IC-570 – Mecânica dos Solos – Prova 1

Questão 1

Solo natural: γ = 16 kN/m³; w = 10%

Solo compactado: $$\gamma_{dmax} = 18 kN/m^{3}$$; $$w_{ot} = 17\,\%$$; CR = 100%

 

a) Solo natural: $$\gamma_{d} = \frac{\gamma}{1+w} \longrightarrow \gamma_{d} = \frac{16}{1+0,1} \longrightarrow \gamma_{d} = 14,55\, kN/m^{3}$$

$$\gamma_{d} = \frac{P_{s}}{V_{c}} \longrightarrow 14,55 = \frac{P_{s}}{V_{c}} \longrightarrow P_{s} = 14,55V_{c}$$

Solo compactado: $$\gamma_{dmax} = \frac{P_{s}}{V_{a}} \longrightarrow 18 = \frac{P_{s}}{V_{a}} \longrightarrow P_{s} = 18V_{a}$$

$$14,55V_{c} = 18V_{a}$$

$$\frac{7}{100} = \frac{x}{y} \longrightarrow y = \frac{100x}{7}$$

$$V_{a} = \frac{50xy}{2} \longrightarrow V_{a} = 100\cdot x^{2}\cdot 25$$

$$V_{c} = \frac{(7-x)(100-y)50}{2} \longrightarrow V_{c} = 25\frac{4900 – 1400x – 100x^{2}}{7}$$

$$25\frac{14,55}{7}(4900 – 1400x – 100x^{2}) = 25\cdot 18\cdot 100x^{2} \longrightarrow x = 1,64\, m$$

$$V_{c} = 25\frac{4900 – 1400\cdot 1,64 – 100\cdot 1,64^{2}}{7} \longrightarrow V_{c} = 8339,43\, m^{3}$$

 

b) $$P_{s} = 14,55 V_{c} \longrightarrow P_{s} = 14,55\cdot 8339,43 \longrightarrow P_{s} = 121338,71\, kN$$

Solo natural: $$w = \frac{P_{wc}}{P_{s}} \longrightarrow 0,1 = \frac{P_{wc}}{121338,71} \longrightarrow P_{wc} = 12133,87\, kN$$

Solo compactado: $$w_{ot} = \frac{P_{wa}}{P_{s}} \longrightarrow 0,17 = \frac{P_{wa}}{121338,71} \longrightarrow P_{wa} = 20627,58\, kN$$

$$\Delta V = \frac{P_{wa} – P_{wc}}{\gamma_{w}} \longrightarrow \Delta V = \frac{20627,58 – 12133,87}{10} \longrightarrow \Delta V = 849,37\, m^{3}$$

 

c) x = 1,64 m, como calculado no item (a).

Questão 2

300x250-728_90-livros2

Curva de compactação para certo nível de energia

a) Curva de compactação para determinada umidade

 

b) Curva de saturação para o item (a)

c) Curva de compactação para determinado peso específico seco

Questão 3

BF4

Ponto A: γ = 19 kN/m³

Para A1:

$$\sigma_{z1} = \frac{\bar{q}}{b\pi}\frac{b(b-x)z + x((b-x)^{2}+z^{2})}{(b-x)^{2} + z^{2}}(\arctan(\frac{b-x}{z}) + \arctan(\frac{x}{z}))$$

$$\sigma_{z1} = \frac{142,5}{7,5\pi}\frac{7,5(7,5-7,5)7,5 + 7,5((7,5-7,5)^{2} +7,5^{2})}{(7,5-7,5)^{2} + 7,5^{2}}(\arctan(\frac{7,5-7,5}{7,5}) + \arctan(\frac{7,5}{7,5}))$$

$$\sigma_{z1} = 35,63\, kN/m^{2}$$

Para A2:

$$\alpha = \arctan(\frac{b}{z}) \longrightarrow \alpha = \arctan(\frac{10}{7,5})$$

β = 0

$$\sigma_{z2} = \frac{q}{\pi}(\alpha + \sin\alpha\cos(\alpha + 2\beta))$$

$$\sigma_{z2} = \frac{142,5}{\pi}(\arctan(\frac{10}{7,5}) + \sin(\arctan(\frac{10}{7,5}))\cos(\arctan(\frac{10}{7,5}) + 2\cdot 0))$$

$$\sigma_{z2} = 63,83\, kN/m^{2}$$

Para A3:

$$\sigma_{z3} = \frac{142,5}{7,5\pi}\frac{7,5(7,5-17,5)7,5 + 17,5((7,5-17,5)^{2} +7,5^{2})}{(7,5-17,5)^{2} + 7,5^{2}}(\arctan(\frac{7,5-17,5}{7,5}) + \arctan(\frac{17,5}{7,5}))$$

$$\sigma_{z3} = 20,06\, kN/m^{2}$$

$$\sigma_{zT} = \sigma_{z1} + \sigma_{z2} + \sigma_{z3} \longrightarrow \sigma_{zT} = 35,63 + 63,83 + 20,06 \longrightarrow \sigma_{zT} = 119,52\, kN/m^{2}$$

Ponto B: γ = 19 kN/m³

Para A1/A3:

$$\sigma_{z1} = \sigma_{z3} = \frac{142,5}{7,5\pi}\frac{7,5(7,5-12,5)7,5 + 12,5((7,5-12,5)^{2}+7,5^{2})}{(7,5-12,5)^{2} + 7,5^{2}}(\arctan(\frac{7,5-12,5}{7,5}) + \arctan(\frac{12,5}{7,5}))$$

$$\sigma_{z1} = \sigma_{z3} = 24,18\, kN/m^{2}$$

Para A2:

$$\alpha = 2\arctan(\frac{b}{2z}) \longrightarrow \alpha = 2\arctan(\frac{10}{2\cdot 7,5}) \longrightarrow \alpha = 2\arctan(\frac{10}{15})$$

$$\sigma_{z2} = \frac{q}{\pi}(\alpha + \sin\alpha)$$

$$\sigma_{z2} = \frac{142,5}{\pi}(2\arctan(\frac{10}{15}) + \sin(2\arctan(\frac{10}{15}))$$

$$\sigma_{z2} = 95,21\, kN/m^{2}$$

$$sigma_{zT} = \sigma_{z1} + \sigma_{z2} + \sigma_{z3} \longrightarrow \sigma_{zT} = 2\cdot 24,18 + 95,21 \longrightarrow \sigma_{zT} = 143,57\, kN/m^{2}$$

Questão 4

Solo 1: $$e = \frac{\gamma_{s}(1+w)}{\gamma} – 1 \longrightarrow 0,72 = \frac{27(1+0,16)}{\gamma} – 1 \longrightarrow \gamma = 18,21\, kN/m^{3}$$

Solo 1 (saturado): $$S = \frac{w\gamma_{s}}{e\gamma_{w}} \longrightarrow 1 = \frac{w\cdot 27}{0,72\cdot 10} \longrightarrow w = 0,27$$

$$\gamma_{d} = \frac{\gamma}{1+w} \longrightarrow \gamma_{d} = \frac{18,21}{1+0,27} \longrightarrow \gamma_{d} = 14,34\, kN/m^{3}$$

$$n = 1 – \frac{\gamma}{\gamma_{s}(1+w)} \longrightarrow n = 1 – \frac{18,21}{27(1+0,27)} \longrightarrow n = 0,47$$

$$\gamma_{sat} = \gamma_{d} + n\gamma_{w} \longrightarrow \gamma_{sat} = 14,34 + 0,47\cdot 10 \longrightarrow \gamma_{sat} = 19,04\, kN/m^{3}$$

Solo 2: $$\gamma_{sat} = \gamma_{d} + n\gamma_{w} \longrightarrow \gamma_{sat} = 15 + 0,17\cdot 10 \longrightarrow \gamma_{sat} = 16,7\, kN/m^{3}$$

 

Tensões totais verticais

$$\sigma_{zA} = 80\, kN/m^{2}$$

$$\sigma_{zB} = 80 + 18,21\cdot 1,5 + 19,04\cdot 1,5 \longrightarrow \sigma_{zB} = 135,88\, kN/m^{2}$$

$$\sigma_{zC} = 80 + 18,21\cdot 1,5 + 19,04\cdot 1,5 + 16,7\cdot 4,5 \longrightarrow \sigma_{zC} = 211,03\, kN/m^{2}$$

Tensões efetivas verticais

$$\sigma_{zA}’ = 80\, kN/m^{2}$$

$$\sigma_{zB}’ = 80 + 18,21\cdot 1,5 + (19,04 – 10)\cdot 1,5 \longrightarrow \sigma_{zB}’ = 120,88\, kN/m^{2}$$

$$\sigma_{zC}’ = 80 + 18,21\cdot 1,5 +(19,04 – 10)\cdot 1,5 + (16,7 – 10)\cdot 4,5 \longrightarrow \sigma_{zC}’ = 151,03\, kN/m^{2}$$

Tensões efetivas horizontais

$$\sigma_{xA}’ = 0,5\cdot 80 \longrightarrow \sigma_{xA}’ = 40\, kN/m^{2}$$

$$\sigma_{xB}’ = 0,5\cdot 120,88 \longrightarrow \sigma_{xB}’ = 60,44\, kN/m^{2}$$

$$\sigma_{xC}’ = 0,4\cdot 151,03 \longrightarrow \sigma_{xC}’ = 60,41\, kN/m^{2}$$

Tensões totais horizontais

$$\sigma_{xA} = 0,5\cdot 80 \longrightarrow \sigma_{xA} = 40\, kN/m^{2}$$

$$\sigma_{xB} = 0,5\cdot 120,88 + 10\cdot 1,5 \longrightarrow \sigma_{xB} = 75,44\, kN/m^{2}$$

$$\sigma_{xC} = 0,4\cdot 151,03 + 10\cdot 6 \longrightarrow \sigma_{xC} = 120,41\, kN/m^{2}$$

Banner 300x250 Telefonia
buenos-aires_300x250.jpg