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
$${Thank}\:{you}\:{so}\:{much}\:{MrW}.{After}\:{looking} \\$$$${at}\:{the}\:{solvings}\:{you}\:{made}\:{I}\:{went}\:{further} \\$$$${to}\:{practice}\:{and}\:{now}\:{I}\:{understand}\:{to}\:{a} \\$$$${good}\:{extent}. \\$$