(6) − 18(1.5) − 18(4.5) = 0 → 6R_C = 27 + 81 = 108 → R_C = 18 kN ↑ ΣF_y = 0: R_Ay = 36 − 18 = 18 kN ↑ ΣF_x = 0: R_Ax = 0 CONVENTION: Assume all unknowns in tension. STEP 3 — JOINT A (unknowns: F_AD, F_AB): ΣF_y = 0: F_AD sin60° + 18 = 0 F_AD (0.866) = −18 F_AD = −20.78 kN → COMPRESSION (20.8 kN C) ΣF_x = 0: F_AD cos60° + F_AB = 0 (−20.78)(0.5) + F_AB = 0 F_AB = 10.39 kN → TENSION (10.4 kN T) STEP 4 — JOINT D (unknowns: F_BD, F_DE; known: F_AD = −20.78 kN, load = 18 kN ↓): Note: F_AD at D acts at 60° below-left (compression pushes D away from A direction). ΣF_y = 0: F_AD_y_at_D + F_BD sin60° − 18 = 0 AD is in compression → it pushes D in the direction away from A, i.e., upward and to the right at 60°. F_AD_y at D = +20.78 sin60° = +18.0 kN (upward) 18.0 + F_BD(0.866) − 18 = 0 → F_BD = 0 (zero-force) ΣF_x = 0: F_AD_x at D + F_DE + F_BD cos60° = 0 F_AD_x at D = +20.78(0.5) = +10.39 (pointing right, away from A) 10.39 + F_DE + 0 = 0 → F_DE = −10.39 kN → COMPRESSION (10.4 kN C) STEP 5 — By symmetry: F_CE = F_AD = 20.8 kN (Compression) F_BC = F_AB = 10.4 kN (Tension) F_BE = F_BD = 0 (Zero-force) FINAL SUMMARY: AD = 20.8 kN C | BC = 10.4 kN T CE = 20.8 kN C | AB = 10.4 kN T BD = 0 (zero-force) | BE = 0 (zero-force) DE = 10.4 kN C

Question Type

numerical

Answer Structure

• Step 1: Determinacy check [0.5 mark]
• Step 2: Support reactions (symmetry argument acceptable) [1 mark]
• Step 3: Joint A → F_AD and F_AB [1 mark]
• Step 4: Joint D → F_BD and F_DE [1.5 marks]
• Step 5: Symmetry argument for remaining members [0.5 mark]
• Step 6: Complete summary table with T/C designations [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Incorrect angle — using 30° instead of 60° for the member angle from horizontal
• Not recognizing BD and BE as zero-force members and wasting time on complex equations
• Missing the symmetry argument — resolving all five joints individually wastes time and risks errors
• Presenting F_AD = −20.78 kN without the Compression label

Key Phrases To Include

• equilateral triangle
• sin60°
• cos60°
• symmetry
• zero-force
• compression
• tension
• summary table

Marks

2

Topic

Zero-Force Members

Difficulty

easy

Template Id

T11

Examiner Tip

The word 'unloaded' is the hinge of both rules. Examiners specifically check whether you have included this condition. Its omission reduces each rule to a half-mark.

Model Answer

Rule 1 (Two-member joint): If only two non-collinear members meet at an unloaded joint, both members are zero-force members. Rule 2 (Three-member joint): If three members meet at an unloaded joint and two of them are collinear, the third non-collinear member is a zero-force member. The two collinear members continue to carry the chord force.

Question Type

very_short_answer

Answer Structure

• Rule 1: Two non-collinear members at unloaded joint → both are zero-force [1 mark]
• Rule 2: Three members, two collinear, unloaded joint → non-collinear is zero-force [1 mark]

Scoring Breakdown

Common Mark Deductions

• Omitting 'unloaded joint' condition — the rule fails if there is a load at the joint
• Applying Rule 1 to collinear members (two collinear members at a joint is trivially a straight member, not a truss joint)
• Saying the collinear members also become zero-force in Rule 2

Key Phrases To Include

• non-collinear
• unloaded joint
• two members
• three members
• collinear
• zero-force

Marks

3

Topic

Truss Determinacy

Difficulty

medium

Template Id

T12

Examiner Tip

Part (b) is a higher-order thinking question — the examiner rewards the qualifier (about zero-force members). Including it demonstrates you understand the limitations of the determinacy equation, which is a PRC board exam differentiator.

Model Answer

(a) DETERMINACY CHECK: m + r = 13 + 3 = 16 2j = 2(8) = 16 Since m + r = 2j = 16, the truss is STATICALLY DETERMINATE (assuming proper member arrangement). (b) EVALUATION OF THE CLAIM: If one member is removed: m becomes 12. m + r = 12 + 3 = 15 < 2j = 16 Since m + r < 2j, the modified structure has fewer equations solvable than unknowns — it becomes a MECHANISM (geometrically unstable). The colleague's claim is CORRECT, provided the removed member is not a redundant (zero-force) member and the geometry does not provide an alternative load path. Note: If a zero-force member is removed, the truss may still satisfy the count but could become geometrically unstable if it was providing bracing — the geometric stability must be checked physically.

Question Type

short_answer

Answer Structure

• Part (a): Compute m + r and 2j, compare, state determinate [1.5 marks]
• Part (b): Recompute with m−1, compare to 2j, evaluate claim with reasoning [1.5 marks]

Scoring Breakdown

Common Mark Deductions

• Saying 'unstable' without the word 'mechanism' or 'geometrically unstable'
• Not providing the nuanced qualification about zero-force members
• Arithmetic error in 2j

Key Phrases To Include

• m + r = 2j
• statically determinate
• m + r < 2j
• mechanism
• geometrically unstable
• zero-force member

Marks

5

Topic

Method of Sections

Difficulty

hard

Template Id

T13

Examiner Tip

Recognizing that two cut forces are parallel (both chords horizontal) and switching to ΣF_y instead of ΣM is a key board exam skill. Examiners specifically test this — showing you know when moments won't work earns a mark by itself.

Model Answer

GIVEN: Span 16 m, 4 panels at 4 m, height 3 m. Loads: 20 kN ↓ at B(4,0), C(8,0), D(12,0). Total load = 60 kN. Pin at A(0,0), roller at E(16,0). STEP 1 — DETERMINACY: m = 13, r = 3, j = 8 → 13+3=16=2(8) ✓ STEP 2 — REACTIONS: ΣM_A = 0: R_E(16) − 20(4) − 20(8) − 20(12) = 0 16R_E = 80 + 160 + 240 = 480 R_E = 30 kN ↑ ΣF_y = 0: R_Ay = 60 − 30 = 30 kN ↑ ΣF_x = 0: R_Ax = 0 VERIFICATION: 30 + 30 = 60 ✓ STEP 3 — GEOMETRY OF DIAGONAL CF: C is at (8,0), F is at (4,3). Δx = 4−8 = −4 m, Δy = 3−0 = 3 m Length L_CF = √(16+9) = 5 m cos θ = 4/5 = 0.800 (angle from horizontal) sin θ = 3/5 = 0.600 STEP 4 — SECTION CUT: Cut plane between panels B–F and C–G, cutting: • Top chord FG (horizontal at y = 3) • Diagonal CF (from C down-right to F, or F up-left to C) • Bottom chord BC (horizontal at y = 0) Consider LEFT portion (A–B–C–F): forces acting on it: • R_Ay = 30 kN ↑ at A • 20 kN ↓ at B • 20 kN ↓ at C • F_FG (unknown, horizontal, at y = 3) • F_CF (unknown, along CF) • F_BC (unknown, horizontal, at y = 0) STEP 5 — FIND F_CF: To eliminate F_FG and F_BC, take moments about their intersection. F_FG acts along y = 3 → passes through F(4,3). F_BC acts along y = 0 → passes through B(4,0) and C(8,0). Both F_FG and F_BC pass through point G(8,3)? No — take ΣM about point G(8,3) where F_FG's line of action intersects. Correct approach: F_FG acts horizontally at y = 3 and F_BC acts horizontally at y = 0; neither passes through the same point. Instead, find intersection: F_FG is along y = 3; F_BC is along y = 0. These are parallel — they do NOT intersect at a finite point. For parallel cuts: Use ΣF_x or ΣF_y to isolate F_CF: ΣF_y = 0 (left portion): R_Ay − 20 − 20 + F_CF sin θ_CF_y = 0 Note: CF goes from C(8,0) to F(4,3). Assuming tension → F_CF pulls C toward F, i.e., upward-left at the cut. The y-component of F_CF (assuming tension, arrow at C pointing toward F): Direction from C to F: (−4, +3)/5 → sin θ = 3/5 = 0.6 ΣF_y = 0: 30 − 20 − 20 + F_CF(0.600) = 0 −10 + 0.6 F_CF = 0 F_CF = 10/0.6 = 16.67 kN F_CF = 16.7 kN TENSION (Positive result confirms tension — consistent with a Pratt truss diagonal under downward loading.) FINAL ANSWER: F_CF = 16.7 kN (Tension)

Question Type

numerical

Answer Structure

• Step 1: Determinacy check [0.5 mark]
• Step 2: Reactions R_Ay = 30 kN, R_E = 30 kN with verification [1 mark]
• Step 3: Geometry of diagonal CF — compute angle and trig ratios [0.5 mark]
• Step 4: Identify three members cut and set up left FBD [1 mark]
• Step 5: Recognize parallel chords → use ΣF_y; solve F_CF [1.5 marks]
• Step 6: Final answer 16.7 kN Tension with sign interpretation [0.5 mark]

Scoring Breakdown

Common Mark Deductions

• Attempting to take moments about an intersection of two parallel forces (impossible — no finite intersection)
• Forgetting the 20 kN load at C in the ΣF_y of the left portion
• Wrong direction for F_CF y-component in ΣF_y

Key Phrases To Include

• parallel chords
• ΣF_y = 0
• left portion
• sin θ = 3/5
• F_CF = 16.7 kN
• Tension
• section cut

Marks

2

Topic

Zero-Force Members

Difficulty

medium

Template Id

T14

Examiner Tip

This is a conceptual 2-mark question with exactly two expected points. Structure your answer as a numbered list — it signals to the examiner that you are giving two distinct, complete reasons.

Model Answer

Zero-force members are retained in real trusses for two principal reasons: 1. Stability under alternate load cases: A member that carries zero force under one specific loading configuration will carry force under a different loading (e.g., live load in a different position, wind load, or asymmetric loading). Removing it would make the structure inadequate for those cases. 2. Buckling prevention (bracing): Long compression members can buckle. Zero-force members that are attached at intermediate points of these long members reduce the unsupported length, thereby increasing the buckling capacity of the primary members. This is particularly important for chord members of roof and bridge trusses under NSCP 2015 provisions.

Question Type

short_answer

Answer Structure

• Point 1: Alternate load cases — zero-force under one case, non-zero under another [1 mark]
• Point 2: Bracing against buckling — reduces unsupported length of compression members [1 mark]

Scoring Breakdown

Common Mark Deductions

• Giving only one reason when two are expected for a 2-mark question
• Vague answer like 'for safety' without specifying the mechanism
• Not mentioning buckling by name

Key Phrases To Include

• alternate load cases
• different loading
• buckling
• bracing
• unsupported length
• compression member

Marks

3

Topic

Method of Joints

Difficulty

medium

Template Id

T15

Examiner Tip

Verification questions test whether you can correctly interpret the physical meaning of compression vs. tension. The key insight — compression pushes the joint, tension pulls — must be explicitly stated to earn the first mark.

Model Answer

GIVEN at Joint C: Load = 20 kN ↓ Member AC: F_AC = 15.63 kN Compression → pushes C away from A direction. Member BC: F_BC = 15.63 kN Compression → pushes C away from B direction. GEOMETRY: θ = tan⁻¹(2.5/3) = 39.81°; cos θ = 0.768, sin θ = 0.640 AC goes from A(0,0) to C(3,2.5) → unit vector at C pointing away from A: direction from A to C = (+3,+2.5)/3.905 = (+0.768, +0.640) BC goes from B(6,0) to C(3,2.5) → unit vector at C pointing away from B: direction from B to C = (−3,+2.5)/3.905 = (−0.768, +0.640) Since AC is in compression, it pushes C in the direction away from A (i.e., toward (+0.768, +0.640) as seen from A, but at joint C the push is OUTWARD from C, i.e., in the direction from A to C extended, or the direction (+0.768, +0.640)). Wait — re-clarify: A member in compression PUSHES the joint. The push direction on joint C from member AC = direction FROM A TO C (away from the member body toward C) = (+0.768, +0.640). But this is UPWARD, not downward. Let me re-set: At joint C, member AC in compression exerts force on C in the direction AWAY from the member, i.e., in the direction from A toward C extended beyond C: (+0.768, +0.640). This is upward and to the right. ΣF_x at C: F_AC_x + F_BC_x = +15.63(0.768) + 15.63(−0.768) = +12.0 − 12.0 = 0 ✓ ΣF_y at C: F_AC_y + F_BC_y − 20 = +15.63(0.640) + 15.63(0.640) − 20 = 10.0 + 10.0 − 20 = 0 ✓ Both equilibrium equations are satisfied at Joint C. The solution is verified.

Question Type

numerical

Answer Structure

• Step 1: State the forces acting at C: two member forces + applied load [0.5 mark]
• Step 2: Determine correct direction of compression force on joint C [1 mark]
• Step 3: ΣF_x = 0 equation and verification [0.5 mark]
• Step 4: ΣF_y = 0 equation and verification [1 mark]

Scoring Breakdown

Common Mark Deductions

• Applying tension convention (pull) to a compression member — getting ΣF_y wrong
• Not clearly stating the direction of each force on joint C
• Simply plugging numbers without showing the x and y components

Key Phrases To Include

• compression pushes
• ΣF_x = 0
• ΣF_y = 0
• equilibrium satisfied
• 0.640
• 0.768