Design of restrained steel beam to Eurocode 3 (EC3) New terminologies Permanent Action (gk) - Dead Load (gk) Variable Action (qk) - Live Load (qk) •
Select the lightest suitable section in S275 steel required for a simply in S275 steel loaded by uniformly distributed loading UDL gk = 9 kN/m and qk = 7.5 kN/m as shown below. Point Load Gk = 10 kN and Qk = 15 kN placed mid span Assume that the beam is fully laterally restrained.
Step 1 – Calculate the Design Action (FEd) For UDL
For Point Load
Step 2 – Calculate the Design Bending Moment (MEd)
Step 3 – Choose a suitable section size Calculate the Minimum Plastic Modulus (Wpl,y) about y-y axis (Horizontal Axis this is x-x axis in BS5950)
MEd = Design Bending Moment (Nmm) = 259.2 kNm = 259.2 x 106 Nmm
γ
M0
= partial factor of safety for the resistance = 1.05 (this is a constant value)
fy = yield strength of the grade of steel = 275N/mm2 (for S275)
From Section tables (page 4 on handout Universal Beams) Use 356 x 171 x 57 UB Wpl,y = 1010 cm3 Step 4 – Strength Classification FOR FLANGE (
FOR WEB
)
(
)
for S275 steel t < 16mm (tf = 8.1)
for S275 steel t < 16mm (tf = 8.1)
fy = 275 N/mm2
fy = 275 N/mm2 (
)
(
)
For Class 1
For Class 1
from steel table, (page 3 of Handout Universal Beams)
from steel table, (page 3 of Handout Universal Beams)
= 5.53
= 38.5
9ε = 9 x 0.92 = 8.28
72ε = 72 x 0.92 = 66.24
5.53 < 8.28
38.5 < 66.24
FLANGE is Class 1 PLASTIC
WEB is Class 1 PLASTIC
Therefore Section is Class 1 (Plastic)
Step 5 – Check the resistance of the cross section to bending Show that: Mpl,Rd > MEd Mpl,Rd = resistance of cross section to bending MEd = design bending moment For Class 1 Sections
Wpl,y (plastic modulus about y-y axis from steel table, page 4) = 1010 cm3 = 1010 x 103 mm3 fy (for steel < 16mm)= 275 N/mm2 γM0 = 1.05
MEd = 259.2kNm Since: Mpl,Rd (264.5 kNm) > MEd (259.2 kNm) The beam is satisfactory in bending
Step 6 – Check the resistance of the cross section to shear Show that: Vpl,Rd > VEd Vpl,Rd = resistance of cross section to shear VEd = design shear force = (
√ )
Av (Shear area from steel table, page ) = 31.5 cm2 = 31.5 x 102 mm2 fy for steel < 16mm = 275 N/mm2 √ = 1.732
Since: Vpl,Rd (476.3 kN) > VEd (111.6 kN) The beam is satisfactory in shear
Step 7 – Check the Deflection Show that: Maximum Permissible Deflection > Actual Deflection Max. Perm. Def. = Max. Perm. Def. = Max. Perm. Def. = 22.22 mm Act. Def. = qk (UDL) = 7.5 kN/m = 7.5 N/mm (remember, only use Variable Action (Live Load)) Qk (Point Load) = 15 kN = 15 x 103 N L = Span = 8 m = 8000 mm E (for steel) = 210 x 103 N/mm2 I = second moment of area about y-y axis (from steel table) = 16038 cm4 = 16038 x 104 mm4 Act. Def. =
Act. Def. = Act. Def. = 11.88 + 4.75 Act. Def. = 16.63 mm Since: Max. Perm. Def. (22.22) > Act. Def. (16.63) The beam is satisfactory is deflection
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