Question 20.2: Use the Mueller-Breslau principle to determine the shape of ...

In Fig. 20.4(b) we impose a unit shear displacement at the section C. In effect we are removing the resistance to shear of the beam at \mathrm{C} by cutting the beam at \mathrm{C}. We then apply positive shear forces to the two faces of the cut section in accordance with the sign convention of Section 3.2. Thus the beam to the right of \mathrm{C} is displaced downwards while the beam to the left of \mathrm{C} is displaced upwards. Since the slope of the influence line is the same on each side of \mathrm{C} we can determine the ordinates of the influence line by geometry. Hence, in Fig. 20.4(b)

\frac{c_{1} \mathrm{e}}{c_{1} a_{1}}=\frac{c_{1} \mathrm{f}}{c_{1} b_{1}}

Therefore

\mathrm{c}_{1} \mathrm{e}=\frac{\mathrm{c}_{1} \mathrm{a}_{1}}{\mathrm{c}_{1} \mathrm{~b}_{1}} \mathrm{c}_{1} \mathrm{f}=\frac{1}{2} \mathrm{c}_{1} \mathrm{f}

Further, since

\begin{gathered}\mathrm{c}_{1} \mathrm{e}+\mathrm{c}_{1} \mathrm{f}=1 \\\mathrm{c}_{1} \mathrm{e}=\frac{1}{3} \quad \mathrm{c}_{1} \mathrm{f}=\frac{2}{3}\end{gathered}

as before. The ordinate \mathrm{d}_{1} \mathrm{~g}\left(=\frac{1}{3}\right) follows.

In Fig. 20.4(c) we have, from the geometry of a triangle,

\alpha+\beta=1 \quad \text { (external angle }=\text { sum of opposite internal angles) }

Then, assuming that the angles \alpha and \beta are small so that their tangents are equal to the angles in radians

\frac{c_{2} h}{c_{2} a_{2}}+\frac{c_{2} h}{c_{2} b_{2}}=1

or

\mathrm{c}_{2} \mathrm{~h}\left(\frac{1}{2}+\frac{1}{4}\right)=1

whence

\mathrm{c}_{2} \mathrm{~h}=\frac{4}{3}

as in Fig. 20.2(d). The ordinate \mathrm{d}_{2} \mathrm{i}\left(=\frac{2}{3}\right) follows from similar triangles.