CELE Strength of Materials — Stresses in BeamsRevision Notes

Formulas

Exam Tips

• Memorize S_rect = bh²/6 and S_circle = πd³/32 — these appear in almost every board exam set.
• When a problem says 'maximum bending stress,' go directly to σ_max = M/S; you don't need to compute y separately.
• For T-beams, the board exam almost always wants you to find the centroid and I first — make a table with A, ȳ, and Ī + Ad² for each sub-area.
• Sagging moment → bottom tension: for reinforced-concrete beams (ACI 318 / NSCP 2015 Section 406), reinforcement goes at the bottom.
• If the problem provides a W-section table (e.g., AISC/NSCP Table), read S_x directly; no need to calculate I and c separately.

Key Points

• Plane sections remain plane after bending (Bernoulli–Euler assumption) — strain is linear through the depth.
• The neutral axis (NA) passes through the centroid of the cross-section; bending stress is zero there.
• Stress varies linearly: σ = My/I, where y is measured from the NA.
• Maximum bending stress occurs at the extreme fiber (y = c): σ_max = Mc/I = M/S.
• For a sagging (positive) moment: bottom fiber is in tension, top fiber is in compression.
• For a hogging (negative) moment: top fiber is in tension, bottom fiber is in compression.
• Units must be consistent: if M is in N·mm and I is in mm⁴, then σ is in N/mm² = MPa.
• For an unsymmetric section (e.g., T-beam), c_top ≠ c_bot, so σ_top ≠ σ_bot for the same M.
• Linearly elastic, isotropic, homogeneous material is assumed (valid for steel, timber; not directly for concrete).

Definitions

Section Title

1. The Flexure (Bending) Formula

Common Mistakes

• Using the full depth h instead of c (= h/2 for symmetric sections) in the formula σ = Mc/I.
• For T-sections or other unsymmetric sections: assuming c_top = c_bot (they are different — always locate the centroid first).
• Mixing units: M in kN·m but I in mm⁴ without converting M to N·mm (multiply by 10⁶).
• Dividing M by I instead of M by S — or forgetting to multiply by c when using σ = Mc/I.
• Forgetting that S = I/c applies per axis; for bending about the weak axis, use the weak-axis I and corresponding c.

Formulas

Exam Tips

• Set up a systematic table: Column headers = Sub-area, A (mm²), ȳ from top (mm), Aȳ (mm³), Ī = bh³/12 (mm⁴), d = ȳ_sub − ȳ_NA (mm), Ad² (mm⁴), I_total contribution.
• For symmetric I-sections (W-shapes), I can be computed as I_outer_rectangle − 2×I_void (each void is a rectangle at the same height).
• Double-check: ȳ (from top) + ȳ (from bottom) must equal the total depth of the section.
• On the board exam, T-beam problems are common in combined flexure topics — always expect an unsymmetric section.

Key Points

• For composite (built-up) sections, find the centroid first: ȳ = ΣAᵢȳᵢ / ΣAᵢ.
• Compute I about the NA using the parallel-axis theorem: I_NA = Σ(Ī_i + Aᵢdᵢ²), where dᵢ is the distance from each sub-area's centroid to the overall NA.
• For a T-beam: flange (wide, thin) on top + rectangular web below — treat as two rectangles.
• For a hollow section: compute I of the outer rectangle minus I of the void (same NA for symmetric voids).
• For a circular section: I = πd⁴/64.
• Always set up a table (sub-area, A, ȳ from reference, Aȳ, Ī, Ad², I contribution) to avoid errors.
• The NA of a T-beam is closer to the flange (larger area) — expect it to be in the upper third of the total depth.
• For the board exam, the most common composite sections are: T-beam, I-section (built-up), box section, and circle with hole.

Definitions

Section Title

2. Moment of Inertia and Centroid of Composite Sections

Common Mistakes

• Measuring ȳ for sub-areas from different references (inconsistency in datum) — always use the same reference line (usually the top fiber).
• Forgetting the Ad² term when applying the parallel-axis theorem — leads to gross underestimate of I.
• For a hollow section, subtracting the wrong I (should subtract I of the void about the same NA, using parallel axis if needed).
• For T-beams, ignoring that the NA shifts toward the flange — do NOT assume it is at mid-depth.

Formulas

Exam Tips

• τ_max = 1.5 V/A for rectangles appears in almost every board exam — commit it to memory.
• For a T-beam shear problem, you will usually be asked for τ at the NA or at the flange-web junction — compute a separate Q for each location.
• When the problem gives you a W-section (steel I-beam), use τ ≈ V/A_web unless asked for the exact formula (VQ/Ib).
• Check units before substituting: V in N, Q in mm³, I in mm⁴, b in mm → τ in MPa.
• At the flange-web junction, Q is the first moment of the flange area only (the area above the junction).

Key Points

• Transverse shear force V in a beam produces horizontal shear stress (by complementary shear, also equal to vertical shear stress) at every horizontal plane.
• The shear stress formula: τ = VQ/(Ib), where Q is the first moment of area of the partial section beyond the plane of interest.
• Shear stress distribution is parabolic for a rectangle — maximum at the NA, zero at the top and bottom fibers.
• This is the exact opposite of bending stress distribution (bending stress is max at fibers, zero at NA).
• For a rectangular section: τ_max = 3V/(2A) = 1.5 V/A — a direct formula to memorize.
• For a solid circular section: τ_max = 4V/(3A) — also at the NA.
• For wide-flange (I-beam) sections: ~90–95% of shear is carried by the web; approximate τ_max ≈ V/A_web.
• The shear stress at a specific level (not just the NA) requires computing Q for the area above (or below) that level.
• Units: if V is in N, Q in mm³, I in mm⁴, b in mm, then τ is in N/mm² = MPa.

Definitions

Section Title

3. Horizontal Shear Stress — τ = VQ/(Ib)

Common Mistakes

• Computing Q as the first moment of the ENTIRE section instead of only the portion above (or below) the cut level.
• Using the wrong b — for T-beams, b changes at the flange-web junction; always use the width at the exact level of the cut.
• Forgetting that τ = 0 at the top and bottom fibers (Q = 0 there); shear stress is max at the NA.
• Confusing the 1.5 factor (rectangle) with the 4/3 factor (circle) — memorize both separately.
• For I-sections, the shear stress jumps discontinuously at the flange-web junction because b changes abruptly — be careful at that interface.

Formulas

Exam Tips

• The board exam usually gives a simple rectangular built-up section; Q of the added board = A_board × d_board_to_NA.
• Memorize: q = VQ/I (shear flow), then s = F/q (spacing). Two steps, done.
• If two identical nails are at each cross-section (one on each side), the effective F doubles: s = 2F/q.
• Watch out: some problems ask for the maximum allowable shear V given nail spacing s. Rearrange: V = qI/Q = (F/s)(I/Q).

Key Points

• A built-up beam is fabricated from individual pieces connected by nails, bolts, welds, or adhesive (e.g., plank-and-joist sections, plate girders).
• Shear flow q = VQ/I is the shear force per unit length that the connectors must resist.
• Shear flow has units of N/mm (or kN/m).
• Connector spacing s = F/q, where F is the allowable load per connector (nail, bolt, weld per unit length).
• If there are n connectors per row, the effective capacity per spacing is nF: s = nF/q.
• Q is the first moment of the connected (attached) piece(s) about the NA of the full section.
• Shear flow is constant along the length between connections if V is constant; spacing can be increased where V (and q) is small.
• AISC 360 Section E6 / NSCP 2015 addresses built-up column connectors; similar principles apply to built-up beams.

Definitions

Section Title

4. Shear Flow in Built-Up Beams

Common Mistakes

• Using Q of the full section instead of Q of only the attached (tributary) piece being connected.
• Forgetting to account for the number of connectors per row (if two nails per row, the spacing s = 2F/q, not F/q).
• Treating shear flow as shear stress — they have different units (N/mm vs. MPa); do not confuse.
• Using V_max everywhere — in practice, spacing can be increased at low-shear regions of the beam.

Formulas

Exam Tips

• Always compute M = Pe first, then treat as a bending moment superimposed on the axial stress.
• Quick kern check: compute e and compare to h/6. If e > h/6 → tension occurs — no need to calculate exact stress for this yes/no answer.
• For compression members: use σ = −P/A ± Mc/I (negative P/A since it's compressive).
• This topic bridges with foundation engineering: footing pressure q = P/A ± Mc/I is tested in soil mechanics exams too.
• Be alert for problems that ask 'find the eccentricity such that no tension occurs' — set σ_min = 0 and solve for e.

Key Points

• When a member carries both an axial force P and a bending moment M (from eccentricity e = M/P), the stresses are superimposed.
• σ = P/A ± Mc/I — algebraically add or subtract the two contributions at each extreme fiber.
• P/A is uniform (same everywhere); Mc/I is linear (opposite signs on opposite fibers).
• Maximum fiber stress: σ_max = P/A + Mc/I (fiber on the side toward eccentricity).
• Minimum fiber stress: σ_min = P/A − Mc/I (fiber away from eccentricity).
• Sign convention: compressive P/A is negative; tensile P/A is positive. Be consistent.
• The kern (core) of a section is the region where the eccentric load can act without causing tension anywhere.
• For a rectangle (b × h): kern extends h/6 from the centroid along each axis (middle-third rule).
• For a solid circle of diameter d: kern radius = d/8.
• If e > h/6 (for rectangle), the load is outside the kern → tensile stress develops on one side.
• This is critical for masonry piers, unreinforced concrete footings, and gravity dam design where tension is undesirable.

Definitions

Section Title

5. Combined Axial and Bending Stress (Eccentric Loading)

Common Mistakes

• Not converting M = Pe to N·mm before substituting (especially when P is in kN and e is in mm).
• Getting the sign of P/A wrong — compressive load gives negative (compressive) uniform stress; tensile load gives positive.
• Forgetting that both fibers must be checked: σ_max uses + and σ_min uses − (or vice versa, depending on bending direction).
• Applying the kern rule for a circle (d/8) to a rectangle, or vice versa.
• In combined loading, bending about two axes (biaxial bending): σ = P/A ± M_x c_y / I_x ± M_y c_x / I_y — easy to miss the second moment term.

Formulas

Exam Tips

• Design problems almost always specify b as a fraction of h (e.g., b = 0.5h). Substitute, then solve the cubic: h³ = 6S_req/k where b = kh.
• For rectangular sections, both flexure and shear checks can be done with two simple formulas — S = bh²/6 and τ_max = 1.5V/A.
• Round UP dimensions after solving (e.g., h = 362 mm → use 365 mm or 370 mm). Never round down for design.
• For steel W-sections, check the NSCP/AISC section table in the exam annex — simply pick the lightest W with S_x ≥ S_req.
• If the problem says 'lightest section,' it means the one with the smallest weight (smallest area) that still satisfies S_req.

Key Points

• A complete beam design must satisfy BOTH flexure (bending stress ≤ σ_allow) and shear (τ ≤ τ_allow) requirements.
• Step 1: Construct the SFD/BMD to find V_max and M_max.
• Step 2 (Flexure): S_req = M_max / σ_allow → select a section with S ≥ S_req.
• Step 3 (Shear): Check τ_max = VQ/(Ib) ≤ τ_allow (or use 1.5V/A for rectangles).
• Step 4 (Serviceability): Check deflection (Chapter 5 — not covered here but always required in full design).
• Governing mode: Short, heavily loaded beams → shear governs. Long, moderately loaded beams → flexure (or deflection) governs.
• For timber beams, NSCP 2015 Section 616 provides allowable bending (F_b) and shear (F_v) stresses.
• For steel beams, AISC 360 / NSCP 2015 Section 506 uses F_y-based nominal strengths with φ factors (LRFD) or Ω factors (ASD).
• If b = kh for a rectangular section, substituting into S_req = bh²/6 gives a cubic equation in h — solve for h, then b.

Definitions

Section Title

6. Beam Design for Strength (Flexure and Shear)

Common Mistakes

• Sizing the beam for flexure only and forgetting to check shear — especially for short deep beams.
• For b = 0.5h (or other ratio) problems: setting up S = bh²/6 correctly but making an algebra error when solving for h.
• Using allowable stress for one material (e.g., steel 165 MPa) on a problem that specifies timber (F_b = 10–16 MPa) — always note material.
• Neglecting self-weight of the beam in the moment and shear calculations when it is significant.