Question

Problem #2: For the beam shown, plot the shear-force and bending-moment diagrams using singularity functions. Label the diagrams properly and provide values at all key points. y 500 lbf 40 lbffin A BIIIIIII

Answer 1

The solution can be calculated as follows,

Shear force (SF) at c = 0 Sf at any distance x to the left of c= 40xx (tve) (tve due to sign convention of SF). a Sf at B, SF2 = 40x6 =240 1 bf next lood that is Now SF will be constant until the at A. Sf = (500 + 240) Ibf. = 740 lbf. This sf will not vary until point o, where a sof of magnitude 740 lbf will act upward at o g So S.F diagrams will come down to zero value 1740 lbf. Bending Moment Diagram Take any section x-x at a distance of a from 'C' - Let 'w be the unifooms lood acting per length. na to da I. .. at x-x, wx is the lood acting So Moment ( M E -XXX [-ve sign due bo sign convention, distanse is the distance at which the effective of a unition load will a - M = - 40* 6* ç = -720 lbf in? (W = 40 lbf len & maximums x = 6 in) , R After point B take a section x-x again before 'A. - 240 *(x+3) 240 I I ME - 240x673)=-2160 lbf live? (bfint

Now, take a section byron between 0 a A. soo abf ,240 dbf effective). X-X M =-500x c) + 240 [x+6+3] - - [800x + 240x + 2160] = - [ 740x + 21607 - MOE - [740% (3) + 2160] = -8080. Hoftiz (bf in?

500 lbf 40 lbf/in 1o 8 in A6 in B6 in 740 lbf 240 lbf 720 lbf in^2 2160 lbf in^2 8080 lbf in^2

The second figure is the shear force diagram and the last one is the bending moment diagram.

Note that in the bending moment diagram the two straight lines have different slopes.