Worked examples

Beam example 2: trace a library-floor load into beam B3

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Design the fully restrained supporting beam B3 in the library floor plan. Characteristic dead and imposed floor loads are $6$ and $4 \frac{kN}{m^{2}}$ ; the dead load includes the self-weight allowances used by the example. Use S355 and the lecturer’s simply supported member model.

Original source: LectureNotes/Ch 3_Beam.pdf — p. 22, p. 23, p. 24, p. 25, p. 26. 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.

Read the diagram and collect the data

Feature	Meaning for loading
Slab span marks crossing the B1 lines	Slab load reaches the parallel B1 beams over a $1.25 m$ tributary width.
B1 above and below each interior B2	Each B2 loading point receives two B1 end reactions.
B2 on both sides of the interior B3	B3 receives two B2 end reactions at its centre.
B3 also directly borders slab strips	Interior B3 carries its own $1.25 m$ tributary slab strip, represented as a UDL.
Given and assumed	All self-weight is already allowed for in the supplied dead intensity. B3 fully restrained; lecturer assumes brittle finishes for serviceability.

Before calculating: recognition and strategy

Work upstream before downstream: slab → B1 → B2 → B3. A supporting beam receives a point load equal to the end reaction of the supported beam, not that beam’s entire load. Keep dead and imposed reactions separate until the B3 loading is established. Do not treat the $5 m$ horizontal bay width as the B3 span.

1. Slab to B1

B1 span

= 3 m

; tributary width

= 1.25 m

g_{B 1} = 6 \frac{kN}{m^{2}} \times 1.25 m = 7.5 \frac{kN}{m}

q_{B 1} = 4 \times 1.25 = 5 \frac{kN}{m}

Dead-load reaction at each end:

\frac{7.5 \times 3}{2} = 11.25 kN

Imposed-load reaction at each end:

\frac{5 \times 3}{2} = 7.5 kN

The division by two comes from a uniformly loaded simply supported span. These are characteristic reactions, before the ultimate factors.

Animation labFollow the floor load in 3D2 concepts · 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 floor carries pressure in $\frac{kN}{m ^{2}}$ . The highlighted strip belongs to one secondary beam.
Multiply pressure by tributary width: $w = q \times b$ . The illustration uses $q = 5 \frac{kN}{m ^{2}}$ .
A primary beam receives the secondary beam reaction at their connection, not a new full-span UDL.
Trace reactions down to columns and foundations. Count each loaded area once.

2. B1 reactions to B2

Each internal B1 line has two spans joining B2:

G_{point, B 2} = 2 \times 11.25 = 22.5 kN

Q_{point, B 2} = 2 \times 7.5 = 15 kN

In the interval

5 m

B2 span has three equally spaced point loads at

1.25, 2.5, 3.75 m

. B2 dead-load reaction at each end:

\frac{3 \times 22.5}{2} = 33.75 kN

B2 imposed-load reaction at each end:

\frac{3 \times 15}{2} = 22.5 kN

There are three internal point loads, not four: four slab strips are bounded by three interior B1 lines and the two B3 boundaries. Symmetry of the three loads gives half the total at each B2 end. Loads arriving directly at column-supported beam ends do not add span bending.

Animation labFollow the floor load in 3D2 concepts · 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 floor carries pressure in $\frac{kN}{m ^{2}}$ . The highlighted strip belongs to one secondary beam.
Multiply pressure by tributary width: $w = q \times b$ . The illustration uses $q = 5 \frac{kN}{m ^{2}}$ .
A primary beam receives the secondary beam reaction at their connection, not a new full-span UDL.
Trace reactions down to columns and foundations. Count each loaded area once.

3. Assemble B3 point load and direct slab UDL

Simple explanation: How a floor load reaches a beam

Each beam collects the load from its own strip of floor.

Find the tributary width from the actual plan.
Area load × tributary width gives load per beam length.
A supporting beam receives the other beam’s end reaction.

Remember: A reaction becomes a point load, not automatically a UDL.

Related concept and full method

Midspan dead point load:

2 \times 33.75 = 67.5 kN

Midspan imposed point load:

2 \times 22.5 = 45 kN

Direct dead UDL:

6 \times 1.25 = 7.5 \frac{kN}{m}

Direct imposed UDL:

4 \times 1.25 = 5 \frac{kN}{m}

Ultimate point load:

P = 1.4 \times 67.5 + 1.6 \times 45 = 94.5 + 72 = 166.5 kN

Ultimate UDL:

w = 1.4 \times 7.5 + 1.6 \times 5 = 10.5 + 8 = 18.5 \frac{kN}{m}

The factor of two at the central point represents the B2 on each side of B3. It does not double B3’s $6 m$ span or its $1.25 m$ tributary strip.

Animation labFollow the floor load in 3D2 concepts · 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 floor carries pressure in $\frac{kN}{m ^{2}}$ . The highlighted strip belongs to one secondary beam.
Multiply pressure by tributary width: $w = q \times b$ . The illustration uses $q = 5 \frac{kN}{m ^{2}}$ .
A primary beam receives the secondary beam reaction at their connection, not a new full-span UDL.
Trace reactions down to columns and foundations. Count each loaded area once.

4. Reactions, shear and moment of B3

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

Characteristic dead-load reaction:

\frac{67.5 + 7.5 \times 6}{2} = 56.25 kN

Characteristic imposed-load reaction:

\frac{45 + 5 \times 6}{2} = 37.5 kN

Ultimate reaction:

1.4 \times 56.25 + 1.6 \times 37.5 = 138.75 kN

Direct check:

\frac{166.5 + 18.5 \times 6}{2} = 138.75 kN

Shear immediately left of midspan:

138.75 - 18.5 \times 3 = 83.25 kN

Immediately right of midspan:

83.25 - 166.5 = - 83.25 kN

M_{\max} = \frac{166.5 \times 6}{4} + \frac{18.5 \times 6^{2}}{8}

= 249.75 + 83.25 = 333 kN \cdot m

Maximum shear occurs at the supports:

| V |_{\max} = 138.75 kN

Animation labBalance reactions and moments2 concepts · 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.

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.

5. Select $457 \times 152 \times 52$ UB

Trial

p_{y} = 355 \frac{N}{{mm}^{2}}

S_{required} = \frac{333 \times 10^{6}}{355} = 938,028.17 {mm}^{3} = 938.028 {cm}^{3}

Selected

S_{x} = 1,096 {cm}^{3} > 938.028

Data File p.9 row $457 \times 152 \times 52$ gives $D = 449.8$ , $t = 7.6$ , $T = 10.9$ , $r = 10.2$ , $d = 407.6 mm$ , $\frac{b}{T} = 6.99$ and $\frac{d}{t} = 53.6$ . Page 10, the same designation, gives $I_{x} = 21,370 {cm}^{4}$ , $Z_{x} = 950 {cm}^{3}$ , $S_{x} = 1,096 {cm}^{3}$ . $T = 10.9 \leq 16$ confirms pᵧ=355. No interpolation is used for section properties: read the exact row.

Animation labRead a table without losing the keys3 concepts · 12 source expressions

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

Name the required property: material strength, section property, buckling strength or a moment factor.
Keep section size, steel grade, thickness band, curve and axis as separate lookup keys.
$cm^{2}$ , $cm^{3}$ and $cm^{4}$ require different conversion powers. Do not use adjacent columns interchangeably.
Use bracketing rows within the same valid column. The original page remains the source of all table values.

6. Classify the trial section

ε = \sqrt{\frac{275}{355}} = 0.880141

\frac{b}{T} = 6.99 < 9 ε = 7.92127

The flange is Class 1.

\frac{d}{t} = 53.6 < 80 ε = 70.4113

The web in bending is Class 1; overall Class 1.

\frac{d}{t} = 53.6 < 70 ε = 61.6099

No separate web shear-buckling check is needed.

Animation labWhy thin elements buckle locally2 concepts · 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.

7. Check shear resistance

A_{v} = t D = 7.6 \times 449.8 = 3,418.48 {mm}^{2}

V_{c} = \frac{355 \times 3,418.48}{\sqrt{3} \times 1,000} = 700.649 kN

138.75 < 700.649

: shear 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.

8. Check moment resistance

Simple explanation: Why one flange squeezes and the other stretches

Bending makes opposite sides of the section do opposite jobs.

Find the moment from the actual loads and supports.
Choose the resistance formula allowed by the section class.
Check whether the coexistent shear changes that formula.

Remember: Use shear at the location being checked, not an unrelated maximum.

Related concept and full method

Shear at the same midspan section

= 83.25 kN

0.6 V_{c} = 420.390 kN > 83.25

is low shear.

p_{y} S_{x} = \frac{355 \times 1,096}{1,000} = 389.08 kN \cdot m

1.2 p_{y} Z_{x} = \frac{1.2 \times 355 \times 950}{1,000} = 404.7 kN \cdot m

M_{c} = 389.08 kN \cdot m > 333 kN \cdot m

Passes.

The source’s requested B3 design demonstrates shear, moment and deflection. It does not specify a bearing geometry for a numerical local-web design here. Do not import the $160 mm$ bearing from Example 1.

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.

9. Check B3 serviceability

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 imposed load

P = 45,000 N

w = 5 \frac{N}{mm}

, and

L = 6,000 mm

I = 21,370 \times 10^{4} = 213,700,000 {mm}^{4}

δ_{P} = \frac{45,000 \times {6,000}^{3}}{48 \times 205,000 \times 213,700,000} = 4.62239 mm

δ_{w} = \frac{5 \times 5 \times {6,000}^{4}}{384 \times 205,000 \times 213,700,000} = 1.92600 mm

δ = 4.62239 + 1.92600 = 6.54839 mm

\frac{L}{360} = \frac{6,000}{360} = 16.66667 mm

6.54839 < 16.66667

: passes under the assumed brittle-finish condition.

Animation labSee stiffness and deflection1 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.

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

B1 characteristic reactions $G$ / $Q$ = $11.25$ / $7.5 kN$ . B2 has three $22.5$ / $15 kN$ points; end reactions $33.75$ / $22.5 kN$ . B3 central $G$ / $Q$ = $67.5$ / $45 kN$ plus $7.5$ / $5 \frac{kN}{m}$ . Ultimate $P = 166.5 kN$ , $w = 18.5 \frac{kN}{m}$ , $V_{\max} = 138.75 kN$ , $M_{\max} = 333 kN \cdot m$ . Select $457 \times 152 \times 52$ UB S355, Class 1: $V_{c} = 700.649 kN$ , $M_{c} = 389.08 kN \cdot m$ , $δ = 6.548 mm < 16.667 mm$ . The demonstrated checks pass.

Mistakes to avoid

Do not make floor pressure a beam UDL without multiplying by tributary width.
Do not transfer a supported beam’s entire load to each end.
Include B3’s directly supported slab strip once.
Do not add self-weight already included by the problem.

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: only the characteristic floor imposed pressure rises from $4$ to $5 \frac{kN}{m^{2}}$ . Find the revised B3 ultimate $P$ and $w$ , and imposed deflection; retain the original dead loading.

Reveal answer and reasoning

All imposed effects multiply by $\frac{5}{4}$ . B3 imposed point $= 56.25 kN$ ; imposed UDL $= 6.25 \frac{kN}{m}$ . $P = 1.4 \times 67.5 + 1.6 \times 56.25 = 184.5 kN$ ; $w = 1.4 \times 7.5 + 1.6 \times 6.25 = 20.5 \frac{kN}{m}$ . $M = \frac{184.5 \times 6}{4} + \frac{20.5 \times 36}{8} = 369 kN \cdot m < 389.08$ . Imposed deflection= $6.54839 \times 1.25 = 8.18549 mm < 16.66667$ . Maximum shear= $\frac{184.5 + 20.5 \times 6}{2} = 153.75 kN < 700.649$ ; these checks still pass.

Animation labFollow the floor load in 3D5 concepts · 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.

The floor carries pressure in $\frac{kN}{m ^{2}}$ . The highlighted strip belongs to one secondary beam.
Multiply pressure by tributary width: $w = q \times b$ . The illustration uses $q = 5 \frac{kN}{m ^{2}}$ .
A primary beam receives the secondary beam reaction at their connection, not a new full-span UDL.
Trace reactions down to columns and foundations. Count each loaded area once.

Beam example 2: trace a library-floor load into beam B3

Read the diagram and collect the data

Before calculating: recognition and strategy

1. Slab to B1

2. B1 reactions to B2

3. Assemble B3 point load and direct slab UDL

4. Reactions, shear and moment of B3

5. Select 457×152×52457×152×52 UB

6. Classify the trial section

7. Check shear resistance

8. Check moment resistance

9. Check B3 serviceability

Compact exam answer

Mistakes to avoid

Procedure for an unfamiliar variant

Independent self-check

5. Select $457 \times 152 \times 52$ UB