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CADmantra Technologies Pvt. Ltd. is one of the best Cad training company in northern zone in India . which are provided many types of courses in cad field i.e AUTOCAD,SOLIDWORK,CATIA,CRE-O,Uniraphics-NX, CNC, REVIT, STAAD.Pro. And many courses
Contact: www.cadmantra.com
www.cadmantra.blogspot.com
www.cadmantra.wix.com
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17-Examples of Beams (Steel Structural Design & Prof. Shehab Mourad)
1. 1 Prepared by Prof. Shehab Mourad – Department of Civil Eng. - KSU
Examples on beam design
A simple steel beam (A36) has a span of 9m and is supporting a uniform service dead
load = 12 kN/m and service live load = 21 kN/m,
Find the lightest W- Shape section for the following Cases :
1) The top flange in laterally supported continuously with RC Slab
2) The top flange is supported at (2.25m) apart
3) The top flange is only supported at mid span
4) The top flange is supported at only ends.
Solution:-
Fy = 250 MPa
wult = 1.2 D + 1.6 L = 1.2 x 12 + 1.6 x 21 = 48 kN/m
Mult max = wL2
/8 = 48 (9)2
/8 = 486 kN.m
Qult max = w L /2 = 48 x 9 /2 = 216 kN
Case 1:
Top flange is continuously laterally supported .
Lb = 0 < Lp
Look for the lightest W- Shape , so that
Ø Mp > 486 kN.m (Zone 1) of curve
From LRFD tables Choose W 610 x 82
Mp= 495 kN.m , Lp= 1.7m, L r = 4.99 m BF= 57.8
For zone 1 of curve
ØMp= Fy.Zx = 495 kN.m
Ø Mn least of
Ø 1.5 Fy Sx =0.9(1.5) 250 x 1870x103
x 10-6
= 631.12 kN.m
Therefore Ø Mn = 495 kN.m > M ult max = 486 kN.m
W 610 x 82 is Safe
486 kN.m
+
+
-
216 k N
216 k N
w
L = 9 m
2. 2 Prepared by Prof. Shehab Mourad – Department of Civil Eng. - KSU
Check shear strength
For W 610 x 82 h = 534 mm , tw = 10 mm, d = 599 mm
h = 534 = 53.4
tw 10
2.45 E = 2.45 2x 105
= 69.30 , h < 2.45 E
Fy 250 tw Fy
Ø Vn= (0.90) 0.6 x 250 x (599 x 10) x 10-3
= 808.65 kN > Vult = 216 kN
The W 610 x 82 section is Safe in shear
Case 2
Top flange is laterally @ 2.25m Lb=2.25m
Try to use same section W 610x82
Mp= 495 kN.m , Lp= 1.7m, Lr = 4.99 m, BF= 57.8
Lb> Lp beam is located in zone 2 on curve
Therefore, Ø Mn = Cb [Ø Mp – BF (L b - L p) ]
Check for part 2 of beam :-
Cb2 = 12.5 M max
2.5 Mmax+ 3MA+4MB+3MC
Cb2 = 12.5 x 486 = 1.061
2.5 x 486 + 3 x 417.7 + 4 x 455.6 + 3 x 478.4
The factored resistance of W 610x82, for Lb = 2.25 m
Ø Mn = 1.061 [495 - 57.8 (2.25-1.70) ] = 491.46 kN. m < Ø Mp= 495 kN.m
Ø Mn = 491.46 kN.m > Mult max = 486 kN.m
Lb= 2.25 Lb= 2.25
Critical part, Cb in
less, and maximum
moment is applied
417.7
455.6
478.4
486
364.5
W610 x 82 is safe for unsupported length of compression flange = 2.25 m
3. 3 Prepared by Prof. Shehab Mourad – Department of Civil Eng. - KSU
Case 3 : Top flange is laterally supported at mid span (4.5m)
Lb = 4.5m and most probably will be greater than Lp
Therefore the beam is located at zone 2 of beam curve
Cb = 12.5 Mmax
2.5M max + 3MA + 4MB + 3Mc
Cb = 12.5 x 486 = 1.30
2.5 x 486 + 3 x 212.6 + 4 x 364.5 + 3 x 455.6
Enter beam graphs and look for the lightest section with,
Lb=4.5 m, and factored resistance not less than Mmax / Cb = 458/1.30 = 373.8 kN.m
From graph, the lightest section is W 610 x 92
From beam tables, for W 610 x 92, Ø Mp= 565 kN.m , Lp = 1.74 m, BF = 62.1
Ø Mn = 1.30 [ 565 -62.1( 4.5-1.74)] = 511.68 < 565 kN.m
Ø Mn= 511.68 kN.m > M ult max = 486 kN.m
Case 4 : The top flange is only laterally supported at ends
Lb= 9m.
Cb = 12.5x486 = 1.136
2.5 x 486 + 3 x 364.5 + 4 x 486 +3x364.56
Look for a lightest section in zone 2 in beam graphs, with Lb =9.0 m,
With a factored resistance not less than Mmax / Cb = 486 /1.136 = 427.7 kN.m
From graph choose, W 460 x 128,
From Tables Ø Mp = 686 kN.m , Ø Mr = 445 kN.m, Lp = 3.33 m , Lr = 10.8 m, BF = 32.3
Ø Mn = 1.136 [ 686 -32.3 (9-3.33)] = 571 kN.m < Ø Mp
Ø Mn = 571 kN.m. > M ult max = 486 kN.m
MA =364.5 MC = 364.5
MB= M max =486
Lb= 9.00
MB =364.5
MC = 455.6
M max =486
MA = 212.6
Lb= 4.50 Lb= 4.50
W 610 x 92 is Safe for Lb= 4.50 m
W 460 x 128 is Safe for Lb= 9.0 m
4. 4 Prepared by Prof. Shehab Mourad – Department of Civil Eng. - KSU
Check deflection due to live load W 460 x 128
From W-shape tables , Ix = 6.37x108
mm4
E = 2 x 105
MPa (N/mm2
)
L = 9000 mm
Service live load = w = 21 kN/m = 21x1000 N = 21 N /mm
1000 mm
For uniform load on simply supported beam, the max deflection = 5 w L 4
384 EI
r max = 5 x 21 x (9000)4
= 14.09 mm
384 x 2 x 105
x 6.37 x 108
r all = L = 9000 = 25 mm
360 360
r max < r all
W 460 x 128 satisfies the deflection limits as well as strength requirements for
unsupported compression flange = 9.0 m