A simply supported beam ABC of length l carries a concentrated load P ...
Solution:
Understanding the problem
A simply supported beam ABC of length l carries a concentrated load P at an intermediate point B. The slope at A is 0.75 times the slope at C. We have to find the length of portion AB.
Assumptions
We assume that the beam is uniform and the cross-section of the beam is constant.
Free Body Diagram
We draw the free body diagram of the beam.
We take moments about A.
∑MA = 0
P(l/2 - AB) - RA * AB = 0
RA = P(l/2 - AB) / AB
Now, we take moments about C.
∑MC = 0
P(l/2 - BC) - RC * BC = 0
RC = P(l/2 - BC) / BC
Slope
The slope at A is given by:
θA = RA * AB2 / (2EI)
The slope at C is given by:
θC = RC * BC2 / (2EI)
We know that θA = 0.75 * θC
Substituting the values of RA and RC, we get:
AB3 / BC3 = 3 * 0.75 * (l - BC) / (2 * BC * (l - AB))
Solving for AB
We simplify the above equation to get:
AB3 = 2.25 * BC3 - 3 * BC2 * AB + 1.5 * l * BC2 - 2.25 * l * BC * AB
We solve this equation to get the value of AB.
Final Answer
The length of portion AB is equal to the value of AB obtained from the above equation.