23-Design of Column Base Plates (Steel Structural Design & Prof. Shehab Mourad)

1. 1 Prepared by Prof. Shehab Mourad – Department of Civil Eng. - KSU
Design of Column Base Plates
v Column bases frequently are designed to resist bending moments as
well as axial load.
v An axial load causes compression between a base plate and the
supporting footing ,while a moment increases the compression on
one side and decreases it on the other side .
A = B x H
I = B x H3
/ 12
y = H / 2
e = M / P
For no tensile stress H > 6 e
M
P
P/A
M. y / I
B
H
e

2. 2 Prepared by Prof. Shehab Mourad – Department of Civil Eng. - KSU
v for small moment the forces may be transferred to the footing
through flexure of the base plate.
v For higher moments ,stiffened or booted connections may be used
Design requirements:
1- Choose a base plate dimension ( H x B) so that the maximum
compression stress should not exceed the maximum bearing stress on
concrete
fc fb = 0.60 x 1.7 f 'c , where f 'c = compressive strength of concrete
2- Thickness of base plate t ³
Where; M pu = moment acting at base plate at center of compression flange
fb = 0.90
6 M p u
fb Fy

3. 3 Prepared by Prof. Shehab Mourad – Department of Civil Eng. - KSU
Example :
Design a moment resisting base plate to support a W 530 x 150 column with an
axial load of Pu = 1735 kN and bending moment Mu = 200 kN.m. Use A36 steel
with Fy = 250 MPa and concrete footing strength = 35 MPa.
Solution
For W530x150 , d = 543 mm , tw = 12.7mm , bf = 312 mm , tf =20.3 mm
Eccentricity , e = (Mu x1000) /Pu = (200x 1000) / 1735 = 115.27 mm
For no tensile stress on base plate, H ³ 6 e = 6 x 115.27 mm = 691.6 mm
Then choose H = 750 mm and B = 500 mm
Check stresses on concrete footing
f = - Pu / A ± Mu .y / I =
- 1735 x 103
/ ( 750 x 500) ± 200 x 106
x 375 / (500 x 7503
/12)
f = - 4.60 ± 4.30
f max = -8.90 MPa < fc Pu = 0.6 x 1.7 x 35 = 35.7 MPa O.K
f min = -0.30 MPa ( still compression) O.K
taking moments at center of compression flange per unit width (sec I-I)
M pu I-I = ( 7.60 x 113.65 2
/ 2 ) + (8.90-7.60) (113.65 2
) /3 = 54679 N.mm
t ³ = ( 6 x 54679/ 0.9x 250) 0.5
= 38.20 mm
Choose t = 40 mm
checking bending in transverse direction (sec II-II)
/2 = 125.2 mmx = ( 500 – 0.8 x 312)
Average stress f = (0.3+8.9)/2 = 4.6 MPa
Mpu II-II = (4.60) (125.2)2
/2 = 36053 N.mm < 54679 N.mm (O.K)
Use base plate 750mm x 500 mm x 40mm
6 M p u
fb Fy
B = 500
H = 750
d = 543
103.5103.5
113.65
-8.90
-0.30
-7.60
x
x
I
I
IIII