A plane frame PQR (fixed at 𝑷 and free at 𝑹) is shown in the figure. Both members (PQ and QR) have length, 𝑳, and flexural rigidity, 𝑬𝑰. Neglecting the effect of axial stress and transverse shear, the horizontal deflection at free end, 𝑹, is

The correct option/answer is:
$\frac{4F{L}^3}{3EI}$

Problem Analysis:
The provided diagram shows a plane frame PQR, which is fixed at support P and has a free end at R. The frame consists of two members:
1. A horizontal member PQ of length L.
2. A vertical member QR of length L.
Both members have a constant flexural rigidity EI. A horizontal force F acts to the right at the free end R. We are tasked with finding the horizontal deflection at the free end R, neglecting axial stress and transverse shear effects.

We can determine the horizontal deflection at the free end R using Castigliano's Theorem. According to Castigliano's Theorem, the deflection at any point in the direction of an applied force is equal to the partial derivative of the total strain energy of the structure with respect to that force:
${\delta }_R=\frac{\partial U}{\partial F}={\left(\frac{\partial U}{\partial F}\right)}_{RQ}+{\left(\frac{\partial U}{\partial F}\right)}_{QP}$
Where the deflection contribution from a member of length L under bending is given by:
$\frac{\partial U}{\partial F}={\int }_0^L\frac{M\left(x\right)·\frac{\partial M\left(x\right)}{\partial F}}{EI}dx$

Step 1: Deflection Contribution from Member RQ (Vertical Member)
Let us choose the origin at the free end R, with the coordinate x running vertically upwards along the member RQ from $x=0$ to $x=L$.
The bending moment at a section at distance x from R is:
$M_x=F·x$
Taking the partial derivative with respect to F:
$\frac{\partial M_x}{\partial F}=x$
Substituting these into the strain energy derivative formula for member RQ:
${\left(\frac{\partial U}{\partial F}\right)}_{RQ}={\int }_0^L\frac{\left(Fx\right)\left(x\right)}{EI}dx=\frac{F}{EI}{\left[\frac{x^3}{3}\right]}_0^L=\frac{F{L}^3}{3EI}$

Step 2: Deflection Contribution from Member QP (Horizontal Member)
Let us choose the origin at joint Q, with the coordinate x running horizontally from Q to P from $x=0$ to $x=L$.
The horizontal force F acting at R has a perpendicular distance of L (the height of member QR) to member QP. Therefore, the bending moment at any section of member QP is uniform and equal to:
$M_x=F·L$
Taking the partial derivative with respect to F:
$\frac{\partial M_x}{\partial F}=L$
Substituting these into the strain energy derivative formula for member QP:
${\left(\frac{\partial U}{\partial F}\right)}_{QP}={\int }_0^L\frac{\left(FL\right)\left(L\right)}{EI}dx=\frac{F{L}^2}{EI}{\left[x\right]}_0^L=\frac{F{L}^3}{EI}$

Step 3: Total Horizontal Deflection at R
Summing the individual deflection contributions from both members:
${\delta }_R=\frac{F{L}^3}{3EI}+\frac{F{L}^3}{EI}$
Combining the terms by finding a common denominator:
${\delta }_R=\left(\frac{1}{3}+1\right)\frac{F{L}^3}{EI}=\frac{4F{L}^3}{3EI}$