Load Sharing Configuration Example

This example appears in the System Analysis Reference book.

A component has five possible failure modes, $$A\,\!$$,  $${{B}_{A}}\,\!$$ ,  $${{B}_{B}}\,\!$$ ,  $${{B}_{C}}\,\!$$  and  $$C\,\!$$. The $$B\,\!$$  modes are interdependent. In addition:


 * The system will fail if mode $$A\,\!$$  occurs, mode  $$C\,\!$$  occurs or two out of the three  $$B\,\!$$  modes occur.


 * Modes $$A\,\!$$  and  $$C\,\!$$  each have a Weibull distribution, with a  $$\beta =2\,\!$$  and  $$\eta =10,000\,\!$$  and $$15,000\,\!$$, respectively.


 * Events $${{B}_{A}}\,\!$$,  $${{B}_{B}}\,\!$$  and  $${{B}_{C}}\,\!$$  each have an exponential distribution with a mean of 10,000 hours.


 * If any $$B$$  event occurs (i.e.,  $${{B}_{A}}\,\!$$,  $${{B}_{B}}\,\!$$  or  $${{B}_{C}}\,\!$$ ), the remaining  $$B\,\!$$  events are more likely to occur.  Specifically, the mean times of the remaining  $$B\,\!$$  events are halved.

Determine the reliability at 1,000 hours for this component.

Solution

The first step is to create the RBD. Modes $$A\,\!$$,  $$C\,\!$$  and a load sharing container with the  $${{B}_{i}}\,\!$$  modes must be drawn in series, as illustrated next.



The next step is to define the properties for each block, including those for the container. Based on the problem statement, the $$B\,\!$$  modes are in a 2-out-of-3 load sharing redundancy. When all three are working (i.e., when no $$B\,\!$$  mode has occurred), each block has an exponential distribution with $$\mu=10,000\,\!$$.

The next step is to set the weight proportionality factor. This factor defines the portion of the load that the particular item carries while operating, as well as the load that shifts to the remaining units upon failure of the item. To illustrate, assume three units (1, 2 and 3) are in a load sharing container with weight proportionality factors of 1, 2 and 3 respectively (and a 1-out-of-3 requirement).


 * Unit 1 carries $$\left( \tfrac{1}{1+2+3} \right)=0.166$$  or 16.6% of the total load.


 * Unit 2 carries $$\left( \tfrac{2}{1+2+3} \right)=0.333$$  or 33.3% of the total load.


 * Unit 3 carries $$\left( \tfrac{3}{1+2+3} \right)=0.50$$  or 50% of the total load.

The actual load on each unit then becomes the product of the entire load defined for the container times the portion carried by that unit. For example, if the container load is 100 lbs, then the portion assigned to Unit 1 will be $$100\cdot 0.166=16.6\,\!$$  lbs.

In the current example, all units share the same load and thus have equal weight proportionality factors. Because these factors are relative, if the same number is used for all three items then the results will be the same. Thus, weight proportional factor is set equal to 1 for each item.

The last properties that need to be defined are the total load and the 2-out-of-3 redundancy. The total load is dependent on how the parameters were computed. In this case, total load was assumed to be 3 when the parameters were computed (i.e., the load per item was 1 when all worked and 1.5 when two worked).

The following figure shows the Block Properties window of the load sharing container. When all of the parameters have been specified in BlockSim, the reliability at 1,000 hours can be determined. From the Analytical QCP, this is found to be 98.1048%.