## Textbooks & Solution Manuals

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

## Tip our Team

Our Website is free to use.
To help us grow, you can support our team with a Small Tip.

## Holooly Tables

All the data tables that you may search for.

## Holooly Help Desk

Need Help? We got you covered.

## Holooly Arabia

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

Products

## Textbooks & Solution Manuals

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

## Holooly Arabia

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

## Holooly Help Desk

Need Help? We got you covered.

## Q. 10.22

A cantilever beam is shown Figure 10.40. Find the free end deflection and rotation of the beam. ## Verified Solution

(a) Free end deflection: Since we have been asked to determine free-end deflection (vertical) and there is no point load acting on the free-end, we therefore assume a dummy load acting at that point.
Then we calculate the strain energy with the dummy load present on the beam. Finally, after calculating deflection using Castigliano’s theorem, we set dummy load to be 0. Similar approach is taken to find rotation at the free-end. In that case, we have to apply a dummy moment. Rest of the calculation is the same as discussed in the previous solved example. Accordingly, we assume a dummy load at B and calculate the strain energy of the system:

From Figure 10.41, bending moment at any distance x from end A is

$M_x=-\left\lgroup P_x+\frac{w_{ o } x^2}{2} \right\rgroup$

Strain energy of the beam is

$U=\left\lgroup \frac{1}{2 E I} \right\rgroup \int_0^L M_x^2 d x$

Differentiating with respect to P, we get

$\frac{\partial U}{\partial P}=\left\lgroup \frac{1}{E I} \right\rgroup \int_0^L M_x \frac{\partial M_x}{\partial P} d x$

By Castigliano’s second theorem (thus, we tacitly assume the beam to be linearly elastic), we get by setting P = 0:

$\delta_{ B }=\underset{P \rightarrow 0}{Lt} \left\lgroup\frac{\partial U}{\partial P} \right\rgroup$

$=\underset{P \rightarrow 0}{Lt }\left\lgroup\frac{1}{E I} \right\rgroup \int_0^L \left\lgroup\frac{\partial M_x}{\partial x} \right\rgroup d x$

$=\underset{P \rightarrow 0}{Lt}\left\lgroup \frac{1}{E I} \right\rgroup \int_0^L\left\lgroup P x+\frac{w_0 x^2}{2} \right\rgroup \left(P+w_{ o } x\right) d x$

$=\left\lgroup \frac{1}{E I} \right\rgroup \int_0^L \frac{w_{ o }^2}{2} x^3 d x=\frac{w_{ o } L^4}{8 E I}$

Thus,

$\delta_{ B }=\frac{w_{ o } L^4}{8 E I}(\downarrow)$

(b) Rotation of free end: We put a dummy moment at the free end of the beam as shown in Figure 10.42.

Now bending moment at a distance x from end A is

$M_x=-\left\lgroup M_{ o }+\frac{w_{ o } x^2}{2} \right\rgroup \Rightarrow \frac{\partial M_x}{\partial M_{ o }}=(-1)$

Thus,

$\theta_{ B }=\underset{M_{ o } \rightarrow 0} {Lt}\frac{\partial U}{\partial M_{ o }}=\left\lgroup \frac{1}{E I} \right\rgroup \underset{M_{ o } \rightarrow 0} {Lt}\int_0^L M_x \frac{\partial M_x}{\partial x} d x$

Therefore,

$\theta_{ B }=\left\lgroup \frac{1}{E I} \right\rgroup \underset{M_{ o } \rightarrow 0}{Lt} \int_0\left\lgroup M_{ o }+\frac{w_{ o } x}{2} \right\rgroup d x$

$=\left\lgroup \frac{1}{E I} \right\rgroup \int_0^L \frac{w_{ o }}{2} x^2 d x=\frac{w_{ o } L^3}{6 E I}$

Free-end rotation of the beam is

$\theta_{ B }=\frac{w_{ o } L^3}{6 E I}$  