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Shear Stress in Beams – Chapter 6

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Strength of Materials · Chapter 6

Shear Stress in Beams

Shear Stress Formula · Rectangular · Triangular · Circular · I-Section · Other Sections · Composite Beams

τ = SAȳ/Ib τ_max(rect) = 3S/2bd τ_max(circ) = 4S/3πR² τ_max(tri) = 3S/2bh at h/2

6.1 Shear Stress Distribution in Beams — Derivation of General Formula

Assumptions (from source):

Material is homogeneous and elastic Material obeys Hooke’s law Shear stress is assumed constant along width; variation along depth only
Figure 6.1 — Derivation of Shear Stress Formula (from source)
Shear stress derivation Beam Element (length dx) NA Shaded area A above y y b (width at y) ȳ x₁ x₂ dx τ acts on bottom face of shaded element dx Key Variables: M = BM at x₁-x₁ M+dM = BM at x₂-x₂ S = SF at x₁-x₁ b = width of section at y from NA Derivation Steps By bending equation at section x₁-x₁: σ = M·y/I Force on strip dA at x₁-x₁: F = M·y·dA/I Force on strip dA at x₂-x₂: F+dF = (M+dM)y·dA/I Net unbalance force on strip: dF = (dM/I)·y·dA Total unbalance force on layers ABCD: F = (dM/I)∫y·dA = (dM/I)·Aȳ ∫y·dA represents moment of area under consideration about the neutral axis = Aȳ This unbalance force is resisted by shear stress acting along length dx: F = τ·b·dx Equating: (dM/I)·Aȳ = τ·b·dx ∴ τ = (dM/dx)·(Aȳ)/(I·b) But dM/dx = S (shearing force at section) General Shear Stress Formula τ = SAȳ/Ib S = Shear force at section A = Area of section above y ȳ = dist of centroid of A from NA I = MOI of whole section about NA b = width of section at distance y Also written as: τ = S·Q / (I·b) where Q = Aȳ = first moment of area above y about NA Shear stress on one face = complementary
shear stress on perpendicular face

Fig. 6.1 — Derivation of τ = SAȳ/Ib. At surface y = ±d/2: τ = 0. Maximum τ occurs at neutral axis (y=0) for most sections.

NOTE (from source): The shear stress on one side of the element is accompanied by complementary shear stress of equal magnitude acting on the perpendicular face — hence horizontal and vertical shear stresses at any level must be equal. Also, shear stresses at top and bottom are zero because adjacent faces are free from any shear stress.

6.1.1 Shear Stress Distribution in Rectangular Section

Consider a beam of rectangular section width b and depth d. Shear stress for an element at distance y above NA:

Derivation from source (pg 255)
A = b(d/2 − y)    ȳ = (y + d/2)/2
I = bd³/12    b = b (constant)
τ = S·b(d/2−y)·(y+d/2)/2 / (bd³/12·b)
τ = 6S/(bd³) · (d²/4 − y²)
(parabolic variation)
Key Results (from source)
At y = ±d/2 (surface): τ = 0
For τ_max: dτ/dy = 0 → y = 0 (NA)
τ_max = 6S/(bd³) · d²/4 = 3S/2bd
τ_max = 3/2 · τ_avg
where τ_avg = S/bd
Location where τ = τ_avg
τ = τ_avg when: (d²/4 − y²) = d²/6
y′ = ± d/2√3 = ± d/(2√3)
Figure 6.2 — Rectangular Section & Shear Stress Distribution
Rectangular section shear stress distribution NA y ȳ A (above y) d/2 d/2 b d Cross-section τ=0 τ=0 τ_max = 3S/2bd = 3/2 · τ_avg τ=τ_avg τ=τ_avg NA Shear stress (parabolic) variation
Shear Stress Variation
PARABOLIC
τ = 6S(d²/4 − y²)/bd³
Maximum Shear Stress
τ_max = 3S/2bd
occurs at Neutral Axis (y=0)
Ratio
τ_max = 3/2 · τ_avg
τ_avg = S/bd

6.1.2 Shear Stress Distribution in Triangular Section

Consider a triangular section of base b and height h. Shear stress at layer DE at a distance x from apex (from source pg 257–258):

Derivation (from source)
Width at x from apex: b’ = bx/h
Area ADE = (1/2)·b’·x = bx²/2h
ȳ = 2h/3 − (2/3)x = 2(h−x)/3
Aȳ = (bx²/2h)·2(h−x)/3
I = bh³/36    b_at_section = bx/h
τ = 12S/(bh³)·(hx − x²)
(parabolic variation)
Key Results (from source)
At x=0 (apex) and x=h (base): τ=0
dτ/dx = 0 → x = h/2 (from apex)
τ_max = 3S/2bh (at x = h/2)
τ_max = 3/2 · τ_avg
τ_max NOT at neutral axis!
(NA is at h/3 from base = 2h/3 from apex)
Shear Stress at NA (from source)
At NA: x = 2h/3 from apex
τ_NA = 8S/3bh = 4/3 · τ_avg
(less than τ_max at h/2)
Figure 6.3 — Triangular Section & Shear Stress Distribution
Triangular section shear stress distribution NA (h/3 from base) h/2 from apex (τ_max) h/2 h/2 b h Cross-section τ=0 τ=0 τ_max = 3S/2bh at h/2 from apex τ_NA = 4/3·τ_avg (less than τ_max) Shear stress distribution
Key Distinction (from source): For triangular section, τ_max = 3/2·τ_avg (same ratio as rectangle) but τ_max occurs at h/2 from apex, NOT at neutral axis. The shear stress at neutral axis = 4/3·τ_avg, which is less than τ_max. This is a very common GATE question!

6.1.3 Shear Stress Distribution in Circular Section

Consider a circular section of radius R. Shear stress at layer EF at a distance y from NA (from source pg 259):

Derivation (from source)
Width: b = 2√(R²−y²)
Aȳ = ∫_y^R y·b·dy = ∫_y^R y·2√(R²−y²)·dy
On integrating:
Aȳ = (2/3)(R²−y²)^(3/2)
τ = S·(2/3)(R²−y²)^(3/2) / [πR⁴/4·2√(R²−y²)]
τ = 4S/3πR² · (R²−y²)
(parabolic variation)
Key Results (from source)
At y = ±R (surface): τ = 0
For τ_max: dτ/dy = 0 → y = 0 (NA)
τ_max = 4S/3πR² = 4S/3A
τ_max = 4/3 · τ_avg
τ_avg = S/πR²
Location where τ = τ_avg (from source)
τ = τ_avg when (R²−y²) = 3πR²/4
wait: 4S(R²−y²)/3πR² = S/πR²
y′ = ± R/2
Figure 6.4 — Circular Section & Shear Stress Distribution
Circular section shear stress distribution NA EF (width b) y R Cross-section (R = radius) b = 2√(R²−y²) τ=0 τ=0 τ_max = 4S/3πR² = 4/3 · τ_avg τ=τ_avg (y=R/2) τ=τ_avg (y=R/2) NA Shear stress (parabolic)
Thin Circular Tube (from source, hollow circle):
For hollow circle with inner radius R₁ and outer radius R₂:
τ_max = (4S/3) · (R₂²+R₁R₂+R₁²)/(R₂²+R₁²)·A
For thin tube (R₁≈R₂=R): τ_max = 2·τ_avg
i.e., τ_max/τ_avg = 2 for thin circular tube

6.1.4 Shear Stress Distribution in I-Section

For an I-section (flanges: width B, depth (D−d)/2 each; web: width b, depth d), shear stress distribution in the flange and web are calculated separately.

Figure 6.5 — I-Section, Dimensions and Shear Stress Distribution
I-section shear stress distribution NA Flange (width B) Web (width b) Flange (width B) B b D d D/2−y d/2 d/2 Shear stress distribution NA τ=0 τ=0 τ just above jxn (flange) = SB(D²−d²)/8BI τ just below jxn (web) = SB(D²−d²)/8bI Jump: B/b × (stress in flange at jxn) τ_max at NA = SB(D²−d²)/8bI + Sd²/8I I-section (flanges wide, web narrow) I-Section Formulae (from source) (i) In FLANGE [d/2 ≤ y ≤ D/2]: A = B(D/2 − y) ȳ = (D/2 + y)/2,   b = B τ = S(D²/4 − y²) / 2I (Parabolic, but b=B so small stress) (ii) In WEB [0 ≤ y ≤ d/2]: Aȳ = Aȳ(flange) + Aȳ(web) = B(D²−d²)/8 + b(d²/4−y²)/2 τ = SB(D²−d²)/8bI + S(d²/4−y²)/2I (Parabolic in web) At y=0 (NA), τ is maximum: τ_max = SB(D²−d²)/8bI + Sd²/8I Jump at junction = B/b × τ(in flange at jxn)

Fig. 6.5 — I-section shear stress distribution. Parabolic in both flange and web. Sudden JUMP (discontinuity) occurs at flange-web junction because width changes from B (flange) to b (web). Most shear force is carried by the web.

Important (from source): In I-sections, practically most of the shear force is resisted by the web (narrow but deep). The flanges carry most of the bending moment. The sudden jump in shear stress at the flange-web junction is because the same Aȳ acts over a much narrower width b (web) compared to B (flange).

6.1.5 Shear Stress in Some Other Sections

Figure 6.6 — Shear Stress Distributions in Various Sections (from source)
Shear stress distribution in various sections Cross Section Section Shear Stress Dist. Parabolic in each web/flange Jump at junctions Hollow Circle Section (hollow) Shear Stress Dist. τ_max = 2τ_avg (thin tube) NA Parabolic; for thin tube τ_max/τ_avg = 2 T-Section NA NA not at mid-depth Shear Stress τ_max at NA Jump at jxn Asymmetric — NA not at mid-depth for T-section Diamond (sq. on diag.) Square section kept along diagonal NA τ_max = 9S/4bh at y = h/4 from NA τ_max = 9/8 · τ_avg

Fig. 6.6 — Shear stress distributions from source (pg 262). Cross, T, L-section: parabolic with jumps at junctions. Hollow circle: parabolic (thin tube τ_max = 2τ_avg). Diamond: max occurs at h/4 from NA (not at NA).

6.2 Shear Stresses in Composite Sections

Analysis of composite sections is done by considering the equivalent section of the composite section in terms of any one material. Then shear stress formula is applied to the equivalent section:

Composite Section Shear Stress Formula
τ = S·(Aȳ)_eq / (I_eq · b_eq)
I_eq = Equivalent MOI about NA | b_eq = Width of equivalent section at the layer under consideration
(Aȳ)_eq = Moment of area of the portion above the layer under consideration about NA (in equivalent section)
Procedure (from source)
  1. Convert composite section into equivalent homogeneous section using modular ratio m = E₁/E₂
  2. Equivalent width of material with modular ratio m = m × original width
  3. Find NA of equivalent section
  4. Calculate I_eq about NA
  5. Calculate b_eq at layer under consideration
  6. Apply τ = S·(Aȳ)_eq / (I_eq · b_eq)
Composite section equivalent section Original Steel (E_s) Wood (E_w) NA Section → m=E_s/E_w Equiv. (wood) Steel → width = m×b Wood New NA Equiv. section Shear stress in equiv. section
Key Note (from source): For composite beam section, shear stress distribution will be similar to the shear stress distribution obtained for the equivalent section. The actual shear stress in each material is the same as in the equivalent section at that level.

Summary — All Sections: Shear Stress Comparison

Summary from Source (pg 269)
  • In general, shear stress is assumed constant along width and variation is considered only along depth
  • Shear stress at any section is given by τ = SAȳ/Ib
  • In rectangular beam section, maximum shear stress occurs at neutral axis and τ_max = 3/2 · τ_avg
  • In triangular section, τ_max does NOT occur at neutral axis — it occurs at h/2 from the apex and τ_max = 3/2 · τ_avg
  • In circular section, τ_max occurs at neutral axis and τ_max = 4/3 · τ_avg
  • In composite section, shear stress is the same as the shear stress given by the equivalent transformed section
Section Shear Stress Formula Variation Shape Location of τ_max τ_max in terms of τ_avg
Rectangle (b×d) τ = 6S(d²/4−y²)/bd³ Parabolic At Neutral Axis (y=0) 3/2 · τ_avg
Triangle (b×h) τ = 12S(hx−x²)/bh³ Parabolic At h/2 from apex (NOT NA!)
τ_NA = 4/3·τ_avg (less than τ_max)		 3/2 · τ_avg
Circle (radius R) τ = 4S(R²−y²)/3πR⁴ Parabolic At Neutral Axis (y=0) 4/3 · τ_avg
Thin Circular Tube τ_max ≈ 2S/A Parabolic At Neutral Axis 2 · τ_avg
I-Section (web) τ = SB(D²−d²)/8bI + S(d²/4−y²)/2I Parabolic (web)
Parabolic (flange)		 At Neutral Axis (y=0)
Jump at flange-web junction		 τ_max = SB(D²−d²)/8bI + Sd²/8I
Diamond (sq.diagonal) τ = 4S(3h−4y)·y/bh³ Parabolic At 3h/8 from top (h/4 from NA) 9/8 · τ_avg
Chapter 6 — Key Formula Reference
General Formula
τ = SAȳ / Ib
S=SF, A=area above y, ȳ=centroid dist, I=MOI, b=width
Rectangle τ_max
3S/2bd = 3/2 · τ_avg
At Neutral Axis | τ variation: parabolic
Triangle τ_max
3S/2bh = 3/2 · τ_avg
At h/2 from apex (NOT at NA!) | τ_NA = 4/3·τ_avg
Circle τ_max
4S/3πR² = 4/3 · τ_avg
At Neutral Axis | τ_avg = S/πR²
Thin Tube τ_max
2 · τ_avg
At Neutral Axis | τ_max/τ_avg = 2
Composite Sections
τ = S·(Aȳ)_eq / I_eq·b_eq
Use equivalent section; same stress distribution

Chapter 6: Shear Stress in Beams — Strength of Materials · MADE EASY Postal Study Package 2019
All technical data, formulae and distributions from source material only

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