CELE Strength of Materials — Stresses in BeamsDetailed Explanation

Examples

Applications

• Sizing timber floor joists and roof rafters per NSCP 2015 Section 616 (allowable bending stress check: fb ≤ Fb).
• Checking steel beams for yielding in flexure per AISC 360-16 Chapter F (Mn = Mp = FyZ for compact sections).
• Determining required reinforcement in RC beams: tension stress in concrete exceeds its modulus of rupture, so steel carries the tension (ACI 318-19 / NSCP 2015 Section 406).
• Bridge girder design — finding the governing fiber stress under truck live loads.
• Foundation beam / grade beam stress checks under eccentric column loads.

Misconceptions

• WRONG: 'The neutral axis is always at mid-depth.' CORRECT: NA is at mid-depth ONLY for symmetric sections. For T, L, or composite sections, compute ȳ = ΣAy/ΣA.
• WRONG: 'I is the moment of inertia about any axis.' CORRECT: In the flexure formula, I must be taken about the CENTROIDAL axis (the NA) of the section.
• WRONG: 'σ_max = Mc/I and σ_max = M/S always give different answers.' CORRECT: They are identical because S is defined as I/c.
• WRONG: 'Bending stress is maximum at the neutral axis.' CORRECT: Bending stress is ZERO at the NA; shear stress is maximum at the NA.
• WRONG: 'For a T-beam under sagging, the top is in tension.' CORRECT: Sagging → top is COMPRESSION, bottom is TENSION.

Related Concepts

• Shear stress distribution (τ = VQ/Ib)
• Section modulus S = I/c
• Moment of inertia and parallel-axis theorem
• Shear Force Diagram (SFD) and Bending Moment Diagram (BMD)
• Euler-Bernoulli beam theory
• Stress-strain relationship (Hooke's Law)
• Kern (core) of a section

Common Exam Questions

Key Points To Remember

• σ = My/I — stress is linear through the depth; zero at the NA, maximum at extreme fibers.
• The neutral axis passes through the CENTROID of the cross-section for homogeneous beams.
• Use the parallel-axis theorem: I_total = Σ(I_own + Ad²) when computing I for composite sections.
• For symmetric sections (rectangle, circle, I-beam), c_top = c_bot = c, so a single σ_max applies.
• For unsymmetric sections (T, L, channel), compute c_top and c_bot separately — they give different stresses.
• Always use consistent units: if M is in N·mm and I is in mm⁴, stress σ comes out in MPa (N/mm²).
• A sagging (positive) moment → bottom in tension, top in compression; hogging → reversed.
• The bottom fiber of a simply supported beam governs for positive bending (tension-critical for concrete).

Examples

Applications

• Rapid selection of W-sections in steel beam design by comparing tabulated S_x to S_req = M_max/φFy (AISC 360 / NSCP).
• Timber beam design per NSCP 2015: fb = M/S ≤ Fb (adjusted allowable bending stress).
• Comparing the efficiency of cross-sections — larger S per unit area means a lighter, more efficient beam.
• Optimization problems: finding minimum-area section (e.g., b:h ratio) for a given S_req.

Misconceptions

• WRONG: 'A wider beam is always stronger than a deeper beam.' CORRECT: S = bh²/6 grows with h², so a deeper section is far more efficient per unit of material.
• WRONG: 'S is the same as I.' CORRECT: S = I/c; they differ by the factor c (extreme fiber distance).
• WRONG: 'For circular sections, S = πd⁴/64.' CORRECT: That formula is I, not S. For circles, S = πd³/32 = I/c where c = d/2.

Related Concepts

• Flexure formula σ = My/I
• Moment of inertia I and parallel-axis theorem
• Plastic section modulus Z (for LRFD / strength design of steel)
• Allowable bending stress (NSCP timber, steel design)

Common Exam Questions

Key Points To Remember

• S = I/c is the section modulus; σ_max = M/S is the simplest form of the flexure formula.
• For rectangles: S = bh²/6. For circles: S = πd³/32. Memorize these.
• S_req = M_max / σ_allow — the fundamental flexural design equation.
• Depth is the most efficient dimension to increase for bending — S scales with h².
• For unsymmetric sections, compute S_top and S_bot separately; the smaller one governs.
• Standard steel sections have tabulated S_x in AISC/NSCP tables — use them directly in steel design.
• Units check: M in N·mm ÷ S in mm³ = σ in N/mm² = MPa.

Examples

Applications

• Timber horizontal shear check: fv = 1.5V/A ≤ Fv per NSCP 2015 Table 616.3-1 (Fv typically 0.5–1.2 MPa for Philippine timber species).
• RC beam shear design: nominal shear Vn = Vc + Vs where Vc is concrete contribution and Vs is from stirrups — concept rooted in shear stress distribution (ACI 318-19 / NSCP 2015 Section 422).
• Steel web shear check: AISC 360-16 Chapter G — web shear yielding: Vn = 0.6 Fy Aw.
• Finding shear stress at flange-web junction of I-beams — critical for checking weld capacity in plate girders.
• Glued joint design in built-up timber sections.

Misconceptions

• WRONG: 'Q is the moment of inertia.' CORRECT: Q is the FIRST moment of area (= A'ȳ'), while I is the SECOND moment of area. They are completely different quantities.
• WRONG: 'Shear stress is maximum at the extreme fibers.' CORRECT: Shear stress is ZERO at the extreme fibers and MAXIMUM at the neutral axis.
• WRONG: 'b in τ = VQ/Ib is the total width of the section.' CORRECT: b is the width AT THE SPECIFIC LEVEL where τ is computed. For an I-beam, b equals the web thickness at the web but the flange width at the flange.
• WRONG: 'Shear stress distribution in a rectangular beam is uniform.' CORRECT: It is PARABOLIC — maximum at the center, zero at the top and bottom.
• WRONG: 'τ_max = V/A for rectangles.' CORRECT: τ_max = 1.5V/A (the factor 1.5 is critical and commonly missed).

Related Concepts

• Flexure formula σ = My/I
• Shear flow q = VQ/I in built-up beams
• Principle of complementary shear stress
• Shear Force Diagram (V diagram)
• Principal stresses and Mohr's Circle (at points with combined τ and σ)

Common Exam Questions

Key Points To Remember

• τ = VQ/Ib — memorize this formula and know each variable's meaning.
• Q is the FIRST MOMENT of area of the part BEYOND (above or below) the cut level about the NA. Q is NOT the moment of inertia.
• At extreme fibers (top, bottom): Q = 0, τ = 0. At the NA: Q = Q_max, τ = τ_max.
• For rectangles: τ_max = 1.5 V/A (at the NA). For circles: τ_max = (4/3) V/A.
• b in the formula is the WIDTH AT THE LEVEL where you compute τ — this changes at flange-web junctions in I-beams.
• Shear stress distribution is PARABOLIC for rectangular sections — it is NOT uniform.
• For timber, horizontal shear (fv = VQ/Ib) must satisfy fv ≤ Fv per NSCP 2015 Table 616.3-1.
• Always confirm V is the shear force AT THE SECTION being checked, from the SFD.

Examples

Applications

• Nailed built-up timber beams in Philippine residential construction — sizing nail spacing per NSCP 2015.
• Determining weld size and spacing in steel plate girders (q = VQ/I → weld force per unit length → select fillet weld size per AISC 360).
• Shear stud spacing in composite steel-concrete beams per AISC 360 Chapter I.
• Glulam beam design — checking adequacy of glue-line shear.
• Plywood-web box beams — a common alternative to standard lumber in Philippine timber construction.

Misconceptions

• WRONG: 'Shear flow q = VQ/Ib (same as shear stress).' CORRECT: Shear flow is q = VQ/I — there is NO 'b'. Shear flow is force per length; shear stress is force per area.
• WRONG: 'Q in the shear flow formula is for the whole section.' CORRECT: Q is ONLY for the CONNECTED AREA (the component being attached to the rest).
• WRONG: 'Connector spacing is constant along the beam length.' CORRECT: Spacing varies because V varies along the span — connectors must be closer together where V is larger (near supports for UDL beams).

Related Concepts

• Shear stress formula τ = VQ/Ib
• Moment of inertia (parallel-axis theorem)
• Shear force diagram (V)
• Nail/bolt design capacity (NSCP 2015 Chapter 624)
• Composite beam design (AISC 360 Chapter I)

Common Exam Questions

Key Points To Remember

• Shear flow q = VQ/I — note: NO 'b' in the denominator (unlike shear stress).
• q has units of N/mm (force per unit length), not MPa.
• The spacing of connectors: s = nF/q, where n = number of connectors per spacing, F = capacity per connector.
• Q is computed for the CONNECTED AREA only (the piece being attached), not the whole section.
• If a component is connected on two sides (e.g., top and bottom nails on a box beam), account for n correctly.
• Compute I for the ENTIRE composite section including all components.
• A higher V (near supports on a UDL beam) means smaller connector spacing — nails are closer together near the ends.

Examples

Applications

• Short columns with eccentric loads — foundations, retaining wall stems.
• Masonry wall design: eccentricity check to avoid tension (unreinforced masonry cannot take tension per NSCP 2015 Chapter 7).
• Footing design: ensuring soil pressure is entirely compressive (no uplift) — eccentricity must stay within the kern of the footing.
• Crane girders subjected to wheel loads + self-weight axial effects.
• Wind-loaded columns: axial gravity load + lateral bending moment from wind.

Misconceptions

• WRONG: 'P/A is always compressive.' CORRECT: If P is a tensile force, P/A is positive (tensile). Sign depends on the direction of P.
• WRONG: 'The kern applies only to the x-direction.' CORRECT: The kern is a two-dimensional region. For a rectangle, it is a rhombus extending ±h/6 in one direction and ±b/6 in the other.
• WRONG: 'Superposition does not work here because the stresses are different types.' CORRECT: Superposition is valid for linear elastic materials — both σ = P/A and σ = Mc/I are normal stresses acting on the same fiber, so they add directly.
• WRONG: 'e = h/6 means no stress at the extreme fiber.' CORRECT: e = h/6 means the minimum fiber stress is ZERO (just touches zero, not negative). Load within the kern → all fibers in compression.

Related Concepts

• Flexure formula σ = My/I
• Axial stress σ = P/A
• Principle of superposition
• Short column design (NSCP 2015 Section 410 / ACI 318-19 Chapter 22)
• Soil bearing pressure distribution under eccentric footings
• Biaxial bending (bending about both axes simultaneously)

Common Exam Questions

Key Points To Remember

• Superpose: σ = P/A ± Mc/I. The sign of P/A is positive for tension, negative for compression.
• For eccentric axial load: M = Pe (multiply eccentricity by axial force to get the induced moment).
• Middle-third rule for rectangles: if e > h/6, tension occurs on one face.
• Kern is the 'safe zone' for compressive loads in masonry, unreinforced concrete, and footings.
• Both σ_max and σ_min must be checked: σ_max against compressive strength, σ_min for possible tension.
• If σ_min is negative (tension) and the material is tension-weak (masonry, plain concrete), redesign is needed.
• Always identify WHICH fiber is at +c and which is at −c from the bending moment direction.

Problem

PROBLEM 1 — Flexure formula, symmetric section: A W-section steel beam has I_x = 178 × 10⁶ mm⁴ and a total depth of 400 mm. It is subjected to a bending moment M = 250 kN·m. Find the maximum bending stress.

Solution

c = 400/2 = 200 mm S_x = I/c = 178 × 10⁶ / 200 = 890,000 mm³ = 890 × 10³ mm³ σ_max = M/S = 250 × 10⁶ / 890,000 = 280.9 MPa Answer: σ_max ≈ 281 MPa

Problem

PROBLEM 2 — Shear stress at a specific level: A 250 mm × 400 mm rectangular beam carries V = 80 kN. Find the shear stress at a level 100 mm above the neutral axis.

Solution

I = 250(400)³/12 = 1,333.33 × 10⁶ mm⁴ Area above the cut (at y = 100 mm from NA, so 100 mm up from center, 100 mm from top): The cut is at y = 100 mm from NA, so the strip from y = 100 to y = 200 (top fiber) has height = 100 mm. A' = 250 × 100 = 25,000 mm² Centroid of A' from NA: ȳ' = 100 + 100/2 = 150 mm Q = A' × ȳ' = 25,000 × 150 = 3,750,000 mm³ τ = VQ/(Ib) = 80,000 × 3,750,000 / (1,333.33 × 10⁶ × 250) τ = 300 × 10⁹ / 333.33 × 10⁹ = 0.90 MPa Answer: τ = 0.90 MPa at y = 100 mm from NA. (Compare with τ_max = 1.5 × 80,000/100,000 = 1.20 MPa at the NA — confirming τ at y = 100 mm < τ_max.)

Problem

PROBLEM 3 — Design for section modulus: A simply supported beam spans 6 m and carries a central concentrated load P = 30 kN. Select a rectangular timber section with b = h/2 if the allowable bending stress is σ_allow = 12 MPa. Check also the allowable shear stress τ_allow = 0.90 MPa.

Solution

Step 1 — Internal forces: V_max = P/2 = 30/2 = 15 kN M_max = PL/4 = 30 × 6/4 = 45 kN·m = 45 × 10⁶ N·mm Step 2 — Required section modulus: S_req = M_max/σ_allow = 45 × 10⁶/12 = 3.75 × 10⁶ mm³ Step 3 — Solve for dimensions (b = h/2): S = bh²/6 = (h/2)h²/6 = h³/12 h³ = 12 × 3.75 × 10⁶ = 45 × 10⁶ h = (45 × 10⁶)^(1/3) = 355.7 mm → use h = 360 mm, b = 180 mm Step 4 — Shear check: A = 180 × 360 = 64,800 mm² τ_max = 1.5V/A = 1.5 × 15,000/64,800 = 0.347 MPa ≤ 0.90 MPa ✓ Flexure governs. Use 180 mm × 360 mm section.

Problem

PROBLEM 4 — Built-up beam nail spacing: A built-up box beam section has I = 4.0 × 10⁷ mm⁴. The top cover plank (A = 6,000 mm², centroid at 80 mm from the beam NA) is nailed with two nails (F = 500 N each) at each spacing. The beam carries V = 5 kN. Find the required nail spacing.

Solution

Step 1 — First moment of area of connected component: Q = A × ȳ = 6,000 × 80 = 480,000 mm³ Step 2 — Shear flow: q = VQ/I = 5,000 × 480,000 / 4.0 × 10⁷ q = 2.4 × 10⁹ / 4.0 × 10⁷ = 60 N/mm Step 3 — Nail spacing (n = 2 nails per spacing): s = nF/q = 2 × 500/60 = 16.67 mm Use s = 15 mm (round down for safety). Answer: s = 15 mm

Problem

PROBLEM 5 — Combined axial and bending stress (eccentric load): A 300 mm × 300 mm short masonry column carries a compressive load of 500 kN at an eccentricity of 60 mm from the centroidal axis. Find the maximum and minimum fiber stresses and determine whether tension occurs. The kern of the section is also to be evaluated.

Solution

Section properties: A = 300 × 300 = 90,000 mm² I = 300(300)³/12 = 675 × 10⁶ mm⁴ c = 150 mm Direct axial stress: σ_direct = P/A = 500,000/90,000 = 5.556 MPa (compression) Bending moment: M = Pe = 500,000 × 60 = 30 × 10⁶ N·mm Bending stress at extreme fiber: σ_bending = Mc/I = 30 × 10⁶ × 150/675 × 10⁶ = 6.667 MPa Combined stresses: σ_max = σ_direct + σ_bending = 5.556 + 6.667 = 12.22 MPa (compression, on load side) σ_min = σ_direct − σ_bending = 5.556 − 6.667 = −1.111 MPa (TENSION on far side) Kern check: e_kern = h/6 = 300/6 = 50 mm e = 60 mm > 50 mm → outside the kern → tension OCCURS. Conclusion: Tension of 1.11 MPa occurs on the far face. For unreinforced masonry (which has negligible tensile strength), this is structurally unacceptable. The load must be moved within 50 mm of the centroid, OR the section must be enlarged, OR steel reinforcement must be added per NSCP 2015 Section 706.