Consider a case where the load \( Q \) for a ship structure has only statistical uncertainties. The probability density function of the load \( p_Q(x) \) is shown in the figure. The characteristic limit value of the load \( Q_L \) is 1.5 and the factor of safety is 1. Which of the following probability of exceedance value(s) of the load will lead to a safe design?
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To calculate the exceedance probability, integrate the probability density function from the given limit value to the maximum value of the distribution.
Step 1: Understanding the problem. The characteristic limit value of the load is \( Q_L = 1.5 \), and the factor of safety \( FS = 1 \), so the design value of the load is: \[ Q_{{design}} = \frac{Q_L}{FS} = 1.5. \] Step 2: Exceedance probability. The exceedance probability for the design load value is the area under the probability density function \( p_Q(x) \) to the right of \( Q_{{design}} \), which is \( Q = 1.5 \). Step 3: Analyzing the given distribution. From the probability density function, we see that the triangular distribution has a peak at \( x = 1 \) and linearly decreases to zero at \( x = 2 \). The cumulative area to the right of \( x = 1.5 \) gives the exceedance probability. By calculating the area under the curve from 1.5 to 2, we find the exceedance probability is between 0.15 and 0.20.
