Problem 6.9. Displacement of a frame

Problem a)

Support displacement, e [mm]=

Problem b)

Support displacement, e [mm]=

Problem a)

Step 1.  Draw moment diagrams from the uniformly distributed vertical load, and from a unit horizontal force placed at right support of the frame.

Check moment diagramsHide moment diagrams

Step 2.  Calculate the displacement with the aid of Castigliano’s II theorem. 

Check deflectionHide deflection

See Footnote j in Section 6.4.

$$\begin{gathered} \mathit{e}=\frac{1}{EI}{\int }_{}{M}_p{M}_1\mathrm{d}x=\frac{1}{EI}\left(\frac{2}{3}\frac{p{h}^2}{2}h\frac{5}{8}h+\frac{1}{2}\frac{p{h}^2}{2}lh\right)= \\ =\frac{1}{2.5\times {10}^5}\left(\frac{2}{3}\frac{5}{8}\frac{3.0\times 4.0^4}{2}+\frac{3.0\times 4.0^3\times 6.0}{4}\right)=\frac{448}{EI}=0.00179 \mathrm{m}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{79}\mathbf{mm} \end{gathered}$$

Problem b)

Step 1.  Draw moment diagrams from the concentrated horizontal load, and from a unit horizontal force placed at right support of the frame.

Check moment diagramsHide moment diagrams

Step 2. Calculate the displacement with the aid of Castigliano’s II theorem. 

Check displacementHide displacement

See Footnote j in Section 6.4.

$$\begin{gathered} \mathit{e}=\frac{1}{EI}{\int }_{}{M}_p{M}_1\mathrm{d}x=\frac{1}{EI}\left(A_1{h}_1+A_2{h}_2+A_3{h}_3\right)=\frac{1}{EI}\left(\frac{F{h}^2}{2}\frac{2}{3}h+Fhlh+\frac{F{h}^2}{2}\frac{2}{3}h\right)= \\ =\frac{F}{EI}\left(2\frac{4.0^3}{3}+4.0^2\times 6.0\right)=\frac{138.7F}{EI}=\frac{138.7\times 1000}{2.5\times {10}^5}=0.000555 \mathrm{m}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{555}\mathbf{ }\mathbf{mm} \end{gathered}$$

Problem a)

Moment diagrams are given in the Figure: from the uniformly distributed vertical load and from a unit horizontal force placed at right support.

The displacement is calculated with the aid of Castigliano’s II theorem. 

See Footnote j in Section 6.4.

$$\begin{gathered} \mathit{e}=\frac{1}{EI}{\int }_{}{M}_p{M}_1\mathrm{d}x=\frac{1}{EI}\left(\frac{2}{3}\frac{p{h}^2}{2}h\frac{5}{8}h+\frac{1}{2}\frac{p{h}^2}{2}lh\right)= \\ =\frac{1}{2.5\times {10}^5}\left(\frac{2}{3}\frac{5}{8}\frac{3.0\times 4.0^4}{2}+\frac{3.0\times 4.0^3\times 6.0}{4}\right)=\frac{448}{EI}=0.00179 \mathrm{m}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{79}\mathbf{mm} \end{gathered}$$

Problem b)

Moment diagrams are given in the Figure: from the concentrated horizontal load and from a unit horizontal force placed at right support of the frame.

The displacement is calculated with the aid of Castigliano’s II theorem. 

See Footnote j in Section 6.4.

$$\begin{gathered} \mathit{e}=\frac{1}{EI}{\int }_{}{M}_p{M}_1\mathrm{d}x=\frac{1}{EI}\left(A_1{h}_1+A_2{h}_2+A_3{h}_3\right)=\frac{1}{EI}\left(\frac{F{h}^2}{2}\frac{2}{3}h+Fhlh+\frac{F{h}^2}{2}\frac{2}{3}h\right)= \\ =\frac{F}{EI}\left(2\frac{4.0^3}{3}+4.0^2\times 6.0\right)=\frac{138.7F}{EI}=\frac{138.7\times 1000}{2.5\times {10}^5}=0.000555 \mathrm{m}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{555}\mathbf{ }\mathbf{mm} \end{gathered}$$