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Substituting in Eq. (13.78) gives - ks y - L - S1 - k f ( y - yc ) = m Solving for cam displacement, yc = L + S1 k f + ks m d 2 y + + . kf kf k f dt 2 (13.79) d2y . dt 2
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1. FCM is a steel member in tension whose failure is likely to cause the entire bridge to collapse. These members are highly vulnerable to a fatigue type failure. The three conditions for identifying fracture critical members are steel material, tensile stress, and vulnerability to collapse due to simple con guration. 2. Redundancy helps to delay fracture and increase fracture toughness. Redundancy is a structural condition resulting when there is more than the minimum number of structural elements for stability. The framing or con guration of the bridge such as number of girders, number of continuous supports, or multiple numbers of members forming a connection would increase redundancy. Load path redundancy: When three or more main load carrying members are present between substructure units and if one member is close to failure, load can be safely distributed to other members. Structural redundancy: A con guration which provides continuity of load path to adjacent spans due to redistribution of moments. Only internal spans, but not the end spans, will provide structural redundancy.
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#include <stdio.h> #include <conio.h> int main(void) { char ch; ch = getch(); putch(ch); // ungetch(ch); ch = getch(); putch(ch); // return 0; } // get keypress show the key // return to buffer // get same key again show the key again
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EXAMPLE An eccentric circle cam drives a translating roller follower. This circle has an eccentricity of 0.500 in. and a diameter of 2.500 in. and rotates at 180 rpm. The roller follower has a diameter of 0.750 in. Find (1) the characteristics of the follower motion after the cam has rotated 30 from its lowest position and (2) the maximum pressure angle. Solution The equivalent mechanism has a crank length E equal to 0.500 in. The connecting rod length equals the eccentric circle radius plus the roller radius, giving
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// An override of GetHashCode(). public override int GetHashCode() { return Val.GetHashCode(); } } class CompareDemo { // Require IEquatable<T> interface. public static bool IsIn<T>(T what, T[] obs) where T : IEquatable<T> { foreach(T v in obs) if(v.Equals(what)) // Uses Equals() return true; return false; } // Require IComparable<T> interface. This method assumes // a sorted array. It returns true if what is inside the range // of elements passed to obs. public static bool InRange<T>(T what, T[] obs) where T : IComparable<T> { if(what.CompareTo(obs[0]) < 0 || what.CompareTo(obs[obs.Length-1]) > 0) return false; return true; } // Demonstrate comparisons. static void Main() { // Use IsIn() with int. int[] nums = { 1, 2, 3, 4, 5 }; if(IsIn(2, nums)) Console.WriteLine("2 is found."); if(IsIn(99, nums)) Console.WriteLine("This won t display.");
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1. Create a file called QSDemo.cs. 2. Create the Quicksort class shown here: // A simple version of the Quicksort. using System; class Quicksort { // Set up a call to the actual quicksort method. public static void QSort(char[] items) { if(items.Length == 0) return; qs(items, 0, items.Length-1); } // A recursive version of quicksort for characters. private static void qs(char[] items, int left, int right) { int i, j; char x, y; i = left; j = right; x = items[(left+right)/2]; do { while((items[i] < x) && (i < right)) i++; while((x < items[j]) && (j > left)) j--; if(i <= j) { y = items[i]; items[i] = items[j]; items[j] = y; i++; j--; } } while(i <= j);
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