Engineering Mechanics

Chapter 15 : Equilibrium of Strings „„„„„ 325

Example 15.1. Three loads of 10 kN, 15 kN and 20 kN are suspended from a string AB as
shown in Fig. 15.3.

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.
Solution. Given. Span (l) = 12 m

(i) Vertical reactions at A and B
Taking moments about A and equating the same,
VB × 12 = (10 × 3) + (15 × 6) + (20 × 9) = 300

∴^300 25 kN
B 12
V == Ans.

and VA= (10 + 15 + 20) – 25 = 20 kN Ans.

(ii) Horizontal thrusts at A and B
Let H= Horizontal thrusts at A and B.
Consider equilibrium of the string ACD. Taking moments of the forces acting in the string
ACD about D and equating the same,
H × 3 = (20 × 6) – (10 × 3) = 90

∴

90

30 kN

3

H== Ans.

(iii) Sag of points C and E

Let yC= Sag of point C, and

yE = Sag of point E.

Taking moments of the forces acting in the string AC about C and equating the same,

30 × yC= 20 × 3 = 60

∴

60

2m

C 30

y == Ans.

Similarly, taking moments of the forces acting in the string EB about E and equating the same,

30 × yE = 25 × 3 = 75

∴

(^75) 2.5 m
E 30
y == Ans.
(iv) Tensions in all the segments of the string
First of all, draw the space diagram and name all the forces and tensions as per Bow’s notations
as shown in Fig. 15.4 (a).