Question:
A plug dam is to be constructed in a mine gallery of size 3.5 m \(\times\) 2.5 m. The interfacial shear strength between the surrounding rock and plug dam is 1.0 MPa. The minimum thickness of the dam, in m, to withstand a water pressure of 10.0 MPa, is . (rounded off to two decimal places)
A plug dam is to be constructed in a mine gallery of size 3.5 m \(\times\) 2.5 m. The interfacial shear strength between the surrounding rock and plug dam is 1.0 MPa. The minimum thickness of the dam, in m, to withstand a water pressure of 10.0 MPa, is . (rounded off to two decimal places)
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Balance the total water force on the dam face against the shear resistance developed around its perimeter.
Updated On: Jul 27, 2026
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Correct Answer: 7.29
Solution and Explanation
Step 1: Find the total force the water pressure pushes on the dam face.
The plug dam seals a gallery that is 3.5 m wide and 2.5 m high, so the face area the water pushes against is
\[ A = 3.5 \times 2.5 = 8.75 \text{ m}^2 \]
Water pressure acts uniformly over this face, so the total pushing force is pressure times area. Since 1 MPa equals 1 MN per square meter, this comes out directly in MN.
\[ F = P \times A = 10.0 \times 8.75 = 87.5 \text{ MN} \]
Step 2: Find how the dam resists this force.
The dam does not resist the water by its own strength alone, it resists through shear that develops all around its edge, where the dam's outer surface grips the surrounding rock. That gripping surface runs around the full perimeter of the gallery and extends back into the rock over the dam's thickness \(t\).
Perimeter of the gallery:
\[ p = 2(3.5+2.5) = 12 \text{ m} \]
Shear resisting force equals shear strength times the contact area (perimeter times thickness):
\[ F_{resist} = \tau \times p \times t = 1.0 \times 12 \times t \]
Step 3: Set the resisting force equal to the water force and solve for \(t\).
For the minimum safe thickness, the resisting shear force must just balance the water force.
\[ 12t = 87.5 \]
\[ t = \frac{87.5}{12} = 7.29 \text{ m} \]
Final Answer:
The plug dam needs a minimum thickness of about 7.29 m to hold back the water. \[ \boxed{7.29 \text{ m}} \]
The plug dam seals a gallery that is 3.5 m wide and 2.5 m high, so the face area the water pushes against is
\[ A = 3.5 \times 2.5 = 8.75 \text{ m}^2 \]
Water pressure acts uniformly over this face, so the total pushing force is pressure times area. Since 1 MPa equals 1 MN per square meter, this comes out directly in MN.
\[ F = P \times A = 10.0 \times 8.75 = 87.5 \text{ MN} \]
Step 2: Find how the dam resists this force.
The dam does not resist the water by its own strength alone, it resists through shear that develops all around its edge, where the dam's outer surface grips the surrounding rock. That gripping surface runs around the full perimeter of the gallery and extends back into the rock over the dam's thickness \(t\).
Perimeter of the gallery:
\[ p = 2(3.5+2.5) = 12 \text{ m} \]
Shear resisting force equals shear strength times the contact area (perimeter times thickness):
\[ F_{resist} = \tau \times p \times t = 1.0 \times 12 \times t \]
Step 3: Set the resisting force equal to the water force and solve for \(t\).
For the minimum safe thickness, the resisting shear force must just balance the water force.
\[ 12t = 87.5 \]
\[ t = \frac{87.5}{12} = 7.29 \text{ m} \]
Final Answer:
The plug dam needs a minimum thickness of about 7.29 m to hold back the water. \[ \boxed{7.29 \text{ m}} \]
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