Worked examples

Assignment 2 Q2: an intermediate restraint at an off-centre load

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For $356 \times 171 \times 67$ UB S355, assumed Class 1, find support reactions and draw annotated SFD/BMD under ultimateP60 kN at $5 m$ from A plus full-span $4 \frac{kN}{m}$ UDL over $8 m$ . Check segmentAB LTB using accurateu andv, and discuss removing restraintB.

Original source: Assignment/AY2627s 1-CON4334-Assignment 2.pdf — p. 2. 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 pp.9–10, exact $356 \times 171 \times 67$ 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	D363.4, web t 9.1, flange T15.7, root r 10.2, clear d 311.6, all $mm$
Local ratios	$\frac{b}{T} = 5.52$ ； $\frac{d}{t} = 34.2$ 。
Major-axis properties	$I_{x} = 19460 {cm}^{4} = 1.946 \times 10^{8} {mm}^{4}$ ; $Z_{x} = 1071 {cm}^{3}$ ; $S_{x} = 1211 {cm}^{3}$ .
LTB properties	$r_{y} = 3.99 cm = 39.9 mm$ ; $u = 0.886$ ; torsional index $x = 24.4$ .

Before calculating: recognition and strategy

Vertical equilibrium gives one simply supported $8 m$ beam. Lateral stability is assessed over shorter segments. Keep these two models distinct: $5 m$ is the AB stability length, not a $5 m$ simply supported vertical-load span.

(a) Support reactions

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

Total downward force:

4 \times 8 + 60 = 92 kN

UDL resultant

32 kN

acts at distance

4 m

from A. Taking moments about A:

8 R_{C} = 32 \times 4 + 60 \times 5 = 128 + 300 = 428

R_{C} = \frac{428}{8} = 53.5 kN

R_{A} = 92 - 53.5 = 38.5 kN

Check moments about C:

38.5 \times 8 = 32 \times 4 + 60 \times 3 = 308 kN \cdot m

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.

(b) Annotated shear and bending diagrams

x

is in metres from A; sagging moment is positive.

0 \leq x \leq 5

V = 38.5 - 4 x kN

M = 38.5 x - 2 x^{2} kN \cdot m

5 \leq x \leq 8

V = 38.5 - 4 x - 60 kN

M = 38.5 x - 2 x^{2} - 60 (x - 5) kN \cdot m

Shear end values:

V (A^{+}) = 38.5

V (B^{-}) = 18.5

V (B^{+}) = - 41.5

V (C^{-}) = - 53.5 kN

Moment end values:

M (A) = 0

M (B) = 38.5 \times 5 - 2 \times 25 = 142.5 kN \cdot m

M (C) = 0

Shear changes sign at B, so the maximum there is

M = 142.5 kN \cdot m

Calculated shear and bending diagrams with labelled support/load ordinates; positive values upward. All values are repeated in the working. — This is a teaching diagram drawn from the calculation, not an original source image. Positions are measured in $m$ ; $V$ is measured in $kN$ ; $M$ is measured in $kN \cdot m$ .Open full-size image.

The SFD slopes downward at $4 kN$ per metre on both intervals; its jump atB is $60 kN$ . The BMD is parabolic between loads and has a change of slope atB. It has no jump because there is no applied point couple.

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.

(c) AB lateral-torsional buckling using accurate parameters

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 = 15.7 mm

, giving

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

ε = \sqrt{\frac{275}{355}} = 0.880141

Flange:

\frac{b}{T} = 5.52 < 9 ε = 7.921268

Web in bending:

\frac{d}{t} = 34.2 < 80 ε = 70.411267

Both are Class 1, so

β_{W} = 1

; plastic bending resistance is permitted.

\frac{d}{t} = 34.2 < 70 ε = 61.609858

No separate shear-buckling calculation is needed under the course limit.

Within the $5 m$ AB segment, quarter points are 1.25,2.5 and 3.75 m fromA. The B point load acts at the segment end; it is accounted for throughRA and the maximum moment.

M_{2} = 38.5 \times 1.25 - 2 \times {1.25}^{2} = 45 kN \cdot m

M_{3} = 38.5 \times 2.5 - 2 \times {2.5}^{2} = 83.75 kN \cdot m

M_{4} = 38.5 \times 3.75 - 2 \times {3.75}^{2} = 116.25 kN \cdot m

m_{LT} = 0.2 + \frac{0.15 \times 45 + 0.5 \times 83.75 + 0.15 \times 116.25}{142.5}

= 0.663596 > 0.44

Equivalent demand

94.562500 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{5000}{39.9} = 125.313283

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

= {[1 + 0.05 {(\frac{125.313283}{24.4})}^{2}]}^{- \frac{1}{4}} = 0.810370

λ_{LT} = u v λ \sqrt{β_{W}}

= 0.886 \times 0.810370 \times 125.313283 \times 1 = 89.973415

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

85

row gives

175

;

90

row gives

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

. Interpolation fraction:

\frac{89.973415 - 85}{90 - 85} = 0.994683

p_{b} = 175 + 0.994683 \times (162 - 175) = 162.069120 \frac{N}{{mm}^{2}}

M_{b} = \frac{p_{b} S_{x}}{1,000} = \frac{162.069120 \times 1211}{1,000} = 196.265704 kN \cdot m

Equivalent uniform moment demand:

m_{LT} M_{\max} = 94.562500 kN \cdot m

Utilisation:

\frac{94.562500}{196.265704} = 0.481809

Passes.

AB passes under the stated normal-load/LE5 m assumption. The u value is 0.886 from this exact UB row; $v$ is calculated, not assumed $1$ .

Animation labA beam bends sideways and twists5 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.

(d) Removing B changes stability but not vertical actions

Simple explanation: A support stops only certain movements

Holding a ruler up does not necessarily stop it sliding sideways or twisting.

Identify vertical support, lateral restraint and torsional restraint separately.
Find which flange is in compression in each region.
Use only restraints actually provided by the question.

Remember: An extra side restraint does not create a new vertical reaction.

Related concept and full method

Without the effective lateral restraint atB, the unrestrained segment becomes the full $8 m$ betweenA andC. The point load remains at $5 m$ , so the reactions and vertical SFD/BMD remain unchanged. Longer $L_{E}$ increases slenderness and reduces buckling resistance.

Full-span quarter points:

x = 2, 4, 6 m

M (2) = 38.5 \times 2 - 2 \times 2^{2} = 69 kN \cdot m

M (4) = 38.5 \times 4 - 2 \times 4^{2} = 122 kN \cdot m

M (6) = 38.5 \times 6 - 2 \times 6^{2} - 60 \times 1 = 99 kN \cdot m

m_{LT} = 0.2 + \frac{0.15 \times 69 + 0.5 \times 122 + 0.15 \times 99}{142.5} = 0.804912

New equivalent demand

114.700000 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{8000}{39.9} = 200.501253

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

= {[1 + 0.05 {(\frac{200.501253}{24.4})}^{2}]}^{- \frac{1}{4}} = 0.691395

λ_{LT} = u v λ \sqrt{β_{W}}

= 0.886 \times 0.691395 \times 200.501253 \times 1 = 122.822299

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

120

row gives

104

;

125

row gives

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

. Interpolation fraction:

\frac{122.822299 - 120}{125 - 120} = 0.564460

p_{b} = 104 + 0.564460 \times (97 - 104) = 100.048781 \frac{N}{{mm}^{2}}

M_{b} = \frac{p_{b} S_{x}}{1,000} = \frac{100.048781 \times 1211}{1,000} = 121.159074 kN \cdot m

Equivalent uniform moment demand:

m_{LT} M_{\max} = 114.700000 kN \cdot m

Utilisation:

\frac{114.700000}{121.159074} = 0.946689

Passes.

The member remains a numerical narrow pass under the sameLE=actual/normal-load assumptions, but stability margin is substantially reduced. It is wrong to say that removing a restraint always forces failure; it always requires a new check. Other beam checks are outside the requested LTB calculation.

Animation labA beam bends sideways and twists5 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.

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

$R_{A} = 38.5$ , $R_{C} = 53.5 kN$ . Shear values: $+ 38.5, + 18.5, - 41.5, - 53.5$ ; moment values: $0, 142.5, 0$ . AB: $λ = 125.313283$ , $v = 0.810370$ , $λ_{LT} = 89.973415$ , $p_{b} = 162.069120$ , $M_{b} = 196.265704 kN \cdot m$ ; demand $94.5625$ : passes. Remove B: $λ = 200.501253$ , $λ_{LT} = 122.822299$ , $M_{b} = 121.159074 kN \cdot m$ ; full-span demand $114.7$ : still just passes. These LTB results assume normal loading and that $L_{E}$ equals the distance between restraints.

Mistakes to avoid

Do not make B a vertical support.
Do not use full-span quarter positions for AB.
Do not discard the UDL in the moment ordinates.
Do not state failure after restraint removal without recalculating.

Procedure for an unfamiliar variant

Solve the original vertical system first.
Mark lateral restraint positions separately.
Find each segment’s maximum and local quarter moments.
Readu, $x$ , $r_{y}$ and calculatev without early rounding.
Change both effective length and moment factor when a restraint is removed.

Independent self-check

Try it yourself. Invented variant: all ultimate loads increase $5 %$ after restraintB is removed. Does the full-span LTB check still pass?

Reveal answer and reasoning

Yes, narrowly. Demand becomes $1.05 \times 114.7 = 120.435 kN \cdot m$ , below $121.159074$ . At $6 %$ increase, demand $121.582 kN \cdot m$ exceeds $121.159074$ and fails. Maximum proportional increase $\approx \frac{121.159074}{114.7} - 1 = 5.6313 %$ . This comparison uses the original section and the stated stability assumptions.

Animation labA beam bends sideways and twists2 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.

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.