1. The document provides examples of checking the strength of beams and columns. 2. In the first example, the beam section W 310 x 97 is checked to resist ultimate loads and is found to be safe. 3. In the second example, the safety of column section W 360 x 72 is checked for a given load of 250 kN when laterally supported at mid-height. It is found to be unsafe by about 8% and requires a larger section.

1. 1 Prepared by Prof. Shehab Mourad – Department of Civil Eng. - KSU
Example 1
Solution:
1- Beam effect (beam strength)
Lb = 9m
For W 310 x 97, f Mp = 358 kN.m, f Mr = 235 kN.m, Lp = 3.83 m, Lr = 13.6 m ,
BF = 12.6
Lr > Lb > Lp beam strength is located in zone 2
Obtain Cb = 1.14 for uniform load on simply supported beam that is laterally
supported at ends only.
f Mn = 1.14 [ 358 – 13.2 (9 – 3.83) ] = 330 kN.m < f Mp
f Mnx = 330 kN.m
2- Beam-Column Effect
The beam is braced at ends,
Maximum moment = Mnt = w L2
/8 = 10 . (9)2
/8 = 101.25 kN.m
B1 = Cm
1 - Pu /Pe1
Cm = 1.0 , for hinged-hinged beam with uniform transverse loads
Pe1 = p2
E I = (3.14)2
(2 x 105
) (222 x 106
) x 10-3
= 5410.00 kN
(KL)2
(9000)2
B1 = 1.0 = 1.15 ³ 1.0 B1 = 1.15
1 – 700/5410
Ultimate maximum moment at mid span = Mux
= B1 x Mnt = 1.15 x 101.25 = 116.44 kN.m
W u = 10 kN/mPu = 700 kN
9 m
Pu = 700 kN
Check the safety of the
shown beam to resist the
given ultimate loads, if the
beam section is W 310 x 97
(A36) and the compression
flange of the beam is only
laterally supported at ends.

2. 2 Prepared by Prof. Shehab Mourad – Department of Civil Eng. - KSU
3- Column Effect
For W 310 x 97,
bf =305 mm, tf = 15.4 mm, h = 241 mm, tw = 9.9 mm,
A= 12300 mm2
, ry = 77mm
Klx = Kly = 9000 mm KLy /ry is critical
ly = (KL/ry) Fy = 9000 / 77 250 = 1.32
p E 3.14 2 x 105
Check local buckling:
For flange, ( bf/2) / tf = (305/ 2) / 15.4 = 9.90 < 0.56 E = 15.80 o.k
Fy
For web, h/ tw = 241 / 9.9 = 24.13 < 1.49 E = 42.10 o.k
Fy
No local buckling in column section
ly = 1.31 < 1.50
Fcr = (0.658) (1.32)
x 250 = 120.6 MPa
f Pn = 0.85 (12300) ( 120.6) x 10-3
= 1260.8 kN
4- Check Strength Equation
Pu = 700 = 0.555 > 0.20
f Pn 1260.8
Pu + 8 ( Mux) = 700 + 8 116.44 = 0.87 < 1.0 o.k
f Pn 9 f Mnx 1260.8 9 330
Section W 310 x 97 is safe to resist the given ultimate loads
2

3. 3 Prepared by Prof. Shehab Mourad – Department of Civil Eng. - KSU
Example 2
Solution:
- Column buckling length (inplane)
Ic/ Ib = 1.75
G1 = 10 (for hinge)
G2 = I/5 = 1.14
1.75 I/10
Braced case
K = 3 x 10 x 1.14 + 1.4 (10 + 1.14) + 0.64 = 0.87
3 x 10 X 1.14 + 2 (10 +1.14) + 1.28
Unbraced case
K = 1.6 x 10 x 1.14 + 4 (10 + 1.14) + 7.5 = 1.94
10 + 1.14 + 7.5
10 m
5 m
W = 50 kN/m
H
W 530 x 66
W 360 x 72 W 360 x 72 3
2
1
Check the safety of column (3), if
Pu = 250 kN and is laterally supported
at mid height. The resulted bending
moment diagrams are given for both
cases, braced and unbraced. All steel is
A36
For W 360x72
A= 9110 mm2
, d= 350 mm, tw=8.6 mm
bf =204 mm, tf = 15.1 mm, h =280 mm
Ix = 201 x 106
mm4
, rx = 149 mm,
ry = 48.5 mm
For W 530 x 66
I x = 351 x 106
mm4
W = 50 kN/m
180 kN.m
180 kN.m
Mnt diagram
(braced case)
100 kN.m
100 kN.mH
Mlt diagram
(unbraced case)

4. 4 Prepared by Prof. Shehab Mourad – Department of Civil Eng. - KSU
1- Beam effect
Lb = 5.0 / 2 = 2.5 m
For W 360 x 72; f Mp = 288 kN.m, f Mr = 187 kN.m,
Lp = 2.41 m, Lr = 7.98 m , BF = 18.1
Lr > Lb > Lp beam strength is located at zone 2
Cb = 12.5 M = 1.25
2.5 M + 3 x 0.63 M + 4 x 0.75 M + 3 x 0.88 M
f Mn = 1.25 [ 288 – 18.1 (2.5 – 2.41)] = 357 kN.m > f Mp = 288 kN.m
f Mn = f Mp = 288 kN.m
2 – Beam – Column effect
(braced case)
B1 = Cm
1 - Pu /Pe1
where Cm = 0.6 – 0.4 ( 0 ) = 0.6
180
Pe1 = p2
E I = (3.14)2
(2 x 105
) (201 x 106
) x 10-3
= 20946 kN
(KLx)2
(0.87 x 5000)2
B1 = 0.6 = 0.61 < 1.0
1 – ( 250/20946)
B1 = 1.0
(unbraced case)
B2 = 1 > 1.0
1 - ∑Pu /∑Pe2
∑Pu = 50 x 10 = 500 kN
Pe2 = p2
E I = (3.14)2
(2 x 105
) (201 x 106
) x 10-3
= 4212.5 kN
(KLx)2
(1.94 x 5000)2
B2 = 1 = 1.06
1 – [500 / (2 x 4212.5)]
Mux = B1 Mnt + B2 Mlt = 1 x 180 + 1.06 x 100 = 286.3 kN.m
Lb = 2.5 m
Lb = 2.5 m
0.5 M
0.63 M
0.75 M
0.88 M
M

5. 5 Prepared by Prof. Shehab Mourad – Department of Civil Eng. - KSU
3- Column Effect (unbraced case)
KLx/rx = 1.94 x 5000 / 149 = 65, KLy /ry = 1 x 2500/ 48.5 = 51.5
KLx/rx controls,
lx = (KL/rx) Fy = 65 250 = 0.736 < 1.50
p E 3.14 2 x 105
Check local buckling:
For flange, ( bf/2) / tf = (204/ 2) / 15.1 = 6.75 < 0.56 E = 15.80 o.k
Fy
For web, h/ tw = 280 / 8.6 = 32.6 < 1.49 E = 42.10 o.k
Fy
There is no local buckling
Fcr = (0.658) (0.736)
x 250 = 199.3 MPa
f Pn
=
0.85 x 199.3 x 9110 x 10-3
= 1543 kN
5- Check beam –column strength
Pu = 250 / 1543 = 0.16 < 0.20
f Pn
Pu + ( Mux) = 250 + 286.3 = 1.08 > 1.0 NOT o.k
2 f Pn f Mnx 2 x 1543 288
W 360 x 72 is unsafe to be used a column by about 8 %, and
need to changed to a larger section
2