# 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

Discussion:

9 comments Page 1 of 1.
Jinish said:
3 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.

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.

Arsalan Wazir said:
4 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.

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.

Govinda said:
4 years ago

Thanks to all for explaining the answer.

Rohan said:
4 years ago

Good thanks @Asif Ali Mundrani.

Amar said:
5 years ago

Thanks to all for the explanation.

Asif Ali Mundrani said:
6 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).

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).

Baloch said:
6 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).

= bd^3/12 divided by 2bd^3/12.

OR

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

Bhavsingh said:
7 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)).

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)).

Mounika Reddy said:
9 years 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.

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

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

So it is 1/2.

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