Products
Rewards
from HOLOOLY

We are determined to provide the latest solutions related to all subjects FREE of charge!

Enjoy Limited offers, deals & Discounts by signing up to Holooly Rewards Program

HOLOOLY

HOLOOLY
TABLES

All the data tables that you may search for.

HOLOOLY
ARABIA

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

HOLOOLY
TEXTBOOKS

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

HOLOOLY
HELP DESK

Need Help? We got you covered.

## Q. 7.13

Determine the horizontal deflection at joint $C$ of the frame shown in Fig. $7.18(\mathrm{a})$ including the effect of axial deformations, by the virtual work method.

 Table 7.11 x Coordinate Segment Origin Limits (ft) M (k-ft) F(k) $M_{v} \left(k-ft\right)$ $F_{v} \left(k\right)$ AB A 0-15 -1.67x -12.50 $\frac{x}{2}$ $\frac{3}{4}$ BC B 0-20 $-25+ 12.5x-{x^{2} }$ -11.67 $7.5-\frac{3}{4} x$ $\frac{1}{2}$ DC D 0-15 11.67x -27.50 $\frac{x}{2}$ $-\frac{3}{4}$

## Verified Solution

The real and virtual systems are shown in Fig. 7.18(b) and (c), respectively. The $x$ coordinates used for determining the bending moment equations for the three members of the frame, $A B, B C$, and $C D$, are also shown in the figures. The equations for $M$ and $M_{v}$ obtained for the three members are tabulated in Table $7.11$ along with the axial forces $F$ and $F_{v}$ of the members. The horizontal deflection at joint $C$ of the frame can be determined by applying the virtual work expression given by Eq. (7.35):

\begin{aligned} 1\left(\Delta\right)=& \sum F_{v}\left(\frac{F L}{A E}\right)+\sum \int \frac{M_{v} M}{E I} d x \end{aligned}     (7.35)

\begin{aligned} 1\left(\Delta_{C}\right)=& \sum F_{v}\left(\frac{F L}{A E}\right)+\sum \int \frac{M_{v} M}{E I} d x \\ 1\left(\Delta_{C}\right)=& \frac{1}{A E}\left[\frac{3}{4}(-12.5)(15)+\frac{1}{2}(-11.67)(20)-\frac{3}{4}(-27.5)(15)\right] \\ &+\frac{1}{E I}\left[\int_{0}^{15} \frac{x}{2}(-1.67 x) d x\right.\\ &\left.+\int_{0}^{20}\left(7.5-\frac{3}{4} x\right)\left(-25+12.5 x-x^{2}\right) d x+\int_{0}^{15} \frac{x}{2}(11.67 x) d x\right] \\ (1 \mathrm{k}) \Delta_{C}=& \frac{52.08 \mathrm{k}^{2}-\mathrm{ft}}{A E}+\frac{9,375 \mathrm{k}^{2}-\mathrm{ft}^{3}}{E I} \end{aligned}

Therefore,

\begin{aligned} \Delta_{C} &=\frac{52.08 \mathrm{k}-\mathrm{ft}}{A E}+\frac{9,375 \mathrm{k}-\mathrm{ft}^{3}}{E I} \\ &=\frac{52.08}{(35)(29,000)}+\frac{9,375(12)^{2}}{(29,000)(1,000)} \\ &=0.00005+0.04655 \\ &=0.0466 \mathrm{ft}=0.559 \mathrm{in} \\ \Delta_{C} &=0.559 \mathrm{in} . \rightarrow \end{aligned}

Note that the magnitude of the axial deformation term is negligibly small as compared to that of the bending deformation term.