Structville

A hub for civil engineering related designs, analyses, discussion, information, and knowledge...

Trending Posts:

Search This Blog

Tuesday, August 1, 2017

Solved Example on Design of Steel Beams According to BS 5950 - 1: 2000


Introduction
Universal beam sections are normally employed in buildings to carry load. Loads on beams may include the load from slab, walls, building services, and their own self weight. It is necessary for structural beams to satisfy ultimate and serviceability limit state requirements.
This post gives a solved design example of a laterally restrained beam according to BS 5950.

Design example
A laterally restrained beam 9m long that is simply supported at both ends supports a dead uniformly distributed load of 15 KN/m and an imposed load uniformly distributed load of 5KN/m. It also carries a dead load of 20KN at distance of 2.5m from both ends. Provide a suitable UB to satisfy ultimate and serviceability limit state requirements (Py = 275 N/mm2).


Solution
At ultimate limit state;
Concentrated dead load = 1.4Gk = 1.4 × 20 = 28 KN
Uniformly distributed load = 1.4Gk + 1.6Qk = 1.4(15) + 1.6(5) = 29 KN/m


Support Reactions
Let ∑MB = 0; anticlockwise negative
(9 × Ay) – (29 × 9 × 4.5) – (28 × 6.5) – (28 × 2.5) = 0
Ay = 158.5 KN

Let ∑MA = 0; clockwise negative
(9 × By) – (29 × 9 × 4.5) – (28 × 6.5) – (28 × 2.5) = 0
By = 158.5 KN

Internal Stresses
Moment
MA = 0 (Hinged support)
MC = MD = (158.5 × 2.5) – (29 × 2.5 × 1.25) = 305.625 KNm
Mmidspan = (158.5 × 4.5) – (29 × 4.5 × 2.25) – (28 ×2) = 363.625 KNm

Shear
QA = Ay = 158.5 KN
QCL = 158.5 – (29 × 2.5) = 86 KN
QCR = 158.5 – (29 × 2.5) - 28 = 58 KN
QDL = 158.5 – (29 × 6.5) – 28 = -58 KN
QDR = 158.5 – (29 × 6.5) – 28 - 28 = - 86 KN
QB = 158.5 – (29 × 9) – 28 - 28 = - 158.5 KN

Internal Forces Diagram



Structural Design to BS 5950
Py = 275 N/mm2

Initial selection of section
Moment Capacity of section Mc = PyS ------- (1)

Where S is the plastic modulus of the section
Which implies that S = Mc/Py = (363.625 × 106)/275 = 1320963.636 mm3 = 1320.963 cm3

With this we can go to the steel sections table and select a section that has a plastic modulus that is slightly higher than 1320.963 cm3

Try section UB 457 × 191 × 67 (S = 1470 cm3)

Properties of the section;
Ixx = 29400 cm4
Zxx = 1300 cm3
Mass per metre = 67.1 kg/m
D = 453.4mm
B = 189.9mm
t = 8.5mm
T = 12.7mm
r = 10.2mm
d = 407.6mm

Strength classification
Since T = 12.7mm < 16mm, Py = 275 N/mm2
Hence ε = √(275/Py) = √(275/275) = 1.0

Section classification
Flange
b/T = 7.48 < 9ε; Flange is plastic class 1
Web
d/t = 48 < 80ε; Web is also plastic class 1

Shear Capacity
As d/t = 48 < 70ε, shear buckling need not be considered (clause 4.4.4)

Shear Capacity Pv = 0.6PyAv  ----- (2)

P= 0.6PytD = 0.6 × 275 × 8.5 × 453.4 = 635893.5 N = 635.89 kN

But design shear force Fv = 158.5 KN. Since Fv(158.5) < Pv(635.89), section is ok for shear.

Now, 0.6Pv = 0.6 × 635.89 = 381.54 kN
Since Fv(158.5KN) < 0.6Pv(381.54 KN), we have low shear load.



Moment Capacity
Design Moment = 363.625 KNm

Moment capacity of section UB 457 × 191 × 67 (S = 1470 cm3 = 1470 × 103 mm3)

Mc = PyS = 275 × 1470 × 103 = 404.25 × 106 N.mm = 404.25 KNm
1.2PyZ = 1.2 × 275 × 1300 × 103 = 429 × 106 N.mm = 429.00 KNm

Mc (404.25 KNm) < 1.2PyZ (429.00 KNm) Hence section is ok

Evaluating extra moment due to self weight of the beam
Self-weight of the beam Sw = 67.1 kg/m = 0.658 KN/m (UDL on the beam)
Moment due to self weight (Msw) = (ql2)/8 = (0.658 × 92)/8 = 6.66 KNm

(363.625 + 6.66) < Mc (404.25 KNm) < 1.2PyZ (429.00 KNm) Hence section is ok for moment resistance.

Deflection Check
We check deflection for the unfactored imposed load; E = 205 KN/mm2 = 205 × 106 KN/m2; Ixx = 29400 cm4 = 29400 × 10-8 m4



The maximum deflection for this structure occurs at the midspan and it is given by;
δ = (5ql4)/384EI = (5 × 5 × 94) / (384 × 205 × 106 × 29400 × 10-8) = 7.087 × 10-3 m = 7.087mm

Permissible deflection; L/360 = 9000/360 = 25mm
7.087mm < 25mm. Hence deflection is satisfactory.

Web bearing
According to Clause 4.5.2 of BS 5950-1:2000, the bearing resistance Pbw is given by:

Pbw = (b1 + nk)tPyw   ----- (3)

Where;
b1 is the stiff bearing length
n = 5 (at the point of concentrated loads) except at the end of a member and n = 2 + 0.6be/k ≤ 5 at the end of the member
be is the distance to the end of the member from the end of the stiff bearing
k = (T + r) for rolled I- or H-sections
T is the thickness of the flange
t is the web thickness
Pyw is the design strength of the web


Web bearing at the supports
Let us assume that beam sits on 200mm bearing, and be = 20mm

k = (T + r) = 12.7 + 10.2 = 22.9mm;
Hence n = 2 + 0.6(20/22.9) = 2.52mm < 5mm.
Pbw = (b1 + nk)tPyw
Pbw = [200 + 2.52(22.9)] × 8.5 × 275 = 602392.45 N = 602.392 KN
Pbw (602.392 KN) > Fv (158.5 KN) Hence it is ok

Contact stress at supports
Pcs = [b1 × 2(r +T )]Py = [200 × 2(22.9)] × 275 = 2519000N = 2519 KN
Pcs (2519 KN) > Fv (158.5 KN) Hence contact stress is ok

Web buckling

According to clause 4.5.3.1 of BS 5950, provided the distance αe from the concentrated load or reaction to the nearer end of the member is at least 0.7d, and if the flange through which the load or reaction is applied is effectively restrained against both;

(a) rotation relative to the web
(b) lateral movement relative to the other flange

The buckling resistance of an unstiffened web is given by;

Px = [25εt/√(b1 + nk)d] Pbw ----- (4)

When αe < 0.7d, the buckling resistance of an unstiffened web is given by;

Px = [(αe + 0.7d)/1.4d] × [25εt/√(b1 + nk)d)] Pbw ------------- (5)

Therefore, αe = 20mm + (200/2) = 120mm
0.7d = 0.7 × 407.6 = 285.32mm
αe(120mm) < 0.7d(285.32mm). Hence equation (5) applies

Px = [(120 + 285.32)/(1.4 × 407.6)] × [(25 × 1 × 8.5 )/√((200 + 2.52 × 22.9) × 407.6)] × 602.392 = 280.538KN

Px(280.538 KN) > Fv(158.5 KN) Hence it is ok.

Thank you for reading, and feel free to share.

Like Our Facebook page
www.facebook.com/structville


1 comment: