Question

5. A beam has the cross-section as shown in the figure. If the shear force at a section is 24.6 KN, find the maximum shear stress and its location. 100 mm 20 mm T 100 mm 20mm

Answer 1

shear stress in I Section Solution : Maximum 100mm 20mm - loomno - 2017m. Т. section Find Centroid of given a, y, t9242 17 a,taz 2 0 Loomm - 20 X 100 2000 mm ta 20mm 2000mm Az - 20 X 100 20mm Toomm 100 = 50mm oy, 2 = llomm +42 = 100 + 20 = 100 +10 2 V (2DOO)(50) + (2000) I10) (2000 + 2000)

17 il Ý lcomm 300h, 100,000 + 220,000 4000 = 80 mm Th₂ Xt 100 I commi 20mm W. Y, = 110mm. 0 ŭ = 80mm Y=50 Romm about Next, Calculate Moment of Inesha T- section x-axis Ixxi + Iz -> Ixx goedia Ixoc, I, + bhita ta, h, ² (5-4,3 o I loommah 12 201() (-10) (46-501 20mm=b Iso = 3.46 x 100 mm²

1.86 x 10 mm 4 Similarly 2 Izt az ha Ixxa 2 bh²+ a2 (92-ý 12 (100)(2037 2000 (110-80) E 12 Toaz in o and Ixxz Put Ixoxo Inca - [3.46x100 + 1.86x10 5.32 x 10 Ixx = Shear Maximum Now, For given T-section will be the Neutral Axis (NA) at Stress FA t max I b F = shear force below the Neutral axis. A= Area below the

lomm 20mm Nallyul axil wob 26) mm Email 8: 8cm 42.82 My mo? 20 mm A = 80 x 20 = lboomm? 9 = P Previous 80 Homm 2 2 b = 20 mm = width of web F = 24 ob kN all values in Tmax equahon Putting 1.I - Ixx IT max FAY I b F = 24-6 IN = 24.6x103N (2462) C16oo)*(H0) C 1.86X106) (20) EN/mm2 С 342.32 N max Maximum shear stress value and it's Location 18 at Neutral alais.