Question:
At a site, Static Cone Penetration Test was carried out. The measured point (tip) resistance \( q_c \) was 1000 kPa at a certain depth. The friction ratio (\( f_r \)) was estimated as 1 % at the same depth.
The value of sleeve (side) friction (in kPa) at that depth was \(\underline{\hspace{2cm}}\) . (in integer)
At a site, Static Cone Penetration Test was carried out. The measured point (tip) resistance \( q_c \) was 1000 kPa at a certain depth. The friction ratio (\( f_r \)) was estimated as 1 % at the same depth.
The value of sleeve (side) friction (in kPa) at that depth was \(\underline{\hspace{2cm}}\) . (in integer)
The value of sleeve (side) friction (in kPa) at that depth was \(\underline{\hspace{2cm}}\) . (in integer)
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The friction ratio in static cone penetration tests is used to estimate the sleeve friction based on the tip resistance.
Updated On: Jan 11, 2026
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Correct Answer: 10
Solution and Explanation
The friction ratio \( f_r \) is defined as the ratio of sleeve friction \( f_s \) to tip resistance \( q_c \):
\[ f_r = \frac{f_s}{q_c} \times 100. \] Given \( f_r = 1% \) and \( q_c = 1000 \, \text{kPa} \), we can solve for \( f_s \):
\[ 1 = \frac{f_s}{1000} \times 100. \] Thus, \( f_s = 10 \, \text{kPa} \).
\[ f_r = \frac{f_s}{q_c} \times 100. \] Given \( f_r = 1% \) and \( q_c = 1000 \, \text{kPa} \), we can solve for \( f_s \):
\[ 1 = \frac{f_s}{1000} \times 100. \] Thus, \( f_s = 10 \, \text{kPa} \).
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