Chapter 4, Problem 1P

The figure shows a torsion bar OA fixed at O, simply supported at A, and connected to a cantilever AB. The spring rate of the torsion bar is k_T, in newton-meters per radian, and that of the cantilever is k_l, in newtons per meter. What is the overall spring rate based on the deflection y at point B?

To determine

The overall spring rate based on the deflection $y$ at point B.

Answer to Problem 1P

The overall spring rate based on the deflection $y$ at point B is $\underset{\_}{\frac{k_T\cdot k_l}{l^2{k}_l+k_T}}$.

Explanation of Solution

Write the expression for the stiffness of the torsion bar.

$\begin{array}{l}{k}_T=\frac{T}{\theta }\\ \theta =\frac{T}{k_T}\end{array}$ (I)

Here, the torsional stiffness of the bar is $k_T$, applied torque is $T$, and the angular deformation is $\theta$.

Write the expression of the torque on the torsional bar.

$T=Fl$ (II)

Here, the length of the bat is $l$ and the applied force is $F$.

Since the force $F$ produce a bending moment on the cantilever AB and torsional moment on the portion AO.

Substitute $Fl$ for $T$ in Equation (I).

$\theta =\frac{Fl}{k_T}$ (III)

Write the expression for stiffness of the cantilever portion AB.

$\begin{array}{c}{k}_l=\frac{F}{\delta }\\ \delta =\frac{F}{k_l}\end{array}$ (IV)

Here, the lateral stiffness of the bar is $k_l$ and the linear deformation is $\delta$.

Write the expression for net deflection of a combined system.

$y=\frac{F}{k}$ (V)

Here, total deflection is $y$, applied load is $F$, and the total stiffness of the bar is $k$.

The net defection of the combined system is the combination of the torsional deflection and lateral deflection.

Write the equation of net deflection of the combined system.

$y=l\theta +\delta$ (VI)

Conclusion:

Substitute the value of $\frac{Fl}{k_T}$ for $\theta$ and $\frac{F}{k_l}$ for $\delta$ in Equation (VI).

$y=\frac{F{l}^2}{k_T}+\frac{F}{k_l}$ (VII)

Substitute $\frac{F}{k}$ for $y$ in Equation (VII).

$\begin{array}{l}\frac{F}{k}=\frac{F{l}^2}{k_T}+\frac{F}{k_l}\\ \frac{1}{k}=\frac{l^2}{k_T}+\frac{1}{k_l}\\ \frac{1}{k}=\frac{l^2{k}_l+k_T}{k_T\cdot k_l}\\ k=\frac{k_T\cdot k_l}{l^2{k}_l+k_T}\end{array}$

Thus, the overall spring rate based on the deflection $y$ at point B is $\underset{\_}{\frac{k_T\cdot k_l}{l^2{k}_l+k_T}}$.