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What is steel plate deflection in mm when you apply 15ton of force on top?


Consider under the steel plate is soil with allowable soil bearing pressure is 3ton/sq.m.


Steel plate size: 1.5m x 1.3m x 0.025m


Load size : 0.500m diameter

What is steel plate deflection in mm when you apply 15ton of force on top Consider under the steel plate is soil with allowable soil

Answer :

Final answer:

To calculate the deflection of a steel plate under a given force, use the formula: Deflection = (Force * Length * Width)^3 / (192 * Elastic Modulus * Thickness^3). Given the force, plate dimensions, and soil bearing pressure, substitute the values into the formula to find the deflection.

Explanation:

To calculate the deflection of a steel plate under a given force, we need to analyze the plate's structural properties and the soil's bearing capacity. In this case, the deflection can be determined using the formula:

Deflection = (Force * Length * Width)^3 / (192 * Elastic Modulus * Thickness^3)

Given the force, plate dimensions, and soil bearing pressure, we can substitute the values into the formula to find the deflection in millimeters.

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Final answer:

To calculate the deflection of the steel plate, the stress applied to the plate and its mechanical properties must be considered. With only the applied force and plate dimensions provided, this needs more information. The stress calculation shows a result of about 76625 kPa, but without details on the nature of the loading and the boundary conditions, calculating the deflection isn't possible.

Explanation:

In order to calculate the deflection of the steel plate, we must consider the stress applied to the plate and its mechanical properties including its modulus of elasticity and Poisson's ratio, assuming that the plate is isotropic and homogeneous. Since only the applied force and the plate dimensions are provided, a full solution of this problem requires additional information.

The applied stress on the plate can be calculated by dividing the force by the area it's applied to. In this case, the force is 15 tons (or 15000 kg) and the area is the area of the load (which can be approximated as a circle with a diameter of 0.5m), that is π*(0.5/2)² = 0.196 m², so the stress is about 76625 kPa.

Now, to calculate the deflection, we need to apply the formula for the maximum deflection of a simply supported rectangular plate under a uniform load. This formula varies depending on the specific cases, such as whether the plate has free or clamped edges, and whether the load is uniform or concentrated in the middle. As crucial information on the nature of the loading and the boundary conditions of the steel plate are missing from the provided details, the actual calculation of the steel plate deflection is limited, implying that additional information or simplifying assumptions would be required to proceed further.

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