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Question Number 97953 by I want to learn more last updated on 10/Jun/20

Commented by mr W last updated on 10/Jun/20

M=T×h=2×2=4 KN  ΔV=(M/l)=(4/4)=1 KN  ΔV_A =+ΔV=+1 KN (increasing)  ΔV_B =ΔV_C =−((ΔV)/2)=−0.5 KN (decreasing)

M=T×h=2×2=4KNΔV=Ml=44=1KNΔVA=+ΔV=+1KN(increasing)ΔVB=ΔVC=ΔV2=0.5KN(decreasing)

Commented by mr W last updated on 10/Jun/20

Commented by I want to learn more last updated on 10/Jun/20

Thanks sir, i appreciate

Thankssir,iappreciate

Answered by Rio Michael last updated on 10/Jun/20

Here′s another method i love. Keep in mind  it is thesame as Mr W method.   if we equate the sum of the moments about the x−axis   we can the determine ΔN_A  the additional reaction force  ⇒ ΣM_x  = 0 ⇔ 4ΔN_A  −2.0 × 2.0kN = 0    ΔN_A  = 1.0 kN  if we equate the sum of the moments about the x−axis parrallel  to the y−axis which goes through point B  to zero. we get  the addition reaction force ΔN_C    ⇒Σ M_y ^B  = 0 ⇔ 4.8ΔN_C  + 2.4N_A  = 0     N_C  = −0.5 kN  if we equate the sum in the z−direction to zero , we can determine  the additional ΔN_B  force   ⇒ ΣF_z  = 0 ⇔ ΔN_A  + ΔN_B  + ΔN_C  = 0   N_B  = −0.5 kN.

Heresanothermethodilove.KeepinminditisthesameasMrWmethod.ifweequatethesumofthemomentsaboutthexaxiswecanthedetermineΔNAtheadditionalreactionforceΣMx=04ΔNA2.0×2.0kN=0ΔNA=1.0kNifweequatethesumofthemomentsaboutthexaxisparralleltotheyaxiswhichgoesthroughpointBtozero.wegettheadditionreactionforceΔNCΣMyB=04.8ΔNC+2.4NA=0NC=0.5kNifweequatethesuminthezdirectiontozero,wecandeterminetheadditionalΔNBforceΣFz=0ΔNA+ΔNB+ΔNC=0NB=0.5kN.

Commented by Rio Michael last updated on 10/Jun/20

Commented by mr W last updated on 11/Jun/20

ΔF_B =0  ΔF_C =2 KN

ΔFB=0ΔFC=2KN

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