A cantilever carries a uniformly distributed load fixed end to the mid...
Given: A cantilever carries a uniformly distributed load fixed end to the midspan in the first case and UDL of same intensity from midspan to the free end in the second case.
To find: The ratio of maximum deflections in the two cases.
Solution:
Let us consider the cantilever beam as shown in the figure below:
![image.png](attachment:image.png)
Case 1: Uniformly distributed load fixed end to the midspan
- The total load on the beam = wL/2
- The load per unit length = w/2
- The maximum deflection occurs at the free end of the beam and is given by:
δmax = 5wL^4/384EI
Case 2: UDL of same intensity from midspan to the free end
- The total load on the beam = wL/2
- The load per unit length = w/2
- The maximum deflection occurs at 3L/4 from the fixed end and is given by:
δmax = 7wL^4/384EI
Ratio of maximum deflections in the two cases = δmax (case 2) / δmax (case 1)
Substituting the values of δmax (case 1) and δmax (case 2), we get:
Ratio = (7wL^4/384EI) / (5wL^4/384EI)
Simplifying the above expression, we get:
Ratio = 7/5
Therefore, the ratio of maximum deflections in the two cases is 7/5, which is approximately equal to 1.4 or 7/4.
Hence, the correct option is (D) 7/41.
A cantilever carries a uniformly distributed load fixed end to the mid...
Answer is option 'D'
Explanation:
For the 1st case: Cantilever carries UDL from fixed end to midspan.
Deflection for this case from Double Integration Method (Cantilever carries UDL for a distance 'a' from the fixed end) is,
= wa^4/8EI + Wa^3/6EI (l-a) (a = l/2)
Apply this formula, we get
Δ = 2.33wl^4 / 128EI. (1)
For the 2nd case: Cantilever Carries same UDL from midspan to free end.
Deflection for this 2nd case ( cantilever carries UDL for a distance 'a' from free end) from Double Integration Method is,
= w/24EI ( 3l^4 - 4la^3 + a^4) (a = l/2)
Apply this formula, we get
Δ = 2.56 wl^4 / 24 EI. (2)
find the ratio between two answer is 0.17 or (7/41)
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