1. 7.4 Cables
Flexible cables and chains are used to support and transmit loads from one member to another
In suspension bridges and trolley wheels, they carry majority of the load
In force analysis, weight of cables is neglected as it is small compared to the overall weight
Consider three cases: cable subjected to concentrated loads, subjected to distributed load and subjected to its own weight

2. 7.4 Cables
Assume that the cable is perfectly flexible and inextensible
Due to its flexibility, the cables offers no resistance to bending and therefore, the tensile force acting in the cable is always tangent to the points along its length
Being inextensible, the length remains constant before and after loading, thus can be considered as a rigid body

3. 7.4 Cables Cable Subjected to Concentrated Loads
For a cable of negligible weight supporting several concentrated loads, the cable takes the form of several straight line segments, each subjected to constant tensile force

4. 7.4 Cables Cable Subjected to Concentrated Loads
Known: h, L1, L2, L3 and loads P1 and P2
Form 2 equations of equilibrium at each point A, B, C and D
If the total length L is given, use Pythagorean Theorem to relate the three segmental lengths
If not, specify one of the sags, yC and yD and from the answer, determine the other sag and hence, the total length L

5. 7.4 Cables Example 7.13 Determine the tension in each segment of
the cable.

6. 7.4 Cables
Solution
FBD for the entire cable

7. 7.4 Cables
Solution

8. 7.4 Cables Solution Consider leftmost section which cuts cable BC
since sag yC = 12m

9. 7.4 Cables
Solution

10. 7.4 Cables
Solution
Consider point A, C and E,

11. 7.4 Cables
Solution
Point A

12. 7.4 Cables
Solution
Point C

13. 7.4 Cables
Solution
Point E

14. 7.4 Cables
Solution
By comparison, maximum cable tension is in segment AB since this segment has the greatest slope
For any left hand side segment, the horizontal component Tcosθ = Ax
Since the slope angles that the cable segment make with the horizontal have been determined, the sags yB and yD can be determined using trigonometry

15. 7.4 Cables Cable Subjected to a Distributed Load
Consider weightless cable subjected to a loading function w = w(x) measured in the x direction

16. 7.4 Cables Cable Subjected to a Distributed Load
For FBD of the cable having length ∆

17. 7.4 Cables Cable Subjected to a Distributed Load
Since the tensile force in the cable changes continuously in both the magnitude and the direction along the cable’s length, this change is denoted on the FBD by ∆T
Distributed load is represented by its resultant force w(x)(∆x) which acts a fractional distance k(∆x) from point O where o < k < 1

18. 7.4 Cables
Cable Subjected to a Distributed Load

19. 7.4 Cables Cable Subjected to a Distributed Load
Divide by ∆x and taking limit,
Integrating,

20. 7.4 Cables Cable Subjected to a Distributed Load Integrating,
Eliminating T,
Perform second integration,

21. 7.4 Cables
Example 7.14
The cable of a suspension bridge supports half of
the uniform road surface between the two columns
at A and B. If this distributed loading wo, determine
the maximum force developed in the cable and the
cable’s required length. The span length L and, sag
h are known.

22. 7.4 Cables Solution Note w(x) = wo Perform two integrations
Boundary Conditions at x = 0

23. 7.4 Cables Solution Therefore, Curve becomes
This is the equation of a parabola
Boundary Condition at x = L/2

24. 7.4 Cables Solution For constant, Tension, T = FH/cosθ
Maximum tension occur at point B for 0 ≤ θ ≤ π/2

25. 7.4 Cables Solution Slope at point B Or Therefore
Using triangular relationship

26. 7.4 Cables Solution For a differential segment of cable length ds
Determine total length by integrating
Integrating yields

27. 7.4 Cables Cable Subjected to its Own Weight
When weight of the cable is considered, the loading function becomes a function of the arc length s rather than length x
For loading function w = w(s)
acting along the cable,

28. 7.4 Cables Cable Subjected to its Own Weight
FBD of a segment of the cable

29. 7.4 Cables Cable Subjected to its Own Weight
Apply equilibrium equations to the force system
Replace dy/dx by ds/dx for direct integration

30. 7.4 Cables Cable Subjected to its Own Weight Therefore
Separating variables and integrating

31. 7.4 Cables Example 7.15 Determine the deflection curve, the length,
and the maximum tension in the uniform
cable. The cable weights wo = 5N/m.

32. 7.4 Cables
Solution
For symmetry, origin located at the center of the cable
Deflection curve expressed as y = f(x)
Integrating term in the denominator

33. 7.4 Cables
Solution
Substitute
So that
Perform second integration or

34. 7.4 Cables Solution Evaluate constants or
dy/dx = 0 at s = 0, then C1 = 0
To obtain deflection curve, solve for s

35. 7.4 Cables Solution Hence Boundary Condition y = 0 at x = 0
For deflection curve,
This equations defines a catenary curve

36. 7.4 Cables Solution Boundary Condition y = h at x = L/2
Since wo = 5N/m, h = 6m and L = 20m,
By trial and error,

37. 7.4 Cables Solution For deflection curve,
x = 10m, for half length of the cable
Hence
Maximum tension occurs when θ is maximum at
s = 12.1m

38. 7.4 Cables
Solution

39. Chapter Summery Internal Loadings
If a coplanar force system acts on a member, then in general a resultant normal force N, shear force V and bending moment M will act at any cross-section along the member
These resultants are determined using the method of sections
The member is sectioned at the point where the internal loadings are to be determined

40. Chapter Summery Internal Loadings
A FBD of one of the sectioned parts is drawn
The normal force is determined from the summation of the forces normal to the cross-section
The shear force is determined from the summation of the forces tangent to the cross section
The bending moment is found from the summation of the moments about the centroid of the cross-sectional area

41. Chapter Summery Internal Loadings
If the member is subjected to 3D loading, in general, a torisonal loading will also act on the cross-section
It can be determined by the summation of the moments about an axis that is perpendicular to the cross section and passes through its centroid

42. Chapter Summery Shear and Moment Diagrams as Functions of x
Section the member at an arbitrary point located distance x from one end to construct the shear and moment diagrams for a member
Unknown shear and moment are indicated on the cross-section in the positive direction according to the established sign convention
Applications of the equilibrium equations will make these loadings functions of x which then, can be plotted

43. Chapter Summery Shear and Moment Diagrams as Functions of x
If the external loadings or concentrated forces and couple moments act on the member, then different expressions of V and M must be determined within regions between these regions
Graphical Methods for Establishing Shear
and Moment Diagrams
Plot the shear and moment diagrams using differential relationships that exist between the distributed loading w and V and M

44. Chapter Summery Graphical Methods for Establishing Shear
and Moment Diagrams
Slope of the shear diagram = distributed loading at any point
dV/dx = -w
Slope of the moment diagram = shear at any point
V = dM/dx
Change in shear between an y two points = area under the shear diagram between points, ∆M = ∫Vdx

45. Chapter Summery Cables
When a flexible and inextensible cable is subjected to a series of concentrated forces, the analysis of the cable can be performed by using the equations of equilibrium applied to the FBD of wither segments or points of application of the loading
If the external distributed loads or weight of the cables are considered, forces and shape of the cable must be determined by analysis of the forces on a differential segment of the cable and integrating the result

46. Chapter Review

47. Chapter Review

48. Chapter Review

49. Chapter Review

50. Chapter Review