Problem 5.9. Creep deflection of timber beam

Final deflection, v₂ [mm]=

Step 1. Determine the creep coefficient.

Check creep coefficientHide creep coefficient

See Figure 5.15.

Creep coefficient varies with the time. Its values at times t₁ and t₂ are

$$\begin{gathered} {\phi }_1=\phi \left(t_1\right)=\frac{v_1–v_0}{v_0}=\frac{10–8}{8}=0.25 \\ {\phi }_2=\phi \left(t_2\right)=\frac{v_2–v_0}{v_0}=\frac{12–8}{8}=0.5 \end{gathered}$$

We may assume the creep function in the following form

$\phi ={\phi }_0\left(1–e^{–c_{1}t}\right)$

The constants (φ₀ and c₁)  can be expressed from the above two equations. Note however to solve this problem constants are not needed.

 

Step 2. Place half of the load on the beam at time t₀. Calculate final deflection from this part of the load.

Check final deflection from half load Hide deflection

See Figure 5.16a.

$v_{2,1}=v_0\left(1+{\phi }_2\right)=4\left(1+0.5\right)=6 \mathrm{mm}$

Step 3. Place other half of the load on the beam at time t₁. Calculate final deflection from this part of the load.

Check final deflection from other half load Hide deflection

See Figure 5.16b and Eq.(5-43).

$v_{2,2}=v_0\left(1+({\phi }_2–{\phi }_1\right)=4\left(1+(0.5–0.25)\right)=5 \mathrm{mm}$

Step 4. Summarize final deflection from both load parts.

Check final deflectionHide deflection

See Figure 5.16c.

Linear creep theory allows superposition, the final deflection results in

${\mathit{v}}_{\mathbf{2}}=v_{2,1}+v_{2,2}=6+5\mathbf{=}\mathbf{11}\mathbf{ }\mathbf{mm}$

Creep coefficient varies with the time. Its values at times t₁ and t₂ are

$$\begin{gathered} \phi \left(t_1\right)={\phi }_1=\frac{v_1–v_0}{v_0}=\frac{10–8}{8}=0.25 \\ \phi \left(t_2\right)={\phi }_2=\frac{v_2–v_0}{v_0}=\frac{12–8}{8}=0.5 \end{gathered}$$

See Figure 5.15.

We may assume the creep function in the following form

$\phi ={\phi }_0\left(1–e^{–c_{1}t}\right)$

The constants (φ₀ and c₁)  can be expressed from the above two equations. Note however to solve this problem constants are not needed.

Half of the load is placed on the beam at time t₀. Final deflection from this part of the load is

See Figure 5.16a.

$v_{2,1}=v_0\left(1+{\phi }_2\right)=4\left(1+0.5\right)=6 \mathrm{mm}$

Other half of the load is placed on the beam at time t₁. Final deflection from this part of the load is

See Figure 5.16b and Eq.(5-43).

$v_{2,2}=v_0\left(1+({\phi }_2–{\phi }_1\right)=4\left(1+(0.5–0.25)\right)=5 \mathrm{mm}$

Linear creep theory allows superposition, the final deflection results in

See Figure 5.16c.

${\mathit{v}}_{\mathbf{2}}=v_{2,1}+v_{2,2}=6+5\mathbf{=}\mathbf{11}\mathbf{ }\mathbf{mm}$