Question

The beam of an overall depth 250 mm (shown below) is used in a buildingsubjected to twodifferent thermal environments. The temperatures at the top and bottom surfaces of the beam are36c and 72c respectively. Considering coefficient of thermal expansion () as 1.5010 5 per c,the vertical deflection of the beam (in mm) at its mid-span due to temperature gradient is _

Answer 1

The vertical deflection of the beam its mid-span due to temperature gradient is 250 mm.

Given:

Depth of the beam = h = 250 mm

Temperature of top surface = 36°C

Temperature of bottom surface = 72°C

Coefficient of thermal expansion $\bold{=\alpha=1.50\times10^{-5}/ \° C}$

To find:

Vertical deflection of the beam = ?

Solution:

When the beam is vertically deflected, then it forms a semi circle.

From the property of the circle is:

$\bold{(2R-\delta)\delta=\frac{L}{2}\times\frac{L}{2}}$

Where,

$\bold{\delta}$ = Deflection

R = Resistance of the beam

L = Length of the beam

Since, the deflection is very small, $\bold{\delta}$ is neglected.

$\bold{\Rightarrow 2R\delta=\frac{L^2}{4}}$

$\bold{\therefore \delta = \frac{L^2}{8R}}$

Now, the resistance of the beam is given by the formula:

$\bold{R=\frac{h}{\alpha T}}$

On substituting the resistance formula in deflection formula, we get,

$\bold{\delta=\frac{L^2\times \alpha T}{8\times h}}$

On substituting the given values, we get,

$\bold{\delta=\frac{1.5\times10^{-5}\times3^2\times(72^{\circ}-36^{\circ})}{8\times250\times10^{-3}}}$

$\bold{\therefore \delta = 250 \ mm}$