Question-159297

Question Number 159297 by rs4089 last updated on 15/Nov/21

Commented by rs4089 last updated on 15/Nov/21

$${how}\:{can}\:{i}\:{find}\:{slope}\:{and}\:{deflection} \\ $$$$\:{of}\:{this}\:{cantilever}\:{beam}\:{at}\:{free}\:{end}\: \\ $$$${point}.\:{by}\:{using}\:{double}\:{integral}\:{method} \\ $$

Answered by mr W last updated on 15/Nov/21

Commented by mr W last updated on 15/Nov/21

M_1 =P_1 a M_2 =P_1 (a+b)+P_2 b EIf_A =((M_1 a)/2)×((2a)/3)+M_1 b×(a+(b/2))+(((M_2 −M_1 )b)/2)×(a+((2b)/3)) EIf_A =((M_1 a^2 )/3)+((M_1 b(2a+b))/2)+(((M_2 −M_1 )b(3a+2b))/6) EIf_A =(((a^3 +3a^2 b+3ab^2 +b^3 )P_1 )/3)+(((3a+2b)b^2 P_2 )/6) EIf_A =(((a+b)^3 P_1 )/3)+(((3a+2b)b^2 P_2 )/6) ⇒f_A =(((a+b)^3 P_1 )/(3EI))+(((3a+2b)b^2 P_2 )/(6EI)) EIϕ_A =((M_1 a)/2)+(((M_1 +M_2 )b)/2) EIϕ_A =((P_1 a^2 )/2)+((P_1 ab+P_1 ab+P_1 b^2 +P_2 b^2 )/2) EIϕ_A =(((a+b)^2 P_1 +P_2 b^2 )/2) ⇒ϕ_A =(((a+b)^2 P_1 +b^2 P_2 )/(2EI))

$${M}_{\mathrm{1}} ={P}_{\mathrm{1}} {a} \\ $$$${M}_{\mathrm{2}} ={P}_{\mathrm{1}} \left({a}+{b}\right)+{P}_{\mathrm{2}} {b} \\ $$$${EIf}_{{A}} =\frac{{M}_{\mathrm{1}} {a}}{\mathrm{2}}×\frac{\mathrm{2}{a}}{\mathrm{3}}+{M}_{\mathrm{1}} {b}×\left({a}+\frac{{b}}{\mathrm{2}}\right)+\frac{\left({M}_{\mathrm{2}} −{M}_{\mathrm{1}} \right){b}}{\mathrm{2}}×\left({a}+\frac{\mathrm{2}{b}}{\mathrm{3}}\right) \\ $$$${EIf}_{{A}} =\frac{{M}_{\mathrm{1}} {a}^{\mathrm{2}} }{\mathrm{3}}+\frac{{M}_{\mathrm{1}} {b}\left(\mathrm{2}{a}+{b}\right)}{\mathrm{2}}+\frac{\left({M}_{\mathrm{2}} −{M}_{\mathrm{1}} \right){b}\left(\mathrm{3}{a}+\mathrm{2}{b}\right)}{\mathrm{6}} \\ $$$${EIf}_{{A}} =\frac{\left({a}^{\mathrm{3}} +\mathrm{3}{a}^{\mathrm{2}} {b}+\mathrm{3}{ab}^{\mathrm{2}} +{b}^{\mathrm{3}} \right){P}_{\mathrm{1}} }{\mathrm{3}}+\frac{\left(\mathrm{3}{a}+\mathrm{2}{b}\right){b}^{\mathrm{2}} {P}_{\mathrm{2}} }{\mathrm{6}} \\ $$$${EIf}_{{A}} =\frac{\left({a}+{b}\right)^{\mathrm{3}} {P}_{\mathrm{1}} }{\mathrm{3}}+\frac{\left(\mathrm{3}{a}+\mathrm{2}{b}\right){b}^{\mathrm{2}} {P}_{\mathrm{2}} }{\mathrm{6}} \\ $$$$\Rightarrow{f}_{{A}} =\frac{\left({a}+{b}\right)^{\mathrm{3}} {P}_{\mathrm{1}} }{\mathrm{3}{EI}}+\frac{\left(\mathrm{3}{a}+\mathrm{2}{b}\right){b}^{\mathrm{2}} {P}_{\mathrm{2}} }{\mathrm{6}{EI}} \\ $$$${EI}\varphi_{{A}} =\frac{{M}_{\mathrm{1}} {a}}{\mathrm{2}}+\frac{\left({M}_{\mathrm{1}} +{M}_{\mathrm{2}} \right){b}}{\mathrm{2}} \\ $$$${EI}\varphi_{{A}} =\frac{{P}_{\mathrm{1}} {a}^{\mathrm{2}} }{\mathrm{2}}+\frac{{P}_{\mathrm{1}} {ab}+{P}_{\mathrm{1}} {ab}+{P}_{\mathrm{1}} {b}^{\mathrm{2}} +{P}_{\mathrm{2}} {b}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${EI}\varphi_{{A}} =\frac{\left({a}+{b}\right)^{\mathrm{2}} {P}_{\mathrm{1}} +{P}_{\mathrm{2}} {b}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\Rightarrow\varphi_{{A}} =\frac{\left({a}+{b}\right)^{\mathrm{2}} {P}_{\mathrm{1}} +{b}^{\mathrm{2}} {P}_{\mathrm{2}} }{\mathrm{2}{EI}} \\ $$

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