Problem 5.2. Linearly varying temperature change

Deflection of beam end, v(L) [mm]=

Rotation of beam end, φ(L) [rad] =

Step 1. Determine the curvature from the linearly varying temperature change.

Check curvatureHide curvature

Eq.(5-12)

$\kappa =\alpha ∆T_1=\alpha \frac{∆T}{h}=–1.2\times {10}^{–5}\frac{50}{20}=–3.00\times {10}^{–5}\frac{1}{\mathrm{mm}}$

Elongation at the top results in negative curvature, which is uniform along the beam’s length.

Step 2. Derive rotation and deflection functions from the curvature.

Check curvatureHide curvature

See Table 3.1.

The displacement functions can be obtained by the integration of the uniform curvature function:

$$\begin{gathered} \kappa =–\frac{d^{2}v}{d{x}^2} \\ \phi =\frac{dv}{dx}=–{\int }_{}\kappa \mathrm{d}x=\alpha \frac{∆T}{h}x+C_1 \\ v=–\iint \kappa \mathrm{d}x\mathrm{d}x=\alpha \frac{∆T}{h}\frac{x^2}{2}+C_{1}x+C_2 \end{gathered}$$

The constants, C₁ and C₂ are determined from the boundary conditions.

Table 3.2.

$$\begin{gathered} x=0: \frac{dv}{dx}=0 \to C_1=0 \\ v=0 \to C_2=0 \end{gathered}$$

The rotation and deflection functions are

$$\begin{gathered} \phi \left(x\right)=\alpha \frac{∆T}{h}x=3\times {10}^{–5}x \\ v\left(x\right)=\alpha \frac{∆T}{h}\frac{x^2}{2}=1.5\times {10}^{–5}{x}^2 \end{gathered}$$

Step 3. Calculate the displacements of the beam end.

Check end displacementsHide end displacements

The curvature arises from the linearly varying temperature change is constant along the beam’s length.

Eq.(5-12)

$\kappa =\alpha ∆T_1=\alpha \frac{∆T}{h}=–1.2\times {10}^{–5}\frac{50}{20}=–3.00\times {10}^{–5}\frac{1}{\mathrm{mm}}$

Elongation at the top results in negative curvature, which is uniform along the beam’s length. The displacement functions can be obtained by the integration of the uniform curvature function:

$$\begin{gathered} \kappa =–\frac{d^{2}v}{d{x}^2} \\ \phi =\frac{dv}{dx}=–{\int }_{}\kappa \mathrm{d}x=\alpha \frac{∆T}{h}x+C_1 \\ v=–\iint \kappa \mathrm{d}x\mathrm{d}x=\alpha \frac{∆T}{h}\frac{x^2}{2}+C_{1}x+C_2 \end{gathered}$$

See Table 3.1.

The constants, C₁ and C₂ are determined from the boundary conditions.

$$\begin{gathered} x=0: \frac{dv}{dx}=0 \to C_1=0 \\ v=0 \to C_2=0 \end{gathered}$$

Table 3.2.

The rotation and deflection functions are

$$\begin{gathered} \phi \left(x\right)=\alpha \frac{∆T}{h}x=3\times {10}^{–5}x \\ v\left(x\right)=\alpha \frac{∆T}{h}\frac{x^2}{2}=1.5\times {10}^{–5}{x}^2 \end{gathered}$$

The displacements of the beam end are$$\begin{gathered} \mathit{\phi }\mathbf{\left(}\mathit{L}\mathbf{\right)}=\alpha \frac{∆T}{h}L=3\times {10}^{–5}\times 600\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{018}\mathbf{ }\mathbf{rad}\mathbf{ }\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{03}\mathbf{°} \\ \mathit{v}\mathbf{\left(}\mathit{L}\mathbf{\right)}=\alpha \frac{∆T}{h}\frac{L^2}{2}=1.5\times {10}^{–5}\times {600}^2\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{40}\mathbf{ }\mathbf{mm} \end{gathered}$$