Strength of Materials — Beams and Columns
🟢 Lite — Quick Review (1h–1d)
Rapid summary for last-minute revision before your exam.
Shear Force and Bending Moment
- SFD/SMD conventions: SF positive when acting downward on left face; BM positive when causing sagging (concave upward)
- Relationship: w = −dV/dx; V = dM/dx; w = −d²M/dx²
- Point of contraflexure: Where M = 0 (bending moment changes sign)
Bending and Shear Stress
- Bending stress: σ = My/I (M = moment, y = distance from NA, I = moment of inertia)
- Section modulus: Z = I/y_max; σ_max = M/Z
- Shear stress: τ = VQ/(It) (Q = first moment, t = web thickness)
Column Buckling
- Euler’s load: P_Euler = π²EI/(KL)²
- Slenderness ratio: λ = KL/r (r = √(I/A))
- Rankine’s formula: 1/P = 1/P_Euler + 1/P_c (for intermediate columns)
Key Formulas Table
| Parameter | Formula |
|---|---|
| Bending Stress | σ = My/I |
| Shear Stress | τ = VQ/(It) |
| Euler Load | P_cr = π²EI/(KL)² |
| Slenderness Ratio | λ = KL/r |
| Slope (Cantilever) | y” = M/(EI) |
| Deflection | y = ∫∫ M/(EI) dx dx |
⚡ GATE Tip: Draw SFD and BMD first for any beam problem. Most mistakes come from wrong sign conventions or missing reactions.
🟡 Standard — Regular Study (2d–2mo)
Standard content for students with a few days to months.
Shear Force and Bending Moment Diagrams
The Shear Force Diagram (SFD) and Bending Moment Diagram (BMD) are essential tools for beam analysis. They are related to the distributed load w(x) through:
dV/dx = −w(x) and dM/dx = V(x)
Sign Conventions
| Convention | Shear Force | Bending Moment |
|---|---|---|
| Positive | ↓ on left face (acts downward on left portion) | Sagging (concave upward, compression at top) |
| Negative | ↑ on left face | Hogging (concave downward, compression at bottom) |
For a simply supported beam with UDL w over entire span L:
- Maximum SFD at supports: R_A = R_B = wL/2
- Maximum BM at midspan: M_max = wL²/8
Bending Stress in Beams
The bending stress formula derives from Euler-Bernoulli beam theory:
σ = My/I
where:
- M = bending moment at the section
- y = perpendicular distance from neutral axis
- I = moment of inertia about neutral axis (I = bd³/12 for rectangle; I = πd⁴/64 for circle)
Neutral Axis (NA): The axis where σ = 0. For symmetric sections it passes through the centroid.
Shear Stress in Beams
The shear stress distribution across a beam depth is:
τ = VQ/(It)
For a rectangular section (width b, depth h):
τ_max = 3V/(2bh) at the neutral axis
For an I-beam, most shear is carried by the web. τ_web = V/(tw·h_web).
⚡ GATE Watch: Q = ∫y dA (first moment of area above/below the point), not the section modulus. Students often confuse Q with Z.
Deflection by Integration
From beam theory: EI d²y/dx² = M(x)
Integrate twice to get slope and deflection. Boundary conditions determine constants.
Standard cases:
- Cantilever, tip load P: δ_max = PL³/(3EI)
- Cantilever, UDL w: δ_max = wL⁴/(8EI)
- Simply supported, point load at center: δ_max = PL³/(48EI)
Moment-Area Method
Two theorems for slope and deflection between two points A and B on a beam:
- θ_B − θ_A = (area under M/EI diagram between A and B)
- Deflection of B relative to tangent at A = (moment of M/EI diagram area about B)
Useful for cantilever deflection at the free end.
Example Problem
A simply supported beam of span 6 m carries a point load of 20 kN at midspan. EI = 300 kN·m². Find maximum BM and midspan deflection.
Solution: R_A = R_B = 20/2 = 10 kN M_max at midspan = R_A × 3 = 10 × 3 = 30 kN·m δ_max = PL³/(48EI) = 20×6³/(48×300) = 4320/14400 = 0.3 m = 300 mm
🔴 Extended — Deep Study (3mo+)
Comprehensive coverage for students on a longer study timeline.
Euler’s Column Theory and Slenderness
Column Buckling
When a compression member is slender, it fails by buckling before material reaches compressive strength. The Euler critical load is:
P_cr = π²EI/(KL)²
where K is the effective length factor:
| End Conditions | K |
|---|---|
| Both ends pinned | 1.0 |
| One end fixed, one free | 2.0 |
| Both ends fixed | 0.5 |
| One end fixed, one pinned | 0.7 |
Slenderness Ratio
λ = KL/r where r = √(I/A) (radius of gyration)
- Short columns (λ < π√(E/σ_y)): Fail by compression (material crushing)
- Long columns (λ > π√(E/σ_y)): Fail by buckling (Euler)
- Intermediate columns: Use Rankine’s formula
Rankine’s Formula (Intermediate Columns)
1/P_R = 1/P_Euler + 1/P_c where P_c = σ_c × A
Equivalent form:
P_R = P_Euler × P_c / (P_Euler + P_c)
This gives a smooth transition between short-column crushing and long-column buckling. Many GATE questions use Rankine’s formula for column design.
Slenderness Ratio and Section Selection
For minimum weight design, the optimal column has a slenderness ratio around 80–100. Sections with high I/A (wide flanges, tubes) are more efficient. The most efficient section for a given area has the largest possible r.
Bending Moment Diagram — Common Patterns
| Loading Type | BMD Shape |
|---|---|
| Cantilever + UDL | Parabolic (positive/hogging) |
| Simply supported + UDL | Parabolic (sagging, max at midspan) |
| Cantilever + point load at tip | Triangular (linear) |
| Simply supported + point load at center | Triangular (max at load) |
| Overhanging beam | May have negative BM region (hogging) |
⚡ Point of Contraflexure: Where BMD crosses zero (M = 0). At this section, bending stress is zero — important for reinforcement design in concrete.
Shear Stress Distribution — I-Beam Details
In an I-beam:
- Flange: carries mostly bending stress
- Web: carries most of the shear
- τ_max occurs at the neutral axis (centroidal axis)
- Q for the web = area of flange × distance from NA to flange centroid
Example — Column Design
A steel column (E = 200 GPa, σ_y = 250 MPa) with both ends pinned, length 4 m, diameter 80 mm solid circular. Find critical load and slenderness ratio.
Solution: I = πd⁴/64 = π(0.08)⁴/64 = 2.01×10⁻⁶ m⁴ A = πd²/4 = π(0.08)²/4 = 5.027×10⁻³ m² r = √(I/A) = √(2.01×10⁻⁶/5.027×10⁻³) = 0.02 m = 20 mm KL = 1.0 × 4 = 4 m λ = KL/r = 4/0.02 = 200 P_cr = π²EI/(KL)² = π²×200×10⁹×2.01×10⁻⁶/16 = 248 kN Check: λ > π√(E/σ_y) = π√(200×10⁹/250×10⁶) = 89 → Long column (Euler applies)
GATE Previous Year Pattern
| Year | Topic | Marks |
|---|---|---|
| 2023 | BMD/SFD — overhanging beam | 5 |
| 2022 | Column buckling — Euler load | 2 |
| 2021 | Deflection by integration | 5 |
| 2020 | Shear stress in I-beam | 2 |
| 2019 | Moment-area method | 2 |
⚡ Common Mistakes: (1) Wrong sign convention for BMD, (2) Forgetting KL factor in Euler formula, (3) Using diameter instead of radius in τ = VQ/(It), (4) Confusing slenderness ratio with flexibility factor.
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