Find analytically the support reaction at B and the load P, for the beam shown in the figure below if the reaction of support A is zero.
∑Fy=0→RA−10+RB−36−P=0→RB=46−P\sum F_y=0\to R_A-10+R_B-36-P=0\to R_B=46-P∑Fy=0→RA−10+RB−36−P=0→RB=46−P
∑MA=0→6RB−10⋅2−20−36⋅5−7P=0→\sum M_A=0\to 6R_B-10\cdot2-20-36\cdot5-7P=0\to∑MA=0→6RB−10⋅2−20−36⋅5−7P=0→
6RB−7P=2206R_B-7P=2206RB−7P=220
So, we have
6⋅(46−P)−7P=220→P=56 (kN)6\cdot(46-P)-7P=220\to P=56\ (kN)6⋅(46−P)−7P=220→P=56 (kN) . Answer
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments