6-footing Design

DESIGN OF FOOTING Design Constants fc’

- 20.7 MPa

fy

- 276 MPa

β

- 0.85

DESIGN OF FOOTING:

Loads to be carried:

qA

= 250 kPa

Dead Load

γ Soil

= 15.6 kPa

Beam Reactions

= 409.073 kN

Effective Bearing Capacity:

Column weight

= 23.54(0.5)(0.5)(12)

qE

PU

= 250 – 15.6(1.4) – 23.54(0.4)

= 409.073 + 70.62 qE

= 479.693 kN

(NSCP 2015 Table 305-1) – Minimum Requirements for Foundations

= qA - γ Soil (1.5) – 23.54(0.4)

= 217.184 kPa

Area Required: Effective Bearing Capacity (qE) 479.693 217.184

A

=

A

= 2.209 m2

Assuming Square Footing: L2

= √ 2.209

Assumptions: Protective Covering

= 75 mm→100mm

Depth of Footing (from top of footing to N.G.L.) = 1500 mm

L

= 400 – 100 = 300 mm

Allowable Foundation Pressure = 100 kPa (Sandy Gravel and/or Gravel at depth of 300 mm) an increase of 20% is allowed for additional 300 mm depth (NSCP 2015, Table 304-1) qA

= Allowable Foundation Pressure = 75 + 0.2(75) (5) +50 (additional 300mm)

= 1.486

→ 1.50 m

Trial Area A

Thickness of Footing (t) = 400 mm Effective Depth (d)

=A

= L2 = 1.52

A

= 2.25 m2

qU

= Ultimate Soil Pressure =

PU A

=

479.693 2.25

=

P A

qU

= 213.197 kPa

Beam Shear

Punching Shear

ν allow = 0.75(Vc) = 0.75(0.17 λ √ fc ' bwd )

ν p =ν allow

= 0.75(0.17√ 20.7)(1500) (d)

where:

ν allow = 870.135d

ν allow = 0.75( Vc) = 0.75(0.17 λ √ fc ' bwd )

VU

= Beam Shear Force = qU AC =

VU

213.197(1500)(550 – d ) 1000

= 0.75(

1 √ 20.7) 3

ν allow = 1.137 MPa

= 319.7955 (550 – d) νp =

νu =ν allow 319.7955 (550 – d) = 870.135d

d = 147.813 mm (minimum depth required for beam shear)

Vp Punching Shear Area

where VP

= qU A C = 213.197 ¿ ¿

VP

=

213.197(15002−(400+ d)2 ) 1000

where: APS

= Punching Shear Area

APS

= 4(400 + d) (d)

APS

= 4d (400 + d)

Reinforcement Bars

νu =ν allow 213.197(15002−(400+ d)2 ) = 1.137 (4d (400 + d)) 1000

d = 161.515 mm minimum depth required for punching shear.

Since (d punch = 161.515 mm) > (d beam shear = 147.813 mm), d punch is the governing d.

MU

= Ultimate Moment = qU (AC) (0.275) = 213.197(0.55x1.500) (0.275)

Assume 16 mm ϕ to be used: Thickness of footing = 400 mm Protective Covering

MU

= 48.369 kN – m

= 75 + 16 + 8 = 99 mm → 100 mm

Actual d

= 400 – 100

Actual d

= 300 mm

Solving for RN: MU

48.369(10002) = 0.9(1500) (3002) RN RN

∴ Since (actual d = 300 mm) > (governing d = 206.003 mm), it is safe.

= ∅ b d2 RN

= 0.398

Solving for ω: ω

=

√ √

2.36 R N fc ' 1.18

1− 1−

1− 1−

ω

=

ω

= 0.0195

2.36(0. 398) 20.7 1.18

Solving for ρ: ω

=

0.0195 = ρ

ρ( fy) fc ' ρ(276) 20.7

= 0.0015

Since (ρ = 0.0015) < ( ρmin = 0.00507), use ρ = ρmin . Solving for the Area of the Steel As bd

ρ

=

AS

= 0.00507(1500) (300)

AS

= 2281.5 mm2

Using 16 mm ϕ reinforcement bars: n

=

AS = 11.34 A 12

→ 12 bars

∴Therefore use 12 – 16 mm ϕ bars both ways.

DET