Parabolic & Catenary Cables PDF

Week 8 Parabolic cables are a common structural element in various engineering applications due to their unique properties and advantages. They are particularly well-suited for structures that require a high degree of flexibility and load-bearing capacity, such as suspension bridges, cable-stayed br...

Week 8 Parabolic cables are a common structural element in various engineering applications due to their unique properties and advantages. They are particularly well-suited for structures that require a high degree of flexibility and load-bearing capacity, such as suspension bridges, cable-stayed bridges, and certain types of roofs. Key Characteristics of Parabolic Cables: Shape: When subjected to a uniformly distributed load, parabolic cables assume a parabolic shape. This shape allows for efficient distribution of forces and minimizes bending moments. Flexibility: Parabolic cables are highly flexible, enabling them to adapt to changes in load or environmental conditions without significant stress or deformation. Strength: When properly designed and constructed, parabolic cables can be extremely strong and capable of supporting heavy loads. Efficiency: The parabolic shape of the cable allows for optimal use of material, resulting in a more efficient and cost-effective structure. Parabolic Cables Ex. The maximum allowable tension in a cable is 90,000 lbs. It is to carry a load of 200 lb/horizontal foot when suspended between level supports 300 ft apart. What is the minimum permissible sag? Note: Since the load as given is horizontally distributed, thus the cable is a parabolic cable. Also the biggest tension in the cable is at the support. Ex. A transmission wire 230 m long having a mass of 0.97 kg per linear meter is suspended between towers of equal elevation 229 m apart. Find the maximum tension. Catenary cables are another type of cable commonly used in structural engineering, often in conjunction with parabolic cables or as standalone elements. While both types share similarities, there are key differences in their shapes and properties. Shape and Geometry: Catenary Curve: The most distinctive feature of catenary cables is their characteristic catenary shape. This curve is defined by the equation y = a * cosh(x/a), where a is a constant related to the cable's properties and the distance between the supports. Sag: Catenary cables typically have a greater sag or dip between their supporting points compared to parabolic cables of similar span and load. Span and Height: The span of a catenary cable is the horizontal distance between its supporting points. The height is the vertical distance between the lowest point of the cable and the supporting points. Mechanical Properties: Flexibility: Catenary cables are generally more flexible than parabolic cables, especially when subjected to concentrated loads or unevenly distributed weights. Tension Distribution: The catenary shape allows for a more uniform distribution of tension along the cable, which can be advantageous in certain applications. Load-Bearing Capacity: Catenary cables can be designed to support heavy loads, making them suitable for various structural applications. Ex. A cable which has a mass of 0.6 kg/m and is 240 m long is to be suspended with a sag of 24 m. Determine the maximum span of the cable. Ex. A cable 60 m long weighs 25 N/m. The tensions at both ends of the cable are 1500 N and 1650 N, respectively. Determine the distance of the lowest point of the cable from the ground. Coulomb’s Theory of Dry Friction 1. Static Case - Coulomb proposed the following law: If there is no relative motion between two surfaces that are in contact, the normal force N and the friction force F satisfy the following relationship. ≤ = where is the maximum static friction force that can exist between the contacting surfaces and is known as coefficient of static friction. The coefficient of static friction is an experimental constant that depends on the composition and roughness of the contacting surfaces. This equation states simply that the friction force that exists under static conditions (no relative motion) has an upper limit that is proportional to the normal force. Coulomb’s Theory of Dry Friction 2. Impending sliding – Consider the static case in which the friction force equals its limiting value; that is = = For this condition, the surfaces are on the verge of sliding, a condition known as impending sliding. When sliding impends, the surfaces are at rest relative to each other. However, any change that would require an increase in the friction force would cause sliding. The direction for can be determined from the observation that g 3. Dynamic Case: = is referred to as the kinetic, or dynamic friction force. The coefficient of kinetic friction is usually smaller than its static counterpart. As in the static case,. Note: All real surfaces also provide a force component that is tangent to the surface, called the friction force, that resists sliding. - In many situations, friction forces are helpful. For example, friction enables you to walk without slipping, screws in place, and it allows us to transmit power by means of clutches and belts. - On the other hand, friction can also be detrimental: It causes wear in machinery and reduces efficiency in the transmission of power by converting mechanical energy into heat. Limitations - Engineers must rely on empirical constants, such as the coefficient of friction. Handbook values for the coefficients of friction should be treated as approximate values. - The theory of dry friction is applicable only to surfaces that are dry or that contain only a small amount of lubricant. If there is relative motion between the surfaces of contact, the theory is valid for low speeds only. Ropes and Flat Belts Example: As shown in figure, a flexible belt placed around a rotating drum of 4-inch radius acts as a brake when the arm ABCD is pulled down by the force P. The coefficient of kinetic friction between the belt and the drum is 0.2. Determine the force P that would result in a braking torque of 400-lb·in., assuming that the drum is rotating counterclockwise. Neglect the weight of the brake arm Given:  0 = 0 + : − − = Eq. 1 Eq. 1 − − = = ℎ : = 0.2 Eq. 1 & 2: > =. = 240° = 1.33 =. = 0.2 1.33 =. Eq. 2 = 0 + 2 60 + 6 60 − 8 = 0 2 76.34 60 + 6 176.34 60 = 8 P=. Bending-moment Diagram (BMD) is a graphical representation that shows how the bending moment varies along the length of a structural element, such as a beam, under a load. The bending moment at any point along the beam is a measure of the internal moment or torque that causes the beam to bend. The diagram helps engineers assess where the maximum bending occurs and aids in the design and analysis of structural components to ensure safety and strength. Remember: In constructing the bending-moment diagram, start at the left end of the beam (where x = 0) and proceed across the beam to the right. Each bending-moment diagram should start at M = 0 at the left end of the beam and return to M = 0 at the right end. This always occurs in a properly constructed diagram. If the moment diagram does not return to M = 0 at the right end of the beam, check your equilibrium calculations to make sure that you've correctly determined the beam reactions. Ex 1: Ex 1: − 10 6 − = 0 = 60 − 10 Ex 1: Ex 2: Centroids and Center of Gravity Definitions: Centroid – the geometric center of the region. Centroids and mass centers coincide only if the distribution of mass is uniform—that is, if the body is homogeneous. Center of gravity – occurs between the contacting surfaces of bodies when there is no lubricating fluid. Center of mass – center of mass of a body is the idealized location where all its mass can be thought to be concentrated. First moment of the area - The first moment of area of a shape, about a certain axis, equals the sum over all the infinitesimal parts of the shape of the area of that part times its distance from the axis. Second moment of the area/Moment of inertia – is a geometrical property of an area which reflects how its points are distributed with regard to an arbitrary axis. Formula for Centroid Location of Common Shapes and Lines Formula for Centroid Location of Common Shapes and Lines Example 1 (Centroid of Areas): Using the method of composite areas, determine the location of the centroid of the shaded area shown in the figure below. Sol’n: The area can be viewed as a rectangle, from which a semicircle and a triangle have been removed. Sol’n: Example 1 (Centroid of Lines): Using the method of composite curves, determine the centroidal coordinates of the line in the that consists of the circular arc 1, and the straight lines 2 and 3. Sol’n: Sol’n: Inertia - the natural tendency of an object to remain at rest when it is at rest or in motion to continue moving at constant speed Moment - (torque) is the cross product of force and the perpendicular distance to which the force is up being applied Moment of Inertia - the natural tendency of the body to rotate or tend to rotate due to distribution of area, volume or mass elements of the body. - a.k.a Second Moment Parallel-Axis Theorem The moment of inertia of the body at a certain axis is equal to the sum of the movement of inertia with respect to the centroidal axis parallel to it and the product of the area and the square of the shortest distance between the two parallel axis = + = − MI w/ respect to x-axis − centroidal MI A – Area d – distance k – radius of gyration Example (Centroid of Areas): The centroid of the plane region is located at C. If the area of the region is 2000mm2 and its moment of inertia about the x-axis is Ix =40 × 106 mm4 , determine Iu. Ix = Ixo + Ad2 Iu = Ixo + Ad2 Ixo = Ix − Ad2 Iu = 23.8 × 106 + 2000 70 2 Ixo = 40 × 106 − 2000 90 2 =. × Ixo = 40 × 106 − 2000 90 2 =. × Example 3 (Product of Inertia): Calculate the product of inertia I x y for the angle shown in Fig. (a) by the method of composite areas. Sol’n: Ixy = Ix + A Ixy = 0 + 140 × 20 10 70 =. × Ixy = Ix + A Ixy = 0 + 160 × 20 100 10 =. × = =. × +. × =. × Thank you…