Solve the problem in Ex. 17.11 by the direct method.

Given: $E I=52.6 \times 10^{4} kN - m ^{2}, d=80 cm , \gamma=17.5 kN / m ^{3}, e=5 m , L=9 m , c=60 kN / m ^{2}$ and $P_{t}=80$ kN.

Question

Solve the problem in Ex. 17.11 by the direct method.

Given: [latex]E I=52.6 imes 10^{4} kN - m ^{2}, d=80 cm , \gamma=17.5 kN / m ^{3}, e=5 m , L=9 m , c=60 kN / m ^{2}[/latex] and [latex]P_{t}=80[/latex] kN.

Accepted Answer

Groundline deflection

From Eq. (16.30 ) for piers in clay

[latex]n_{h}=\frac{125 c^{1.5} \sqrt{E I \gamma d} /(1+e / d)^{1.5}}{P_{e}^{1.5}}[/latex] (16.30)

[latex]n_{h}=\frac{125 c^{1.5} \sqrt{E I \gamma d}}{1+\frac{e}{d}^{1.5} imes P_{e}^{1.5}}[/latex]

Substituting and simplifying

[latex]n_{h}=\frac{125 imes 60^{1.5} \sqrt{52.6 imes 10^{4} imes 17.5 imes 0.8}}{1+\frac{5}{0.8}^{1.5} imes P_{e}^{1.5}}=\frac{808 imes 10^{4}}{P_{e}^{1.5}} kN / m ^{3}[/latex] (a)

Step 1

Assume [latex]P_{e}=P_{t}=80 kN[/latex],

From Eq. (a), [latex]n_{h}=11,285 kN / m ^{3}[/latex] and

[latex]T=\frac{E I}{n_{h}}^{0.2}=\frac{52.6 imes 10^{4}}{11,285}^{0.2}=2.16 m[/latex]

Step 2

From Eq. (16.22) [latex]P_{e}=P_{t} imes 1+0.67 \frac{e}{T}=801+0.67 imes \frac{5}{2.16}=204 kN[/latex]

[latex]P_{e}=P_{t}\left(1+0.67 \frac{e}{T} ight)[/latex] (16.22)

FromEq. (a), [latex]n_{h}=2772 kN / m ^{3} ext {, hence } T=2.86 m[/latex]

Step 3

Continue the above process till convergence is reached. The final values are

[latex]P_{e}=177 kN , n_{h}=3410 kN / m ^{3} ext { and } T=2.74 m[/latex]

For [latex]P_{e}=190 kN ext {, we have } n_{h}=8,309 kN / m ^{3} ext { and } T=2.29 m[/latex]

Step 4

FromEq. (17.46)

[latex]y_{t}( ext { combined })=2.43 \frac{P_{t} T^{3}}{E I}+1.62 \frac{M_{t} T^{2}}{E I}[/latex] (17.46)

[latex]y_{t}=\frac{2.43 imes 177 imes(2.74)^{3}}{52.6 imes 10^{4}} imes 1000=16.8 mm[/latex]

By Duncan et al, method [latex]y_{t}=9.6[/latex] mm

Maximum moment from Eq. (16.12)

[latex]V=\left[P_{t} ight] A_{v}+\left[\frac{M_{i}}{T} ight] B_{v}[/latex] (6.12)

[latex]M=\left[P_{t} T ight] A_{m}+\left[M_{t} ight] B_{m}=[80 imes 2.74] A_{m}+[400] B_{m}=219.2 A_{m}+400 B_{m}[/latex]

Depth x/T = Z	[latex]A_{m}[/latex]	[latex]B_{m}[/latex]	[latex]M_{1}[/latex]	[latex]M_{1}[/latex]	M (kN-m)
0	0	1	0	400	400
0.4	0.379	0.99	83	396	479
0.5	0.46	0.98	101	392	493
0.6	0.53	0.96	116	384	500
0.7	0.6	0.94	132	376	508 (max)
0.8	0.65	0.91	142	364	506
The maximum bending moment occurs at x/T=0.1 or [latex]x=0.7 imes 2.74=1.91[/latex] m (6.26 ft). As per Duncan et al., method M(max) = 470.5 kN-m. This occurs at a depth of 1.3 m.

Question 17.12: Solve the problem in Ex. 17.11 by the direct method. Given: ...

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