Question:
A stable drop structure is designed with a base length of 5.7 m. The sum of vertical forces excluding uplift is 92.5 kN, sum of uplift forces is 72.5 kN, and sum of moments of all the horizontal and vertical forces about an axis passing through mid-section of base length is 60 kN.m. The eccentricity denoting the longitudinal distance between the centroid of base area and the point of application of resultant vertical load is _________ m. (Rounded off to 2 decimal places)
A stable drop structure is designed with a base length of 5.7 m. The sum of vertical forces excluding uplift is 92.5 kN, sum of uplift forces is 72.5 kN, and sum of moments of all the horizontal and vertical forces about an axis passing through mid-section of base length is 60 kN.m. The eccentricity denoting the longitudinal distance between the centroid of base area and the point of application of resultant vertical load is _________ m. (Rounded off to 2 decimal places)
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The eccentricity is calculated by dividing the sum of moments by the vertical force, and this helps determine the location of the resultant vertical load.
Updated On: Apr 14, 2025
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Solution and Explanation
The eccentricity (\(e\)) can be calculated using the formula:
\[
e = \frac{M}{F_v},
\]
Where:
\(M\) is the sum of the moments of all the horizontal and vertical forces about an axis passing through the mid-section of the base (60 kN.m),
\(F_v\) is the sum of the vertical forces excluding uplift (92.5 kN). Substitute the values: \[ e = \frac{60}{92.5} \approx 0.649 \, {m}. \] Thus, the eccentricity is approximately 1.18 m.
\(M\) is the sum of the moments of all the horizontal and vertical forces about an axis passing through the mid-section of the base (60 kN.m),
\(F_v\) is the sum of the vertical forces excluding uplift (92.5 kN). Substitute the values: \[ e = \frac{60}{92.5} \approx 0.649 \, {m}. \] Thus, the eccentricity is approximately 1.18 m.
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