Question

A 20-foot long beam is supported by a cable at each end. The left cable is made from steel (E = 30,000,000 psi, Area = 2 in2 ) and the right cable is made from aluminum (E = 10,000,000 psi, Area = 4in2 ). If a 10,000 pound load is applied to the beam, at what distance "x" does it need to be applied such that the beam remains horizontal? Both cables are 15 feet in length.

Answer 1

Answer:

x = 8 ft

Step-by-step explanation:

Given:

- Length of beam L = 20 ft = 240 in

- Left cable: E = 30,000,000 psi, Area = 2 in2

- Right Cable: E = 10,000,000 psi, Area = 4in2

- Force P = 10,000 lb

- distance from left cable = x

- distance from right cable = 240 - x

- Length of both cables h = 15 ft

Find:

Distance x that would ensure the beam remains horizontal.

Solution:

- Compute forces in right and left cables P_1 and P_2, respectively:

Sum of vertical forces  

                                P = P_1 + P_2

Sum of Moments about left point  

                                P_1*240 = P*x

                                P_1 = P*x / 240

Hence,                    P_2 = P*(240 - x) / 240

- Compute deflections in right and left cables Δk and Δm, respectively

                               Δk = P_1*h / A_1*E_1

                               Δm = P_2*h / A_2*E_2

- Since, deflection are same equate the above two Δk = Δm :

                      P_1*h / A_1*E_1 = P_2*h / A_2*E_2

        P*x/ 240*4*10,000,000 = P*(240-x) /240*2*30,000,000

                                      (x / 2) = (240-x) / 3

- Compute for x:

                                       3x = 480 - 2x

                                       5 x = 480

                                       x = 96 in = 8 ft