Finite Element Method Examples 1-5

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}$$