Civil Engineering - Strength of Materials - Discussion


If the length of a cantilever carrying an isolated load at its free end is doubled, the deflection of the free end will increase by

[A]. 8
[B]. 1/8
[C]. 1/3
[D]. 2
[E]. 3

Answer: Option A


No answer description available for this question.

Shraddha said: (Sep 22, 2015)  
Answer: 8 times.

Deflection at the free end due to point load = Wl*L*L/3EI.

Put 2L in place of L, it becomes 8 times.

Subhra Nandi said: (Sep 24, 2015)  
No answer is wrong its will be ANS-A (8).

Ashish Gusain said: (Jul 9, 2016)  
Option B is wrong.

Deflection = Wl.l.l/3EI,

Given that L = 2l,

Then deflection = W.2l.2l.2l/EI,

=> 8x W.l.l.l/3EI.

So option A is right.

Shweta Puri said: (Aug 28, 2016)  
Read the question carefully, they are not telling about length they said if the load at the free end is double it mean W = 2W. The deflection of the free end is WL^3/3EI when we put W = 2w it become1/2 which is not given in the answer but the right answer is 1/2.

Anni said: (Sep 23, 2016)  

I think, your answer is wrong.

Baloch said: (Nov 6, 2016)  
It's not doubling the load, it's doubling the length.

L^3, 2L^3 = 8, And is C.

Surjya said: (Jan 15, 2017)  
It will be option A i.e. 8.

Don Dai said: (Feb 9, 2017)  
More W & L, more deflection.
More B & D, less Deflection.

That's why 8 times.

Minesh Rathore said: (Jun 6, 2017)  
I think the Answer should be A.

Aaqib said: (Apr 20, 2018)  
Wl^3÷3Ei÷w (2l) ^3÷3Ei,
=1/8 Answer.

I think option B, please correct me if I m wrong.

Sushil Thakur said: (Sep 30, 2018)  
Right @Aaqib.

But it is not ask about ratio they ask about how much increase in deflection.

So, x1/x2 =1/8,
i.e x2=8x1 (answer).
I hope you understand.

Vishal Amit Patil said: (Jul 1, 2019)  

Here, it is mentioned to double the load not length.

Jinish said: (Aug 14, 2019)  
Deflection 1= wl^3/ 3EI and,
If l = 2 then deflection 2 = 8wl^3/3 EI.
So deflection = deflection 1/ deflection 2 = 1/8.
Correction answer is 1/8.
Here, mentioned length is to doubled l = 2l.

Dhru said: (Apr 21, 2020)  
Wl^3/3EI upon W(2l^3) /3EI.

Solve then the answer will be 1/8.

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