Practice

Graduated practice and unfamiliar variants

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

All questions on this page are invented practice, not extra past papers or predictions. Work on paper. First identify the mechanism, then select the formula. A correct number with an unexplained wrong method does not count as independent mastery.

Stage 1: essential arithmetic and recognition

P1 · units

Try it yourself. A section has $I_{x} = 19460 {cm}^{4}$ , $S_{x} = 1211 {cm}^{3}$ and $r_{y} = 3.99 cm$ . Convert each to $mm$ units.

Hint

Powers of length scale by the corresponding power of $10$ .

Reveal answer and reasoning

$I_{x} = 19460 \times 10^{4} = 194600000 {mm}^{4}$ ; $S_{x} = 1211 \times 1000 = 1211000 {mm}^{3}$ ; $r_{y} = 3.99 \times 10 = 39.9 mm$ . Do not apply the same factor to all three.

Animation labUnits and powers1 concept · 3 source expressions

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

Force, length and stress must use compatible units before substitution.
$1 cm = 10 mm$ . An area has two length factors, so $1 cm^{2} = 100 mm^{2}$ .
A section modulus has $length^{3}$ ; second moment of area has $length^{4}$ . Use $1 0^{3}$ and $1 0^{4}$ for cm to mm.
$\frac{N}{mm ^{2}} \times mm^{3} = N \cdot mm$ . Divide by $1 0^{6}$ to express that moment in $kN \cdot m$ .

P2 · area to line load

Try it yourself. A slab has $G = 4 \frac{kN}{m^{2}}$ , $Q = 3 \frac{kN}{m^{2}}$ and beam tributary width $2 m$ . Beam self-weight is $0.5 \frac{kN}{m}$ . Find design line load.

Hint

Keep $G$ and $Q$ separate; self-weight is dead load.

Reveal answer and reasoning

$w_{G} = 4 \times 2 + 0.5 = 8.5 \frac{kN}{m}$ ； $w_{Q} = 3 \times 2 = 6 \frac{kN}{m}$ 。 $w_{u} = 1.4 \times 8.5 + 1.6 \times 6 = 11.9 + 9.6 = 21.5 \frac{kN}{m}$

Animation labFollow the floor load in 3D2 concepts · 3 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.

P3 · reaction and moment

Try it yourself. The P2 beam is simply supported over $6 m$ . Find reactions and peak bending moment.

Hint

Symmetry gives equal reactions; the shear changes sign at midspan.

Reveal answer and reasoning

Total= $21.5 \times 6 = 129 kN$ ; $R_{A} = R_{B} = \frac{129}{2} = 64.5 kN$ . $M_{\max} = \frac{w L^{2}}{8} = \frac{21.5 \times 36}{8} = 96.75 kN \cdot m$ . Check $R_{A} + R_{B} = 129 kN$ and $R_{B} \times 6 = 129 \times 3$ .

Animation labBalance reactions and moments2 concepts · 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.

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.

P4 · method selection

Try it yourself. A vertical support is present at each end of an $8 m$ beam and a lateral restraint at $3 m$ . How many vertical spans and LTB segments exist?

Hint

Separate sideways restraint from vertical support.

Reveal answer and reasoning

One $8 m$ vertical span; two LTB segments, $3 m$ and $5 m$ . Analyze vertical moments over the $8 m$ beam first. Moment at the lateral restraint is not necessarily zero.

P5 · strength row

Try it yourself. S355 flange thickness is $20 mm$ , $\frac{b}{T} = 8.5$ and web $\frac{d}{t} = 40$ under bending. Classify the section.

Hint

Use py 345, then compare flange with $9 ε$ and $10 ε$ and web with $80 ε$ .

Reveal answer and reasoning

$ε = \sqrt{\frac{275}{345}} = 0.892804$ . Flange $8.5 > 9 ε = 8.035235$ , but $8.5 < 10 ε = 8.928039$ , Class 2. Web $40 < 80 ε = 71.424314$ , Class 1. Overall Class 2; capped plastic moment resistance is permitted at low shear.

Animation labWhy thin elements buckle locally2 concepts · 4 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.

Animation labUnits and powers7 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.

Force, length and stress must use compatible units before substitution.
$1 cm = 10 mm$ . An area has two length factors, so $1 cm^{2} = 100 mm^{2}$ .
A section modulus has $length^{3}$ ; second moment of area has $length^{4}$ . Use $1 0^{3}$ and $1 0^{4}$ for cm to mm.
$\frac{N}{mm ^{2}} \times mm^{3} = N \cdot mm$ . Divide by $1 0^{6}$ to express that moment in $kN \cdot m$ .

Stage 2: choose and apply a resistance

P6 · threaded double shear

Try it yourself. FourM20 Grade 8.8 bolts carry $300 kN$ through a symmetric double-cover splice. Threads cross both shear planes. Is bolt shear adequate?

Hint

Per bolt there are two planes; the full joint force is divided among four bolts on ONE transfer half.

Reveal answer and reasoning

$A_{t} = 245 {mm}^{2}$ , $p_{s} = 375$ . One plane $P_{s} = \frac{245 \times 375}{1000} = 91.875 kN$ ; two planes $183.75 kN$ ; four bolts $735 kN > 300$ . Demand per bolt $75 kN$ ; per symmetric shear plane $37.5$ . This checks bolt shear only; plates and layout require separate checks.

Animation labCount the bolt shear planes1 concept · 4 source expressions

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

Load must cross an interface between the connected plates.
A lap joint gives one shear plane through a bolt.
A symmetric double-cover joint may provide two shear planes. Count load-transfer interfaces, not just visible plates.
Use the source’s area and shear strength. Then check bearing, plate resistance and detailing separately.

P7 · bearing selection

Try it yourself. OneM20 Grade 8.8 bolt bears on an $8 mm$ S355 plate with standard $22 mm$ holes, pitch $60$ and end $30 mm$ . Calculate the three connected-part limits using pbs 550,Us 510,Ub 800.

Hint

Use d 20 for bearing area, but pitch minus hole $22$ for clear ligament.

Reveal answer and reasoning

Diameter term $\frac{20 \times 8 \times 550}{1000} = 88 kN$ . End-distance term $\frac{0.5 \times 30 \times 8 \times 550}{1000} = 66 kN$ . Clear ligament between holes $l_{c} = 60 - 22 = 38 mm$ ; tearing limit: $\frac{\min (1.5 \times 38 \times 8 \times 510, 2 \times 20 \times 8 \times 800)}{1000} = \min (232.56, 256) = 232.56 kN$ Governing value $66 kN$ . $30 mm$ end distance may also fail the sheared-edge minimum of $34 mm$ , so state the edge preparation before accepting the layout.

Animation labBearing and the remaining ligament1 concept · 4 source expressions

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

Force transfers through contact between bolt and connected plate.
The plate around the hole carries bearing stress. Bolt bearing and plate bearing are separate checks.
A short ligament can tear out towards the end. The direction of force determines the relevant edge.
Evaluate every specified bearing and ligament bound for each layer and retain the smallest applicable resistance.

P8 · bolt interaction

Try it yourself. AnM20 Grade 8.8 bolt carriesFs 40 kN andFt 90 kN. Apply the course combined rule.

Hint

Ps 91.875; Pt 137.2; Pnom 109.76 kN. Individual checks remain necessary.

Reveal answer and reasoning

$\frac{40}{91.875} = 0.435374$ , $\frac{90}{109.76} = 0.819971$ ; sum $1.255345 < 1.4$ . Individual shear $40 < 91.875$ and tension $90 < 137.2$ both pass (also $90 < 109.76$ ). Do not reject it simply because the interaction value exceeds $1$ .

Animation labCount the bolt shear planes3 concepts · 6 source expressions

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

Load must cross an interface between the connected plates.
A lap joint gives one shear plane through a bolt.
A symmetric double-cover joint may provide two shear planes. Count load-transfer interfaces, not just visible plates.
Use the source’s area and shear strength. Then check bearing, plate resistance and detailing separately.

P9 · weld length and detail

Try it yourself. A direct $140 kN$ force is carried equally by two straight S355/Class 42 fillets. Leg $6 mm$ , plates $10$ and $12 mm$ , available run length $100 mm$ each. Check strength and min/max leg.

Hint

For each terminated run, effective length=physical− $2 s$ .

Reveal answer and reasoning

$q = \frac{0.7 \times 6 \times 250}{1000} = 1.05 \frac{kN}{mm}$ . Each weld $L_{eff} = 100 - 12 = 88 mm$ ; total resistance $2 \times 88 \times 1.05 = 184.8 kN > 140$ . Required $L_{eff} = \frac{140}{2 \times 1.05} = 66.666667 mm$ ; actual length $\geq 78.666667 mm$ . Thicker plate $12 mm$ requires minimum weld leg $5$ ; $10 mm$ plate-edge maximum is $10 - 2 = 8$ ; $6$ lies between the two limits. Effective length $88 \geq \max (24, 40)$ . Lap/end returns and transverse spacing still need checking against the actual geometry.

Animation labFrom fillet leg to effective throat1 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.

An equal-leg fillet between perpendicular plates has an approximately right-triangular section.
For this geometry, throat $a \approx 0.7 s$ . It is shorter than the leg.
Effective resisting area = $a \times L_{eff}$ . With $p_{w} = 220 \frac{N}{mm ^{2}}$ here, capacity per length is $0.7 s \times 220 \frac{N}{mm}$ .
Use the course end allowances, minimum size and length rules; increasing the geometric length alone does not resolve every detailing check.

P10 · interpolation

Independent practice.Curve c, $p_{y} = 345$ : $60$ and $62$ rows give $p_{c}$ respectively as $241$ and $236 \frac{N}{{mm}^{2}}$ . If $λ = 61.2$ , $A_{g} = 11300 {mm}^{2}$ , find $P_{c}$ .

Hint

The fraction is $\frac{1.2}{2}$ , not $\frac{61.2}{62}$ .

Reveal answer and reasoning

$p_{c} = 241 + \frac{61.2 - 60}{62 - 60} \times (236 - 241) = 241 - 3 = 238 \frac{N}{{mm}^{2}}$ . $P_{c} = \frac{238 \times 11300}{1000} = 2689.4 kN$ . It lies between $2723.3$ and $2666.8 kN$ , the capacities at the bounding rows.

Animation labEffective length and buckling axes2 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 column can bow sideways before its section reaches the crushing resistance.
Each axis has its own radius of gyration and restraint spacing.
$λ = \frac{L _{E}}{r}$ , with compatible length units. A tie affects only the directions it actually restrains.
Select each buckling curve and compressive strength before forming Pc. The governing axis is determined by resistance, not slenderness alone.

Animation labCount the bolt shear planes7 concepts · 3 source expressions

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

Load must cross an interface between the connected plates.
A lap joint gives one shear plane through a bolt.
A symmetric double-cover joint may provide two shear planes. Count load-transfer interfaces, not just visible plates.
Use the source’s area and shear strength. Then check bearing, plate resistance and detailing separately.

Stage 3: unfamiliar variants and diagnosis

P11 · off-centre load

Try it yourself. A6 m simple beam carries a $150 kN$ design point load $2 m$ from A. Ignore self-weight. Find reactions, maximum moment and full-span quarter-point $m_{LT}$ .

Hint

The quarter points are $1.5$ , $3$ , $4.5 m$ . Derive their moments rather than using a central-load shortcut.

Reveal answer and reasoning

$R_{B} \times 6 = 150 \times 2$ givesRB50 kN; RA100 kN. Mmax atx 2 is $100 \times 2 = 200 kN \cdot m$ . $M_{1.5} = 150$ ; $M_{3} = 100 \times 3 - 150 \times 1 = 150$ ; $M_{4.5} = 450 - 150 \times 2.5 = 75 kN \cdot m$ . $m_{LT} = \max [0.44, 0.2 + \frac{0.15 \times 150 + 0.5 \times 150 + 0.15 \times 75}{200}] = 0.74375$ . Equivalent demand $148.75 kN \cdot m$ . The $0.85$ central-point shortcut does not match this load position.

Animation labBalance reactions and moments3 concepts · 6 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.

P12 · high-shear variant

Independent practice.A section has $V = 0.8 V_{c}$ , $p_{y} = 355$ , $S = 1200000$ , $Z = 1050000$ , $S_{v} = 250000 {mm}^{3}$ . It is Class 1. Find the reduced $M_{c}$ .

Hint

Useρ and both reduced expressions including the $1.2$ ceiling.

Reveal answer and reasoning

$ρ = {(2 \times 0.8 - 1)}^{2} = 0.36$ . First resistance term: $\frac{355 (1200000 - 0.36 \times 250000)}{10^{6}} = 394.05 kN \cdot m$ Upper bound: $\frac{1.2 \times 355 (1050000 - \frac{0.36 \times 250000}{1.5})}{10^{6}} = 421.74 kN \cdot m$ $M_{c} = \min (394.05, 421.74) = 394.05 kN \cdot m$ . Demand $400 kN \cdot m$ fails, even though the original low-shear resistance was $426 kN \cdot m$ .

Animation labHigh shear reduces bending resistance2 concepts · 4 source expressions

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

Part of the web contributes to bending while also carrying shear.
The course low-shear route applies at $∣ V ∣ \leq 0.6 V_{c}$ . Above it, follow the applicable reduction expression.
The highlighted web contribution is reduced, not the complete cross-section by an arbitrary percentage.
Retain the section-class-specific formula and all bounds shown in the original calculation.

P13 · load increase

Independent practice.Imposed-load-only deflection is $17 mm$ , limit $\frac{L}{360} = 20 mm$ . Only $Q$ increases by $25 %$ ， $G$ is unchanged. A student multiplies all ULS internal forces by $1.25$ . Correct the calculation.

Hint

Elastic deflection is linear in its load, but the ULS total contains an unchanged dead component.

Reveal answer and reasoning

New deflection $δ = 1.25 \times 17 = 21.25 mm > 20$ , so SLS fails. New ULS actions $1.4 G + 1.6 \times 1.25 Q$ , not $1.25 (1.4 G + 1.6 Q)$ . Separate the $G$ ／ $Q$ contributions and recalculate reactions and moments.

Animation labSee stiffness and deflection3 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.

P14 · unknown restraint

Try it yourself. A beam’s strength calculation passes only ifLE=L. The drawing shows end supports but no lateral detail or load-height statement. What is a complete conclusion?

Hint

An assumption is not a dimension inferred from a line drawing.

Reveal answer and reasoning

State the calculated result conditional on the normal-load/end-restraint modelLE=L. Identify the needed lateral/torsional restraint and load-height information. RecalculateLE from the applicable course condition if those differ. Do not invent a restraint and report unconditional adequacy.

P15 · weak-axis tie

Try it yourself. AUC hasrx 120 mm,ry 60 mm, actual length $4 m$ . Both ends pinned; a midheight tie restrains only weak-axis lateral displacement. Findλx andλy.

Hint

A tie changes only the restrained buckling direction.

Reveal answer and reasoning

LEx 4000,L Ey 2000 mm; $λ_{x} = \frac{4000}{120} = 33.333333$ , $λ_{y} = \frac{2000}{60} = 33.333333$ . Equal slenderness does not prove equalpc because the selected curves may differ. Read the correct curve for each axis.

Animation labEffective length and buckling axes1 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.

A column can bow sideways before its section reaches the crushing resistance.
Each axis has its own radius of gyration and restraint spacing.
$λ = \frac{L _{E}}{r}$ , with compatible length units. A tie affects only the directions it actually restrains.
Select each buckling curve and compressive strength before forming Pc. The governing axis is determined by resistance, not slenderness alone.

P16 · rectangle becomes a U

Try it yourself. A four-sided rectangular weld loses its top horizontal run. Can you use the same centroid andIx+Iy?

Hint

Changing effective weld geometry changes both direct stress and torsion.

Reveal answer and reasoning

No. Total length reduces and the length-weighted centroid moves toward the remaining bottom run. Recalculate centroidΣLi yi/ΣLi; calculate each remaining line inertia about that centroid using parallel-axis terms; recompute $e$ from the force line to the new centroid; combine direct and torsional components at candidate extreme points. The four-sided formula is inapplicable.

Animation labBalance reactions and moments8 concepts · 6 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.

How to know you can apply the method independently

Pass a stage only after solving its calculations with units and explaining each applicability choice without revealing the answer. After a failed attempt, read one relevant worked example, then solve a different question rather than immediately copying the same one. Use the timed mock only after you can complete Stage 3 without hidden assumptions.

Take the full timed mock · Return to every lecture/tutorial example

Animation labFollow the calculation sequence1 concept

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

Locate the load, supports, connection geometry and any stated assumptions.
Keep given values, table lookups and calculated values distinct; reconcile their units.
The calculation player steps through the existing expressions in their original order.
Compare demand with resistance or the relevant limit. Keep missing inputs and conditional conclusions explicit.