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

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This document discusses the design of column base plates to resist both axial loads and bending moments. It provides equations to calculate stresses on the base plate and footing. It then gives an example of designing a base plate for a column supporting an axial load of 1735 kN and bending moment of 200 kN.m. The design process involves calculating eccentricity, base plate dimensions, stresses on the footing, required plate thickness, and checking bending in two directions. The example concludes by specifying a base plate of dimensions 750mm x 500mm x 40mm that satisfies all design requirements.

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 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 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

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Lec03 Earthquake Force Using Response Specturum Method (1) (Earthquake Engine...

Hossam Shafiq II

Lec02 Earthquake Damage to Concrete Structure (Earthquake Engineering هندسة ا...

Hossam Shafiq II

Lec01 Design of RC Structures under lateral load (Earthquake Engineering هندس...

Hossam Shafiq II

Lec13 Continuous Beams and One Way Slabs(3) Footings (Reinforced Concrete Des...

Hossam Shafiq II

Lec12 Continuous Beams and One Way Slabs(2) Columns (Reinforced Concrete Desi...

Hossam Shafiq II

Lec11 Continuous Beams and One Way Slabs(1) (Reinforced Concrete Design I & P...

Hossam Shafiq II

The document discusses reinforced concrete continuity and analysis methods for continuous beams and one-way slabs. It describes how steel reinforcement must extend through members to provide structural continuity. The ACI/SBC coefficient method of analysis is summarized, which uses coefficient tables to determine maximum shear forces and bending moments for continuous beams and one-way slabs under various loading conditions in a simplified manner compared to elastic analysis. Requirements for applying the coefficient method include having multiple spans with ratios less than 1.2, prismatic member sections, and live loads less than 3 times dead loads.

Lec10 Bond and Development Length (Reinforced Concrete Design I & Prof. Abdel...

Hossam Shafiq II

This document discusses bond and development length in reinforced concrete. It defines bond as the adhesion between concrete and steel reinforcement, which is necessary to develop their composite action. Bond is achieved through chemical adhesion, friction from deformed bar ribs, and bearing. Development length refers to the minimum embedment length of a reinforcement bar needed to develop its yield strength by bonding to the surrounding concrete. The development length depends on factors like bar size, concrete strength, bar location, and transverse reinforcement. It also provides equations from design codes to calculate the development length for tension bars, compression bars, bundled bars, and welded wire fabric. Hooked bars can be used when full development length is not available, and the document discusses requirements for standard hook geome

Lec09 Shear in RC Beams (Reinforced Concrete Design I & Prof. Abdelhamid Charif)

Hossam Shafiq II

This document discusses shear in reinforced concrete beams. It covers shear stress and failure modes, shear strength provided by concrete and steel stirrups, design according to code provisions, and critical shear sections. Key points include: transverse loads induce shear stress perpendicular to bending stresses; shear failure is brittle and must be designed to exceed flexural strength; nominal shear strength comes from concrete and steel stirrups according to code equations; design requires checking section adequacy and providing minimum steel area and maximum stirrup spacing. Critical shear sections for design are located a distance d from supports.

Lec07 Analysis and Design of Doubly Reinforced Beam (Reinforced Concrete Desi...

Hossam Shafiq II

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