Question

4force 000 lbf 001 0.002 ,sec 1000lbf 5. A load cell (second-order model) has the specifications as follows Undamped natural frequency 1000Hz Damping Ratiob-0.707 Sensitivity-10mv/lbf Stiffness 100000 lbfin If a dynamic force as shown below is applied to the load cell, determine (a) the steady- state output voltage as a function of time, and (b) the useful dynamic frequency range ( 5%).

Answer 1

• project network diagram of above project is shown below.

  

  path 1 time 98 10 6 25 path 2 time 90 95 20 12 30 15 7 end 4 1 2 start 5 30 15 10 15 5 critical path - path 3 -> 1->2->5->8->10->11 path 3 time 115

  The project has 3 paths.

  1. Start -> 1 -> 2 -> 3 -> 6 -> 9 -> 11 -> End, time: 98 days.
  2. Start -> 1 -> 2 -> 4 -> 7 -> 9 -> 11 -> End, time: 90 days.
  3. Start -> 1 -> 2 -> 5 -> 8 -> 10 -> 11 -> End, time: 115 days.

  c)

  The critical path is the path has longest path . In the above network 3 rd path is the critical path with time = 115

  The critical path of the project => start -> 1 -> 2 -> 5 -> 8 -> 10 -> 11 -> end => time = 115.

  b) slack (SL) or float of the path is defined as delaying time to complete the waiting time.i.e difference between critical path time and time for the corresponding path.

  -> SL of first path = critical path time - first  path time = 115 - 98 = 17

  -> SL of second path = critical path time - second path time = 115 - 90 = 25

  -> SL of critical path always zero.

  Calculating Early Start (ES), Early Finish (EF), Late Start (LS), and Late Finish (LF) for each node in the network

  Calculating Early Start (ES), Early Finish (EF) for each node in the path 1

  -> Using password pass, ES and EF are calculated from starting node.

  ES of initial node always starts with 1 ,because every project starts from first day.

  So , ES of node 1 = 1

  formula for ES = EF of previous node + 1 (from second node onwards)

  formula for EF =  Activity time + ES of the node - 1 .

  S0, EF of node 1 = 30 + 1` - 1 = 30.

  formula for ES = EF of previous node + 1

  ES of node 2 = 30 +1 =31

  EF of node 2 = 20 + 31` - 1 = 50.

  ES of node 3 = 50 +1 =51

  EF of node 3 = 25 + 51` - 1 = 75.

  ES of node 6 = 75 +1 =76

  EF of node 6 = 10 + 76` - 1 = 85.

  ES of node 9 = 85 +1 =86

  EF of node 9 = 8 + 86` - 1 = 93

  ES of node 11 = 93 +1 =94

  EF of node 11 = 5 + 94` - 1 = 98.

  Calculating Early Start (ES), Early Finish (EF) for each node in the path 2

  -> Using password pass, ES and EF are calculated from starting node.

  ES of initial node always starts with 1 ,because every project starts from first day.

  So , ES of node 1 = 1

  formula for ES = EF of previous node + 1 (from second node onwards)

  formula for EF =  Activity time + ES of the node - 1 .

  S0, EF of node 1 = 30 + 1` - 1 = 30.

  formula for ES = EF of previous node + 1

  ES of node 2 = 30 +1 =31

  EF of node 2 = 20 + 31` - 1 = 50.

  ES of node 4 = 50 +1 =51

  EF of node 4 = 15 + 51` - 1 = 65.

  ES of node 7 = 65 +1 =66

  EF of node 7 = 12 + 66` - 1 = 77.

  ES of node 9 = 77 +1 =78

  EF of node 9 = 8 + 78` - 1 = 85

  ES of node 11 = 85 +1 =86

  EF of node 11 = 5 + 86` - 1 = 90.

  Calculating Early Start (ES), Early Finish (EF) for each node in the path 3

  -> Using password pass, ES and EF are calculated from starting node.

  ES of initial node always starts with 1 ,because every project starts from first day.

  So , ES of node 1 = 1

  formula for ES = EF of previous node + 1 (from second node onwards)

  formula for EF =  Activity time + ES of the node - 1 .

  S0, EF of node 1 = 30 + 1` - 1 = 30.

  formula for ES = EF of previous node + 1

  ES of node 2 = 30 +1 =31

  EF of node 2 = 20 + 31` - 1 = 50.

  ES of node 5 = 50 +1 =51

  EF of node 5 = 30 + 51` - 1 = 80.

  ES of node 8 = 80 +1 =81

  EF of node 8 = 15 + 81` - 1 = 95.

  ES of node 10 = 95 +1 =96

  EF of node 10 = 15 + 96` - 1 = 110

  ES of node 11 = 110 +1 =111

  EF of node 11 = 5 + 111` - 1 = 115

  Calculating Late Start (LS), Late Finish (LF) for each node in the path 1

  -> Using backword pass, LS andLF are calculated from ending node.

  LF of last node always starts with critical path time ,because every project not completes before critical time.

  So , LF of node 11 = 115

  formula for LF = LS of successor activity – 1 (from next node onwards)

  formula for LS =  LF of activity – activity time + 1 .

  So, LS of node 11 = 115 - 5 + 1` = 111

  LF of node 9 = 111 - 1 = 110

  LS of node 9 = 110 - 8 + 1 = 103

  LF of node 6 = 103 - 1 = 102

  LS of node 6 = 102 - 10 + 1 = 93

  LF of node 3 = 93 - 1 = 92

  LS of node 3 = 92 - 10 + 1 = 83

  -> node 2 has 3 successor nodes , So which node has least LS will be taken as LF of successor activity.

  since edge 2<->4 is least .so Late start of node 2 is 30 + 20 + 15 +1 = 66 (it will be got LS from path 2)

  LF of node 2 = 66 - 1 = 65

  LS of node 2 = 65 - 15 + 1 = 51

  LF of node 1 = 51 - 1 = 50

  LS of node 1 = 50 - 20 + 1 = 31

  => similarily for path2 and path3 can be calculated.

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