Problem 3.22. Unkown load function

Derive load function, t (x)=

Check resultHide result

Step 1. Give the Saint- Venant torque.

Show Saint-Venant torqueHide Saint Venant torque

Eq.(3-115)

$T_{SV}=G{I}_{\mathrm{t}}\frac{d\psi }{dx}=G{I}_{\mathrm{t}}\frac{T_0}{G{I}_t\mu }\left(\mu -\mu e^{\mu x}\right)=T_0\left(1-e^{\mu x}\right)$

Step 2. Determine the bimoment function.

Show bimomentHide bimoment

Eq.(3-121)

$M_{\omega }=-E{I}_{\omega }\frac{d^2\psi }{d{x}^2}=-E{I}_{\omega }\frac{T_0}{G{I}_t\mu }\left(-{\mu }^2{e}^{\mu x}\right)=T_0\frac{e^{\mu x}}{\mu }$

Step 3. Express restrained warping induced torque from the bimoment.

Show restrained warping induced torqueHide restrained warping induced torque

Eq.(3-125)

$T_{\omega }=\frac{M_{\omega }}{dx}=T_0{e}^{\mu x}$

Step 4. Write the total torque function.

Show total torqueHide total torque

Eq.(3-126)

$T=T_{SV}+T_{\omega }=T_0\left(1-e^{\mu x}\right)+T_0{e}^{\mu x}=T_0$

Step 5. Show the load function.

Show loadHide load

Equilibrium is given by Eq.3-127

$t\left(x\right)=-\frac{d{T}_{SV}}{dx}-\frac{d{T}_{\omega }}{dx}=\mu T_0{e}^{\mu x}-\mu T_0{e}^{\mu x}=0$

The beam is unloaded along its length. 

The end loads can be determined from the boundary conditions. Check which boundary conditions are satisfied by the displacement functions.

See Table 3.11.

At x = 0 

$$\begin{gathered} \psi \left(0\right)=\frac{T_0}{G{I}_{\mathrm{t}}\mu }\left(1+\mu \times 0-e^{\mu \times 0}\right)=0 \\ \frac{d\psi \left(0\right)}{dx}=\frac{T_0}{G{I}_{\mathrm{t}}}\left(1-e^{\mu \times 0}\right)=0 \end{gathered}$$

Thus the x = 0 end is built-in.

At x = L

$$\begin{gathered} \psi \left(L\right)=\frac{T_0}{G{I}_{\mathrm{t}}\mu }\left(1+\mu L-e^{\mu \times L}\right)\ne 0 \\ {M}_{\omega }\left(L\right)=T_0\frac{e^{\mu L}}{\mu } \\ {T}_{\omega }\left(L\right)=T_0 \end{gathered}$$

This end must be connected to an other beam, which transmits the above end loads.

The total torque is the sum of the Saint- Venant torque and the restrained warping induced torque.

The Saint Venant torque is

Eq.(3-115)

The restrained warping induced torque is expressed from the bimoment function:

Eq.(3-121)

$M_{\omega }=-E{I}_{\omega }\frac{d^2\psi }{d{x}^2}=-E{I}_{\omega }\frac{T_0}{G{I}_t\mu }\left(-{\mu }^2{e}^{\mu x}\right)=T_0\frac{e^{\mu x}}{\mu }$

Eq.(3-125)

$T_{\omega }=\frac{M_{\omega }}{dx}=T_0{e}^{\mu x}$

The total torque function is

Equation 3-126

The load function is determined from the equilibrium:

See Eq.3-127

Thus the beam is unloaded along its length. 

The end loads can be determined from the boundary conditions. Let us check which boundary conditions are satisfied by the displacement functions.

See Table 3.11.

$$\begin{gathered} \psi \left(0\right)=\frac{T_0}{G{I}_{\mathrm{t}}\mu }\left(1+\mu \times 0-e^{\mu \times 0}\right)=0 \\ \frac{d\psi \left(0\right)}{dx}=\frac{T_0}{G{I}_{\mathrm{t}}}\left(1-e^{\mu \times 0}\right)=0 \end{gathered}$$

Thus the x = 0 end is built-in.

At x = L

$$\begin{gathered} \psi \left(L\right)=\frac{T_0}{G{I}_{\mathrm{t}}\mu }\left(1+\mu L-e^{\mu \times L}\right)\ne 0 \\ {M}_{\omega }\left(L\right)=T_0\frac{e^{\mu L}}{\mu } \\ {T}_{\omega }\left(L\right)=T_0 \end{gathered}$$

This end must be connected to an other beam, which transmits the above end loads.