Example 1 Truss Theory - Pile Cap Design Verification

This example is taken from Reinforced Concrete Design to BS 8110 Simply Explained page 189, Example 16.4.

Pile/Pile Cap Parameters

Pile Diameter, d_p= 450mm
Column Loads:
P_service = 1750 kN
P_u = 2800 kN
Column Dimensions, a = 500mm, b = 500mm
Concrete Strength, f'_c= 35 N/mm2
Reinforcing Steel Strength, f_y= 460 N/mm2
Pile Embedment into Cap = 35mm
Concrete Clear Cover = 40mm
Pile Spacing = 3d_p = 1350mm
Pile Edge distance = 150 + d_p/2 = 375mm
Assume total depth = Pile Spacing / 2 + Pile Embedment + Concrete Cover = 1350 / 2 + 35 + 40 = 750mm
Average Bar Depth d = 650mm

For a 4 pile - pile cap:

A_{s} = 2 \frac{P_{u}}{24 l d} (3 l^{2} - b^{2}) \frac{1}{0.95 f_{y}}

A_{s} = 2 \frac{2, 800 {(10)}^{3}}{24 (1, 350) (650)} [3 {(1, 350)}^{2} - 500^{2}] \frac{1}{0.95 (450)} = 2, 345 {mm}^{2}

V_u = 2,800 kN

V_{c} = 0.8 \sqrt{35} (4) (500) (650) = 6, 152 kN > V_{u}

V_u = 2,800 / 2 = 1,400 kN

a_v = 675 - 250 - 225 + 460/5 = 292 mm

\frac{100 A_{s}}{b d} = \frac{100 (3, 245)}{2, 100 (650)} = 0.24

v_c = 0.40(35/25)^1/3 = 0.45 per BS8110-97 Table 3.8

Shear Enhancement = 2d/a_v = 1,300/290 = 4.48

V_c = 0.45(4.48)(2,100)(650) = 2,571 kN > V_u, OK

If this footing had more than 5 piles, the following procedure would have been followed:

Flexure

Bending moment at column face = 1,400(675 - 250)/1,000 = 595 kN-m

\frac{M}{b d^{2}} = \frac{595 {(10)}^{6}}{2, 100 {(650)}^{2}} = 0.67

\frac{100 A_{s}}{b d} = \frac{100 (2, 400)}{2, 100 (650)} = 0.176

Try A_s = 2,400 mm²

Note that the required reinforcement is much less than with the truss theory.

Shear

Shear capacity based on flexural reinforcement based on truss theory.

\frac{100 A_{s}}{b d} = \frac{100 (2, 400)}{2, 100 (650)} = 0.176

v_c = 0.35(35/25)^1/3 = 0.39 per BS8110-97 Table 3.8

V_c = 0.39(4.48)(2,100)(650) = 2,385 kN > V_u, OK

Note the shear capacity is less due to the smaller flexural reinforcement when compared to the truss method tension reinforcement.