Engineering Mechanics

(^324) „„„„„ A Textbook of Engineering Mechanics
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 veiw :

1. String carrying point loads.


2. String carrying uniformly distributed load.


3. String supported at different levels.

15.4.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 W 1 , W 2 and W 2 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 T 1 = Tension in the string AC,

T 2 = Tension in the string CD,

T 3 = Tension in the string DE, and

T 4 = Tension in the string EB.

Since all the points of the cable are in equilibrium, therefore vector diagram with the help of

loads W 1 , W 2 , W 3 as well as tensions T 1 , T 2 , T 3 and T 4 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 :

1. Select some suitable point p and draw a vertical line pq equal to the load W 1 to some
   suitable scale,


2. Similarly, draw qr, rs equal to loads W 2 and W 3 to the scale.


3. Through p, draw a line parallel to AC and through q draw a line parallel to CD, meeting
   the first line at o.


4. Join or and os. Now the vector diagram is given by the figure pqrsop.


5. 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.


6. Now the tensions in the cable AC (T 1 ), CD (T 2 ), DE (T 3 ) and EB (T 4 ) 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.
   Note. If the exact positions of the points C, D, E are not given then the position of point o in
   the vector diagram is obtained by locating the point m such that pm is equal to the vertical reaction at
   A and ms is equal to the vertical reaction at B and then cutting mo equal to the horizontal thrust to the
   scale.