Problem 5.3. Temperature change of statically indeterminate structure

Reaction force of the hinged support, A [kN]=

Draw moment and deflection diagrams.

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Maximum deflection, v_max [mm]=

Similar problem is discussed in Figures 5.3 and 5.4.

Apply the force method.

Force method is presented in Section 9.1

Step 1. Choose a primary structure.

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Step 2. Determine the deflection and moment functions of the primary structure from the temperature load and from the redundant, respectively.

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From the linearly varying temperature change κ = αΔT/h curvature arise.

Eq.(5-12)

Bending moment and displacement function from the temperature change are:

See Problem 5.2

$$\begin{gathered} M_0\left(x\right)=0 \\ {v}_0\left(x\right)=\alpha \frac{∆T}{h}\frac{x^2}{2} \end{gathered}$$

Bending moment and displacement function from the unit load are:

$$\begin{gathered} M_1=(L–x) \\ {v}_1\left(x\right)=–\frac{L}{EI}\left(\frac{x^2}{2}–\frac{x^3}{6L}\right) \end{gathered}$$

Step 3. Determine redundant from the compatibility condition.

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Step 4. Calculate the support reactions.

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Step 5. Draw moment and deflection diagrams.

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Step 6. Calculate the maximum deflection.

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Similar problem is discussed in Figures 5.3 and 5.4.

We apply the force method.

Force method is presented in Section 9.1

The primary structure is chosen to be a cantilever, thus the redundant is the support reaction of the hinge.

The deflection and moment functions of the primary structure from the temperature load and from the redundant are shown in the Figure below.

From the linearly varying temperature change κ = αΔT/h curvature arise.

Eq.(5-12)

Bending moment and displacement function from the temperature change are:

See Problem 5.2

$$\begin{gathered} M_0\left(x\right)=0 \\ {v}_0\left(x\right)=\alpha \frac{∆T}{h}\frac{x^2}{2} \end{gathered}$$

Bending moment and displacement function from the unit load are:

$$\begin{gathered} M_1=(L–x) \\ {v}_1\left(x\right)=–\frac{L}{EI}\left(\frac{x^2}{2}–\frac{x^3}{6L}\right) \end{gathered}$$

The redundant is determined from the compatibility condition. Compatibility condition is written at the end of the cantilever, where there is no deflection of the original structure:

$$\begin{gathered} a_0=v_0\left(L\right), a_1=v_1\left(L\right) \\ {a}_0+a_{1}X=0 \to X=–\frac{a_0}{a_1}=\frac{\alpha {\displaystyle \frac{∆T{L}^2}{2h}}}{{\displaystyle \frac{L^3}{3EI}}}=\frac{3\alpha ∆TEI}{2hL} \end{gathered}$$

The support reaction at the hinge is equal to the redundant:

$\mathit{A}=X=\frac{3\alpha ∆TEI}{2hL}=\frac{3\times 1.2\times {10}^{–5}\times 50\times 210\times {10}^3\times {\displaystyle \frac{30\times {20}^3}{12}}}{2\times 20\times 600}\mathbf{=}\mathbf{315}\mathbf{ }\mathbf{N}\mathbf{ }$

The reactions at the built-in support are

$$\begin{gathered} B=–A=315 \mathrm{N} \\ M=AL=315\times 0.6=189 \mathrm{Nm} \end{gathered}$$

Moment and deflection functions are:

$$\begin{gathered} M\left(x\right)=A(L–x) \\ v\left(x\right)=\frac{\alpha ∆T}{h}\frac{x^2}{2}–\frac{AL}{EI}\left(\frac{x^2}{2}–\frac{x^3}{6L}\right) \end{gathered}$$

The deflection function reaches its maximum where the first derivative of the function is zero:

$$\begin{gathered} \frac{dv}{\mathrm{d}x}=\frac{{\displaystyle \alpha ∆T}}{{\displaystyle h}}x–\frac{{\displaystyle AL}}{{\displaystyle EI}}\left(x–\frac{{\displaystyle x^2}}{{\displaystyle 2L}}\right)=0 \to x=\frac{2}{3}L \\ {\mathit{v}}_{\mathbf{max}}=v\left(\frac{2L}{3}\right)=\frac{\alpha ∆T}{h}\frac{{\left(2/3L\right)}^2}{2}–\frac{AL}{EI}\left(\frac{{\left(2/3L\right)}^2}{2}–\frac{{\left(2/3L\right)}^3}{6L}\right)\mathbf{=}\mathbf{–}\mathbf{0}\mathbf{.}\mathbf{400}\mathbf{ }\mathbf{mm} \end{gathered}$$