Stresses in Beams - Detailed Explanation
Stresses in beams form the analytical backbone of structural design. Once the internal forces — bending moment M and shear force V — are determined from equilibrium (SFD/BMD), the next step is translating those forces into actual stresses that the material must resist. This chapter covers the flexure formula (σ = My/I), the shear stress formula (τ = VQ/Ib), section modulus, shear flow in built-up sections, and combined axial-plus-bending stress. These topics appear consistently in the PRC Civil Engineer Licensure Examination under Strength of Materials (formerly Engineering Mechanics — Mechanics of Deformable Bodies) and underpin the reinforced concrete (NSCP 2015 / ACI 318-19), structural steel (NSCP / AISC 360-16), and timber design topics on the same board exam. Mastery here is non-negotiable.
The Flexure (Bending) Formula
Examples
A simply supported timber beam 150 mm wide × 300 mm deep spans 4 m and carries a UDL of 6 kN/m. Determine the maximum bending stress.
Solution
Step 1 — Maximum bending moment:
M_max = wL²/8 = 6(4)²/8 = 12 kN·m = 12 × 10⁶ N·mm
Step 2 — Section modulus for rectangle (b = 150 mm, h = 300 mm):
S = bh²/6 = 150(300)²/6 = 2.25 × 10⁶ mm³
Step 3 — Maximum bending stress:
σ_max = M/S = 12 × 10⁶ / 2.25 × 10⁶ = 5.33 MPa
This is a straightforward flexure formula application. Always verify units first. The 12 kN·m moment must be converted to N·mm before dividing by S in mm³ to obtain stress in MPa. The result (5.33 MPa) is well within typical allowable bending stress for structural timber (≈ 8–12 MPa per NSCP 2015 Table 616.3-1), so the beam is adequate in flexure.
A T-beam has a 200 mm wide × 50 mm thick flange on top of a 50 mm wide × 200 mm deep web (total depth 250 mm). For a sagging moment M = 30 kN·m, find the top and bottom fiber stresses.
Solution
Step 1 — Locate the neutral axis (measure ȳ from top):
A_flange = 200 × 50 = 10,000 mm² at ȳ₁ = 25 mm
A_web = 50 × 200 = 10,000 mm² at ȳ₂ = 50 + 200/2 = 150 mm
ȳ = (10,000 × 25 + 10,000 × 150) / 20,000 = 87.5 mm from the top
Step 2 — Extreme fiber distances:
c_top = 87.5 mm
c_bot = 250 − 87.5 = 162.5 mm
Step 3 — Moment of inertia about NA (parallel-axis theorem):
Flange: I_f = (200 × 50³)/12 + 10,000 × (87.5 − 25)² = 2.083 × 10⁶ + 39.063 × 10⁶ = 41.146 × 10⁶ mm⁴
Web: I_w = (50 × 200³)/12 + 10,000 × (150 − 87.5)² = 33.333 × 10⁶ + 39.063 × 10⁶ = 72.396 × 10⁶ mm⁴
I_total = 41.146 × 10⁶ + 72.396 × 10⁶ = 113.54 × 10⁶ mm⁴
Step 4 — Fiber stresses (M = 30 × 10⁶ N·mm, sagging):
σ_top = M × c_top / I = 30 × 10⁶ × 87.5 / 113.54 × 10⁶ = 23.1 MPa (compression)
σ_bot = M × c_bot / I = 30 × 10⁶ × 162.5 / 113.54 × 10⁶ = 42.9 MPa (tension)
The T-section has its centroid shifted toward the flange (only 87.5 mm from the top vs. 162.5 mm to the bottom). As a result, the bottom tension stress (42.9 MPa) is almost double the top compression stress (23.1 MPa). This is why concrete T-beams need significant tensile steel at the bottom — the concrete alone cannot resist that tension. This scenario is a perennial board-exam favorite.
When a beam is subjected to a bending moment M, the cross-section rotates about the neutral axis (NA). The fundamental assumption — plane sections remain plane after bending (Euler-Bernoulli hypothesis) — means that strain varies linearly through the depth. For a linearly elastic material, stress is proportional to strain (Hooke's Law), so bending stress σ also varies linearly through the depth:
σ = My / I
where:
- σ = bending (flexural) stress at distance y from the neutral axis (MPa)
- M = bending moment at the section (N·mm)
- y = distance from the neutral axis to the fiber of interest (mm)
- I = second moment of area (moment of inertia) of the entire cross-section about the neutral (centroidal) axis (mm⁴)
The neutral axis passes through the centroid of the cross-section. At the NA, y = 0 and σ = 0. At the extreme fiber (top or bottom), y = c (the maximum distance from NA to the outermost fiber), so:
σ_max = Mc / I
Sign convention (sagging positive moment):
- Bottom fiber → TENSION (positive y below NA)
- Top fiber → COMPRESSION (negative in beam convention)
For an UNSYMMETRIC section (e.g., T-beam), the distance from the NA to the top fiber (c_top) differs from the distance to the bottom fiber (c_bot). Therefore, σ_top ≠ σ_bot — this is a classic board-exam scenario.
The neutral axis location is found by computing the centroid:
ȳ = ΣAᵢyᵢ / ΣAᵢ
The moment of inertia is computed using the parallel-axis theorem:
I_NA = Σ(I_own + Aᵢdᵢ²)
where dᵢ is the distance from each sub-area centroid to the section NA.
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.
Common 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
Example
A W200×100 steel section (S_x = 990 × 10³ mm³) carries M = 120 kN·m. σ_max = 120 × 10⁶ / 990 × 10³ = 121 MPa.
Approach
- Find M_max from SFD/BMD. 2. Compute S = I/c (or use formulas: S_rect = bh²/6; S_circle = πd³/32). 3. σ_max = M/S. Watch units.
Example
T-beam with unequal flanges — compute centroid, then top and bottom stresses separately.
Approach
- Divide section into rectangles. 2. Compute ȳ = ΣAy/ΣA from a reference edge. 3. Compute I_NA using parallel-axis. 4. Find c_top, c_bot. 5. Apply σ = Mc/I separately for each fiber.
Example
For M_max = 40 kN·m and σ_allow = 10 MPa: S_req = 4.0 × 10⁶ mm³. For b = h/2: solve h³ = 12 × 4.0 × 10⁶ → h = 363 mm.
Approach
- Establish S_req = M_max / σ_allow. 2. Express S in terms of unknown dimension (e.g., S = bh²/6). 3. Solve for the dimension.
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).
Section Modulus
Examples
Select the minimum dimensions of a rectangular timber beam (b = h/3) for M_max = 54 kN·m and σ_allow = 12 MPa.
Solution
Step 1 — Required section modulus:
S_req = M_max / σ_allow = 54 × 10⁶ / 12 = 4.5 × 10⁶ mm³
Step 2 — Express S in terms of h (with b = h/3):
S = bh²/6 = (h/3)(h²)/6 = h³/18
Step 3 — Solve for h:
h³ = 18 × 4.5 × 10⁶ = 81 × 10⁶
h = (81 × 10⁶)^(1/3) = 433 mm
b = h/3 = 144 mm
Round up to standard timber dimensions: b = 150 mm, h = 450 mm (check new S ≥ S_req):
S_actual = 150(450)²/6 = 5.0625 × 10⁶ mm³ > 4.5 × 10⁶ ✓
This is a typical design problem. Express S as a function of the unknown dimension, set equal to S_req, and solve algebraically. Always round up to the next standard size and verify.
A solid circular timber beam is needed for M_max = 9 kN·m and σ_allow = 8 MPa. Find the required diameter d.
Solution
Step 1 — Required section modulus:
S_req = 9 × 10⁶ / 8 = 1.125 × 10⁶ mm³
Step 2 — Section modulus for circle:
S = πd³/32
Step 3 — Solve for d:
d³ = 32 × 1.125 × 10⁶ / π = 11.459 × 10⁶
d = (11.459 × 10⁶)^(1/3) = 225.6 mm
Use d = 230 mm (next standard size).
For circular sections, always use S = πd³/32. Cube both sides after isolating d³. A 225 mm diameter round timber is a common answer in PRC review materials.
The section modulus S = I/c is a single geometric property that consolidates the flexural efficiency of a cross-section. It eliminates the need to track both I and c separately when computing maximum bending stress:
σ_max = M/S
For a RECTANGULAR section (width b, depth h):
I = bh³/12, c = h/2
S = bh²/6
For a SOLID CIRCULAR section (diameter d):
I = πd⁴/64, c = d/2
S = πd³/32
For STANDARD STEEL SECTIONS (W, S, C shapes), S_x values are tabulated in AISC Steel Construction Manual (also NSCP Part II steel tables). The Philippine Board Exam often provides these directly in the problem.
Why S matters for design:
Given an allowable bending stress σ_allow (from material specifications or NSCP), the required section modulus is:
S_req = M_max / σ_allow
Then select any cross-section whose actual S ≥ S_req. This is the most efficient beam-design shortcut on the exam.
Importance of depth:
For a rectangle, S = bh²/6 grows with the SQUARE of depth h. Doubling the depth (while keeping b constant) quadruples S — which is why increasing beam depth is far more effective than increasing width. This is why I-beams concentrate material at the flanges (maximum y from the NA), making them highly efficient.
Note on unsymmetric sections:
For sections not symmetric about the bending axis (e.g., T-beam), there are TWO section moduli:
S_top = I / c_top and S_bot = I / c_bot
The smaller one (for the more critical fiber) governs design.
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.
Common 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
Example
Rectangular beam 100 × 200 mm, M = 6 kN·m: S = 100(200)²/6 = 666,667 mm³; σ_max = 6 × 10⁶/666,667 = 9 MPa.
Approach
Identify cross-section shape → use S formula → σ_max = M/S.
Example
b = 2h/3, M = 20 kN·m, σ_allow = 10 MPa → S_req = 2.0 × 10⁶ mm³ → (2h/3)h²/6 = h³/9 = 2 × 10⁶ → h = 261 mm.
Approach
S_req = M/σ_allow → express S in terms of unknown → solve algebraically → round up.
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.
Horizontal Shear Stress in Beams
Examples
For the same 150 mm × 300 mm timber beam carrying 6 kN/m on 4 m span, find the maximum horizontal shear stress.
Solution
Step 1 — Maximum shear force:
V_max = wL/2 = 6 × 4/2 = 12 kN = 12,000 N
Step 2 — Cross-sectional area:
A = 150 × 300 = 45,000 mm²
Step 3 — Maximum shear stress (rectangular formula):
τ_max = 3V / (2A) = 3(12,000) / (2 × 45,000) = 36,000 / 90,000 = 0.40 MPa
Verification using VQ/Ib:
I = 150(300)³/12 = 337.5 × 10⁶ mm⁴
Q_max = A'ȳ' = [150 × 150] × 75 = 1,687,500 mm³ (area above NA × its centroid distance from NA)
τ_max = VQ/(Ib) = 12,000 × 1,687,500 / (337.5 × 10⁶ × 150) = 0.40 MPa ✓
Both methods give the same answer. Use τ_max = 1.5V/A for rectangles as a quick check — but you must be comfortable with VQ/Ib for non-rectangular sections and for shear stress AT SPECIFIC LEVELS (not just the maximum).
A 250 mm wide × 400 mm deep rectangular beam carries V = 80 kN. Find the shear stress at (a) the neutral axis, and (b) 100 mm from the neutral axis (i.e., at the mid-depth of either half).
Solution
Given: b = 250 mm, h = 400 mm, V = 80,000 N
I = 250(400)³/12 = 1,333.33 × 10⁶ mm⁴
A = 250 × 400 = 100,000 mm²
(a) At the neutral axis (y = 0 from NA):
Area above NA: A' = 250 × 200 = 50,000 mm²
Centroid of A' from NA: ȳ' = 100 mm
Q_NA = 50,000 × 100 = 5,000,000 mm³
τ_NA = VQ/(Ib) = 80,000 × 5,000,000 / (1,333.33 × 10⁶ × 250) = 1.20 MPa
Check: τ_max = 1.5V/A = 1.5 × 80,000/100,000 = 1.20 MPa ✓
(b) At 100 mm from the NA (y = 100 mm — i.e., mid-depth of the upper half):
Area above this cut: A'' = 250 × (200 − 100) = 250 × 100 = 25,000 mm²
Centroid of A'' from NA: ȳ'' = 100 + 50 = 150 mm
Q = 25,000 × 150 = 3,750,000 mm³
τ = 80,000 × 3,750,000 / (1,333.33 × 10⁶ × 250) = 0.90 MPa
Note: τ at y=100 mm (0.90 MPa) < τ at NA (1.20 MPa), confirming the parabolic distribution.
Part (b) illustrates the parabolic shear stress variation. The shear stress reduces as you move away from the NA. This is a common board-exam scenario where shear stress at an intermediate level is requested, not just the maximum.
When a beam carries a transverse shear force V, shear stresses develop on both the transverse (vertical) and longitudinal (horizontal) planes — they are equal in magnitude at any point (principle of complementary shear). The shear stress formula is:
τ = VQ / (Ib)
where:
- V = shear force at the section (N)
- Q = first moment of area of the cross-sectional area ABOVE (or BELOW) the level of interest, computed about the neutral axis (mm³)
- I = moment of inertia of the ENTIRE cross-section about the NA (mm⁴)
- b = width of the section AT the level where τ is computed (mm)
- τ = shear stress at that level (MPa)
How to compute Q:
For any area A' above (or below) the cut level:
Q = A' × ȳ'
where ȳ' is the distance from the centroid of A' to the neutral axis of the entire section.
Shear stress distribution:
- At the top and bottom EXTREME FIBERS: Q = 0, so τ = 0
- At the NEUTRAL AXIS: Q is maximum, so τ is maximum
- Distribution is PARABOLIC for a rectangular section
Special cases:
- RECTANGULAR section (b × h):
τ_max = 3V / (2A) = 1.5 V/A (at the NA)
This factor 1.5 must be memorized for boards.
- SOLID CIRCULAR section (diameter d, area A):
τ_max = 4V / (3A) ≈ 1.333 V/A (at the NA)
- WIDE-FLANGE (I-beam) section:
Nearly all shear is carried by the WEB. A useful approximation:
τ_avg,web ≈ V / A_web where A_web = t_w × d
Note on WHERE shear governs:
- Short, heavily loaded beams → shear critical
- Long beams with moderate loads → flexure or deflection critical
- Timber beams: horizontal shear (τ = VQ/Ib) is the primary shear check per NSCP 2015 Section 616.3 because wood is weak in shear parallel to grain.
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.
Common 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
Example
V = 60 kN, circular solid section d = 200 mm. A = π(200)²/4 = 31,416 mm². τ_max = (4/3)(60,000)/31,416 = 2.55 MPa.
Approach
Use τ_max = 1.5V/A (rectangle) or τ_max = (4/3)V/A (circle). Identify V from SFD.
Example
I-beam: find τ at flange-web junction. A' = flange area; ȳ' = distance from flange centroid to beam NA; b = web thickness.
Approach
- Identify the cut level (distance from NA). 2. Compute A' and ȳ' for the area beyond the cut. 3. Q = A'ȳ'. 4. τ = VQ/(Ib). Use b at the cut level.
Example
Timber beam: τ_max = 0.40 MPa ≤ Fv = 0.70 MPa (for Apitong per NSCP). Safe in shear.
Approach
Compute τ_max = VQ/Ib (or 1.5V/A for rect.) and compare with τ_allow. If τ_max > τ_allow, the section fails in shear.
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.
Shear Flow in Built-Up Beams
Examples
A built-up T-beam consists of a 100 mm × 25 mm flange nailed to a 25 mm × 150 mm web. The beam is loaded so that V = 5 kN. Each nail has a shear capacity of 400 N. Find (a) the shear flow at the flange-web junction, and (b) the required nail spacing.
Solution
Step 1 — Locate the NA of the entire section (ȳ from top):
A_flange = 100 × 25 = 2,500 mm² at ȳ₁ = 12.5 mm
A_web = 25 × 150 = 3,750 mm² at ȳ₂ = 25 + 75 = 100 mm
ȳ = (2,500 × 12.5 + 3,750 × 100) / 6,250 = (31,250 + 375,000) / 6,250 = 64.5 mm from top
Step 2 — Moment of inertia about NA:
Flange: I_f = (100 × 25³)/12 + 2,500(64.5 − 12.5)² = 130,208 + 6,760,000 = 6,890,208 mm⁴
Web: I_w = (25 × 150³)/12 + 3,750(100 − 64.5)² = 7,031,250 + 4,725,938 = 11,757,188 mm⁴
I_total = 6,890,208 + 11,757,188 = 18,647,396 mm⁴ ≈ 18.647 × 10⁶ mm⁴
Step 3 — Q for the flange (connected area about the whole NA):
Q = A_flange × (ȳ_NA − ȳ_flange centroid) = 2,500 × (64.5 − 12.5) = 2,500 × 52 = 130,000 mm³
Step 4 — Shear flow:
q = VQ/I = 5,000 × 130,000 / 18.647 × 10⁶ = 34.86 N/mm
Step 5 — Nail spacing (n = 1 nail per spacing):
s = F/q = 400/34.86 = 11.5 mm
Note: This is very close spacing — likely two nails per spacing would give s = 2 × 400/34.86 = 23 mm. In practice, minimum spacing per NSCP 2015 Section 624 governs if the calculated spacing is impractically small.
The shear flow formula is the key. Note that Q is only for the flange (the piece being connected), not for the whole section. The very small spacing indicates the beam experiences high shear — a good reason to check whether the design load is realistic or whether the section needs to be enlarged.
A box beam is made of four 50 × 200 mm planks. For V = 6 kN, Q = 1.5 × 10⁵ mm³, I = 4.0 × 10⁷ mm⁴, and each nail capacity F = 500 N, find the nail spacing if two nails are used at each joint (n = 2).
Solution
q = VQ/I = 6,000 × 150,000 / 40,000,000 = 22.5 N/mm
s = nF/q = 2 × 500 / 22.5 = 44.4 mm
Use s = 40 mm (round down for safety).
With two nails per spacing, the spacing doubles compared to one nail. Always round DOWN the computed spacing (smaller spacing = more connectors = safer). Check also minimum and maximum nail spacing per NSCP 2015 Section 624.
When a beam is fabricated by joining two or more pieces (planks nailed together, a plate welded to a W-section, a box beam glued from lumber), the connectors (nails, bolts, welds, glue) at the joints must transfer the horizontal shear forces between the components. The intensity of this horizontal shear per unit length is called SHEAR FLOW, q:
q = VQ / I (in N/mm or N/m)
where:
- V = shear force at the section (N)
- Q = first moment of area of the CONNECTED COMPONENT about the NA of the whole section (mm³)
- I = moment of inertia of the ENTIRE built-up section about the NA (mm⁴)
Note the difference from the shear stress formula: shear flow q does not have b in the denominator because it is a force per unit LENGTH (not a stress).
Connector spacing:
If each connector (nail, bolt) has an allowable shear capacity F (N), and there are n connectors per pitch spacing s, the shear flow compatibility gives:
s = nF / q
For a single row of connectors (n = 1):
s = F / q
For two connectors per spacing (n = 2, e.g., two nails at each connection point on a box beam with two sides):
s = 2F / q
Practical rules:
- Compute I of the entire section (using parallel-axis).
- Identify the connected component (the piece being attached).
- Compute Q = A_connected × ȳ_connected-centroid (from the whole section's NA).
- q = VQ/I.
- s = nF/q.
This principle applies to:
- Nailed/bolted timber built-up beams (common in Philippine construction)
- Welded plate girders (weld size from shear flow per AISC 360)
- Glued laminated timber (glulam) joints
- Composite steel-concrete beams (shear stud spacing per AISC 360 Chapter I)
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.
Common 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
Example
Board-exam style: 'A T-section built-up beam is loaded with V = 8 kN. Q = 2 × 10⁵ mm³, I = 5 × 10⁷ mm⁴, nail capacity = 600 N. Find nail spacing.' → q = 8,000 × 200,000/50,000,000 = 32 N/mm; s = 600/32 = 18.75 mm → use 18 mm.
Approach
- Compute I of entire section. 2. Compute Q for connected component. 3. q = VQ/I. 4. s = nF/q. Round down.
Example
s = 50 mm, n = 1, F = 800 N, I = 3 × 10⁷ mm⁴, Q = 1.2 × 10⁵ mm³. q = 800/50 = 16 N/mm. V = 16 × 3 × 10⁷/1.2 × 10⁵ = 4,000 N = 4 kN.
Approach
Reverse of above: given s and F, find q = nF/s. Then V = qI/Q.
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.
Combined Flexure and Axial Load (Eccentric Loading)
Examples
A short rectangular post 200 mm × 200 mm carries P = 300 kN applied 40 mm off-center. Find the extreme fiber stresses. Does tension occur?
Solution
Step 1 — Section properties:
A = 200 × 200 = 40,000 mm²
I = 200(200)³/12 = 133.33 × 10⁶ mm⁴
c = 100 mm
Step 2 — Induced moment:
M = Pe = 300,000 × 40 = 12 × 10⁶ N·mm
Step 3 — Direct axial stress (compression):
σ_axial = P/A = 300,000/40,000 = 7.5 MPa (compression)
Step 4 — Bending stress at extreme fiber:
σ_bending = Mc/I = 12 × 10⁶ × 100/133.33 × 10⁶ = 9.0 MPa
Step 5 — Combined stresses:
σ_max = 7.5 + 9.0 = 16.5 MPa (compression, on the side of load application)
σ_min = 7.5 − 9.0 = −1.5 MPa (TENSION on the far side)
Step 6 — Kern check:
h/6 = 200/6 = 33.3 mm
Since e = 40 mm > 33.3 mm (kern limit), tension IS expected — confirmed by σ_min = −1.5 MPa.
The 7.5 MPa axial compression is not enough to overcome the 9.0 MPa bending tension on the far side, resulting in net tension. If this post were unreinforced concrete or masonry, this would be unacceptable. Moving the load within the kern (e ≤ 33.3 mm) would eliminate tension.
A 300 mm × 300 mm short concrete column carries 500 kN at an eccentricity of 60 mm. Find extreme fiber stresses and state whether tension occurs.
Solution
A = 300 × 300 = 90,000 mm²
I = 300(300)³/12 = 675 × 10⁶ mm⁴
c = 150 mm
M = Pe = 500,000 × 60 = 30 × 10⁶ N·mm
σ_axial = 500,000/90,000 = 5.56 MPa (compression)
σ_bending = 30 × 10⁶ × 150/675 × 10⁶ = 6.67 MPa
σ_max = 5.56 + 6.67 = 12.23 MPa (compression)
σ_min = 5.56 − 6.67 = −1.11 MPa (TENSION)
Kern check: h/6 = 300/6 = 50 mm < e = 60 mm → tension confirmed.
For plain concrete, this is unacceptable. Reinforcing steel is needed on the tension face (ACI 318-19 / NSCP 2015 Section 409).
The problem directly mirrors a common board-exam question on eccentric columns. The kern check is a quick preliminary indicator: if e > h/6, expect tension without reinforcement.
Many real structural members carry BOTH an axial force P and a bending moment M simultaneously. The stress at any fiber is the algebraic sum (superposition) of the uniform axial stress and the linearly varying bending stress:
σ = P/A ± Mc/I
where:
- P/A = direct (axial) stress — uniform over the entire cross-section (tension +, compression −)
- Mc/I = bending stress — linear, maximum at extreme fibers, zero at NA
- The ± applies to the two extreme fibers: use + for the fiber on the tension side of bending, − for the compression side
For ECCENTRIC LOADING (load P applied at eccentricity e from the centroidal axis, so M = Pe):
σ_max = P/A + Mc/I (most compressed or most tensile fiber)
σ_min = P/A − Mc/I
The KERN (CORE) of a section:
The kern is the central region within which a compressive load can act without producing TENSION anywhere in the cross-section. Tension is problematic in masonry walls, unreinforced concrete footings, and soil under a footing.
For a RECTANGULAR section (b × h):
Kern extends ±h/6 in the h-direction and ±b/6 in the b-direction — the 'MIDDLE-THIRD RULE':
- If eccentricity e ≤ h/6 → no tension in the section
- If e > h/6 → tension occurs
For a CIRCULAR section (diameter d):
Kern radius = d/8 (middle quarter).
Mathematical check: tension occurs if |P/A| < |Mc/I|, i.e., if the bending stress exceeds the uniform axial compression stress.
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.
Common 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
Example
P = 200 kN at e = 30 mm on 150×150 mm post: σ_axial = 8.89 MPa; σ_bending = 8.89 × (30/(150/6)) = → Mc/I approach gives the exact values.
Approach
- Compute A, I, c. 2. M = Pe. 3. σ_axial = P/A. 4. σ_bending = Mc/I. 5. σ = σ_axial ± σ_bending. State tension/compression for each.
Example
For a 200×400 mm rectangle, kern in the h-direction: e_max = 400/6 = 66.7 mm. Load must be within 66.7 mm of the centroid.
Approach
Set σ_min = 0: P/A − Mc/I = 0 → P/A = P·e·c/I → e = I/(Ac) = S/A (for one direction). For rectangle: e_max = h/6.
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.
Practice problems
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
For a symmetric W-section, c = d/2 where d is the total depth. The result (281 MPa) exceeds F_y = 248 MPa (A36 steel) — this section would yield. A deeper or heavier section is needed for A36, or use A572 Grade 345 steel (F_y = 345 MPa > 281 MPa). This is a common context in board exam steel design problems.
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.)
The trick in this type of problem is correctly identifying A' (the area ABOVE the cut level) and its centroid distance from the NA. The shear stress here (0.90 MPa) is 75% of the maximum (1.20 MPa), consistent with the parabolic distribution — at the quarter-depth level, τ/τ_max = 0.75.
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.
This is a complete beam design: determine M_max and V_max, size for flexure first, then verify shear. For long beams with moderate loads, shear rarely governs for timber — but always check. The factor 1.5 in the shear formula is easily forgotten under exam pressure.
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
The small spacing (15 mm) reflects the very high shear flow relative to nail capacity. In practice, larger-capacity connectors (bolts, threaded rods) or a smaller Q (lighter top component) would be considered. The board exam will often give you Q and I directly and ask only for the spacing — plug-and-solve.
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.
This problem combines four key concepts: direct stress, bending stress, superposition, and the kern. Always evaluate the kern first as a quick check — if e > h/6, tension is guaranteed. In the Philippine context, unreinforced masonry walls and plain concrete footings must remain within the kern under service loads.
Exam preparation tips
- MEMORIZE the formula cluster: σ = My/I, σ_max = Mc/I = M/S, S_rect = bh²/6, S_circle = πd³/32, τ = VQ/Ib, τ_max,rect = 1.5V/A, τ_max,circle = (4/3)V/A, q = VQ/I, σ = P/A ± Mc/I. Write them on one index card and drill daily.
- ALWAYS draw a cross-section sketch and label the NA, c_top, c_bot, and the level of interest before solving any shear or flexure problem — this prevents using the wrong y or c value.
- CHECK UNITS before substituting: use N and mm consistently. M in N·mm, I in mm⁴, c in mm → σ in MPa. Mixing kN·m with mm⁴ is the most common arithmetic error on the board exam.
- For T-sections and other unsymmetric sections: ALWAYS locate the centroid first using ȳ = ΣAy/ΣA before computing I. The neutral axis is NOT at mid-depth for these sections.
- Distinguish Q from I: Q = A'ȳ' is the FIRST moment of area (units: mm³). I is the SECOND moment of area (units: mm⁴). Many examinees confuse these — a fatal error in shear flow or shear stress calculations.
- The factor 1.5 for rectangular maximum shear stress (τ_max = 1.5V/A) is frequently required on the board. Similarly, (4/3) for circular sections. Don't use V/A alone — it underestimates τ by 33–50%.
- For built-up beam problems: nail spacing s = nF/q (NOT s = F/(q×b)). Shear FLOW q = VQ/I has NO 'b' in the denominator.
- For the kern / middle-third rule: memorize e_max = h/6 for rectangles. Load outside the kern → tension; load inside → pure compression. This directly connects to masonry wall and footing design questions.
- In combined axial + bending problems: treat compressive P/A as NEGATIVE (or carefully define your sign convention and stick to it throughout). Inconsistent sign handling causes the most errors.
- For the PRC board, the Strength of Materials / Mechanics of Deformable Bodies component often includes 15–20 questions. Of these, bending and shear stress problems typically account for 5–8 questions. Prioritize this topic over more obscure topics.
- Practice solving problems under time pressure: each board problem should take 2–4 minutes. If a problem takes more than 5 minutes, skip it and return — time management is critical.
- Review NSCP 2015 allowable stresses for timber (Section 616) and reference the concept of adjusted design values — the board may provide Fb and ask whether a beam is adequate in bending.
- When the problem asks 'which fiber governs?', the answer is the fiber with the LARGER value of |Mc/I| — for unsymmetric sections, the fiber farther from the NA governs.
Diagrams
Stresses in Beams — Master Concept Map
A comprehensive mind map showing all major topics in the Stresses in Beams chapter: the flexure formula, section modulus, shear stress, shear flow in built-up beams, combined axial-bending stress, and the overall design process. Use this map to orient yourself before studying individual concepts.
Step-by-Step Beam Stress Analysis Procedure
A comprehensive decision-tree flowchart that guides examinees through the correct procedure for any beam stress problem: starting from identifying the type of stress required, through section property calculation, to the final stress computation. Covers flexure, shear stress, shear flow, and combined axial-bending stress.
Stress Distribution in Beams — Visual Comparison
Side-by-side comparison of three stress distributions in beams: (1) bending stress — linear, zero at NA, max at fibers; (2) shear stress — parabolic, zero at fibers, max at NA; (3) combined axial-bending — shifted linear distribution that may produce tension on one face. This visual highlights the key contrast that board exams test.
Computing Q — First Moment of Area Step by Step
A step-by-step flowchart for computing Q (first moment of area), the quantity that most examinees confuse or compute incorrectly. Shows how to identify A-prime, find its centroid distance from the NA, compute Q = A'ȳ', and then apply it to either the shear stress formula (with b) or the shear flow formula (without b).
Kern of a Rectangular Section — Middle-Third Rule
A decision flowchart illustrating the kern (core) concept for rectangular sections: how to determine whether an eccentric compressive load causes tension, using the middle-third rule (e ≤ h/6), and what design actions to take if tension occurs. Relevant to masonry, plain concrete footings, and short column design per NSCP 2015.
Beam Stress Analysis — Historical Development and Code References
A timeline tracing the development of beam stress theory from Galileo (1638) through Euler-Bernoulli beam theory, Saint-Venant's shear formula, and modern codes — culminating in the Philippine National Structural Code (NSCP 2015) and RA 544, which are directly relevant to the PRC Civil Engineer Licensure Examination.
In summary
Stresses in beams is one of the highest-yield topics on the PRC Civil Engineer Licensure Examination. The core toolkit — the flexure formula σ = My/I, the section modulus S = I/c, the shear stress formula τ = VQ/Ib, shear flow q = VQ/I, and combined stress σ = P/A ± Mc/I — is compact but demands precision in application. The most common board-exam pitfalls are: confusing Q with I, forgetting the 1.5 factor for rectangular maximum shear, failing to locate the correct neutral axis for unsymmetric (T, L) sections, and making unit errors by mixing kN·m with mm⁴.
In practice, these formulas underpin every major design code used in the Philippines. The NSCP 2015 timber design tables require fv = VQ/Ib ≤ Fv and fb = M/S ≤ Fb checks. Steel beam design per AISC 360-16 / NSCP 2015 begins with S_x ≥ M_max/φFy or M_max/Ω·Fb. Reinforced concrete design (ACI 318-19 / NSCP 2015 Section 406) is motivated by the fact that concrete's tensile stress at the bottom fiber exceeds its modulus of rupture — hence tensile reinforcement. The kern concept directly governs footing design (soil must stay in compression) and masonry wall design (no tension in unreinforced masonry per NSCP 2015 Chapter 7).
For exam success: internalize the formula set, practice computing centroids and moments of inertia for composite and T-sections under timed conditions, and develop the habit of always checking both the flexure and shear conditions. Remember that RA 544 (Republic Act No. 544, the Philippine Civil Engineering Act) mandates that only licensed civil engineers may sign and seal structural plans — the knowledge in this chapter is not just exam content; it is the professional foundation you will carry throughout your career.