← Strength of Materials · GSSSB Work Assistant Civil

Chapter 11 of 17

Bending and Shear Stress

In the GSSSB Work Assistant Civil syllabus under Strength of Materials · 2 parts

Bending Stresses in Beams · Shear Stresses in Beams

CIVILGYAN TEST SERIES · civilgyantests.com — free study notes for civil engineering exams

📑 Contents (21 sections)

Part 1 of 2

Bending Stresses in Beams

Last reviewed 16 Sept 2026 · 8 min read

Pure bending

A beam segment carrying a bending moment with no shear force is in pure (simple) bending — for example, the middle portion of a simply supported beam with two equal loads placed symmetrically. The theory below is exact for pure bending and is used, with negligible error, where shear is also present.

When a beam sags, its top fibres shorten and its bottom fibres lengthen. Somewhere between lies a layer that neither shortens nor lengthens: the neutral layer. Its intersection with a cross-section is the neutral axis (NA).

Assumptions of the theory of simple bending

The material is homogeneous, isotropic and obeys Hooke's law; stresses are within the elastic limit.
$E$ is the same in tension and compression.
Plane sections remain plane after bending (Bernoulli–Euler hypothesis).
The beam is initially straight and every layer is free to expand or contract independently.
The radius of curvature is large compared with the depth.
The loads act in a plane of symmetry of the section (so the beam bends without twisting).

The flexure formula

Consider a layer at distance $y$ from the neutral layer, which bends to radius $R$ . Its original length equals the neutral-layer length $R d θ$ ; after bending it is $(R + y) d θ$ . So its strain is

ε = \frac{( R + y ) d θ - R d θ}{R d θ} = \frac{y}{R} \Rightarrow σ = E ε = \frac{E}{R} y

Stress is proportional to the distance from the neutral axis.

Force equilibrium (no net axial force): $\int σ d A = \frac{E}{R} \int y d A = 0$ → $\int y d A = 0$ → the neutral axis passes through the centroid of the section.
Moment equilibrium: $M = \int σ y d A = \frac{E}{R} \int y^{2} d A = \frac{E}{R} I$ .

FormulaBending equation

\frac{M}{I} = \frac{σ}{y} = \frac{E}{R}

$M$ = bending moment, $I$ = second moment of area about the neutral axis, $σ$ = bending stress at distance $y$ from the NA, $E$ = Young's modulus, $R$ = radius of curvature of the neutral layer. $E I$ is the flexural rigidity.

FigureBending stress distribution across a rectangular section (sagging)

Linear distribution: zero at the neutral axis, maximum at the extreme fibres.

Section modulus and moment of resistance

The maximum stress occurs at the extreme fibre, $y_{ma x}$ :

σ_{ma x} = \frac{M}{I} y_{ma x} = \frac{M}{Z}, Z = \frac{I}{y _{ma x}}

$Z$ is the section modulus. For a permissible stress $σ$ , the moment of resistance is $M_{r} = σ Z$ . The beam is safe if $M \leq M_{r}$ . A larger $Z$ means a stronger section in bending.

Section	$I$ about centroidal axis	$Z$
Rectangle $b \times d$ (bending about the $b$ side)	$b d^{3} /12$	$b d^{2} /6$
Square of side $a$	$a^{4} /12$	$a^{3} /6$
Square with diagonal vertical	$a^{4} /12$	$a^{3} / (62)$
Solid circle, diameter $d$	$π d^{4} /64$	$π d^{3} /32$
Hollow circle $D$ , $d$	$π (D^{4} - d^{4}) /64$	$π (D^{4} - d^{4}) / (32 D)$
Hollow rectangle $B \times D$ outer, $b \times d$ inner	$(B D^{3} - b d^{3}) /12$	$(B D^{3} - b d^{3}) / (6 D)$
Triangle, base $b$ , height $h$ (base horizontal)	$b h^{3} /36$	$b h^{2} /24$ (to apex), $b h^{2} /12$ (to base)

Exam TipDoubling depth versus doubling width

For a rectangle $Z = b d^{2} /6$ . Doubling the width doubles the strength; doubling the depth makes it four times stronger (and eight times stiffer, since $I \propto d^{3}$ ). This is why beams are placed with the longer side vertical.

Economical sections

Most of the bending resistance comes from material far from the neutral axis, where stress is high. Material near the NA is lightly stressed. I-sections and box sections put most material in the flanges, giving a large $Z$ for a small area — the most economical shapes for steel beams.

Comparing shapes of equal area

Square vs circle (same area): $Z_{s q u a r e} / Z_{c i r c l e} = 1.18$ — the square is stronger.
Square with a side horizontal vs diagonal vertical: $Z$ ratio $= 2 = 1.414$ — the square resting on a side is stronger.
I-section vs rectangle of the same area and depth: the I-section is several times stronger.

Strongest and stiffest rectangular beam cut from a round log of diameter $D$

With $b^{2} + d^{2} = D^{2}$ :

Strongest (maximum $Z = b d^{2} /6$ ): $b = D / 3$ , $d = D 2/3$ , so $d / b = 2$ .
Stiffest (maximum $I = b d^{3} /12$ ): $b = D /2$ , $d = 3 D /2$ , so $d / b = 3$ .

Unsymmetrical sections about the neutral axis

For a T-section, channel with web vertical, or an inverted T, the NA (through the centroid) is not at mid-depth. Then $y_{t o p} \neq = y_{b o tt o m}$ and there are two section moduli:

σ_{t o p} = \frac{M y _{t o p}}{I}, σ_{b o tt o m} = \frac{M y _{b o tt o m}}{I}

The fibre farther from the NA has the larger stress. For materials weak in tension (cast iron), the section is arranged so that the tension face is nearer the NA — e.g. a cast-iron T-beam with its flange on the tension side.

Composite (flitched) beams

A flitched beam is timber strengthened by steel plates bolted to it, bending together. Plane sections still remain plane, so at a given $y$ both materials have the same strain; their stresses are in the ratio of their moduli.

Take modular ratio $m = E_{s} / E_{t}$ .
Replace the steel by an equivalent timber area, $m$ times as wide (at the same depth).
Analyse the transformed timber section: $σ_{t} = M y / I_{e q}$ ; the steel stress at the same level is $σ_{s} = m σ_{t}$ .
Check both materials against their permissible stresses; the one reaching its limit first governs.

Beams of uniform strength

A beam of uniform strength has the same maximum bending stress at every section, so no material is wasted. Since $σ = M / Z$ , $Z$ must vary with $M$ .

Simply supported beam, central load $W$ , constant depth: width must vary linearly from zero at the supports to maximum at the centre (a rhombus in plan) — $b_{x} = \frac{3 W x}{σ d ^{2}}$ .
Same beam with constant width: depth varies parabolically, $d_{x} = 3 W x / (σ b)$ .

Leaf springs and tapered cantilever brackets use this idea.

Worked examples

Worked ExampleExample 1 — rectangular timber beam

A timber beam 150 mm wide and 300 mm deep carries a bending moment of 30 kN·m. Find the maximum bending stress.

Solution. $Z = \frac{150 \times 30 0 ^{2}}{6} = 2.25 \times 1 0^{6}$ mm³

$σ_{ma x} = \frac{30 \times 1 0 ^{6}}{2.25 \times 1 0 ^{6}} = 13.33 N/m m^{2}$ (compression at top, tension at bottom for sagging)

Worked ExampleExample 2 — I-section

A symmetrical I-section has flanges 200 mm × 20 mm and a web 10 mm × 360 mm (overall depth 400 mm). It carries $M$ = 150 kN·m. Find the maximum bending stress and the stress at the junction of flange and web.

Solution. $I = \frac{200 \times 40 0 ^{3} - 190 \times 36 0 ^{3}}{12} = \frac{12.800 \times 1 0 ^{9} - 8.865 \times 1 0 ^{9}}{12} = 3.279 \times 1 0^{8}$ mm⁴

$σ_{ma x} = \frac{150 \times 1 0 ^{6} \times 200}{3.279 \times 1 0 ^{8}} = 91.5 N/m m^{2}$

At the junction ( $y = 180$ mm): $σ = 91.5 \times \frac{180}{200} = 82.3 N/m m^{2}$

Worked ExampleExample 3 — T-section

A T-section has a flange 100 mm × 20 mm and a web 20 mm × 100 mm (overall depth 120 mm). It carries a sagging moment of 4 kN·m. Find the stresses at the top and bottom.

Solution. Areas: flange 2000 mm² (centroid 10 mm from top), web 2000 mm² (centroid 70 mm from top).

$\overset{y}{ˉ} = \frac{2000 \times 10 + 2000 \times 70}{4000} = 40$ mm from top → $y_{t o p}$ = 40 mm, $y_{b o tt o m}$ = 80 mm.

$I = \frac{100 \times 2 0 ^{3}}{12} + 2000 \times 3 0^{2} + \frac{20 \times 10 0 ^{3}}{12} + 2000 \times 3 0^{2} = 5.333 \times 1 0^{6}$ mm⁴

$σ_{t o p} = \frac{4 \times 1 0 ^{6} \times 40}{5.333 \times 1 0 ^{6}} = 30 N/m m^{2}$ (compression)

$σ_{b o tt o m} = \frac{4 \times 1 0 ^{6} \times 80}{5.333 \times 1 0 ^{6}} = 60 N/m m^{2}$ (tension)

Worked ExampleExample 4 — flitched beam

A timber beam 100 mm × 200 mm has a steel plate 10 mm × 200 mm bolted on each side. $E_{s} / E_{t}$ = 20. Permissible stresses: timber 8 N/mm², steel 150 N/mm². Find the moment of resistance.

Solution. Equivalent timber width $= 100 + 20 \times (10 + 10) = 500$ mm.

$Z_{e q} = \frac{500 \times 20 0 ^{2}}{6} = 3.333 \times 1 0^{6}$ mm³

If timber reaches 8 N/mm², steel (same extreme fibre) would be $20 \times 8 = 160$ N/mm² > 150 — not allowed.

So steel governs: $σ_{t} = 150/20 = 7.5$ N/mm² → $M_{r} = 7.5 \times 3.333 \times 1 0^{6} = 25 kN \cdot m$

Worked ExampleExample 5 — strongest beam from a log

Find the dimensions of the strongest rectangular beam that can be cut from a log 300 mm in diameter.

Solution. $b = D / 3 = 300/1.732 = 173.2 mm$ ; $d = 2 b = 244.9 mm$

Frequently tested points

$M / I = σ / y = E / R$ ; stress varies linearly, zero at the NA.
NA passes through the centroid (for elastic bending without axial force).
$Z = I / y_{ma x}$ ; rectangle $b d^{2} /6$ ; circle $π d^{3} /32$ .
For equal area: square is 1.18 times as strong as a circle; square on side is $2$ times as strong as on diagonal.
Strongest beam from a log: $d / b = 2$ ; stiffest: $d / b = 3$ .
Flitched beams: equal strain at a level; $σ_{s} = m σ_{t}$ .
Radius of curvature $R = E I / M$ ; curvature $1/ R = M / E I$ .

Common MistakeCommon mistakes

Using $b d^{3} /12$ with $b$ and $d$ swapped for a beam bent about its weak axis.
Taking $y$ as half the depth for T and channel sections — find the centroid first.
In composite beams, assuming both materials reach their permissible stress together.
Calling $Z$ "moment of inertia" — $Z$ has units of mm³, $I$ has mm⁴.

Revision SummaryChapter summary

Plane sections remain plane → strain and stress vary linearly with distance from the NA.
$M / I = σ / y = E / R$ ; NA through the centroid.
$σ_{ma x} = M / Z$ ; strength depends on section modulus, stiffness on $I$ .
I and box sections are economical; strongest log beam $d / b = 2$ , stiffest $3$ .
Unsymmetrical sections have different top and bottom stresses; flitched beams use the modular ratio.

Part 2 of 2

Shear Stresses in Beams

Last reviewed 16 Sept 2026 · 7 min read

Shear in beams — the idea

A beam carrying transverse loads has a shear force $V$ at most sections. That force is resisted by shear stresses spread over the cross-section. Unlike bending stress, shear stress is not uniform: it is zero at the top and bottom surfaces and largest near the neutral axis.

Transverse shear always comes with horizontal shear between layers. Stack loose planks and load them — they slide over one another. Glue them and they act as one deep beam; the glue now carries horizontal shear. By complementary shear, the horizontal and vertical shear stresses at any point are equal.

Shear stress formula

Consider a beam element $d x$ long. The bending moment changes from $M$ to $M + d M$ across it. Isolate the part of the element above a level at distance $y$ from the neutral axis. The bending stresses on its two ends differ, and the difference must be balanced by a horizontal shear force on the cut face.

Net horizontal force on the isolated part $= \frac{d M}{I} \int y d A = \frac{d M}{I} A \overset{y}{ˉ}$
This equals $τ b d x$ , where $b$ is the width at the cut.

Since $d M / d x = V$ :

FormulaShear stress at a level in a beam

τ = \frac{V A y ˉ}{I b}

$V$ = shear force at the section, $A$ = area of the section beyond the level considered (above or below it), $\overset{y}{ˉ}$ = distance of the centroid of that area from the neutral axis, $I$ = moment of inertia of the whole section about the NA, $b$ = width of the section at the level considered.

$A \overset{y}{ˉ}$ is the first moment of area, often written $Q$ .

Assumptions: the shear stress is uniform across the width at a given level, and the bending formula holds. The result is exact for narrow rectangles and a good approximation for most practical sections.

Rectangular section ( $b \times d$ )

At distance $y$ from the NA: $A = b (\frac{d}{2} - y)$ and $\overset{y}{ˉ} = \frac{1}{2} (\frac{d}{2} + y)$ , so

τ = \frac{V}{2 I} (\frac{d ^{2}}{4} - y^{2})

The distribution is parabolic.
$τ = 0$ at the top and bottom ( $y = \pm d /2$ ).
Maximum at the NA: $τ_{ma x} = \frac{V d ^{2}}{8 I} = \frac{3}{2} \frac{V}{b d}$ .

Remember

For a rectangle, $τ_{ma x} = 1.5 τ_{a v g}$ , where $τ_{a v g} = V / (b d)$ .

Circular section (diameter $d$ )

The distribution is also parabolic, with maximum at the NA:

τ_{ma x} = \frac{4}{3} \frac{V}{π d ^{2} /4} = \frac{4}{3} τ_{a v g}

Triangular section (base $b$ , height $h$ , base horizontal)

Here the width changes with depth. The maximum shear stress is not at the neutral axis but at mid-height:

τ_{ma x} = 1.5 τ_{a v g} at h /2 from the apex, τ_{N A} = \frac{4}{3} τ_{a v g}

The neutral axis lies at $2 h /3$ from the apex.

Summary of maximum-to-average ratios

Section	Location of $τ_{ma x}$	$τ_{ma x} / τ_{a v g}$
Rectangle	Neutral axis	1.5
Solid circle	Neutral axis	4/3 ≈ 1.33
Triangle	Mid-height (not the NA)	1.5 (at NA 4/3)
Thin-walled circular tube	Neutral axis	2.0
Square with a diagonal vertical (diamond)	$d /8$ from the NA, above and below	9/8 = 1.125
I-section	Neutral axis, in the web	web carries most of the shear

I-section

The flanges are wide and the web thin. Because $τ = V A \overset{y}{ˉ} / (I b)$ has $b$ in the denominator:

In the flange, $b$ = flange width, so $τ$ is small.
At the flange–web junction, the same $A \overset{y}{ˉ}$ is divided by the web thickness instead of the flange width, so the stress jumps by the ratio $B / t_{w}$ .
In the web, $τ$ varies parabolically, maximum at the NA.

The web carries almost all (typically 90–95%) of the shear force, while the flanges carry most of the bending moment. Hence a quick design check for steel beams: average shear stress $= V / (t_{w} d)$ over the web.

FigureShear stress distribution in an I-section

T-section

The NA is closer to the flange. Maximum shear stress is at the NA (in the web). There is a jump at the flange–web junction, as in the I-section, but the distribution is not symmetrical.

Shear flow

The shear flow $q$ is the horizontal shear force per unit length of the beam at a given level:

q = τ b = \frac{V A y ˉ}{I}

Its unit is N/mm. It is independent of the width at the level, which makes it the natural quantity for connections in built-up beams:

spacing of fasteners s = \frac{capacity of one fastener (or row)}{q}

Applications: nails in a built-up timber beam, bolts or welds connecting a cover plate to a steel beam, shear connectors in composite steel–concrete beams, and the calculation of the shear centre of open thin-walled sections.

Worked examples

Worked ExampleExample 1 — rectangular beam

A simply supported beam of 100 mm × 200 mm section carries a shear force of 20 kN. Find the maximum and average shear stresses and the shear stress 50 mm above the NA.

Solution. $τ_{a v g} = \frac{20 000}{100 \times 200} = 1.0 N/m m^{2}$ ; $τ_{ma x} = 1.5 \times 1.0 = 1.5 N/m m^{2}$

$I = \frac{100 \times 20 0 ^{3}}{12} = 6.667 \times 1 0^{7}$ mm⁴

At $y$ = 50 mm: $τ = \frac{V}{2 I} (\frac{d ^{2}}{4} - y^{2}) = \frac{20 000}{2 \times 6.667 \times 1 0 ^{7}} (10000 - 2500) = 1.125 N/m m^{2}$

Worked ExampleExample 2 — I-section

An I-section has flanges 150 mm × 20 mm, web 10 mm × 260 mm (overall depth 300 mm). Shear force = 100 kN. Find the shear stress at the bottom of the top flange, at the top of the web, and at the NA.

Solution. $I = \frac{150 \times 30 0 ^{3} - 140 \times 26 0 ^{3}}{12} = \frac{4.050 \times 1 0 ^{9} - 2.461 \times 1 0 ^{9}}{12} = 1.324 \times 1 0^{8}$ mm⁴

Flange area above the junction $= 150 \times 20 = 3000$ mm², centroid at 140 mm from NA → $A \overset{y}{ˉ} = 4.20 \times 1 0^{5}$ mm³.

Flange just above junction ( $b$ = 150): $τ = \frac{100 000 \times 4.20 \times 1 0 ^{5}}{1.324 \times 1 0 ^{8} \times 150} = 2.11 N/m m^{2}$
Web just below junction ( $b$ = 10): $τ = 2.11 \times 15 = 31.7 N/m m^{2}$
At NA: add the web above the NA, area $10 \times 130 = 1300$ mm² at 65 mm → $A \overset{y}{ˉ} = 4.20 \times 1 0^{5} + 0.845 \times 1 0^{5} = 5.045 \times 1 0^{5}$ mm³ $τ_{ma x} = \frac{100 000 \times 5.045 \times 1 0 ^{5}}{1.324 \times 1 0 ^{8} \times 10} = 38.1 N/m m^{2}$

The simple average over the web, $V / (t_{w} d) = 100 000/ (10 \times 300) = 33.3$ N/mm², is close to the exact peak of 38.1 N/mm² — which is why steel design checks shear on the web area.

Worked ExampleExample 3 — nail spacing in a built-up beam

A T-beam is made by nailing a 150 mm × 50 mm plank (flange) on top of a 50 mm × 200 mm plank (web). Each nail can carry 1.0 kN in shear. For $V$ = 3 kN, find the nail spacing.

Solution. Flange: 7500 mm² at 25 mm from top. Web: 10 000 mm² at 150 mm from top.

$\overset{y}{ˉ} = \frac{7500 \times 25 + 10 000 \times 150}{17 500} = 96.4$ mm from top.

$I = \frac{150 \times 5 0 ^{3}}{12} + 7500 \times 71. 4^{2} + \frac{50 \times 20 0 ^{3}}{12} + 10000 \times 53. 6^{2} = 1.019 \times 1 0^{8}$ mm⁴

The nails carry the shear flow at the flange–web joint: flange centroid is $96.4 - 25 = 71.4$ mm above the NA.

$q = \frac{V A y ˉ}{I} = \frac{3000 \times 7500 \times 71.4}{1.019 \times 1 0 ^{8}} = 15.8$ N/mm

Spacing $= \frac{1000}{15.8} = 63$ mm → adopt 60 mm.

Frequently tested points

Shear stress: $τ = V A \overset{y}{ˉ} / (I b)$ ; zero at extreme fibres.
Rectangle: parabolic, $τ_{ma x} = 1.5 τ_{a v g}$ at NA.
Circle: $τ_{ma x} = \frac{4}{3} τ_{a v g}$ at NA.
Triangle: maximum at mid-height, $1.5 τ_{a v g}$ ; at NA $\frac{4}{3} τ_{a v g}$ .
Thin circular tube: $τ_{ma x} = 2 τ_{a v g}$ .
I-section: web carries most shear; sudden jump at flange–web junction.
Shear flow $q = V Q / I$ decides the spacing of nails, bolts and welds.

Common MistakeCommon mistakes

Taking $b$ as the flange width when computing web stress (or vice versa) at a junction.
Using the area of the whole section as $A$ — it is only the part beyond the level.
Assuming maximum shear is always at the NA (false for triangles and diamonds).

Revision SummaryChapter summary

Transverse shear produces equal horizontal shear between layers.
$τ = V A \overset{y}{ˉ} / (I b)$ , parabolic for rectangles and circles.
Max/average: rectangle 1.5, circle 4/3, thin tube 2, triangle 1.5 at mid-height.
In I-sections the web resists shear; stress jumps at the flange–web junction.
Shear flow $q = V A \overset{y}{ˉ} / I$ designs connectors in built-up beams.

Downloaded from civilgyantests.com · Not for resale. Verify every code clause and date against the official source.

Finished reading? Test yourself.

Practise Strength of Materials with the GSSSB Work Assistant Civil test series.

Open GSSSB Work Assistant Civil tests →

Bending Stresses in Beams

Pure bending

Assumptions of the theory of simple bending

The flexure formula

Section modulus and moment of resistance

Economical sections

Comparing shapes of equal area

Strongest and stiffest rectangular beam cut from a round log of diameter D

Unsymmetrical sections about the neutral axis

Composite (flitched) beams

Beams of uniform strength

Worked examples

Frequently tested points

Shear Stresses in Beams

Shear in beams — the idea

Shear stress formula

Rectangular section (b×d)

Circular section (diameter d)

Triangular section (base b, height h, base horizontal)

Summary of maximum-to-average ratios

I-section

T-section

Shear flow

Worked examples

Frequently tested points

Finished reading? Test yourself.

Strongest and stiffest rectangular beam cut from a round log of diameter $D$

Rectangular section ( $b \times d$ )

Circular section (diameter $d$ )

Triangular section (base $b$ , height $h$ , base horizontal)