Q. 300U ID(X) - 200 ID/TT (X squared) - 9850 ID-TI 33. The distance between point A and the shear force VAB=0 * O a. 6 feet b. 8 feet c. 10 feet O d. 12 feet 34. The degree of bending moment equations for segments AB and BC. * O a. 1st degree and 2rd degree O b. 3rd degree and 2nd degree c. 2nd degree and 3rd degree O d. None of the above. Back Next Page 2 of 4 Clear formProblem Solving No.2 ............................. B. Derive the shear force equation,bending moment equation for the beam * shown. Neglect the weight of the beam. 200 lb/fl 12001b'ft 6ft~L4ft~l'4ft>| Determine the following: 26. Reaction support at point B. @ a. 1350 lb (Upward Direction) 0 b. 1380 lleownward Direction) 0 o. 1400 lb (Upward Direction) 0 d. 1420 lb (Downward Direction) 30. Bending moment equation for segment AB. * O a.-5.40 psf X cube O b. 5.45 psf X cube O c. -5.55 psf X cube O d. 6.55 psf X cube 31. Shear force equation for segment BC. * O a. 2000 lb - 200 lb/ft ( X) b. 2100 lb - 300 lb/ft ( X) O c. 2150 lb - 300 lb/ft ( X) O d. 3000 lb - 200 lb/ft ( X) 32. Bending moment equation for segment BC. * O a. 2000 lb(X) - 100 1b/ft (X squared) - 9600 lb-ft b. 2100 lb(X) - 120 lb/ft (X squared) - 9700 lb-ft O c. 21500 lb(X) - 200 lb/ft (X squared) - 9800 lb-ft d. 3000 lb(X) - 200 lb/ft (X squared) - 9850 lb-ft2?. Reaction support at point D. * O a. one |b [Upward Direction) 0 b. mo lb [Downward Direction] 0 :2. Bill] lb {Upward Direction} (E) d. 90D lb [Downward Direction] 23. Centroid of the triangular load. * O a. 588 lb (Downward Direction) 0 b. 593 lb [Upward Direction} 0 :1. sun lb{[}ownward Direction] (E) d. (Upward Direction) 29. Shear force equation for segment AB. * O a_1b.65 psf I squared O b. 16.55 psf X squared @ 11-15.?5 psi): squared O d.1b_?5 psf X squared