Deflect1

Rev. 1.0 - 'E' for steel corrected. Coefficients remain unchanged- 23/01/02

Steel Beam Deflection Coefficients ("E" has been taken as = 2.05E5 N/mm²)

Table 1 - ''I' required = K₁.W.L²	'K₁' values
Type of Loading and Support Condition	L/200	L/300	L/360	L/400	L/500	L/600
	1.27	1.91	2.29	2.54	3.18	3.81
	2.03	3.05	3.66	4.07	5.08	6.10
	12.2	18.3	22.0	24.4	30.5	36.6
	32.5	48.8	58.5	65.0	81.3	97.6
	0.254	0.381	0.457	0.508	0.635	0.762
	0.508	0.762	0.915	1.02	1.27	1.52
	0.527	0.791	0.949	1.05	1.32	1.58
	0.909	1.36	1.64	1.82	2.27	2.73
	0.893	1.34	1.61	1.79	2.23	2.68
	1.47	2.20	2.64	2.94	3.67	4.40

Table 2 - ''I' required = K₂.M.L	'K₂' values
Type of Loading and Support Condition	L/200	L/300	L/360	L/400	L/500	L/600
	10.20	15.20	18.30	20.30	25.40	30.50
	8.13	12.20	14.60	16.30	20.30	24.40
	24.40	36.60	43.90	48.80	61.00	73.20
	32.50	48.80	58.50	65.00	81.30	97.60
	3.05	4.57	5.49	6.10	7.62	9.15
	4.07	6.10	7.32	8.13	10.20	12.20
	4.22	6.33	7.59	8.44	10.50	12.70
	4.85	7.27	8.73	9.70	12.10	14.50
	9.52	14.30	17.10	19.00	23.80	28.60
	7.22	10.80	13.00	14.40	18.10	21.70

Units :-

I = cm⁴......... W = KN......... L = m........... M = kN.m

Example on the Use of Deflection Coefficients

CADOSS ^(R)	Project Useful Data			Job Ref WS2001.001
	Part of Structure Restrained Beam - Abridged Design			Calc Sheet No. / Rev Beam / 1
	Drawing Ref n / a	Calc By M.G.D	Date Dec 2001	Check By	Date

Fully Restrained Beam

Span = 10m

Centres = 2.5m

U.D. Load 'W'	Service Load (kN/m²)	g_f	Factored Load (kN/m²)
Dead	5.0	1.4	7.0
Imposed	5.0	1.6	8.0
Total	10.0		15.0

W (service imposed load)	= 5 kN/m²x 2.5m x 10m	= 125 kN
W (service total load)	= 10 kN/m² x 2.5m x 10m	= 250 kN
W (factored total load)	= 15 kN/m² x 2.5m x 10m	= 375 kN

Fv	= 375kN / 2	= 187.5 kN
Mx	= 375kN x 10m / 8	= 468.8 kN.m
I req'd	= 250kN x 10² x 1.59	= 39750 cm⁴(for D=l/250 under total load)
I req'd	= 125kN x 10²x 2.29	= 28625 cm⁴(for D=l/360 due to imposed load)

From Capacity Tables: - Try 533 UB x 82^kg Grade S275
Mcx	= 566 kN.m	> Mx
Ix	= 47500 cm⁴	> I req'd
Pv	= 837 kN	> Fv
Pbw (web bearing)	= 269 kN	> Fv {assumes 50mm of stiff bearing at end supports}
Px (web buckling)	= 168 kN	< Fv {Web fails if unstiffened but with web cleats, welded end plates or web stiffeners the web will be OK for buckling resistance}

Use 533 x 210 UB x 82 kg Grade S275 (No Camber)

With web stiffened against buckling.

Reactions:

Factored total load = 187.5 kN

Service dead load = 62.5 kN

Service Imposed load = 62.5kN

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Notes

1. BS5950 gives some suggested limits for beam deflections based on imposed loads. It is considered here that dead load or total load deflections are equally as important as imposed load deflections as the above example attempts to demonstrate. Had the code suggested limit, of D = l/200 due to imposed load, been used in the above example then a total load deflection of 100mm could have been the end result!

2. If a cambered beam had been chosen for the above example then a beam with an Inertia of => 28625 cm⁴ may have proved adequate. This could have resulted in a lighter section being used especially if steel of grade S355 was also employed. Be careful though, when deflections become high there is a possibility that vibrations during service may become unacceptable.

3. Using Table 1 you will notice that the beam I required is calculated, from the beam span and applied load, before the beam size is selected. This is considered more efficient than selecting a beam and subsequently checking its deflection only to find that it fails!

4. Table 2 is very efficient to use when a beam has complex multiple loads applied. Using Table 2 the maximum moment due to service loads is used to derive I required. If you look at the difference in the coefficients for UDL and Point loads you will see that using Table 2, rather than Table 1, will enable a fairly close estimate to be made for the coefficient required for complex loading.

Document (No. 20011229/WS/001/R1.0)