Example 7.1
Derive the stiffness
matrix of the tapered bar element (which deforms in the axial direction) shown
in next figure. The diameter of the bar decreases from D to d over its length.
As diameter h(x)=D at
x=0 and h(x)=d at x=l, we have
$$\begin{align}
& h(x)=D+\left( \frac{d-D}{l} \right)x \\
& I)\text{Stiffness matrix:} \\
& V(t)=\text{strain energy of element}=\frac{1}{2}{{\int\limits_{0}^{l}{EA\left( \frac{\partial u}{\partial x} \right)}}^{2}}dx \\
& u(x,t)=\left( 1-\frac{x}{l} \right){{u}_{1}}(t)+\left( \frac{x}{l} \right){{u}_{2}}\left( t \right) \\
& A\left( x \right)=\frac{\pi {{h}^{2}}}{4}=\frac{\pi }{4}\left[ {{D}^{2}}+2D\left( \frac{d-D}{l} \right)x+{{\left( \frac{d-D}{l} \right)}^{2}}{{x}^{2}} \right] \\
& \text{Thus the strain energy expression becomes:} \\
& V=\frac{\pi E}{24l}\left( {{D}^{2}}+{{d}^{2}}+dD \right)\left( u_{1}^{2}+u_{2}^{2}-2{{u}_{1}}{{u}_{2}} \right) \\
& V=\frac{1}{2}{{\overrightarrow{u}}^{T}}\left[ k \right]\overrightarrow{u}=\frac{1}{2}\left( {{u}_{1}}\text{ }{{u}_{2}} \right)\left[ \begin{matrix}
{{k}_{11}} & {{k}_{12}} \\
{{k}_{21}} & {{k}_{22}} \\
\end{matrix} \right]\left\{ \begin{matrix}
{{u}_{1}} \\
{{u}_{2}} \\
\end{matrix} \right\} \\
& \text{This gives the element matrix as:} \\
& \left[ k \right]=\frac{\pi E}{12l}\left( {{D}^{2}}+{{d}^{2}}+dD \right)\left[ \begin{matrix}
1 & -1 \\
-1 & 1 \\
\end{matrix} \right] \\
& II)\text{Consistent mass matrix:} \\
& T(t)=\frac{1}{2}{{\int\limits_{0}^{l}{\rho A\left( \frac{\partial u}{\partial t} \right)}}^{2}}dx \\
& \dot{u}(x,t)=\frac{\partial u}{\partial t}\left( x,t \right)=\left( 1-\frac{x}{l} \right){{{\dot{u}}}_{1}}\left( t \right)+\left( \frac{x}{l} \right){{{\dot{u}}}_{2}}\left( t \right) \\
& T=\frac{1}{2}\int\limits_{0}^{l}{\rho \frac{\pi }{4}\left\{ {{D}^{2}}+2D\left( \frac{d-D}{l} \right)x+{{\left( \frac{d-D}{l} \right)}^{2}}{{x}^{2}} \right\}}\left\{ \dot{u}_{1}^{2}+x\left( -\frac{2}{l}\dot{u}_{1}^{2}-\frac{2}{l}{{{\dot{u}}}_{1}}{{{\dot{u}}}_{2}} \right) \right. \\
& +{{x}^{2}}\left. \left( \frac{\dot{u}_{1}^{2}}{{{l}^{2}}}+\frac{\dot{u}_{2}^{2}}{{{l}^{2}}}-\frac{2{{{\dot{u}}}_{1}}{{{\dot{u}}}_{2}}}{{{l}^{2}}} \right) \right\}dx \\
& T=\frac{1}{2}\left( {{{\dot{u}}}_{1}}\text{ }{{{\dot{u}}}_{2}} \right)\left[ \begin{matrix}
{{m}_{11}} & {{m}_{12}} \\
{{m}_{21}} & {{m}_{22}} \\
\end{matrix} \right]\left\{ \begin{matrix}
{{{\dot{u}}}_{1}} \\
{{{\dot{u}}}_{2}} \\
\end{matrix} \right\} \\
& \left[ {{m}_{c}} \right]=\frac{\pi \rho l}{4}\left[ \begin{matrix}
\left( \frac{{{D}^{2}}}{5}+\frac{{{d}^{2}}}{30}+\frac{Dd}{10} \right) & \left( \frac{{{D}^{2}}}{20}+\frac{{{d}^{2}}}{20}+\frac{Dd}{15} \right) \\
\left( \frac{{{D}^{2}}}{20}+\frac{{{d}^{2}}}{20}+\frac{Dd}{15} \right) & \left( \frac{{{D}^{2}}}{30}+\frac{{{d}^{2}}}{5}+\frac{Dd}{10} \right) \\
\end{matrix} \right] \\
\end{align}$$
$$\begin{align}
& III)\text{Lumped mass matrix:} \\
& M=\frac{\pi \rho }{4}\int\limits_{0}^{l}{{{h}^{2}}}\left( x \right)dx \\
& M=\frac{\pi \rho l}{4}\left( {{D}^{2}}+{{d}^{2}}Dd \right) \\
& M=\frac{\pi \rho l}{4}\left[ \begin{matrix}
\left( {{D}^{2}}+{{d}^{2}}+Dd \right) & 0 \\
0 & \left( {{D}^{2}}+{{d}^{2}}+Dd \right) \\
\end{matrix} \right] \\
\end{align}$$
Example 7.2
Derive the stiffness
matrix of the bar element in longitudinal vibration whose cross-sectional area
varies as A(x) = A0Exp(-(x/t)) where A0 is the area at the root.
Assume linear
displacement model
$$\begin{align}
& u(x)={{\alpha }_{1}}+{{\alpha }_{2}}x={{U}_{1}}+\left( \frac{{{U}_{2}}-{{U}_{1}}}{l} \right)x \\
& {{\varepsilon }_{x}}=\frac{\partial u}{\partial x}=\frac{{{U}_{2}}-{{U}_{1}}}{l} \\
& {{\sigma }_{x}}=E{{\varepsilon }_{x}}=E\left( \frac{{{U}_{2}}-{{U}_{1}}}{l} \right) \\
& V(x)=\iiint\limits_{{{V}^{(e)}}}{\frac{1}{2}}{{\sigma }_{x}}{{\varepsilon }_{x}}dV=\frac{1}{2}\int\limits_{x=0}^{l}{E}{{\left( \frac{{{U}_{2}}-{{U}_{1}}}{l} \right)}^{2}}A(x)dx \\
& V(x)=\frac{E}{2{{l}^{2}}}{{\left( {{U}_{2}}-{{U}_{1}} \right)}^{2}}l{{A}_{0}}\left( 1-\frac{1}{e} \right) \\
& V(x)=\frac{1}{2}{{\overrightarrow{U}}^{T}}\left[ K \right]\overrightarrow{U}=\frac{1}{2}\left( {{U}_{1}}\text{ }{{U}_{2}} \right)\left[ k \right]\left\{ \begin{matrix}
{{U}_{1}} \\
{{U}_{2}} \\
\end{matrix} \right\} \\
& \left[ k \right]=\frac{E{{A}_{0}}}{l}\left( 0.6321 \right)\left[ \begin{matrix}
1 & -1 \\
-1 & 1 \\
\end{matrix} \right] \\
\end{align}$$
Example 7.3
The tapered cantilever
beam shown in next figure is used as a spring to carry a load P. Derive the
stiffness matrix of the beam using a one-element idealization. Assume b = 10
cm, t = 2.5 cm, l = 2 m, E = 2.07 * 1011 N/m2, and P = 1000 N.
$$\begin{align}
& w(x)=B-\left( \frac{B-b}{l} \right)x \\
& I(x)=\frac{1}{12}w(x){{t}^{3}}=\frac{B{{t}^{3}}}{12}-\left( \frac{\left( B-b \right){{t}^{3}}}{12l} \right)x={{a}_{1}}-{{a}_{2}}x \\
& {{a}_{1}}=\frac{B{{t}^{3}}}{12};{{a}_{2}}=\frac{\left( B-b \right){{t}^{3}}}{12l}; \\
& \text{Deflection of beam:} \\
& w(x)=\sum\limits_{i=1}^{4}{{{N}_{i}}\left( x \right)}{{W}_{i}} \\
& {{N}_{1}}(x)=1-3{{\left( \frac{x}{l} \right)}^{2}}+2{{\left( \frac{x}{l} \right)}^{3}} \\
& {{N}_{2}}(x)=x-2l{{\left( \frac{x}{l} \right)}^{2}}+l{{\left( \frac{x}{l} \right)}^{3}} \\
& {{N}_{3}}(x)=3{{\left( \frac{x}{l} \right)}^{2}}-2{{\left( \frac{x}{l} \right)}^{3}} \\
& {{N}_{3}}(x)=-l{{\left( \frac{x}{l} \right)}^{2}}+l{{\left( \frac{x}{l} \right)}^{3}} \\
& w(x)={{N}_{1}}(x)+{{N}_{2}}(x)+{{N}_{3}}(x)+{{N}_{4}}(x) \\
& w(x)=1+x-3l{{\left( \frac{x}{l} \right)}^{2}}+2l{{\left( \frac{x}{l} \right)}^{3}} \\
& \text{Strain energy of element is given by} \\
& V=\frac{1}{2}{{\int\limits_{0}^{l}{EI(x)\left( \frac{{{d}^{2}}w(x)}{d{{x}^{2}}} \right)}}^{2}}dx \\
& \frac{{{d}^{2}}{{N}_{1}}(x)}{d{{x}^{2}}}=-\frac{6}{{{l}^{2}}}+\frac{12}{{{l}^{3}}}x,\frac{{{d}^{2}}{{N}_{2}}(x)}{d{{x}^{2}}}=-\frac{4}{l}+\frac{6}{{{l}^{2}}}x, \\
& \frac{{{d}^{2}}{{N}_{3}}(x)}{d{{x}^{2}}}=\frac{6}{{{l}^{2}}}+\frac{12}{{{l}^{3}}}x,\frac{{{d}^{2}}{{N}_{4}}(x)}{d{{x}^{2}}}=-\frac{2}{l}+\frac{12}{{{l}^{2}}}x, \\
& {{\left( \frac{{{d}^{2}}w(x)}{d{{x}^{2}}} \right)}^{2}}=c_{1}^{2}+c_{2}^{2}+2{{c}_{1}}{{c}_{2}}x={{\left( {{c}_{1}}+{{c}_{2}}x \right)}^{2}} \\
& {{c}_{1}}=-\frac{6}{{{l}^{2}}}{{W}_{1}}-\frac{4}{l}{{W}_{2}}+\frac{6}{{{l}^{2}}}{{W}_{3}}-\frac{2}{l}{{W}_{4}} \\
& {{c}_{2}}=\frac{12}{{{l}^{3}}}{{W}_{1}}+\frac{6}{{{l}^{2}}}{{W}_{2}}-\frac{12}{{{l}^{3}}}{{W}_{3}}+\frac{6}{{{l}^{2}}}{{W}_{4}} \\
\end{align}$$
$$\begin{align}
& V=\frac{1}{2}E\left\{ W_{1}^{2}\left[ \frac{36}{{{l}^{4}}}\left( {{a}_{1}}l-{{a}_{2}}\frac{{{l}^{2}}}{2} \right)-\frac{72}{{{l}^{5}}} \right. \right.\left. +\frac{144}{{{l}^{6}}}\left( {{a}_{1}}\frac{{{l}^{3}}}{3}-{{a}_{2}}\frac{{{l}^{4}}}{4} \right) \right]+W_{2}^{2}\left[ \frac{16}{{{l}^{2}}}\left( {{a}_{1}}l-{{a}_{2}}\frac{{{l}^{2}}}{2} \right) \right. \\
& \left. -\frac{24}{{{l}^{4}}}\left( {{a}_{1}}{{l}^{2}}-2{{a}_{2}}\frac{{{l}^{3}}}{3} \right)+\frac{36}{{{l}^{4}}}\left( {{a}_{1}}\frac{{{l}^{3}}}{3}-{{a}_{2}}\frac{{{l}^{4}}}{4} \right) \right]+W_{3}^{2}\left[ \frac{36}{{{l}^{4}}} \right.\left( {{a}_{1}}l-{{a}_{2}}\frac{{{l}^{2}}}{2} \right)-\frac{72}{{{l}^{5}}}\left( {{a}_{1}}{{l}^{2}}-2{{a}_{2}}\frac{{{l}^{3}}}{3} \right) \\
& \left. +\frac{144}{{{l}^{6}}}\left( {{a}_{1}}\frac{{{l}^{3}}}{3}-{{a}_{2}}\frac{{{l}^{4}}}{4} \right) \right]+W_{4}^{2}\left[ \frac{4}{{{l}^{2}}} \right.\left( {{a}_{1}}l-{{a}_{2}}\frac{{{l}^{2}}}{2} \right)\left. -\frac{12}{{{l}^{3}}}\left( {{a}_{1}}{{l}^{2}}-2{{a}_{2}}\frac{{{l}^{3}}}{3} \right)+\frac{36}{{{l}^{4}}}\left( {{a}_{1}}\frac{{{l}^{3}}}{3}-{{a}_{2}}\frac{{{l}^{4}}}{4} \right) \right] \\
& +2{{W}_{1}}{{W}_{2}}\left[ \frac{24}{{{l}^{3}}}\left( {{a}_{1}}l-{{a}_{2}}\frac{{{l}^{2}}}{2} \right) \right.-\frac{42}{{{l}^{4}}}\left( {{a}_{1}}{{l}^{2}}-2{{a}_{2}}\frac{{{l}^{3}}}{3} \right)\left. +\frac{72}{{{l}^{5}}}\left( {{a}_{1}}\frac{{{l}^{3}}}{3}-{{a}_{2}}\frac{{{l}^{4}}}{4} \right) \right] \\
& +2{{W}_{1}}{{W}_{3}}\left[ -\frac{36}{{{l}^{4}}}\left( {{a}_{1}}l-{{a}_{2}}\frac{{{l}^{2}}}{2} \right)+\frac{72}{{{l}^{5}}}\left( {{a}_{1}}{{l}^{2}}-2{{a}_{2}}\frac{{{l}^{3}}}{3} \right)-\frac{144}{{{l}^{6}}}\left( {{a}_{1}}\frac{{{l}^{3}}}{3}-{{a}_{2}}\frac{{{l}^{4}}}{4} \right) \right] \\
& +2{{W}_{1}}{{W}_{4}}\left[ \frac{12}{{{l}^{3}}}\left( {{a}_{1}}l-{{a}_{2}}\frac{{{l}^{2}}}{2} \right) \right.-\frac{30}{{{l}^{4}}}\left( {{a}_{1}}{{l}^{2}}-2{{a}_{2}}\frac{{{l}^{3}}}{3} \right)\left. +\frac{72}{{{l}^{5}}}\left( {{a}_{1}}\frac{{{l}^{3}}}{3}-{{a}_{2}}\frac{{{l}^{4}}}{4} \right) \right] \\
& +2{{W}_{2}}{{W}_{3}}\left[ -\frac{24}{{{l}^{3}}}\left( {{a}_{1}}l-{{a}_{2}}\frac{{{l}^{2}}}{2} \right) \right.+\frac{42}{{{l}^{4}}}\left( {{a}_{1}}{{l}^{2}}-2{{a}_{2}}\frac{{{l}^{3}}}{3} \right)\left. -\frac{72}{{{l}^{5}}}\left( {{a}_{1}}\frac{{{l}^{3}}}{3}-{{a}_{2}}\frac{{{l}^{4}}}{4} \right) \right] \\
& +2{{W}_{2}}{{W}_{4}}\left[ \frac{8}{{{l}^{2}}} \right.\left( {{a}_{1}}l-{{a}_{2}}\frac{{{l}^{2}}}{2} \right)\left. -\frac{18}{{{l}^{3}}}\left( {{a}_{1}}{{l}^{2}}-2{{a}_{2}}\frac{{{l}^{3}}}{3} \right)+\frac{36}{{{l}^{4}}}\left( {{a}_{1}}\frac{{{l}^{3}}}{3}-{{a}_{2}}\frac{{{l}^{4}}}{4} \right) \right] \\
& \left. +2{{W}_{3}}{{W}_{4}}\left[ -\frac{12}{{{l}^{3}}}\left( {{a}_{1}}l-{{a}_{2}}\frac{{{l}^{2}}}{2} \right) \right.+\frac{30}{{{l}^{4}}}\left( {{a}_{1}}{{l}^{2}}-2{{a}_{2}}\frac{{{l}^{3}}}{3} \right)\left. -\frac{72}{{{l}^{5}}}\left( {{a}_{1}}\frac{{{l}^{3}}}{3}-{{a}_{2}}\frac{{{l}^{4}}}{4} \right) \right] \right\} \\
& Bywriting:V=\frac{1}{2}{{\overrightarrow{W}}^{T}}\left[ k \right]\overrightarrow{W} \\
& \overrightarrow{W}=\left\{ \begin{matrix}
{{W}_{1}} \\
{{W}_{2}} \\
{{W}_{3}} \\
{{W}_{4}} \\
\end{matrix} \right\},\text{thestiffness matrix can be identified}\text{.} \\
\end{align}$$
$$\begin{align}
& \text{Defining }{{\text{a}}_{1}}=\frac{B{{t}^{3}}}{12},{{d}_{1}}=\frac{\left( B-b \right){{t}^{3}}}{12},{{a}_{2}}=\frac{{{d}_{1}}}{l},{{a}_{2}}l={{d}_{1}} \\
& \text{the elements of }\left[ k \right]\text{can be expressed as:} \\
& {{k}_{11}}=E\left\{ {{a}_{1}}\left( \frac{12}{{{l}^{3}}} \right)-{{d}_{1}}\left( \frac{6}{{{l}^{3}}} \right) \right\},{{k}_{22}}=E\left\{ {{a}_{1}}\left( \frac{4}{l} \right)-{{d}_{1}}\left( \frac{1}{l} \right) \right\}, \\
& {{k}_{33}}=E\left\{ {{a}_{1}}\left( \frac{12}{{{l}^{3}}} \right)-{{d}_{1}}\left( \frac{6}{{{l}^{3}}} \right) \right\},{{k}_{11}}=E\left\{ {{a}_{1}}\left( \frac{4}{l} \right)-{{d}_{1}}\left( \frac{3}{l} \right) \right\}, \\
& {{k}_{12}}=E\left\{ {{a}_{1}}\left( \frac{6}{{{l}^{2}}} \right)-{{d}_{1}}\left( \frac{2}{{{l}^{2}}} \right) \right\},{{k}_{13}}=E\left\{ {{a}_{1}}\left( -\frac{12}{{{l}^{3}}} \right)+{{d}_{1}}\left( \frac{6}{{{l}^{3}}} \right) \right\}, \\
& {{k}_{14}}=E\left\{ {{a}_{1}}\left( \frac{6}{{{l}^{2}}} \right)-{{d}_{1}}\left( \frac{4}{{{l}^{2}}} \right) \right\},{{k}_{23}}=E\left\{ {{a}_{1}}\left( -\frac{6}{{{l}^{2}}} \right)+{{d}_{1}}\left( \frac{2}{{{l}^{2}}} \right) \right\}, \\
& {{k}_{24}}=E\left\{ {{a}_{1}}\left( \frac{2}{l} \right)-{{d}_{1}}\left( \frac{1}{l} \right) \right\},{{k}_{34}}=E\left\{ {{a}_{1}}\left( -\frac{6}{{{l}^{2}}} \right)+{{d}_{1}}\left( \frac{4}{{{l}^{2}}} \right) \right\}, \\
& b)Stressesinducedinthebeam: \\
& B=0.25m \\
& b=0.10m \\
& t=0.025m \\
& l=2\text{m} \\
& E=2.07\cdot {{10}^{11}}\text{N/}{{\text{m}}^{\text{2}}} \\
& P=1000\text{N} \\
& {{a}_{1}}=32552.0833\cdot {{10}^{-11}} \\
& {{d}_{1}}19531.25\cdot {{10}^{-11}} \\
& \text{Stiffness matrix can be computed as:} \\
& \left[ k \right]=\begin{matrix}
\begin{matrix}
{{W}_{1}}\text{ } & {{W}_{2}}\text{ } & {{W}_{3}}\text{ } & {{W}_{4}} \\
\end{matrix} \\
\left[ \begin{matrix}
70750 & 80860 & -70750 & 60640 \\
80860 & 114600 & -80860 & 47170 \\
-70750 & -80860 & 70750 & -60640 \\
60640 & 47170 & -60640 & 74120 \\
\end{matrix} \right] \\
{} \\
\end{matrix}\begin{matrix}
{{W}_{1}} \\
{{W}_{2}} \\
{{W}_{3}} \\
{{W}_{4}} \\
\end{matrix} \\
& \text{Equilibrium equations:} \\
& \left[ k \right]\overrightarrow{W}=\overrightarrow{F} \\
& \left[ \begin{matrix}
70750 & -60640 \\
-60640 & 74120 \\
\end{matrix} \right]\left\{ \begin{matrix}
{{W}_{3}} \\
{{W}_{4}} \\
\end{matrix} \right\}=\left\{ \begin{matrix}
0 \\
1000 \\
\end{matrix} \right\} \\
& \text{The solution of these equations is:} \\
& {{W}_{3}}=0.03871\text{ m} \\
& {{W}_{4}}=0.04516\text{ m} \\
& \text{stress at root:} \\
& {{\left. {{\sigma }_{\max }} \right|}_{x=0}}={{\left. \frac{M{{y}_{\max }}}{I} \right|}_{x=0}}={{\left. \frac{EI(x)\frac{{{d}^{2}}w(x)}{d{{x}^{2}}}{{y}_{\max }}}{I(x)} \right|}_{x=0}}=3.3392\cdot {{10}^{7}}\text{N/}{{\text{m}}^{\text{2}}} \\
\end{align}$$
Example 7.4
For a dynamical system shown
in next figure derive the local stiffness matrices of elements.
$$\begin{align}
& {{A}^{\left( i \right)}}=13\cdot {{10}^{-4}}{{m}^{2}},i=1,2,3,4 \\
& E=200\cdot {{10}^{9}}N/{{m}^{2}} \\
& {{l}^{\left( 1 \right)}}=\sqrt{{{1.25}^{2}}+{{2.50}^{2}}}=2.795085m \\
& {{l}^{\left( 2 \right)}}=\sqrt{{{1.25}^{2}}+{{2.50}^{2}}}=2.795085m \\
& {{l}^{\left( 1 \right)}}=\sqrt{{{3.75}^{2}}+{{1.25}^{2}}}=3.952847m \\
& {{l}^{\left( 1 \right)}}=\sqrt{{{2.5}^{2}}+{{3.75}^{2}}}=4.506939m \\
& \left[ {{k}^{\left( i \right)}} \right]=\frac{{{E}^{\left( i \right)}}{{A}^{\left( i \right)}}}{{{l}^{\left( i \right)}}}\left[ \begin{matrix}
1 & -1 \\
-1 & 1 \\
\end{matrix} \right] \\
& \left[ {{\overline{k}}^{\left( i \right)}} \right]={{\left[ {{\lambda }^{\left( i \right)}} \right]}^{T}}\left[ {{k}^{\left( i \right)}} \right]\left[ {{\lambda }^{i}} \right] \\
& \left[ {{\lambda }^{i}} \right]=\left[ \begin{matrix}
\cos {{\theta }_{i}} & \sin {{\theta }_{i}} & 0 & 0 \\
0 & 0 & \cos {{\theta }_{i}} & \sin {{\theta }_{i}} \\
\end{matrix} \right] \\
& \left[ {{\overline{k}}^{\left( i \right)}} \right]=\frac{{{E}^{\left( i \right)}}{{A}^{\left( i \right)}}}{{{l}^{\left( i \right)}}}\left[ \begin{matrix}
{{\cos }^{2}}{{\theta }_{i}} & \cos {{\theta }_{i}}\sin {{\theta }_{i}} & -{{\cos }^{2}}{{\theta }_{i}} & -\cos {{\theta }_{i}}\sin {{\theta }_{i}} \\
\cos {{\theta }_{i}}\sin {{\theta }_{i}} & {{\sin }^{2}}{{\theta }_{i}} & -\cos {{\theta }_{i}}\sin {{\theta }_{i}} & -{{\sin }^{2}}{{\theta }_{i}} \\
-{{\cos }^{2}}{{\theta }_{i}} & -\cos {{\theta }_{i}}\sin {{\theta }_{i}} & {{\cos }^{2}}{{\theta }_{i}} & \cos {{\theta }_{i}}\sin {{\theta }_{i}} \\
-\cos {{\theta }_{i}}\sin {{\theta }_{i}} & -{{\sin }^{2}}{{\theta }_{i}} & \cos {{\theta }_{i}}\sin {{\theta }_{i}} & {{\sin }^{2}}{{\theta }_{i}} \\
\end{matrix} \right] \\
& {{\theta }_{1}}={{63.4349}^{\circ }} \\
& {{\theta }_{2}}={{116.5651}^{\circ }} \\
& {{\theta }_{3}}={{18.4350}^{\circ }} \\
& {{\theta }_{4}}={{56.3099}^{\circ }} \\
& \frac{{{E}^{\left( 1 \right)}}{{A}^{\left( 1 \right)}}}{{{l}^{\left( 1 \right)}}}=\frac{200\cdot {{10}^{9}}\cdot 13\cdot {{10}^{-4}}}{2.795085}=93.0204\cdot {{10}^{6}}N/m, \\
& cos{{\theta }_{1}}=0.4472, \\
& \sin {{\theta }_{1}}=0.8944, \\
& \left[ {{\overline{k}}^{\left( 1 \right)}} \right]={{10}^{6}}\begin{matrix}
\begin{matrix}
{{U}_{1}}\text{ } & \text{ }{{U}_{2}}\text{ } & {{U}_{3}}\text{ } & {{U}_{4}} \\
\end{matrix} \\
\left[ \begin{matrix}
18.6041 & 37.2082 & -18.6041 & -37.2082 \\
37.2082 & 74.4162 & -37.2082 & -74.4162 \\
-18.6041 & -37.2082 & -18.6041 & 37.2082 \\
-37.2082 & -74.4162 & 37.2082 & 74.4162 \\
\end{matrix} \right] \\
{} \\
\end{matrix}\begin{matrix}
{{U}_{1}} \\
{{U}_{2}} \\
{{U}_{3}} \\
{{U}_{4}} \\
\end{matrix} \\
\end{align}$$
$$\begin{align}
& \frac{{{E}^{\left( 2 \right)}}{{A}^{\left( 2 \right)}}}{{{l}^{\left( 2 \right)}}}=\frac{200\cdot {{10}^{9}}\cdot 13\cdot {{10}^{-4}}}{2.795085}=93.0204\cdot {{10}^{6}}N/m, \\
& cos{{\theta }_{2}}=-0.4472, \\
& \sin {{\theta }_{2}}=0.8944, \\
& \left[ {{\overline{k}}^{\left( 2 \right)}} \right]={{10}^{6}}\begin{matrix}
\begin{matrix}
{{U}_{5}}\text{ } & \text{ }{{U}_{6}}\text{ } & {{U}_{3}}\text{ } & {{U}_{4}} \\
\end{matrix} \\
\left[ \begin{matrix}
18.6041 & 37.2082 & -18.6041 & 37.2082 \\
-37.2082 & 74.4162 & -37.2082 & -74.4162 \\
-18.6041 & 37.2082 & 18.6041 & -37.2082 \\
37.2082 & -74.4162 & -37.2082 & 74.4162 \\
\end{matrix} \right] \\
{} \\
\end{matrix}\begin{matrix}
{{U}_{5}} \\
{{U}_{6}} \\
{{U}_{3}} \\
{{U}_{4}} \\
\end{matrix} \\
& \frac{{{E}^{\left( 3 \right)}}{{A}^{\left( 3 \right)}}}{{{l}^{\left( 3 \right)}}}=\frac{200\cdot {{10}^{9}}\cdot 13\cdot {{10}^{-4}}}{3.952847}=65.7754\cdot {{10}^{6}}N/m, \\
& \cos {{\theta }_{3}}=0.9487, \\
& \sin {{\theta }_{3}}=0.3162, \\
& \left[ {{\overline{k}}^{\left( 3 \right)}} \right]={{10}^{6}}\begin{matrix}
\begin{matrix}
{{U}_{3}}\text{ } & \text{ }{{U}_{4}}\text{ } & {{U}_{7}}\text{ } & {{U}_{8}} \\
\end{matrix} \\
\left[ \begin{matrix}
59.1978 & 19.7326 & -59.1978 & -19.7326 \\
-19.7326 & 6.57757 & -19.7326 & -6.57757 \\
-59.1978 & -19.7326 & 59.1978 & -19.7326 \\
-19.7326 & -6.57757 & 19.7326 & 6.57757 \\
\end{matrix} \right] \\
{} \\
\end{matrix}\begin{matrix}
{{U}_{3}} \\
{{U}_{4}} \\
{{U}_{7}} \\
{{U}_{8}} \\
\end{matrix} \\
& \frac{{{E}^{\left( 4 \right)}}{{A}^{\left( 4 \right)}}}{{{l}^{\left( 4 \right)}}}=\frac{200\cdot {{10}^{9}}\cdot 13\cdot {{10}^{-4}}}{4.506939}=57.6888\cdot {{10}^{6}}N/m, \\
& \cos {{\theta }_{4}}=0.5547, \\
& \sin {{\theta }_{4}}=0.8321, \\
& \left[ {{\overline{k}}^{\left( 4 \right)}} \right]={{10}^{6}}\begin{matrix}
\begin{matrix}
{{U}_{5}}\text{ } & \text{ }{{U}_{6}}\text{ } & {{U}_{7}}\text{ } & {{U}_{8}} \\
\end{matrix} \\
\left[ \begin{matrix}
17.7504 & 26.6256 & -17.7504 & -26.6256 \\
26.6256 & 6.57757 & -26.6256 & -6.57757 \\
-17.7504 & -26.6256 & 17.7504 & -26.6256 \\
-26.6256 & -6.57757 & 26.6256 & 6.57757 \\
\end{matrix} \right] \\
{} \\
\end{matrix}\begin{matrix}
{{U}_{5}} \\
{{U}_{6}} \\
{{U}_{7}} \\
{{U}_{8}} \\
\end{matrix} \\
\end{align}$$
Example 7.5
Find the stresses in
the stepped beam shown in next figure when a moment of 1000 N-m is applied at
node 2 using a two-element idealization. The beam has a square cross section
50x50 mm between nodes 1 and 2 and 25x25 mm between nodes 2 and 3. Assume the Young
s modulus as 2.1 * 10^11 Pa.
$$\begin{align}
& Element1: \\
& I=\frac{1}{12}\left( \frac{50}{1000} \right){{\left( \frac{50}{1000} \right)}^{3}}=0.5208\cdot {{10}^{-6}}{{m}^{4}} \\
& \frac{EI}{{{l}^{3}}}=\frac{2.1\cdot {{10}^{11}}\cdot 0.5208\cdot {{10}^{-6}}}{{{0.25}^{3}}}=0.7\cdot {{10}^{7}} \\
& \left[ {{k}^{\left( 1 \right)}} \right]=0.7\cdot {{10}^{7}}\begin{matrix}
\begin{matrix}
{{w}_{1}}\text{ } & {{w}_{2}}\text{ } & {{w}_{3}}\text{ } & {{w}_{4}} \\
\end{matrix} \\
\left[ \begin{matrix}
12 & 1.5 & -12 & 1.5 \\
1.5 & 0.25 & -1.5 & 0.125 \\
-12 & -1.5 & 12 & -1.5 \\
1.5 & 0.125 & -1.5 & 0.25 \\
\end{matrix} \right] \\
{} \\
\end{matrix}\begin{matrix}
{{w}_{1}} \\
{{w}_{2}} \\
{{w}_{3}} \\
{{w}_{4}} \\
\end{matrix}= \\
& 0.7\cdot {{10}^{7}}\begin{matrix}
\begin{matrix}
{{w}_{3}}\text{ } & {{w}_{4}} \\
\end{matrix} \\
\left[ \begin{matrix}
12 & -1.5 \\
-1.5 & 0.25 \\
\end{matrix} \right] \\
{} \\
\end{matrix}\begin{matrix}
{{w}_{3}} \\
{{w}_{4}} \\
\end{matrix} \\
\end{align}$$
$$\begin{align}
& Element2: \\
& I=\frac{1}{12}\left( \frac{25}{1000} \right){{\left( \frac{25}{1000} \right)}^{3}}=0.3255\cdot {{10}^{-7}}{{m}^{4}} \\
& \frac{EI}{{{l}^{3}}}=\frac{2.1\cdot {{10}^{11}}\cdot 0.3255\cdot {{10}^{-7}}}{{{0.4}^{3}}}=10.6805\cdot {{10}^{4}} \\
& \left[ {{k}^{\left( 2 \right)}} \right]=10.6805\cdot {{10}^{4}}\begin{matrix}
\begin{matrix}
{{w}_{3}}\text{ } & {{w}_{4}}\text{ } & {{w}_{5}}\text{ } & {{w}_{6}} \\
\end{matrix} \\
\left[ \begin{matrix}
12 & 2.4 & -12 & 2.4 \\
2.4 & 0.64 & -2.4 & 0.32 \\
-12 & -2.4 & 12 & -2.4 \\
2.4 & 0.32 & -2.4 & 0.64 \\
\end{matrix} \right] \\
{} \\
\end{matrix}\begin{matrix}
{{w}_{3}} \\
{{w}_{4}} \\
{{w}_{5}} \\
{{w}_{6}} \\
\end{matrix} \\
& \left[ {{k}^{\left( 2 \right)}} \right]=10.6805\cdot {{10}^{4}}\begin{matrix}
\begin{matrix}
{{w}_{3}}\text{ } & {{w}_{4}} \\
\end{matrix} \\
\left[ \begin{matrix}
12 & 2.4 \\
2.4 & 0.64 \\
\end{matrix} \right] \\
{} \\
\end{matrix}\begin{matrix}
{{w}_{3}} \\
{{w}_{4}} \\
\end{matrix} \\
\end{align}$$
$$\begin{align}
& \text{Assembled stiffness matrix:} \\
& \left[ K \right]=\left[ \begin{matrix}
8.4\cdot {{10}^{7}}+0.1282\cdot {{10}^{7}} & -1.05\cdot {{10}^{7}}+0.0256\cdot {{10}^{7}} \\
-1.05\cdot {{10}^{7}}+0.0256\cdot {{10}^{7}} & 0.175\cdot {{10}^{7}}+0.0068\cdot {{10}^{7}} \\
\end{matrix} \right] \\
& \text{Equilibrium equations:} \\
& {{10}^{7}}\left[ \begin{matrix}
8.5282 & -1.0244 \\
-1.0244 & 0.1818 \\
\end{matrix} \right]\left\{ \begin{matrix}
{{W}_{3}} \\
{{W}_{4}} \\
\end{matrix} \right\}=\left\{ \begin{matrix}
0 \\
{{10}^{3}} \\
\end{matrix} \right\} \\
& {{W}_{3}}=2.0446\cdot {{10}^{-4}}\text{m} \\
& {{W}_{4}}=1.7021\cdot {{10}^{-3}}\text{m} \\
& \text{Stresses in elements:} \\
& M=EI\frac{{{d}^{2}}w(x)}{d{{x}^{2}}} \\
& w(x)=\sum\limits_{i=1}^{4}{{{W}_{i}}{{N}_{i}}\left( x \right)} \\
& {{\sigma }_{\max }}=\frac{Mc}{I}=\frac{Mh}{2I}=\frac{Eh}{2}\frac{{{d}^{2}}w(x)}{d{{x}^{2}}} \\
& {{\sigma }_{\max }}=\frac{Eh}{2}\left[ \frac{{{W}_{1}}}{{{l}^{3}}}\left( 12x-6l \right)+\frac{{{W}_{2}}}{{{l}^{2}}}\left( 6x-4l \right)+\frac{{{W}_{3}}}{{{l}^{3}}}\left( 6l-12x \right)+\frac{{{W}_{4}}}{{{l}^{2}}}\left( 6x-2l \right) \right] \\
& \text{For element 1:} \\
& {{W}_{1}}=0,{{W}_{2}}=0,{{W}_{3}}=2.0446\cdot {{10}^{-4}},{{W}_{4}}=1.7021\cdot {{10}^{-3}}, \\
& l=0.25,h=0.05,E=2.1\cdot {{10}^{11}} \\
& {{\left. {{\sigma }_{\max }} \right|}_{\text{fixed end}}}={{\sigma }_{\max }}\left( x=0 \right)=3.1560\cdot {{10}^{7}}\text{N/}{{\text{m}}^{2}} \\
& {{\left. {{\sigma }_{\max }} \right|}_{\text{loaded end}}}={{\sigma }_{\max }}\left( x=0.25 \right)=3.9929\cdot {{10}^{7}}\text{N/}{{\text{m}}^{2}} \\
& \text{For element 2:} \\
& {{W}_{1}}=2.0446\cdot {{10}^{-4}},{{W}_{2}}=1.7021\cdot {{10}^{-3}},{{W}_{3}}=0,{{W}_{4}}=0, \\
& l=0.4,h=0.025,E=2.1\cdot {{10}^{11}} \\
& {{\left. {{\sigma }_{\max }} \right|}_{\text{loaded end}}}={{\sigma }_{\max }}\left( x=0 \right)=-6.4807\cdot {{10}^{7}}\text{N/}{{\text{m}}^{2}} \\
& {{\left. {{\sigma }_{\max }} \right|}_{\text{fixed end}}}={{\sigma }_{\max }}\left( x=0.4 \right)=4.2467\cdot {{10}^{7}}\text{N/}{{\text{m}}^{2}} \\
\end{align}$$
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