Question

The timber beam shown has the cross section shown. The moment of inertia about the z axis is 1,184 in.4. If P = 6360 lb, determine the normal stress Oy acting at point H. 2 in. 3.25 in. H 8 in. 48 in 48 in. 36 in 2 in. 2 in OIN 2 in. -193 psi - 143 psi C-206 psi -222 psi -268 psi

Answer 1

6360 lb { H - A RA 48in 36 in RB =2312:7 eb = above Neto Neutral Axis ENA) and tension Step2 = Calculations for Bending Moment at to section of location of Ho NA- A 48 in = 4047. 2746 Mx=36in RB X 36 Sign convention: (2) Sagging Moment Hogging at section, moment any section. 23/2.73x36 lb-in Mx Step 3: Stress calwlation at H. is of of sagging nature at section X-X; cross-section will face compression at any My = 83258.28 lb-in As Mx below NA

Due to symmetry about Z-axis, NA will lie middepth & either from top bottom. *H 2.75in NAI bin From the concept of Flexeral formula, blo ISH لا أنا where I R M= BM at any section I= MOI about the bending Axis. Bending Stress acting, at distance y from NA. &= Modulus of Elasticity Ra Radous of curvature MY T flence on My Y Iz

- 67 83258.28 * 2.75 Morinxin 1184164 193.379 lb 193 Psi TH in2 M Compressive stresse of are take compressive stress as fine and tensile stress tive OM 192 Psi Note - Bending Stress is also a normal stress only acting acting due to Bending Moment and not due to Axial loads,