Question

The simply support beam below is loaded as shown. = 45 75 kip 30 kip/ft 500 kip-ft mo 1 ft 3 ft 1.5 ft 2.5 ft 2.5ft The exact value of the normal force (N), shear force (V), and bending moment (M) at a distance of x=7 feet. Be sure to indicate both magnitude and direction with arrows. (a) FBD and support reactions (b) FBD to find internal forces (c) Calculate N, V, and M at 7 feet

Answer 1

Free Body Diagram Sookip-f4 75kip Bokin-ff- hus V VIVE Apply Equilibrium equations, ki*3-*15*2.5*2.5' 'D VE VA I EFx=0 HA - 75 Cos45=0 HA = 53.03 kip GEMA=0 (VE X8 )-(30x5x8) - ( 15xsin 45°x4)-500 - O VE = 239.02 kpl +1 Sty=0 VA + VE 75sin45 (3085)=0 VA+VE = 203.03 VA +239,02 = 203.03 IVA = -35.98 kip b) FBD to find internal forces at x=7 ft. 500 kipff. 75 KIP. 30 kpft 53.03k I J45 2 . K1-+ 3' * 1.5 * 1.5 -1.512 -35.98 KIP V x=7ft

c) Apply Equilibrium equation to above FBD, - Sfx 53.03 - N=0 Normal Force, IN = 53.03 kip/(k +12 Fy=0 - 35.98 - (75x sin65) - (30x1.5) +V=0 Y = 134.01 kip (D (+ ²M=0 -M-500 + (35.98877+ (75xsing 583) + (30x1.5x1.5 5X1,5) -M-500+ 444.71=0 M =-55.29 kip-ft. Negative Sign indicates Counter clock wise direction. ..M = 55.29 kip-ft (CCW)
If you have any doubts, feel free to ask me. Don't forget to rate me.