Compute Qu and Qa for the pier given in Ex. 17.5 by the following methods. 1 . Use the SPT value [Eq. ( 1 5 .48) j for bored piles 2. Use the Tomlinson method of estimating Qb and Table 15.2 for estimating Qf. Compare the results of the various methods.

Question

Compute [latex]Q_{u} ext { and } Q_{a}[/latex] for the pier given in Ex. 17.5 by the following methods.

1 . Use the SPT value [Eq. ( 1 5 .48) j for bored piles

[latex]Q_{u}=133 N_{c o r} A_{b}+0.67 \vec{N}_{c o r} A_{s}[/latex] (15.48)

2. Use the Tomlinson method of estimating [latex]Q_{b}[/latex] and Table 15.2 for estimating [latex]Q_{f}[/latex]. Compare the results of the various methods.

Table 15.2 Values of [latex]\overline{ K }_{s} ext { and } \delta[/latex](Broms, 1966)
		Values of [latex]\bar{K}_{s}[/latex]
Pile material	[latex]\delta[/latex]	Low [latex]D_{r}[/latex]	High [latex]D_{r}[/latex]
Steel	20°	0.5	1
Concrete	[latex]3 / 4 \phi[/latex]	1	2
Wood	[latex]2 / 3 \phi[/latex]	1.5	4

Accepted Answer

Use of the SPT value [Mey erhof Eq. (15.48)]

[latex]\begin{aligned}&q_{b}=133 N_{c o r}=133 imes 30=3,990 kN / m ^{2} \&Q_{b}=\frac{3.14 imes(1.5)^{2}}{4} imes 3,990=7,047 kN \&f_{s}=0.67 N_{c o r}=0.67 imes 30=20 kN / m ^{2} \&Q_{f}=3.14 imes 1.5 imes 25 imes 20=2,355 kN \&Q_{u}=7,047+2,355=9,402 kN \&Q_{a}=\frac{9,402}{2.5}=3,760 kN\end{aligned}[/latex]

Tomlinson Method for [latex]Q_{b}[/latex]

For a driven pile

From Fig. 15.9 [latex]N_{q}=65 ext { for } \phi=36^{\circ} ext { and } \frac{L}{d}=\frac{25}{1.5} \approx 17[/latex]

Hence [latex]q_{b}=q_{o}^{\prime} N_{q}=437.5 imes 65=28,438 kN / m ^{2}[/latex]

For bored pile

[latex]q_{b}=\frac{1}{3} q_{b} ext { (driven pile) }=\frac{1}{3} imes 28,438=9,479 kN / m ^{2}[/latex]

[latex]Q_{b}=A_{b} q_{b}=1.766 imes 9,479=16.740 kN[/latex]

[latex]Q_{f}[/latex] from Table 15.2

For [latex]\phi=36^{\circ}, \delta=0.75 imes 36=27, ext { and } \bar{K}_{s}=1.5[/latex] (for medium dense sand).

[latex]f_{s}=\bar{q}_{o}^{\prime} \bar{K}_{s} an \delta=\frac{437.5}{2} imes 1.5 an 27^{\circ}=167 kN / m ^{2}[/latex]

As per Tomlinson (1986) [latex]f_{s}[/latex] is limited to 110 [latex]kN / m ^{2} ext {. Use } f_{s}=110 kN / m ^{2}[/latex].

Therefore [latex]Q_{f}=3.14 imes 1.5 imes 25 imes 110=12,953 kN[/latex]

[latex]Q_{u}=Q_{b}+Q_{f}=16,740+12,953=29,693 kN[/latex]

[latex]Q_{a}=\frac{29,693}{2.5}=11,877 kN[/latex]

Comparison of estimated results [latex]\left(F_{s}=2.5 ight)[/latex]
Example No	Name of method	[latex]Q_{b}( kN )[/latex]	[latex]Q_{f}( kN )[/latex]	[latex]Q_{u}( kN )[/latex]	[latex]Q_{a}( kN )[/latex]
17.5	Vesic	19.429	12.953	32.382	12.953
17.5	O'Neill and Reese, for [latex]Q_{b}[/latex] and Vesic for [latex]Q_{f}[/latex]	3.046	12.953	15.999	6.4
17.6	MeyerhofEq. (15.49)	7.047	2.355	9.402	3.76
17.6	Tomlinson for [latex]Q_{b}[/latex] (Fig. 15.9) Table 15.2 for [latex]Q_{f}[/latex]	16.74	12.953	29.693	11.877

Which method to use

The variation in the values of [latex]Q_{b} ext { and } Q_{f}[/latex], are very large between the methods. Since the soils encountered in the field are generally heterogeneous in character no theory holds well for all the soil conditions. Designers have to be practical and pragmatic in the selection of any one or combination of the theoretical approaches discussed earlier.

Question 17.6: Compute Qu and Qa for the pier given in Ex. 17.5 by the foll...

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