Question

Pargas MD RMN3553 2.MP4 2013 Mchar om COCO GRMN3553.MP4 A bear is loaded and supported as shown, with M = 3180 ft·lb, w - 190 lb/ft and W2 = 450 lb/ft. GRMN3552.MP4 The dimensions on the figure are a = 5 ft and b = 5.7 ft. Draw and label shear and bending moment diagrams for this beam. From your diagrams, enter the following information. Limits of an interval are defined anywhere there is a change in the loading or the shear diagram becomes zero. The first interval is on the left and subsequent intervals are in order moving to the right. GRMN3554.MP4 Reactions: RA GRMN3551.MP4 MA= Shear and Bending Moment Diagram Data: tres xright Interval Vesi Vright function type curvature Mert Mrg RMN3553 2.mov he

Answer 1

Before moving into the problem let me explain the sign convention of shear force and bending moment used in this problem. Convention of shear force If in a beam the shear force about a section (here, Section X-X and yy) is such that it creates a clockwise rotation of the beam then the shear force is positive, but if the direction of shear force is such that it causes an auticlock wise moment then the shear force is negative. Hence here, Y, and Va are positive and Y₃ and Yq are negative. Siqu Couvention of bending moment 13 " Sagging moment about a section is taken as positive while hogging moment is taken as negative. So here M, and Mz are posit while M3 and My are negative.

| 450 Lb/ft - > 519045/17 5ft 157ft Es The above diagram is the free body diagram of the beem. At the support the reaction force is purely vertical This is because there is no horizontal force acting on the beam, so by equilibrium of horizontal forces, the horizontal reaction at support (A) is zero. Take the equilibrium of vestical forces, EV=O Ra + 160 15- 450 X 5-7 =0 Rp = 1615 lb Take the equilibrium of moment about point A, EMA =O (considering clockwise moment as positive) -MA – 190x5x5 - M + 450X57*(5+ 53) =0 MA= 450 X5.7x7.85–190x5x 2.5— 3180 Mar 14580.25 lb.ft

at distance of Consides a section P-p between A and B ips from A. The shear force at P.P is given by the equation, Vp-p = RA + 180xp Vpp = 1615+180xp At A, X-0, P-O, VA = 16151 At B, X=5ft, p=5ft, VB= 25651b The shear force rasies linearly from Ato B. The bending moment at P.P is given by the equation, Mpp = - MA + Rex p + 180x pxş Mp-p = -14580.25 + 1615xp +95xp? At A, x=0, P=O, MA= -14580-25 lb.ft To just left of B, X= 5ft, p=5ft, MB left = -4130.25 1657 The bending moment shows parabolic variation from A to B. Consider a section Q-Q between B and C at a distance of qe from C. The shear force at Q-Q is given by the equation, Vq-Q = 450x9

At B, X = 55+, V=5.7ft, VB= 2565lb At C, x= 10.7ft, q=oft, &c=0 The shear force shows linear variation from B to C. The bending moment at Q-Q is given by the equation, Ma-Q = -450x ex♡ Mo-a = 22582 To just right of B, x=5ft, q= 5.757, MB sight -7310-25 165+ At C, x= 10.750 q=oft, Mc=0 The bending moment shows parabolic variation from B toc. Shear force Diagram. 1256516 ТЫѕ ь 1615 db i دی (همین (0 . 7ft = Bending Moment Diagrem. (x-2) (2=5ft) . (2=10.7ft) 14130.25lb ft 7310.25 lb ft. 14580-251b.ft