Equilibrium of Strings Chapter 15 Equilibrium of Strings

Learning Objectives Introduction Shape of a Loaded String Tension in a String Tension in a String Carrying Point Loads Tension in a String Carrying Uniformly Distributed Load Tension in a String when the Two Supports are at Different Levels Length of a String Length of a String when the Supports are at the Same Level Length of a String when the Supports are at Different Levels The Catenary

Shape of a Loaded String Introduction A string (or rope), in its theoretical sense, is absolutely flexible, light (i.e. its weight is neglected) and inextensible. It is capable of offering only tensile resistance. The slope of a loaded string depends upon its length and the loads supported by it. It will be interesting to know that if the loads carried by the string are changed (in magnitude or position) its shape will also change. Shape of a Loaded String Consider a string or cable suspended at two points A and B at the same level, and carrying a uniformly distributed load over its horizontal span as shown in Fig 15.1. Fig. 15.1.

Let w = Uniformly distributed load per unit length, l = Span of the cable, and yc = Central dip of the cable. Now consider any point (P) on the string. Let the coordinates of this point be x and y with respect to C, the lowest point of the string as origin. Now draw the tangent at P. Let q be the inclination of the tangent with the horizontal as shown in the figure. We know that the portion CP of the string is in equilibrium under the action of the following forces : Load (w.x) acting vertically downwards, Horizontal pull (H) acting horizontally at C, and Tension (T) acting at P along the tangent. Resolving the forces vertically and horizontally,

Tension in a String The determination of tension in the string or cable is one of the important criterion for its design. As a matter of fact, the tension in a string depends upon the magnitude and type of loading as well as levels of the two supports. Though there are many types of strings and loadings, yet the following are important from the subject point of view : String Carrying Point Loads String Carrying Uniformly Distributed Load String Supported at Different Levels

Tension in a String Carrying Point Loads Fig. 15.2 Consider a string or cable suspended at two points A and B at the same level and carrying point loads W1, W2 and W2 at C, D and E respectively. Let us assume the weight of the string to be negligible as compared to the point loads and the cable to take the shape as shown in Fig. 15.2. Let T1 = Tension in the string AC, T2 = Tension in the string CD, T3 = Tension in the string DE, and T4 = Tension in the string EB. Since all the points of the cable are in equilibrium, therefore vector diagram with the help of loads W1, W2, W3 as well as tensions T1, T2, T3 and T4 in the cable must close. Now draw the vector diagram for the given loads and tensions as shown in Fig. 15.2.(b) and as discussed below :

Select some suitable point p and draw a vertical line pq equal to the load W1 to some suitable scale, Similarly, draw qr, rs equal to loads W2 and W3 to the scale. Through p, draw a line parallel to AC and through q draw a line parallel to CD, meeting the first line at o. Join or and os. Now the vector diagram is given by the figure pqrsop. Through o, draw om perpendicular to the load line pqrs. The vertical reactions at A and B are given by pm and ms respectively to the scale. Now the tensions in the cable AC (T1), CD (T2), DE (T3) and EB (T4) are given by the lengths op, oq, or and os respectively to the scale. And the horizontal thrust is given by om to the scale. Example Three loads of 10 kN, 15 kN and 20 kN are suspended from a string AB as shown in Fig. 15.3. If the point D is at a depth of 3 m from the supports, find (i) vertical reactions at A and B ; (ii) horizontal thrusts at A and B ; (iii) sag of points C and E ; and (iv) tensions in all the segments of the string.

Tension in a String Carrying Uniformly Distributed Load Consider a string or cable suspended at two points A and B at the same level and carrying a uniformly distributed load over the horizontal span of the cable as shown in Fig. 15.5.

Example A suspension bridge of 40 m span with 1.5 m wide platform is subjected to an average load of 20 kN/m2. The bridge is supported by a pair of cables having a central dip of 5 m. Find the necessary cross sectional area of the cable, if the maximum permissible strees in the cable material is not to exceed 1050 N/mm2.

Tension in a String Supported at Different Levels Fig. 15.6. Tension in string ACB. Consider a string or cable ACB, supported at different levels at A and B, and carrying a uniformly distributed load as shown in Fig 15.6. Let C be the lowest point of the cable. Let w = Uniformly distributed load per unit length of the span, l = Span of the string, yc = Depth of the lowest point of the string C, from the lower support B, d = Difference between the levels of the two supports, l1 = Horizontal length between A and C, and l2 = Horizontal length between C and B.

Since the string is supporting vertical loads only, therefore the horizontal thrust at A, must be equal to the horizontal thrust at B. In order to locate position of the lowest point C, let us imagine the portion CB of the string to be extended to CB1, such that the new support B1 is at the same level as that of A. Similarly, imagnie the portion AC of the string to be cut short to A1C, such that the new support A1 is at the same level as that of B. From the geometry of the figure, we find that the string ACB1 has a span of 2l1 and a central dip of (yc + d) ; whereas A1CB has a span of 2l2 and a central dip of yc. Now in the string ACB1 the horizontal thrust

Length of a String It means the actual length of a string or cable required between two supports, when it is loaded when it is loaded with a uniformly distributed load and hangs in the form of a parabola. Here we shall discuss the following two cases : When the supports are at the same level When the supports are at different levels Length of a String when the Supports are at the Same Level Consider a string ACB supported at A and B at the same level, and carrying a uniformly distributed load as shown in Fig 15.9. Let C be the lowest point of the cable.

Example A steel wire, of uniform section, is hung in the form of a parabola. Find the maximum horizontal span, if the central dip is 1/12th of the span and the strees in steel wire is not to exceed 120 N/mm2. Take mass density of the steel as 7800 kg/m3.