Question Number 68222 by necxxx last updated on 07/Sep/19 | ||
$${Sketch}\:{the}\:{shear}\:{and}\:{moment}\:{diagrams} \\ $$$${of}\:{a}\:{simply}\:{supported}\:{beam}\:{of}\:\mathrm{6}{m}.{The} \\ $$$${load}\:{on}\:{the}\:{beam}\:{consists}\:{of}\:{UDL}\:{of} \\ $$$$\mathrm{15}{KN}/{m}\:{over}\:{the}\:{left}\:{half}\:{of}\:{the}\:{span}. \\ $$$$ \\ $$ | ||
Commented by necxxx last updated on 07/Sep/19 | ||
$${please}\:{help}\:{me}\:{with}\:{this}\:{problem}.\:{I}'{ve} \\ $$$${tried}\:{solving}\:{but}\:{my}\:{result}\:{isnt}\:{in}\: \\ $$$${accordance}\:{with}\:{the}\:{source}.\:{Please}\:{post} \\ $$$${steps}\:{and}\:{diagrams}\:{where}\:{necessary}. \\ $$$${I}'{ll}\:{also}\:{post}\:{the}\:{steps}\:{I}\:{took}\:{so}\:{you}\:{all}\:{can}\: \\ $$$${help}\:{me}\:{with}\:{the}\:{amendments}. \\ $$$${Thanks}\:{in}\:{advance}. \\ $$ | ||
Commented by necxxx last updated on 07/Sep/19 | ||
Commented by necxxx last updated on 07/Sep/19 | ||
Commented by necxxx last updated on 07/Sep/19 | ||
Commented by necxxx last updated on 07/Sep/19 | ||
$${here}\:{is}\:{the}\:{diagram}\:{given} \\ $$$$ \\ $$ | ||
Answered by mr W last updated on 07/Sep/19 | ||
Commented by necxxx last updated on 07/Sep/19 | ||
$${Thank}\:{you}\:{so}\:{much}\:{sir}\:{but}\:{this}\:{SFD} \\ $$$${and}\:{BMD}\:{do}\:{not}\:{tally}\:{with}\:{the}\:{answer} \\ $$$${given}\:{in}\:{the}\:{reference}\:{material}\:{I}\:{used}. \\ $$$$ \\ $$ | ||
Commented by mr W last updated on 07/Sep/19 | ||
$${the}\:{values}\:{i}\:{gave}\:{are}\:\mathrm{100\%}\:{corrrect}, \\ $$$${that}'{s}\:{for}\:{sure}. \\ $$$${your}\:{book}\:{might}\:{use}\:{different}\:{signs}. \\ $$$${therefore}\:{i}\:{haven}'{t}\:{given}\:{any}\:{sign}\:{for} \\ $$$${the}\:{values}. \\ $$ | ||
Commented by necxxx last updated on 07/Sep/19 | ||
Commented by mr W last updated on 07/Sep/19 | ||
$${if}\:{the}\:{maximum}\:{of}\:{moment}\:{is}\:{needed}, \\ $$$${it}\:{is} \\ $$$$\frac{\mathrm{135}}{\mathrm{4}}×\frac{\mathrm{9}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{15}×\left(\frac{\mathrm{9}}{\mathrm{4}}\right)^{\mathrm{2}} =\frac{\mathrm{1215}}{\mathrm{32}}=\mathrm{37}.\mathrm{97}\:{KNm} \\ $$$${which}\:{is}\:{there}\:{where}\:{the}\:{shear}\:{force} \\ $$$${is}\:{zero}. \\ $$ | ||
Commented by mr W last updated on 07/Sep/19 | ||
$${in}\:{fact}\:{i}\:{drawed}\:{the}\:{sketches}\:{without} \\ $$$${any}\:{calculation}. \\ $$ | ||
Commented by necxxx last updated on 07/Sep/19 | ||
$${ok}.{Thank}\:{you}\:{so}\:{much}.{I}'{ll}\:{try}\:{to}\:{study} \\ $$$${it}\:{more} \\ $$ | ||
Commented by mr W last updated on 07/Sep/19 | ||
$${the}\:{value}\:{given}\:{in}\:{your}\:{book}\:{for}\:{M}_{{max}} \\ $$$${is}\:{wrong}.\:{it}\:{is}\:{not}\:\mathrm{37}.\mathrm{37}\:{KNm},\:{but} \\ $$$$\mathrm{37}.\mathrm{97}\:{KNm}.\:{other}\:{values}\:{in}\:{your} \\ $$$${book}\:{are}\:{correct}\:{and}\:{tally}\:{with}\:{my} \\ $$$${results}. \\ $$ | ||
Commented by necxxx last updated on 08/Sep/19 | ||
$${please}\:{how}\:{did}\:{you}\:{do}\:{that}.{Really}\:{I}\:{would}\: \\ $$$${like}\:{to}\:{know}\:{how}\:{you}\:{solved}\:{it} \\ $$$${comprehensively}. \\ $$$$ \\ $$$$ \\ $$$${Thanks} \\ $$ | ||
Commented by necxxx last updated on 09/Sep/19 | ||
$${Thank}\:{you}\:{so}\:{much}\:{MrW}.{After}\:{looking} \\ $$$${at}\:{the}\:{solvings}\:{you}\:{made}\:{I}\:{went}\:{further} \\ $$$${to}\:{practice}\:{and}\:{now}\:{I}\:{understand}\:{to}\:{a} \\ $$$${good}\:{extent}. \\ $$ | ||