Friday, January 11, 2013

Design Example: Reinforced Pad Base.

The axially loaded pad base in Design Example 2 is to be redesigned as a reinforced base, founded in the weathered sandstone. Assuming settlements have been judged to be satisfactory, the base will have an allowable bearing pressure, na = 550 kN/m2.


Since dead and imposed loads are approximately equal, a
combined partial load factor of γP = 1.5 will be used.

Area of base

Adopt a 3.0 m × 3.0 m square base, i.e. L = B = 3.0 m (see Fig. 11.25). Reactive design pressure on base for concrete design

Reinforced pad base design example.
Fig. 11.25 Reinforced pad base design example.

Depth of base
The base and its reinforcement must be capable of resisting bending, beam shear and punching shear. At first glance it is not always possible to judge which is critical. The process of selecting a suitable depth for the base is simplified by use of the charts for estimating effective depth. The effective depth will be checked for each case, assuming a typical reinforcement percentage
of between 0.25% and 0.50%. The results are shown in  Table 11.1.

 Table 11.1 Estimating effective depth for reinforced pad base design example
Estimating effective depth for reinforced pad base design example

This indicates that bending is critical, i.e. it requires the greatest effective depth, for low percentages of reinforcement.

For this particular example an average effective depth in both directions of d = 600 mm will be selected.

Overall depth of base is, h = 600 + 25 (bar diameter) 
                                        + 50 (cover)
                                        = 675 mm
From Fig. 11.25, the cantilever moment at face of base  plate is

The base should be checked for both beam shear and punching shear, since either may be critical. Grade C40  concrete has been specified.

Local shear at column face
The shear at the face of the column should be checked.

Allowable concrete shear stress, vc = 0.57 N/mm2

From BS 8110: Part 1:, the critical location for beam shear is at a distance 2d = 2 × 600 = 1200 mm from the face  of the load (i.e. from the edge of the base plate in this  example). The shear force acting across this failure plane is

Vbeam = (design pressure) × (area of base beyondcritical location)

Punching shear
The critical location for punch- ing shear for a square load is a square perimeter a distance 1.5d = 1.5 × 600 = 900 mm from the face of the load.

The length of one side of this perimeter is

Area of base outside of perimeter

Comparison with vbeam = 0.12 N/mm2 indicates that, in this instance, punching shear is more critical than beam shear.

This is normally the case with square pad foundations. If however a foundation size of say 2 m × 4 m had been chosen in this example, beam shear may well become critical.

Local bond
Although not covered by BS 8110, local bond can be a problem in foundation design, and should therefore be checked at sections with high shear stress. Local bond is given by

where ∑ us is the sum of the bar perimeters at the section being considered.
Punching shear, Vu = 2979 kN

The length of the punching shear is u = 8800 mm.
T25 bars @ 175 centres each way are proposed. The total number of bars crossing the shear perimeter is u/175 = 50.
The local bond stress is

where la is the lever arm which CP 110 approximates to the effective depth d.

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