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Keith_Richards [23]
4 years ago
5

A 650-kN column load is supported on a 1.5 m square, 0.5 m deep spread footing. The soil below is a well-graded, normally consol

idated sand with 훾 = 18.0 kN/m3 and the following SPT N60 values:
Engineering
1 answer:
insens350 [35]4 years ago
6 0

<u>Explanation:</u>

Determine the weight of footing

W_{f}=\gamma(L)(B)(D)

Where W_{f} is the weight of footing, γ is the unit weight of concrete,  L is the length of footing is the width of footing, and D is the depth of footing

Substitute 2 m \text { for } L, 1.5 m \text { for } B, 0.5 m \text { for } D \text { and } 23.6 kN / m ^{3} for γ in the equation

\begin{aligned}W_{f} &=\left(23.6 kN / m ^{3}\right)(2 m )(1.5 m )(0.5 m ) \\&=35.4 kN\end{aligned}

Therefore, the weight of the footing is 35.4 kN

Determine the initial vertical effective stress.

\sigma_{z p}^{\prime}=\gamma(D+B)-u

Here,   \sigma_{z^{p}}^{\prime} is initial vertical stress at a depth below ground surface  γ is the unit weight of soil, D is depth and u is pore water pressure.

Substitute 18 kN / m ^{3} \text { for } \gamma, 1.5 m \text { for } B, 0.5 m \text { for } D \text { and } 0 for u in the equation

\begin{aligned}\sigma_{z p}^{\prime} &=\left(18 kN / m ^{3}\right)(1.5+0.5) m -0 \\&=36 kPa\end{aligned}

Therefore, the initial vertical stress is 36 kPa

Determine the vertical effective stress.

\sigma_{z D}^{\prime}=\gamma D

Here,   \sigma_{z^{p}}^{\prime} is initial vertical stress at a depth below ground surface  γ is the unit weight of soil, D is depth and u is pore water pressure.

Substitute \(18 kN / m ^{3}\) for \(\gamma, 0.5 m\) for \(D\) and 0 for \(u\) in the equation.

\begin{aligned}\sigma_{z b}^{\prime} &=\left(18 kN / m ^{3}\right)(0.5 m )-0 \\&=9 kPa\end{aligned}

Therefore, the vertical stress at a depth below the ground surface is

9 kPa

Determine the influence factor at the midpoint of soil layer,

I_{e p}=0.5+0.1 \sqrt{\frac{q-\sigma_{s 0}^{\prime}}{\sigma_{z p}^{\prime}}}

Here I_{e p} is the influence factor at the midpoint of soil layer  \sigma_{z^{p}}^{\prime} is initial vertical stress, \sigma_{z^{p}}^{\prime} is vertical effective stress, and Q is bearing pressure

Substitute 36 kPa for \(\sigma_{z p}^{\prime}, 228.47\) kPa for \(q,\) and 9 kPa for \(\sigma_{z D}^{\prime}\) in the equation\begin{aligned}I_{\epsilon P} &=0.5+0.1 \sqrt{\frac{228.47 kPa -9 kPa }{36 kPa }} \\&=0.75\end{aligned}

Therefore the influence factor at the midpoint of the soil layer is 0.693

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Dividing both the numerator and denominator by cosθ, we have:

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3 years ago
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Answer:

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                     J_cd = pi/2 (c_cd1^4 - c_cd2^4)= pi/2(1.5^4-1.25^4) = 4.1172 in^4

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                    T_d = 4.1172*5.6*10^6*0.006545 / 8

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                    T_max = min ( 12.63 , 19.21 , 18.86 )

                    T_max = 12.63 kip.in

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                    Ф_a = Ф_a/d = Ф_a/b + Ф_c/d:

                    T_max* ( L_ab / J_ab*G_ab + L_cd / J_cd*G_cd)

                    12.63*(12/0.92*11.2*10^6 + 8/4.117*5.6*10^6)

                    0.01908 rads = 1.093°

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