Shear Force and Bending Moment Diagrams: A Worked Steel Beam Assignment Example

This is a worked shear force and bending moment diagram example of the kind set in a first structures or mechanics module: one simply supported steel floor beam, 6.0 m long, carrying a uniformly distributed load and a point load. Every number is computed on the page and every line carries its unit.

It is a worked example for study, not a design to build: the checks show how the clauses are applied, not advice about any real structure. Our engineering assignment help covers structures, mechanics and the engineering maths behind them, and more worked samples sit in computer science assignment samples, which includes our engineering samples, and in maths and science samples.

What Is the Beam Problem in This Worked Example?

A 6.0 m simply supported steel beam carries a UDL and a point load at 1.5 m from A, factored to 12.0 kN/m and 24.0 kN. The reactions are 54.0 and 42.0 kN, the maximum moment is 73.5 kN m at 2.5 m, and a 254 x 146 x 31 UB in S275 passes every check in this exercise.

The question as set (characteristic loads)
Beam: simply supported, span L = 6.0 m, pin at A (x = 0), roller at B (x = 6.0 m)
UDL over the full span: permanent g_k = 5.0 kN/m, variable q_k = 3.5 kN/m
Point load from a secondary beam, 1.5 m from A: permanent G_k = 10 kN, variable Q_k = 7 kN
Trial section: 254 x 146 x 31 UB, grade S275

The beam's own weight, 31.1 kg/m or about 0.31 kN/m, is inside g_k. The sheet asks for:

1. ULS design loads, BS EN 1990 expression 6.10;
2. support reactions;
3. shear force and bending moment diagrams with key values;
4. maximum bending moment and its position;
5. bending and shear checks, BS EN 1993-1-1;
6. a deflection check for a brittle finish.

One assumption comes before any calculation: the floor restrains the top flange laterally along its length, which is why lateral-torsional buckling is not checked below. A beam without that restraint also needs clause 6.3.2 of BS EN 1993-1-1, and an answer that never states the assumption leaves the examiner guessing whether the writer knew it mattered.

How Do You Work Out the Design Loads and the Support Reactions?

Factor the loads, then take moments about one support. Expression 6.10 of BS EN 1990 (1.35 on permanent, 1.5 on variable actions) gives a design UDL of 12.0 kN/m and a design point load of 24.0 kN. Moments about A give R_B = 42.0 kN; vertical equilibrium gives R_A = 54.0 kN, and together they carry the 96.0 kN applied.

The factors are the UK National Annex values in Table NA.A1.2(B) of BS EN 1990, as printed on SteelConstruction.info's design codes page, which also lists the alternative pair 6.10a and 6.10b; the sheet names 6.10, and your answer should say which expression it used.

Design loads (ULS)
w_Ed = 1.35 × 5.0 + 1.5 × 3.5 = 6.75 + 5.25 = 12.0 kN/m
P_Ed = 1.35 × 10 + 1.5 × 7 = 13.5 + 10.5 = 24.0 kN
Total distributed load = 12.0 × 6.0 = 72.0 kN, acting at midspan, 3.0 m from A

Reactions. The free body has R_A upward at x = 0, R_B upward at x = 6.0 m, 72.0 kN downward at 3.0 m and 24.0 kN downward at 1.5 m. Taking moments about A, clockwise positive:

R_B × 6.0 = 72.0 × 3.0 + 24.0 × 1.5 = 216.0 + 36.0 = 252.0 kN m
R_B = 252.0 / 6.0 = 42.0 kN
Vertical equilibrium: R_A = 72.0 + 24.0 − 42.0 = 54.0 kN

Write the check as a sentence, not bare arithmetic: R_A + R_B = 96.0 kN, the total applied load, and moments about B give R_A × 6.0 = 72.0 × 3.0 + 24.0 × 4.5 = 324.0 kN m, so R_A = 54.0 kN again. A wrong reaction is the costliest error on a beam question, because it is carried into every shear value, every moment value and both resistance checks.

How Do You Draw the Shear Force Diagram?

Start at A with V = R_A = 54.0 kN, fall at 12.0 kN per metre under the UDL, drop by 24.0 kN at the point load and arrive at −42.0 kN (minus R_B) at B. The shear crosses zero at x = 2.5 m. Join the values with straight lines, because a uniform load gives a linear shear.

Shear force diagram for the 6.0 m beam, V(x) in kN

Chart data

Point (m)	Shear force V
0	54 kN
0.5	48 kN
1	42 kN
1.5 (just left)	36 kN
1.5 (just right)	12 kN
2	6 kN
2.5	0 kN
3	-6 kN
3.5	-12 kN
4	-18 kN
4.5	-24 kN
5	-30 kN
5.5	-36 kN
6	-42 kN

The sign convention comes from Udoeyo's Structural Analysis, section 1.4: "a shear force that tends to move the left of the section upward or the right side of the section downward will be regarded as positive", and a bending moment is positive when it causes sagging. Under this convention dV/dx = −w and dM/dx = V, and every expression below obeys it.

Cut the beam at a distance x from A and sum the vertical forces to the left of the cut. There are two regions, because the point load changes the expression at 1.5 m.

Region 1, 0 ≤ x < 1.5 m
V(x) = R_A − w_Ed x = 54.0 − 12.0x
V(0) = 54.0 kN
V just left of the load, x = 1.5 m: 54.0 − 18.0 = 36.0 kN

Region 2, 1.5 m < x ≤ 6.0 m
V(x) = 54.0 − 12.0x − 24.0 = 30.0 − 12.0x
V just right of the load, x = 1.5 m: 30.0 − 18.0 = 12.0 kN
V(6.0) = 30.0 − 72.0 = −42.0 kN, which is −R_B, so the diagram closes
Zero shear: 30.0 − 12.0x = 0, so x = 2.5 m from A

The drop at 1.5 m is exactly the point load, 36.0 − 12.0 = 24.0 kN, which is the second check on this section: every vertical step in a shear diagram must equal a point load or a reaction, and the total of the steps and slopes must return the diagram to zero at the far end.

How Do You Draw the Bending Moment Diagram?

Take moments of everything to the left of the cut, sagging positive. Left of the point load, M = 54.0x − 6.0x², which gives 67.5 kN m at the load. Beyond it the load's own moment is subtracted, the curve peaks at 73.5 kN m at 2.5 m where the shear is zero, and returns to zero at B.

Bending moment diagram for the 6.0 m beam, M(x) in kN m

Chart data

Point (m)	Bending moment M
0	0 kN m
0.5	25.5 kN m
1	48 kN m
1.5	67.5 kN m
2	72 kN m
2.5	73.5 kN m
3	72 kN m
3.5	67.5 kN m
4	60 kN m
4.5	49.5 kN m
5	36 kN m
5.5	19.5 kN m
6	0 kN m

Region 1, 0 ≤ x ≤ 1.5 m
M(x) = R_A x − w_Ed x² / 2 = 54.0x − 6.0x²
M(0) = 0
M(1.5) = 81.0 − 13.5 = 67.5 kN m

Region 2, 1.5 m ≤ x ≤ 6.0 m
M(x) = 54.0x − 6.0x² − 24.0(x − 1.5)
M(2.5) = 135.0 − 37.5 − 24.0 = 73.5 kN m
M(3.0) = 162.0 − 54.0 − 36.0 = 72.0 kN m
M(6.0) = 324.0 − 216.0 − 108.0 = 0, so the diagram closes at B

Working from B gives the same peak: with 3.5 m of beam to the right of 2.5 m, M(2.5) = R_B × 3.5 − 12.0 × 3.5² / 2 = 147.0 − 73.5 = 73.5 kN m. The area rule in Baker and Haynes, Engineering Statics, section 8.5 ("the change in the moment value between two points is the area under the shear curve between those points") gives a third route: the trapezium under the shear from 0 to 1.5 m, (54.0 + 36.0) / 2 × 1.5 = 67.5 kN m, is M(1.5), and the triangle to 2.5 m adds 12.0 × 1.0 / 2 = 6.0 kN m, so M(2.5) = 73.5 kN m.

The shape follows from the same rule. Where the shear is a sloping line the moment is a parabola; where the shear steps down by 24.0 kN the slope of the moment diagram changes by 24.0 kN m per metre, which is the kink at 1.5 m; and where the shear passes through zero the moment is flat, which is the peak. A moment diagram that fails to close to zero at a roller has an arithmetic error somewhere upstream.

The chart draws sagging above the axis, as Udoeyo does ("positive bending moments are drawn above the x-centroidal axis"). Some modules draw on the tension side instead, which puts sagging below; follow your module and label the diagram.

Where Is the Maximum Bending Moment, and Why Is It Where the Shear Force Is Zero?

At 2.5 m from A, not at midspan, with M_max = 73.5 kN m. Because dM/dx = V, the moment is stationary wherever the shear passes through zero, and on a sagging beam that point is the maximum. Solve V(x) = 30.0 − 12.0x = 0 rather than assuming the middle: midspan gives 72.0 kN m, 2 percent low.

Udoeyo's section 1.4 states the rule directly: "maximum bending moment occurs where the shearing force equals zero". The point load pulls the crossing from midspan towards A by its moment about A over the total UDL, P_Ed a / (w_Ed L) = 24.0 × 1.5 / (12.0 × 6.0) = 0.5 m, from 3.0 m to 2.5 m, where a is its distance from A. A heavier load, or the same load further from A, pulls it further, until the crossing reaches the load (at a = 2.25 m for these loads). Beyond that the shear jumps across zero at the load without touching it, and the maximum sits under the load: read it from the moment expression rather than solving V = 0.

Does a 254 x 146 x 31 UB in S275 Pass the Bending and Shear Checks?

It passes both in this exercise. The section is Class 1, so its plastic modulus of 393 cm³ applies and M_c,Rd = 108.1 kN m against M_Ed = 73.5 kN m, a utilisation of 0.68. The plastic shear resistance is 260.3 kN against V_Ed = 54.0 kN, below half of V_pl,Rd, so the moment resistance needs no reduction for shear.

Section check results, 254 x 146 x 31 UB in S275

Design moment M_Ed

73.5 kN m At 2.5 m from A; 72.0 kN m at midspan.

Moment resistance M_c,Rd

108.1 kN m Class 1, so W_pl,y = 393 cm³; W_el,y would give 96.5 kN m.

Bending utilisation

68% M_Ed / M_c,Rd = 0.68.

Shear resistance V_pl,Rd

260.3 kN A_v = 1,639 mm²; design shear V_Ed = 54.0 kN at support A.

Shear utilisation

21% V_Ed / V_pl,Rd = 0.21.

Half of V_pl,Rd

130.1 kN V_Ed is below it, so M_c,Rd is not reduced for shear (clause 6.2.8).

Section properties, from British Steel's universal beams datasheet
Mass 31.1 kg/m; depth h = 251.4 mm; width b = 146.1 mm
Web t_w = 6.0 mm; flange t_f = 8.6 mm; root radius r = 7.6 mm
Depth between fillets d = 219.0 mm; area A = 39.7 cm²
Second moment of area I_y = 4,413 cm⁴
Elastic modulus W_el,y = 351 cm³; plastic modulus W_pl,y = 393 cm³

Material, from SteelConstruction.info's steel material properties and member design pages
S275 with t_f = 8.6 mm ≤ 16 mm, so f_y = 275 N/mm² (BS EN 10025-2); E = 210,000 N/mm²; γ_M0 = 1.00 (UK National Annex)

Classification, Table 5.2 of BS EN 1993-1-1 (Class 1 limits: 9ε for an outstand flange in compression, 72ε for a web in bending)
ε = √(235 / f_y) = √(235 / 275) = 0.924
Flange outstand c = (b − t_w − 2r) / 2 = (146.1 − 6.0 − 15.2) / 2 = 62.45 mm; c / t_f = 62.45 / 8.6 = 7.26 ≤ 9ε = 8.32, so the flange is Class 1
Web c = d = 219.0 mm; c / t_w = 219.0 / 6.0 = 36.5 ≤ 72ε = 66.6, so the web is Class 1 in bending
The section is Class 1, and W_pl,y applies

The datasheet prints B/2T = 8.49, which is the whole half-flange over its thickness; Table 5.2 measures the outstand from the root fillet, so 7.26 is the figure compared with the limit. A section that came out Class 3 would use W_el,y instead, and here that would give 351 × 10³ mm³ × 275 N/mm² / 1.00 = 96.5 × 10⁶ N mm = 96.5 kN m, a utilisation of 0.76 rather than 0.68. The class decides the modulus; the modulus is not a free choice.

Bending resistance, clause 6.2.5
W_pl,y = 393 cm³ = 393 × 10³ mm³
M_c,Rd = W_pl,y f_y / γ_M0 = 393 × 10³ mm³ × 275 N/mm² / 1.00 = 108.1 × 10⁶ N mm
10⁶ N mm = 1 kN m, so M_c,Rd = 108.1 kN m
M_Ed / M_c,Rd = 73.5 / 108.1 = 0.68, which is less than 1.0

Shear resistance, clause 6.2.6
Shear area for a rolled I-section loaded parallel to the web, clause 6.2.6(3)(a), with A = 39.7 cm² = 3,970 mm²:
A_v = A − 2 b t_f + (t_w + 2r) t_f = 3,970 − 2 × 146.1 × 8.6 + (6.0 + 15.2) × 8.6 = 3,970 − 2,512.9 + 182.3 = 1,639 mm²
The clause sets a minimum of η h_w t_w; with η = 1.0, the conservative value, and h_w = 251.4 − 2 × 8.6 = 234.2 mm, that is 1.0 × 234.2 × 6.0 = 1,405 mm², so 1,639 mm² stands
Shear buckling: h_w / t_w = 234.2 / 6.0 = 39.0 ≤ 72ε / η = 66.6 (55.5 if η = 1.2), so clause 6.2.6(6) calls for no shear buckling check
V_pl,Rd = A_v (f_y / √3) / γ_M0 = 1,639 mm² × (275 / 1.732) N/mm² / 1.00 = 1,639 × 158.8 = 260,300 N = 260.3 kN
V_Ed = R_A = 54.0 kN, the largest shear on the beam; V_Ed / V_pl,Rd = 0.21
0.5 V_pl,Rd = 130.1 kN, and 54.0 kN is below it

The M_c,Rd and V_pl,Rd expressions are as printed on the SteelConstruction.info member design page, with the clause 6.2.8 rule that shear below 0.5 V_pl,Rd may be neglected "except where shear buckling reduces the section resistance". The h_w / t_w line closes that exception, and at 2.5 m, where the moment peaks, the shear is zero. The same page says "beams with sufficient restraint to the compression flange are not susceptible to lateral-torsional buckling"; without the floor, the beam needs a clause 6.3.2 check, named here and not computed.

Does the Beam Pass the Deflection Check?

It does, in this exercise. Under unfactored variable actions alone, the 3.5 kN/m UDL deflects the beam 6.37 mm at midspan and the 7 kN point load adds 2.34 mm there: 8.7 mm in total against the span/360 limit of 16.7 mm under a brittle finish. The UK National Annex leaves the permanent actions out.

The limits are the suggested values in the UK National Annex to BS EN 1993-1-1, as tabled on SteelConstruction.info's design codes page: span/360 under plaster or another brittle finish, span/200 for other beams, length/180 for cantilevers, all under variable actions only; a client can agree others for a project. Deflection is a serviceability check, so no partial factors apply.

Values for the check
q_k = 3.5 kN/m = 3.5 N/mm; Q_k = 7 kN = 7,000 N; L = 6,000 mm
E = 210,000 N/mm²; I_y = 4,413 cm⁴ = 4.413 × 10⁷ mm⁴
E I_y = 210,000 × 4.413 × 10⁷ = 9.267 × 10¹² N mm²

The formulas below come from the American Wood Council's Design Aid 6. They are elastic results, valid for any material in consistent units, and follow from integrating EI d²y/dx² = M twice (Udoeyo, section 1.7).

UDL at midspan (Design Aid 6, Figure 1)
δ₁ = 5 q_k L⁴ / (384 E I) = 5 × 3.5 × 6,000⁴ / (384 × 9.267 × 10¹²) = 6.37 mm

Point load, 1.5 m from A (Design Aid 6, Figure 8)
Figure 8 gives the maximum for a > b, with x and a measured from the same support, so read this beam from B: a = 4,500 mm (B to the load), b = 1,500 mm (load to A), a + 2b = 7,500 mm
At midspan x = 3,000 mm, which is less than a: δ₂ = Q_k b x (L² − b² − x²) / (6 E I L) = 7,000 × 1,500 × 3,000 × (6,000² − 1,500² − 3,000²) / (6 × 9.267 × 10¹² × 6,000) = 2.34 mm
The load is off-centre, so its own maximum is not at midspan: δ_max = Q_k a b (a + 2b) √(3a(a + 2b)) / (27 E I L) = 7,000 × 4,500 × 1,500 × 7,500 × √(3 × 4,500 × 7,500) / (27 × 9.267 × 10¹² × 6,000) = 2.38 mm
It occurs at x = √(a(a + 2b) / 3) = √(4.5 × 7.5 / 3) = 3.35 m from B, which is 2.65 m from A; the midspan value is 1.6 percent below it

Total and limit
Superposition applies because the equations are linear: Roylance, section 4.3, "the response to a combination of loads is the sum of the responses that would be generated by each separate load acting alone"
At midspan: δ = 6.37 + 2.34 = 8.71 mm
The two maxima are 0.35 m apart, so their sum, 6.37 + 2.38 = 8.75 mm, is an upper bound on the deflection anywhere on the span
Limit for a brittle finish: L / 360 = 6,000 / 360 = 16.7 mm; 8.7 mm is 52 percent of it
For comparison, L / 200 = 30.0 mm for a beam with no brittle finish

A common mistake on this question is running the deflection with the ULS loads. Put 12.0 kN/m and 24.0 kN into the same midspan formulas and the answer is 29.9 mm, which fails the 16.7 mm limit and would send the student off to a heavier section for no reason.

What Loses Marks in a Shear Force and Bending Moment Assignment?

The mistakes we correct most often are unchecked reactions, mixed sign conventions and dropped units. A reaction that is 6 kN out puts every later number out; a shear formula from one textbook and a moment relation from another flip the sign of the diagram; a modulus left in cm³ against N/mm² gives a resistance a thousand times wrong.

What loses the mark and what earns it, step by step

Point of comparison	What loses the mark	What earns it
Free-body diagram	What loses the mark No sketch; the reactions appear from nowhere	What earns it Beam, supports, factored loads and both reactions drawn and labelled before the first equation
Reactions	What loses the mark R_A found from one equation and never checked	What earns it R_A + R_B = 96.0 kN stated in a sentence, and R_A confirmed by moments about B
Sign convention	What loses the mark A shear formula from one textbook and a moment relation from another	What earns it One convention named once (left side up is positive shear, sagging is positive moment, dM/dx = V) and kept to the end
Units	What loses the mark kN mixed with N, cm³ with mm³, no unit on the final answer	What earns it Every line carries its unit; cm³ to mm³ and N mm to kN m converted where they happen
Section modulus	What loses the mark W_el used for a Class 1 section, or W_pl for a Class 3	What earns it Class found from Table 5.2 first, then the modulus that class allows
Deflection	What loses the mark Factored ULS loads with the permanent actions included (29.9 mm here)	What earns it Unfactored variable actions only (8.7 mm here), against the limit the finish requires
Assumptions and labels	What loses the mark Restraint never stated; diagrams drawn with no values on them	What earns it Restraint assumption stated up front; every key value written on both diagrams

The sign convention trap is set by good books. Roylance's Mechanics of Materials, section 4.1, counts shear as positive when it points upward on a face whose outward normal points along +x, the reverse of Udoeyo's left-side-up rule, so its relations read dV/dx = −q and dM/dx = −V; both books call sagging positive. Each set is correct inside its own convention. Take the shear expression from one and the moment relation from the other and the bending moment diagram comes out upside down while every number in it looks plausible, which is why the convention belongs in the first line of the answer.

Units are the other quiet failure. Each conversion in this example (cm² to mm², cm³ to mm³, N mm to kN m, cm⁴ to mm⁴) is written on the line where it happens. A line with no unit on it is a line the reader cannot check.

How Do You Repeat This Method for Your Own Beam?

Sketch the free body, factor the loads with expression 6.10 of BS EN 1990, find and check the reactions, write one shear and one moment expression for each region between load changes, and set V = 0 to locate M_max. Then classify the section with Table 5.2 of BS EN 1993-1-1 and run the resistance and deflection checks.

The method this example followed, in seven steps

1. 1 Sketch the beam and its free body Span, supports, every load with its position, and the two unknown reactions drawn in. State the restraint assumption here.
2. 2 Factor the loads BS EN 1990 expression 6.10 with 1.35 on permanent and 1.5 on variable actions for the ULS checks; keep the unfactored variable actions aside for deflection.
3. 3 Find the reactions and check them Moments about one support for the other reaction, vertical equilibrium for the first, then the sum against the total load and moments about the second support.
4. 4 Write a shear expression for each region One expression between each pair of load changes; evaluate at both ends of every region and just either side of every point load.
5. 5 Write a moment expression for each region Moments of everything to the left of the cut, sagging positive; check that the moment is zero at a pinned or roller end.
6. 6 Set V = 0 and find M_max Solve the shear expression for x, substitute into the moment expression, and confirm the value from the other support.
7. 7 Run the resistance and deflection checks Class, M_c,Rd, V_pl,Rd, the 0.5 V_pl,Rd rule, LTB if the flange is unrestrained, then deflection under variable actions against the National Annex limit.

On a cantilever, Udoeyo's procedure lets you skip the reactions by starting at the free end; the moment is hogging, negative under this convention, and peaks at the fixed support, where the shear is largest. The V = 0 rule finds turning points inside a span, and a cantilever's maximum sits at its boundary instead. An overhanging beam has a hogging maximum over the support and a sagging maximum in the span: check each against its own resistance. Wherever the moment hogs, the bottom flange is in compression and a floor on the top flange does not restrain it, so that region needs the clause 6.3.2 lateral-torsional buckling check (M_b,Rd) even when the span does not.

Need help with a structures or mechanics assignment? Send us the beam, frame or truss problem set on WhatsApp with the section tables you have been given and your deadline, and our engineering assignment help team works it the way this page does, with every step shown and every line carrying its unit.

The same show-every-step working runs through our Boolean algebra worked example and probability and statistics worked example; if this beam leads to a final-year build, see our final year project ideas.

Sources

All sources were opened on 24 September 2026. BS EN 1990 and BS EN 1993-1-1 with their UK National Annexes are BSI standards and not free to read, so rule wording is quoted from SteelConstruction.info; clause and table numbers are the standards' own.

• Udoeyo, F. (n.d.) Structural Analysis, section 1.4, Internal Forces in Beams and Frames. Engineering LibreTexts, CC BY-NC-ND 4.0. eng.libretexts.org. The sign convention, dV/dx = −w and dM/dx = V, the zero-shear rule, positive moment drawn above the axis, and the procedure behind steps 1 and 3 to 6, including starting a cantilever at its free end.
• Udoeyo, F. (n.d.) Structural Analysis, section 1.7, Deflection of Beams: Geometric Methods. Engineering LibreTexts. eng.libretexts.org. The elastic curve equation EI d²y/dx² = M and double integration.
• Baker, D. W. and Haynes, W. (n.d.) Engineering Statics: Open and Interactive, section 8.5, Relations between Loading, Shear and Moment. engineeringstatics.org. The slope and area rules behind the M(1.5) and M(2.5) checks and both diagram shapes.
• Roylance, D. (n.d.) Mechanics of Materials, section 4.1, Shear and Bending Moment Diagrams. Engineering LibreTexts, CC BY-NC-SA 4.0. eng.libretexts.org. Its shear convention is the reverse of Udoeyo's (positive shear upward on a positive face), so its moment relation reads dM/dx = −V; it also calls sagging positive. No formula is taken from it.
• Roylance, D. (n.d.) Mechanics of Materials, section 4.3, Beam Displacements. Engineering LibreTexts. eng.libretexts.org. Superposition of deflections.
• American Wood Council (2007) Design Aid No. 6: Beam Design Formulas with Shear and Moment Diagrams. American Forest & Paper Association. awc.org (PDF). Figure 1 (uniform load) and Figure 8 (concentrated load at any point); its notation is in pounds and inches.
• British Steel (2024) Sections: Universal beams (UB), dimensions and properties, datasheet ref. CUBD:ENG:072024. britishsteel.co.uk (PDF). Every property of the 254 x 146 x 31 UB.
• SteelConstruction.info (n.d.) Member design. Steel Construction Institute. steelconstruction.info. Wording of the classification rule, the M_c,Rd, V_pl,Rd and shear-area expressions, the 0.5 V_pl,Rd rule and its shear-buckling exception, γ_M0 = 1.00 (UK National Annex), and the sentences on restrained compression flanges and M_b,Rd; the page gives no clause numbers.
• SteelConstruction.info (n.d.) Design codes and standards. Steel Construction Institute. steelconstruction.info. The partial factors 1.35 and 1.5 from Table NA.A1.2(B) of the UK National Annex to BS EN 1990 (expressions 6.10, 6.10a, 6.10b) and the suggested deflection limits under variable actions only.
• SteelConstruction.info (n.d.) Steel material properties. Steel Construction Institute. steelconstruction.info. f_y = 275 N/mm² for S275 at t ≤ 16 mm (BS EN 10025-2) and E = 210,000 N/mm².
• BSI. BS EN 1990, Eurocode: Basis of structural design, with the UK National Annex; and BS EN 1993-1-1, Eurocode 3: Design of steel structures, Part 1-1: General rules and rules for buildings, with the UK National Annex. The standards whose expression 6.10, Table 5.2 (limits 9ε and 72ε) and clauses 6.2.5, 6.2.6, 6.2.6(3)(a), 6.2.6(6), 6.2.8 and 6.3.2 are applied above.