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Question-68418

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

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