Engineering Mechanics — Statics

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Equilibrium and Free Body Diagrams

• Equilibrium: ΣF_x = 0, ΣF_y = 0, ΣM = 0 (2D problems)
• Support reactions: Roller → 1 reaction (perp. to surface); Pin → 2 reactions (F_x, F_y); Fixed → 3 reactions (F_x, F_y, M)
• Static determinacy: 3 unknowns for 2D → statically determinate; >3 → indeterminate

Truss Analysis

• Method of joints: Equilibrium at each joint; use ΣF_x = 0, ΣF_y = 0
• Method of sections: Cut through ≤3 members; ΣM = 0 about a point
• Zero-force members: If two members meet at a joint with no external force → both are zero-force; if one member and no force → that member is zero-force

Friction and Centroids

• Friction force: F ≤ μN (static); F = μN (impending motion)
• Centroid: x̄ = ∫x dA/A; ȳ = ∫y dA/A
• Moment of inertia: I = ∫y² dA about axis; I = I_G + Ad² (parallel axis theorem)

Pappus-Guldinus: Surface area = (generating curve length) × (distance traveled by centroid); Volume = (generating area) × (distance traveled by centroid)

⚡ GATE Tip: Always draw the FBD before writing equilibrium equations. Most statics errors come from omitting forces or mislabeling directions.

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🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Free Body Diagrams and Equilibrium

A Free Body Diagram (FBD) is a sketch showing all forces and moments acting on a body, isolated from its surroundings. Every force must be accounted for.

Support Reactions

Support Type	Reactions	Statically Determinate?
Roller (smooth)	1 perpendicular	Yes
Hinge/Pin	2 (F_x, F_y)	Yes
Fixed/Built-in	3 (F_x, F_y, M)	No (indeterminate)
Slider	2 (perpendicular to slot + moment if rotation prevented)	Yes

⚡ GATE Watch: A fixed support in 2D gives 3 unknowns (H, V, M). A cantilever is fixed at one end — statically determinate. A propped cantilever (fixed at one end, roller at the other) has 4 unknowns → statically indeterminate to 1°.

Equilibrium Equations

For a 2D problem:

ΣF_x = 0 ΣF_y = 0 ΣM_O = 0 (moments about any point O)

For a 3D problem, add ΣF_z = 0 and ΣM_x = ΣM_y = ΣM_z = 0.

Procedure:

1. Draw FBD showing all known and unknown forces
2. Choose coordinate axes (horizontal and vertical usually suffice)
3. Write equilibrium equations
4. Solve for unknowns
5. Check: Substitute back; verify moment equilibrium about a different point

Friction

Static friction: F_s ≤ μ_s N (force needed to start motion equals μ_s N) Kinetic friction: F_k = μ_k N (during motion) Impending motion: F = μ_s N (at the point of slipping)

⚡ GATE Tip: When solving friction problems with multiple bodies, check if motion is impending or not. If unsure, assume impending and check consistency.

Friction on inclined planes: For a block on incline with angle θ: F = W sin θ, N = W cos θ. Sliding begins when tan θ = μ_s.

Truss Analysis

Method of Joints

Each joint is a pin joint in equilibrium under two-force members. At each joint:

ΣF_x = 0 and ΣF_y = 0

Always start at a joint with ≤2 unknowns (usually a support joint).

Sign convention: Members in tension are pulled apart at the joint; members in compression are pushed together.

Zero-Force Members

These situations produce zero-force members:

1. Two members at a joint with no external force → both are zero-force
2. Three members at a joint with one external force collinear with one member → the other two are zero-force
3. Four members at a joint forming an X with no external force → diagonal members are zero-force (if symmetric and no deformation)

⚡ GATE Watch: GATE frequently asks “identify zero-force members” — knowing these rules saves time.

Method of Sections

Cut the truss through members you want to find (≤3 unknowns), then apply ΣM = 0 about a point to isolate individual member forces.

ΣM_point = 0 → eliminates the force through that point, solving for perpendicular forces directly.

This method is faster than joints when you need forces in specific members (not the whole truss).

Centroids and Moment of Inertia

Centroid of a composite area:

x̄ = Σx_i A_i / ΣA_i; ȳ = Σy_i A_i / ΣA_i

Common centroids:

Shape	x̄	ȳ
Rectangle	b/2	h/2
Triangle	b/3 from vertex	h/3 from base
Semicircle	0	4r/(3π) from base
Quarter circle	4r/(3π)	4r/(3π)

Moment of Inertia

I_x = ∫y² dA; I_y = ∫x² dA

Parallel Axis Theorem:

I = I_G + Ad²

where I_G is about centroidal axis, A is area, d is distance between parallel axes.

Radius of gyration:

r = √(I/A)

Pappus-Guldinus Theorems

Theorem 1 (Surface area):

S = L × 2πr̄

The surface area generated by rotating a plane curve of length L about an axis equals L times the distance traveled by its centroid.

Theorem 2 (Volume):

V = A × 2πr̄

The volume generated by rotating a plane area A about an external axis equals A times the distance traveled by its centroid.

⚡ GATE Watch: The axis of rotation must NOT intersect the curve/area being rotated. Both theorems require the centroid’s path to be a circle.

Example Problem

A truss has a 10 kN load at joint C. Find forces in members BC, AC, and AD using method of joints. (Truss: A is pinned support, D is roller support, B is directly below A on horizontal line AD, C is midpoint of BD.)

Solution (joint A): R_A + R_D = 10 kN (vertical equilibrium) ΣM_A: R_D × 6 = 10 × 3 → R_D = 5 kN, R_A = 5 kN

Joint D: Only members AD and BD meet. R_D = 5 kN ↑. Since no horizontal force at D, member AD has zero force. BD carries 5 kN (compression).

Joint B: Members BD (5 kN compression), BC, AB. ΣF_y = 0: BC sin θ = 0 → BC = 0 kN (zero-force member). ΣF_x = 0: AB = 5 kN (tension).

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Advanced Truss Analysis

Compound and Complex Trusses

Compound truss types:

• Type 1: Three肝炎 connected by pins (simplest truss)
• Type 2: Multiple simple trusses connected by joints + 1 member or 3 members
• Type 3: Complex trusses (non-Houston type) — must analyze by method of joints or computer methods

Determinacy of plane trusses:

m + r = 2j (determinate); m + r > 2j (indeterminate) where m = number of members, r = number of reactions, j = number of joints.

Space Trusses

For 3D trusses:

ΣF_x = ΣF_y = ΣF_z = ΣM_x = ΣM_y = ΣM_z = 0 (6 equations per joint)

A space truss with m = 3j − 6 is statically determinate (just rigid).

Friction Applications

Wedge Friction

A wedge under a block introduces additional force directions. Equilibrium requires:

• ΣF_x = 0, ΣF_y = 0 at each contacting surface
• Friction force at each contact ≤ μN

For a self-locking wedge: tan α ≤ μ where α is the wedge angle.

Screw Jack

A screw thread with lead angle λ and friction angle φ acts like an inclined plane:

T = W r tan(λ + φ) (raising the load) T = W r tan(λ − φ) (lowering, if λ > φ)

If λ < φ, the screw is self-locking (load won’t fall).

Belt Friction

For a flat belt wrapping around a pulley:

T₁/T₂ = e^(μθ) (Euler-Euler belt friction formula)

where T₁ = tight side tension, T₂ = slack side tension, θ = angle of contact in radians.

⚡ GATE Watch: For V-belts, use T₁/T₂ = e^(μθ/sin(β)) where β = V-groove half-angle. Most flat belt problems assume flat surfaces.

Virtual Work Method for Trusses

For a determinate truss, the deflection at a joint can be found using virtual work:

δ = Σ(F_i × (∂F_i/∂P) × L_i / (A_i × E))

where F_i is member force due to unit load at the point/direction of desired displacement. This is the unit load method — particularly useful when only one displacement is needed.

Example — Moment of Inertia (Composite Section)

Find I_NA about the neutral axis of an I-section: Flange 150×20 mm, Web 20×100 mm, all about the horizontal centroidal axis.

Solution: Convert to all dimensions in mm. Total height = 20 + 100 + 20 = 140 mm. A_total = 150×20×2 + 20×100 = 8000 mm²

Centroid ȳ = (150×20×10×2 + 20×100×70)/8000 = (60000 + 140000)/8000 = 25 mm (from bottom)

I_about_bottom = I_flange×2 + A_flange×d²_flange×2 + I_web + A_web×d²_web = (150×20³/12)×2 + 3000×65²×2 + (20×100³/12) + 2000×45² = (150×8000/12)×2 + 3000×4225×2 + (20×1000000/12) + 2000×2025 = 100000 + 25350000 + 1666667 + 4050000 = 31,066,667 mm⁴

I_NA = I_about_bottom − A_total × ȳ² = 31,066,667 − 8000×25² = 31,066,667 − 5,000,000 = 26,066,667 mm⁴

Friction Example — Block on Inclined Plane

A 100 N block rests on a 30° incline with μ_s = 0.4. (a) Will it slide? (b) Minimum force P parallel to incline needed to start upward motion.

Solution: (a) Component down plane: W sin 30° = 100 × 0.5 = 50 N Normal force: N = W cos 30° = 100 × 0.866 = 86.6 N μ_s N = 0.4 × 86.6 = 34.6 N < 50 N → Block slides down

(b) To start upward: P ≥ μ_s N + W sin 30° = 34.6 + 50 = 84.6 N

GATE Previous Year Pattern

Year	Topic	Marks
2023	Method of sections — truss	5
2022	FBD + equilibrium (beam)	5
2021	Friction — wedge	2
2020	Centroid + moment of inertia	5
2019	Method of joints + zero-force	2

⚡ Common Mistakes: (1) Omitting self-weight or reactions on FBD, (2) Wrong direction for friction force, (3) Misidentifying zero-force members, (4) Using parallel axis theorem with wrong distance d, (5) Applying Pappus-Guldinus when axis intersects the generating shape.

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