Civil Engineering - Strength of Materials - Discussion

The correct answer is: 1/2 times.

The deflection of a simply supported beam with a point load at the centre is given by the formula:
δ = (W * L³) / (48 * E * I).

For a rectangular beam, the moment of inertia (I) is given by:
I = (b * d³) / 12

If the width (b) is doubled (to 2b), the new moment of inertia (I') becomes:
I' = (2b * d³) / 12 = 2 * (b * d³) / 12 = 2I.

Now, let's look at how this affects the deflection. The new deflection (δ') with the doubled width is:
δ' = (W * L³) / (48 * E * I')
= (W * L³)/(48 * E * 2I)
= (1/2) * (W * L³)/(48 * E * I) = (1/2) * δ.

Therefore, if the width of the beam is doubled, the deflection is changed by *1/2 times*.

It should be 1/2.
If b = 2b.

Then deflection 1/2.
If d = 2d.

Then deflection is 1/8
Because:
∆ = WL^3/48EI.
I = bd^3/12.

1/2 is the correct answer.

For width it is 1/2.

For depth it is 1/8.

It should be 1/2.

The given answer is wrong.

In a simply supported beam when a) length is double deflection is double b) width is double then deflection is 1/2 and when depth is double then deflection 8 times.

So here is the explanation:

For b, it will be 1/2 and for d it will b 1/8.
Deflection = Wl3/48EI.
I = BD3/12.

And inverse i = 12/BD3.
For D it will be a cube
For B it will b as such.
If D is doubled (2)3 8.
Id B doubled then only (2).

1/2 is the right answer.

1/2 will be the answer.

Deflection =wl3/48EI,
I=bd3/12.

Width is (b) is doubled = 2b.
Then I=2bd3/12, I= bd3/6.

Deflection = wl3/48Ebd3/6.
= wl3/8Ebd3.

I agree with the given answer.