Question and Answers Forum

All Questions      Topic List

Mechanics Questions

Previous in All Question      Next in All Question      

Previous in Mechanics      Next in Mechanics      

Question Number 68418 by necxxx last updated on 10/Sep/19

Commented by necxxx last updated on 10/Sep/19

The UDL is 2.5KN/m. The first point load is 35KN. I′ve tried solving but I encountered some issues with the diagram. I′ll be really grateful to get this solved with the possible diagrams.

$${The}\:{UDL}\:{is}\:\mathrm{2}.\mathrm{5}{KN}/{m}.\:{The}\:{first}\:{point} \\ $$$${load}\:{is}\:\mathrm{35}{KN}.\:{I}'{ve}\:{tried}\:{solving}\:{but}\:{I} \\ $$$${encountered}\:{some}\:{issues}\:{with}\:{the}\:{diagram}. \\ $$$${I}'{ll}\:{be}\:{really}\:{grateful}\:{to}\:{get}\:{this}\:{solved} \\ $$$${with}\:{the}\:{possible}\:{diagrams}. \\ $$

Commented by mr W last updated on 10/Sep/19

point loads are 35 (not 85), 25, 5 KN?

$${point}\:{loads}\:{are}\:\mathrm{35}\:\left({not}\:\mathrm{85}\right),\:\mathrm{25},\:\mathrm{5}\:{KN}? \\ $$

Commented by necxxx last updated on 10/Sep/19

exactly sir.I also stated that the UDL is 2.5KN/m

$${exactly}\:{sir}.{I}\:{also}\:{stated}\:{that}\:{the}\:{UDL} \\ $$$${is}\:\mathrm{2}.\mathrm{5}{KN}/{m} \\ $$

Answered by mr W last updated on 10/Sep/19

Commented by mr W last updated on 10/Sep/19

12R_2 =5×14+25×9+2.5×6×6+35×3=490 ⇒R_2 =((245)/6) 12R_1 +5×2=35×9+2.5×6×6+25×3=480 ⇒R_1 =((480−10)/(12))=((235)/6) check: R_1 +R_2 =((245+235)/6)=80 35+2.5×6+25+5=80 ⇒ok

$$\mathrm{12}{R}_{\mathrm{2}} =\mathrm{5}×\mathrm{14}+\mathrm{25}×\mathrm{9}+\mathrm{2}.\mathrm{5}×\mathrm{6}×\mathrm{6}+\mathrm{35}×\mathrm{3}=\mathrm{490} \\ $$$$\Rightarrow{R}_{\mathrm{2}} =\frac{\mathrm{245}}{\mathrm{6}} \\ $$$$\mathrm{12}{R}_{\mathrm{1}} +\mathrm{5}×\mathrm{2}=\mathrm{35}×\mathrm{9}+\mathrm{2}.\mathrm{5}×\mathrm{6}×\mathrm{6}+\mathrm{25}×\mathrm{3}=\mathrm{480} \\ $$$$\Rightarrow{R}_{\mathrm{1}} =\frac{\mathrm{480}−\mathrm{10}}{\mathrm{12}}=\frac{\mathrm{235}}{\mathrm{6}} \\ $$$${check}:\:{R}_{\mathrm{1}} +{R}_{\mathrm{2}} =\frac{\mathrm{245}+\mathrm{235}}{\mathrm{6}}=\mathrm{80} \\ $$$$\mathrm{35}+\mathrm{2}.\mathrm{5}×\mathrm{6}+\mathrm{25}+\mathrm{5}=\mathrm{80}\:\Rightarrow{ok} \\ $$

Commented by mr W last updated on 10/Sep/19

Commented by mr W last updated on 10/Sep/19

to determine M_(max) position with shear force =0: distance to point load 35 KN: ((25)/(25+65))×3=(5/6)m M_(max) =((235)/6)×(3+(5/6))−35×(5/6)−((2.5)/2)×((5/6))^2 =((5765)/(48))=120.1 KNm

$${to}\:{determine}\:{M}_{{max}} \\ $$$${position}\:{with}\:{shear}\:{force}\:=\mathrm{0}: \\ $$$${distance}\:{to}\:{point}\:{load}\:\mathrm{35}\:{KN}: \\ $$$$\frac{\mathrm{25}}{\mathrm{25}+\mathrm{65}}×\mathrm{3}=\frac{\mathrm{5}}{\mathrm{6}}{m} \\ $$$${M}_{{max}} =\frac{\mathrm{235}}{\mathrm{6}}×\left(\mathrm{3}+\frac{\mathrm{5}}{\mathrm{6}}\right)−\mathrm{35}×\frac{\mathrm{5}}{\mathrm{6}}−\frac{\mathrm{2}.\mathrm{5}}{\mathrm{2}}×\left(\frac{\mathrm{5}}{\mathrm{6}}\right)^{\mathrm{2}} \\ $$$$=\frac{\mathrm{5765}}{\mathrm{48}}=\mathrm{120}.\mathrm{1}\:{KNm} \\ $$

Commented by necxxx last updated on 10/Sep/19

Thank you so much mrW.I′m very grateful

$${Thank}\:{you}\:{so}\:{much}\:{mrW}.{I}'{m}\:{very}\:{grateful} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com