si vuole sapere \( \displaystyle \Delta{w}_{{c}},\phi_{{a}} \). La forza è inclinata di 45°
con metodi analitici risulta \( \displaystyle \Delta{w}_{{c}}=-\frac{\sqrt{{2}}}{{2}}\frac{{{F}{{l}}^{{3}}}}{{{e}{i}}},\phi_{{a}}=-\frac{{7}}{{12}}\sqrt{{2}}\frac{{{F}{{l}}^{{2}}}}{{{e}{i}}}\lt{b}\frac{{r}}{\gt}{c}{o}{n}{i}{l}{m}{e}\to{d}{o}\partial{l}{a}{c}{o}{m}{p}{o}{s}{i}{z}{i}{o}\ne{c}\in{e}{m}{a}{t}{i}{c}{a}{h}{o},{p}{e}{r}{s}{o}{v}{r}{a}{p}{p}{o}{s}{i}{z}{i}{o}\ne{d}{e}{g{{l}}}{i}{e}{f{{f{{e}}}}}{\mathtt{{i}}}, \)Deltaw_c=Deltaw_c^(AB)+Deltaw_c^(BC)+Deltaw_c^(CD), phi_a=phi_a^(AB)+phi_a^(BC)+phi_a^(CD)\( \displaystyle ,{d}{o}{v}{e}{l}'{a}\pi{c}{e}{s}{i}{g{{n}}}{\quad\text{if}\quad}{i}{c}{a}{c}{h}{e}è{d}{e}{f{{\quad\text{or}\quad}}}{m}{a}{b}{i}\le{s}{o}{l}{o}{q}{u}{e}{l}{t}{r}{a}{\mathtt{\odot}}\lt{b}\frac{{r}}{\gt}{S}{i}{a}{a}{l}{l}{\quad\text{or}\quad}{a}{d}{e}{f{{\quad\text{or}\quad}}}{m}{a}{b}{i}\le{s}{o}{l}{a}{m}{e}{n}{t}{e}{A}{B}:{l}{a}{d}{e}{f{{\quad\text{or}\quad}}}{m}{a}{t}{a}{q}{u}{a}{l}{i}{t}{a}{t}{i}{v}{a}\partial{l}{a}{s}{t}{r}{u}{\mathtt{{u}}}{r}{a}{s}{a}{r}à:\lt{b}\frac{{r}}{\gt}\lt{b}\frac{{r}}{\gt}\lt{a}{h}{r}{e}{f{=}}\text{http://img820.imageshack.us/i/deformata1.png/}{c}{l}{a}{s}{s}=\text{postlink}\gt\lt{i}{m}{g{{s}}}{r}{c}=\text{http://img820.imageshack.us/img820/5671/deformata1.png}{a}\leq\frac{\text{Immagine}}{\gt}\frac{\lt}{{a}}\gt\lt{b}\frac{{r}}{\gt}{i}{l}{t}{r}{a}{\mathtt{{o}}}{r}{i}{g{{i}}}{d}{o}{s}{i}{c}{o}{m}{p}{\quad\text{or}\quad}{t}{a}{c}{o}{m}{e}{s}{e}{a}{v}{e}{s}{s}{e}{u}{n}{c}{a}{r}{r}{e}{l}{l}{o}:\lt{b}\frac{{r}}{\gt}\lt{a}{h}{r}{e}{f{=}}\text{http://img40.imageshack.us/i/trattorigido1.png/}{c}{l}{a}{s}{s}=\text{postlink}\gt\lt{i}{m}{g{{s}}}{r}{c}=\text{http://img40.imageshack.us/img40/7538/trattorigido1.png}{a}\leq\frac{\text{Immagine}}{\gt}\frac{\lt}{{a}}\gt\lt{b}\frac{{r}}{\gt}{p}{e}{r}{i}{l}{t}{r}{a}{\mathtt{{o}}}{d}{e}{f{{\quad\text{or}\quad}}}{m}{a}{b}{i}\le{a}{\mathbf{{i}}}{a}{m}{o}{c}{h}{e}{b}{n}{o}{n}{p}{u}ò{t}{r}{a}{s}{l}{a}{r}{e}{v}{e}{r}{t}{i}{c}{a}{l}{m}{e}{n}{t}{e}{e}{n}{o}{n}{p}{u}ò{r}{u}{o}{t}{a}{r}{e}\lt{b}\frac{{r}}{\gt}{p}{e}{r}{c}{i}ò{l}{o}{s}{c}{h}{e}{m}{a}\equiv{a}\le{n}{t}{e}{p}{e}{r}{A}{B}è:\lt{b}\frac{{r}}{\gt}\lt{a}{h}{r}{e}{f{=}}\text{http://img684.imageshack.us/i/schemaequivalente1.png/}{c}{l}{a}{s}{s}=\text{postlink}\gt\lt{i}{m}{g{{s}}}{r}{c}=\text{http://img684.imageshack.us/img684/7571/schemaequivalente1.png}{a}\leq\frac{\text{Immagine}}{\gt}\frac{\lt}{{a}}\gt\lt{b}\frac{{r}}{\gt}{c}{o}{n}{\mathfrak{{o}}}{n}{t}{a}\to{c}{o}{n}{l}{o}{s}{c}{h}{e}{m}{a}{d}{i}{r}{\quad\text{if}\quad}{e}{r}{i}{m}{e}{n}\to,{d}{o}{v}{e}\ne{l}{p}{u}{n}\to\in{c}{u}{i}è{a}{p}{p}{l}{i}{c}{a}{t}{a}{l}{a}{f{{\quad\text{or}\quad}}}{z}{a} \)v=(Fl^3)/(48EI)\( \displaystyle {e}{d}{a}{l}{l}{a}{p}{a}{r}{t}{e}\partial{l}{a}{c}{e}{r}{n}{i}{e}{r}{a} \)phi=-(Fl^2)/(16EI)
con \( \displaystyle {F}'=\frac{\sqrt{{2}}}{{2}}{F},{l}'={2}{l} \) trovo \( \displaystyle {{w}_{{b}}^{{{A}{B}}}}={{w}_{{{{c}}^{{-}}}}^{{{A}{B}}}}=\frac{\sqrt{{2}}}{{12}}\frac{{{F}{{l}}^{{3}}}}{{{E}{I}}},{\phi_{{a}}^{{{A}{B}}}}=-\frac{\sqrt{{2}}}{{8}}\frac{{{F}{{l}}^{{2}}}}{{{E}{I}}}\lt{b}\frac{{r}}{\gt}\lt{b}\frac{{r}}{\gt}{s}{i}{a}{\quad\text{or}\quad}{a}{d}{e}{f{{\quad\text{or}\quad}}}{m}{a}{b}{i}\le{i}{l}{t}{r}{a}{\mathtt{{o}}}{B}{C}:{l}{a}{d}{e}{f{{\quad\text{or}\quad}}}{m}{a}{t}{a}{q}{u}{a}{l}{i}{t}{a}{t}{i}{v}{a}{s}{a}{r}à:\lt{b}\frac{{r}}{\gt}\lt{a}{h}{r}{e}{f{=}}\text{http://img824.imageshack.us/i/deformata2.png/}{c}{l}{a}{s}{s}=\text{postlink}\gt\lt{i}{m}{g{{s}}}{r}{c}=\text{http://img824.imageshack.us/img824/7218/deformata2.png}{a}\leq\frac{\text{Immagine}}{\gt}\frac{\lt}{{a}}\gt\lt{b}\frac{{r}}{\gt}\lt{b}\frac{{r}}{\gt}{B}{e}{C}{h}{a}\cap{o}{c}{o}{m}{e}{u}{n}{i}{c}{a}{r}{e}{s}{t}{r}{i}{z}{i}{o}\ne{c}\in{e}{m}{a}{t}{i}{c}{a}{l}{a}{t}{r}{a}{s}{l}{a}{z}{i}{o}\ne{v}{e}{r}{t}{i}{c}{a}\le{i}{m}{p}{e}{d}{i}{t}{a},{p}{e}{r}{c}{i}ò{t}{r}{a}{s}{c}{u}{r}{\quad\text{and}\quad}{o}{i}{l}{m}{o}\to{r}{i}{g{{i}}}{d}{o},{p}{e}{r}{i}{l}{t}{r}{a}{\mathtt{{o}}}{A}{B}{l}{o}{s}{c}{h}{e}{m}{a}\equiv{a}\le{n}{t}{e}è:\lt{b}\frac{{r}}{\gt}\lt{b}\frac{{r}}{\gt}\lt{a}{h}{r}{e}{f{=}}\text{http://img820.imageshack.us/i/schemaequivalente2.png/}{c}{l}{a}{s}{s}=\text{postlink}\gt\lt{i}{m}{g{{s}}}{r}{c}=\text{http://img820.imageshack.us/img820/3069/schemaequivalente2.png}{a}\leq\frac{\text{Immagine}}{\gt}\frac{\lt}{{a}}\gt\lt{b}\frac{{r}}{\gt}\lt{b}\frac{{r}}{\gt}{m}{e}{n}{t}{r}{e}{i}{l}{m}{i}{o}{s}{c}{h}{e}{m}{a}{d}{i}{r}{\quad\text{if}\quad}{e}{r}{i}{m}{e}{n}\toè:\lt{b}\frac{{r}}{\gt}\lt{a}{h}{r}{e}{f{=}}\text{http://img34.imageshack.us/i/schemadifirerimento2.png/}{c}{l}{a}{s}{s}=\text{postlink}\gt\lt{i}{m}{g{{s}}}{r}{c}=\text{http://img34.imageshack.us/img34/2850/schemadifirerimento2.png}{a}\leq\frac{\text{Immagine}}{\gt}\frac{\lt}{{a}}\gt\lt{b}\frac{{r}}{\gt}{c}{o}{n}{p}{e}{r}{l}{a}{c}{e}{r}{n}{i}{e}{r}{a} \)phi=-(ml)/(6EI)\( \displaystyle ,{p}{e}{r}{i}{l}{c}{a}{r}{r}{e}{l}{l}{o} \)phi=(ml)/(3EI)
concludo \( \displaystyle {\phi_{{B}}^{{{B}{C}}}}=-\frac{\sqrt{{2}}}{{6}}\frac{{{F}{{l}}^{{2}}}}{{{E}{I}}}={\phi_{{a}}^{{{B}{C}}}},{{w}_{{c}}^{{{B}{C}}}}={{w}_{{b}}^{{{B}{C}}}}={l}{\left|{\phi_{{b}}^{{{B}{C}}}}\right|}=\frac{\sqrt{{2}}}{{6}}\frac{{{F}{{l}}^{{3}}}}{{{E}{I}}}\lt{b}\frac{{r}}{\gt}\lt{b}\frac{{r}}{\gt}{s}{e}è{d}{e}{f{{\quad\text{or}\quad}}}{m}{a}{b}{i}\le{s}{o}{l}{o}{C}{D},{l}{a}{d}{e}{f{{\quad\text{or}\quad}}}{m}{a}{t}{a}{q}{u}{a}{l}{i}{t}{a}{t}{i}{v}{a}{s}{a}{r}à\lt{b}\frac{{r}}{\gt}\lt{b}\frac{{r}}{\gt}\lt{a}{h}{r}{e}{f{=}}\text{http://img412.imageshack.us/i/deformata3.png/}{c}{l}{a}{s}{s}=\text{postlink}\gt\lt{i}{m}{g{{s}}}{r}{c}=\text{http://img412.imageshack.us/i/deformata3.png/}{a}\leq\frac{\text{Immagine}}{\gt}\frac{\lt}{{a}}\gt\lt{b}\frac{{r}}{\gt}\lt{b}\frac{{r}}{\gt}{i}{l}{m}{i}{o}{s}{c}{h}{e}{m}{a}\equiv{a}\le{n}{t}{e}è:\lt{b}\frac{{r}}{\gt}\lt{b}\frac{{r}}{\gt}\lt{a}{h}{r}{e}{f{=}}\text{http://img822.imageshack.us/i/schemaequivalente3.png/}{c}{l}{a}{s}{s}=\text{postlink}\gt\lt{i}{m}{g{{s}}}{r}{c}=\text{http://img822.imageshack.us/img822/7613/schemaequivalente3.png}{a}\leq\frac{\text{Immagine}}{\gt}\frac{\lt}{{a}}\gt\lt{b}\frac{{r}}{\gt}\lt{b}\frac{{r}}{\gt}{c}{o}{n}{c}{l}{u}{d}{o} \)v_(c^+)^(CD)=w_(c^-)^(CD)=sqrt2/6 (Fl^3)/(EI)=l|phi_a^(CD)|
\( \displaystyle {\phi_{{a}}^{{{C}{D}}}}=-\frac{\sqrt{{2}}}{{6}}\frac{{{F}{{l}}^{{2}}}}{{{E}{I}}}\lt{b}\frac{{r}}{\gt}\lt{b}\frac{{r}}{\gt}\in{d}{e}{f{\in}}{i}{t}{i}{v}{a} \)phi_a=-11/24sqrt2(Fl^2)/(EI), Deltaw_c=-5/12sqrt2(Fl^3)/(EI)$, che è diverso da quello ottenuto analiticamente
