Explanation Given condition: The maximum tension in the cord equals the cord’s total weight. Given information: A uniform cord is suspended between the two points having the same elevation. Show the free-body diagram of the cable as in Figure (1). Using Figure (1), Write the equation for deflection curve as follows: (Equation 7.15). x = ∫ d s [ 1 + 1 F H 2 ( ∫ w ( s ) d s ) 2 ] 1 / 2 (I) Integrate Equation (I). x = ∫ d s [ 1 + ( 1 F H 2 ) ( w 0 s + C 1 ) 2 ] 1 / 2 (II) Take ( 1 F H ) ( w 0 s + C 1 ) as u , so that d u = ( w 0 F H ) d s . Integrate Equation (II). x = F H w 0 ( sinh − 1 u + C 2 ) (III) Substitute ( 1 F H ) ( w 0 s + C 1 ) for u in Equation (III). x = F H w 0 ( sinh − 1 ( ( w 0 s + C 1 ) F H ) + C 2 ) (IV) Determine the constants using the equation below (Equation 7.14). d y d x = 1 F H ∫ w 0 d s = 1 F H ( w 0 s + C 1 ) (V) Condition 1: d y d x = 0 at s = 0 : Substitute 0 for d y d x and 0 for s in Equation (V). 0 = 1 F H ( w 0 ( 0 ) + C 1 ) C 1 = 0 Substitute and 0 for C 1 in Equation (V). d y d x = 1 F H ( w 0 s + 0 ) d y d x = w 0 s F H (VI) Condition 2: s = 0 at x = 0 : Substitute 0 for x , 0 for s , and 0 for C 1 in Equation (IV). 0 = F H w 0 ( sinh − 1 ( ( w 0 × 0 + 0 ) F H ) + C 2 ) sinh − 1 ( 0 F H ) + C 2 = 0 C 2 = 0 Substitute 0 for C 1 and 0 for C 2 in Equation (IV). x = F H w 0 ( sinh − 1 ( ( w 0 s + 0 ) F H ) + 0 ) x = F H w 0 ( sinh − 1 ( w 0 s F H ) ) s = F H w 0 sinh ( w 0 x F H ) (VII) Substitute F H w 0 sinh ( w 0 x F H ) for s in Equation (VI). d y d x = w 0 ( F H w 0 sinh ( w 0 x F H ) ) F H = sinh ( w 0 x F H ) (VIII) Integrate the Equation (VIII). y = F H w 0 cosh ( w 0 x F H ) + C 3 (IX) Condition 3: y = 0 at x = 0 : Substitute 0 for y and 0 for x in Equation (IX)

13th Edition

ISBN: 9780132915540

Author: Russell C. Hibbeler

Publisher: Prentice Hall

See similar textbooks

Question

Chapter 7.4, Problem 122P

To determine

The sag to span ratio.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 7.4, Problem 121P

Chapter 7.4, Problem 123P

Students have asked these similar questions

A gas mixture with a molar analysis of 40% CH4 (methane) and 60% air enters a control volume operating at steady state at location 1 with a mass flow rate of 5 kg/min, as shown in the figure below. Air enters as a separate stream at 2 and dilutes the mixture. A single stream exits with a mole fraction of methane of 5%. Assume air has a molar analysis of 21% O2 and 79% N2. (CH4, Air) m₁ = = 5 kg/min Air (21% O2, 79% N₂) 3 + (5% CH4, 95% Air)

Argon (Ar), at T₁ = 350 K, 1 bar with a mass flow rate of m₁ 3 kg/s enters the insulated mixing chamber shown in the figure below and mixes with carbon dioxide (CO2) entering as a separate stream at 575 K, 1 bar with a mass flow rate of 0.5 kg/s. The mixture exits at 1 bar. Assume ideal gas behavior with k = 1.67 for Ar and k = 1.25 for CO2. Argon (Ar) P₁ = 1 bar mT For steady-state operation, determine: (a) the molar analysis of the exiting mixture. (b) the temperature of the exiting mixture, in K. (c) the rate of entropy production, in kW/K. Insulation 3 + Mixture exiting P3 = 1 bar 2+ Carbon dioxide (CO2) T₂ = 575 K P2 = 1 bar m2 = 0.5 kg/s

Chapter 7 Solutions

Engineering Mechanics: Statics

Ch. 7.1 - Determine the normal force, shear force, and...Ch. 7.1 - Determine the normal force, shear force, and...Ch. 7.1 - Determine the normal force, shear force, and...Ch. 7.1 - Determine the normal force, shear force, and...Ch. 7.1 - Determine the normal force, shear force, and...Ch. 7.1 - Assume A is pinned and B is a roller. Prob. F7-6 Ch. 7.1 - Assume the support at B is a roller. Point C is...Ch. 7.1 - Determine the shear force and moment at points C...Ch. 7.1 - If the suspended load has a weight of 2 kN and a...Ch. 7.1 - Prob. 4P

Knowledge Booster