CELE Strength of Materials — Stresses in BeamsStudy Notes

The flexure formula is the foundation of beam design. It relates the internal bending moment M to the bending stress σ at any point within a cross-section: σ = (M × y) / I where: • σ = normal (flexural) bending stress at distance y from the neutral axis (in MPa or N/mm²) • M = internal bending moment at the section (in N·mm or kN·m, converted consistently) • y = perpendicular distance from the neutral axis to the point of interest (in mm) • I = second moment of area (moment of inertia) about the neutral (centroidal) axis (in mm⁴) KEY PRINCIPLE: Plane sections remain plane. This assumption (valid for linearly elastic, homogeneous materials like timber, steel, and concrete in the uncracked state) means strain varies linearly through the depth. Since stress = E × strain, stress is also linear. The neutral axis (NA) passes through the centroid of the cross-section and is the surface of zero stress and zero strain. For a SAGGING (positive) moment (concave-up curvature): • The bottom fiber is in TENSION (maximum tensile stress at the extreme bottom fiber) • The top fiber is in COMPRESSION (maximum compressive stress at the extreme top fiber) For a HOGGING (negative) moment (concave-down curvature), the stresses reverse. Maximum bending stress occurs at the extreme fiber, distance c from the neutral axis: σ_max = (M × c) / I This is rewritten as: σ_max = M / S where S = I / c is the SECTION MODULUS (a single descriptor of bending efficiency, in mm³). IMPORTANT: The neutral axis is located at the CENTROID for symmetric and unsymmetric sections alike. For unsymmetric sections (T-beams, L-beams, angles), the centroid is NOT at mid-depth, so the extreme fiber distances differ (c_top ≠ c_bottom), meaning max tensile and max compressive stresses are unequal even for the same moment magnitude.

The section modulus S = I / c encapsulates a cross-section's resistance to bending. It is the KEY quantity for flexural design: σ_max = M / S ⟹ S_required = M_max / σ_allow Once the required section modulus is known, you select a standard section or size a custom one. This principle underpins design tables in AISC 360 (steel), ACI 318 (concrete), and timber design standards. COMMON SECTION MODULI: 1. RECTANGLE (width b, depth h): I = bh³ / 12 c = h / 2 S = bh² / 6 ← Most-used formula on the board exam Note: S depends on depth SQUARED. Doubling the depth increases S by 4×, explaining why beams are "tall and narrow" for bending efficiency. 2. SOLID CIRCLE (diameter d): I = πd⁴ / 64 c = d / 2 S = πd³ / 32 3. HOLLOW CIRCLE (outer diameter D, inner diameter d): I = π(D⁴ – d⁴) / 64 S = πd⁴ – d⁴) / (32 × D/2) [use the outer radius for c] 4. WIDE-FLANGE I-SECTION (W-section, standard in steel): I and S are provided in design tables (e.g., AISC Steel Manual). For hand calculations, approximate I ≈ A × d² / 12 if flanges dominate. 5. T-BEAM and L-BEAM (unsymmetric): Must locate the centroid first, then compute I about the centroid, then S = I / c_top and S = I / c_bot separately (different values). DESIGN PROCEDURE FOR FLEXURE: 1. Analyze the beam under given loads to find M_max. 2. Calculate S_required = M_max / σ_allow. 3. If cross-section is to be designed: a. Assume a shape (rectangle, circle, etc.). b. Express S in terms of unknown dimension(s). c. Solve for the dimension(s). 4. If selecting a standard section: a. Look up sections in design tables (AISC, timber species tables). b. Choose the smallest section with S ≥ S_required (to minimize weight/cost). IMPORTANT: Always check that your choice also satisfies shear requirements (Section 3), deflection limits (Chapter 5), and any other serviceability criteria per NSCP 2015.

Transverse shear forces (V) produce shear stresses on both horizontal and vertical planes. By the principle of complementary shear, if there is vertical shear stress on a horizontal plane, there must be equal horizontal shear stress on a vertical plane. Unlike bending stress, shear stress distribution is NOT linear through the cross-section. THE HORIZONTAL SHEAR STRESS FORMULA: τ = (V × Q) / (I × b) where: • τ = horizontal (or vertical) shear stress at the level of interest (in MPa or N/mm²) • V = internal shear force at the section (in N or kN, consistent units) • Q = first moment of area (first moment of inertia) of the part of the cross-section above (or below) the level where τ is evaluated, computed about the neutral axis (in mm³) • I = second moment of area (moment of inertia) of the entire cross-section about the neutral axis (in mm⁴) • b = width (thickness) of the cross-section at the level where τ is calculated (in mm) DISTRIBUTION OF SHEAR STRESS: Unlike bending stress (which is maximum at the outer fibers and zero at the NA), shear stress is: • ZERO at the top and bottom extreme fibers (free surfaces, no traction) • MAXIMUM at the neutral axis (where Q is maximum) • Varies parabolically through the depth for a rectangular section For rectangular sections, this produces a parabolic profile, so max shear stress ≠ average shear stress. The maximum shear stress is 1.5 times the average: τ_max = (3V) / (2A) = 1.5 × (V / A), where A = b × h For circular sections (solid): τ_max = (4V) / (3A) For I-beams and similar shapes with a thin web: • Nearly all shear is carried by the web. • Useful approximate formula: τ_max ≈ V / A_web (web area only). • The flanges contribute very little; their large width (b) at the flange levels means Q is small near the top and bottom, so τ remains small there. DESIGN FOR SHEAR: After checking flexure (bending), verify that the maximum shear stress does not exceed the allowable shear stress per the applicable code: τ_max ≤ τ_allow For timber (NSCP 2015): τ_allow is typically 0.5–1.5 MPa, depending on species and grade. For steel (AISC 360): τ_allow is typically 0.4 × F_y (yield strength). For concrete (ACI 318): τ limit and reinforcement are more complex (nominal shear strength V_n); concrete carries part, shear reinforcement (stirrups) carries the rest. SHORT, HEAVILY LOADED BEAMS → often shear-governed. LONG, LIGHTLY LOADED BEAMS → usually flexure or deflection-governed.

A built-up beam is fabricated by joining separate pieces (planks, plates, etc.) using fasteners (nails, bolts, screws, welds, or adhesive). Examples include laminated timber joists, plate girders, and composite beams. The question is: how many fasteners, and at what spacing, are needed to transmit shear between the connected parts? SHEAR FLOW (q): The shear flow is the shear force per unit length along the beam: q = (V × Q) / I [units: N/mm or kN/m] Note: This is similar to the shear stress formula but WITHOUT the factor of 1/b. The shear flow represents the total horizontal shear that must be transferred across a given longitudinal plane per unit length. DESIGN OF FASTENER SPACING: If one fastener (nail, bolt, weld, etc.) can carry a capacity F_fastener (in N), the spacing s between fasteners is: s = F_fastener / q Example: If q = 5 N/mm and each nail can carry 500 N, then s = 500 / 5 = 100 mm (nails spaced 100 mm apart along the beam). APPLICATIONS: 1. TIMBER BEAMS with side planks or flange plates nailed or bolted on. 2. STEEL PLATE GIRDERS: flanges welded to the web; the weld size and spacing transmit shear flow. 3. COMPOSITE BEAMS (timber + steel, or precast + cast-in-place concrete): fasteners or mechanical interlock must transfer the shear flow. 4. BOX BEAMS and other fabricated sections. PRACTICAL CONSIDERATIONS: • Fasteners must be distributed along the beam, placed closer together where V (and thus q) is large. • Often, fasteners near the supports are more closely spaced (large V); fewer fasteners are needed near midspan (small V). • Per NSCP 2015 (for timber) and similar codes, minimum edge distance and end distance are specified for fasteners. • For welded connections, weld size (leg length) and spacing are designed so the weld strength ≥ required shear flow.

Many structural members carry both an axial force P and a bending moment M simultaneously. This occurs in: • Short columns with eccentric loads (P applied off-center at distance e, creating M = P × e). • Members in frames under combined axial compression and lateral loads. • Footings under eccentric loads. STRESS SUPERPOSITION: For a linearly elastic material, the axial stress and bending stress superpose (add algebraically): σ = σ_axial + σ_bending = (P / A) ± (M × c / I) Sign convention: • Compression: P/A is positive (or negative, depending on convention; here we use positive for compression). • If M is sagging (concave up) and c is at the bottom fiber, then M × c / I is tensile (subtracted from axial). More commonly written as: σ = (P / A) ± (M / S) The ± accounts for the two extreme fibers (top and bottom). For eccentric loading (P at distance e from the centroid), M = P × e: σ = (P / A) ± (P × e × c / I) = (P / A) ± (P × e / S) EXTREME FIBER STRESSES: σ_max = (P / A) + (M × c / I) [top, if M is sagging] σ_min = (P / A) – (M × c / I) [bottom, if M is sagging] If σ_min becomes negative (tension), the member develops tension on one side. THE KERN (CORE) OF A SECTION: The kern is the region within which an axial load can be applied without producing tension anywhere in the cross-section. If a load is applied outside the kern, one extreme fiber will see tension. For a RECTANGULAR section: • Distance from centroid to extreme fiber: c = h / 2 • Kern boundary: e_kern = I / (A × c) = (A × c² / 12) / (A × c) = c / 6 = h / 12 • Kern extends from –h/6 to +h/6 from the centroid → the middle-third rule (load must be in the middle third of the depth). For a CIRCULAR section: • Kern radius: r_kern = I / (A × c) = (πd⁴ / 64) / (πd² / 4 × d/2) = d / 8. For a TRIANGULAR section (apex up): • Kern extends from the centroid to h / 9 outward (h / 18 on each side of centroid, but the calculation is more complex). APPLICATIONS: 1. MASONRY AND CONCRETE FOOTINGS: Tension is undesirable (mortar and concrete are weak in tension). The design ensures the resultant load stays within the kern, avoiding tension. 2. SHORT COLUMNS: Eccentric loads can cause tension on the side away from the load. 3. ECCENTRICALLY LOADED BRACKETS: Combined bending and axial load. DESIGN PROCEDURE: 1. Compute σ at both extreme fibers. 2. If tension is unacceptable (masonry, concrete), ensure σ_min ≥ 0, i.e., load within the kern. 3. If tension is acceptable (steel, timber), design for σ_max (tension or compression, whichever is larger in magnitude).

Designing a beam for strength involves multiple checks. Here is a systematic approach aligned with NSCP 2015, ACI 318, and AISC 360: DESIGN CHECKLIST: 1. LOAD ANALYSIS: Draw shear and bending-moment diagrams. Identify M_max and V_max. 2. FLEXURAL DESIGN (usually governs): a. Calculate S_required = M_max / σ_allow. b. Select or size a section to meet S_required. c. Record S_actual ≥ S_required. 3. SHEAR DESIGN: a. Calculate τ_max at the section of maximum shear (usually at supports). b. Check τ_max ≤ τ_allow. c. If V-governs (unlikely in typical beams), redesign with a wider or deeper section, focusing on increasing I or decreasing Q (less likely). 4. DEFLECTION CHECK (Chapter 5): Deflection δ_max should not exceed span/240 or span/360 (per NSCP 2015 or ACI 318). Formula (simply supported, UDL): δ_max = (5 × w × L⁴) / (384 × E × I) If δ exceeds the limit, increase I (deeper section, higher E material). 5. CONNECTION/FASTENER DESIGN (if built-up): Verify bolts, welds, or nails carry the shear flow. 6. SERVICEABILITY CHECKS: • Ponding (roof beams): ensure drainage or provide stiffness. • Vibration (floor beams): check frequency if dynamic loads present. • Cracking (concrete beams): limit crack width per ACI 318. COMMON BOARD-EXAM PITFALLS: • **Pitfall 1: Using the wrong I formula.** Students confuse I for different shapes. Use I = bh³ / 12 ONLY for rectangles with the axis parallel to b and perpendicular to h. For other shapes, look up or derive carefully (parallel-axis theorem for composite sections). • **Pitfall 2: Forgetting to locate the centroid for unsymmetric sections.** For a T-beam or L-beam, if the centroid is not at mid-depth, then c_top ≠ c_bot, and the max stresses differ even for the same M. Many students calculate S using the wrong c. • **Pitfall 3: Applying the wrong Q in τ = VQ / (Ib).** The first moment Q is the moment of the area ABOVE (or BELOW) the level, NOT INCLUDING the level itself. At the neutral axis, Q_at_NA = (half the cross-section area) × (distance from NA to its own centroid), not the full area. • **Pitfall 4: Using the 1.5 factor for non-rectangular sections.** The τ_max = 1.5 × V/A formula applies ONLY to rectangles. For circles, use 4V/(3A). For I-beams or other shapes, use the general formula τ = VQ / (Ib). • **Pitfall 5: Unit inconsistency.** If M is in kN·m, convert to N·mm (1 kN·m = 10⁶ N·mm). If I is in mm⁴, then σ comes out in N/mm² (MPa). Mixing units (e.g., M in kN·m and I in cm⁴) leads to errors by factors of 10 or 100. • **Pitfall 6: Sign confusion in combined axial + bending.** Keep track of which fiber is in tension and which in compression. A well-drawn stress diagram (linear through depth) prevents errors. • **Pitfall 7: Forgetting to check BOTH flexure and shear.** Some students design for flexure and assume shear is OK. In short beams or beams with concentrated loads near supports, shear can govern. • **Pitfall 8: Misunderstanding the kern rule.** The kern/middle-third rule applies to sections where you want to ensure NO TENSION ANYWHERE (masonry, unreinforced concrete). For reinforced concrete, the kern doesn't constrain design; instead, the reinforcement is sized for the actual tension. • **Pitfall 9: Over-designing or under-designing.** Select the SMALLEST section meeting S_required (and other checks) to minimize cost and weight. However, ensure all design criteria are met; safety margin comes from factors of safety in the allowable stresses, not from excess section size. • **Pitfall 10: Neglecting lateral-torsional buckling of steel beams.** While outside the scope of "stresses in beams" per se, slender beams can fail by buckling before reaching the yield stress. AISC 360 Section F provides limits on unbraced length. REAL-WORLD DESIGN EXAMPLE (TIMBER BEAM UNDER NSCP 2015): Design a simply supported timber beam to carry a floor live load of 2.4 kPa (typical office) over a 5 m span, with 3 m spacing between beams (tributary width = 3 m). • Distributed load: w = (dead load 1.2 kPa + live 2.4 kPa) × 3 m = 10.8 kN/m • M_max = wL² / 8 = 10.8 × 25 / 8 = 33.75 kN·m ≈ 34 kN·m • V_max = wL / 2 = 10.8 × 5 / 2 = 27 kN • Allowable stress (Grade 1 softwood): σ_b = 10 MPa (adjust for moisture, load duration per NSCP) • Allowable shear: τ = 1.2 MPa Step 1: Size for flexure. S_req = 34 × 10⁶ N·mm / 10 N/mm² = 3.4 × 10⁶ mm³ For a rectangular section b × h with b = 100 mm (standard dimension): S = 100 × h² / 6 = 3.4 × 10⁶ ⟹ h² = 204,000 ⟹ h = 452 mm Choose h = 450 mm (nearest standard size below) or 475 mm (if available). With 100 × 450 mm: S_actual = 100 × 450² / 6 = 100 × 202,500 / 6 = 3.375 × 10⁶ mm³ ✓ (close to requirement) Step 2: Check shear. A = 100 × 450 = 45,000 mm² τ_max = 1.5 × 27,000 / 45,000 = 0.9 MPa < 1.2 MPa ✓ Step 3: Check deflection (preliminary; detailed in Chapter 5). δ_max = 5 × 10.8 × 5000⁴ / (384 × 8,000 × 3.375 × 10⁶) = (5 × 10.8 × 6.25 × 10¹¹) / (384 × 2.7 × 10¹⁰) = 3.375 × 10¹³ / 1.037 × 10¹³ ≈ 32.5 mm L / 240 = 5,000 / 240 = 20.8 mm < 32.5 mm (FAILS) L / 360 = 5,000 / 360 = 13.9 mm < 32.5 mm (FAILS) Deflection governs! Increase depth. With h = 550 mm: I ∝ h³, so I_new / I_old = (550/450)³ ≈ 1.35 δ_new ≈ 32.5 / 1.35 ≈ 24 mm > L/240 (still marginal). Try h = 600 mm. With h = 600 mm: (600/450)³ ≈ 1.78, so δ ≈ 32.5 / 1.78 ≈ 18.3 mm < L/240 ✓ FINAL SELECTION: 100 × 600 mm timber beam (or nearest available size: 4 × 24 inch nominal = 100 × 600 mm approx). This example shows that DEFLECTION often governs long timber spans, not flexure or shear.