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Question Number 169448 by BHOOPENDRA last updated on 30/Apr/22

Commented by BHOOPENDRA last updated on 30/Apr/22

calculation sir

$${calculation}\:{sir} \\ $$

Commented by BHOOPENDRA last updated on 30/Apr/22

thankyou sir

$${thankyou}\:{sir} \\ $$

Commented by BHOOPENDRA last updated on 01/May/22

Commented by BHOOPENDRA last updated on 01/May/22

Commented by BHOOPENDRA last updated on 01/May/22

sir please help me where i m doing wrong can you please help me in the solution

$${sir}\:{please}\:{help}\:{me}\:{where}\:{i}\:{m}\:{doing}\: \\ $$$${wrong}\:{can}\:{you}\:{please}\:{help}\:{me}\:{in}\:{the} \\ $$$${solution} \\ $$

Commented by mr W last updated on 01/May/22

there are many methods to solve. tell what you get for M_A ,M_B ,M_C ,M_D , maybe i can check if i can get the same.

$${there}\:{are}\:{many}\:{methods}\:{to}\:{solve}. \\ $$$${tell}\:{what}\:{you}\:{get}\:{for}\:{M}_{{A}} ,{M}_{{B}} ,{M}_{{C}} ,{M}_{{D}} , \\ $$$${maybe}\:{i}\:{can}\:{check}\:{if}\:{i}\:{can}\:{get}\:{the} \\ $$$${same}. \\ $$

Commented by BHOOPENDRA last updated on 01/May/22

i was helping somebody sir i have done this type of question 3,4 year ago now i dont remeber any method properly so please help me out this

$${i}\:{was}\:{helping}\:{somebody}\:{sir}\:{i}\:{have}\:{done} \\ $$$${this}\:{type}\:{of}\:{question}\:\mathrm{3},\mathrm{4}\:{year}\:{ago} \\ $$$${now}\:{i}\:{dont}\:{remeber}\:{any}\:{method} \\ $$$${properly}\:{so}\:{please}\:{help}\:{me}\:{out}\:{this} \\ $$

Commented by mr W last updated on 01/May/22

For calculation by hand we apply usually Method I: we take the moments at the supports as unknowns ⇒system with 4 unknowns Method II: we take the slopes at the supports B,C as unknowns ⇒system with 2 unknowns both are easy, but it needs time to work out...

$${For}\:{calculation}\:{by}\:{hand}\:{we}\:{apply} \\ $$$${usually} \\ $$$${Method}\:{I}:\: \\ $$$${we}\:{take}\:{the}\:{moments}\:{at}\:{the}\:{supports} \\ $$$${as}\:{unknowns} \\ $$$$\Rightarrow{system}\:{with}\:\mathrm{4}\:{unknowns} \\ $$$${Method}\:{II}: \\ $$$${we}\:{take}\:{the}\:{slopes}\:{at}\:{the}\:{supports}\:{B},{C} \\ $$$${as}\:{unknowns} \\ $$$$\Rightarrow{system}\:{with}\:\mathrm{2}\:{unknowns} \\ $$$$ \\ $$$${both}\:{are}\:{easy},\:{but}\:{it}\:{needs}\:{time}\:{to} \\ $$$${work}\:{out}... \\ $$

Commented by BHOOPENDRA last updated on 01/May/22

thankyou sir i did roughly but i did not get the ans.so can you please calculate it sir

$${thankyou}\:{sir}\:{i}\:{did}\:{roughly}\:{but}\:{i}\:{did}\:{not}\:{get} \\ $$$${the}\:{ans}.{so}\:{can}\:{you}\:{please}\:{calculate}\:{it} \\ $$$${sir} \\ $$

Answered by mr W last updated on 01/May/22

Moment under loads: M_(A,r,0) =−((1.5×4^2 )/(12))=−2 KNm M_(B,l,0) =((1.5×4^2 )/(12))=2 KNm M_(B,r,0) =−((7×4)/8)=−3.5 KNm M_(C,l,0) =((7×4)/8)=3.5 KNm M_(C,r,0) =−((8×1×3^2 )/4^2 )=−4.5 KNm M_(D,l,0) =((8×3×1^2 )/4^2 )=1.5 KNm Moment under 𝛉_1 =slope at B×EI=1: M_(A,r,1) =(2/4)=0.5 KNm M_(B,l,1) =(4/4)=1 KNm M_(B,r,1) =(4/4)=1 KNm M_(C,l,1) =(2/4)=0.5 KNm M_(C,r,1) =0 KNm M_(D,l,1) =0 KNm Moment under 𝛉_2 =slope at C×EI=1: M_(A,r,2) =0 KNm M_(B,l,2) =0Nm M_(B,r,2) =(2/4)=0.5 KNm M_(C,l,2) =(4/4)=1 KNm M_(C,r,2) =(4/4)=1 KNm M_(D,l,2) =(2/4)=0.5 KNm ΣM_B =(1+1)θ_1 +0.5θ_2 +2−3.5=0 ⇒4θ_1 +θ_2 =3 ...(i) ΣM_C =0.5θ_1 +(1+1)θ_2 +3.5−4.5=0 ⇒θ_1 +4θ_2 =2 ...(ii) solving (i) and (ii) we get θ_1 =(2/3) θ_2 =(1/3) M_(A,r) =−2+0.5×((2/3))=−(5/3) KNm M_(B,l) =2+1×((2/3))+0×((1/3))=(8/3) KNm M_(B,r) =−3.5+1×((2/3))+0.5×((1/3))=−(8/3) KNm M_(C,l) =3.5+0.5×((2/3))+1×((1/3))=((25)/6) KNm M_(C,r) =−4.5+0×((2/3))+1×((1/3))=−((25)/6) KNm M_(D,l) =1.5+0×((2/3))+0.5×((1/3))=(5/3) KNm

$$\boldsymbol{{Moment}}\:\boldsymbol{{under}}\:\boldsymbol{{loads}}: \\ $$$${M}_{{A},{r},\mathrm{0}} =−\frac{\mathrm{1}.\mathrm{5}×\mathrm{4}^{\mathrm{2}} }{\mathrm{12}}=−\mathrm{2}\:{KNm} \\ $$$${M}_{{B},{l},\mathrm{0}} =\frac{\mathrm{1}.\mathrm{5}×\mathrm{4}^{\mathrm{2}} }{\mathrm{12}}=\mathrm{2}\:{KNm} \\ $$$${M}_{{B},{r},\mathrm{0}} =−\frac{\mathrm{7}×\mathrm{4}}{\mathrm{8}}=−\mathrm{3}.\mathrm{5}\:{KNm} \\ $$$${M}_{{C},{l},\mathrm{0}} =\frac{\mathrm{7}×\mathrm{4}}{\mathrm{8}}=\mathrm{3}.\mathrm{5}\:{KNm} \\ $$$${M}_{{C},{r},\mathrm{0}} =−\frac{\mathrm{8}×\mathrm{1}×\mathrm{3}^{\mathrm{2}} }{\mathrm{4}^{\mathrm{2}} }=−\mathrm{4}.\mathrm{5}\:{KNm} \\ $$$${M}_{{D},{l},\mathrm{0}} =\frac{\mathrm{8}×\mathrm{3}×\mathrm{1}^{\mathrm{2}} }{\mathrm{4}^{\mathrm{2}} }=\mathrm{1}.\mathrm{5}\:{KNm} \\ $$$$ \\ $$$$\boldsymbol{{Moment}}\:\boldsymbol{{under}}\:\boldsymbol{\theta}_{\mathrm{1}} =\boldsymbol{{slope}}\:\boldsymbol{{at}}\:\boldsymbol{{B}}×\boldsymbol{{EI}}=\mathrm{1}: \\ $$$${M}_{{A},{r},\mathrm{1}} =\frac{\mathrm{2}}{\mathrm{4}}=\mathrm{0}.\mathrm{5}\:{KNm} \\ $$$${M}_{{B},{l},\mathrm{1}} =\frac{\mathrm{4}}{\mathrm{4}}=\mathrm{1}\:{KNm} \\ $$$${M}_{{B},{r},\mathrm{1}} =\frac{\mathrm{4}}{\mathrm{4}}=\mathrm{1}\:{KNm} \\ $$$${M}_{{C},{l},\mathrm{1}} =\frac{\mathrm{2}}{\mathrm{4}}=\mathrm{0}.\mathrm{5}\:{KNm} \\ $$$${M}_{{C},{r},\mathrm{1}} =\mathrm{0}\:{KNm} \\ $$$${M}_{{D},{l},\mathrm{1}} =\mathrm{0}\:{KNm} \\ $$$$ \\ $$$$\boldsymbol{{Moment}}\:\boldsymbol{{under}}\:\boldsymbol{\theta}_{\mathrm{2}} =\boldsymbol{{slope}}\:\boldsymbol{{at}}\:\boldsymbol{{C}}×\boldsymbol{{EI}}=\mathrm{1}: \\ $$$${M}_{{A},{r},\mathrm{2}} =\mathrm{0}\:{KNm} \\ $$$${M}_{{B},{l},\mathrm{2}} =\mathrm{0}{Nm} \\ $$$${M}_{{B},{r},\mathrm{2}} =\frac{\mathrm{2}}{\mathrm{4}}=\mathrm{0}.\mathrm{5}\:{KNm} \\ $$$${M}_{{C},{l},\mathrm{2}} =\frac{\mathrm{4}}{\mathrm{4}}=\mathrm{1}\:{KNm} \\ $$$${M}_{{C},{r},\mathrm{2}} =\frac{\mathrm{4}}{\mathrm{4}}=\mathrm{1}\:{KNm} \\ $$$${M}_{{D},{l},\mathrm{2}} =\frac{\mathrm{2}}{\mathrm{4}}=\mathrm{0}.\mathrm{5}\:{KNm} \\ $$$$ \\ $$$$\Sigma{M}_{{B}} =\left(\mathrm{1}+\mathrm{1}\right)\theta_{\mathrm{1}} +\mathrm{0}.\mathrm{5}\theta_{\mathrm{2}} +\mathrm{2}−\mathrm{3}.\mathrm{5}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{4}\theta_{\mathrm{1}} +\theta_{\mathrm{2}} =\mathrm{3}\:\:\:...\left({i}\right) \\ $$$$\Sigma{M}_{{C}} =\mathrm{0}.\mathrm{5}\theta_{\mathrm{1}} +\left(\mathrm{1}+\mathrm{1}\right)\theta_{\mathrm{2}} +\mathrm{3}.\mathrm{5}−\mathrm{4}.\mathrm{5}=\mathrm{0} \\ $$$$\Rightarrow\theta_{\mathrm{1}} +\mathrm{4}\theta_{\mathrm{2}} =\mathrm{2}\:\:\:...\left({ii}\right) \\ $$$${solving}\:\left({i}\right)\:{and}\:\left({ii}\right)\:{we}\:{get} \\ $$$$\theta_{\mathrm{1}} =\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\theta_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$ \\ $$$${M}_{{A},{r}} =−\mathrm{2}+\mathrm{0}.\mathrm{5}×\left(\frac{\mathrm{2}}{\mathrm{3}}\right)=−\frac{\mathrm{5}}{\mathrm{3}}\:{KNm} \\ $$$${M}_{{B},{l}} =\mathrm{2}+\mathrm{1}×\left(\frac{\mathrm{2}}{\mathrm{3}}\right)+\mathrm{0}×\left(\frac{\mathrm{1}}{\mathrm{3}}\right)=\frac{\mathrm{8}}{\mathrm{3}}\:{KNm} \\ $$$${M}_{{B},{r}} =−\mathrm{3}.\mathrm{5}+\mathrm{1}×\left(\frac{\mathrm{2}}{\mathrm{3}}\right)+\mathrm{0}.\mathrm{5}×\left(\frac{\mathrm{1}}{\mathrm{3}}\right)=−\frac{\mathrm{8}}{\mathrm{3}}\:{KNm} \\ $$$${M}_{{C},{l}} =\mathrm{3}.\mathrm{5}+\mathrm{0}.\mathrm{5}×\left(\frac{\mathrm{2}}{\mathrm{3}}\right)+\mathrm{1}×\left(\frac{\mathrm{1}}{\mathrm{3}}\right)=\frac{\mathrm{25}}{\mathrm{6}}\:{KNm} \\ $$$${M}_{{C},{r}} =−\mathrm{4}.\mathrm{5}+\mathrm{0}×\left(\frac{\mathrm{2}}{\mathrm{3}}\right)+\mathrm{1}×\left(\frac{\mathrm{1}}{\mathrm{3}}\right)=−\frac{\mathrm{25}}{\mathrm{6}}\:{KNm} \\ $$$${M}_{{D},{l}} =\mathrm{1}.\mathrm{5}+\mathrm{0}×\left(\frac{\mathrm{2}}{\mathrm{3}}\right)+\mathrm{0}.\mathrm{5}×\left(\frac{\mathrm{1}}{\mathrm{3}}\right)=\frac{\mathrm{5}}{\mathrm{3}}\:{KNm} \\ $$

Commented by BHOOPENDRA last updated on 01/May/22

thankyou sir

$${thankyou}\:{sir} \\ $$

Commented by mr W last updated on 01/May/22

Commented by mr W last updated on 01/May/22

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