A linear variation/expansion occurs in all types of beams when it is exposed to increasing temperatures. Most of the time, we consider there is a uniform temperature change in the beam, which leads to linear bending deflection. But, theoretically, temperature changes in non-uniform, the temperature of the beam across the depth will be different.

The deflection and slope of any beam (not particularly a simply supported one) primary depend on the load case it is subjected upon. If the load case varies, its deflection, slope, shear force and bending moment get changed.

This article will help you find the deflection and slope developed at any point of simply supported and cantilever beams, subjected to temperature change.

## Uniform Temperature Change in Simply Supported Beams

The simply supported beam is one of the most modest structures. The configuration of a simply supported beam is so simple having one hinge support at an end and roller support at the other end. With this setup the beam can only rotate horizontally, any vertical moment is restained.

In a case of uniform temperature change, the linear change in dimension is,

Change = L α (ΔT)

Where, L: Original span of the beam

α: Coefficient of Thermal Expansion

ΔT: Temperature Change

So, if the temperature change in a beam is uniform, we can directly apply the above formula to find the deflection due to temperature change. But if the temperature changes in non-uniform, the temperature of the beam across the depth will be different. Let us see that case as well.

## Non-Uniform Temperature Change in Simply Supported Beams

Consider, T0: Initial temperature of the beam

T1: Final temperature of the top of the beam

T2: Final temperature of the bottom of the beam

Assume, T2 > T1.

Take a small part **dx **for analysis.

Now, let us find the change in length at the top and bottom of the beam, considering both are two separate cases.

For the bottom part, the change in length = α dx (T2-T0). So, the new bottom length, Lb = dx + α dx (T2-T0)

For the bottom part, the change in length = α dx (T1-T0). So, the new bottom length, Lt = dx + α dx (T1-T0)

Due to different expansion in two sides of the beam, the beam will rotate and form an angle dΘ.

So, we can write:

dΘ x h = Lb - Lt = α dx (ΔT)

⇒ (dΘ/dx) = α ΔT/h (Equation 1)

According to the Double Integration formula of beam deflection,

d²y/dx² = (dΘ/dx) = M/EI (Equation 2)

Now, comparing equations 1 and 2,

**d****²****y/dx****² = α ΔT/h **(Equation 3)

This is the final formula, we use to get the slope and deflection of the beam due to nonuniform temperature variation in the beam.

Hope you are very clear with the concept. So what to wait then? Let's solve a question on a similar concept. Don't scroll the blog further, solve by yourself, then match your explanation/answer with mine!

## Example

Find the deflection at the centre of the beam, due to temperature variation at the bottom and top of the beam as shown.

### Step1: Finding the deflection function, y

According to the equation 3, d²y/dx² = α ΔT/h

Integrating once with respect to x, dy/dx = α ΔTx/h + C1

Integrating once again, y = α ΔTx²/2h + xC1 + C2

### Step 2: Apply Boundary Conditions

Applying the boundary/end conditions

(i) At x=0, y=0, so C2=0

(ii) At x=L, y=0, C1=(-α ΔTL/2h)

### Step 3: Solve the deflection function

Now putting the values of C1 and C2, in the deflection function,

y = α ΔTx²/2h - xα ΔTL/2h

As the question asks to find out the deflection at the centre of the beam, Put x = L/2

**So, y = (-α ΔTL²/8h) [DOWNWARD]**

All Clear now? Solve one more question.

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