# Civil Engineering - Strength of Materials - Discussion

Discussion Forum : Strength of Materials - Section 1 (Q.No. 32)
32.
If the width of a simply supported beam carrying an isolated load at its centre is doubled, the deflection of the beam at the centre is changed by
1/2
1/8
2
8
4
Explanation:
No answer description is available. Let's discuss.
Discussion:
9 comments Page 1 of 1.

Asif Ali Mundrani said:   7 years ago
If width=b,
Then, I=bh^3/12.
So,
(Def)max=PL^3/48EI

Now,
If width=b'=2b
Then,
I'=b'h^3/12=(2b)h^3/12=2(bh^3/12)
I'=2I

So,
(Def)'max = PL^3/48EI'
= PL^3/48E(2I)
= 1/2[PL^3/48EI]
(Def)'max = 1/2(Def)max

Hence Proved That,

If the width of a simply supported beam carrying an isolated load at its centre is doubled, the deflection of the beam at the centre is changed by (1/2).
(2)

Jinish said:   5 years ago
Deflection at centre = wl^3/48EI where I= bd^3/12.
Now b = 2b so I= 2bd^3/12 so I=bd^3/6.
Now deflection= WL^3/48E*(bd^3/6).
= WL^3* (6)/ bd^3 * (48).
= 1/8.
Now changed of beam = deflection normal/ deflection due to doubled width.
= 1/(1/8)
= 8.
So the centre is changed by 8 times.
(3)

Arsalan Wazir said:   5 years ago
Deflection has an inverse relation with square of the area of the beam .
deflection = 1/(A)2 ie A=bd.

So if we increase the width of the beam. Then;
Deflection =1/(2bd)2.
Deflection= 1/4(bd)2.
Deflection will be decreased by 1/4.

Bhavsingh said:   9 years ago
Deflection = wl^3/48EI.

So, from data b = 2b (double).

MOI = bd^3/12 = (2bd^3/12).

y = (wl^3)/48E (2bd^3)/12).

y = 1/2 ((wl^3)/(48EI)).
(2)

Baloch said:   8 years ago
Only by using 2nd moment,

= bd^3/12 divided by 2bd^3/12.
OR
bd^3/12 X 12/2bd^3 (after canceling what left behind is 1/2).
(1)

Mounika Reddy said:   1 decade ago
y = (wl^3)/(48*EI) FOR WIDTH b.

If it is 2b, I = 2bh^3/12.

y' = (wl^3)/(48*E2I).

So it is 1/2.

Govinda said:   5 years ago
Thanks to all for explaining the answer.

Amar said:   6 years ago
Thanks to all for the explanation.

Rohan said:   5 years ago
Good thanks @Asif Ali Mundrani.