Worked examples

Tutorial 3 Q4: an end couple reverses the bending diagram

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Check $533 \times 210 \times 82$ UB S355 against LTB for a $350 kN \cdot m$ design end moment and $180 kN$ design point load. Supports A and C are $10.5 m$ apart; point B is $6 m$ from A and $4.5 m$ from C. A, B and C have lateral restraints. Normal loads; effective lengths equal actual segment lengths.

Original source: LectureNotes/Ch 3_Beam.pdf — p. 35, p. 36. 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

Input	Given/lookup
Loads	Already design values M350 kNm and P180 kN; no $1.4$ / $1.6$ refactoring.
Spans	Dimension chain $6 m + 4.5 m = 10.5 m$ ; B is a load/restraint point, not a support.
LTB lengths	AB6,000 mm; BC4,500 mm under the explicit effective-length assumption.

Lookup: Data File pp.9–10, exact $533 \times 210 \times 82$ 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	D528.3, web t 9.6, flange T13.2, root r 12.7, clear d 476.5, all $mm$
Local ratios	$\frac{b}{T} = 7.91$ ； $\frac{d}{t} = 49.6$ 。
Major-axis properties	$I_{x} = 47540 {cm}^{4} = 4.754 \times 10^{8} {mm}^{4}$ ; $Z_{x} = 1800 {cm}^{3}$ ; $S_{x} = 2059 {cm}^{3}$ .
LTB properties	$r_{y} = 4.38 cm = 43.8 mm$ ; $u = 0.864$ ; torsional index $x = 41.6$ .

Before calculating: recognition and strategy

An applied couple contributes to moment equilibrium but does not add vertical force. It produces a jump in the bending diagram at the end. Find the signed internal end moments for each segment, then use the end-moment LTB factor; the bending diagram is linear between the discrete loads.

1. Find support reactions with the end couple included

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

Upwards is positive; applied anticlockwise moment is positive.

Σ V = 0 : R_{A} + R_{C} - 180 = 0

Σ M_{A} = 0 : R_{C} \times 10.5 + 350 - 180 \times 6 = 0

R_{C} = \frac{1,080 - 350}{10.5} = 69.523810 kN

R_{A} = 180 - 69.523810 = 110.476190 kN

Check moments about C:

180 \times 4.5 + 350 - 110.476190 \times 10.5 = 0

Treating B as a vertical support would create a different statically indeterminate problem. Its× symbol only restrains lateral movement for the stated LTB model.

Animation labBalance reactions and moments1 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.

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.

2. Construct signed shear and bending ordinates

Internal sagging moment is positive;

x

is measured from A in

m

. At

0 < x < 6

V = + 110.476190 kN

M (x) = - 350 + 110.476190 x kN \cdot m

In the interval

6 < x < 10.5

V = 110.476190 - 180 = - 69.523810 kN

M (x) = - 350 + 110.476190 x - 180 (x - 6)

M (A^{+}) = - 350 kN \cdot m

M (B) = - 350 + 110.476190 \times 6 = + 312.857143 kN \cdot m

M (C) = 312.857143 - 69.523810 \times 4.5 = 0

Contraflexure point within AB:

0 = - 350 + 110.476190 x

x = \frac{350}{110.476190} = 3.168103 m

from A.

Location	Shear immediately right	Bending moment
$A$	$+ 110.476190 kN$	$- 350 kN \cdot m$ (end-couple jump)
$B^{-}$	$+ 110.476190 kN$	$+ 312.857143 kN \cdot m$
$B^{+}$	$- 69.523810 kN$	$+ 312.857143 kN \cdot m$
$C^{-}$	$- 69.523810 kN$	$0$
$C^{+}$	$0$	$0$

Draw horizontal SFD segments at the listed levels, a downward $180 kN$ jump at B, and upward $69.523810 kN$ at C. Draw straight BMD lines joining $- 350$ at A to $+ 312.857143$ at B to $0$ at C. The zero crossing at $3.168103 m$ is not a physical lateral restraint.

Animation labBuild the shear and moment diagrams1 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 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.

3. Check the $6 m$ AB 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

S355, flange

T = 13.2 mm

, so

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

ε = \sqrt{\frac{275}{355}} = 0.880141

Flange:

\frac{b}{T} = 7.91 < 9 ε = 7.921268

. Web in bending:

\frac{d}{t} = 49.6 < 80 ε = 70.411267

. Both are Class 1, so

β_{W} = 1

; plastic bending resistance is permitted.

\frac{d}{t} = 49.6 < 70 ε = 61.609858

; no shear-buckling calculation is required under the course limit.

AB has reversing moment: maximum magnitude $350$ occurs at A. The smaller signed end ordinate is $+ 312.857143$ opposite to $- 350$ , so the ratio is negative.

β = \frac{- 312.857143}{350} = - 0.893878

Table 8.4a:

m_{LT} = \max (0.44, 0.6 + 0.4 β)

= \max (0.44, 0.242449) = 0.44

Equivalent demand:

0.44 \times 350 = 154 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}{43.8} = 136.986301

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

= {[1 + 0.05 {(\frac{136.986301}{41.6})}^{2}]}^{- \frac{1}{4}} = 0.897360

λ_{LT} = u v λ \sqrt{β_{W}}

= 0.864 \times 0.897360 \times 136.986301 \times 1 = 106.208098

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

105

row gives

129

;

110

row gives

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

. Interpolation fraction:

\frac{106.208098 - 105}{110 - 105} = 0.241620

p_{b} = 129 + 0.241620 \times (120 - 129) = 126.825424 \frac{N}{{mm}^{2}}

M_{b} = \frac{p_{b} S_{x}}{1,000} = \frac{126.825424 \times 2059}{1,000} = 261.133547 kN \cdot m

Equivalent uniform demand:

m_{LT} M_{\max} = 154.000000 kN \cdot m

Utilisation:

\frac{154.000000}{261.133547} = 0.589737

Passes.

Animation labA beam bends sideways and twists6 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.

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.

4. Check the $4.5 m$ BC segment and conclude

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

The BC moment diagram is triangular, from

+ 312.857143

0

β = 0

; Table 8.4a:

m_{LT} = 0.6 + 0.4 \times 0 = 0.6

Equivalent demand:

0.6 \times 312.857143 = 187.714286 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{4500}{43.8} = 102.739726

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

= {[1 + 0.05 {(\frac{102.739726}{41.6})}^{2}]}^{- \frac{1}{4}} = 0.935620

λ_{LT} = u v λ \sqrt{β_{W}}

= 0.864 \times 0.935620 \times 102.739726 \times 1 = 83.052333

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

80

row gives

190

;

85

row gives

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

. Interpolation fraction:

\frac{83.052333 - 80}{85 - 80} = 0.610467

p_{b} = 190 + 0.610467 \times (175 - 190) = 180.843001 \frac{N}{{mm}^{2}}

M_{b} = \frac{p_{b} S_{x}}{1,000} = \frac{180.843001 \times 2059}{1,000} = 372.355739 kN \cdot m

Equivalent uniform demand:

m_{LT} M_{\max} = 187.714286 kN \cdot m

Utilisation:

\frac{187.714286}{372.355739} = 0.504126

Passes.

Both required segment LTB checks pass. As a scope check, the peak shear and largest moment magnitude also satisfy the local section checks:

A_{v} = t D = 9.6 \times 528.3 = 5071.68 {mm}^{2}

V_{c} = \frac{p_{y} A_{v}}{\sqrt{3}} = \frac{355 \times 5071.68}{\sqrt{3} \times 1,000} = 1039.488214 kN

V_{\max} = 110.476190 kN < V_{c}

: shear passes.

0.6 V_{c} = 623.692928 kN > V_{\max}

; even peak shear is low, so low-shear bending applies throughout.

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

= \min (\frac{355 \times 2059}{1,000}, \frac{1.2 \times 355 \times 1800}{1,000})

= \min (730.945, 766.8) = 730.945 kN \cdot m

| M |_{\max} = 350.000000 kN \cdot m

: bending passes.

A numerical support bearing check or imposed-load deflection cannot be derived from this source because bearing dimensions and characteristic load components are not supplied. They are not part of Q4’s stated LTB-only request.

Animation labA beam bends sideways and twists4 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.

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 exam answer

$R_{A} = 110.47619 kN$ , $R_{C} = 69.52381 kN$ . $M_{A} = - 350$ , $M_{B} = + 312.85714$ , $M_{C} = 0 kN \cdot m$ . AB: $L_{E} = 6 m$ , $β = - 0.893878$ , $m_{LT} = 0.44$ ; demand $154 < M_{b} = 261.134 kN \cdot m$ . BC: $L_{E} = 4.5 m$ , $β = 0$ , $m_{LT} = 0.6$ ; demand $187.714 < M_{b} = 372.356 kN \cdot m$ . S355 $533 \times 210 \times 82$ UB passes LTB in both segments.

Mistakes to avoid

Do not factor already-design loads again.
Do not omit the $350 kN \cdot m$ couple in support equilibrium.
Do not use absolute-value end moments to form a same-side $β$ when the BMD reverses.
Do not insert a restraint at the moment-zero point.

Procedure for an unfamiliar variant

Name supports and load positions using the source dimension chains.
Choose a sign convention; include applied couples in moment equilibrium.
Find reactions, then piecewise shear and bending moment.
Locate extrema where shear is zero and at load/support discontinuities.
Divide only at actual effective lateral restraints; classify and read the exact section row.
Calculate segment moment factors and LTB resistance, retaining UDL curvature where present.
Check shear, local bending and serviceability; state any missing bearing/detail inputs.

Independent self-check

Try it yourself. Invented variant: remove the lateral restraint at B but keep the same vertical loads. Can the two segment pass results still be used?

Reveal answer and reasoning

No. The unrestrained length becomes $10.5 m$ . Recalculate the whole-span quarter-point BMD factor and $λ_{LT}$ with $10,500 mm$ . The vertical reactions and BMD are unchanged; the stability resistance changes because B is no longer a restraint.

Tutorial 3 Q4: an end couple reverses the bending diagram

Read the diagram and collect the data

Before calculating: recognition and strategy

1. Find support reactions with the end couple included

2. Construct signed shear and bending ordinates

3. Check the 6m6m AB segment

4. Check the 4.5m4.5m BC segment and conclude

Compact exam answer

Mistakes to avoid

Procedure for an unfamiliar variant

Independent self-check

3. Check the $6 m$ AB segment

4. Check the $4.5 m$ BC segment and conclude