Lecture Notes on Problem 1-37

Problem Statement

• Given a plate with a width of 0.5 meters.
• Stress distribution at the support varies as described.
• Objective: Determine the load P applied to the plate and the distance D from one end where it is applied.

Details of the Problem

• Stress formula:
  • Stress ( ( \sigma ) = 15 ( X^{1/2} ) Mega Pascal
• Width of the plate: 0.5 meters

Free Body Diagram

• Draw a diagram of the plate with the following elements:
  • Load P acting at a distance D from one end.
  • Stress direction: X direction.
  • Varying load represented as a function of X.

Differential Force Calculation

• Consider a small element of the plate:
  • Differential area ( dA = width \cdot dx = 0.5 \cdot dx )
  • Differential force ( dF = \sigma \cdot dA )
  • Substitute stress value:
    • ( dF = 15 \cdot X^{1/2} \cdot 10^{6} \cdot (0.5 , dx) )
    • Result: ( dF = 7.5 \cdot 10^{6} \cdot X^{1/2} , dx )

Equilibrium Condition for Forces

• Apply the first equilibrium condition:
  • ( \sum F_y = 0 )
  • Forces acting:
    • Load P (downward) = negative
    • Integral of differential forces (upward) = positive
• Equation setup:
  • ( -P + \int_{0}^{4} 7.5 \cdot 10^{6} \cdot X^{1/2} , dx = 0 )
• Solve integral:
  • ( P = 40 \cdot 10^{6} , N ightarrow 40 , MegaN )

Equilibrium Condition for Moments

• Apply the second equilibrium condition:
  • ( \sum M_O = 0 )
  • Moments:
    • Moment due to load P: ( -P imes D ) (clockwise)
    • Moment due to differential force ( dF ): ( dF imes X ) (counterclockwise)
• Set up the equation:
  • ( -P imes D + \int_{0}^{4} dF imes X = 0 )
• Substitute values and solve:
  • Result: ( D = 2.40 , meters )

Conclusion

• Load P = 40 Mega Newton
• Distance D = 2.40 meters from one end where load is applied.

Final Remarks

• The problem was solved methodically using equilibrium principles and integration.
• Hope you enjoyed the explanation!