Worked examples

Beam example 3: three restrained segments and an omitted deflection solution

Open “Animation lab” beside a teaching step for a visual explanation or a walkthrough of its original expressions. Models are illustrative; source answers remain unchanged.

← Read this lecture example beside its chapter concepts

Chinese–English terminology

Need a simpler picture? Open “Simple explanation” beside a difficult step. These optional notes do not replace the full solution.

Check/design the S355 simply supported $9 m$ beam for shear and bending with lateral restraints at A, B, C and D. $AB = BC = CD = 3 m$ . At B: dead $40 kN$ and imposed $60 kN$ ; at C: dead $20 kN$ and imposed $30 kN$ . Dead self-weight is $3 \frac{kN}{m}$ . Effective segment lengths equal actual lengths. Also work through the lecturer’s requested follow-up web and deflection checks.

Original source: LectureNotes/Ch 3_Beam.pdf — p. 27, p. 28, p. 29, p. 30. Values tagged given are in the question or diagram; lookup values come from a named table; calculated values follow from the working; assumptions are stated explicitly.

Ch 3 p.27: only A and D have vertical support symbols. Crosses at A, B, C and D identify lateral restraints. B and C carry downward point loads; the small arrows over the entire span represent the specified

3 \frac{kN}{m}

self-weight. DL denotes dead load, LL imposed load, S/W self-weight, Shear shear force, and BM bending moment. From left to right, the original shear diagram prints

146, 133, 19, 32, 108, 120 kN

. The first two labels are above zero and the last four below. The source prints only magnitudes for those last four; under this lesson's sign convention they are negative shear values, not four positive loads. At B and C, the original moment diagram labels

419

and

342 kN \cdot m

. These are rounded values on the lecturer's diagram. The adjacent working recalculates from the given loads at higher precision and does not treat the diagram results as additional inputs.Open full-size image

Read the diagram and collect the data

Input	Exact diagram or table source
Vertical span	Three dimension strings of $3 m$ give $A = 0, B = 3, C = 6, D = 9 m$ .
Lateral model	Crosses define three $3 m$ unbraced segments; question sets Lᴇ=3,000 mm.
Trial section	The source selects $457 \times 191 \times 74$ UB. Data pp.9–10 give $D = 457$ , $t = 9$ , $T = 14.5$ , $r = 10.2$ , $d = 407.6 mm$ ; $\frac{b}{T} = 6.57$ , $\frac{d}{t} = 45.3$ ; $r_{y} = 4.20 cm$ ; $I_{x} = 33,320 {cm}^{4}$ ; $Z_{x} = 1,458$ , $S_{x} = 1,653 {cm}^{3}$ ; $u = 0.877$ , $x = 33.9$ .
Missing local-web data	No stiff bearing lengths, distances to member ends or complete local flange-restraint details are given. Crosses alone identify lateral restraint, not both web-buckling conditions.

Before calculating: recognition and strategy

Analyse the continuous $9 m$ beam on its two vertical supports. Use that BMD to examine the three separate lateral-buckling segments. First reproduce the lecturer’s rounded, straight-line BC approximation, then retain the self-weight curvature with Table 8.4b to check every segment. Deflection uses the imposed loads $60$ and $30 kN$ at their actual off-centre positions; a central-point formula cannot be substituted.

1. Factored loads

P_{B} = 1.4 \times 40 + 1.6 \times 60 = 56 + 96 = 152 kN

P_{C} = 1.4 \times 20 + 1.6 \times 30 = 28 + 48 = 76 kN

w = 1.4 \times 3 = 4.2 \frac{kN}{m}

Total load:

152 + 76 + 4.2 \times 9 = 265.8 kN

Animation labFrom characteristic to design load1 concept · 1 source expressions

Supplement to the original lesson. Enable JavaScript to play, step through calculations and rotate 3D models. The following explanation remains readable offline.

G is permanent load; Q is imposed load. A surface load and a line load also have different units.
This illustration uses the course gravity case $1.4 G + 1.6 Q$ . Other combinations in the original text retain their own factors.
For illustrative $G = 5 \frac{kN}{m ^{2}}$ , change Q and watch each separate contribution.
Do not carry this ULS total automatically into deflection. Follow the stated SLS load case.

2. Reactions without premature rounding

Simple explanation: Why reactions balance the loads

Think of a seesaw that must neither fall nor turn.

Upward and downward forces must balance.
Take moments about a support to eliminate its reaction.
Use perpendicular distance from the point to the force line.

Remember: A lateral restraint is not automatically a vertical support.

Related concept and full method

Take moments about D: each downward load contributes its distance to D; the UDL resultant $4.2 \times 9$ acts at midspan, $4.5 m$ from D.

R_{A} \times 9 = 152 \times 6 + 76 \times 3 + (4.2 \times 9) \times 4.5

= 912 + 228 + 170.1 = 1,310.1 kN \cdot m

R_{A} = \frac{1,310.1}{9} = 145.566667 kN

R_{D} = 265.8 - 145.566667 = 120.233333 kN

Check by moments about A: $120.233333 \times 9 = 152 \times 3 + 76 \times 6 + 37.8 \times 4.5 = 1,082.1 kN \cdot m$ .

Animation labBalance reactions and moments1 concept · 2 source expressions

Supplement to the original lesson. Enable JavaScript to play, step through calculations and rotate 3D models. The following explanation remains readable offline.

This demonstrator is a simply supported 6 m beam with a 60 kN point load; it is not the page’s original loading diagram.
$R_{B} \times L = P \times a$ . Moving the load towards B increases $R_{B}$ .
$R_{A} = P - R_{B}$ . The two upward reactions must sum to P.
With the load at midspan, $R_{A} = R_{B} = 30 kN$ . End couples, UDLs and overhangs require their own equilibrium terms.

3. Construct the SFD and BMD from the left-hand free body

For shear (

kN

), start with

V (x) = 145.566667 - 4.2 x

; when

x > 3

, add

- 152

; when

x > 6

, also add

- 76

. Negative signs mean the point loads cause downward shear jumps.
For moment (

kN \cdot m

), start with

M (x) = 145.566667 x - \frac{4.2 x^{2}}{2}

; when

x > 3

, add

- 152 (x - 3)

; when

x > 6

, also add

- 76 (x - 6)

. Each negative term is a point load times its lever arm to the section; include it only after the section has passed that load.

Read each “when” term as zero before its load position and include it afterwards. $x$ is in metres from A. The shorthand is often written with positive-part brackets $⟨ x - a ⟩$ ; it is a beam-position coordinate, unrelated to the torsional index.

Position	Shear just left / right ( $kN$ )	Moment ( $kN \cdot m$ )
$A$	$0$ ／ $+ 145.5667$	$0$
B, $x = 3$	$+ 132.9667$ ／ $- 19.0333$	$145.566667 \times 3 - 2.1 \times 9 = 417.8$
C, $x = 6$	$- 31.6333$ ／ $- 107.6333$	$145.566667 \times 6 - 2.1 \times 36 - 152 \times 3 = 341.8$
$D$	$- 120.2333$ ／ $0$	$0$

B is the maximum because shear changes from positive to negative there. BC and CD have negative shear throughout, so their bending moment decreases. Draw straight sloping shear segments, vertical jumps at the point loads, and curved bending segments continuous at B and C.

Animation labBuild the shear and moment diagrams1 concept · 5 source expressions

Supplement to the original lesson. Enable JavaScript to play, step through calculations and rotate 3D models. The following explanation remains readable offline.

For the illustrative 60 kN point load on a 6 m simply supported beam, balance the reactions before making a cut.
$V = R_{A}$ and $M = R_{A} x$ before the point load. Moment grows linearly.
Shear jumps down by P. To its right, $V = R_{A} - P$ and $M = R_{A} x - P (x - a)$ .
For this case, moment is continuous and returns to zero at B. A concentrated applied couple instead creates a moment jump.

4. Confirm strength and section class

T = 14.5 mm \leq 16 \to p_{y} = 355 \frac{N}{{mm}^{2}}

ε = \sqrt{\frac{275}{355}} = 0.880141

\frac{b}{T} = 6.57 < 9 ε = 7.92127; \frac{d}{t} = 45.3 < 80 ε = 70.4113

The section is Class 1;

\frac{d}{t} = 45.3 < 70 ε = 61.6099

, so no separate web shear-buckling calculation is needed.

Animation labWhy thin elements buckle locally1 concept · 1 source expressions

Supplement to the original lesson. Enable JavaScript to play, step through calculations and rotate 3D models. The following explanation remains readable offline.

The flange outstand and web have different widths, thicknesses and edge support conditions.
A thinner plate can wrinkle locally before the complete member loses stability.
Class 1 allows plastic rotation; Class 2 reaches plastic resistance; Class 3 reaches elastic resistance; Class 4 requires effective properties.
Check every relevant compression element with the supplied limits and stress distribution. The deformation shown is qualitative.

5. Shear resistance

A_{v} = t D = 9 \times 457 = 4,113 {mm}^{2}

V_{c} = \frac{355 \times 4,113}{\sqrt{3} \times 1,000} = 842.9978 kN

V_{\max} = 145.5667 < 842.9978

Passes.

Animation labSee shear in the web1 concept · 1 source expressions

Supplement to the original lesson. Enable JavaScript to play, step through calculations and rotate 3D models. The following explanation remains readable offline.

Internal shear keeps the two sides of the cut in vertical equilibrium.
For the course’s common I-section case, the web provides the principal shear area; use the specified definition.
A slender web may require a different shear-buckling route before a simple shear-resistance formula is used.
Use $V_{c} = \frac{p _{y} A _{v}}{3}$ where applicable in the course. Convert N to kN before comparing with design shear.

6. Actual section moment resistance

Maximum shear coexisting with the moment at B

= 132.9667 kN

0.6 V_{c} = 505.7987 kN > 132.9667

Use the low-shear branch.

p_{y} S_{x} = \frac{355 \times 1,653}{1,000} = 586.815 kN \cdot m

1.2 p_{y} Z_{x} = \frac{1.2 \times 355 \times 1,458}{1,000} = 621.108 kN \cdot m

M_{c} = 586.815 kN \cdot m > 417.8

Section bending resistance passes.

Animation labCompression and tension across a section1 concept · 1 source expressions

Supplement to the original lesson. Enable JavaScript to play, step through calculations and rotate 3D models. The following explanation remains readable offline.

For sagging, the top flange is in compression and the bottom in tension; hogging reverses this.
Elastic bending stress varies with distance from the neutral axis: $σ = \frac{M y}{I}$ .
The section class governs whether elastic, plastic or effective properties may be used.
Use the shear at the section under examination; the largest shear elsewhere is not automatically coexistent.

7. Check every unbraced segment

Simple explanation: A beam can escape sideways

The compressed flange can move sideways while the section twists.

Divide the beam at effective lateral restraints.
Use each segment’s effective length to obtain its buckling resistance.
Compare that resistance with the segment’s equivalent moment demand.

Remember: A section bending check alone does not check lateral-torsional buckling.

Related concept and full method

All three segments have the same section and $3 m$ effective length, hence the same buckling resistance. Use the weak-axis radius $4.20 cm = 42 mm$ .

λ = \frac{3,000}{42} = 71.428571

\frac{λ}{x} = \frac{71.428571}{33.9} = 2.10704

v = {[1 + 0.05 {(\frac{71.428571}{33.9})}^{2}]}^{- 0.25} = 0.951117

For Class 1,

β_{w} = 1

λ_{LT} = 0.877 \times 0.951117 \times 71.428571 \times \sqrt{1} = 59.580686

Table 8.3a,

p_{y} = 355

column:

55

row gives

274

，

60

row gives

257 \frac{N}{{mm}^{2}}

f = \frac{59.580686 - 55}{60 - 55} = 0.916137

p_{b} = 274 + 0.916137 (257 - 274) = 258.425666 \frac{N}{{mm}^{2}}

M_{b} = \frac{258.425666 \times 1,653}{1,000} = 427.177626 kN \cdot m

Lecturer route: approximate BC as straight, $β = \frac{342}{419} \approx 0.816$ ; read mᴸᵀ≈0.93, giving $0.93 \times 419 = 389.67 kN \cdot m$ . Rounded λᴸᵀ≈59.5 gives pᵦ≈259 and Mᵦ≈428 kNm. It passes. To include the UDL exactly, use quarter-point magnitudes from $M (x)$ above:

Segment	Quarter positions from A ( $m$ )	$M_{2}$ 、 $M_{3}$ 、 $M_{4}$ （ $kN \cdot m$ ）	$M_{\max}$ （ $kN \cdot m$ ）	$m^{LT} M_{\max}$ （ $kN \cdot m$ ）
$AB$	$0.75, 1.5, 2.25$	$107.99375, 213.625, 316.89375$	$417.8$	$254.105625$
$BC$	$3.75, 4.5, 5.25$	$402.34375, 384.525, 364.34375$	$417.8$	$390.825625$
$CD$	$6.75, 7.5, 8.25$	$259.89375, 175.625, 88.99375$	$341.8$	$208.505625$

Segment BC:

M (3.75) = 145.566667 \times 3.75 - 2.1 \times {3.75}^{2} - 152 \times 0.75 = 402.34375

M (4.5) = 145.566667 \times 4.5 - 2.1 \times {4.5}^{2} - 152 \times 1.5 = 384.525

M (5.25) = 145.566667 \times 5.25 - 2.1 \times {5.25}^{2} - 152 \times 2.25 = 364.34375

m_{LT} = 0.2 + \frac{[0.15 \times 402.34375 + 0.5 \times 384.525 + 0.15 \times 364.34375]}{417.8}

= 0.935437 > 0.44

Equivalent moment demand:

0.2 \times 417.8 + 0.15 \times 402.34375 + 0.5 \times 384.525 + 0.15 \times 364.34375

= 390.825625 kN \cdot m < 427.177626

Passes.

For AB insert its three ordinates into the same expression: $0.2 \times 417.8 + 0.15 \times 107.99375 + 0.5 \times 213.625 + 0.15 \times 316.89375 = 254.105625$ . For CD: $0.2 \times 341.8 + 0.15 \times 259.89375 + 0.5 \times 175.625 + 0.15 \times 88.99375 = 208.505625$ . Their $m$ factors also exceed $0.44$ . BC governs and all three pass.

Animation labA beam bends sideways and twists1 concept · 7 source expressions

Supplement to the original lesson. Enable JavaScript to play, step through calculations and rotate 3D models. The following explanation remains readable offline.

For the illustrated sagging beam the top flange is compressed.
The unrestrained compression flange can move sideways while the complete cross-section twists.
Only effective restraints divide the member into unbraced segments. They do not automatically add vertical supports.
Use the segment effective length, section properties and matching moment factor. This is an exaggerated mode shape, not a calculated displacement.

8. The omitted local web checks: what can and cannot be concluded

Simple explanation: A concentrated force can hurt one small region

A narrow contact presses much harder locally than the same force spread over a wider contact.

Find the actual stiff bearing length and end position.
Check local crushing and local web buckling separately.
Use the restraint conditions required by each expression.

Remember: Missing contact dimensions cannot be guessed from drawing scale.

Related concept and full method

Page 30 asks students to check bearing and web buckling, but neither the support bearing geometry nor the load-introduction details at B and C are supplied. A lateral-restraint cross does not specify the stiff bearing length or rotational restraint of the loaded flange relative to the web. Therefore the numerical local-web follow-up remains incomplete from the available source data.

Known section values:

t = 9 mm

p_{yw} = 355

k = T + r = 14.5 + 10.2 = 24.7 mm

. At A:

n = \min (2 + \frac{0.6 b_{e, A}}{24.7}, 5)

P_{bw, A} = \frac{(b_{1, A} + 24.7 n) \times 9 \times 355}{1,000}

= 3.195 (b_{1, A} + 24.7 n) kN

Require

\geq 145.5667 kN

. At D, use its actual bearing inputs and require

\geq 120.2333 kN

. If B/C are internal bearing locations,

n = 5

, giving

P_{bw} = 3.195 (b_{1} + 123.5) kN

; compare with the point force actually transferred at that contact.

Once $b_{1}$ , $b_{e}$ , $a_{e}$ and both local restraint conditions are known, evaluate $P_{x}$ using $d = 407.6 mm$ and the correct end/interior branch in the web-buckling procedure. If local restraint is absent, Lᴱ,web is also needed. Do not copy Example 1’s $160 mm$ , $80 mm$ and $0 mm$ values into this different diagram. The source’s overall “adequate” statement is supported for shear, section bending and LTB, but local-web adequacy is not established by the supplied information.

Animation labSpread a concentrated force into the web3 concepts · 2 source expressions

Supplement to the original lesson. Enable JavaScript to play, step through calculations and rotate 3D models. The following explanation remains readable offline.

A concentrated reaction first enters through the bearing/contact region.
The flange and root geometry spread the force before it enters the web.
A wider effective bearing region can reduce local stress for the same force.
End distance, stiff bearing length and restraint conditions must come from the original question. This slider is illustrative only.

9. Complete the omitted off-centre deflection calculation

Simple explanation: How much does the beam sag?

Strength asks whether it fails; deflection asks how far it moves.

Use the serviceability load case specified by the course question.
Choose the expression matching the support and load positions.
Use consistent units for load, length, E and I.

Remember: The largest deflection is not always at midspan.

Related concept and full method

Use only the locations $x = 3 m$ and their imposed loads $60 kN$ , and $x = 6 m$ of $30 kN$ ; there is no imposed UDL. This calculation uses $N$ , $mm$ , constant $E = 205,000$ , and exact section-table value $I_{x} = 33,320 \times 10^{4} = 333,200,000 {mm}^{4}$ . Imposed-load reactions are $R_{A} = \frac{60 \times 6 + 30 \times 3}{9} = 50 kN$ , $R_{D} = 40 kN$ .

The general method starts from the curvature relationship. Taking downward deflection $v$ as positive, $E I v^{″} = - M$ . Integrate the bending moment twice. Define ${⟨ x - a ⟩}^{n}$ as follows: when $x < a$ , its value is zero; when $x \geq a$ , its value is ${(x - a)}^{n}$ . Use the support conditions $v (0) = v (9,000) = 0$ to determine the integration constants.

E I v (x) = C x - \frac{50,000 x^{3}}{6} + \frac{60,000 {⟨ x - 3,000 ⟩}^{3}}{6} + \frac{30,000 {⟨ x - 6,000 ⟩}^{3}}{6}

C = \frac{[50,000 \times {9,000}^{3} - 60,000 \times {6,000}^{3} - 30,000 \times {3,000}^{3}]}{6 \times 9,000}

= 420,000,000,000 N \cdot {mm}^{2}

E I v^{'} (x) = C - \frac{50,000 x^{2}}{2} + \frac{60,000 {⟨ x - 3,000 ⟩}^{2}}{2} + \frac{30,000 {⟨ x - 6,000 ⟩}^{2}}{2}

Maximum downward deflection occurs where slope $v^{'} = 0$ . It lies between B and C: slope numerator at $3,000$ is $+ 195 \times 10^{9}$ ; at $6,000$ it is $- 210 \times 10^{9}$ , and the positive bending moment makes slope decrease continuously.

In the interval

3,000 < x < 6,000

0 = 420 \times 10^{9} - 25,000 x^{2} + 30,000 {(x - 3,000)}^{2}

= 5,000 x^{2} - 180,000,000 x + 690 \times 10^{9}

x = \frac{[180,000,000 - \sqrt{{180,000,000}^{2} - 4 \times 5,000 \times 690 \times 10^{9}}]}{2 \times 5,000}

= 4,361.818 mm

The other root is outside this interval and is rejected.

v_{\max} = \frac{[420 \times 10^{9} \times 4,361.818 - \frac{50,000 \times {4,361.818}^{3}}{6} + \frac{60,000 \times {1,361.818}^{3}}{6}]}{205,000 \times 333,200,000}

= 17.06545 mm

At midspan the deflection is $17.04645 mm$ , close but not the exact maximum. The source does not specify the finish category for this example. If the brittle-finish $\frac{L}{360}$ criterion used in preceding examples is adopted explicitly, the limit is $\frac{9,000}{360} = 25 mm$ and $17.06545 < 25$ passes. The calculated deflection is determined; the acceptance statement is conditional on that chosen serviceability limit.

Animation labSee stiffness and deflection1 concept · 15 source expressions

Supplement to the original lesson. Enable JavaScript to play, step through calculations and rotate 3D models. The following explanation remains readable offline.

Use the specified SLS load, span and support arrangement. The demonstrator has a full-span UDL.
The loaded beam bends; the deformation is exaggerated so its shape can be seen.
For a simply supported full-span UDL, $δ = \frac{5 w L ^{4}}{384 E I}$ . Double L with w, E and I unchanged: δ becomes 16 times as large.
The readout uses $w = 10 \frac{N}{mm}$ , $E = 205000 \frac{N}{mm ^{2}}$ and $I = 200000000 mm^{4}$ . Select the finish/support-specific limit from the original table.

Compact exam answer

$457 \times 191 \times 74$ UB S355 is Class 1. Rₐ/Rᴅ= $145.5667$ / $120.2333 kN$ ; Mᴮ/Mᶜ= $417.8$ / $341.8 kN \cdot m$ . $V_{c} = 842.998 kN$ ; $M_{c} = 586.815 kN \cdot m$ . All segments Lᴱ=3 m: λᴸᵀ=59.5807,pᵦ=258.426,Mᵦ=427.178 kNm. Exact equivalent demands AB/BC/CD $= 254.106$ / $390.826$ / $208.506 kN \cdot m$ ; all pass, BC controls. Imposed $δ_{\max} = 17.065 mm$ at $4.362 m$ ; passes if $\frac{L}{360}$ applies. Local bearing/web-buckling numerical checks are unresolved because bearing and local-restraint data are missing.

Mistakes to avoid

Do not turn B and C into vertical supports.
Do not use $3 m$ as the span for the whole-beam deflection.
Do not omit BC end moments when evaluating its quarter points.
Do not present assumed bearing dimensions as given.

Procedure for an unfamiliar variant

Identify the vertical supports, span, lateral restraints and individual load positions.
Keep dead and imposed loads separate; form the required ultimate and serviceability cases.
Find reactions and the maximum shear/moment by equilibrium, with units.
Choose a section-table row, confirm thickness-dependent strength, and classify both flange and web.
Check shear and bending; add segment LTB where restraint is discrete.
Complete the requested web and deflection checks; state the governing result and any missing data.

Independent self-check

Try it yourself. Invented variant: with all loads unchanged, the lateral restraint at C is removed. Which parts of this solution must be recalculated?

Reveal answer and reasoning

Vertical reactions, whole-beam SFD/BMD, section shear/bending and elastic deflection do not change. LTB segments become $AB = 3 m$ and $BD = 6 m$ . For BD use Lᴱ=6,000 mm (if the same effective-length rule is stipulated), recompute $λ$ , $v$ ,λᴸᵀ,pᵦ and Mᵦ, and use quarter positions $4.5, 6, 7.5 m$ with the true BD maximum atB. The point load atC remains even though its lateral restraint is removed.

Animation labSee stiffness and deflection4 concepts · 2 source expressions

Supplement to the original lesson. Enable JavaScript to play, step through calculations and rotate 3D models. The following explanation remains readable offline.

Use the specified SLS load, span and support arrangement. The demonstrator has a full-span UDL.
The loaded beam bends; the deformation is exaggerated so its shape can be seen.
For a simply supported full-span UDL, $δ = \frac{5 w L ^{4}}{384 E I}$ . Double L with w, E and I unchanged: δ becomes 16 times as large.
The readout uses $w = 10 \frac{N}{mm}$ , $E = 205000 \frac{N}{mm ^{2}}$ and $I = 200000000 mm^{4}$ . Select the finish/support-specific limit from the original table.