Answer:
The change in length of the circular strut DC = 0.0028 in.
The vertical displacement of the rigid bar at point B = 0.00378 in.
Explanation:
We have the following parameters or information in the question given above:
=> The outer diameter = 15 in., the inner diameter = 14.4 in., the modulus elasticity of E = 29,000 ksi, and the Point load P = 5kips.
The diagram showing the rigid bar ACB is supported by an elastic cir-cular strut DC is given in the attached picture below.
According to Newton's law of motion, it can be seen that the force on CD, that is FCD is equal and opposite to ACD. Hence, FCD = ACD.
Where FCD = p × [4 + 5] ÷ [sin Ф × 4].
kindly note that from the diagram sin Ф = 3/5, cos Ф = 4/5 and tan Ф = 3/4. Also p =5.
Hence, FCD =[ 5 × 9] ÷ [3/5 × 4] = 18.75 kip. So FCD = ACD.
The next thing here is to determine the area and length of CD, say the area of CD is G, thus, G = π/4 × [ 15² - 14.4²] = 13.854 in².
The lenght of CD is = √[4² + 3²] = √[16 + 9] = 5ft. Thus, 5 × 12 = 60in.
Hence, the change in length of the circular strut DC = [18.75 × 60] ÷ 13.854 × 29000 = 0.0028 in.
The vertical deflection of CD = 0.0028 × 3/5 = 0.00168 in.
We have that; 4 /CV = 9BV. Hence, BV = 9/4× CV.
(CV = vertical deflection of CD).
The vertical displacement of the rigid bar at point B = 9/ 4 × 0.00168 in = 0.00378 in.
