CELE Engineering Mechanics — Frames, Machines and CablesExam Answer Templates

Marks

1

Topic

Frames — Member Classification

Difficulty

easy

Template Id

T1

Examiner Tip

One crisp sentence with the phrase 'more than two points' plus a valid example is sufficient for full marks. Do not over-explain in a 1-mark item.

Model Answer

A multi-force member is a structural member that is subjected to forces (or moments) at more than two points, developing internal shear and bending in addition to axial force. Example: the horizontal beam of a portal frame loaded at its midspan.

Question Type

very_short_answer

Answer Structure

• Sentence 1: Define 'multi-force member' with the key phrase 'more than two points' [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing 'a member that carries many forces' without specifying 'at more than two points' — too vague for the mark.
• Giving a two-force member (e.g., a diagonal strut loaded only at its pin ends) as the example.

Key Phrases To Include

• more than two points
• shear and bending
• multi-force member

Marks

1

Topic

Frames — Two-Force Members

Difficulty

easy

Template Id

T2

Examiner Tip

The word 'collinear' is the key examiner trigger word. Include it explicitly.

Model Answer

A two-force member: (1) has forces applied at exactly two points only, and (2) carries no couple or distributed load — so the resultant force at each end must be equal, opposite, and collinear (directed along the line joining the two points).

Question Type

very_short_answer

Answer Structure

• Condition 1: Forces applied at exactly two points only [½ mark]
• Condition 2: No intermediate loads or couples; force is axial (along the member axis) [½ mark]

Scoring Breakdown

Common Mark Deductions

• Stating only one condition and omitting the collinearity or 'no intermediate load' requirement.
• Confusing 'two-force' with 'two-member' — a common mix-up.

Key Phrases To Include

• two points only
• collinear
• along the line joining
• no distributed load or couple

Marks

2

Topic

Frames vs. Machines

Difficulty

easy

Template Id

T3

Examiner Tip

Structure this as a two-row comparison. Use the exact phrase 'transmit or modify forces' for the machine definition — it is the textbook phrasing examiners expect.

Model Answer

A frame is a stationary structure designed to support loads; it contains at least one multi-force member but does not have moving parts. A machine, on the other hand, contains moving parts and is designed to transmit or modify forces (e.g., pliers, a toggle clamp, a crane hook assembly). Both are analyzed by dismembering into individual members and applying equilibrium to each part, but the machine analysis additionally yields a mechanical advantage (M.A. = load / effort).

Question Type

short_answer

Answer Structure

• Line 1: Define 'frame' — stationary, supports loads, multi-force member, no moving parts [1 mark]
• Line 2: Define 'machine' — moving parts, transmits/modifies force, yields mechanical advantage [1 mark]

Scoring Breakdown

Common Mark Deductions

• Saying a frame 'has no multi-force members' — incorrect; a truss has no multi-force members, but a frame does.
• Omitting mechanical advantage in the machine definition — loses the second mark.

Key Phrases To Include

• stationary structure
• moving parts
• transmit or modify forces
• mechanical advantage
• multi-force member

Marks

2

Topic

Machines — Mechanical Advantage

Difficulty

easy

Template Id

T4

Examiner Tip

Always verify: M.A. > 1 means mechanical advantage (load > effort). If M.A. < 1, re-check which arm is which.

Model Answer

Given: effort arm a = 0.5 m; load arm b = 0.08 m; effort F_e = 40 N. (a) Mechanical Advantage: M.A. = effort arm / load arm = 0.5 / 0.08 = 6.25 (b) Load balanced: Load F_L = effort × M.A. = 40 × 6.25 = 250 N Answer: M.A. = 6.25 (dimensionless); Load = 250 N.

Question Type

numerical

Answer Structure

• Step 1: Write M.A. formula and substitute [1 mark]
• Step 2: Compute load using F_L = F_e × M.A. with correct units [1 mark]

Scoring Breakdown

Common Mark Deductions

• Inverting the ratio (load arm / effort arm) — gives M.A. < 1, physically unreasonable for this lever.
• Omitting units on the load (N).
• Not stating M.A. is dimensionless.

Key Phrases To Include

• M.A. = effort arm / load arm
• mechanical advantage
• 250 N
• 6.25

Marks

2

Topic

Cables — Concentrated Loads

Difficulty

easy

Template Id

T5

Examiner Tip

The phrase 'horizontal component H is constant throughout' is the most tested concept in cable problems. State it first, before any formula.

Model Answer

Property: The horizontal component of tension H is constant throughout the entire length of a cable, regardless of the number or magnitude of concentrated loads applied. This follows from horizontal equilibrium of any cable segment (no horizontal loads are applied to the cable itself). Formula: At any segment inclined at angle θ to the horizontal, T = H / cos θ or equivalently, H = T cos θ, V = T sin θ where V is the vertical component of tension at that segment.

Question Type

short_answer

Answer Structure

• Statement: H is constant throughout the cable [1 mark]
• Formula: T = H / cosθ (or H = T cosθ) with definition of θ [1 mark]

Scoring Breakdown

Common Mark Deductions

• Writing T = H cosθ (inverted — gives T < H, which is impossible since T is the resultant).
• Not defining what θ represents.

Key Phrases To Include

• horizontal component H is constant
• T = H / cosθ
• angle of inclination
• no horizontal loads applied to cable

Marks

3

Topic

Cables — Parabolic (Uniform Horizontal Load)

Difficulty

medium

Template Id

T6

Examiner Tip

Write both cable formulas at the top of your solution before substituting. Examiners award a formula-recall mark even if your arithmetic contains a minor error.

Model Answer

Given: L = 30 m, w = 5 kN/m, d = 3 m. For a uniformly horizontally loaded cable, the shape is parabolic. (a) Horizontal tension H: H = wL² / (8d) H = (5)(30)² / (8 × 3) H = (5 × 900) / 24 H = 4500 / 24 H = 187.5 kN (b) Total vertical load = wL = 5 × 30 = 150 kN Vertical reaction at each support = wL/2 = 75 kN Maximum tension at the support: T_max = √(H² + (wL/2)²) T_max = √(187.5² + 75²) T_max = √(35156.25 + 5625) T_max = √40781.25 T_max = 201.9 kN Answer: H = 187.5 kN; T_max = 201.9 kN (at the supports, where the cable is steepest).

Question Type

numerical

Answer Structure

• Step 1: Identify loading type → parabolic cable, state formula H = wL²/8d [1 mark]
• Step 2: Substitute and compute H = 187.5 kN [1 mark]
• Step 3: Apply T_max formula, substitute, and compute T_max = 201.9 kN [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using d = L/2 = 15 m instead of the given sag d = 3 m — catastrophic substitution error.
• Forgetting to square inside the square root — computing √(H + wL/2) instead of √(H² + (wL/2)²).
• Not stating that T_max occurs at the supports (location marks are often separate).
• Omitting units (kN) on both answers.

Key Phrases To Include

• parabolic cable
• H = wL²/8d
• T_max = √(H² + (wL/2)²)
• maximum tension at supports
• 187.5 kN
• 201.9 kN

Marks

3

Topic

Frames — Analysis Procedure

Difficulty

medium

Template Id

T7

Examiner Tip

Numbered steps with keywords are faster to write and easier for examiners to mark than flowing paragraphs. Use exactly this format under timed conditions.

Model Answer

Step-by-step procedure for frame analysis by dismembering: Step 1 — Identify member types: Examine the frame and label each member as either a two-force member (loaded at two pin ends only, force acts along the member) or a multi-force member (loaded at more than two points or carrying a moment). Step 2 — Analyze the whole frame: Draw the FBD of the entire frame as a rigid body. Apply ΣFx = 0, ΣFy = 0, and ΣM = 0 to determine the external support reactions (pin components and roller reactions). Step 3 — Dismember and draw individual FBDs: Separate each member. At every shared pin, introduce the pin force components (Cx, Cy, etc.) on each member. By Newton's Third Law, the force on Member A at pin C is equal in magnitude but opposite in direction to the force on Member B at the same pin. Step 4 — Apply equilibrium to each member: Write ΣFx = 0, ΣFy = 0, ΣM = 0 for each multi-force member, choosing the moment center to eliminate as many unknowns as possible. Two-force members immediately give the direction of the pin force. Step 5 — Check: Verify that the pin forces on mating members are equal and opposite, and that the overall equilibrium of the reassembled frame is satisfied.

Question Type

short_answer

Answer Structure

• Step 1: Identify two-force vs. multi-force members [1 mark]
• Step 2: FBD of whole frame → find external reactions [1 mark]
• Step 3 & 4: Dismember, apply Newton's Third Law at pins, write equilibrium for each member [1 mark]

Scoring Breakdown

Common Mark Deductions

• Starting with individual members before finding external reactions — leads to more unknowns than equations.
• Not mentioning Newton's Third Law at pin connections — a fundamental concept mark is lost.
• Omitting the check step (not penalized heavily, but shows incomplete understanding).

Key Phrases To Include

• two-force member
• multi-force member
• FBD of the whole frame
• dismember
• Newton's Third Law
• equal and opposite
• ΣFx = 0, ΣFy = 0, ΣM = 0

Marks

3

Topic

Cables — Concentrated Loads

Difficulty

medium

Template Id

T8

Examiner Tip

Always draw the FBD at the loaded joint showing the two cable segments pulling away and the applied load pulling down. This ensures the factor of 2 is not missed.

Model Answer

Given: Span L = 12 m; load P = 10 kN at midspan; sag d = 1.5 m. By symmetry, the load divides the cable into two equal segments of horizontal projection 6 m each. (a) Geometry of cable segment: tanθ = sag / horizontal projection = 1.5 / 6 = 0.25 θ = arctan(0.25) = 14.04° sinθ = 0.2425, cosθ = 0.9701 (b) Vertical equilibrium at load point (two symmetrical cable segments pulling upward): ΣFy = 0: 2T sinθ = P 2T(0.2425) = 10 T = 10 / (2 × 0.2425) T = 20.6 kN (c) Horizontal component: H = T cosθ = 20.6 × 0.9701 = 20.0 kN Verification: H is constant throughout both segments (no horizontal load applied to cable). ✓ Answer: T = 20.6 kN; H = 20.0 kN.

Question Type

numerical

Answer Structure

• Step 1: Draw FBD at load point; compute geometry (tanθ, sinθ, cosθ) [1 mark]
• Step 2: Apply ΣFy = 0 → solve for T = 20.6 kN [1 mark]
• Step 3: Compute H = T cosθ = 20.0 kN; verify H is constant [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using tanθ = 1.5/12 (using full span instead of half-span for the horizontal projection) — geometry error.
• Writing T sinθ = P (forgetting the factor of 2 for two symmetric cable segments).
• Not computing H at all — loses the third mark entirely.

Key Phrases To Include

• tanθ = 1.5/6
• 2T sinθ = P
• T = 20.6 kN
• H = T cosθ
• H = 20.0 kN
• H is constant

Marks

2

Topic

Cables — Concentrated Loads / Funicular Polygon

Difficulty

medium

Template Id

T9

Examiner Tip

The trap is saying 'tension is constant.' Memorize: tension T varies, but horizontal component H is constant. This distinction is a favorite board-exam trick question.

Model Answer

A funicular polygon is the shape adopted by a perfectly flexible, inextensible cable subjected to a series of concentrated (point) loads. Between each pair of adjacent loads, the cable segment is straight (a chord), so the entire loaded cable forms a series of straight line segments — a polygon — that is in equilibrium with the applied loads. The polygon is called 'funicular' (from Latin: funis = rope) because it is the natural hanging shape. The quantity that remains constant throughout every segment of the cable is the horizontal component of tension, H. Since no horizontal external forces act on the cable between load points, horizontal equilibrium requires H to be the same in every segment.

Question Type

short_answer

Answer Structure

• Sentence 1-2: Define funicular polygon — straight segments between loads forming a polygon [1 mark]
• Sentence 3: State that H (horizontal tension component) is constant throughout [1 mark]

Scoring Breakdown

Common Mark Deductions

• Saying the 'tension' is constant — tension T varies from segment to segment; only H is constant.
• Describing the catenary (self-weight) instead of the polygon (concentrated loads).

Key Phrases To Include

• funicular polygon
• straight segments between loads
• horizontal component H is constant
• no horizontal external forces

Marks

2

Topic

Cables — Parabola vs. Catenary

Difficulty

medium

Template Id

T10

Examiner Tip

Remember: suspension bridge → parabola (deck load per horizontal meter). Power line sagging under its own weight → catenary. This real-world hook helps you recall which is which.

Model Answer

Parabolic cable: Occurs when the load is uniformly distributed per unit horizontal length (e.g., the deck load on a suspension bridge, where the cable supports a horizontal roadway). The cable equation is y = wx²/(2H), a parabola. Analysis uses H = wL²/8d. Catenary cable: Occurs when the load is the cable's own self-weight, uniformly distributed per unit length along the cable itself (not along the horizontal). The cable hangs in the shape of a hyperbolic cosine: y = H/w_c [cosh(w_c x/H) − 1], where w_c is the cable weight per unit length. Key distinction: Load per unit horizontal distance → parabola; load per unit cable length (self-weight) → catenary. For cables with small sag-to-span ratios (d/L < 1/8), the two shapes are nearly identical, and the parabolic formula is a good engineering approximation for the catenary.

Question Type

short_answer

Answer Structure

• Parabola: loading condition + equation or formula [1 mark]
• Catenary: loading condition + shape description; note the approximation for shallow sags [1 mark]

Scoring Breakdown

Common Mark Deductions

• Saying both are parabolas — loses the catenary mark.
• Confusing the loading direction: 'per unit horizontal' vs. 'per unit cable length' is the critical distinction.

Key Phrases To Include

• uniform load per unit horizontal length
• parabola
• self-weight per unit cable length
• catenary
• hyperbolic cosine
• shallow sag approximation

Marks

5

Topic

Frames — Dismembering Method

Difficulty

hard

Template Id

T11

Examiner Tip

Always take moments about the pin with the most unknowns (usually the wall pin A) to get F_CD directly in one equation. This is the single most efficient move in a 5-mark frame problem.

Model Answer

Given: Beam AB pinned at A(0,0); load P = 12 kN ↓ at B(6,0); strut CD from C(3,0) to D(0,−3); the strut CD is a two-force member (pin-connected at both ends, no intermediate load). Step 1 — Geometry of strut CD: CD runs from C(3,0) to D(0,−3): Δx = −3 m, Δy = −3 m Length of CD = √(3² + 3²) = 3√2 = 4.243 m Angle of CD below horizontal: tanα = 3/3 = 1 → α = 45° The strut force acts along CD. On the beam at C, the strut pushes upward and to the right at 45° (in compression). F_CD,x = F_CD cos45° = 0.7071 F_CD (→ positive x) F_CD,y = F_CD sin45° = 0.7071 F_CD (↑ positive y) Step 2 — FBD of beam AB: Forces: A_x (→), A_y (↑) at A; F_CD,x (→) and F_CD,y (↑) at C; 12 kN (↓) at B. Step 3 — Moment about A (eliminates A_x and A_y): ΣM_A = 0: +(F_CD sin45°)(3) − 12(6) = 0 0.7071 F_CD × 3 = 72 2.121 F_CD = 72 F_CD = 72 / 2.121 F_CD = 33.95 kN ≈ 34.0 kN (compression in strut) Step 4 — ΣFy = 0 on beam AB: A_y + F_CD sin45° − 12 = 0 A_y + 0.7071(33.95) − 12 = 0 A_y + 24.0 − 12 = 0 A_y = −12.0 kN (i.e., 12.0 kN downward at A) Step 5 — ΣFx = 0 on beam AB: A_x + F_CD cos45° = 0 A_x + 0.7071(33.95) = 0 A_x + 24.0 = 0 A_x = −24.0 kN (i.e., 24.0 kN directed toward the wall, in −x direction) Step 6 — Resultant at A: R_A = √(A_x² + A_y²) = √(24.0² + 12.0²) = √(576 + 144) = √720 R_A = 26.8 kN Direction: θ_A = arctan(12/24) = arctan(0.5) = 26.6° below the −x axis (toward wall and downward) Newton's Third Law check: The strut pushes beam at C with 34.0 kN at 45° up-right; the beam pushes the strut at C with 34.0 kN at 45° down-left (equal and opposite). ✓ Answers: (a) F_CD = 34.0 kN (compression) (b) A_x = 24.0 kN ← (toward wall); A_y = 12.0 kN ↓ (c) R_A = 26.8 kN at 26.6° below horizontal

Question Type

numerical

Answer Structure

• Step 1: Identify strut CD as two-force member; compute geometry (angle = 45°) [1 mark]
• Step 2: Draw FBD of beam AB with all forces labeled [1 mark]
• Step 3: Apply ΣM_A = 0 → F_CD = 34.0 kN [1 mark]
• Step 4 & 5: Apply ΣFy = 0 → A_y = 12.0 kN ↓; ΣFx = 0 → A_x = 24.0 kN ← [1 mark]
• Step 6: Compute R_A = 26.8 kN; invoke Newton's Third Law at pin C as verification [1 mark]

Scoring Breakdown

Common Mark Deductions

• Not identifying the strut as a two-force member — leads to incorrectly assumed force direction and extra unknowns.
• Taking moments about C instead of A — while valid, it does not eliminate both A_x and A_y simultaneously, leading to simultaneous equations and higher error risk.
• Sign errors in A_y: students often write A_y = +12 kN (upward) without checking the equilibrium direction.
• Skipping the resultant computation or Newton's Third Law check — loses the fifth mark.

Key Phrases To Include

• two-force member
• 45° angle
• ΣM_A = 0
• F_CD = 34.0 kN compression
• A_x = 24.0 kN
• A_y = 12.0 kN
• R_A = 26.8 kN
• Newton's Third Law

Marks

5

Topic

Cables — Parabolic, Full Analysis

Difficulty

hard

Template Id

T12

Examiner Tip

Part (d) arc length is a bonus-type sub-question. If time is short, secure parts (a), (b), and (c) first — they carry 3 of the 5 marks.

Model Answer

Given: L = 40 m, w = 6 kN/m (horizontal), d = 4 m. The cable shape is parabolic (load is uniform per unit horizontal length). (a) Horizontal tension H: H = wL² / (8d) H = (6)(40)² / (8 × 4) H = (6 × 1600) / 32 H = 9600 / 32 H = 300 kN (b) Total vertical load = wL = 6 × 40 = 240 kN Vertical reaction at each support = wL/2 = 120 kN (by symmetry) Maximum tension: T_max = √(H² + (wL/2)²) T_max = √(300² + 120²) T_max = √(90000 + 14400) T_max = √104400 T_max = 323.0 kN (at the supports) (c) Cable slope at support: tanθ_s = (wL/2) / H = 120 / 300 = 0.40 θ_s = arctan(0.40) = 21.8° (This confirms T_max occurs at the support where the cable is steepest.) (d) Approximate cable length: S ≈ L[1 + (8/3)(d/L)²] d/L = 4/40 = 0.10 (d/L)² = 0.01 S ≈ 40[1 + (8/3)(0.01)] S ≈ 40[1 + 0.02667] S ≈ 40 × 1.02667 S ≈ 41.07 m Answers: (a) H = 300 kN (b) T_max = 323.0 kN at the supports (c) θ_s = 21.8° (d) S ≈ 41.07 m

Question Type

numerical

Answer Structure

• Step 1: State parabolic cable; compute H = wL²/8d = 300 kN [1 mark]
• Step 2: Compute T_max = √(H² + (wL/2)²) = 323.0 kN [1 mark]
• Step 3: Compute support slope θ_s = arctan(wL/2H) = 21.8° [1 mark]
• Step 4: Apply arc length formula S ≈ 41.07 m [1 mark]
• Step 5: Correct units, logical consistency, and noting T_max location [1 mark]

Scoring Breakdown

Common Mark Deductions

• Using d = L/2 = 20 m in the H formula — a catastrophic error yielding H = 30 kN (off by an order of magnitude).
• Computing T_max at midspan (where T = H, the minimum, not maximum) — conceptual error.
• Forgetting (8/3) coefficient in the arc length formula or using d/L = 4 instead of 0.10.
• Not stating units for arc length S (meters).

Key Phrases To Include

• parabolic cable
• H = wL²/8d
• 300 kN
• T_max = √(H² + (wL/2)²)
• 323.0 kN
• at the supports
• θ_s = 21.8°
• S ≈ 41.07 m

Marks

5

Topic

Cables — Multiple Concentrated Loads

Difficulty

hard

Template Id

T13

Examiner Tip

For third-point loading with equal loads, exploit symmetry immediately. This reduces the problem to finding one unknown (H) from a single moment equation on the left sub-FBD.

Model Answer

Setup: Let A be the left support, B be the right support (span AB = 9 m, horizontal). Load P₁ = 8 kN at point C (3 m from A), load P₂ = 8 kN at D (6 m from A). By symmetry, A and B have equal vertical reactions. Step 1 — External reactions: ΣM_A = 0: B_y(9) − 8(3) − 8(6) = 0 9 B_y = 24 + 48 = 72 B_y = 8 kN ↑ ΣFy = 0: A_y + B_y = 16 → A_y = 8 kN ↑ (confirmed by symmetry) Step 2 — Horizontal component H (constant throughout): Consider the FBD of the LEFT segment AC (from A to load C). At C, sag = 1.5 m below the chord AB (assumed supports at the same level). Horizontal projection of AC = 3 m; vertical drop at C = 1.5 m. ΣM_C (for segment AC) = 0: H(1.5) = A_y(3) H × 1.5 = 8 × 3 H = 24 / 1.5 H = 16 kN Step 3 — Tension in segment AC (left outer segment, slope = 1.5/3 = 0.5, θ_AC = arctan(0.5) = 26.57°): T_AC = √(H² + A_y²) = √(16² + 8²) = √(256 + 64) = √320 = 17.89 kN Step 4 — Tension in segment CD (middle segment, horizontal by symmetry): At load C, vertical equilibrium: ΣFy = 0 at C: T_AC sinθ_AC − 8 − T_CD sinθ_CD = 0 8 kN (upward from AC) − 8 kN (load) = T_CD sinθ_CD T_CD sinθ_CD = 0 → θ_CD = 0° This means segment CD is horizontal (sag at C = sag at D = 1.5 m, symmetric profile → chord CD is horizontal). T_CD = H = 16 kN (horizontal segment) Step 5 — Tension in segment DB (right outer segment, symmetric to AC): T_DB = T_AC = 17.89 kN (by symmetry) Summary of answers: T_AC = T_DB = 17.89 kN ≈ 17.9 kN (outer segments) T_CD = 16.0 kN (middle horizontal segment) H = 16.0 kN (constant horizontal component — verified: T_CD = H ✓)

Question Type

numerical

Answer Structure

• Step 1: External reactions by ΣM_A = 0 → A_y = B_y = 8 kN [1 mark]
• Step 2: H from ΣM_C = 0 on left sub-FBD → H = 16 kN [1 mark]
• Step 3: T_AC = √(H² + A_y²) = 17.89 kN [1 mark]
• Step 4: Middle segment CD is horizontal by symmetry → T_CD = H = 16 kN [1 mark]
• Step 5: T_DB = T_AC = 17.89 kN by symmetry; verify H is constant [1 mark]

Scoring Breakdown

Common Mark Deductions

• Not using a sub-FBD to find H — instead trying to find H from geometry alone without moment equation.
• Failing to recognize the middle segment CD is horizontal when loads are equal and symmetric.
• Computing T = H + V instead of T = √(H² + V²) — algebraic error.
• Not verifying H is the same in all three segments.

Key Phrases To Include

• A_y = B_y = 8 kN
• H = 16 kN
• constant horizontal component
• T_AC = 17.89 kN
• middle segment horizontal
• T_CD = 16 kN
• symmetry

Marks

3

Topic

Frames/Machines — Newton's Third Law at Pins

Difficulty

medium

Template Id

T14

Examiner Tip

A short labeled sketch of two members meeting at a pin, with arrows showing (C_x, C_y) on one member and (−C_x, −C_y) on the other, earns full marks faster than three paragraphs of prose.

Model Answer

Newton's Third Law (action-reaction principle) states that when two bodies interact, the force exerted by body A on body B is equal in magnitude, opposite in direction, and collinear to the force exerted by body B on body A. In the dismembering method, when a frame is separated into its individual members at a shared pin connection, unknown internal pin forces are introduced on the FBD of each member. Newton's Third Law mandates that if pin C exerts a force (C_x, C_y) on Member 1, then Member 1 exerts a force of (−C_x, −C_y) on Member 2 — exactly equal in magnitude but reversed in direction. This is essential because: 1. Consistency: Without applying Newton's Third Law, the force on one member would be assumed independently from the force on the adjacent member, leading to two separate unknowns for what is physically one force pair — the system becomes over-specified and the solution inconsistent. 2. Equilibrium: When all individual members are reassembled into the whole frame, the internal pin forces (which are internal forces to the assembly) cancel in pairs, and only the external loads and reactions remain in the global equilibrium — this is the mathematical self-consistency check. In practice: always assign one direction (e.g., C_x to the right) on Member 1's FBD and write C_x to the left on Member 2's FBD. If the solved value of C_x is positive, the assumed direction on Member 1 is correct; it is reversed on Member 2.

Question Type

short_answer

Answer Structure

• Sentence 1: State Newton's Third Law precisely [1 mark]
• Paragraph 2: Apply to pin connections — force on Member 1 is equal and opposite to force on Member 2 [1 mark]
• Paragraph 3: Explain why essential — consistency and global equilibrium check [1 mark]

Scoring Breakdown

Common Mark Deductions

• Stating Newton's Third Law correctly but not connecting it specifically to the pin forces in a frame.
• Not explaining why the law is essential — describing it without its consequence loses the third mark.

Key Phrases To Include

• equal in magnitude, opposite in direction
• collinear
• shared pin connection
• action-reaction
• internal forces cancel
• global equilibrium

Marks

2

Topic

Cables — Location of Maximum Tension

Difficulty

medium

Template Id

T15

Examiner Tip

Board exams frequently ask 'where is tension maximum?' as a trap. The answer is always the supports — where the cable is steepest. Memorize the justification: steepest slope → smallest cosθ → largest T = H/cosθ.

Model Answer

(a) Single midspan concentrated load: Maximum tension occurs in the cable segments at the supports (on either side of the load point), NOT at the load point. At the supports, the cable is steepest (largest inclination angle θ), giving the largest resultant T = H/cosθ. Since H is constant and cosθ is smallest (θ is largest) at the supports, T is greatest there. (b) Uniform horizontal load (parabolic cable): Maximum tension also occurs at the supports. As the cable rises toward the supports, its slope θ increases continuously from zero at midspan (where the cable is horizontal, T_min = H) to a maximum at the supports, where T_max = √(H² + (wL/2)²). This formula shows that T_max > H always, confirming the support location.

Question Type

short_answer

Answer Structure

• Part (a): Supports for concentrated load + justification via θ and T = H/cosθ [1 mark]
• Part (b): Supports for parabolic cable + formula T_max = √(H² + (wL/2)²) [1 mark]

Scoring Breakdown

Common Mark Deductions

• Saying maximum tension is 'at the load point' or 'at midspan' — the most common conceptual error in cable problems.
• Not providing justification — answering 'at the supports' without explaining why loses a half-mark.

Key Phrases To Include

• at the supports
• steepest inclination
• T = H/cosθ
• smallest cosθ
• T_max = √(H² + (wL/2)²)