As shown, beam ABC is supported by the roller at A and pin at C . The geometry o
ID: 1828779 • Letter: A
Question
As shown, beam ABC is supported by the roller atA and pin at C. The geometry of the beam is given by a=4.0 ft, b=6.5 ft, and c=9.00 ft. The applied forces are F1=1.25kip andF2=1.25 kip. Force F1 is applied at an angletheta=55? with the horizontal. Neglect the weight of the beam.
Find the vertical reaction at A, and find the vertical and horizontal reaction at C.
As shown, beam ABC is supported by the roller atA and pin at C. The geometry of the beam is given by a=4.0 ft, b=6.5 ft, and c=9.00 ft. The applied forces are F1=1.25 kip andF2=1.25 kip. Force F1 is applied at an angletheta=55? with the horizontal. Neglect the weight of the beam. Find the vertical reaction at A, and find the vertical and horizontal reaction at C.Explanation / Answer
1st Step: Draw a FBD (The custom diagram isnt working, but it looks almost the same as the problem except F1 would be broken into x and y componets and your reactions will be at A and C.)
2nd: Find the x and y componets of F1 F1x = 1.25k(cos55) F1y= 1.25k(sin55)
3rd: Summation of forces in the X direction
?Fx=0
-F1x + Cx = 0
-1.25k(cos55) + Cx = 0
Cx = 1.25k(cos55) = 0.716k
4th: sum the moments at C
?Mc = 0 (clockwise is positive)
-F2(9'-4') + Ay(9'+6.5') =0
-1.25k(6') + Ay(15.5') = 0
Ay = [1.25k(6')] / (15.5') = 0.48387k
5th: Now that you have Ay you can sum the moments in the Y direction to find Cy
?Fy = 0
Ay + Cy - F2 - F1y = 0
Cy = -Ay + 1.25k + 1.25k(sin55) = 1.7900k
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