Question

4. SHEAR AND MOMENT DIAGRAMS Determine the shear and moment diagrams for the beam shown below. There is a pin at A and a rocker at B. a. There are 3 sections. Draw the FBD for each section. Also, give the shear and moment equation for each section. у 125 lb/ft 1000 lb Х A B 10 ft 6 ft 10 ft

Answer 1

We have to start by finding the support reactions. Then we can write the shear force and bending moment equations for all the sections seperately. Please find the attached solution.

Prob. 1000 lb 125 lblft PI С. 13 777 loft 6ft loft RB RA Eryzo RA+Rg = 1000+ 125X10 RATRBF = 2250 EMATO RBX 26 1ooox lot 125 X10 X 21 RB 1394.23 lb RA = 855.7 of lb X sign, Shear force 125 db/ft 13 Itue love Section BC 1394.23 F = - 1394.23 + 1250

at i FB = - 1394.23 lb P-2 at co (x = loft.) fo -1394.23 + 1250 For a 144.23 1000 lb. I Section CDI С. B Fx = -1394.23 +1250 [constant - 144.23+ luoo fo= 855. Obi 1000 lb Section DA B 855.77 x Fe = ass. lb (constant) FA= 855.11- 855.4 0.

SFD. 855.7 P-3 + D C B 144.23 1394.23 Bending moment Directions. True] G-ve) section Bei Ma 1394.23xx 2 125.2 Parabolic) at B MB=0 250 at ci Me= 1692.3 eb.ft. 21 =10 Section CD. 125* 10 x (X-5) Mx = (1394.23) x Mr = 1394.23X - 1250 x + 6250 Man 144.23 x + 6250 (Linear)

so at point Di = 16 ft. MD= 144.23X16 + 6250 P-4 Mo= 8557.68 eb.ft Section DA Mat (1394.23) 1250 (0-5) - 1000(X-16) -855.777 X + 22250 (linear) at point a 126 MA=0 8557.68 Ibofti BMD linear 1692.3 lb ft. + Parabolic. linear.