Question

Question 1 10 points Save Answer Determine the shear force 5.94 ft to the right of point A for the beam shown below if p = 965 lb, W = 332 lb/ft, T = 4,028 lb ft, a = 3.82 ft and b = 5.59 ft. P W T в! E D leathe 2 ft2 ft b | 24 |

Answer 1

Question.1: Shear Force at a section;

P=965 l. A la с X a = By Ey ve Question 1 W = 332 lb/ft. T=4028 lb-ft. E Be of @ 5.94.47 = ? * - 3-82f bo 5.59 ft. " 244 24t.1 Sigm Comention Step:/- find Support Reactions: 9, ntve t, {Horizontal forces =0 Bx=0 lb. Evertical forces =0 By + Ey = 2020.88 lb. {moment at point B =0 → 6-965 * 3-82) + ( 332 *5.59 * 5.9 5.99)+(4028) +t- Ey *9.59) = 576.526 lb ; By = 22 44.354 lb. To find Shear force at any section stand at that section and look either left (or) light. The summation of forces either to the left (100) sufight echt will give liequired shear force Nous, Shear force @ 5.94 It from point to Othe Sf = -965 +2244.354-332*2.12 965 lb. 332 lb ft. If=575.514 lb. K3.82- 2.12 2244. 354 lb.

We have to determine the shear force at a section which is 5.94 ft. to the right of point A. To find the shear force we either take the summation of vertical forces of the left side of the section or to the right of the section.

Shear Force @ 5.94 ft. from point A = 575.514 lb.

(Positive sign indicates the upward shear force as per sign convention)

Question.2: Bending Moment at a section;

Question.a. P= 1029 lb. W= 326 lb lift T= 5,603.0bft A B Br X By Byt Ey = 1029 +1669-12 0698.12 lb. BM @ 6:7 ft..? 'A= 3-25ft) b= 5.12 oft.“ 244 244 Sign Cornertieren alep: 1. To find support leactions ? a Altre E Horizonted forces =0 → By = 0 lb. dia Evertical forces=0 By + Ey= & Moment at point B-0 + (-102943-25) + (326 * 5-12 * 5:22) +(5603) + Ey *9.12) = 0 Ey = 716.195 lb. depid. To find moment at my section stand at that section and look either left (oo) liight, summation of all the moments to the left (ory to the right of the section will give the required bending moment. Now, BM @ bil ft from point 326 Ulft. X BM = (-1029* 6-1) +(1981.925*2.85) 3.25 ft. & 2.85 ft.t +(-386*2.85* 2:25 4:05) -1952.38145 lb-ft. . By = 1981.925 lb. > - 1029 lb. + X 1981.925 IBM =

We have to determine the bending moment at a section which is 6.1 ft. from the point A. To find the bending moment we either take the summation of all the moments of the left side or of the right side of the section.

BM @ 6.1 ft. from point A = -1952.38125 lb-ft.

(Negative sign indicates the anti-clockwise moment as per the sign convention)