Question Number 68222 by necxxx last updated on 07/Sep/19
![Sketch the shear and moment diagrams of a simply supported beam of 6m.The load on the beam consists of UDL of 15KN/m over the left half of the span.](https://www.tinkutara.com/question/Q68222.png)
$${Sketch}\:{the}\:{shear}\:{and}\:{moment}\:{diagrams} \\ $$$${of}\:{a}\:{simply}\:{supported}\:{beam}\:{of}\:\mathrm{6}{m}.{The} \\ $$$${load}\:{on}\:{the}\:{beam}\:{consists}\:{of}\:{UDL}\:{of} \\ $$$$\mathrm{15}{KN}/{m}\:{over}\:{the}\:{left}\:{half}\:{of}\:{the}\:{span}. \\ $$$$ \\ $$
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.](https://www.tinkutara.com/question/Q68223.png)
$${please}\:{help}\:{me}\:{with}\:{this}\:{problem}.\:{I}'{ve} \\ $$$${tried}\:{solving}\:{but}\:{my}\:{result}\:{isnt}\:{in}\: \\ $$$${accordance}\:{with}\:{the}\:{source}.\:{Please}\:{post} \\ $$$${steps}\:{and}\:{diagrams}\:{where}\:{necessary}. \\ $$$${I}'{ll}\:{also}\:{post}\:{the}\:{steps}\:{I}\:{took}\:{so}\:{you}\:{all}\:{can}\: \\ $$$${help}\:{me}\:{with}\:{the}\:{amendments}. \\ $$$${Thanks}\:{in}\:{advance}. \\ $$
Commented by necxxx last updated on 07/Sep/19
![](https://www.tinkutara.com/question/9204.png)
Commented by necxxx last updated on 07/Sep/19
![](https://www.tinkutara.com/question/9205.png)
Commented by necxxx last updated on 07/Sep/19
![](https://www.tinkutara.com/question/9206.png)
Commented by necxxx last updated on 07/Sep/19
![here is the diagram given](https://www.tinkutara.com/question/Q68228.png)
$${here}\:{is}\:{the}\:{diagram}\:{given} \\ $$$$ \\ $$
Answered by mr W last updated on 07/Sep/19
![](https://www.tinkutara.com/question/9213.png)
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.](https://www.tinkutara.com/question/Q68247.png)
$${Thank}\:{you}\:{so}\:{much}\:{sir}\:{but}\:{this}\:{SFD} \\ $$$${and}\:{BMD}\:{do}\:{not}\:{tally}\:{with}\:{the}\:{answer} \\ $$$${given}\:{in}\:{the}\:{reference}\:{material}\:{I}\:{used}. \\ $$$$ \\ $$
Commented by mr W last updated on 07/Sep/19
![the values i gave are 100% corrrect, that′s for sure. your book might use different signs. therefore i haven′t given any sign for the values.](https://www.tinkutara.com/question/Q68249.png)
$${the}\:{values}\:{i}\:{gave}\:{are}\:\mathrm{100\%}\:{corrrect}, \\ $$$${that}'{s}\:{for}\:{sure}. \\ $$$${your}\:{book}\:{might}\:{use}\:{different}\:{signs}. \\ $$$${therefore}\:{i}\:{haven}'{t}\:{given}\:{any}\:{sign}\:{for} \\ $$$${the}\:{values}. \\ $$
Commented by necxxx last updated on 07/Sep/19
![](https://www.tinkutara.com/question/9214.png)
Commented by mr W last updated on 07/Sep/19
![if the maximum of moment is needed, it is ((135)/4)×(9/4)−(1/2)×15×((9/4))^2 =((1215)/(32))=37.97 KNm which is there where the shear force is zero.](https://www.tinkutara.com/question/Q68251.png)
$${if}\:{the}\:{maximum}\:{of}\:{moment}\:{is}\:{needed}, \\ $$$${it}\:{is} \\ $$$$\frac{\mathrm{135}}{\mathrm{4}}×\frac{\mathrm{9}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{15}×\left(\frac{\mathrm{9}}{\mathrm{4}}\right)^{\mathrm{2}} =\frac{\mathrm{1215}}{\mathrm{32}}=\mathrm{37}.\mathrm{97}\:{KNm} \\ $$$${which}\:{is}\:{there}\:{where}\:{the}\:{shear}\:{force} \\ $$$${is}\:{zero}. \\ $$
Commented by mr W last updated on 07/Sep/19
![in fact i drawed the sketches without any calculation.](https://www.tinkutara.com/question/Q68252.png)
$${in}\:{fact}\:{i}\:{drawed}\:{the}\:{sketches}\:{without} \\ $$$${any}\:{calculation}. \\ $$
Commented by necxxx last updated on 07/Sep/19
![ok.Thank you so much.I′ll try to study it more](https://www.tinkutara.com/question/Q68253.png)
$${ok}.{Thank}\:{you}\:{so}\:{much}.{I}'{ll}\:{try}\:{to}\:{study} \\ $$$${it}\:{more} \\ $$
Commented by mr W last updated on 07/Sep/19
![the value given in your book for M_(max) is wrong. it is not 37.37 KNm, but 37.97 KNm. other values in your book are correct and tally with my results.](https://www.tinkutara.com/question/Q68254.png)
$${the}\:{value}\:{given}\:{in}\:{your}\:{book}\:{for}\:{M}_{{max}} \\ $$$${is}\:{wrong}.\:{it}\:{is}\:{not}\:\mathrm{37}.\mathrm{37}\:{KNm},\:{but} \\ $$$$\mathrm{37}.\mathrm{97}\:{KNm}.\:{other}\:{values}\:{in}\:{your} \\ $$$${book}\:{are}\:{correct}\:{and}\:{tally}\:{with}\:{my} \\ $$$${results}. \\ $$
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](https://www.tinkutara.com/question/Q68262.png)
$${please}\:{how}\:{did}\:{you}\:{do}\:{that}.{Really}\:{I}\:{would}\: \\ $$$${like}\:{to}\:{know}\:{how}\:{you}\:{solved}\:{it} \\ $$$${comprehensively}. \\ $$$$ \\ $$$$ \\ $$$${Thanks} \\ $$
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.](https://www.tinkutara.com/question/Q68345.png)
$${Thank}\:{you}\:{so}\:{much}\:{MrW}.{After}\:{looking} \\ $$$${at}\:{the}\:{solvings}\:{you}\:{made}\:{I}\:{went}\:{further} \\ $$$${to}\:{practice}\:{and}\:{now}\:{I}\:{understand}\:{to}\:{a} \\ $$$${good}\:{extent}. \\ $$