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Tutorial 4 Q5: opposite reactions and equal moment sharing

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Check $254 \times 254 \times 89$ UC S355 below a simple-construction floor joint. Effective length $5.1 m$ ; actual height $6 m$ . Upper/lower column stiffnesses are equal. All shown loads are first-order ultimate values: central $1,200 kN$ ; Fₓ₁150 at $220 mm$ left, Fₓ₂100 at $110 mm$ right; Fᵧ₁180 at $230 mm$ above, Fᵧ₂300 at $270 mm$ below in plan. Amplification $1.15$ .

Original source: LectureNotes/Ch 4_Column.pdf — p. 41. 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

Lookup: Data File p.11, exact row $254 \times 254 \times 89$ UC. Read the dimensions/local ratios table and the properties table separately. The $x - x$ axis crosses the web horizontally; $y - y$ passes vertically through its centre in the table sketch.

Property	Value and units
Flange/web/root-to-root web depth	$T = 17.3 mm$ 、 $t = 10.3 mm$ 、 $d = 200.3 mm$ 。
Local slenderness	$\frac{b}{T} = 7.41$ 、 $\frac{d}{t} = 19.4$ 。
Radii (converted from $cm$ )	$r_{x} = 11.2 \times 10 = 112 mm$ ； $r_{y} = 6.55 \times 10 = 65.5 mm$ 。
Area	$A = 113 {cm}^{2} = 11300 {mm}^{2}$ 。
Elastic moduli	$Z_{x} = 1096 {cm}^{3}$ ； $Z_{y} = 379 {cm}^{3}$ 。
Plastic moduli	$S_{x} = 1224 {cm}^{3}$ ； $S_{y} = 575 {cm}^{3}$ 。
LTB parameters	$u = 0.85$ , $x = 14.5$ ; both dimensionless

Before calculating: recognition and strategy

Resolve two separate balances at the floor joint: the column below carries all downward force, while net moment is shared equally between upper and lower columns by the given equal stiffness. Use effective $5.1 m$ for compression and actual $6 m$ for the special simple-construction LTB expression.

1. Sum compression and opposing moments

F_{lower} = 1,200 + 150 + 100 + 180 + 300 = 1,930 kN

Net joint moment magnitude:

| M_{x, joint} | = \frac{| 300 \times 270 - 180 \times 230 |}{1,000}

= \frac{| 81,000 - 41,400 |}{1,000} = 39.6 kN \cdot m

| M_{y, joint} | = \frac{| 150 \times 220 - 100 \times 110 |}{1,000}

= \frac{| 33,000 - 11,000 |}{1,000} = 22 kN \cdot m

The central $1,200 kN$ is described as axial force above, including self-weight; do not add self-weight again. The lower segment takes the full $1,930 kN$ . The upper segment is not assigned half of that load.

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. Section interaction

At a cross section, compression and bending share the material. Use amplified moments and capped plastic resistances. The total must not exceed $1$ ; all terms below are dimensionless.

A_{g} p_{y} = \frac{113 \times 100 \times 345}{1,000} = 3898.5 kN

M_{x, amp} = 1.15 \times 19.8 = 22.77 kN \cdot m

M_{y, amp} = 1.15 \times 11 = 12.65 kN \cdot m

M_{cx} = \min (\frac{345 \times 1224}{1,000}, \frac{1.2 \times 345 \times 1096}{1,000})

= \min (422.28, 453.744) = 422.28 kN \cdot m

M_{cy} = \min (\frac{345 \times 575}{1,000}, \frac{1.2 \times 345 \times 379}{1,000})

= \min (198.375, 156.906) = 156.906 kN \cdot m

Section ratio:

\frac{1930}{3898.5} + \frac{22.77}{422.28} + \frac{12.65}{156.906} = 0.629605

Passes.

Animation labA strong slice can belong to an unstable member2 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.

Combine axial compression and the two bending demands using the specified section resistances.
The whole member adds effective-length and buckling-curve effects.
This uses its own moment factor and bending resistance; it is not a copy of the section check.
Elastic, plastic and buckling resistances are not interchangeable. Read the three original expressions and their first-order/amplified moments.

4. Flexural member interaction using $5.1 m$

Simple explanation: The column needs more than one pass

A slice can be strong while the whole member still buckles.

Check cross-section compression plus bending.
Then check the separate member-buckling expressions.
Keep each moment, factor and resistance in its specified expression.

Remember: The three checks do not share interchangeable denominators.

Related concept and full method

Table 8.7: hot-rolled H-section (UC), maximum thickness $\leq 40 mm$ , about the $x$ axis use curve $b$ , about the $y$ axis use curve $c$ . Use the Data File p.8 $p_{y} = 345 \frac{N}{{mm}^{2}}$ column. The two axes use different curves, so slenderness alone cannot identify the governing axis.

L_{Ex} = L_{Ey} = 5100 mm

is the given effective length.

λ_{x} = \frac{5100}{112} = 45.535714

λ_{y} = \frac{5100}{65.5} = 77.862595

x

axis, curve b:

44

row gives

302

;

46

row gives

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

. Interpolation fraction:

\frac{45.535714 - 44}{46 - 44} = 0.767857

p_{c} = 302 + 0.767857 \times (298 - 302) = 298.928571 \frac{N}{{mm}^{2}}

y

axis, curve c:

76

row gives

196

;

78

row gives

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

. Interpolation fraction:

\frac{77.862595 - 76}{78 - 76} = 0.931298

p_{c} = 196 + 0.931298 \times (191 - 196) = 191.343511 \frac{N}{{mm}^{2}}

P_{cx} = \frac{A p_{cx}}{10} = \frac{113 \times 298.928571}{10} = 3377.892857 kN

P_{cy} = \frac{113 \times 191.343511}{10} = 2162.181679 kN

P_{c} = \min (P_{cx}, P_{cy}) = 2162.181679 kN

Member interaction uses elastic moment denominators pᵧZ, even when the cross-section check used plastic moduli. Use amplified moments here.

M_{ex} = \frac{345 \times 1096}{1,000} = 378.12 kN \cdot m

M_{ey} = \frac{345 \times 379}{1,000} = 130.755 kN \cdot m

\frac{F}{P_{c}} + \frac{m_{x} M_{x, amp}}{M_{ex}} + \frac{m_{y} M_{y, amp}}{M_{ey}}

= \frac{1930}{2162.181679} + \frac{1 \times 22.77}{378.12} + \frac{1 \times 12.65}{130.755}

= 1.049582

Fails.

Animation labA strong slice can belong to an unstable member5 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.

Combine axial compression and the two bending demands using the specified section resistances.
The whole member adds effective-length and buckling-curve effects.
This uses its own moment factor and bending resistance; it is not a copy of the section check.
Elastic, plastic and buckling resistances are not interchangeable. Read the three original expressions and their first-order/amplified moments.

5. Simple-construction LTB using actual $6 m$

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

Simple-construction special rule: use actual storey length $L$ in 0.5L/rᵧ; the axial effective length is a different quantity.

λ_{LT} = \frac{0.5 \times 6000}{65.5} = 45.801527

Read Data File p.5 Table 8.3a, pᵧ345 column:

45

row gives

302

;

50

row gives

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

. Interpolation fraction:

\frac{45.801527 - 45}{50 - 45} = 0.160305

p_{b} = 302 + 0.160305 \times (285 - 302) = 299.274809 \frac{N}{{mm}^{2}}

M_{b} = p_{b} S_{x} = \frac{299.274809 \times 1224}{1,000} = 366.312366 kN \cdot m

Course Eq.8.81 uses first-order minor-axis moment in its last term. Mᴸᵀ is the specified amplified major-axis value; do not amplify it twice. The axial denominator is $P_{cy}$ .

\frac{F}{P_{cy}} + \frac{m_{LT} M_{LT}}{M_{b}} + \frac{m_{y} M_{y, first}}{M_{ey}}

= \frac{1930}{2162.181679} + \frac{1 \times 22.77}{366.312366} + \frac{1 \times 11}{130.755}

= 1.038904

Fails.

$p_{y} = 345 \frac{N}{{mm}^{2}}$ ; $P_{cx} = 3377.893 kN$ , $P_{cy} = 2162.182 kN$ ; $M_{cx} = 422.280 kN \cdot m$ , $M_{cy} = 156.906 kN \cdot m$ . $M_{b} = 366.312 kN \cdot m$ . Ratios: section $0.629605$ , flexural buckling $1.049582$ , axial force/LTB $1.038904$ . At least one required ratio exceeds $1$ , so the column fails.

Animation labA strong slice can belong to an unstable member5 concepts · 9 source expressions

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

Combine axial compression and the two bending demands using the specified section resistances.
The whole member adds effective-length and buckling-curve effects.
This uses its own moment factor and bending resistance; it is not a copy of the section check.
Elastic, plastic and buckling resistances are not interchangeable. Read the three original expressions and their first-order/amplified moments.

Compact exam answer

Lower F1,930 kN. Joint Mx 39.6/My 22 kNm; half-shares $19.8$ / $11$ ; amplified $22.77$ /12.65. LE5,100 mm for axial buckling; L6,000 mm for λᴸᵀ=0.5L/rᵧ. All $m = 1$ .

Mistakes to avoid

Do not use $355 \frac{N}{{mm}^{2}}$ for a flange thicker than $16 mm$ .
Do not substitute plastic moduli in the member elastic denominators.
Do not amplify an already amplified Mᴸᵀ twice.
Do not treat an axial resistance pass as proof of combined-load adequacy.
Do not halve the axial force when halving joint moment.
Do not add magnitudes of opposite-side eccentric moments.
Do not use $5.1 m$ in the actual-height LTB rule.

Procedure for an unfamiliar variant

Identify construction type, axis, actual length and effective length.
Read thickness, grade and the exact UC row. Classify before using plastic moduli.
Sum vertical forces and derive signed eccentric moments; share moments only when the joint has two columns.
Apply specified amplification once. Check the capped section interaction.
Select curves $b$ / $c$ for these rolled H sections, interpolate both strengths and check flexural interaction with elastic moduli.
Use the appropriate simple/continuous LTB slenderness rule and the course first-order minor-axis term.
Report every utilisation and let any failed check govern.

Independent self-check

Try it yourself. Invented variant: the lower column has twice the upper column’s $\frac{I}{L}$ stiffness, with all loads unchanged. Find its first-order moments.

Reveal answer and reasoning

Lower fraction= $\frac{2}{2 + 1} = \frac{2}{3}$ . $M_{x} = \frac{39.6 \times 2}{3} = 26.4 kN \cdot m$ ; $M_{y} = \frac{22 \times 2}{3} = 14.666667 kN \cdot m$ . Amplified moments $30.36$ and $16.866667 kN \cdot m$ . Axial load remains $1,930 kN$ . All interactions must be recalculated; the stiffer lower column attracts more moment.

Animation labEccentric reactions and stiffness sharing3 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.

A beam reaction can act away from the column centre even at a nominally pinned beam connection.
$M = R e$ . Opposing reactions can cancel part of the signed moment, while both still add compression.
The course simple model distributes the joint moment in proportion to $\frac{I}{L}$ of the columns above and below.
Equal relevant stiffness gives half each. A roof joint with no upper column is a different case.

Tutorial 4 Q5: opposite reactions and equal moment sharing

Read the diagram and collect the data

Before calculating: recognition and strategy

1. Sum compression and opposing moments

2. Share joint moments, amplify and classify

3. Section interaction

4. Flexural member interaction using 5.1m5.1m

5. Simple-construction LTB using actual 6m6m

Compact exam answer

Mistakes to avoid

Procedure for an unfamiliar variant

Independent self-check

4. Flexural member interaction using $5.1 m$

5. Simple-construction LTB using actual $6 m$