Question

4. (25 pt.) The beam subjected to a uniform distributed load as shown in Figure 4(a) has a triangular cross-section as shown in Figure 4(b). 1) (6 pt.) Determine mathematical descriptions of the shear force function V(x) and the moment function M(x). 2) (6 pt.) Draw the shear and moment diagrams for the beam. 3) (5 pt.) What is the maximum internal moment Mmar in the beam? Where on the beam does it occur? 4) (8 pt.) Determine the absolute maximum bending stress in the beam. The neutral axis passes through the centroid of the cross-sectional area. Refer to Figure 4(e) for the location of the centroid and the formula for the moment of inertia of a triangular area. 4 kN/m 30 cm Triangular area 6 m 40 cm 1,- (a) (b) (c) Figure 4

Answer 1

4 k nim mano 30cm zb Ant B Flour RBV -6m 12 kw RB va 12 kw Rout=12 kw om Iz bhs Efuzo Rave RBV a 4x6 =24 1 Rover RBV > 24 kw ] 76 uox(30)3 x 10-8 -(0) - 2 36 EMAzo TI = 3x10 mg 24x32 Roux 6 [REV 2 12 kN I RAV , 12 KW 12 kW Shear force diagram ILN Bending moment diagram Mman

No ന് 6w! 12 kw e 12kw (S. f)e-x 2 Shear force in Section *-* [(Sf)x.x = 12- 4* 7 = (PM) x-x 2 Bending moment in section *-* 2 Rovex - 4xxx 8 Hx-x 2 123 - 2x2 72 M® To calculate Mmay ll.c a (BM)x-X zo dre la - use zo 23m at cenfure of beam (BM)xazm z 12X3 – 273)2 2 36-182 18 kwam distance from neutralasia Mmay ry a Mmas 12h/3) 2 188103 x dx (0.2 E ? (Colony Hou Honeydelse) - Justo y es pot 3xlory z 12X100 Pa ay 12 MPa