Mock solutions

Mock Q4: full solution and invented marking guide

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.

Chinese–English terminology

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

Return to question. All marking guidance is invented.

MOCK Q4 · two equal design point loads; all geometry repeated in the adjacent question. — Invented mock-question diagram: two equal design point loads. The $457 \times 191 \times 67$ UB has self-weight neglected; downward arrows at B and C each represent $140 kN$ . A–B, B–C and C–D are each $2 m$ ; A and D are vertical supports. Schematic, not to scale; stated dimensions govern.Open full-size image.

(a) Equilibrium and diagrams

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

R_{D} \times 6 = 140 \times 2 + 140 \times 4 = 840

Therefore

R_{D} = 140 kN

R_{A} = 280 - 140 = 140 kN

Segment AB

V = + 140

; segment BC

V = 0

; segment CD

V = - 140

.
Segment AB

M = 140 x

; segment BC

M = 280

; segment CD

M = 140 (6 - x)

, where

x

is measured in

m

M_{\max} = 280 kN \cdot m

; end moments are zero.

Calculated shear and bending diagrams with labelled support/load ordinates; positive values upward. All values are repeated in the working. — Teaching diagrams drawn from the calculation, not original source images. Labels identify calculated diagrams, positive ordinates upwards, shear force and bending moment. Positions are in $m$ , $V$ is measured in $kN$ , $M$ is measured in $kN \cdot m$ ; A/B/C/D match the question diagram. Numerical values are also given in the piecewise equations above.Open full-size image.

Animation labBuild the shear and moment diagrams2 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.

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.

(b) End-only lateral restraint

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

Lookup: Data File pp.9–10, exact $457 \times 191 \times 67$ UB row. Dimensions on p.9; properties on p.10. $D$ is overall depth; $d$ is the clear web depth between root fillets, not the nominal designation.

Property	Lookup value / conversion
Dimensions	D453.4, web t 8.5, flange T12.7, root r 10.2, clear d 407.6, all $mm$
Local ratios	$\frac{b}{T} = 7.48$ ； $\frac{d}{t} = 48$ 。
Major-axis properties	$I_{x} = 29380 {cm}^{4} = 2.938 \times 10^{8} {mm}^{4}$ ; $Z_{x} = 1296 {cm}^{3}$ ; $S_{x} = 1471 {cm}^{3}$ .
LTB properties	$r_{y} = 4.12 cm = 41.2 mm$ ; $u = 0.872$ ; torsional index $x = 37.9$ .

S355, flange

T = 12.7 mm

, so

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

ε = \sqrt{\frac{275}{355}} = 0.880141

Flange:

\frac{b}{T} = 7.48 < 9 ε = 7.921268

Web in bending:

\frac{d}{t} = 48 < 80 ε = 70.411267

Both are Class 1, so

β_{W} = 1

; plastic moment resistance is permitted.

\frac{d}{t} = 48 < 70 ε = 61.609858

No shear-buckling calculation is required under the course limit.

A_{v} = t D = 8.5 \times 453.4 = 3853.9 {mm}^{2}

V_{c} = \frac{p_{y} A_{v}}{\sqrt{3}} = \frac{355 \times 3853.9}{\sqrt{3} \times 1,000} = 789.892822 kN

V_{\max} = 140.000000 kN < V_{c}

Shear passes.

0.6 V_{c} = 473.935693 kN > V_{\max}

Even maximum shear is low, so the low-shear bending formula applies throughout the beam.

M_{c} = \min (p_{y} S_{x}, 1.2 p_{y} Z_{x})

= \min (\frac{355 \times 1471}{1,000}, \frac{1.2 \times 355 \times 1296}{1,000})

= \min (522.205, 552.096) = 522.205 kN \cdot m

| M |_{\max} = 280.000000 kN \cdot m

Bending passes.

Full-span quarter points

x = 1.5

3

4.5 m

, giving

M_{2} = 210

M_{3} = 280

M_{4} = 210 kN \cdot m

m_{LT} = 0.2 + \frac{0.15 \times 210 + 0.5 \times 280 + 0.15 \times 210}{280}

= 0.2 + \frac{203}{280} = 0.925

Equivalent demand:

0.925 \times 280 = 259 kN \cdot m

Normal-load effective length is the given segment length. All quantities in $λ$ and λᴸᵀ are dimensionless. This Class 1 section has $β_{W} = 1$ .

λ = \frac{L_{E}}{r_{y}} = \frac{6000}{41.2} = 145.631068

v = {[1 + 0.05 {(\frac{λ}{x})}^{2}]}^{- \frac{1}{4}}

= {[1 + 0.05 {(\frac{145.631068}{37.9})}^{2}]}^{- \frac{1}{4}} = 0.870908

λ_{LT} = u v λ \sqrt{β_{W}}

= 0.872 \times 0.870908 \times 145.631068 \times 1 = 110.596877

Table 8.3a, Data File p.5, pᵧ355 column:

110

gives

120

，

115

gives

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

. Interpolation fraction:

\frac{110.596877 - 110}{115 - 110} = 0.119375

p_{b} = 120 + 0.119375 \times (111 - 120) = 118.925621 \frac{N}{{mm}^{2}}

M_{b} = \frac{p_{b} S_{x}}{1,000} = \frac{118.925621 \times 1471}{1,000} = 174.939589 kN \cdot m

Equivalent uniform moment demand:

m_{LT} M_{\max} = 259.000000 kN \cdot m

Utilisation:

\frac{259.000000}{174.939589} = 1.480511

Fails.

Animation labA beam bends sideways and twists7 concepts · 14 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.

(c) Restraint at both loads

Simple explanation: Which length belongs in the calculation?

A sideways tie can shorten the buckling region without shortening the beam’s vertical span.

Separate vertical-support spacing from lateral-restraint spacing.
Apply the course rule for the actual end restraint and loading.
Keep the original vertical load analysis unless its supports change.

Remember: Restraints in one direction may not restrain the other direction.

Related concept and full method

Vertical supports remain A/D and the moment diagram is unchanged. New LTB segments AB, BC and CD each have $L_{E} = 2000 mm$ . AB/CD moment diagrams are triangular, from $0$ increasing to $280$ or from $280$ decreasing to $0$ , giving $m_{LT} = 0.6$ . BC moment is constant at $280$ , giving $m_{LT} = 1$ . Segment resistances are equal, so BC governs.

AB/CD demand:

0.6 \times 280 = 168 kN \cdot m

BC demand:

1 \times 280 = 280 kN \cdot m

Normal-load effective length is the given segment length. All quantities in $λ$ and λᴸᵀ are dimensionless. This Class 1 section has $β_{W} = 1$ .

λ = \frac{L_{E}}{r_{y}} = \frac{2000}{41.2} = 48.543689

v = {[1 + 0.05 {(\frac{λ}{x})}^{2}]}^{- \frac{1}{4}}

= {[1 + 0.05 {(\frac{48.543689}{37.9})}^{2}]}^{- \frac{1}{4}} = 0.980484

λ_{LT} = u v λ \sqrt{β_{W}}

= 0.872 \times 0.980484 \times 48.543689 \times 1 = 41.503979

Table 8.3a, Data File p.5, pᵧ355 column:

40

gives

325

，

45

gives

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

. Interpolation fraction:

\frac{41.503979 - 40}{45 - 40} = 0.300796

p_{b} = 325 + 0.300796 \times (309 - 325) = 320.187267 \frac{N}{{mm}^{2}}

M_{b} = \frac{p_{b} S_{x}}{1,000} = \frac{320.187267 \times 1471}{1,000} = 470.995470 kN \cdot m

Equivalent uniform moment demand:

m_{LT} M_{\max} = 280.000000 kN \cdot m

Utilisation:

\frac{280.000000}{470.995470} = 0.594486

Passes.

Since the governingBC demand passes andAB/CD are smaller with the same resistance, all three LTB segment checks pass. The unchanged shear and section bending checks remain valid.

Animation labA beam bends sideways and twists4 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.

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.

Compact script and invented marking guide

$R_{A} = R_{D} = 140 kN$ ; Mmax $280 kN \cdot m$ . End-onlymLT0.925, equivalent demand $259 kN \cdot m$ :LTB fails despite section-strength passes. WithB/C restraint eachLE2 m; AB/CD demand $168$ ,BC280 kNm, all pass against the common calculatedMb above.

Invented credit: (a) reactions $2$ , correctly labelledSFD/BMD2; (b) properties/class $4$ , shear/bending $4$ , moment factor $3$ , accurate LTB lookup/check $5$ ; (c) unchanged vertical analysis/segment lengths $2$ , all moment shapes $3$ , common resistance $3$ , conclusions $2$ .

Animation labRead the moment shape within one segment5 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.

Moment ordinates must belong to the same effective unbraced segment.
Same-side and reverse-curvature diagrams have different signed end ratios.
The markers show ¼, ½ and ¾ of this segment, not of the entire beam.
LTB and column flexural factors are different. Preserve their individual bounds and coefficient sets.